Samacheer Kalvi 11th Economics Solutions Chapter 11 Tamil Nadu Economy

Immediately get Chapter Wise Tamilnadu State Board Solutions for 11th Economics to gain more marks and start Quick Revision of all Chapters. All Chapters Pdf is provided along with the Questions and Answers. You Can Dowload Samacheer Kalvi 11th Economics Book Solutions Questions and Answers for Chapter wise are given with a clear explanation. Tamilnadu State Board Solutions for 11th Economics Chapter 11 Tamil Nadu Economy Questions and Answers is for free of cost.

Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 11 Tamil Nadu Economy

Tamilnadu State Board Solutions for 11th Economics Chapter 11 Tamil Nadu Economy Questions and Answers PDF has all given in Chapter Wise Section. Check Out daily basis with Tamilnadu State Board Solutions 11th Economics PDF will help to improve your score. Improve your level of accuracy to answer a question by reading with Samacheer Kalvi 11th Economics Book Solutions Questions and Answers PDF.

Samacheer Kalvi 11th Economics Tamil Nadu Economy Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
In health index, Tamil Nadu is ahead of
(a) Kerala
(b) Punjab
(c) Gujarat
(d) all the above
Answer:
(c) Gujarat

Question 2.
In sex ratio, Tamil Nadu ranks
(a) first
(b) second
(c) third
(d) fourth
Answer:
(c) third

Question 3.
Tamil Nadu is rich in
(a) Forest resource
(b) human resource
(c) mineral resource
(d) all the above
Answer:
(b) human resource

Question 4.
The main source of irrigation in Tamil Nadu is
(a) river
(b) tank
(c) well
(d) canals
Answer:
(c) well

Question 5.
Knitted garment production is concentrated in
(a) Coimbatore
(b) Tiruppur
(c) Erode
(d) Karur
Answer:
(b) Tiruppur

Question 6.
Which of the following is wrongly matched?
(a) Gateway of Tamil Nadu – Thoothukudi
(b) Home textile city – Erode
(c) Steel city – Salem
(d) Pump city – Coimbatore
Answer:
(b) Home textile city – Erode

Question 7.
Which of the following cities does not have international airport?
(a) Madurai
(b) Tiruchirappalli
(c) Paramakudi
(d) Coimbatore
Answer:
(c) Paramakudi

Question 8.
TN tops in the production of the following crops except
(a) Banana
(b) Coconut
(c) plantation crops
(d) cardamom
Answer:
(d) cardamom

Question 9.
Largest area of land is used in the cultivation of
(a) Paddy
(b) sugarcane
(c) Groundnut
(d) Coconut
Answer:
(a) Paddy

Question 10.
In literacy rate, TN ranks
(a) second
(b) fourth
(c) sixth
(d) eighth
Answer:
(d) eighth

Question 11.
In investment proposals filed by MSMEs, TN ranks
(a) I
(b) II
(c) III
(d) IV
Answer:
(a) I

Question 12.
Which district in TN has the highest sex ratio?
(a) Nagapattinam
(b) Nilgiris
(c) Tiruchirapalli
(d) Thanjavur
Answer:
(b) Nilgiris

Question 13.
Which district has the lowest child sex ratio?
(a) Madurai
(b) Theni
(c) Ariyalur
(d) Cuddalore
Answer:
(c) Ariyalur

Question 14.
Which Union Territory has the highest sex ratio?
(a) Chandigarh
(b) Pondicherry
(c) Lakshadeep
(d) Andaman Nicobar
Answer:
(b) Pondicherry

Question 15.
The largest contribution to GSDP in Tamil Nadu comes from
(a) agriculture
(b) industry
(c) mining
(d) services
Answer:
(d) services

Question 16.
In human development index, TN is ranked
(a) Second
(b) fourth
(c) sixth
(d) third
Answer:
(d) third

Question 17.
SPIC is located in
(a) Chennai
(b) Madurai
(c) Tuticorin
(d) Pudukottai
Answer:
(c) Tuticorin

Question 18.
The TICEL park is
(a) Rubber Park
(b) Textile park
(c) Food park
(d) Bio
Answer:
(d) Bio

Question 19.
In India’s total cement production, Tamil Nadu ranks
(a) third
(b) fourth
(c) first
(d) second
Answer:
(a) third

Question 20.
The Headquarters of Southern Railway is at
(a) Tiruchirappalli
(b) Chennai
(c) Madurai
(d) Coimbatore
Answer:
(b) Chennai

Part – B
Answer the following questions in one or two sentences

Question 21.
State any two districts with favourable sex ratio. Indicate the ratios.
Answer:

  1. The Nilgiris (1041 females)
  2. Thanjavur (1031 females)

Question 22.
Define GSDP.
Answer:
The Gross State Domestic Product refers to the total money value of all the goods and services produced annually in the state.

Question 23.
Mention any four food crops which are favourable to Tamil Nadu.
Answer:

  1. Rice
  2. Maize
  3. Groundnut
  4. Oil seeds

Question 24.
What are major ports in Tamil Nadu?
Answer:
Major ports at Chennai, Ennore and Tuticorin Intermediate port in Nagapattinam.

Question 25.
What is heritage tourism?
Answer:
Heritage tourism is travelling to experience the places and activities that authentically represent the stories and people of the past.

Question 26.
What are the nuclear power plants in Tamil Nadu?
Answer:
The Kalpakkam Nuclear power plant and the Koodankulam Nuclear power plant.

Question 27.
Define Micro industry.
Answer:
The enterprises with a capital investment not exceeding 25 lakhs.

Part – C
Answer the following questions in one Paragraph

Question 28.
Write a note on mineral resources in Tamil Nadu.
Answer:
Mineral resources in Tamil Nadu:

  1. Tamil Nadu has a few mining projects based on Titanium, Lignite, Magnesite, Graphite, Limestone, Granite and Bauxite.
  2. The first one is the Neyveli Lignite Corporation that has led development of large industrial complex around Neyveli in Cuddalore district with Thermal Power Plants, Fertilizer and carbonisation plants.
  3. Magnesite mining is at Salem from which mining of Bauxite ores are carried out at Yercaud and this region is also rich in Iron ore at Kanjamalai.
  4. Molybdenum is found in Karadikuttam in Madurai district.

Question 29.
Explain GSDP in Tamil Nadu.
Answer:
GSDP in Tamil Nadu:

  1. GSDP refers to the total money value of all the goods and services produced annually in the state.
  2. According to Tamil Nadu’s directorate of Economics and Statistics, Tamil Nadu is the second largest economy in India with a GSDP of $207.8 billion in 2016 -17
  3. The GSDP of Tamil Nadu is equal to the GDP of Kuwait on nominal terms and GDP of UAE on PPP terms.
  4. Sectoral contribution of GSDP of Tamil Nadu
    • Tertiary sector – 63. 70%
    • Secondary sector – 28.5%
    • Primary sector – 7.76%

Question 30.
Describe development of textile industry in Tamil Nadu.
Answer:
Development of textile industry in Tamil Nadu:

  1. Tamil Nadu is the largest textile hub of India. Tamil Nadu is known as the “Yam Bowl” of the country.
  2. Tamil Nadu accounts for 41% of India’s cotton yam production.
  3. It produce direct employment to 35 million people.
  4. It contributes 4% of GDP and 35% of gross export earnings.
  5. The textile sector contributes to 14% of the manufacturing sector.

Question 31.
Compare productivity of any two food crops between Tamil Nadu and India.
Answer:

  1. Asia result of Government’s efforts Tamil Nadu tops in productivity, in food crops as well as non-food crops, among the states in India.
  2. Among Indian states Tamil Nadu ranks first in maize, Kambu, Groundnut, Oil seeds and Cotton.
  3. Second in rice and coconut.
  4. Third in Sugarcane, Sunflower and Jowar.

Question 32.
Explain the prospect for development of tourism.
Answer:
Prospect for development of tourism:

  1. Tamil Nadu has emerged as one of the leading tourist destinations for both domestic and foreign tourists.
  2. Tourism in Tamil Nadu is promoted by Tamil Nadu Tourism Development Corporation (TTDC), a Government of Tamil Nadu undertaking.
  3. The state currently ranks the highest among Indian states with about 25 crore arrivals.
  4. The annual growth rate of this industry stood at 16 percent.
  5. Approximately 28 lakh foreign and 11 crore domestic tourists visit the state.

Question 33.
What are the renewable sources of power in Tamil Nadu?
Answer:
Renewable sources of power in Tamil Nadu:

  1. Tamil Nadu is the fore front of all other Indian States in installed capacity.
  2. Muppandal wind farm is a renewable energy source, supplying the villagers with electricity for work.
  3. Wind farms were built in Nagercoil and Tuticorin apart from already existing ones around Coimbatore, Pollachi, Dharapuram and Udumalaipettai.
  4. Wind energy contributes 2% of the total power output of India.
  5. Tamil Nadu tops in solar power generation in India.
  6. There are about 20 hydro electric units in Tamil Nadu.

Question 34.
Describe the performance of Tamil Nadu economy in health.
Answer:
Performance of Tamil Nadu economy in health:

  1. Tamil Nadu has a three – tier health infrastructure comprising hospitals, primary health centres, health units, community health centres and sub-centres.
  2. Tamil Nadu has placed third in health index as per the NITIAAYOG report.
  3. The neo natal mortality rate is 14 which is lower than many other states.
  4. The under 5 mortality has dropped from 21 in 2014 to 20 in 2015.

Part – D
Answer the following questions in about a page

Question 35.
Describe the qualitative aspects of population.
Answer:

  1. Population : Tamil Nadu stands sixth in population with 7.21 crore.
  2. Density : Tamil Nadu ranks 12th in density with 555 people per sq.km.
  3. Urban population : Tamil Nadu is one of the most urbanized state with 48.4% of urban population.
  4. Sex ratio : Balanced sex ratio implies improvement in quality of life of female population. The sex ratio of Tamil Nadu is nearing balance with 995 which stands third next to Kerala and Puducherry.
  5. Infant mortality rate : According to NITI AAYOG, the IMR is 17 for Tamil Nadu which is just half of national average of 34 as on 2016.
  6. Maternal mortality rate : Tamil Nadu has a good record of controlling MMR, ranking third with 79 against national average of 159.
  7. Literacy :
    • The literacy rate of Tamil Nadu is higher than in many states.
    • Tamil Nadu has a highest Gross Enrollment Ratio in higher education.

Question 36
Explain the various sources of energy in Tamil Nadu.
Answer:
Tamil Nadu tops in power generation among the southern states. Tamil Nadu is in the forefront of all other Indian states in installed capacity.
Muppandal wind farm is a renewable energy source, Supplying the villagers with electricity for work.

1. Nuclear energy : The Kalpakkam and the Koddankulam Nuclear Power Plants are the major nuclear energy plants for the energy grid.

2. Thermal power: In Tamil Nadu the share of thermal power in total energy sources is very high and the thermal power plants are at Athippattu, Ennore, Mettur, Neyveli and Thoothukudi.

3. Hydel energy : There are about 20 hydro electric units in Tamil Nadu. The prominent units are Hundah, Mettur, Periyar, Maravakandy, Parson valley etc.

4. Solar energy : Tamil Nadu tops in solar power generation in India. Southern Tamil Nadu is considered as one of the most suitable regions the country for developing solar power projects.

5. Wind energy : Tamil Nadu has the highest installed wind energy capacity in India.
The state has very high quality of off shore wind energy potential off the Tirunelveli coast and Southern Thoothukudi and Rameswaram coast.

Question 37.
Explain the public transport system in Tamil Nadu.
Answer:
Tamil Nadu has a well established transportation system that connects all parts of the state. This is partly responsible for the investment in the state.

Road transport :

  • There are 28 national highways in the state, covering a total distance of 5,036 km.
  • The state has a total road length of 1,67,000 km.
  • It ranks second in India with a share of over 20% in total road projects.

Rail transport :

  • Tamil Nadu has a well – developed rail network as part of Southern Railway, Head quartered at Chennai.
  • Tamil Nadu has a total railway track length of 6,693 KM and there are 690 railway stations in the state.
  • The system connects it with most major cities in India.
  • Chennai developed a metro system, which came into operation since May 2017.

Air transport :

  • Tamil Nadu has four major international airports.
  • Chennai, Coimbatore, Tiruchirapalli and Madurai International airports.
  • It also has domestic airports at Tuticorin, Salem and Madurai.

Samacheer Kalvi 11th Economics Tamil Nadu Economy Additional Questions and Answers

Part-A
Choose the best options

Question 1.
Molybdenum is found in _____ district of Tamil Nadu.
(a) Chennai
(b) Madurai
(c) Tirunelveli
(d) Vellore
Answer:
(b) Madurai

Question 2.
Tamil Nadu has _____ Population as against 121 crore of India.
(a) 7.12 crore
(b) 1.72 crore
(c) 7.21 crore
(d) 1.27 crore
Answer:
(c) 7.21 crore

Question 3.
Tamil Nadu’s per capital income is _____
(a) 1100 dollar
(b) 2200 dollar
(c) 3360 dollar
(d) 4400 dollar
Answer:
(d) 4400 dollar

Question 4.
Gross state domestic product is
(a) GSDP
(b) MMR
(c) GDP
(d) IMF
Answer:
(a) GSDP

Question 5.
The detroit of Asia is _____
(a) Bengaluru
(b) Madurai
(c) Coimbatore
(d) Chennai
Answer:
(d) Chennai

Question 6.
_____ is Tamil Nadu’s steel city
(a) Chennai
(b) Karur
(c) Salem
(d) Namakkal
Answer:
(c) Salem

Question 7.
_____ is famous for bus body building.
(a) Karur
(b) Trichy
(c) Vellore
(d) Chennai
Answer:
(a) Karur

Question 8.
_____ is the yarn bowl of India.
(a) Orissa
(b) Kerala
(c) Andhra
(d) Tamil Nadu
Answer:
(d) Tamil Nadu

Question 9.
Gateway of Tamil Nadu is _____
(a) Thoothukudi
(b) Chennai
(c) Coimbatore
(d) Ranipet
Answer:
(a) Thoothukudi

Question 10.
_____ is the knitting city.
(a) Erode
(b) Karur
(c) Namakkal
(d) Tirupur
Answer:
(d) Tirupur

Question 11.
Which of the following is wrongly matched ?
(a) SAIL – Salem
(b) The Pumpcity – Coimbatore
(c) BHEL – Trichy
(d) Knitting City – Karur
Answer:
(d) Knitting City – Karur

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 11 Tamil Nadu Economy
(a) 4 3 2 1
(b) 1 2 3 4
(c) 4 1 2 3
(d) 3 4 2 1
Answer:
(a) 4 3 2 1

Question 2.
Samacheer Kalvi 11th Economics Solutions Chapter 11 Tamil Nadu Economy
(a) 4 3 2 1
(b) 2 1 4 3
(c) 1 2 3 4
(d) 2 1 3 4
Answer:
(b) 2 1 4 3

Choose the incorrect pair

Question 3.
Samacheer Kalvi 11th Economics Solutions Chapter 11 Tamil Nadu Economy
Answer:
(c) Little japan (iii) Coimbatore

Question 4.
Samacheer Kalvi 11th Economics Solutions Chapter 11 Tamil Nadu Economy
Answer:
(d) Steel City (iv) Erode

Choose the correct statement

Question 5.
(a) Tamil Nadu is the third largest contributor to India’s GDP
(b) Tamil Nadu ranks second in coconut production.
(c) Tamil Nadu ranks first in cement production
(d) Chennai is called as Detroit of Asia.
Answer:
(d) Chennai is called as Detroit of Asia.

Question 6.
(a) The population density of T.N is 480 as per 2011 census,
(b) There are 30 national highways in Tamil Nadu.
(c) Tamil Nadu has the highest installed wind energy capacity in India
(d) The headquarters of southern railway is at Trichy.
Answer:
(c) Tamil Nadu has the highest installed wind energy capacity in India

Choose the incorrect pair

Question 7.
Samacheer Kalvi 11th Economics Solutions Chapter 11 Tamil Nadu
Answer:
(b) Artificial Harbour (ii) Chennai

Question 8.
Samacheer Kalvi 11th Economics Solutions Chapter 11 Tamil Nadu Economy
Answer:
(a) Highest sex ratio (i) The Nilgris

Question 9.
Samacheer Kalvi 11th Economics Solutions Chapter 11 Tamil Nadu Economy
Answer:
(a) Rajapalayam (i) Surgical cotton product

Pick the odd one out

Question 10.
(a) Coimbatore
(b) Erode
(c) Madurai
(d) Trichy
Answer:
(b) Erode

Question 11.
(a) Coconut
(b) Groundnut
(c) Rice
(d) Sugarcane
Answer:
(c) Rice

Choose the incorrect statement

Question 12.
(a) Chennai is referred as banking capital of India.
(b) Tamil Nadu ranks first in maize production.
(c) Tamil Nadu ranks 3rd in human development Index.
(d) Tamil Nadu is the third largest contributor to India’s GDP
Answer:
(d) Tamil Nadu is the third largest contributor to India’s GDP

Question 13.
(a) Tamil Nadu has eight agro climatic zones
(b) There are 17 river basins in Tamil Nadu.
(c) The tertiary sector is the major contributor to Tamil Nadu’s GSDP.
(d) Tamil Nadu is the second largest economy in India.
Answer:
(d) Tamil Nadu is the second largest economy in India.

Fill in the blanks with suitable option given below

Question 14.
Tamilnadu has _______ population as against 121 crore of India
(a) 7.12 crore
(b) 1.72 crore
(c) 7.21 crore
(d) 1.27 crore
Answer:
(c) 7.21 crore

Question 15.
Tamil Nadu’s per capita income is ______
(a) 1100 dollar
(b) 2200 dollar
(c) 3360 dollar
(d) 4400 dollar
Answer:
(b) 2200 dollar

Question 16.
Gross state domestic product is ______
(a) GSDP
(b) MMR
(c) GDP
(d) IMF
Answer:
(a) GSDP

Question 17.
IMR is the number of mortality before completing
(a) 1 year of age
(b) 5 years of age
(c) 3 years of age
(d) 10 years of age
Answer:
(a) 1 year of age

Question 18.
The world’s biggest bagasse based paper mill is at ______
(a) Trichy
(b) Karur
(c) Coimbatore
(d) Erode
Answer:
(b) Karur

Question 19.
Every year India international leather fair is hosted at ______
(a) Coimbatore
(b) Madurai
(c) Chennai
(d) Mumbai
Answer:
(c) Chennai

Question 20.
______ wind farm is a renewable energy source
(a) Nagapattinam
(b) Valapandal
(c) Nagercoil
(d) Muppandal
Answer:
(d) Muppandal

Part – B
Answer the following questions in one or two sentences

Question 1.
How can you calculate per capita income?
Answer:
Percapita income = Samacheer Kalvi 11th Economics Solutions Chapter 11 Tamil Nadu Economy

Question 2.
Define infant mortality rate
Answer:
IMR is the number of mortality before completing 1 year of age.

Question 3.
What is cash – deposit ratio?
Answer:
C-D ratio is the ratio of bank advances to deposits.

Question 4.
What is the position of Tamil Nadu in automotives ?
Answer:
Tamil Nadu has 28% share each in automotive and auto components industries, 19% in the trucks segment and 18% each in passenger cars and two wheelers.

Question 5.
Why is Sivakasi called as Little Japan ?
Answer:
Sivakasi is a leader in the areas of printing, fireworks, and safety matches. It contributes 80% of India’s fireworks production and 60% of India’s total offset printing solution. As there are too many industries in Sivakasi it is called as Little Japan.

Question 6.
Name the cement manufacturing places of Tamil Nadu?
Answer:
Ariyalur, Virudhunagar, Coimbatore and Tirunelveli.

Question 7.
What is the measure of unemployment in Tamil Nadu?
Answer:
Tamil Nadu ranks 22nd with unemployment rate of 42 per 1000.

Part – C
Answer the following questions in one Paragraph

Question 1.
Explain about unemployment and poverty in Tamil Nadu?
Answer:
Unemployment: Tamil Nadu ranks 22nd with unemployment rate of 42 per 1000 as against national average of 50.

Poverty :

  1. Tamil Nadu is one of the India’s richest states.
  2. Since 1994, the state has seen a steady decline in poverty.
  3. Tamil Nadu has lower levels of poverty than most other states in the country.

Question 2.
Explain about water resources of Tamil Nadu.
Answer:
Water resources of Tamil Nadu:

  1. Tamil Nadu is not endowed with rich natural resources compared to other states.
  2. It accounts for three percent of water sources, four percent of land area against six percent of population.
  3. North east monsoon is the major source of rainfall followed by south west monsoon.
  4. There are 17 river basins in Tamil Nadu. The main rivers are Palar, Cheyyar, Ponnaiyar, Cauvery, Bhavani, Vaigai, Chittar, Tamiraparani, Vellar, Noyyal, Siruvani, Gundar, Vaipar etc.
  5. Wells are the largest source of irrigation in Tamil Nadu.

Question 3.
Name the water resources of Tamil Nadu.
Answer:
Water resources of Tamil Nadu:

  1. Ranipet, Ambur, Vaniyambadi – Leather
  2. Salem . – Powerlooms, Home textiles, Steel, Sago
  3. Sankagiri – Lorry fleet operators
  4. Tiruchengode – Borewell drilling services
  5. Naakkal – Transportation Poultry
  6. Karur – Coach-building, Powerlooms
  7. Erode – Powerlooms, Turmeric
  8. Coimbatore – Spinning mills, Engineering industries
  9. Tirupur – Knitwear, Readymade garments
  10. Rajapalayam – Surgical cotton products
  11. Sivakasi – Safety matches, Fireworks, Printing

Part – D
Answer the following questions in about a page

Question 1.
Elaborate the highlights of Tamil Nadu economy.
Answer:
Highlights of Tamil Nadu economy:

  1. Growth of GSDP in Tamil Nadu has been among the fastest in India since 2005.
  2. Poverty reduction is faster than that in many other states.
  3. It contains a smaller proportion of India’s poor population.
  4. Second largest contributor to India’s GDP.
  5. 3rd in Human Development Index.
  6. Ranks 3rd in terms of invested capital and value of total industrial output.
  7. Ranks first among the states in terms of invested capital and value of total industrial output.
  8. Third in health index
  9. Has highest Gross Enrollment Ratio in higher education.
  10. Has the largest number of engineering colleges.
  11. Has emerged as a major hub for renewable energy.
  12. Has highest credit deposit ratio in commercial and cooperative banks.
  13. Ranks first on investment proposals filled by MSMES.

Question 2.
Compare the percapital income of Tamil Nadu with other countries.
Answer:

  1. The percapital GSDP of Tamil Nadu is $ 2200 which is higher than that of many other states in India.
  2. Per capital GSDP of Tamil Nadu is nearly 1.75 times higher than the national average, as per 2018 data.
  3. In term of the per capital income in Tamil Nadu was Rs. 1,03,600 in 2010-11 and it has increased to Rs. 1,88,492 in 2017-18 as per the budget figures 2018.

Samacheer Kalvi 11th Economics Solutions Chapter 11 Tamil Nadu Economy

Question 3.
Explain the role of MSMEs in Tamil Nadu.
Answer:
Role of MSMEs in Tamil Nadu:

  1. The MSME enterprises are classified as manufacturing and service enterprises based on the investment in plant, machinery and equipment MSMED act 2006.
  2. Tamil Nadu accounts of 15,07% micro, small and medium enterpriser which is highest in the country.
  3. There are 6.89 lakhs registered MSMEs producing over 8000 varieties of product for investment more than Rs. 32,008 crore.
  4. The prominent products of MSMEs are the engineering, electrical, chemicals, plastics, steel paper, matches, textiles, hosiery and garments sector.
  5. Around 15.61 lakh entrepreneurs have registered, providing employment opportunities to about 99.7 lakhs persons with total investment of Rs. 1,68,331 crore.

Question 4.
Explain about banks in Tamil Nadu.
Answer:
1. In Tamil Nadu, Nationalised banks account for 52% with 5,337 branches, private commercial banks 30% (3060) branches, state bank of India and its associates 13% (1,364) RRBs 5% (537) branches and the remaining 22 foreign bank branches.

Deposits of the banks :

  1. Total deposits of the banks in Tamil Nadu increased by 14.32% by March 2017 and touched Rs. 6,65,068.59 crores.
  2. Total credits of the banks increased by 13.50% by March 2017 and touched
    Rs. 6,95,500.31 crores.

Advances of the banks :

  1. The share of priority sector advances stands at 45.54% as against the national average of 40%
  2. Primary sector advances stands at 45.54%
  3. Agricultural advances stands at 19.81% on March 2017.
  4. Banks in Tamil Nadu has the highest credit deposit ratio of 119.15% in the country.

If you need some more information about Tamilnadu State Board Solutions for 11th Economics Chapter 11 Tamil Nadu Economy Questions and Answers then visit our site frequently and get all the resources that you look for. Also, share this page with your friends and make them learn all the concepts thoroughly before the exams.

Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell: The Unit of Life

Students who are in search of 11th Bio Botany exam can download this Tamilnadu State Board Solutions for 11th Bio Botany Chapter 6 Cell: The Unit of Life from here for free of cost. These cover all Chapter 6 Cell: The Unit of Life Questions and Answers, PDF, Notes, Summary. Download the Samacheer Kalvi 11th Biology Book Solutions Questions and Answers by accessing the links provided here and ace up your preparation.

Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell: The Unit of Life

Kickstart your preparation by using this Tamilnadu State Board Solutions for 11th Bio Botany Chapter 6 Cell: The Unit of Life Questions and Answers get the max score in the exams. You can cover all the topics of Chapter 6 Cell: The Unit of Life Questions and Answers easily after studying the Tamilnadu State Board 11th Bio Botany Textbook solutions pdf.

Samacheer Kalvi 11th Bio Botany Cell: The Unit of Life Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The two subunits of ribosomes remain united at critical ion level of …………… .
(a) Magnesium
(b) Calcium
(c) Sodium
(d) Ferrous
Answer:
(a) Magnesium

Question 2.
Sequences of which of the following is used to know the phylogeny?
(a) mRNA
(b) rRNA
(c) tRNA
(d) Hn RNA
Answer:
(b) rRNA

Question 3.
Many cells function properly and divide mitotically even though they do not have …………… .
(a) Plasma membrane
(b) Cytoskeleton
(c) Mitochondria
(d) Plastids
Answer:
(d) Plastids

Question 4.
Keeping in view the fluid mosaic model for the structure of cell membrane, which one of the following statements is correct with respect to the movement of lipids and proteins from one lipid monolayer to the other…………… .
(a) Neither lipid nor proteins can flip – flop
(b) Both lipid and proteins can flip – flop
(c) While lipids can rarely flip – flop proteins cannot
(d) While proteins can flip – flop lipids cannot
Answer:
(c) While lipids can rarely flip – flop proteins cannot

Question 5.
Match the columns and identify the correct option:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 1
Answer:
(c) iii, iv, i, ii.

Question 6.
Bring out the significance of phase contrast microscopy.
Answer:
Phase contrast microscope is used to observe living cells, tissues and the cells cultured invitro during mitosis.

Question 7.
State the protoplasm theory.
Answer:
Protoplasm theory was proposed by Max Schultze which states that the protoplasm is the living content of cell and is a complex colloidal system.

Question 8.
Distinguish between prokaryotes and eukaryotes.
Answer:
Between prokaryotes and eukaryotes:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 2

Question 9.
Difference between plant and animal ceil.
Answer:

Plant Cell

Animal Cell

1. Usually they are larger than animal cells1. Usually smaller than plant cells
2. Cell wall present in addition to plasma membrane and consists of middle lamellae, primary and secondary walls2. Cell wall absent
3. Plasmodesmata present3. Plasmodesmata absent
4. Chloroplast present4. Chloroplast absent
5. Vacuole large and permanent5. Vacuole small and temporary
6. Tonoplast present around vacuole6. Tonoplast absent
7. Centrioles absent except motile cells of lower plants7. Centrioles present
8. Nucleus present along the periphery of the cell8. Nucleus at the centre of the cell
9. Lysosomes are rare9. Lysosomes present
10. Synthesizes amino acids, coenzymes and vitamins required by them10. Cannot synthesize aminoacids, coenzymes and vitamins required by them
11. Storage material is starch grains11. Storage material is a glycogen granules

Question 10.
Draw the ultra structure of Plant Cell.
Answer:
The ultra structure of Plant Cell:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 3

Samacheer Kalvi 11th Bio Botany Cell: The Unit of Life Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Scientist who named the unicellular particles as ‘animalcules’ …………… .
(a) Aristotle
(b) Robert Brown
(c) Antonie von Leeuwenhoek
(d) Robert Hooke
Answer:
(c) Antonie van Leeuwenhoek

Question 2.
Cell theory was modified by …………… .
(a) Schwann
(b) Schleiden
(c) Virchow
(d) Dutrochet
Answer:
(c) Virchow

Question 3.
Which of the following microscope produce 3D – image?
(a) Phase contrast
(b) TEM
(c) SEM
(d) Dark field
Answer:
(c) SEM

Question 4.
Which of the following electron opaque chemical is used in Electron microscope?
(a) Strontium
(b) Deuterium
(c) Palladium
(d) Uranium
Answer:
(c) Palladium

Question 5.
Medium for electron movement in TEM is …………… .
(a) Air
(b) Oil
(c) Water
(d) Vacuum
Answer:
(d) Vacuum

Question 6.
Resolving power of SEM is …………… .
(a) 5 – 10 Å
(b) 2 – 10 Å
(c) 5 – 20 nm
(d) 5 – 20 m
Answer:
(c) 5 – 20 nm

Question 7.
Which among the following is NOT an exception to cell theory?
(a) Viruses
(b) Viroids
(c) Prions
(d) Fungi
Answer:
(d) Fungi

Question 8.
Scientist who named the cytoplasm as “Sarcode” is …………… .
(a) Dujardin
(b) Corti
(c) Purkinje
(d) Hugo Van Mohl
Answer:
(a) Dujardin

Question 9.
The pH of protoplasm is around …………… .
(a) 6.6
(b) 6.7
(c) 6.8
(d) 6.9
Answer:
(c) 6.8

Question 10.
The refractive index of protoplasm is …………… .
(a) 1.4
(b) 2.4
(c) 3.4
(d) 0.4
Answer:
(a) 1.4

Question 11.
Histone proteins are seen in the DNA of …………… .
(a) Pseudokaryotes
(b) Prokaryotes
(c) Mesokaryotes
(d) Eukaryotes
Answer:
(d) Eukaryotes

Question 12.
Which of the following organelle is believed to be an endosymbiont?
(a) Ribosomes
(b) Mitochondrion
(c) Golgi bodies
(d) Nucleus
Answer:
(b) Mitochondrion

Question 13.
is the living content of the cell …………… .
(a) Cytoplasm
(b) Protoplasm
(c) Nucleoplasm
(d) Nucleus
Answer:
(b) Protoplasm

Question 14.
Fungal cell wall is made of …………… .
(a) Cutin
(b) Chitin
(c) Hemicellulose
(d) Pectin
Answer:
(b) Chitin

Question 15.
…………… acts as a channel between the protoplasm of adjacent cells.
(a) Middle lamellae
(b) Pits
(c) Plasmodesmata
(d) Primary cell wall
Answer:
(c) Plasmodesmata

Question 16.
Fluid mosaic model was proposed by …………… .
(a) Schleiden and Schwann
(b) Singer and Nicolson
(c) Binning and Roher
(d) G. Palade
Answer:
(b) Singer and Nicolson

Question 17.
Which is the largest of the internal membranes?
(a) Golgi bodies
(b) Endoplasmic reticulum
(c) Tonoplast
(d) Nuclear membrane
Answer:
(b) Endoplasmic reticulum

Question 18.
In plant cells, golgi bodies are found as small vesicles called …………… .
(a) Polysomes
(b) Cytosomes
(c) Cytosol
(d) Dictyosomes
Answer:
(d) Dictyosomes

Question 19.
organelle plays a major role in post translation process of protein …………… .
(a) Golgi bodies
(b) Nucleolus
(c) Ribosomes
(d) ER
Answer:
(a) Golgi bodies

Question 20.
Zymogen granules are synthesized in …………… .
(a) Lysosomes
(b) Golgi bodies
(c) Mitochondria
(d) Chloroplast
Answer:
(b) Golgi bodies

Question 21.
Altmann named mitochondrion as …………… .
(a) Apoplast
(b) Elaioplast
(c) Symplast
(d) Bioplast
Answer:
(d) Bioplast

Question 22.
DNA of mitochondrion is …………… .
(a) Helical
(b) Dumb – bell
(c) Circular
(d) Spiral
Answer:
(c) Circular

Question 23.
Mitochondria are inherited from parent …………… .
(a) Male
(b) Female
(c) Both
(d) None
Answer:
(b) Female

Question 24.
F1 particles are also called as …………… .
(a) Polysomes
(b) Glyoxysomes
(c) Peroxisomes
(d) Oxysomes
Answer:
(d) Oxysomes

Question 25.
Elaioplasts store …………… .
(a) Starch
(b) Lipid
(c) Protein
(d) Chlorophyll
Answer:
(b) Lipid

Question 26.
The photosynthetic units are called as …………… .
(a) Oxysomes
(b) Quantosomes
(c) Thylakoids
(d) Chloroplasts
Answer:
(b) Quantosomes

Question 27.
Which organelle is not membrane bound?
(a) Mitochondrion
(b) Golgi bodies
(c) Chloroplast
(d) Ribosomes
Answer:
(d) Ribosomes

Question 28.
Ribosomes of mitochondrion are …………… .
(a) 16 S
(b) 80 S
(c) 70 S
(d) 50 S
Answer:
(c) 70 S

Question 29.
…………… mineral is required for structural cohesion of ribosomes.
(a) Ca2+
(b) H+
(c) Mg2+
(d) Cl
Answer:
(c) Mg2+

Question 30.
Lysosomes originate from …………… .
(a) Mitochondrion
(b) Nucleus
(c) ER
(d) Golgi bodies
Answer:
(d) Golgi bodies

Question 31.
In mammals, peroxisomes are seen in …………… cells.
(a) Brain
(b) Lung
(c) Liver
(d) Heart
Answer:
(c) Liver

Question 32.
Which organelle has a single unit membrane?
(a) Ribosomes
(b) Glyoxysomes
(c) Polysomes
(d) Nucleus
Answer:
(b) Glyoxysomes

Question 33.
The single unit membrane of vacuoles is called as …………… .
(a) Tonoplast
(b) Symplast
(c) Apoplast
(d) Amyloplast
Answer:
(a) Tonoplast

Question 34.
Vacuoles of Apple cells store …………… .
(a) Sucrose
(b) Malic acid
(c) Citrate
(d) Flavanoid
Answer:
(b) Malic acid

Question 35.
Ribosomal biogenesis occur at …………… .
(a) Mitochondrion
(b) Polysomes
(c) Nucleolus
(d) Chromosomes
Answer:
(c) Nucleolus

Question 36.
The term chromosome was introduced by …………… .
(a) Bridges
(b) Strasburger
(c) Waldeyer
(d) Poster
Answer:
(c) Waldeyer

Question 37.
Stability to chromosome is offered by …………… .
(a) Satellite
(b) Telomere
(c) Kinetochore
(d) Nucleolus
Answer:
(b) Telomere

Question 38.
Life span of the cells is determined by …………… .
(a) Kinetochore
(b) Satellite
(c) Chromatin
(d) Telomere
Answer:
(d) Telomere

Question 39.
The metacentric chromosomes are …………… shaped.
(a) L
(b) V
(c) J
(d) I
Answer:
(b) V

Question 40.
Polytene chromosomes are observed in …………… of Drosophila.
(a) Endocrine gland
(b) Gall bladder
(c) Salivary gland
(d) Exocrine gland
Answer:
(c) Salivary gland

Question 41.
Lampbrush chromosomes occur at …………… stage of meiotic Prophase I.
(a) Leptotene
(b) Diplotene
(c) Zygotene
(d) Pachytene
Answer:
(b) Diplotene

Question 42.
Number of basal rings in gram positive bacteria …………… .
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(a) 2

Question 43.
Microtubules are made of …………… .
(a) Dyenin
(b) Tubin
(c) Tubulin
(d) Nexin
Answer:
(c) Tubulin

Question 44.
Cytoplasm is stained …………… by eosin.
(a) Pink
(b) Blue
(c) Greenish blue
(d) Green
Answer:
(a) Pink

Question 45.
Key difference between plant cell & animal cell is …………… .
(a) Ribosomes
(b) Vacuoles
(c) Cell wall
(d) Centrioles
Answer:
(c) Cell wall

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Name the scientist who proposed the cell theory.
Answer:
Matthias Schleiden and Theodor Schwann.

Question 2.
Define resolving power of a microscope.
Answer:
Resolving power or resolution refers to the ability of the lenses to show the details of object lying between two points. It is the finest detail available from an object. It can be calculated using the following formula.
⇒ Resolution = \(\frac {0.61λ}{NA}\),
where λ, = wavelength of the light and
NA is the numerical aperture.

Question 3.
Define magnification. How will you calculate it?
Answer:
The optical increase in the size of an image is called magnification. It is calculated by the following formula
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 4

Question 4.
List out the types of objective lenses used in Bright field microscope.
Answer:
The types of objective lenses used in Bright field microscope:

  1. 5x
  2. 10x
  3. 45x
  4. 100x

Question 5.
In Bright field microscope, where does the primary & secondary magnification occurs?
Answer:
Primary magnification is obtained through, objective lens and secondary magnification is obtained through eye piece lens.

Question 6.
List out the components of Electron microscope.
Answer:
The components of the microscope are as follows:

  1. Electron Generating System
  2. Electron Condensor
  3. Specimen Objective
  4. Tube Lens
  5. Projector

Question 7.
Name the organisms that are exceptions to cell theory.
Answer:
Viruses, Virions and Prions.

Question 8.
Name the types of cells based on nuclear characteristics.
Answer:
The types of cells based on nuclear characteristics:

  1. Prokaryotes
  2. Mesokaryotes and
  3. Eukaryotes.

Question 9.
Define nucleoid.
Answer:
Nucleoid refers to the primitive nucleus which is not bound by nuclear membrane and the DNA is without histone protein.

Question 10.
Name the organelles which are believed to be endosymbiont in Eukaryote cell.
Answer:
Chloroplast and Mitochondrion.

Question 11.
Write a note on endosymbiont theory.
Answer:
Two eukaryotic organelles believed to be the descendants of the endosymbiotic prokaryotes. The ancestors of the eukaryotic cell engulfed a bacterium and the bacteria continued to function inside the host cell.

Question 12.
Point out any four prokaryotes.
Answer:Four prokaryotes:

  1. Bacteria
  2. Blue Green algae
  3. Mycoplasma and
  4. Rickettsiae.

Question 13.
Why spirochaetae is said to be a prokaryote?
Answer:
Spirochaetae is a prokaryote since its DNA lies in nucleoid and not bound by nuclear membrane and also devoid of histone protein.

Question 14.
Name any two unique structures / organelles of a plant cell.
Answer:
Cell wall and Chloroplast.

Question 15.
What are the components of protoplasm?
Answer:
Protoplasm is composed of a mixture of small particles, such as ions, amino acids, monosaccharides, water, macro-molecules like nucleic acids, proteins, lipids and polysaccharides.

Question 16.
What is the cell wall composition of the following organism?
(a) Fungi
(b) Bacteria
(c) Algae
Answer:
(a) Fungi – Chitin and fungal cellulose.
(b) Bacteria – Peptidoglycan
(c) Algae – Cellulose, mannan and galactan.

Question 17.
Which cell wall layer of plant cell is laid during maturation? Mention its role.
Answer:
Secondary wall is laid during maturation. It plays a key role in determining the shape of a cell.

Question 18.
How pits are formed?
Answer:
In plant cell, at few regions the secondary wall layer is laid unevenly, whereas the primary wall and middle lamellae are laid continuously such regions are called pits.

Question 19.
What is the role of plasmodesmata in a plant cell?
Answer:
Plasmodesmata act as a channel between the protoplasm of adjacent cells through which many substances pass through.

Question 20.
Name the two types of Pits.
Answer:
The two types of Pits:

  1. Simple pit and
  2. Bordered pit.

Question 21.
Define flip – flop movement.
Answer:
The movement of membrane lipids from one side of the membrane to the other side by vertical movement is called flip – flopping or flip – flop movement.

Question 22.
Name the two types of protein seen in cell membrane.
Answer:
Two types of protein seen in cell membrane:

  1. Integral proteins and
  2. peripheral proteins.

Question 23.
Define cytoplasmic streaming.
Answer:
Cytoplasmic streaming refers to the movement of the cytoplasm along with the cellular materials inside the cell.

Question 24.
Name any four endomembrane structures.
Answer:
Four endomembrane structures:

  1. Nuclear membrane
  2. endoplasmic reticulum
  3. vascular membrane
  4. golgi bodies.

Question 25.
What are dictyosomes?
Answer:
In plant cells, the golgi bodies are found as small vesicles which are called as dictyosomes.

Question 26.
What are porins?
Answer:
Porins are the proteins in the outer membrane of mitochondrion which forms the channel that allows the free diffusion of molecules smaller than about 1000 daltons.

Question 27.
What are oxysomes?
Answer:
The inner membrane of mitochondrion consists of stalked particles called elementary particles or Fernandez Moran particles. F1 particles or oxysomes.

Question 28.
How mitochondria helps in evolutionary studies?
Answer:
Mitochondrial DNA is used to track and date recent evolutionary time because it mutates 5 to 10 time faster than DNA in the nucleus.

Question 29.
Why mitochondrion is called as semi – autonomous organelle?
Answer:
Since mitochondria has it own DNA, it is called as Organelle.

Question 30.
Why mitochondria are called “Power houses of cell?
Answer:
Mitochondria are called Power house of a cell, as they produce energy rich (Adenosine Triphosphate) ATP.

Question 31.
Classify plastids based on colour.
Answer:
Chloroplast, Phaeoplast and Rhodoplast.

Question 32.
Classify plastids based on storage & mention their storage component.
Answer:

Plastids

Storage Content

1. Amyloplast1. Starch
2. Elaioplast2. Lipids / Oils
3. Aleuroplast or Proteoplast3. Protein

Question 33.
What are Quantosomes?
Answer:
Quantosomes are small, rounded photosynthetic units present in thylakoids.

Question 34.
List out the functions of chloroplast.
Answer:
Photosynthesis, photo respiration.

Question 35.
What is Svedberg unit?
Answer:
Svedberg unit is a measure of a particle size dependent on the speed with which particle sediment in the ultracentrifuge.

Question 36.
Where the biogenesis of ribosomes occur?
Answer:
Biogenesis of ribosome are denova formation, auto replication and nucleolar origin.

Question 37.
What are Polysomes? State its function.
Answer:
During protein synthesis many ribosomes are attached to the single mRNA called polysomes or polyribosomes. The function of polysomes is the formation of several copies of a particular polypeptide during protein synthesis.

Question 38.
Define Autolysis.
Answer:
Autolysis is the process, where the lysosome causes self destruction of cell on insight of disease they destroy the cells.

Question 39.
List out the enzymes of lysosomes.
Answer:
Acid Hydrolases, Nuclease, Proteases, Glycosidases, Lipases, Phosphatases, Sulphatases and Phospholipidases.

Question 40.
Name any two organelles involved in photorespiration.
Answer:
Peroxisomes and Chloroplast.

Question 41.
Name few single unit membrane bound organelles.
Answer:
Peroxisomes, Glyoxysomes and Sphaerosomes.

Question 42.
What are Tonoplast?
Answer:
In plant cells vacuoles are large, bounded by a single unit membrane called Tonoplast.

Question 43.
How vacuoles helps to maintain the structure of a plant cell?
Answer:
The major function of plant vacuole is to maintain water pressure known as turgor pressure, which maintains the plant structure.

Question 44.
What are metachromatic granules?
Answer:
Inorganic inclusions in bacteria are polyphosphate granules (volutin granules) and sulphur granules. These granules are also known as metachromatic granules.

Question 45.
Which is the largest organelle in a cell? State its function.
Answer:
Nucleus. It controls all the cellular activities.

Question 46.
What is a pore complex?
Answer:
The pores are enclosed by circular structures called annuli. The pore and annuli forms the pore complex.

Question 47.
What are nucleolar organizers region?
Answer:
Secondary constrictions contains the genes for ribosomal RNA which induce the formation of nucleoli and are called nucleolar organizer regions.

Question 48.
Draw the types of chromosomes based on centromere position.
Answer:
The types of chromosomes based on centromere position:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 5

Question 49.
What is a telomere?
Answer:
Telomere is the terminal part of chromosome. It offers stability to the chromosome. DNA of the telomere has specific sequence of nucleotides.

Question 50.
Classify chromosomes based on function.
Answer:
Autosomes and sex chromosomes.

Question 51.
Name any two giant chromosomes.
Answer:
Polytene chromosomes and lamp brush chromosomes.

Question 52.
Define proton motive force.
Answer:
The proton motive force is the force derived from the electrical potential and the hydrogen ion gradient across the cytoplasmic membrane.

Question 53.
Name the basal rings found in flagella of gram negative bacteria.
Answer:
The basal rings found in flagella of gram negative bacteria:

  • L – Lipopolysacchride ring
  • P – Peptidoglycan ring
  • S – Super membrane ring and
  • M – membrane ring.

Question 54.
(a) Define histochemistry.
(b) What is Somatic pairing?
Answer:
(a) The technique of staining the cells and tissue is called histochemical staining or histo chemistry.
(b) Maternal and paternal homologues remain associated side by side is called somatic pairing.

III. Short Answer Type Questions (3 Marks)

Question 1.
Explain the principle involved in Scanning Electron Microscope (SEM).
Answer:
In Scanning Electron Microscope (SEM) electrons are focused by means of lenses into a very fine point. The interaction of electrons with the specimen results in the release of different forms of radiation (such as auger electrons, secondary electrons, back scattered electrons) from the surface of the specimen. These radiations are then captured by an appropriate detector, amplified and then imaged on fluorescent screen.

Question 2.
Write a note on solation gelation property of protoplasm.
Answer:
The protoplasm exist either in semisolid (jelly – like) state called ‘gel’ due to suspended particles and various chemical bonds or may be liquid state called ‘sol’. The colloidal protoplasm which is in gel form can change into sol form by solation and the sol can change into gel by gelation. These gel – sol conditions of colloidal system are prime basis for mechanical behaviour of cytoplasm.

Question 3.
Explain the nuclear characters of Mesokaryotes.
Answer:
Mesokaryotes contains well organized nucleus with nuclear membrane and the DNA is organized into chromosomes but without histone protein components divides through amitosis similar with prokaryotes.

Question 4.
List out the unique characters of a Eukaryotic cell.
Answer:
Eukaryotes have true nucleus. The DNA is associated with protein bound histones forming the chromosomes. Membrane bound organelles are present.

Question 5.
Name the chemicals seen in the cell wall of plant cells.
Answer:
Cellulose, hemicellulose, pectin, lignin, cutin, suberin and silica.

Question 6.
Name the 3 distinct regions of plant cell wall.
Answer:
In plant, cell wall shows three distinct regions

  1. Primary wall
  2. Secondary wall and
  3. Middle lamellae.

Question 7.
Explain the role of hemicellulose, pectin & glycoprotein in primary cell wall.
Answer:
Hemicellulose binds the microfibrils with matrix and glycoproteins control the orientation of microfibrils while pectin serves as filling material of the matrix.

Question 8.
In cell membrane, phospholipids undergo flip – flop movement but not the protein. Why?
Answer:
The phospholipids can have flip – flop movement because the phospholipids have smaller polar regions, whereas the proteins cannot flip – flop because the polar region is extensive.

Question 9.
Define signal transduction.
Answer:
The process by which the cells receive information from outside and respond is called signal transduction.

Question 10.
Distinguish between rough endoplasmic reticulum and smooth endoplasmic reticulum.
Answer:
Between rough endoplasmic reticulum and smooth endoplasmic reticulum:

RER

SER

1. Ribosomes are present in the outer surface of membrane1. Ribosomes are absent on the membrane
2. It is involved in protein synthesis2. It is involved in lipid synthesis

Question 11.
Write the major roles of Golgi bodies.
Answer:
Golgi complex plays a major role in post translational modification of proteins and glycosidation of lipids.

Question 12.
Which is the most abundant protein on Earth? Where it is encoded?
Answer:
Rubisco is the abundant protein on Earth. It is encoded by the chloroplast DNA.

Question 13.
Classify ribosomes with an example.
Answer:
Types of Ribosomes
1. 70S Ribosomes (subunit 30S and 50S): 3 RNA molecule

  • 16SrRNA in 30S subunit
  • 23S and 5S in 50S large subunit
    (Prokaryotic cells of Blue green algae, Bacteria, Mitochondria and Chloroplast of many Algae and higher plants)

2. 80S Ribosomes (subunits 40S and 60S): 4 RNA molecule

  • 18Sr RNA in 40S small sub unit
  • 28S, 5.8S and 5S in larger 60S subunit.
    (Eukaryotic cells of plants and animals).

Question 14.
Write a note on Glyoxysomes.
Answer:
Glyoxysome is a single membrane bound organelle. It is a sub cellular organelle and contains enzymes of glyoxylate pathway – oxidation of fatty acid occurs in glyoxysomes of germinating seeds.

Question 15.
What are cell inclusions? Give example.
Answer:
The cell inclusions are the non – living materials present in the cytoplasm. They are organic and inorganic compounds.  Example: Phosphate granules.

Question 16.
Draw and label the structure of Nucleus.
Answer:
The structure of Nucleus:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 6Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 7

Question 17.
Distinguish between Euchromatin & Heterochromatin.
Answer:
The portion of Eukaryotic chromosome which is transcribed into mRNA contains active genes that are not tightly condensed during interphase is called Euchromatin. The portion of a Eukaryotic chromosome that is not transcribed into mRNA which remains condensed during interphase and stains intensely is called Heterochromatin.

Question 18.
Define Kinetochore.
Answer:
The centromere contains a complex system of protein fibres called kinetochore. Kinetochore is the region of chromosome which is attached to the spindle fibre during mitosis.

Question 19.
Write a note on SAT – chromosome.
Answer:
A satellite or SAT – chromosome are short chromosomal segment or rounded body separated from main chromosome by a relatively elongated secondary constriction. It is a morphological entity in certain chromosomes.

Question 20.
How telomeres helps in cancer studies?
Answer:
Maintenance of telomeres appears to be an important factor in determining the life span and reproductive capacity of cells so studies of telomeres and telomerase have the promise of providing new insights into conditions such as ageing and cancer.

Question 21.
Name the three types of centromere in eukaryotes.
Answer:
The three types of centromere in eukaryotes:

  1. Point centromere
  2. Regional centromere and
  3. Holocentromere.

Question 22.
What are giant chromosomes?
Answer:
These chromosomes are larger in size and are called giant chromosomes. In certain plants they are found in the suspensors of the embryo. The polytene chromosome and lamp brush chromosome occur in animals and are also called as giant chromosomes.

Question 23.
How polyteny condition is achieved?
Answer:
Polyteny is achieved by repeated replication of chromosomal DNA several times without nuclear division and the daughter chromatids aligned side by side and do not separate (endomitosis).

Question 24.
What is Microphotography?
Answer:
Images of structures observed through microscopes can be further magnified, projected and saved by attaching a camera to the microscope by a microscope coupler or eyepiece adaptor. Picture taken using a inbuilt camera in a microscope is called microphotography or microphotograph.

Question 25.
Draw the Structure of Peroxisomes.
Answer:
The Structure of Peroxisomes:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 7

IV. Long Answer Type Questions (5 Marks)

Question 1.
Write a detailed account on Dark field microscope.
Answer:
The dark field microscope was discovered by Z. Sigmondy (1905). Here the field will be dark but object will be glistening so the appearance will be bright. A special effect in an ordinary microscope is brought about by means of a special component called “Patch Stop Carrier”. It is fixed in metal ring of the condenser component.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 8
Patch stop is a small glass device which has a dark patch at centre of the disc leaving a small area along the margin through which the light passes. The light passing through the margin will travel oblique like a hollow cone and strikes the object in the periphery, therefore the specimen appears glistening in a dark background.

Question 2.
Compare Transmission Electron Microscope with Scanning Electron Microscope.
Answer:
Transmission Electron Microscope with Scanning Electron Microscope:
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 9

Question 3.
List out the features of Cell Doctrine.
Answer:
The features of cell doctrine are as follows:

  1. All organisms are made up of cells.
  2. New cells are produced from the pre – existing cells.
  3. Cell is a structural and functional unit of all living organisms.
  4. A cell contains hereditary information which is passed on from cell to cell during cell division.
  5. All the cells are basically the same in chemical composition and metabolic activities.
  6. The structure and function of cell is controlled by DNA.
  7. Sometimes the dead cells may remain functional as tracheids and vessels in plants and horny cells in animals.

Question 4.
Enumerate the properties of protoplasm.
Answer:
The properties of protoplasm:

  1. Protoplasm is translucent, odourless and polyphasic fluid.
  2. It is a crystal colloid solution which is a mixture of chemical substances forming crystalloid and colloidal solution.
  3. Protoplasm exhibits three Brownian movement, amoeboid movement and cytoplasmic streaming or cyclosis. Viscosity of protoplasm is 2 – 20 centipoises. The Refractive index of the protoplasm is 1.4.
  4. The pH of the protoplasm is around 6.8, contain 90% water (10% in dormant seeds).
  5. Approximately 34 elements are present in protoplasm but only 13 elements are main or universal elements i.e. C, H, O, N, Cl, Ca, P, Na, K, S, Mg, I and Fe. Carbon, Hydrogen, Oxygen and Nitrogen form the 96% of protoplasm.
  6. Protoplasm is neither a good nor a bad conductor of electricity.
  7. Cohesiveness: Particles or molecules of protoplasm are adhered with each other by forces, such as Van der Waal’s bonds, that hold long chains of molecules together. This property varies with the strength of these forces.
  8. Contractility: The contractility of protoplasm is important for the absorption and removal of water especially stomatal operations.
  9. Surface Tension: The proteins and lipids of the protoplasm have less surface tension, hence they are found at the surface forming the membrane. On the other hand the chemical substances (NaCl) have high surface tension, so they occur in deeper parts of the cell protoplasm.

Question 5.
List out the functions of the Cell Wall.
Answer:
The cell wall plays a vital role in holding several important functions given below.

  1. Offers definite shape and rigidity to the cell.
  2. Serves as barrier for several molecules to enter the cells.
  3. Provides protection to the internal protoplasm against mechanical injury.
  4. Prevents the bursting of cells by maintaining the osmotic pressure.
  5. Plays a major role by acting as a mechanism of defense for the cells.

Question 6.
Explain in detail about Fluid mosaic model.
Answer:
Jonathan Singer and Garth Nicolson (1972) proposed fluid model: It is made up of lipids and proteins together with a little amount of carbohydrate. The lipid membrane is made up of phospholipid. The phospholipid molecule has a hydrophobic tail and hydrophilic head. The hydrophobic tail repels water and water loving polar molecule are called hydrophilic molecule. They have polar phosphate group responsible for attracting water. Water hating non – polar molecule are called as hydrophobic molecule. They have fatty acid which is non – polar which cannot attract water.

Hydrophilic head attracts water. The proteins of the membrane are globular proteins which are found intermingled between the lipid bilayer most of which are projecting beyond the lipid bilayer. These proteins are called as integral proteins. Few are superficially attached on either surface of the lipid bilayer which are called as peripheral proteins. The proteins are involved in transport of molecules across the membranes and also act as enzymes, receptors or antigens.

Question 7.
Describe Endocytosis and Exocytosis.
Answer:
Cell surface membrane are able to transport individual molecules and ions. There are processes in which a cell can transport a large quantity of solids and liquids into cell endocytosis or out of cell exocytosis.

1. Endocytosis: During endocytosis the cell wraps the cell surface membrane around the material and brings it into cytoplasm inside a vesicle. There are two types of endocytosis:

  • Phagocytosis – Particle is engulfed by membrane, which folds around it and forms a vesicle. The enzymes digest the material and products are absorbed by cytoplasm.
  • Pinocytosis – Fluid droplets are engulfed by membrane, which forms vesicles around them.

2. Exocytosis: Vesicles fuse with plasma membrane and eject contents. This passage of material out the cell is known as exocytosis. This material may be a secretion in the case of digestive enzymes, hormones or mucus.

Question 8.
List out the functions of Golgi bodies.
Answer:
Functions of Golgi bodies:

  1. Glycoproteins and glycolipids are produced.
  2. Transporting and storing lipids.
  3. Formation of lysosomes.
  4. Production of digestive enzymes.
  5. Cell plate and cell wall formation
  6. Secretion of carbohydrates for the formation of plant cell walls and insect cuticles.
  7. Zymogen granules (proenzyme / pre – cursor of all enzyme) are synthesized.

Question 9.
Draw & describe the structure of mitochondrion.
Answer:
Mitochondria are ovoid, rounded, rod shape and pleomorphic structures. Mitochondrion consists of double membrane, the outer and inner membrane. The outer membrane is smooth, highly permeable to small molecules and it contains proteins called Porins, which form channels that allows free diffusion of molecules smaller than about 1000 daltons and the inner membrane divides the mitochondrion into two compartments, outer chamber between two membranes and the inner chamber filled with matrix.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 10
The inner membrane is convoluted (infoldings), called crista (plural: cristae). Cristae contain most of the enzymes for electron transport system. Inner chamber of the mitochondrion is filled with proteinaceous material called mitochondrial matrix. The inner membrane consists of stalked particles called elementary particles or Fernandez Moran particles. F1 particles or Oxysomes. Each particle consists of a base, stem and a round head. In the head ATP synthase is present for oxidative phosphorylation. Inner membrane is impermeable to most ions, small molecules and maintains the proton gradient that drives oxidative phosphorylation.

Question 10.
Draw & describe the structure of Nucleus.
Answer:
Nucleus is an important unit of cell which control all activities of the cell. Nucleus holds the hereditary information. It is the largest among all cell organelles. It may be spherical, cuboidal, ellipsoidal or discoidal. It is surrounded by a double membrane structure called nuclear envelope, which has the inner and outer membrane. The inner membrane is smooth without ribosomes and the outer membrane is rough by the presence of ribosomes and is continues with irregular and infrequent intervals with the endoplasmic reticulum.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 6
The membrane is perforated by pores known as nuclear pores which allows materials such as mRNA, ribosomal units, proteins and other macromolecules to pass in and out of the nucleus. The pores enclosed by circular structures called annuli. The pore and annuli forms the pore complex. The space between two membranes is called perinuclear space. Nuclear space is filled with nucleoplasm, a gelatinous matrix has uncondensed chromatin network and a conspicuous nucleoli. The chromatin network is the uncoiled, indistinct and remain thread like during the interphase. It has little amount of RNA and DNA bound to histone proteins in eukaryotic cells.

Question 11.
Write in detail about the 3 types of centromere in eukaryotes.
Answer:
There are three types of centromere in Eukaryotes. They are as follows:

  1. Point Centromere: The type of centromere in which the kinetochore is assembled as a result of protein recognition of specific DNA sequences. Kinetochores assembled on point centromere bind a single microtubule. It is also called as localized centromere. It occurs in budding yeasts.
  2. Regional Centromere: In regional centromere where the kinetochore is assembled on a variable array of repeated DNA sequences. Kinetochore assembled on regional centromeres bind multiple microtubules. It occurs in fission yeast cell, humans and so on.
  3. Holocentromere: The microtubules bind all the along the mitotic chromosome. Example: Caenorbabditis elegans (transparent nematode) and many insects.

Question 12.
List the functions of Nucleus.
Answer:
Functions of the Nucleus:

  1. Controlling all the cellular activities
  2. Storing the genetic or hereditary information
  3. Coding the information in the DNA for the production of enzymes and proteins.
  4. DNA duplication and transcription takes place in the nucleus.
  5. In nucleolus ribosomal biogenesis takes place.

Question 13.
Explain the structure and movement of Eukaryotic flagella.
Answer:
Structure: Eukaryotic Flagella are enclosed by unit membrane and it arises from a basal body. Flagella is composed of outer nine pairs of microtubules with two microtubules in its centre (9 + 2 arrangement). Flagella are microtubule projection of the plasma membrane. Flagellum is longer than cilium (as long as 200 m). The structure of flagellum has an axoneme made up microtubules and protein tubulin.
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 11
Movement: Outer microtubule doublet is associated with axonemal dynein which generates force for movement. The movement is ATP driven. The interaction between tubulin and dynein is the mechanism for the contraction of cilia and flagella. Dynein molecules uses energy from ATP to shift the adjacent microtubules. This movement bends the cilium or flagellum.

Question 14.
Describe the steps involved in cytologieal techniques.
Answer:
There are different types of mounting based on the portion of a specimen to be observed.

  1. Whole mount: The whole organism or smaller structure is mounted over a slide and observed.
  2. Squash: Is a preparation where the material to be observed is crushed/squashed onto a slide so as to reveal their contents. Example: Pollen grains, mitosis and meiosis in root tips and flower buds to observe chromosomes.
  3. Smears: Here the specimen is in the fluid (blood and microbial cultures etc) are scraped, brushed or aspirated from surface of organ. Example: Epithelial cells.
  4. Sections: Free hand sections from a specimen and thin sections are selected, stained and mounted on a slide. Example: Leaf and stem of plants.

Question 15.
Name any 5 common stains their colour & their affinity used in cytologieal studies.
Answer:
Common Stains used in Histochemistry
Samacheer Kalvi 11th Bio Botany Solutions Chapter 6 Cell The Unit of Life 12

V. Higher Order Thinking Skills (HOTs)

Question 1.
What makes the plant cell more rigid than animal cells?
Answer:
Plants cells posses cell wall which provides sufficient rigidity and proper shape to them whereas in case of animal cells, cell wall is totally absent.

Question 2.
Cleaning organelle in the cell – Explain.
Answer:
Lysosomes contains a variety of hydrolytic enzymes, which can digest the materials within the cell. Thus lysosomes act as cleaning organelle of the cell.

Question 3.
Ribosomes are single membrane organelles present in both prokaryotes & eukaryotes. List out the sites where ribosomes are present in plant cell.
Answer:
The sites where ribosomes are present in plant cell:

  1. Cytoplasm
  2. On rough endoplasmic reticulum
  3. Inside mitochondira & chloroplast.

Question 4.
What does ‘S’ refer in a 70 S and an 80 S ribosomes?
Answer:
‘S’ refers to Svedberg units named after Theoder Svedberg. The size of ribosome and their subunits are represented by Svedberg unit.

Question 5.
Briefly give the contributions of the following scientists in the field of cytology.
(a) Schleiden and Schwann
(b) Singer and Nicolson
Answer:
(a) Schleiden and Schwann proposed the cell theory.
(b) Singer & Nicolson proposed the fluid mosaic model of plasmomembrane.

Question 6.
Is extra genomic DNA is present in prokaryotes & Eukaryotes? If yes, locate them in both the types of organisms.
Answer:
Locate them in both the types of organisms:

  1. In prokaryotes like bacteria, plasmids are the extra genomic DNA present in cytoplasm.
  2. In Eukaryotes, the circular DNA present in matrix of mitochondria & chloroplast are extragenomic DNA.

Question 7.
Give possible reasons to call mitochondria & chloroplast as semi – autonomous organelles.
Answer:
Mitochondria & chloroplasts are considered as semi – autonomoius organelle due to following
facts.

  1. Both mitochondria & chloroplasts have their own DNA, which can replicate independently.
  2. They have their own ribosomes by which they self-synthesize same of their proteins without depending on cellular DNA.

Believing that the Samacheer Kalvi 11th Biology Book Solutions Questions and Answers learning resource will definitely guide you at the time of preparation. For more details about Tamilnadu State 11th Bio Botany Chapter 6 Cell: The Unit of Life textbook solutions, ask us in the below comments and we’ll revert back to you asap.

Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy

Immediately get Chapter Wise Tamilnadu State Board Solutions for 11th Economics to gain more marks and start Quick Revision of all Chapters. All Chapters Pdf is provided along with the Questions and Answers. You Can Dowload Samacheer Kalvi 11th Economics Book Solutions Questions and Answers for Chapter wise are given with a clear explanation. Tamilnadu State Board Solutions for 11th Economics Chapter 10 Rural Economy Questions and Answers is for free of cost.

Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy

Tamilnadu State Board Solutions for 11th Economics Chapter 10 Rural Economy Questions and Answers PDF has all given in Chapter Wise Section. Check Out daily basis with Tamilnadu State Board Solutions 11th Economics PDF will help to improve your score. Improve your level of accuracy to answer a question by reading with Samacheer Kalvi 11th Economics Book Solutions Questions and Answers PDF.

Samacheer Kalvi 11th Economics Rural Economy Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
Which is considered as the basic unit for rural areas?
(a) Panchayat
(b) Village
(c) Town
(d) Municipality
Answer:
(b) Village

Question 2.
Which feature is identified with rural areas?
(a) Low population density
(b) High population density
(c) Low natural resources
(d) Low human resources
Answer:
(a) Low population density

Question 3.
Identify the feature of rural economy.
(a) Dependence on agriculture
(b) High population density
(c) Low level of population
(d) Low level of inequality
Answer:
(a) Dependence on agriculture

Question 4.
What percentage of the total population live in rural area, as per 2011 censes?
(a) 40
(b) 50
(c) 68.84
(d) 70
Answer:
(c) 68.84

Question 5.
How do you term people employed in excess over and above the requirements?
(a) Unemployment
(b) Underemployment or Disguised Unemployment
(c) Full employment
(d) Self-employment
Answer:
(b) Underemployment or Disguised Unemployment

Question 6.
What is the term used to denote the coexistence of two different features in an economy?
(a) Technology
(b) Dependency
(c) Dualism
(d) Inequality
Answer:
(c) Dualism

Question 7.
The process of improving the rural areas, rural people and rural living is defined as
(a) Rural economy
(b) Rural economics
(c) Rural employment
(d) Rural development
Answer:
(d) Rural development

Question 8.
Identify the agriculture related problem of rural economy.
(a) Poor communication
(b) Small size of landholding
(c) Rural poverty
(d) Poor banking network
Answer:
(b) Small size of landholding

Question 9.
The recommended nutritional intake per person in rural areas.
(a) 2100 calories
(b) 2100 calories
(c) 2300 calories
(d) 2400 calories
Answer:
(d) 2400 calories

Question 10.
Indicate the cause for rural poverty.
(a) Lack of non-farm employment
(b) High employment
(c) Low inflation rate
(d) High investment.
Answer:
(a) Lack of non-farm employment

Question 11.
What is the other name for concealed unemployment?
(a) Open
(b) Disguised
(c) Seasonal
(d) Rural
Answer:
(b) Disguised

Question 12.
How do you term the employment occurring only on a particular season?
(a) Open
(b) Disguised
(c) Seasonal
(d) Rural
Answer:
(c) Seasonal

Question 13.
Identify an example for rural industries?
(a) Sugar factory
(b) Mat making industry
(c) Cement industry
(d) Paper industry
Answer:
(b) Mat making industry

Question 14.
How much share of rural families in India is in debt?
(a) Half
(b) One fourth
(c) Two third
(d) Three fourth
Answer:
(d) Three fourth

Question 15.
Identify the cause for rural indebtedness in India.
(a) Poverty
(b) High population
(c) High productivity
(d) Full employment
Answer:
(a) Poverty

Question 16.
In which year, Regional Rural Banks came into existence?
(a) 1965
(b) 1970
(c) 1975
(d) 1980
Answer:
(c) 1975

Question 17.
Identify the year of launch of MUDRA Bank?
(a) 1995
(b) 2000
(c) 2010
(d) 2015
Answer:
(d) 2015

Question 18.
Identify the year in which National Rural Health Mission was launched.
(a) 2000
(b) 2005
(c) 2010
(d) 2015
Answer:
(b) 2005

Question 19.
Identify the advantages of rural roads.
(a) Rural marketing
(b) Rural employment
(c) Rural development
(d) All the above
Answer:
(d) All the above

Question 20.
“An Indian farmer is born in debt, lives in debt, dies in debt and bequeaths debt”- who said this?
(a) Adam Smith
(b) Gandhi
(c) Amartya Sen
(d) Sir Malcolm Darling
Answer:
(d) Sir Malcolm Darling

Part – B
Answer the following questions in one or two sentences

Question 21.
Define Rural Economy.
Answer:
Rural economy refers to villages. Rural economics deals with the application of economic principles in understanding and developing rural areas.

Question 22.
What do you mean by Rural Development?
Answer:
According to the world bank, “Rural development is a strategy designed to improve the economic and social life of a specific group of people rural poor”

Question 23.
Rural Poverty – Define.
Answer:
Rural poverty refers to the existence of poverty in rural areas. Poverty in India is the situation in which an individual fails to earn sufficient income to buy the basic minimum of subsistence.

Question 24.
Define Open Unemployment.
Answer:
In open unemployment, unemployed persons are identified as they remain without work. This type of unemployment is found among agricultural labourers, rural artisans and literate persons.

Question 25.
What is meant by Disguised Unemployment?
Answer:
Disguised unemployment is a situation where many are employed below their productive capacity and even if they are withdrawn from work the output will not diminish.

Question 26.
Define Cottage Industry.
Answer:
Cottage industries are generally associated with agriculture and provide both part time and full time jobs in rural areas.

Question 27.
What do you mean by Micro Finance?
Answer:
Micro finance is a financial service that offers loans, savings and insurance to entrepreneurs and small business owners.

Question 28.
State any two causes of housing problem in rural areas.
Answer:

  1. Rapid adaptation of nuclear families.
  2. Problems in provision of proper water supply, good sanitation, proper disposal of sewage.

Question 29.
Define Rural Electrification.
Answer:
Rural electrification refers to providing electrical power to rural areas.

Question 30.
State any two factors hindering Rural Electrification in India.
Answer:

  1. Lack of funds
  2. Inter – state disputes
  3. Uneven terrain

Part – C
Answer the following questions in one Paragraph

Question 31.
State the importance of Rural Development.
Answer:
Importance of Rural Development:

  1. India cannot be developed by retaining rural as backward
  2. As the rural economy supports the urban sector, the backwardness of the rural sector would be a major impediment to the overall progress of the economy.
  3. Improvements in education, health and sanitation in villages can help avoid many urban problems.
  4. To provide gainful employment and improve food production.
  5. The evils of brain drain and rural urban migration can be reduced.
  6. For the better utilization of resources.
  7. To minimize the gap between rural and urban areas.

Question 32.
Explain the causes for Rural Backwardness.
Answer:

  1. Skewed distribution of land.
  2. Lack of non-farm employment.
  3. Lack of public sector investment.
  4. Inflation.
  5. Low productivity.
  6. Unequal benefit of growth.
  7. Low rate of economic growth.
  8. More emphasis on large industries.
  9. Social evils.

Question 33.
Enumerate the remedial measures to Rural Poverty.
Answer:
Creation of employment opportunities would support elimination of poverty. The poverty eradication schemes implemented in India are

  1. 20 point programme.
  2. Integrated Rural Development Programme (IRDP)
  3. Training Rural Youths for Self Employment (TRYSEM)
  4. Food for Work Programme (FWP)
  5. National Rural Employment Programme (NREP)
  6. RLEGP, JRY and MGN REGS

Question 34.
What are the remedial measures for Rural Unemployment?
Answer:

  1. Subsidiary occupation : To reduce the seasonal unemployment rural people should be encouraged to adopt subsidiary occupation.
  2. Rural works programme : Rural works programme should be planned during off season to provide gainful employment.
  3. Irrigation facilities : Irrigation facilities should be expanded to enable the farmers to adopt multiple cropping.
  4. Rural industrialization : New industries should be set up in rural areas.
  5. Technical education : Employment oriented courses should be introduced to literate the youth to start their own units.

Question 35.
Write a note on Regional Rural Banks.
Answer:
Regional rural banks came into existence in 1975. RRBs are recommended with a view to develop rural economy by providing credit and other facilities to the small and marginal farmers, agricultural labourers, artisans and small entrepreneurs.

RRBs are set up by the joint efforts of the centre and state governments and commercial banks. At present there are 64 RRBs in India. RRBs confine their lending’s only to the weaker sections and their lending rates are at par with the prevailing rate of cooperative societies.

Question 36.
Mention the features of SHGs.
Answer:

  1. SHG is an economically homogeneous group formed through a process of self selection based upon the affinity of its members.
  2. Most SHGs are women’s groups with membership ranging between 10 and 20.
  3. SHGs have well-defined rules and by laws, hold regular, meetings and maintain records and savings and credit discipline.
  4. SHGs are self-managed institutions characterized by participatory and collective decision making.

Question 37.
List out the objectives of MUDRA Bank.
Answer:

  1. Regulate the lender and the borrower of micro finance and bring stability to the micro finance system.
  2. Extend finance and credit support to micro finance institutions.
  3. Register all MFIs and introduce a system of performance rating and accreditation for the past time.
  4. Offer a credit guarantee scheme for providing guarantees to loans being offered to micro businesses.
  5. Introduce appropriate technologies to assist in the process of efficient lending, borrowing and monitoring of distributed capital.

Part – D
Answer the following questions in about a page

Question 38.
‘The features of Rural Economy are peculiar’- Argue.
Answer:
The Rural economy has very peculiar features which are :

  1. Village is an institution : The village is a primary institution and it satisfies almost all the needs of the rural community.
  2. Dependence on agriculture : Agriculture and allied activities are the main occupation in rural areas.
  3. Life of rural people: Life styles in villages are very simple. Public services like education, housing,health and sanitation, transport and communication, banking, roads and markets are limited and unavailable.
  4. Population density : Density of population is very low and houses are scattered in the entire villages.
  5. Employment: There exists unemployment, seasonal unemployment and underemployment in rural areas.
  6. Poverty : According to the 2011 – 12 estimates about 22 crores of people in rural areas live below the poverty line.
  7. Indebtedness : Sir Malcolm Darling stated that an Indian farmer is born in debt, lives in debt, dies in debt and bequeaths debt.
  8. Rural income : As the rural economy is not sufficiently vibrant to provide them with jobs or self employment opportunities the income of the rural people is constrained.
  9. Dependency : Rural households are dependent on social grants and remittances from family members working in urban areas and cities.
  10. Dualism : Dualism means the co-existence of two extremely different features which is the common characteristics in rural area.
  11. Inequality : The distribution of income, wealth and assets are highly skewed among rural people.
  12. Migration : Rural people are forced to migrate to urban areas to seek employment which gives rise to the formation of cities. Enmity and lack of basic amenities in rural areas also push the people to migrate. This is called double poisoning by Schumacher.

Question 39.
Discuss the problems of Rural Economy.
Answer:
The problems of rural economy are.
1. People related problems : The problem consists of illiteracy, lack of technical know how, low level of confidence, dependence on sentiments and beliefs etc.

2. Agriculture related problems : This include lack of awareness, knowledge, skill and attitude, unavailability of inputs, poor marketing facility, insufficient extension of staff and services, small size of land holding, absence of infrastructure, primitive technology, reduced public investment and absence of role for farmers in fixing the prices for their own products.

3. Infrastructural related problems : Poor infrastructure facilities like water, electricity, transport, educational institutions communication, health, employment are found in rural areas.

4. Economics related problems : Inability to adopt high cost technology, high cost of inputs, under privileged rural industries, low income, indebtedness and existence of inequality in land holdings and assets.

5. Leadership related problems : Leadership among the hands of inactive and incompetent people, self-interest of leaders, biased political will, less bargaining power and negotiation skills and dominance of political leaders.

6. Administrative problems : Political interference, lack of motivation and interest, low wages in villages, improper utilization of budget and absence of monitoring and implementation of rural development programme.

Question 40.
Analyse the causes for Rural Indebtedness.
Answer:

1. Poverty of farmers : The vicious circle of poverty fore es the farmers to borrow for consumption, cultivation and celebrations. Thus, poverty’ debt and high rates of interest hold the farmer in the grip of money lenders.

2. Failure of monsoon : Frequent failure of monsoon is a curse to the farmers and they have to suffer due to the failure of nature. Therefore, farmers find it difficult to identify good years to repay their debts.

3. Litigation : Due to land disputes litigation in the court compels them to borrow heavily. Being uneducated and ignorant they are caught in the litigation process and dry away their savings and resources.

4. Money lenders and high rate of interest: The rate of interest charged by the local money lenders is very high and the compounding of interest leads to perpetuate indebtedness of the farmer.

Samacheer Kalvi 11th Economics Rural Economy Additional Questions and Answers

Part-A
Choose the best options

Question 1.
According to the 2011 population census, there are ___________ villages in India.
(a) 6,40,867
(b) 3,40,867
(c) 4,40,867
(d) 5,40,867
Answer:
(a) 6,40,867

Question 2.
Of 121 crore total population ___________ percent lives in rural areas.
(a) 64.88
(b) 68.84
(c) 68.48
(d) 88.64
Answer:
(b) 68.84

Question 3.
___________ are geographical areas located outside towns and cities.
(a) Townships
(b) Corporations
(c) Rural areas
(d) None
Answer:
(c) Rural areas

Question 4.
Total percentage of poverty in 2009-2010
(a) 33.80
(b) 38.30
(c) 33.08
(d) 80.33
Answer:
(a) 33.80

Question 5.
In 2009, the total number of micro finance institutions in India was around ___________
(a) 140
(b) 150
(c) 160
(d) 170
Answer:
(b) 150

Question 6.
SHG Bank Linked Programme (SBLP) was started in ___________
(a) 1991
(b) 1990
(c) 1992
(d) 1993
Answer:
(c) 1992

Question 7.
Micro Units Development and Refinance Agency is ___________
(a) IRDB
(b) MUDRA
(c) BUNDAN
(d) ICICI
Answer:
(b) MUDRA

Question 8.
MUDRA Bank was started in the year ___________
(a) 2011
(b) 2012
(c) 2015
(d) 2013
Answer:
(c) 2015

Question 9.
___________ was the first professor of economics at Madras University
(a) Gilbert Slater
(b) Samuelson
(c) Marshall
(d) Manmohan Singh
Answer:
(a) Gilbert Slater

Question 10.
National Rural Health Mission was launched on ___________
(a) 2001
(b) 1991
(c) 2005
(d) 2000
Answer:
(c) 2005

Question 11.
___________ are setup with a view to develop rural economy by providing credit and other facilities to small and marginal farmers.
(a) MUDRA Bank
(b) RRB
(c) SHG
(d) NABARD
Answer:
(b) RRB

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
(a) 1 2 3 4
(b) 2 1 3 4
(c) 4 3 2 1
(d) 3 4 1 2
Answer:
(b) 2 1 3 4

Question 2.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
(a) 4 3 2 1
(b) 2 1 4 3
(c) 3 4 1 2
(d) 1 2 3 4
Answer:
(c) 3 4 1 2

Choose the correct option

Question 3.
National rural health mission was established in
(a) March 12, 2005
(b) June 12, 2005
(c) April 12, 2005
(d) June 13, 2005
Answer:
(c) April 12, 2005

Question 4.
Gross national happiness index is
(a) GDP
(b) HDF
(c) WEI
(d) GNHI
Answer:
(d) GNHI

Question 5.
Poverty line is calculated on the basis of
(a) Based on inflation
(b) Based on poverty
(c) Based on income or consumption
(d) Based on unemployment
Answer:
(c) Based on income or consumption

Fill in the blanks with suitable option given below

Question 6.
Micro credit is also known as
(a) Micro finance
(b) Small finance
(c) Large finance
(d) None
Answer:
(a) Micro finance

Question 7.
The co-existence of two extremely different features is called as
(a) Unity
(b) Dualism
(c) Trism
(d) None
Answer:
(b) Dualism

Question 8.
The first economics professor of madras university was
(a) A.K. Sen
(b) Charan Singh
(c) Manmohan Singh
(d) Gilbert Slater
Answer:
(d) Gilbert Slater

Choose the incorrect statement

Question 9.
(a) Physical quality of life index is PQLI
(b) The existence of poverty in rural areas is called rural poverty
(c) SSI s are also known as micro, small and medium enterprises (MSMEs)
(d) Petter diamond was the first economics professor of madras university
Answer:
(d) Petter diamond was the first economics professor of madras university

Question 10.
(a) MUDRA is micro units development and Refinance Agency Bank
(b) NRHM is National Rural Health Mission.
(c) The length of roads in India as per 2018 is 24 lakh kms.
(d) PURA is provision of urban facilities for rural areas.
Answer:
(c) The length of roads in India as per 2018 is 24 lakh kms.

Choose the incorrect pair

Question 11.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
Answer:
(c) NHRM (iii) Rural health

Question 12.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
Answer:
(d) All Indian Debt and investment Survey (iv) AIDS

Question 13.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
Answer:
(d) National Food Security (iv) 2013

Question 14.
Samacheer Kalvi 11th Economics Solutions Chapter 10 Rural Economy
Answer:
(a) Adarsh Gram Sadak Yojana (i) 2010

Choose the correct statement

Question 15.
(a) Unemployment is a situation in which a person who is unwilling to work employed,
(b) According to 2011 12 estimates 42 crore village population list below the poverty line.
(c) In 2015, 60 crore people lived in villages.
(d) Sir Malcolm Darling stated that ‘An Indian farmer is bom in debt, lives in debt, dies in debt and bequeaths debt’
Answer:
(d) Sir Malcolm Darling stated that ‘An Indian farmer is bom in debt, lives in debt, dies in debt and bequeaths debt’

Question 16.
(a) Number of people living per sq.km is called population scarcity.
(b) There are 5,40,867 villages in India.
(c) The concept of ‘Double poisoning’ was coined by Schumacher.
(d) In 2009-10, the total percentage of poverty in India is 54.10
Answer:
(c) The concept of ‘Double poisoning’ was coined by Schumacher.

Pick the odd one out

Question 17.
(a) Petter Diamond
(b) A.K.Sen
(c) Dale Mortensen
(d) Christopher Pissarides
Answer:
(b) A.K.Sen

Question 18.
(a) NCC
(b) IRDP
(c) TRYSEM
(d) MGNREGS
Answer:
(a) NCC

Analyse the reason for the following

Question 19.
Assertion (A) : Creation of employment opportunities eliminate poverty’in rural areas.
Reason (R) : Rural unemployment and rural poverty are interrelated.
(a) (A) is true; but (R) is false
(b) Both (A) and (R) are false
(c) Both (A) and (R) are true but (R) is not the correct explanation of (A)
(d) Both (A) and (R) are true (R) is the correct explanation of (A)
Answer:
(d) Both (A) and (R) are true (R) is the correct explanation of (A)

Question 20.
Assertion (A) : Agricultural operations are seasonal in nature.
Reason (R) : To reduce the seasonal unemployment rural people should be encouraged to adopt subsidiary occupations.
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) (A) is false but (R) is true.
Answer:
(b) Both (A) and (R) are true, (R) is not the correct explanation of (A)

Part – B
Answer the following questions in one or two sentences

Question 1.
Name some of the economic indicators?
Answer:
Human Development Index (HDI),Women Empowerment Index (WEI), Gender Disparity Index (GDI), Physical Quality of Life Index (PQLI) and Gross National Happiness Index (GNHI).

Question 2.
How can you measure poverty by nutritional index?
Answer:
On the basis of recommended nutritional intake, persons consuming less than 2400 calories per day is rural areas are considered as rural poor.

Question 3.
What is unemployment?
Answer:
Unemployment is a situation in which a person actively searching for employment but unable to find work at the prevailing wage rate.

Question 4.
Write a note on seasonal unemployment ?
Answer:
In seasonal unemployment, employment occurs only in a particular season and workers remain unemployed in the remaining period of a year.

Question 5.
Write a note on NRHM.
Answer:
The National Rural Health Mission (NRHM) was launched on 12th April 2005, to provide accessible, affordable and quality health care to the vulnerable groups in rural areas.

Question 6.
What is population density?
Answer:
Population density is the number of persons living in per square kilometre.

Part – C
Answer the following questions in one Paragraph

Question 1.
What are the types of rural unemployment ?
Answer:

  1. Open unemployment: Unemployed persons are identified as they remain without work.
  2. Concealed unemployment: Many are employed below their productive capacity and even if they are withdrawn from work the output will not diminish. It is also called as disguised unemployment or under employment.
  3. Seasonal unemployment: Employment occurs only on a particular season and the remaining period of a year the rural people are unemployed or partially employed.

Question 2.
Explain about village industries.
Answer:
Village industries are traditional in nature and depend on local raw material. They cater to the needs of local population.
(Eg.) Gur and Khandsari, Cane and Bamboo basket, Shoe making, Pottery and Leather tanning.

Question 3.
Write a note on rural indebtedness.
Answer:
Rural indebtedness refers to the situation of the rural people who are unable to repay the loan accumulated over a period. Existence of the rural indebtedness indicates the weak financial infrastructure of our country in reaching the needy farmers landless people and the agricultural labourers.

Question 4.
Write a note on rural roads.
Answer:
Rural roads constitute the very life line of rural economy. A well-constructed road network in rural area would bring several benefits including the linking of remote villages with urban centres, reduction in cost of transportation of agricultural inputs and promotion of marketing for rural produces.

Part – D
Answer the following questions in about a page

Question 1.
Explain the causes for rural unemployment.
Answer:
1. Absence of skill development and employment generation : Lack of Government initiatives to give required training and then to generate employment opportunities.

2. Seasonal nature of agriculture : Agricultural operations are seasonal in nature, so non-farm employment opportunities must be created to reduce seasonal unemployment.

3. Lack of subsidiary occupation : Rural people are not able to start subsidiary occupations due to shortages of funds for investment and lack of proper marketing arrangements.

4. Mechanization of agriculture : Mechanization of agricultural operations reduces employment opportunities for the farm labour.

5. Capital – Intensive technology : The expanding private industrial sector in urban areas do not create additional employment opportunities due to the application of capital intensive technologies.

6. Defective system of education : The present System of education has also aggravated the rural unemployment problem. Students want to get degrees only, not any skill. Degrees should be awarded only on the basis of skills acquired.

If you need some more information about Tamilnadu State Board Solutions for 11th Economics Chapter 10 Rural Economy Questions and Answers then visit our site frequently and get all the resources that you look for. Also, share this page with your friends and make them learn all the concepts thoroughly before the exams.

Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India

Immediately get Chapter Wise Tamilnadu State Board Solutions for 11th Economics to gain more marks and start Quick Revision of all Chapters. All Chapters Pdf is provided along with the Questions and Answers. You Can Dowload Samacheer Kalvi 11th Economics Book Solutions Questions and Answers for Chapter wise are given with a clear explanation. Tamilnadu State Board Solutions for 11th Economics Chapter 9 Development Experiences in India Questions and Answers is for free of cost.

Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India

Tamilnadu State Board Solutions for 11th Economics Chapter 9 Development Experiences in India Questions and Answers PDF has all given in Chapter Wise Section. Check Out daily basis with Tamilnadu State Board Solutions 11th Economics PDF will help to improve your score. Improve your level of accuracy to answer a question by reading with Samacheer Kalvi 11th Economics Book Solutions Questions and Answers PDF.

Samacheer Kalvi 11th Economics Development Experiences in India Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
Which of the following is the way of Privatisation?
(a) Disinvestment
(b) Denationalization
(c) Franchising
(d) All the above
Answer:
(d) All the above

Question 2.
Countries today are to be ______ for their growth.
(a) Dependent
(b) Interdependent
(c) Free trade
(d) Capitalist
Answer:
(b) Interdependent

Question 3.
The Arguments against LPG is ______
(a) Economic growth
(b) More investment
(c) Disparities among people and regions
(d) Modernization
Answer:
(c) Disparities among people and regions

Question 4.
Expansion of FDI ______
(a) Foreign Private Investment
(b) Foreign Portfolio
(c) Foreign Direct Investment
(d) Forex Private Investment
Answer:
(c) Foreign Direct Investment

Question 5.
India is the largest producer of ______ in the world.
(a) fruits
(b) gold
(c) petrol
(d) diesel
Answer:
(a) fruits

Question 6.
Foreign investment includes ______
(a) FDI only
(b) FPI and FFI
(c) FDI and FPI
(d) FDI and FFI
Answer:
(a) FDI only

Question 7.
The Special Economic Zones policy was announced in ______
(a) April 2000
(b) July 1990
(c) April 1980
(d) July 1970
Answer:
(a) April 2000

Question 8.
Agricultural Produce Market Committee is a ______
(a) Advisory body
(b) Statutory body
(c) Both a and b
(d) none of these above
Answer:
(b) Statutory body

Question 9.
Goods and Services Tax is ______
(a) a multi point tax
(b) having cascading effects
(c) like Value Added Tax
(d) a single point tax with no cascading effects.
Answer:
(d) a single point tax with no cascading effects.

Question 10.
The New Foreign Trade Policy was announced in the year ______
(a) 2000
(b) 2002
(c) 2010
(d) 2015
Answer:
(d) 2015

Question 11.
Financial Sector reforms mainly related to ______
(a) Insurance Sector
(b) Banking Sector
(c) Both a and b
(d) Transport Sector
Answer:
(c) Both a and b

Question 12.
The Goods and Services Tax Act came in to effect on ______
(a) 1st July 2017
(b) 1st July 2016
(c) 1st January 2017
(d) 1st January 2016
Answer:
(a) 1st July 2017

Question 13.
The new economic policy is concerned with the following ______
(a) foreign investment
(b) foreign technology
(c) foreign trade
(d) all the above
Answer:
(d) all the above

Question 14.
The recommendation of Narashimham Committee Report was submitted in the year ______
(a) 1990
(b) 1991
(c) 1995
(d) 2000
Answer:
(b) 1991

Question 15.
The farmers have access to credit under Kisan credit card scheme through the following except ______
(a) co-operative banks
(b) RRBs
(c) Public sector banks
(d) private banks
Answer:
(a) co-operative banks

Question 16.
The Raja Chelliah Committee on Trade Policy Reforms suggested the peak rate on import duties at ______
(a) 25%
(b) 50%
(c) 60%
(d) 100%
Answer:
(b) 50%

Question 17.
The first ever SEZ in India was set up at
(a) Mumbai
(b) Chennai
(c) Kandla
(d) Cochin
Answer:
(c) Kandla

Question 18.
‘The Hindu Rate of Growth’ coined by Raj Krishna refers to ______
(a) low rate of economic growth
(b) high proportion of Hindu population
(c) Stable GDP
(d) none
Answer:
(a) low rate of economic growth

Question 19.
The highest rate of tax under GST ______ is (as on July, 2017)
(a) 18%
(b) 24%
(c) 28%
(d) 32%
Answer:
(c) 28%

Question 20.
The transfer of ownership from public sector to private sector is known as ______
(a) Globalization
(b) Liberalization
(c) Privatization
(d) Nationalization
Answer:
(c) Privatization

Part – B
Answer the following questions in one or two sentences

Question 21.
Why was structural reform implemented in Indian Economy?
Answer:
Indian economy introduced structural reforms to face the economic crisis in the form of balance of payments problem in 1991.

Question 22.
State the reasons for implementing LPG.
Answer:
To correct the weaknesses and rigidities in the various sectors of the economy India implemented LPG which is the triple pillars of New Economic Policy.

Question 23.
State the meaning of Privatization.
Answer:
Privatization means transfer of ownership and management of enterprises from public sector to private sector.

Question 24.
Define disinvestment.
Answer:
Disinvestment means selling of government securities of public sector undertakings to other PSUs or private sector or banks.

Question 25.
Write three policy initiative introduced in 1991 – 1992 to correct the fiscal imbalance.
Answer:

  1. Reduction in fertilizer subsidy.
  2. Abolition of subsidy on sugar.
  3. Disinvestment of a part of the government’s equity holdings in select public sector undertakings.
  4. Expenditures on welfare measures were reduced.

Question 26.
State the meaning of Special Economic Zones.
Answer:
SEZs is an area in which business and trade laws are different from rest of the country mainly aiming at increasing trade, investment and job creation.

Question 27.
State the various components of Central government schemes under post – harvest measures.
Answer:

  1.  Mega food parks, Integrated cold chain, value addition preservation infrastructure, modernization of slaughter house.
  2. Scheme for quality-assurance, codex standards, research and development and other promotional activities.

Part – C
Answer the following questions in one Paragraph

Question 28.
How do you justify the merits of Privatisation?
Answer:
Privatisation was necessitated because of the belief that the private sector was not given enough opportunities to earn more money.

Question 29.
What are the measures taken towards Globalization?
Answer:
Globalization refers to the integration of the domestic economy with the rest of the world. Import liberalization through reduction of tariff and non-tariff barriers, opening the doors to foreign direct investment and foreign portfolio investment are some of the measures towards globalization.

Question 30.
Write a note on Foreign investment policy?
Answer:
The major feature of the economic reform opened gate to foreign investment and foreign technology. Foreign investment including FDI and FPI were allowed.

In 1991, the government announced a specified list of industries where in automatic permission was granted for FDI. Foreign Investment Promotion Board (FIPB) has been set up to negotiate with international firms and approve foreign direct investment in select areas.

Question 31.
Give short note on Cold storage.
Answer:
Problems relating to the marketing of fruits and vegetables is related with their perishability. Perishability is responsible for high marketing costs, market gluts, price fluctuations and other similar problems.

In order to overcome this constraint, the Government of India and the ministry of agriculture promulgated “Cold Storage Order 1964” under section 3 of the Essential Commodities Act, 1955.

However, the cold storage facility is still very poor and highly inadequate.

Question 32.
Mention the functions of APMC.
Answer:

  1. To promote public private partnership in the ambit of agricultural markets.
  2. To provide market led extension services to farmer.
  3. To promote agricultural activities.
  4. To bring transparency in pricing system and market transactions.
  5. To ensure payments to the farmer for the sale on the same day.
  6. To display data on arrivals and rates from time to time in the market.

Question 33.
List out the features of new trade policy.
Answer:
The trade policy of 1 April 1992 freed imports of almost all intermediate and capital goods. Only 71 items remained restricted. This would affect the domestic industries. Rationalization of tariff structure and removal of quantitative restrictions.

The Chelliah Committee’s report had suggested drastic reduction in import duties. As a first step towards a gradual reduction in the tariffs, the 1991 -92 budget had reduced the peak rate of import duty.

Question 34.
What is GST? Write its advantages.
Answer:
GST is a comprehensive indirect tax levied on manufacture, sale and consumption of goods as well as services.

Advantages of GST :

  1. Removing cascading tax effect.
  2. Single point tax.
  3. Higher threshold for registration
  4. Composition scheme for small business.
  5. Online simpler procedure under GST.
  6. Defined treatment for e-commerce.
  7. Increased efficiency in logistics

Question 35.
Explain the objectives and characteristics of SEZs.
Answer:
The Special Economic Zones (SEZs) policy was announced in April 2000.
The major objectives of SEZs are :

  1. To enhance foreign investment especially to attract foreign direct investment (FDI) and thereby increasing GDP.
  2. To increase shares in global export.
  3. To generate additional economic activity.
  4. To create employment opportunities.
  5. To develop infrastructure facilities.
  6. To exchange technology in the global market.

Main characteristics of SEZs :
Geographically demarked area with physical security.

  1. Administrated by single body / authority.
  2. Streamlined procedures.
  3. Having separate custom area
  4. Governed by more liberal economic laws.
  5. Greater freedom to the firms located in SEZS.

Question 37.
Describe the salient features of EXIM policy (2015 – 2020)
Answer:
The new EXIM policy has been formulated focusing on increasing in exports scenario, boosting production and supporting the concepts like Make in India and Digital India.

  1. Reduce export obligations by 25% and give boost to domestic manufacturing supporting the “Make in India” concept.
  2. As a step to Digital India concept, online procedure to upload digitally signed document, and mobile app for filing tax, stamp duty has been developed.
  3. Repated submission of physical copies of documents are not required.
  4. Export obligation period for export items related to defence, military store, aerospace and nuclear energy to be 24 months.
  5. EXIM policy 2015 – 2020 is expected to double the share of India in World Trade from present level of 3% by the year 2020.This appears to the too ambitious.

Samacheer Kalvi 11th Economics Development Experiences in India Additional Questions and Answers

Part-A
Choose the best options

Question 1.
The expansion of FDI is
(a) Foreign comprehensive investment ______
(b) Foreign direct investment
(c) Foreign indirect investment
(d) Foreign private investment
Answer:
(b) Foreign direct investment

Question 2.
India ranked ______ position in Asia’s GDP.
(a) 4
(b) 3
(c) 1
(d) 2
Answer:
(b) 3

Question 3.
The monopoly and Restrictive Trade Practices Act was abolished in ______
(a) 1919
(b) 1971
(c) 1991
(d) 1791
Answer:
(c) 1991

Question 4.
Crop Insurance scheme was launched on
(a) 2009
(b) 2014
(c) 2016
(d) 2012
Answer:
(c) 2016

Question 5.
The expansion of KCC ______
(a) Kisan Crop Card
(b) Kisan Cash Card
(c) Kisan Credit Card
(d) None of the above
Answer:
(c) Kisan Credit Card

Question 6.
EPZ means ______
(a) Export Pay Zone
(b) Export Policy Zone
(c) Export Processing Zone
(d) Special Economic Zone
Answer:
(c) Export Processing Zone

Question 7.
Goods and Services Tax came into effect on ______
(a) 30,h June 2017
(b) 1st July 2016
(c) 1st July 2017
(d) 1st April 2017
Answer:
(c) 1st July 2017

Question 8.
Fiscal deficit should not exceed ______ percent of GDP
(a) 2
(b) 4
(c) 3
(d) 5
Answer:
(c) 3

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India
(a) 1 2 3 4
(b) 2 1 4 3
(c) 3 4 2 1
(d) 1 2 4 3
Answer:
(b) 2 1 4 3

Question 2.

(a) 4 3 2 1
(b) 3 2 4 1
(c) 1 2 3 4
(d) 3 1 4 2
Answer:
(d) 3 1 4 2

Choose the correct option

Question 3.
New economic policy was introduced in
(a) 2010
(b) 1991
(c) 1884
(d) 1894
Answer:
(b) 1991

Question 4.
Kisan card was introduced in
(a) 2000
(b) 2015
(c) 1998
(d) 2017
Answer:
(c) 1998

Question 5.
was one of the first in Asia to recognize the effectiveness of the export processing zone
(a) India
(b) China
(c) America
(d) Soviet Union
Answer:
(a) India

Fill in the blanks with suitable option given below

Question 6.
Disinvestment means
(a) Selling of government securities
(b) Selling of public sectors
(c) Selling of private securities
(d) None
Answer:
(a) Selling of government securities

Question 7.
was set up to encourage FDI
(a) Domestic investment promotion board
(b) Foreign investment promotion board
(c) Both
(d) None
Answer:
(b) Foreign investment promotion board

Question 8.
The share of India to Asia’s GDP is
(a) 7.50%
(b) 6.50%
(c) 8.50%
(d) 9.50%
Answer:
(c) 8.50%

Choose the incorrect statement

Question 9.
(a) Prime minister’s crop insurance scheme was launched on 2016
(b) New industrial policy was established on 1991
(c) Kisan credit card was introduced by RBI and NABARD
(d) Goods and Services Tax was introduced in 2015
Answer:
(d) Goods and Services Tax was introduced in 2015

Question 10.
(a) New foreign trade policy was introduced on 2015
(b) First ever export processing zone was set up by China
(c) Cold storage order was introduced on 1964.
(d) India’s GDP crossed 2 trillion dollars in 2015-16
Answer:
(b) First ever export processing zone was set up by China

Choose the incorrect pair

Question 11.
Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India
(a) 1 2 3 4
(b) 2 1 4 3
(c) 3 4 2 1
(d) 1 2 4 3
Answer:
(b) 2 1 4 3

Question 12.
Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India
(a) 4 3 2 1
(b) 3 2 4 1
(c) 1 2 3 4
(d) 3 1 4 2
Answer:
(c) Chelliah committee (iii) GST

Choose the correct pair

Question 13.
Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India
Answer:
(d) Cash reserve ratio (iv) 1991

Question 14.
Samacheer Kalvi 11th Economics Solutions Chapter 9 Development Experiences in India
Answer:
(a) Dr.Raj Krishna (i) Low rate of economic growth

Pick the odd one out

Question 15.
(a) Liberalization
(b) Industrialization
(c) Privatization
(d) Globalization
Answer:
(b) Industrialization

Question 16.
(a) Crop insurance
(b) Cold storage
(c) Agriculture subsidy reduction
(d) Kisan credit card
Answer:
(c) Agriculture subsidy reduction

Choose the correct statement

Question 17.
(a) China is the largest producer of fruits.
(b) India is the largest producer of vegetables.
(c) Agricultural produce market committee (ADMC) promote agricultural activities.
(d) Tobacco is a plantation crop
Answer:
(c) Agricultural produce market committee (ADMC) promote agricultural activities.

Question 18.
(a) Privatization measures discourages the continuance of monopoly.
(b) Disinvestment process was fully implemented
(c) New industrial policy abolished MRTP act.
(d) Foreign investment includes FDI and FII
Answer:
(c) New industrial policy abolished MRTP act.

Analyse the reason for the following

Question 19.
Assertion (A) : Agriculture in India is highly prone to risks like droughts and floods.
Reason (R) : Agriculture is a seasonal activity.
(a) (A) and (R) are true; (R) is the correct explanation of (A)
(b) (A) and (R) are true. (R) is not the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) (A) is true (R) is false.
Answer:
(a) (A) and (R) are true; (R) is the correct explanation of (A)

Question 20.
Assertion (A) : GST is a comprehensive indirect tax.
Reason (R) : GST is a multi point tax
(a) (A) is true (R) is false.
(b) (A) is false (R) is true
(c) Both (A) and (R) are true.
(d) Both (A) and (R) are false.
Answer:
(a) (A) is true (R) is false.

Part -B
Answer the following questions in one or two sentences

Question 1.
What is industrial delicensing policy or Red Tapism?
Answer:
The new industrial policy of 1991 abolished the procedure of securing licenses (Red Tapism) to start an industry which is called as industrial delicensing policy.

Question 2.
What is liberalization?
Answer:
Liberalization refers to removal of relaxation of governmental restrictions in all stages in industry.

Question 3.
What is Statutory Liquidity Ratio (SLR) ?
Answer:
SLR refers to the amount that the commercial banks require to maintain in the form of cash or gold or government approved securities before providing credit to the customers. Agricultural Produce Market Committee (APMC) is a statutory body constituted by state government in order to trade in agricultural or horticultural or livestock produce.

Question 5.
Write a note on Kisan Credit Card.
Answer:
A Kisan Credit Card (KCC) is a credit delivery mechanism that is aimed at enabling farmers to have quick and timely access to affordable credit. It was launched in 1998 by RBI and NABARD.

Question 6.
What is globalization ?
Answer:
Globalization stand for the consolidation of the various economies of the world.

Question 7.
Explain agrarian crisis after the economic reforms of 1991.
Answer:

  1. High input costs
  2. Cutback in agricultural subsidies
  3. Reduction of import duties
  4. Paucity of credit facilities.

Part – C
Answer the following questions in one Paragraph

Question 1.
What is crop insurance scheme ?
Answer:
To protect the farmers from natural calamities and ensure their credit eligibility for the next season, the Government of India introduced many agricultural schemes throughout the country.

One of the scheme is the Pradhan Mantri Fasal Bima Yojana which was launched on 18 February 2016. It envisages a uniform premium of 2% for Kharif and 1.5% for Rabi and 5% for commercial and horticultural crops.

Question 2.
State the arguments against LPG.
Answer:

  1. Liberalization measures favor the unrestricted entry of foreign companies which prevents the growth of the local manufactures.
  2. Privatization measures favor the continuance of the monopoly power.
  3. Only the developed countries are favored under globalization but the welfare of (he less developed countries will be neglected.

Question 3.
Explain the major changes in India after 1991.
Answer:

  1. Foreign exchange reserves started rising.
  2. There was a rapid industrialization.
  3. The pattern of consumption started improving.
  4. Infrastructure facilities such as express highways, metro rails, flyovers and airports started expanding (but the local people were thrown away).

Part -D
Answer the following questions in about a page

Question 1.
Explain the measures taken in Ranking system under new economic policy 1991.
Answer:

  1. Reduction in statutory liquidity ratio (SLR) and the Cash Reserve Ratio (CRR) were recommended by the Narasimham Committee Report 1991.
  2. Interest rate liberalization.
  3. Greater competition among public, private and foreign banks and elimination of administrative constraints.
  4. Liberalization of bank branch licensing policy.
  5. Banks were given freedom to relocate branches and open specialised branches.
  6. Guidelines for opening new private sector banks.
  7. New accounting norms regarding classification of assets and provisions of bad debt were introduced.

If you need some more information about Tamilnadu State Board Solutions for 11th Economics Chapter 9 Development Experiences in India Questions and Answers then visit our site frequently and get all the resources that you look for. Also, share this page with your friends and make them learn all the concepts thoroughly before the exams.

Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence

Immediately get Chapter Wise Tamilnadu State Board Solutions for 11th Economics to gain more marks and start Quick Revision of all Chapters. All Chapters Pdf is provided along with the Questions and Answers. You Can Dowload Samacheer Kalvi 11th Economics Book Solutions Questions and Answers for Chapter wise are given with a clear explanation. Tamilnadu State Board Solutions for 11th Economics Chapter 8 Indian Economy Before and After Independence Questions and Answers is for free of cost.

Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence

Tamilnadu State Board Solutions for 11th Economics Chapter 8 Indian Economy Before and After Independence Questions and Answers PDF has all given in Chapter Wise Section. Check Out daily basis with Tamilnadu State Board Solutions 11th Economics PDF will help to improve your score. Improve your level of accuracy to answer a question by reading with Samacheer Kalvi 11th Economics Book Solutions Questions and Answers PDF.

Samacheer Kalvi 11th Economics Indian Economy Before and After Independence Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
The arrival of Vasco da Gama in Calicut, India
(a) 1498
(b) 1948
(c) 1689
(d) 1849
Answer:
(a) 1498

Question 2.
In 1614 Sir Thomas Roe was successful in getting permission from
(a) Akbar
(b) Shajakan
(c) Jahangir
(d) Noorjakhan
Answer:
(c) Jahangir

Question 3.
The power for governance of India was transferred from the East India Company (EIC) to the British crown in
(a) 1758
(b) 1858
(c) 1958
(d) 1658
Answer:
(b) 1858

Question 4.
Ryotwari system was initially introduced in _______
(a) Kerala
(b) Bengal
(c) Tamil Nadu
(d) Maharastra
Answer:
(c) Tamil Nadu

Question 5.
First World War started in the year _______
(a) 1914
(b) 1814
(c) 1941
(d) 1841
Answer:
(a) 1914

Question 6.
When did the Government of India declared its first Industrial Policy ?
(a) 1956
(b) 1991
(c) 1948
(d) 2000
Answer:
(c) 1948

Question 7.
The objective of the Industrial Policy 1956 was
(a) Develop heavy industries
(b) Develop agricultural sector only
(c) Develop private sector only
(d) Develop cottage industries only
Answer:
(a) Develop heavy industries

Question 8.
The industry which was de-reserved in 1993 ?
(a) Railways
(b) Mining of copper and zinc
(c) Atomic energy
(d) Atomic minerals
Answer:
(b) Mining of copper and zinc

Question 9.
The father of Green Revolution in India was _______
(a) M.S. Swaminathan
(b) Gandhi
(c) Visweswaraiah
(d) N.R. Viswanathan
Answer:
(a) M.S. Swaminathan

Question 10.
How many commercial banks were nationalised in 1969 ?
(a) 10
(b) 12
(c) 14
(d) 16
Answer:
(c) 14

Question 11.
The main objective of nationalisation of banks was _______
(a) Private social welfare
(b) Social welfare
(c) To earn
(d) Industries monopoly
Answer:
(b) Social welfare

Question 12.
The Planning Commission was setup in the year _______
(a) 1950
(b) 1955
(c) 1960
(d) 1952
Answer:
(a) 1950

Question 13.
In the first five year plan, the top priority was given to _______ Sector.
(a) Service
(b) Industrial
(c) Agriculture
(d) Bank
Answer:
(c) Agriculture

Question 14.
Tenth Five year plan period was _______
(a) 1992-1997
(b) 2002-2007
(c) 2007-2012
(d) 1997-2002
Answer:
(b) 2002-2007

Question 15.
According to HDR (2016), India ranked _______ out of 188 countries.
(a) 130
(b) 131
(c) 135
(d) 145
Answer:
(b) 131

Question 16.
Annual Plans formed in the year _______
(a) 1989-1991
(b) 1990-1992
(c) 2000-2001
(d) 1981-1983
Answer:
(b) 1990-1992

Question 17.
The Oldest large scale industry in India ________
(a) cotton
(b) jute
(c) steel
(d) cement
Answer:
(a) cotton

Question 18.
Human development index (HDI) was developed _________
(a) Jawaharlal Nehru
(b) M.K.Gandhi
(c) Amartiya Sen
(d) Tagore
Answer:
(c) Amartiya Sen

Question 19.
The main theme of the Twelth Five Year Plan __________
(a) faster and more inclusive growth
(b) growth with social Justice
(c) socialistic pattern of society
(d) faster, more inclusive and sustainable growth
Answer:
(d) faster, more inclusive and sustainable growth

Question 20.
The PQLI was developed by _______
(a) Planning Commission
(b) Nehru
(c) Morris
(d) Morris
Answer:
(c) Morris

Part – B
Answer the following questions in one or two sentences

Question 21.
What are the Phases of colonial exploitation of India?
Answer:
The period of merchant capital, the period of industrial capital, the period of finance capital.

Question 22.
Name out the different types of land tenure existed in India before Independence.
Answer:

  1. Zamindari system
  2. Mahalwari system
  3. Ryotwari system

Question 23.
State the features that distinguish a land tenure system from other system.
Answer:
The land tenure system differs as

  1. Who owns the land ?
  2. Who cultivates the land ?
  3. Who is responsible for paying the land revenue to the government ?

Question 24.
List out the weaknesses on Green Revolution.
Answer:

  1. Indian agriculture was still a gamble of the monsoons.
  2. This strategy needed heavy investment.
  3. The income gap has increased
  4. Increased unemployment among agricultural labourers.
  5. Reduced the soil fertility and spoiled human wealth.

Question 25.
What are the objectives of Tenth five year plan ?
Answer:

  1. Double the per capita income in the next 10 years.
  2. Aimed to reduce the poverty ratio to 15% by 2012.
  3. Growth target was 8.0% but it achieved only 7.2%

Question 26.
What is the difference between HDI and PQLI ?
Answer:

  1. Human Development Index : It is constructed based on life expectancy index, Education index and GDP per capita.
  2. Physical quality of life index : The PQLI is a measure to calculate the quality of life.

Question 27.
Mention the indicators which are used to calculate HDI.
Answer:

  1. Life expectancy index.
  2. Educational index.
  3. GDP per capita.

Part – C
Answer the following questions in one Paragraph

Question 28.
Explain about the Period of Merchant Capital.
Answer:

  1. The period of merchant capital was from 1757 to 1813.
  2. The only aim of the East India company was to earn profit by establishing monopoly trade.
  3. During this period, India had been considered as the best hunting ground for capital
  4. The objective of monopoly trade was fulfilled by achieving political control.
  5. The company administration succeeded in generating huge surpluses which were repatriated to England.

Question 29.
The Handicrafts declined in India in British Period. Why?
Answer:

  1. Through discriminatory tariff policy, the British Government purposefully destroyed the handicrafts.
  2. With the disappearance of Nawabs and Kings, there was no one to protect Indian handicrafts.
  3. Indian products could not compete with machine made products.
  4. Introduction of railways in India increased the domestic market for the British goods.

Question 30.
Elucidate the different types of land tenure system in colonial India.
Answer:
The three types of land tenure system are Zamindari, Mahalwari and Ryotwari system.

1. Zamindari system or the land – lords : This system was created in 1793,after the introduction of permanent settlement act. Under this system the landlords were declared as the owners of the land. They were responsible to pay the land revenue. The share of the rent to government is fixed at 10/11th and the balance as zamindar’s remuneration.

2. Mahalwari system or communal farming : The ownership of the land was maintained by the collective body of villagers which served as a unit of management. They distribute the land and collect revenue and pay it to the state.

3. Ryotwari or the owner – cultivator system : The ownership rights of use and control of land were held by the tiller himself.

Question 31.
State the reasons for nationalization of commercial banks.
Answer:

  1. The main objective of the economic planning aimed at social welfare.
  2. Before independence commercial banks were in the private sector.
  3. These commercial banks failed in helping the government to achieve social objectives of planning.
  4. Therefore, the government decided to nationalize 14 major commercial banks on 1969

Question 32.
Write any three objectives of Industrial Policy 1991.
Answer:

  1. Ensuring rapid industrial development in a competitive environment.
  2. Enchancing support to the small scale sector.
  3. Providing more incentives for industrialisation of the backward areas.

Question 33.
Give a note on Twelfth Five Year Plan.
Answer:

  1. Twelfth five year plan (2012 – 17).
  2. Its main theme is “Faster more inclusive and sustainable growth”
  3. Its growth rate target is 8%

Question 34.
What is PQLI ?
Answer:
Morris D Morris developed the physical quality of life index (PQLI). The PQLI is a measure to calculate the quality of life.
PQLI includes three indicators life expectancy, infant mortality rate and literacy rate. A scale of each indicator ranges from the number 1 to 100. Represents the worst performance and 100 is the best performance.
HDI includes income while PQLI do not. PQLI has only the physical aspects of like.

Part – D
Answer the following questions in about a page

Question 35.
Discuss about the Indian economy during British Period.
Answer:
Britain had exploited India over a period of two centuries of its colonial rule. On the basis of the form of colonial exploitation, economic historians have divided the whole period into three phases namely the period of merchant capital, period of industrial capital and the period of finance capital.

1. Period of merchant capital (1757 to 1813):
The only aim of the East India company was to earn profit by establishing monopoly trade. India was considered as the best hunting ground for capital. By attaining political power the objective of monopoly trade was fulfilled. The company administration succeeded in generating huge surpluses which were repatriated to England.

2. Period of industrial capital (1813 to 1858):
India had become a market for British textiles. Indians were exploited by fixing low price for exports and high price for imports. India’s traditional handicrafts were thrown out of gear.

3. Period of financial capital (Late 19th century – 1947) : Finance imperialism began to entrench. Britain decided to make massive investments in various fields by plundering Indian capital.

4. The land tenure system in India : Land tenure refers to the system of land ownership and managements.
The Zamindari system, the Mahalwari system and the Ryotwari system were the systems of land tenure introduced by British.

5. Problems of British rule : Their profit motives led to drain of resources from India. The handicraft industries were collapsed.
The British rule stunted the growth of Indian enterprise. Capital formation in India were retarded.

Question 36.
Explain the role of SSIs in economic development?
Answer:

1. SSIs provide employment:

  • SSIs use labour intensive techniques thus reduce the problem of unemployment to a great extent.
  • They provide employment to people in villages and unorganized sectors.
  • The employment – capital ratio is high for the SSIs.

2. SSIs bring balanced regional development: SSIs remove regional disparities by industrializing rural and backward areas and bring balanced regional development.

3. Help in mobilization of local resources : SSIs help to mobilize and utilize local resources like small savings, entrepreneurial talent etc.,

4. Pave for optimization of capital: SSIs require less capital per unit of output. They function as a stabilizing force by providing output-capital ratio as well as high employment capital ratio.

5. Promote exports : SSIs earn valuable foreign exchange through exports from India.

6. Complement large scale industries : SSIs serve as ancillaries to large scale units.

7. Meet consumer demands : SSIs serves as an anti-inflationary force by providing goods of daily use.

8. Develop entrepreneurship : They promote self-employment and spirit of self-reliance in the society. They help to increase the per capita income. They help in distributing national income in more efficient and equitable manner.

Question 37.
Explain the objectives of nationalization of commercial banks.
Answer:

  1. The main objective of nationalization was to attain social welfare. Sectors such as agriculture, small and village industries were in need of funds for their expansion and further economic development.
  2. It helped to curb private monopolies in order to ensure a smooth supply of credit to socially desirable sections.
  3. To encourage the banking habit among the rural population.
  4. To reduce the regional imbalances where the banking facilities were not available.
  5. After nationalization, new bank branches were opened in both rural and urban areas, and they created credit facilities mainly to the agriculture sector and its allied activities.

Question 38.
Describe the performance of 12th five year plan in India.
Answer:
Economic planning is the process in which the limited natural resources are used skillfully so as to achieve the desired goals.

1. First five year plan (1951 – 56) : Main focus was on the agricultural development it achieved the growth rate of 3.6%.

2. Second five year plan (1956 – 61) : Focus was on the industrial development of the country and achieved the growth rate of 4.1%.

3. Third five year plan (1961 – 66): To make the economy independent and to reach take off.

4. Plan holiday (1966 – 69): Equal priority was given to agriculture, its allied sectors and the industrial sectors during annual plans.

5. Fourth five year plan (1969 – 74): Growth with stability and progressive achievement of self reliance was the goal but the plan failed.

6. Fifth five year plan (1975 – 79): A successful plan prioritized agriculture and then industry and mines.

7. Rolling plan : This plan was started with an annual plan for 1978-79.

8. Sixth five year plan (1980 – 85): Based on investment yojana, its objective was poverty eradication and technological self reliance.

9. Seventh five year plan (1985 – 90): Establishment of the self sufficient economy and opportunities for productive employment. Private sector got the priority over public sector.

10. Annual plans : Two annual plans were formed in 1990 – 91 & 1991 -92.

11. Eighth five year plan (1992 – 97): Priority was given to development of the human resources. New economic policy was introduced.

12. Ninth five year plan (1997 – 02):

  • Aimed to double the percapita income in the next 10 years.
  • Aimed to reduce the poverty ratio to 15% by 2012.

13. Tenth five year plan (2002 – 07): Growth with justice and equity was the focus but the plan failed with growth rate of 5.6%. Aimed to double the percapita income in the next 10 years. Aimed to reduce the poverty ratio to 15% by 2012.

14. Eleventh five year plan (2007 – 12): Main theme was faster and more inclusive growth.

15. Twelfth five year plan (2012 – 17): Main theme was faster more inclusive and sustainable growth. The five year plans played a’very prominent role in the economic development of the country. These plans had guided the government as to how it should utilise scarce resources so that maximum benefits can be gained.

Samacheer Kalvi 11th Economics Indian Economy Before and After Independence Additional Questions and Answers

Part – A
Choose the best options

Question 1.
Lord Cornwallis introduced “Permanent Settlement Act in” _______
(a) 1793
(b) 1794
(c) 1795
(d) 1796
Answer:
(a) 1793

Question 2.
In the year _______ green revolution started in India.
(a) 1950
(b) 1951
(c) 1948
(d) 1960
Answer:
(d) 1960

Question 3.
KGN. Daber established Mumbai’s spinning and weaving co. in _______
(a) 1810
(b) 1854
(c) 1845
(d) 1948
Answer:
(b) 1854

Question 4.
The oil well of Digboi, Assam was dug in.
(a) 1889
(b) 1898
(c) 1988
(d) 1810
Answer:
(a) 1889

Question 5.
The period of twelfth five year plan _______
(a) 2010 – 15
(b) 2011 – 16
(c) 2012 – 17
(d) 2013 – 18
Answer:
(c) 2012 – 17

Question 6.
NITI Aayog replaced the planning commission in _______
(a) 2013
(b) 2015
(c) 2014
(d) 2016
Answer:
(b) 2015

Question 7.
Human development index is _______
(a) HDI
(b) UNDP
(c) PQLI
(d) HID
Answer:
(a) HDI

Question 8.
Physical Quality of Life Index (PQLI) was developed by _______
(a) HDI
(b) PQLI
(c) UNDP
(d) HID
Answer:
(c) UNDP

Question 9.
In _______ system the ownership rights of use and control of land were held by the tiller himself.
(a) Zamindari
(b) Mahalwari
(c) Ryotwari
(d) Green revolution
Answer:
(c) Ryotwari

Question 10.
The period of industrial capital is _______
(a) 1757 – 1813
(b) 1813 – 1858
(c) 1757 – 1858
(d) 19 th century
Answer:
(b) 1813 – 1858

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
(a) 12 3 4
(b) 2 4 3 1
(c) 2 3 1 4
(d) 4 3 2 1
Answer:
(c) 2 3 1 4

Question 2.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
(a) 12 3 4
(b) 2 1 4 3
(c) 4 3 2 1
(d) 2 3 1 4
Answer:
(b) 2 1 4 3

Choose the correct option

Question 3.
The period of merchant capital is _______
(a) 1842 – 1857
(b) 1918 – 1920
(c) 1757 – 181
(d) 1721 – 1838
Answer:
(c) 1757 – 181

Question 4.
How many types of land tenure system were there before independence ?
(a) 2
(b) 3
(c) 1
(d) 5
Answer:
(b) 3

Question 5.
The first Indian modernized cotton cloth mill was established _______
(a) Chennai
(b) Calcutta
(c) Mumbai
(d) Bengaluru
Answer:
(b) Calcutta

Fill in the blanks with suitable option given below

Question 6.
The plan holiday was _______
(a) 1951-53
(b) 1966-69
(c) 2001 -02
(d) 1976-79
Answer:
(b) 1966-69

Question 7.
The planning commission was replaced by _______
(a) NITI Aayog
(b) Planning group
(c) Finance commission
(d) None of the above
Answer:
(a) NITI Aayog

Question 8.
NITI Aayog was formed in _______
(a) 2001
(b) 2014
(c) 2015
(d) 2016
Answer:
(c) 2015

Choose the incorrect statement

Question 9.
(a) Ryotwari system was first established in Tamilnadu.
(b) Zamindari system was established by British East India Company
(c) Mahalwari system was also called as the owner – cultivator system
(d) Zamindari system was created in 1793
Answer:
(c) Mahalwari system was also called as the owner – cultivator system

Question 10.
(a) The investment of small service enterprises should be more than ? 10 lakhs.
(b) The investment of micro service enterprises should not exceed ?10 lakhs.
(c) The investment of medium service enterprises should be more than ?2 crores,
(d) The investment of medium service enterprises should not exceed ?10 crores.
Answer:
(d) The investment of medium service enterprises should not exceed ?10 crores.

Choose the incorrect pair

Question 11.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
Answer:
(c) Vijaya nagar (iii) England

Question 12.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
Answer:
(c) Serampur (iii) Iron and steel

Chooose the correct pair

Question 13.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
Answer:
(a) Industrial policy (i) 1948

Question 14.
Samacheer Kalvi 11th Economics Solutions Chapter 8 Indian Economy Before and After Independence
Answer:
(d) White revolution (iv) Milk

Analyse the reason for the following

Question 15.
Assertion (A) : Before the advent of the British, Indian economy was self sufficient.
Reason (R) : Before the advent of the British, India lived in village.
(a) Both (A) and (R) are true
(b) (A) is true (R) is false
(c) Both (A) and (R) are false.
(d) (A) is false (R) is true.
Answer:
(a) Both (A) and (R) are true

Choose the incorrect statement

Question 16.
(a) The industrial policy resolutions in India ushered a socialistic economy
(b) The high yielding varieties programme was called as pilot project of green revolution
(c) The aim of seventh five year plan is ‘Garibi Hatao’
(d) Finance commission replaced the planning commission.
Answer:
(b) The high yielding varieties programme was called as pilot project of green revolution

Question 17.
(a) Switzerland ranks first in HDI.
(b) PQLI includes income whereas HDI excludes income
(c) The process of transforming private assets to government ownership is called privatization
(d) The third plan was called as ‘Gadgil plan’.
Answer:
(d) The third plan was called as ‘Gadgil plan’.

Pick the odd one out

Question 18.
(a) The period of revenue capital
(b) The period of industrial capital
(c) The period of merchant capital
(d) The period of financial capital
Answer:
(c) The period of merchant capital

Question 19.
(a) Investment system
(b) Zamindari system
(c) Mahalwari system
(d) Ryotwari system
Answer:
(a) Investment system

Question 20.
(a) Black revolution
(b) Red revolution
(c) White revolution
(d) Violet revolution
Answer:
(d) Violet revolution

Part – B
Answer the following questions in one or two sentences

Question 1.
Name the classification of industries in India?
Answer:

  1. Public sector
  2. Public-cum-private sector
  3. Controlled private sector
  4. Private and co-operative sectors.

Question 2.
What is the resolution of industrial policy 1948 ?
Answer:
The first industrial policy was declared on 6th April 1948. Its main importance was introducing the system of mixed economy in India.

Question 3.
What is the resolution of industrial policy 1956 ?
Answer:
The policy was shaped by the mahalanobis model of growth with emphasis on heavy industries which would lead to higher growth path.

Question 4.
Write a note on micro manufacturing enterprises.
Answer:
In micro enterprises, the investment in plant and machinery does not exceed Rs. 25 lakhs.

Question 5.
Write a note on small manufacturing enterprises
Answer:
The investment in plant and machinery is more than twenty five lakh rupees but does not exceed Rs. 5 crores.

Question 6.
Write a note on medium manufacturing enterprises.
Answer:
The investment in plant and machinery is more than Rs. 5 crores but not exceeding Rs. 10 crores.

Question 7.
Name the classification of public sector banks.
Answer:

  1. Nationalised Banks.
  2. State Bank and its associates

Part – C
Answer the following questions in one Paragraph

Question 1.
Describe the problems of British rule in Indian economy ?
Answer:

  1. The British rule stunted the growth of Indian enterprise. It retarded the capital formation in India.
  2. The drain of wealth financed capital development in Britain.
  3. Indian agricultural sector became stagnant and deteriorated.
  4. Indian handicraft industries were collapsed.
  5. The system of capitalist firms with profit motives led to drain of resources from India.

Question 2.
Mention the achievements of green revolution.
Answer:

  1. The production of major cereals wheat and rice were boosted. India became food surplus exporting food grains to European countries.
  2. It was confined to high yielding variety cereals.
  3. Production of commercial crops were increased.
  4. Per hectare productivity of all crops had increased due to better seeds.
  5. The revolution had positive effect on development of industries manufacturing agricultural tools.
  6. It brought prosperity to rural people.
  7. Demand for labour increased.
  8. Financial resources were provided by banks and co-operative societies.

Question 3.
Examine the requirements of second green revolution.
Answer:

  1. Introduction of genetically modified (GM) seeds which double the per average production.
  2. Contribution of private sector to market the usage of GM foods.
  3. Government can play a key role in expediting irrigation schemes and managing water resources.
  4. Linking of rivers to transfer surplus water to deficient areas.

Question 4.
Write a note on iron and steel industry.
Answer:

  1. First steel industry at Kulti, near Jharia, West Bengal – Bengal iron works company in 1870.
  2. First large scale steel plant TISCO at Jamshedpur in 1907.
  3. Steel authority of India Ltd. (SAIL) was established in 1974 and was made responsible for the development of the steel industry.
  4. India is the eighth largest steel producing country in the world.

Question 5.
Write a note on petrol and natural gas.
Answer:

  1. First successful oil well was dug in India in 1889 at Digboi, Assam.
  2. At present a number of regions with oil reserves have been identified and oil is being extracted in these regions.
  3. For exploration purpose, Oil and Natural Gas Commission (ONGC) was established in 1956 at Dehradun, Uttarakhand.

If you need some more information about Tamilnadu State Board Solutions for 11th Economics Chapter 8 Indian Economy Before and After Independence Questions and Answers then visit our site frequently and get all the resources that you look for. Also, share this page with your friends and make them learn all the concepts thoroughly before the exams.

Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Immediately get Chapter Wise Tamilnadu State Board Solutions for 11th Economics to gain more marks and start Quick Revision of all Chapters. All Chapters Pdf is provided along with the Questions and Answers. You Can Dowload Samacheer Kalvi 11th Economics Book Solutions Questions and Answers for Chapter wise are given with a clear explanation. Tamilnadu State Board Solutions for 11th Economics Chapter 7 Indian Economy Questions and Answers is for free of cost.

Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy

Tamilnadu State Board Solutions for 11th Economics Chapter 7 Indian Economy Questions and Answers PDF has all given in Chapter Wise Section. Check Out daily basis with Tamilnadu State Board Solutions 11th Economics PDF will help to improve your score. Improve your level of accuracy to answer a question by reading with Samacheer Kalvi 11th Economics Book Solutions Questions and Answers PDF.

Samacheer Kalvi 11th Economics Indian Economy Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
The main gold mine region in Karnataka is _________
(a) Kolar
(b) Ramgiri
(c) Anantpur
(d) Cochin
Answer:
(a) Kolar

Question 2.
Economic growth of a country is measured by national income indicated by _________
(a) GNP
(b) GDP
(c) NNP
(d) Per capital income
Answer:
(b) GDP

Question 3.
Which one of the following is a developed nations ?
(a) Mexico
(b) Ghana
(c) France
(d) SriLanka
Answer:
(c) France

Question 4.
The position of Indian Economy among the other strongest economies in the world is _______
(a) Fourth
(b) Seventh
(c) Fifth
(d) Tenth
Answer:
(b) Seventh

Question 5.
Mixed economy means _________
(a) Private sectors and banks
(b) Co-existence of Public and Private sectors
(c) Public sectors and banks
(d) Public sectors only
Answer:
(b) Co-existence of Public and Private sectors

Question 6.
The weakness of Indian Economy is _________
(a) Economic disparities
(b) Mixed economy
(c) Urbanisation
(d) Adequate employment opportunities
Answer:
(a) Economic disparities

Question 7.
A scientific study of the characteristics of population is _________
(a) Topography
(b) Demography
(c) Geography
(d) Philosophy
Answer:
(b) Demography

Question 8.
The year 1961 is known as _________
(a) Year of small divide
(b) Year of Population Explosion
(c) Year of Urbanisation
(d) Year of Great Divide
Answer:
(b) Year of Population Explosion

Question 9.
In which year the population of India crossed one billion mark ?
(a) 2000
(b) 2001
(c) 2005
(d) 1991
Answer:
(b) 2001

Question 10.
The number of deaths per thousand population is called as ________
(a) Crude Death Rate
(b) Crude Birth Rate
(c) Crude Infant Rate
(d) Maternal Mortality Rate
Answer:
(a) Crude Death Rate

Question 11.
The number of births per thousand population is called as _________
(a) Crude death rate
(b) Mortality rate
(c) Morbidity rate
(d) Crude Birth Rate
Answer:
(d) Crude Birth Rate

Question 12.
Density of population = _________
(a) Land area / Total Population
(b) Land area / Employment
(c) Total Population / Land area of the region
(d) Total Population
Answer:
(c) Total Population / Land area of the region

Question 13.
Who introduced the National Development Council in India?
(a) Ambedkar
(b) Jawaharlal Nehru
(c) Radhakrishanan
(d) V.K.R.V. Rao
Answer:
(b) Jawaharlal Nehru

Question 14.
Who among the following propagated Gandhian Ecomomic thinkings.
(a) Jawaharlar Nehru
(b) VKRV Rao
(c) JC Kumarappa
(d) A.K.Sen
Answer:
(c) JC Kumarappa

Question 15.
The advocate of democratic socialism was
(a) Jawaharlal Nehru
(b) P.C. Mahalanobis
(c) Dr. Rajendra Prasad
(d) Indira Gandhi
Answer:
(a) Jawaharlal Nehru

Question 16.
Ambedkar the problem studied by in the context of Indian Economy is ________
(a) Small land holdings and their remedies
(b) Problem of Indian Currency
(c) Economics of socialism
(d) All of them
Answer:
(b) Problem of Indian Currency

Question 17.
Gandhian Economics is based on the Principle _________
(a) Socialistic idea
(b) Ethical foundation
(c) Gopala Krishna Gokhale
(d) Dadabhai Naoroji
Answer:
(b) Ethical foundation

Question 18.
V.K.R.V Rao was a student of _________
(a) J.M. Keynes
(b) Colin Clark
(c) Adam smith
(d) Alfred Marshal
Answer:
(a) J.M. Keynes

Question 19.
Amartya Kumara Sen received the Nobel prize in Economics in the year.
(a) 1998
(b) 2000
(c) 2008
(d) 2010
Answer:
(a) 1998

Question 20.
Thiruvalluvar economic ideas mainly dealt with _________
(a) Wealth
(b) Poverty is the curse in the society
(c) Agriculture
(d) All of them
Answer:
(d) All of them

Part – B
Answer the following questions in one or two sentences

Question 21.
Write the meaning of Economic Growth.
Answer:
A country’s economic growth is usually indicated by Gross Domestic Product (GDP). The GDP is the total monetary value of the goods and services produced by that country over a specific period of time, usually one year.

Question 22.
State any two features of developed economy.
Answer:

  1. High National Income
  2. High per capital Income

Question 23.
Write the short note on natural resources
Answer:
Natural resources are stock or reserve that can be drawn from nature. The major natural resources are land, forest, water, mineral and energy.

Question 24.
Point out any one feature of Indian Economy
Answer:
Indian economy is a mixed economy. In India both the private and public sectors coexist.

Question 25.
Give the meaning of non-renewable energy
Answer:
The sources of energy which cannot be renewed or re-used are called non-renewable energy.
(Eg.) Coal, Oil, Gas etc.

Question 26.
Give a short note on Sen’s ‘Choice of Technique’.
Answer:
In a labour surplus economy, generation of employment cannot be increased at the initial stage by the adaptation of capital intensive technique whatsen called as choice of technique.

Question 27.
List out the reasons for low per capital income as given by V.K.R.V. Rao.
Answer:

  1. Uneconomic holdings.
  2. Low levels of water availability for crops.
  3. Excess population pressure on agriculture due to the absence of a large industrial sector.
  4. Absence of capital.
  5. Absence of autonomy in currency policy.

Part – C
Answer the following questions in one Paragraph

Question 28.
Define Economic Development.
Answer:
The level economic development is indicated not just by GDP, but by an increase in citizen’s quality of life or well-being. The quality of life is being assessed by several indices such as Human Development Index (HDI), Physical Quality of Life Index (PQLI) and Gross National Happiness Index (GNHI)

Question 29.
State Ambedkar’s Economic ideas on agricultural economics.
Answer:
Dr. B.R.Ambedkar was a versatile personality. In 1918, he published a paper “Small holding in India and their remedies” citing Adam Smith’s “Wealth of Nations”, he made a fine distinction between “Consolidation of holdings” and “Enlargement of holdings”.

Question 30.
Write on short note on village sarvodhaya.
Answer:
Village sarvodhaya is the concept of Gandhiji on the development of villages. To Gandhi India lives in villages. He was interested in developing the villages as self – sufficient units.

According to him, “Real India was to be found in villages and not in towns or cities”.
So he suggested the development of self-suffficient, self dependent villages.

Question 31.
Write the strategy of Jawahar lal Nehru in India’s planning.
Answer:

  1. Jawahar lal Nehru was responsible for the introduction of planning in our country.
  2. To Nehru, the plan was essentially an integrated approach for development.
  3. He said the essence of planning is to find the best way to utilize all resources of manpower, of money and so on.
  4. Planning for Nehru was essentially linked up with industrialization and eventual self reliance for the country’s economic growth on a self-accelerating growth.
  5. Nehru carried through this basic strategy of planned development.

Question 32.
Write the V.K.R.V.Rao’s contribution on multiplier concept.
Answer:

  1. Rao’s examination of the “interrelation between investment, income and multiplier in an under developed economy (1952)” was his major contribution to macroeconomic theory.
  2. In it he asserts that keynesian multiplier principle remains inoperative in underdeveloped countries.
  3. V.K.R.V. Rao was the best equipped of all keynes’ pupils.

Question 33.
Write a short note on Welfare Economics given by Amartya Sen.
Answer:

  1. Amartya Sen was awarded the nobel prize for his contributions to welfare economics.
  2. Sen’s major point has been that the distribution of income / consumption among the persons below the poverty line is to be taken into account.
  3. Sen has focused on the poor, viewing them not as objects of pity requiring charitable hand-outs, but as disempowered folk needing empowerment in all aspects.

Question 34.
Explain Social infrastructure.
Answer:

  1. Social infrastructure refers to those structures which are improving the quality of man power and contribute indirectly towards the growth of an economy.
  2. These structures are outside the system of production and distribution.
  3. The development of these social structures help in increasing the efficiency and productivity of man power.
    (Eg.) Schools, Colleges, Hospitals

Part – D
Answer the following questions in about a page

Question 35.
Explain strong features Indian economy
Answer:

  1. India has a mixed economy : In India both private and public sectors coexist.
  2. Agriculture plays the key role : Around 60% of the people in India depend upon agriculture for their livelihood.
  3. An emerging market: India has a high potential for prospective growth which attracts FDI and FII.
  4. Emerging economy : As a result of rapid economic growth Indian economy has a place among the G20 countries.
  5. Fast growing economy : India has emerged as the world’s fastest growing economy in 2016-17 with 7.1% GDP next to China.
  6. Fast growing service sector : The service sector, contributes a lion’s share of the GDP in India.
  7. Large domestic consumption : Due to large domestic consumption the standard of living has considerably improved and life style has changed.
  8. paid growth of urban areas : Improved connectivity in transport and communication, education and health have speeded up the pace of urbanization.
  9. Stable macro economy : The current year’s economic survey represents the Indian economy to be a heaven of macroeconomic stability, resilence and optimism.
  10. Demographic dividend : India is a pride owner of the maximum percentage of youth. This has invited foreign investments to the country and outsourcing opportunities.

Question 36.
Write the importance of mineral resources in India.
Answer:

  1. Iron-ore : Hematite iron is mainly found in Chattisgarh, Jharkhand, Odisha, Goa and Karnataka. Magnetite is found in Western Coast of Karnataka. Some deposits of iron ore are also found in Kerala, Tamil Nadu and Andhra Pradesh.
  2. Coal and Lignite : India ranks third in the coal production. The main centres are West-Bengal, Bihar, Madhya Pradesh, Maharashtra, Odisha and Andhra Pradesh. Bulk production comes from Bengal-Jharkhand coal fields. Lignite is found in Neyveli in Tamil Nadu.
  3. Bauxite : Bauxite is a main source of aluminium. Major reserves are found in Odisha and Andhra Pradesh.
  4. Mica : India stand first in sheet mica production. Andhra Pradesh, Jharkhand, Bihar and Rajasthan are major reserves.
  5. Crude Oil : Oil is being explored in India at many places of Assam and Gujarat.
  6. Gold : India possesses only a limited gold reserve.
  7. Diamond : The total reserves of diamond is estimated at around 4582 carats.

Question 37.
Bring out Jawharlal Nehru’s contribution to the idea of economic development.
Answer:
Jawahar Lal Nehru was one of the Chief builders of modem India. He was a great patriot, thinker and statesman. His ideas of economic development are :

1. Democracy : He was a firm believer in democracy. He believed in free speech, civil liberty, adult franchise and the rule of law and parliamentary democracy.

2. Secularism : Secularism is another signal contribution of Nehru to India. There are so many religions in India but there is no domination by religious majority.

3. Planning : Nehru was responsible for the introduction of planning in our country. The plan was essentially an integrated approach for development. Planning for Nehru was essentially linked up with industrialization and eventual self-reliance for the country’s economic growth on a self-accelerating growth.

4. Advancement of Science : Nehru made a great contribution to the advancement of Science, research, technology and industrial development. In his period, many IITs and research institutions were established. He always insisted on scientific temper.

5. Democratic socialism : Nehru put the country on the road towards a socialistic pattern of society. Nehru’s socialism is democratic socialism.

Question 38.
Write a brief note on the Gandhian economic ideas.
Answer:
Gandhian economics is based on ethical foundations.
Salient features of Gandhian economic thought:

  1. Village republics : Gandhi was interested in developing the villages as self-sufficient units.
  2. On machinery : Gandhi described machinery as ‘Great Sin’.
  3. Industrialism : Gandhi considered industrialism as a curse on mankind.
  4. Decentralization : He advocated a decentralized economy.
  5. Village sarvodaya : Gandhi suggested the development of self-sufficient, self-dependent villages.
  6. Bread labour: Gandhi realized the dignity of human labour. Bread labour or body labour was the expression that Gandhi used to mean manual labour.
  7. The doctrine of trusteeship : Trusteeship provides a means of transforming the present capitalist society into an egalitarian one.
  8. On the food problem : Gandhi was against any sort of food controls. Food controls only create artificial scarcity.
  9. On population : Gandhi was in favour of birth control through brahmacharya or self-control.
  10. On prohibition : Gandhi advocated cent per cent prohibition.

Samacheer Kalvi 11th Economics Indian Economy Additional Questions and Answers

Part -A
Choose the best options

Question 1.
Third world country __________
(a) India
(b) China
(c) Bangladesh
(d) Srilanka
Answer:
(c) Bangladesh

Question 2.
__________ is the year of great divide.
(a) 1921
(b) 1918
(c) 1943
(d) 1969
Answer:
(a) 1921

Question 3.
__________ are BIMARU States.
(a) Tamil Nadu, Kerala, Karnataka, Madhya Pradesh
(b) Bihar, Madhya Pradesh, Rajasthan, Uttarpradesh
(c) West Bengal, Himalchalpradesh, Punjab, Delhi
(d) Andhra Pradesh, Odisa, Gujarat, Chattisgarh
Answer:
(b) Bihar, Madhya Pradesh, Rajasthan, Uttarpradesh

Question 4.
In India, Lignite is found in.
(a) New Delhi
(b) Kolkatta
(c) Ranji
(d) Neyveli
Answer:
(d) Neyveli

Question 5.
Indian educational system comprises of __________ stages.
(a) 6
(b) 12
(c) 3
(d) 4
Answer:
(a) 6

Question 6.
__________ is considered as the best pupil of J.M. Keynes
(a) A.K. Sen
(b) J.C.Kumarappa
(c) V.K.R.V. Rao
(d) Manmohan Singh
Answer:
(c) V.K.R.V. Rao

Question 7.
Indian Railways introduced first wifi facility in __________
(a) Chennai
(b) Calcutta
(c) Bengaluru
(d) Hyderabad
Answer:
(c) Bengaluru

Question 8.
National Harbour board was set up in __________
(a) 1947
(b) 1950
(c) 1948
(d) 1951
Answer:
(b) 1950

Question 9.
A study on poverty and famines and concept of entitlements and capability development was done by __________
(a) Adamsmith
(b) A.K.Sen
(c) Marshall
(d) V.K.R.V. Rao
Answer:
(b) A.K.Sen

Question 10.
Percapital income is __________
(a) National income / output
(b) National income / population
(c) National income / percapital output
(d) National income / male population
Answer:
(b) National income / population

Question 11.
‘India lives in villages’ said by __________
(a) V.K.R.V. Rao
(b) J.C. Kumarappa
(c) Mahatma Gandhi
(d) Jawaharlal Nehru
Answer:
(c) Mahatma Gandhi

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
(a) 1 2 3 4
(b) 3 4 2 1
(c) 1 3 4 2
(d) 2 4 3 1
Answer:
(b) 3 4 2 1

Question 2.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
(a) 4 1 2 3
(b) 3 4 2 1
(c) 2 4 3 1
(d) 1 2 4 3
Answer:
(a) 4 1 2 3

Question 3.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
(a) 4 1 2 3
(b) 1 2 3 4
(c) 4 3 2 1
(d) 2 4 1 3
Answer:
(d) 2 4 1 3

Choose the odd one out

Question 4.
(a) Poverty and famine
(b) Choice of technique
(c) Economics of caste
(d) Concept of capability
Answer:
(c) Economics of caste

Question 5.
(a) Budget by assignment
(b) Budget by assigned revenue
(c) Budget by shared revenues
(d) Budget by fiscal revenues
Answer:
(d) Budget by fiscal revenues

Question 6.
(a) Net Domestic Product (NDP)
(b) Human Development Index (HDI)
(c) Physical Quality of Life Index (PQLI)
(d) Gross National Happiness Index (GNHI)
Answer:
(a) Net Domestic Product (NDP)

Choose the correct statement

Question 7.
(a) Indian economy is the seventh largest in the world
(b) Indian economy is a mixed economy
(c) India ranks third in terms of purchasing power parity
(d) The human capital of India is old
Answer:
(d) The human capital of India is old

Question 8.
(a) The annual addition of population in India equals the total population of Australia.
(b) There exists a huge economic disparity in the Indian economy
(c) Kerala has the highest birth rate
(d) Every 6th person in the world is an Indian
Answer:
(c) Kerala has the highest birth rate

Choose the incorrect statement

Question 9.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
Answer:
(b) Demographic transition – (ii) 2001

Question 10.
Samacheer Kalvi 11th Economics Solutions Chapter 7 Indian Economy
Answer:
(c) 1971 – (iii) 1040

Choose the correct statement

Question 11.
(a) Nehru always insisted on scientific temper
(b) Gandhi advocated a centralized economy
(c) Valluvar has recommended an unbalanced budget
(d) V.K.R.V. Rao’s economic thought is coined as Gandhian Economics.
Answer:
(a) Nehru always insisted on scientific temper

Question 12.
(a) Until 1979, Education in India was in state list
(b) Education in India follows the 10+3 pattern
(c) Sex ratio refers to the number of females per 1000 males.
(d) India ranks fourth in coal production of the world.
Answer:
(c) Sex ratio refers to the number of females per 1000 males.

Analyse the reason for the following 

Question 13.
Assertion (A) : There exist a huge economic disparity in the Indian economy.
Reason (R) : The proportion of income and assets owned by top 10% of Indians goes on increasing.
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is the not correct explanation of (A)
(c) (A) is true and (R) are false.
(d) (A) is false, but (R) is true.
Answer:
(a) Both (A) and (R) are true, (R) is the correct explanation of (A)

Question 14.
Assertion (A) : In Kerala the adult sex ratio is 1084 as on 2011.
Reason (R) : Kerala provides better status to women as compared to other states.
(a) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(b) Both (A) and (R) are true, (R) is the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) None of the above
Answer:
(b) Both (A) and (R) are true, (R) is the correct explanation of (A)

Fill in the blanks with suitable option given below

Question 15.
In India, Lignite is found in _____
(a) New Delhi
(b) kolkata
(c) Ranji
(d) Neyveli
Answer:
(d) Neyveli

Question 16.
Indian educational system comprises of _____ stages
(a) 12
(b) 3
(c) 6
(d) 4
Answer:
(c) 6

Question 17.
Indian railways introduced first wifi facility in _____
(a) Chennai
(b) Bengaluru
(c) Calcutta
(d) Hyderabad
Answer:
(b) Bengaluru

Choose the best option 

Question 18.
_____ is considered as the best pupil of J.M. Keynes
(a) A.K. Sen
(b) J.C.Kumarappa
(c) V.K.R.V.Rao
(d) Nehru
Answer:
(c) V.K.R.V.Rao

Question 19.
Third world country is _____
(a) India
(b) China
(c) Srilanka
(d) Bangladesh
Answer:
(d) Bangladesh

Question 20.
Per capital income is _____
(a) National income / output
(b) National income / population
(c) National income / per capital output
(d) National income / male population
Answer:
(b) National income / population

Part – B
Answer the following questions in one or two sentences 

Question 1.
What is Gross National Happiness Index?
Answer:
GNHI is an indicator of progress, which measures sustainable development, environmental conservation, promotion of culture and good governance.

Question 2.
State any two weakness of Indian Economy.
Answer:

  1. Large population.
  2. Inequality and poverty.

Question 3.
What is sex ratio ?
Answer:
Sex ratio refers to the number of females per 1,000 males.

Question 4.
Explain economic infrastructure.
Answer:
Economic infrastructure is the support system which helps in facilitating production

Question 5.
What is Gandhian Economics ?
Answer:
J.C. Kumarappa developed economic theories based on Gandhism. He developed a school of economic thought coined as Gandhian Economics.

Question 6.
Which are “BIMARU” states ?
Answer:

  1. Bihar
  2. Madhya Pradesh
  3. Rajasthan
  4. Uttarpradesh

Question 7.
What is year of small divide ?
Answer:
In 1951, population growth rate has come down from 1.33% to 1.25%. Hence it is known as year of small divide.

Part – C
Answer the following questions in one Paragraph

Question 1.
What are the aspects of demographic trends in India ?
Answer:

  1. Size of population
  2. Rate of growth.
  3. Birth and death rates
  4. Density of population
  5. Sex – ratio
  6. Life – expectancy at birth
  7. Literacy ratio.

Question 2.
Explain Thiruvalluvar’s ideas on agriculture.
Answer:
According to Thiruvalluvar, agriculture is the most fundamental economic activity. They are the axle-pin of the world, for on their prosperity revolves prosperity of other sectors of other sectors of the economy. He says, ploughmen alone is the freemen of the soil. He believes that agriculture is superior to all other occupation.

Question 3.
Write short note on Birth rate and Death rate.
Answer:
Short Note On Birth Rate And Death Rate:

  1. Birth Rate : It refers to the number of births per thousand of population.
  2. Death Rate : It refers to the number of deaths per thousand of population.

Kerala has the lowest birth rate and Uttarpradesh has the highest birth rate. West Bengal has the lowest and Orissa has the highest death rates.

Part – D
Answer the following questions in about a page

Question 1.
Explain about health in India.
Answer:
Health in India is a state government responsibility. The central council of health and welfare formulates the various health care projects and health department reform policies. The administration of health industry is the responsibility of the ministry of health and welfare.

Health care in India has many forms. But, all medical systems are under one ministry AYUSH.
Health status is better in Kerala as compared to other states. India’s health status is poor compared to Sri Lanka.

Question 2.
Explain Thiruvalluvar’s economic ideas.
Answer:
The economic ideas of Thiruvalluvar are found in his immortal work, Thirukkural in its second part porutpal.

  1. Factors of production : Thiruvalluvar has made many passing references about the factors of production, land, labour, capital, organisation.
  2. Agriculture : Agriculture is the most fundamental economic activity. The prosperity of other sectors depends on the prosperity of agriculture. Valluvar believes that agriculture is superior to all other occupation.
  3. Public finance : He elaborately explained public finance under the headings public revenue, financial administration and public expenditure.
  4. Public expenditure : Valluvar has recommended a balanced budget. He advocates
    1. Defence
    2.  Public works
    3. Social Services
  5. External assistance : Valluvar was against seeking external assistance. He advocated a self-sufficient economy.
  6. Poverty and begging : According to him ‘poverty is the root cause of all other evils which would lead to ever-lasting sufferings.
  7. Wealth: Valluvar has regarded wealth as only a means and not an end.
  8. Welfare state : The important elements of a welfare state are
    1. Perfect health of the people without disease
    2. Abundant wealth
    3. Good crop
    4. Prosperity and happiness
    5. Full security for the people.

If you need some more information about Tamilnadu State Board Solutions for 11th Economics Chapter 7 Indian Economy Questions and Answers then visit our site frequently and get all the resources that you look for. Also, share this page with your friends and make them learn all the concepts thoroughly before the exams.

Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis

Immediately get Chapter Wise Tamilnadu State Board Solutions for 11th Economics to gain more marks and start Quick Revision of all Chapters. All Chapters Pdf is provided along with the Questions and Answers. You Can Dowload Samacheer Kalvi 11th Economics Book Solutions Questions and Answers for Chapter wise are given with a clear explanation. Tamilnadu State Board Solutions for 11th Economics Chapter 6 Distribution Analysis Questions and Answers is for free of cost.

Tamilnadu Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis

Tamilnadu State Board Solutions for 11th Economics Chapter 6 Distribution Analysis Questions and Answers PDF has all given in Chapter Wise Section. Check Out daily basis with Tamilnadu State Board Solutions 11th Economics PDF will help to improve your score. Improve your level of accuracy to answer a question by reading with Samacheer Kalvi 11th Economics Book Solutions Questions and Answers PDF.

Samacheer Kalvi 11th Economics Distribution Analysis Text Book Back Questions and Answers

Part – A
Multiple Choice Questions

Question 1.
In Economics, distribution of income is among the
(a) factors of production
(b) individual
(c) firms
(d) traders
Answer:
(a) factors of production

Question 2.
Theory of distribution is popularly known as ………………………..
(a) Theory of product – pricing
(b) Theory of factor – pricing
(c) Theory of wages
(d) Theory of Interest
Answer:
(b) Theory of factor-pricing

Question 3.
Rent is the reward for the use of
(a) capital
(b) labour
(c) land
(d) organization
Answer:
(c) land

Question 4.
The concept of ‘Quasi – Rent’ is associated with ………………………
(a) Ricardo
(b) Keynes
(c) Walker
(d) Marshall
Answer:
(d) Marshall

Question 5.
The Classical Theory of Rent was propounded by
(a) Ricardo
(b) Keynes
(c) Marshall
(d) Walker
Answer:
(a) Ricardo

Question 6.
‘Original and indestructible powers of the soil’ is the term used by ……………………….
(a) J.S.Mill
(b) Walker
(c) Clark
(d) Ricardo
Answer:
(d) Ricardo

Question 7.
The reward for labour is
(a) rent
(b) wage
(c) profit
(d) interest
Answer:
(b) wage

Question 8.
Money wages are also known as ……………………….
(a) Real wages
(b) Nominal wages
(c) Original wages
(d) Transfer wages
Answer:
(b) Nominal wages

Question 9.
Residual Claimant Theory is propounded by
(a) Keynes
(b) Walker
(c) Hawley
(d) Knight
Answer:
(b) Walker

Question 10.
The reward is given for the use of capital ………………………
(a) Rent
(b) Wage
(c) Interest
(d) Profit
Answer:
(c) Interest

Question 11.
Keynesian Theory of interest is popularly known as
(a) Abstinence Theory
(b) Liquidity Preference Theory
(c) Loanable Funds Theory
(d) Agio Theory
Answer:
(b) Liquidity Preference Theory

Question 12.
According to the Loanable Funds Theory, supply of loanable funds is equal to …………………………
(a) S + BC + DH + DI
(b) I + DS + DH + BM
(c) S+ DS + BM+ DI
(d) S + BM + DH + DS
Answer:
(a) S + BC + DH + DI

Question 13.
The concept of meeting unexpected expenditure according to Keynes is
(a) Transaction motive
(b) Precautionary motive
(c) Speculative motive
(d) Personal motive
Answer:
(b) Precautionary motive

Question 14.
The distribution of income or wealth of a country among the individuals are ………………………….
(a) Functional distribution
(b) Personal distribution
(c) Goods distribution
(d) Services distribution
Answer:
(b) Personal distribution

Question 15.
Profit is the reward for
(a) land
(b) organization
(c) capital
(d) labour
Answer:
(b) organization

Question 16.
Innovation Theory of profit was given by …………………………..
(a) Hawley
(b) Schumpeter
(c) Keynes
(d) Knight
Answer:
(b) Schumpeter

Question 17.
Quasi-rent arises in
(a) Man-made appliances
(b) Homemade items
(c) Imported items
(d) None of these
Answer:
(a) Man-made appliances

Question 18.
“Wages as a sum of money are paid under contract by an employer to a worker for service rendered”- Who said this?
(a) Benham
(b) Marshall
(c) Walker
(d) J.S.Mill
Answer:
(a) Benham

Question 19.
Abstinence Theory of Interest was propounded by
(a) Alfred Marshall
(b) N.W Senior
(c) Bohm-Bawerk
(d) Knut Wicksell
Answer:
(b) N.W Senior

Question 20.
Loanable Funds Theory of Interest is called as …………………………
(a) Classical Theory
(b) Modem Theory
(c) Traditional Theory
(d) Neo-Classical Theory
Answer:
(d) Neo-Classical Theory

Part – B
Answer the following questions in one or two sentences

Question 21.
What is meant by distribution?
Answer:
Distribution means division of income among the four factors of production as rent, wage, interest and profit.

Question 22.
Mention the types of distribution?
Answer:
Personal Distribution: Personal Distribution is the distribution of national income among the individuals.
Functional Distribution: Functional Distribution means the distribution of income among the four factors of production namely land, labour, capital and organization for their services in production process.

Question 23.
Define ‘Rent’.
Answer:
Rent is that part of payment made by a tenant to his landlords for the use of land only.

Question 24.
Distinguish between real and money wages.
Answer:
Real wages :
Wages paid in terms of goods and services. It refers to the purchasing power of money wages.

Money wages :
Nominal wages are referred to as wages paid in terms of money.

Question 25.
What do you mean by interest?
Answer:

  1. Interest is the reward paid by the borrower to the lender for the use of capital.
  2. Interest is the price paid for the use of capital in any market.
  3. Generally speaking, interest is a payment made by a borrower to the lender for the money borrowed.

Question 26.
What is profit?
Answer:
Profit is the amount left with the entrepreneur after he has made payments for all the other factors of production.

Question 27.
State the meaning of liquidity preference?
Answer:

  1. Liquidity preference means the preference of the people to hold wealth in the form of liquid cash rather than in other non – liquid assets like bonds, securities, bills of exchange, land, building, gold etc.
  2. “ Liquidity Preference is the preference to have an amount of cash rather than of claims against others”.

Part – C
Answer the following questions in One Paragraph

Question 28.
What are the motives of demand for money?
Answer:
1. Transaction motive :
It relates to the desire of the people to hold cash for the current transactions.
Mt=f (y)

2. Precautionary motive :
It relates to the desire of the people to hold cash to meet unexpected or unforeseen expenditures.
Mp = f (y)

3. Speculative motive :
It relates to the desire of the people to hold cash in order to take advantage of market movements regarding future changes in the price of bonds and securities in the capital market.
Ms = f (i)

Question 29.
List out the kinds of wages:
Answer:
Wages are divided into four types.

  1. Nominal Wages or Money Wages: Nominal wages are referred to the wages paid in terms of money.
  2. Real Wages: Real wages are the wages paid in terms of goods and services. Hence, real wages are the purchasing power of money wages.
  3. Piece Wages: Wages that are paid on the basis of quantum of work done.
  4. Time Wages: Wages that are paid on the basis of the amount of time that the worker works.

Question 30.
Distinguish between rent and quasi-rent
Answer:
Rent:

  1. Rent accrues to land
  2. The supply of land is fixed forever
  3. It enters into price

Quasi – rent:

  1. Quasi-rent accrues to man-made appliances
  2. The supply of man made appliances is fixed for a short period only
  3. It does not enter into price.

Question 31.
Briefly explain the Subsistence Theory of Wages?
Answer:
The subsistence theory of wages:

  1. Subsistence theory of wages is one of the oldest theories of wages.
  2. According to this theory, wage must be equal to the subsistence level of the labourer and his family.
  3. Subsistence means the minimum amount of food, clothing and shelter which workers and their family require for existence.
  4. If workers are paid higher wages than the subsistence level, the workers would be better off and they will have large families.
  5. Hence, the population would increase.
  6. When the population increases, the supply of labourer would increase and therefore, wages will come down.
  7. If wages are lower than the subsistence level, there would be a reduction in population and thereby the supply of labour falls and wages increase to the subsistence level.

Question 32.
State the Dynamic Theory of Profit.
Answer:

  • J.B. Clark propounded this theory in 1900. To him, profit is the difference between price and cost of production of the commodity.
  • Profit is the reward for dynamic changes in society. He points out that, profit cannot arise in a static society. In static society everything remains stationary.
  • The following changes take place in a dynamic society.
    1. Population is increasing.
    2. Volume of capital is increasing.
    3. Methods of production are improving.
    4. Forms of industrial organization are changing.
    5. The wants of consumer are multiplying.

Question 33.
Describe briefly the Innovation Theory of Profit?
Answer:
Innovation Theory of Profit:

  1. The innovation theory of profit was propounded by Joesph. A.Schumpeter.
  2. Schumpeter says an entrepreneur is not only an undertaker of a business but also an innovator in the process of production.
  3. Profit is the reward for “innovation”.
  4. According to Schumpeter, an innovation may consist of the following:
    • Introduction of a new product.
    • Introduction of a new method of production.
    • Opening up of a new market.
    • Discovery of new raw materials
    • Reorganization of an industry / firm.

When any one of these innovations is introduced by an entrepreneur, it leads to reduction in the cost of production and thereby brings profit to an entrepreneur. To obtain profit continuously, the innovator needs to innovate continuously. The real innovators do so. Imitative entrepreneurs cannot innovate.

Question 34.
Write a note on the Risk-bearing Theory of Profit.
Answer:

  • Risk – the bearing theory was propounded by F.B.Hawley in 1907.
  • Profit is the reward for “risk-taking” in business.
  • Every business involves some risks. So risk-taking is an essential function of the entrepreneur and is the basis of profit.
  • Higher the risks, the greater are the profit.
  • Profit induces entrepreneurs to undertake risks.

Part – D
Answer the following questions in about a page

Question 35.
Explain the Marginal Productivity Theory of Distribution.
Answer:
Marginal productivity theory of distribution was developed by Clark, Wickseed and Walras. This theory explains how the prices of various factors of production are determined. This theory is also known as “General theory of distribution” or “National dividend theory of distribution”.
Assumptions:

  1. All the factors of production are homogenous and can be substituted for each other.
  2. There is perfect competition in both factor and product market.
  3. There is perfect mobility and full employment of factors of production.
  4. This theory is applicable only in the long-run.
  5. The entrepreneurs aim at profit maximization.
  6. There is no government intervention and no technological change.

Explanation of the theory :
Each factor is rewarded according to its marginal productivity.

Marginal product:
The marginal product of a factor of production means the addition made to the total product by employment of an additional unit of that factor.

Marginal physical product (MPP) :
The MPP of a factor is the increment in the total product obtained by the employment of an additional unit of that factor.

Value of marginal product (VMP):
VMP = MPP x price

Statement of the theory:

  1. The price of a factor of production depends upon its productivity.
  2. The price of a factor is determined by and will be equal to marginal revenue product of that factor.
  3. Under certain conditions, the price of a factor will be equal to both average and marginal products of that factor.

Marginal productivity under perfect competition:
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 1
X- axis represents factor units and Y-axis represents the factor price and revenue product.
MRP – Marginal Revenue Product Curve
ARP – Average Revenue Product Curve
AFC – Average Factor cost Curve
MFC – Marginal Factor cost Curve
AFC – is horizontal under perfect competition and MFC coincides with it.
When there is perfect competition in the factor market, the firm is in equilibrium only when MFC = MRP
At the point Q by employing ON units of factors and paying OP price (NQ) where MFC = MRP
At Q, MRP = ARP
Price of the factor (NQ) = Marginal revenue
Product (NQ) = Average revenue product (NQ)
There is no exploitation of factors under perfect competition.
Beyond the point Q the price paid to the factor is more than marginal revenue product and average revenue product, so employer do not employ the factors.

Question 36.
Illustrate the Ricardian Theory of Rent.
Answer:
The classical theory of rent is called “Ricardian theory of rent”.
Definition :
Rent is that portion of the produce of the earth which is paid to the landlord for the use of the original and indestructible powers of the soil.

Assumptions :

  1. Land differs in fertility.
  2. The law of diminishing returns operates in agriculture.
  3. Rent depends upon fertility and location of land.
  4. Theory assumes perfect competition and long period.
  5. There is existence of marginal land or no-rent land.
  6. Land has certain “Original and indestructible powers”.
  7. Land, is used for cultivation only.
  8. Most fertile lands are cultivated first.

Statement of the theory with illustration :
Assume that some people settle in newly discovered island. People will first cultivate the most fertile ‘A’ grade land. They produce 40 bags of paddy.

Suppose after some time if another group of people settle down in the same island. They cultivate ‘B’ grade land which produce 30 bags of paddy. Suppose yet another group of people settle down there they cultivate ‘C’ grade land. It produce 20 bags of paddy.

This surplus of ‘A’ grade land is now raised to 20 bags (40 – 20) and it is the ‘Economic Rent’ of ‘A’ grade land. The surplus of ‘B’ grade land is 10 bags (30 – 20). In ‘C’ grade land cost of production is equal to the price of its produce and it does not yield any rent (20 – 20). Hence ‘C’ grade land is called ‘no-rent land’ or marginal land.

The land which yields rent is called “intra-marginal land”.
Rent indicates the differential advantage of the superior land over the marginal land.
Ricardian Theory’ of Rent :

Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 2
Diagrammatic illustration :
X – axis represents various grades of land and Y axis yield per acre (in bags). OA, AB and BC are the ‘A’, ‘B’, ‘C’ grade lands respectively.

The ‘C’ grade land is the no-rent land. A and B grade lands are “intra-marginal lands. The economic rent yielded by ‘A’ and ‘B’ grade lands is equal to the shaded area of their respective rectangles.

Criticisms :

  1. The order of cultivation from most fertile to least fertile lands is historically wrong.
  2. This theory assumes that rent does not enter into price. But in reality, rent enters into price.

Question 37.
Elucidate the Loanable Funds Theory of Interest.
Answer:

  1. The loanable funds theory, also known as the “Neoclassical theory”. This theory was developed by Swedish economists like Wicksell, Bertilohlin, Viner, Gunnarmyrdal and others.
  2. Interest is the price paid for the use of loanable funds.
  3. The rate of interest is determined by the equilibrium between demand for and supply of loanable funds in the credit market.

Demand for loanable funds :

  1. Demand for Investment (I)
  2. Demand for Consumption (C)
  3. Demand for Hoarding (H)

Supply of loanable funds :
1. Savings (S) :
Savings may be of two types, namely.

  1. Savings planned by individuals are “ex-ante savings”. (Eg.) LIC premium
  2. Unplanned savings are called “ex-post savings”

2. Bank credit:
Commercial banks create credit and supply loan able funds to the investors.

3. Dishoarding (DH) :
Dishoarding means bringing out the hoarded money into use and thus it constitutes a source of supply of loanable funds.

4. Disinvestment (DI):
Disinvestment is the opposite of investment. It means not providing sufficient funds for depreciation of equipment.

Equilibrium :
The rate of interest is determined by the equilibrium between the total demand for and the total supply of loanable funds.
Supply of loanable funds = S + BC + DH + DI
Demand for loanable funds = I + C + H
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 3

Explanation :
X-axis represents the demand for and supply of loanable funds, Y axis represents the rate of interest. The LS curve represents the total supply curve ofloanable funds. The LD curve represents the total demand for loanable funds. The LD and LS curves, intersect each other at the point “E” the equilibrium point. At this point OR rate of interest and OM is the amount ofloanable funds.

Criticisms :

  1. Many factors have been included in this theory’. Still there are many more factors like
    • Asymmetric information
    • Moral Hazard.
  2. It is very difficult to combine real factors with monetary factors.

Question 38.
Explain the Keynesian Theory of Interest.
Answer:
According to Keynes, interest is the reward for parting with liquidity’ for a specified period of time.
Meaning of liquidity preference:
Liquidity preference means the preference of the people to hold wealth in the form of liquid cash rather than in other non-liquid assets.
Motives of demand for money:
The three motives of liquidity preference are.

  1. The transaction motive Mt = f (y)
  2. The precautionary motive Mp = f (y)
  3. The speculative motive Ms = f (i)

Determination of Rate of Interest:
According to Keynes, the rate of interest is determined by the demand for and the supply of money. The demand for money is liquidity preference. The supply of money is determined by the policies of the government and the central bank.

Equilibrium between demand and supply of money:
The equilibrium between liquidity preference and demand for money determine the rate of interest. In short-run, the supply of money is assumed to be constant.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 4

LP is the liquidity preference curve. M2 shows the supply curve of money to satisfy speculative motive. Both curves intersect at the point E which is the equilibrium. Here, rate of interest is I.

If liquidity preference increases from LP to L1 P1 the supply of money remains constant, the rate of interest increase from OI to OI1
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 5

Suppose LP remains constant. If the supply of money is OM2, the interest is OI2 and if the supply of money is reduced from OM2 to OM2, the interest would increase from OM2 to OM4. If the supply of money is increased from OM2 to OM4 the interest would decrease from OI2 to Ol4

Criticisms :
1. This theory does not explain the existence of different interest rates prevailing in the market at the same time.
2. It explains interest rate only in the short – run.

Samacheer Kalvi 11th Economics Distribution Analysis Additional Questions and Answers

Part -A
Choose the best options

Question 1.
Marginal Productivity theory is the …………………….. theory of distribution.
(a) Average
(b) Marginal
(c) Liquidity preference theory
(d) General
Answer:
(d) General

Question 2.
“Interest is the price paid for the use of capital in any market” is said by_______
(a) Adam Smith
(b) Torrance
(c) Walker
(d) Marshall
Answer:
(d) Marshall

Question 3.
The standard of living of workers in a country depends upon the ……………………… wages.
(a) Real
(b) Effective
(c) Direct
(d) Elastic
Answer:
(a) Real

Question 4.
The author of agio-theory of interest.
(a) N.W. Senior
(b) Bohm Bawerk
(c) Walker
(d) Marshall
Answer:
(b) Bohm Bawerk

Question 5.
Who propounded the risk bearing theory of profit?
(a) Bohm – Bawerk
(b) F.B Hawley
(c) Marshall
(d) Walker
Answer:
(b) F.B Hawley

Question 6.
Dynamic theory of profit was propounded by_______
(a) Marshall
(b) J.B.Clark
(c) J.M. Keynes
(d) Walker
Answer:
(b) J.B.Clark

Question 7.
What is meant by MPS?
(a) Marginal production supply
(b) Marginal production sale
(c) Marginal production service
(d) Marginal propensity to save
Answer:
(d) Marginal propensity to save

Question 8.
Time preference theory was given by_______
(a) John Ray
(b) Bohm Bawerk
(c) F.B. Hawley
(d) Irving Fisher
Answer:
(d) Irving Fisher

Question 9.
The demand for labour is ………………………
(a) Effective demand
(b) Direct demand
(c) Derived demand
(d) Elastic demand
Answer:
(c) Derived demand

Question 10.
_______ is the author of modern theory of rent
(a) Joan Robinson
(b) Boulding
(c) Both
(d) None
Answer:
(c) Both

Match the following and choose the answer using the codes given below

Question 1.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 6
(a) 3 4 2 1
(b) 4 3 1 2
(c) 1 2 3 4
(d) 2 3 4 1
Answer:
(c) 1 2 3 4

Question 2.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 7
(a) 2 1 4 3
(b) 1 4 2 3
(c) 1 2 3 4
(d) 4 1 3 2
Answer:
(a) 2 1 4 3

Choose the incorrect pair

Question 3.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 8
Answer:
(d) Total demand for money (iv) Mt + Mp

Question 4.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 9
Answer:
(b) Wage is the residual portion (ii) J.S. Mill

Choose the correct statement

Question 5.
(a) The supply of land is fixed forever
(b) Quasi-rent accrues to land
(c) The supply of man-made appliances are fixed
(d) Quasi-rent enters price
Answer:
(a) The supply of land is fixed forever

Question 6.
(a) Rent is the reward for labour
(b) Profit is the reward for labour
(c) Wages are the reward for labour
(d) Interest is the reward for organisation
Answer:
(b) Profit is the reward for labour

Choose the odd one out

Question 7.
(a) Bank credit
(b) Hoarding
(c) Dishoarding
(d) Disinvestment
Answer:
(b) Hoarding

Question 8.
(a) Monopoly profit
(b) Wind fall profit
(c) Functional reward
(d) Reward for labourer
Answer:
(d) Reward for labourer

Question 9.
(a) Nominal wages
(b) Real wages
(c) Direct wages
(d) Piece wages
Answer:
(c) Direct wages

Question 10.
(a) Abstinence theory
(b) Loanable funds theory
(c) Liquidity preference theory
(d) Risk bearing theory
Answer:
(b) Loanable funds theory

Analyse the reason for the following

Question 11.
Assertion (A) : Demand for loanable funds depends on investment, consumption and hoarding.
Reason (R) : Supply of loanable funds depends on hoarding and investment.
(a) (A) and (R) are true, (R) is not the correct explanation of (A)
(b) (A) and (R) are true, (R) is the correct explanation of (A)
(c) (A) is true (R) are false.
(d) (A) is false (R) is true.
Answer:
(c) (A) is true (R) are false.

Question 12.
Assertion (A) : The rate of interest is determined by the demand for money and the supply of money.
Reason (R) : The demand for money is liquidity preference.
(a) Both (A) and (R) are true, (R) is not the correct explanation of (A)
(b) Both (A) and (R) are correct, (R) is the correct explanation of (A)
(c) Both (A) and (R) are false.
(d) (A) is false and (R) is true.
Answer:
(a) Both (A) and (R) are true, (R) is not the correct explanation of (A)

Question 13.
Assertion (A) : Only superior land get rent
Reason (R) : Rent arose on account of differences in the fertility of land.
(a) Both (A) and (R) are true
(b) Both (A) and (R) are false correct explanation of (A)
(c) (A) is true and (R) are false.
(d) (A) is false and (R) is true.
Answer:
(a) Both (A) and (R) are true

Choose the incorrect pair

Question 14.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 10
Answer:
(d) Wage fund theory (iv) J.S. Mill

Question 15.
Samacheer Kalvi 11th Economics Solutions Chapter 6 Distribution Analysis 11
Answer:
(b) Wages (ii) Labour

Fill in the blanks with suitable option given below

Question 16.
F.A. Walker wrote a book ______ in 1875.
(a) Political economy
(b) Social economy
(c) Principles of economics
(d) Wealth of nations
Answer:
(a) Political economy

Question 17.
The author of Agio theory of interest______
(a) N.W. Seniors
(b) Bohm-Bawerk
(c) Walker
(d) Marshall
Answer:
(b) Bohm-Bawerk

Question 18.
______ is the produced means of production.
(a) Land
(b) Labour
(c) Capital
(d) Organisation
Answer:
(c) Capital

Choose the best option

Question 19.
The theory of factor prices is popularly known as the theory of
(a) Distribution
(b) Exchange
(c) Wages
(d) Profit
Answer:
(a) Distribution

Question 20.
______ is the author of the modern theory of rent
(a) Joan Robinson
(b) Boulding
(c) Both
(d) None
Answer:
(c) Both

Part – B
Answer the following questions in one or two sentences

Question 1.
What is the theory of distribution?
Answer:
The theory of factor prices is popularly known as the theory of distribution.

Question 2.
Define VMP?
Answer:

  1. VMP – means Value of Marginal Product.
  2. The Value of Marginal Product is obtained by multiplying the Marginal Physical Product of the factor by the price of product. Symbolically, VMP = MPP x Price.

Question 3.
Define : Marginal productivity theory of distribution.
Answer:
The theory states that price or the reward for any factor of production is equal to the marginal productivity of that factor.

Question 4.
What is marginal product ? Define marginal physical product ?
Answer:
The addition made to the total product by employment of an additional unit of that factor.

Question 5.
Define modern theory of rent.
Answer:
Rent is the difference between the actual earnings of a factor of production and its transfer earning.
Rent = Actual earning – Transfer earning

Question 6.
What is “Wage Fund“.
Answer:
According to Mill, “Every employer will keep a given amount of capital for payment to the workers”. It is known as “Wage Fund”.

Question 7.
Define Quasi Rent.
Answer:
“Quasi rent is the income derived from machines and other appliances made by man
QR = Total revenue – Total variable cost

Part – C
Answer the following questions in one Paragraph

Question 1.
Explain the types of distribution.
Answer:
1. Personal distribution :
It is the distribution of national income among individuals.

2. Functional distribution :
It means the distribution of income among the four factors of production for their services in the production process.

Question 2.
Explain the standard of living theory of wages.
Answer:
This theory of wages was developed by Torrance is an improved version of the subsistence theory of wage. According to this theory, the wage is equal to the standard of living of the workers. If the standard of living is high, wages will be high and vice versa. The standard of a living wage means the amount necessary to maintain the labourer in the standard of life to which he is accustomed.

Question 3.
Briefly explain the Abstinence theory or waiting theory?
Answer:

  1. This theory was propounded by N.W.Senior.
  2. According to Senior, capital is the result of saving.
  3. Saving involves “ abstinence ” or “ Sacrifice
  4. It is possible to save only if one abstains from present consumption.
  5. Interest is the reward or compensation paid to the saver (capitalist) for his “ abstinence ” or “ sacrifice”.
  6. Marshall accepted the Abstinence theory of interest.
  7. According to him, interest is a reward for waiting.
  8. Saving involves waiting.
  9. Therefore, interest is the reward paid to the saver for his “ waiting ”.

Question 4.
State the theories of interest.
Answer:

  1. Abstinence theory or waiting theory.
  2. Agio theory or the psychological theory.
  3. Loanable funds theory or the neoclassical theory.
  4. Liquidity preference theory or the monetary theory.

Question 5.
What are the four sources of loanable funds.
Answer:
The supply of loanable funds depends upon the following four sources.

  1. Savings (S)
  2. Bank Credit (BC)
  3. Dishoarding (DH)
  4. Disinvestment (DI)

Question 6.
What are the kinds of profit.
Answer:
1. Monopoly profit:
Profit earned by the firm because of its monopoly control.

2. Windfall profit:
Some times, profit arises due to changes in price level. Profit is due to unforeseen factors.

3. Profit as functional reward :
Just like rent, wage and interest, profit is earned by the entrepreneur for his entrepreneurial function.

Question 7.
State the residual claimant theory of wage.
Answer:
This theory was propounded by the American economist F.A. Walkar in 1875, in his book political economy. According to this theory, wage is the residual portion after paying the remuneration of all the other three factors namely land capital and organization.

Criticisms :

  1. This theory does not explain the role of trade unions can secure higher wage for workers.
  2. Demand side of labour in the determination of wages needs to be considered.

If you need some more information about Tamilnadu State Board Solutions for 11th Economics Chapter 6 Distribution Analysis Questions and Answers then visit our site frequently and get all the resources that you look for. Also, share this page with your friends and make them learn all the concepts thoroughly before the exams.

Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding

Enhance your subject knowledge with Tamilnadu State Board for Chapter 10 Chemical Bonding and learn all the underlying concepts easily. Make sure to Download Samacheer Kalvi 11th Chemistry Book Solutions Notes Pdf Chapter 10 Chemical Bonding Questions and Answers PDF on a day to day basis and score well in your exams. Are given after enormous research by people having high subject knowledge. You can rely on them and prepare any topic of Chemistry as per your convenience easily.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding

Students looking for Chapter 10 Chemical Bonding Concepts can find them all in one place from our Tamilnadu State Board Chemical Bonding. Simply click on the links available to prepare the corresponding topics of Chemistry easily. Samacheer Kalvi 11th Chemistry Chapter wise Questions and Answers are given to you after sample research and as per the latest edition textbooks. Clarify all your queries and solve different questions to be familiar with the kind of questions appearing in the exam. Thus, you can increase your speed and accuracy in the final exam.

Samacheer Kalvi 11th Chemistry Chapter 10 Chemical Bonding Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Chemical Bonding Multiple Choice Questions

Question 1.
In which of the following compounds does the central atom obey the octet rule?
(a) XeF4
(b) AICI3
(c) SF6
(d) SO2
Answer:
(d) SCl2
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-1
Hence in (d) SCl2 octet rule is followed

Question 2.
In the molecule OA = C = OB the formal charge on OA, C and OB are respectively.
(a) – 1, 0, +1
(b) +1, 0, – 1
(c) – 2, 0, +2
(d) 0, 0, 0
Answer:
(d) 0, 0, 0
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-2
Formal charge of Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-3
Formal charge of C = 4 – \(\left( 0+\frac { 8 }{ 2 } \right)\) = 4 – 4 = 0

Question 3.
Which of the following is electron deficient?
(a) PH3
(b) (CH3)2
(c) BH3
(d) NH3
Answer:
(c) BH3
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-3 – electron rich,
CH3 – CH3 – Covalent neutral molecule,
BH3 – electron deficient

Question 4.
Which of the following molecule contain no π bond?
(a) SO2
(b) NO2
(c) CO2
(d) H2O
Answer:
(d) H2O
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-5
Water (H2O) contains only σ bonds and no π bonds.

Question 5.
The ratio of number of sigma (σ) and pi (π) bonds in 2- butynal is …………..
(a) 8/3
(b) 5/3
(c) 8/2
(d) 9/2
Answer:
(d) 9/2
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-6
no. of σ bonds = 8 [4C – H; 3C – C; 1C— O]
no.of π bonds = 3 [2C – C; 1C – O]
∴ratio = \(\frac { 8 }{ 3 }\)

Question 6.
Which one of the following is the likely bond angles of sulphur tetrafluoride molecule?
(a) 120°, 80°
(b) 109°. 28
(c) 90°
(d) 89°, 117°
Answer:
(d) 89°, 117°
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-7
Solution:
Normal bond angle in regular trigonal bipyramidal are 90° and 120°. Due to l.p – b.p repulsion, bond angle is reduced to 89°, 117° option (d).

Question 7.
Assertion: Oxygen molecule is paramagnetic.
Reason: It has two unpaired electron in its bonding molecular orbital.
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) Both assertion and reason are false
Answer:
(c) assertion is true but reason is false
Correct statement: Oxygen molecule is paramagnetic
Correct Reason: It has two unpaired electrons in its antibonding molecular orbital.

Question 8.
According to Valence bond theory, a bond between two atoms is formed when ……………….
(a) fully filled atomic orbitals overlap
(b) half filled atomic orbitals overlap
(c) non-bonding atomic orbitals overlap
(d) empty atomic orbitals overlap
Answer:
(b) half filled atomic orbitals overlap

Question 9.
In CIF3, NF3 and BF3 molecules the chlorine, nitrogen and boron atoms are …………………….
(a) sp3 hybridised
(b) sp3, sp3 and sp2 respectively
(c) sp2 hybridised
(d) sp3d, sp3 and sp2 hybridised respectively
Answer:
(d) sp3d, sp3 and sp2 hybridised respectively
Solution:
CIF3 – sp3d hybridisation
NF3 – sp3 hybridisation
BF3 – sp2 hybridisation

Question 10.
When one s and three p orbitais hybridise,
(a) four equivalent orbitais at 900 to each other will be formed
(b) four equivalent orbitais at 1090 28’ to each other will be formed.
(c) four equivalent orbitals, that are lying the same plane will be formed
(d) none of these
Answer:
(b) four equivalent orbitals at 109° 28′ to each other will be formed.

Question 11.
Which of these represents the correct order of their increasing bond order?
(a) C2 < C22- < O2
(b) C22- < C2+ < O2 < O22-
(c) O22- < O2+ < O2 < C22-
(d) O22- < C2+ < O2 < C22-
Answer:
(d) O22- < C2+ < O2 < C22-
Solution:
bond order = (nb – na)
bond order of O22- = \(\frac { 1 }{ 2 }\) (8 – 6) = 1
bond order of C2+ = \(\frac { 1 }{ 2 }\) (5 – 2) = 1.5
bond order of O2 = \(\frac { 1 }{ 2 }\) (8 – 4) = 2
bond order of C22- = \(\frac { 1 }{ 2 }\) (8 – 2) = 3

Question 12.
Hybridisation of central atom in PCl5 involves the mixing of orbitais.
(a) s, px, py, dx2, dx2-y2
(b) s, px, py, pxy, dx2-y2
(c) s, px, py, pz, dx2-y2
(d) px, py, pxy, dx2-y2
Answer:
(c) s, px, py, pz, dx2-y2
Solution:
PCl5 – sp3d hybridisation s, px, py, pz, dx2-y2

Question 13.
The correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is ……………..
(a) H2O2 > O3 > O2
(b) O2 > O3 > H2O
(c) O2 > H2O2 > O3
(d) O3 > O2 > H2O2
Answer:
(b) O2 > O3 > H2O2
Solution:
The bond order for O2, O3 and H2O2 decreases in the order 2 > 1.5 > 1

Question 14.
Which one of the following is diamagnetic?
(a) O2
(b) O22-
(c) O22+
(d) None of these
Answer:
(b) O22-
Solution:
O22- is diamagnetic. Additional two electrons are paired in anti-bonding molecular orbits π*2py and π*2pz

Question 15.
Bond order of a species is 2.5 and the number of electrons are in its bonding molecular orbital is found to be 8. The no. of electrons in its anti-bonding molecular orbital is ………………….
(a) three
(b) four
(c) zero
(d) cannot be calculated form the given information.
Answer:
(a) three
Solution:
Bond order = \(\frac { 1 }{ 2 }\) (nb – na)
2.5 = \(\frac { 1 }{ 2 }\) (8 – na)
⇒ 5 = 8
⇒ na = 8 – 5 = 3

Question 16.
Shape and hybridisation of IF5 are ………….
(a) Trigonal bipyramidal, sp3d2
(b) Trigonal bipyramidal, sp3d
(c) Squai2e pyramidal, sp3d2
(d) Octahedral, sp3d2
Answer:
(c) Square pyramidal, sp3d2
Solution:
IF5 – 5 bond pair + 1 lone pair
∴ hybridisation sp3d2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-10

Question 17.
Pick out the incorrect statement from the following.
(a) sp3 hybrid orbitais are equivalent and are at an angle of 1 09°28’ with each other.
(b) dsp2 hybrid orbitais are equivalent and bond angle between any two of them is 900.
(c) All five sp3d hybrid orbitais arc not equivalent. Out of these five sp3d hybrid orbitais, three are at an angle of 120°, remaining two are perpendicular to the plane containing the other three
(d) none of these
Answer:
(c) All five sp3d hybrid orbitals are not equivalent. Out of these five sp3d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three.

Question 18.
The molecules having same hybridisation, shape and number of lone pairs of electrons are ………………..
(a) SeF4, XeO2F2
(b) SF4, XeF2
(c) XeOF4, TeF4
(d) SeCI4, XeF4
Answer:
(a) SeF4, XeO2F2
Solution:
SeF4, XeO2F2 – sp3d hybridisation
T – shaped, one lone pair on central atom.

Question 19.
In which of the following molecules / ions BF3, NO2, H2O the centrai atom is sp2 hybridised?
(a) NO2 and H2O
(b) NO2 and H2O
(c) BF3 and NO2
(d) BF3 and NH2
Answer:
(c) BF3 and NO2
Solution:
H2O – Central atom sp3 hybridised
NO2 – Central atom sp2 hybridised
BF3 – Central atom sp2 hybridised
NH2  – Central atom sp3 hybridised

Question 20.
Some of the following properties of two species, NO3 and H3O+ are described below. Which one of them is correct?
(a) dissimilar in hybridisation for the central atom with different structure.
(b) isostnictural with same hybridisation for the Central atom.
(c) different hybridisation for the central atom with same structure
(d) none of these
Answer:
(a) dissimilar in hybridisation for the central atom with different structure.
Solution:
NO3 – sp2 hybridisation, planar
H3O+ – sp3 hybridisation, pyramidal

Question 21.
The types of hybridisation on the five carbon atom from right to left in the, 2,3 pentadiene.
(a) sp3, sp2, sp, sp2, sp3
(b) sp3, sp, sp, sp, sp3
(c) sp2, sp, sp2, sp , sp3
(d) sp3, sp3, sp2, sp3, sp3
Answer:
(a) sp3, sp2, sp, sp2, sp3
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-11

Question 22.
XeF2 is isostructural with ………..
(a) SbCl2
(b) BaCl2
(c) TeF2
(d) ICl2
Answer:
(d) ICl2
Solution:
XeF2 is isostructural with ICI2

Question 23.
The percentage of s-character of the hybrid orbitais in methane, ethane, ethene and ethyne are respectively ………………..
(a) 25, 25, 33.3, 50
(b) 50, 50, 33.3, 25
(c) 50, 25, 33.3, 50
(d) 50, 25, 25. 50
Answer:
(a) 25, 25, 33.3, 50
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-12

Question 24.
Of the following molecules, which have shape similar to carbon dioxide?
(a) SnCI2
(b) NO2
(c) C2H2
(d) All of these
Answer:
(c) C2H2
Solution:
CO2 – Linear
C2H2 – Linear

Question 25.
According to VSEPR theory, the repulsion between different parts of electrons obey the order …………………
(a) l.p – l.p > b.p – b.p > l.p – b.p
(b) b.p – b.p > b.p – 1.p > l.p – b.p
(c) l.p – l.p > b.p – l.p > b.p – b.p
(d) b.p – b.p > l.p – l.p > b.p – l.p
Answer:
(c) l.p – l.p > b.p – l.p > b.p – b.p

Question 26.
Shape of CIF3 is ……………………..
(a) Planar triangular
(b) Pyramidal
(c) ‘T’ Shaped
(d) none of these
Answer:
(c) ‘T’ Shaped
Solution:
dF3 – sp3d hybridisation

Question 27.
Non- Zero dipole moment is shown by …………………
(a) CO2
(b) p – dichlorobenzene
(c) carbon tetrachloride
(d) water
Answer:
(d) water
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-13

Question 28.
Which of the following conditions is not correct for resonating structures?
(a) the contributing structure must have the same number of unpaired electrons.
(b) the contributing structures should have similar energies.
(c) the resonance hybrid should have higher energy than any of the contributing structure.
(d) none of these
Answer:
(c) the resonance hybrid should have higher energy than any of the contributing structure.
Solution:
Correct statement is – the resonance hybrid should have lower energy than any of the contributing structure.

Question 29.
Among the following, the compound that contains, ionic, covalent and coordinate linkage is …………………
(a) NH4Cl
(b) NH3
(c) NaCl
(d) none of these
Answer:
(a) NH4Cl

Question 30.
CaO and NaCl have the same crystal structure and approximately the same radii. If U is the lattice energy of NaCl, the approximate lattice energy of CaO is ………….
(a) U
(b) 2U
(c) U/2
(d) 4U
Answer:
(d) 4U

Samacheer Kalvi 11th Chemistry Chemical Bonding Short Answer Questions.

Question 31.
Define the following

  1. Bond order
  2. Hybridisation
  3. a- bond

Answer:
1. Bond order:
Bond orderSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-14
The number of bonds formed between the two bonded atoms in a molecule is called bond order.

2. Hybridisation:
It is a process of mixing of atomic orbitais of the same atom with. comparable energy to form equal number of new equivalent orbitais with same energy. The resultant orbitais are called hybridised orbitais and they possess maximum symmetry and definite orientation in space so as to minimise the force of repulsion between their electrons.

3. σ – bond:
When two atomic orbitais overlap linearly along the axis, the resultant bond is called sigma (σ) bond.

Question 32.
What is a pi bond?
Answer:
Pi – bond:
When two atomic orbitals overlap sideways, the resultant covalent bond is called a pi (π) bond.

Question 33.
In CH4, NH3 and H2O, the central atom undergoes sp3 hybridisation-yet their bond angles are different, why?
Answer:
1. In CH4, NH3 and H2O the central atom undergoes sp3 hybridisation. But their bond angles are different due to the presence of lone pair of electrons.

2. It can be explained by VSEPR theory. According to this theory, even though the hybridisation is same, the repulsive force between the bond pairs and lone pairs arc not same.

3. Bond pair-Bond pair < Bond pair – Lone pair < Lone pair – Lone pair So due to the varying repulsive force the bond pairs and lone pairs are distorted from regular geometry and organise themselves in such a way that repulsion will be minimum and stability will be maximum.

4. In case of CH4, there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry. i.e., tetrahedral with bond angle 109° 28’.

5. H2O has 2 bond pairs and 2 lone pairs. There is large repulsion between lp – lp. Again repulsion between lp – bp is more than that of 2 bond pairs. So 2 bonds are more restricted to form inverted V shape (or) bent shape molecule with a bond angle of 104° 35’.

6. NH3 has 3 bond pairs and 1 lone pair. There is repulsion between lp – bp. So 3 bonds are more restricted to form pyramidal shape with bond angle equal to 107° 18’.

Question 34.
Explain sp2 hybridisation in BF3
Answer:
1. sp2 hybridisation in boron trifluoride – Boron atom – B. Electronic configuration [H2]2s22p2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-15

2. In boron, the s orbital and two p orbitals in the valence shell hybridises to generate three equivalent sp2 orbitais. These 3 orbitaIs lie in the same xy plane and the angle between any two orbitals is equal to 120°.

3. The 3 sp2 hybridised orbitais of boron now overlap with the 2pz orbitais of fluorine (3 atoms). This overlap takes place along the axis.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-16
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-17

Question 35.
Draw the M.O diagram for oxygen molecule and calculate its bond order and show that O2 is paramagnetic.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-18

  1. Electronic configuration of O atom is is 1s2 2s2 2P4
  2. Electronic configuration of O, molecule is
    Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-19
  3. Bond order = Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-20
  4. Molecule has two unpaired electrons, hence it is paramagnetic.

Question 36.
Draw MO diagram of CO and calculate its bond order.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-21

  1. Electronic configuration of C atom: 1s2 2s2 2p2
    Electronic configuration of O atom: 1s2 2s2 2p4
  2. Electronic configuration of CO molecule is: σ1s2 σ*1s2 σ2s2 σ*22 π2py2 π2pz2 σ2px2
  3. Bond order Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-22
  4. Molecule has no unpaired electron, hence it is diamagnetic.

Question 37.
What do you understand by Linear combination of atomic orbitais in MO theory.
Answer:
Linear combination of atomic orbitais (LCAO):
1. The wave functions for the molecular orbitais can be obtained by solving Schrodinger wave equation for the molecule. Since solving Schrodinger wave equation is too complex, a most common method linear combination of atomic orbitais (LCAO) is used to obtain wave function for molecular orbitals.

2. Atomic orbitais are represented by wave functions ψ. Consider two atomic orbitals represented by the wave functions ψA and ψB with comparable energy that combines to form two molecular orbitals.

3. One is bonding molecular orbitai (ψ bonding) and the other is anti-bonding molecular orbital (ψ anti-bonding).

4. The wave function for molecular orbitais, ψA and ψB can be obtained by the LCAO as shown below:
ψbonding = ψA + ψB
ψanti-bonding = ψA – ψB

5. The formation of bonding molecular orbital can be considered as the result of constructive interference of the atomic orbitais and the formation of anti-bonding molecular orbital can be the result of the destructive interference of the atomic orbitais.

6. The formation of two molecular orbitals from two is orbitals is show below.

Constructive interaction:
The two is orbitals are in phase and have the same signs.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-23

Destructive interaction:
The two is orbitals are out of phase and have opposite signs
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-24
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-25

Question 38.
Discuss the formation of N2 molecule using MO Theory.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-26

  1. Electronic configuration of N atom 1s2 2s2 2p3.
  2. Electronic configuration of N, molecule is: σ1s2 σ*1s2 σ2s2 σ*22 π2py2 π2pz2 σ2px2
  3. Bond order Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-27
  4. Molecule has no unpaired electrons hence, it is diamagnetic.

Question 39.
What is dipole moment?
Answer:
1. The polarity of a covalent bond can be measured in terms of dipole moment which is defined as: µ = q x 2d, where µ is the dipole moment, q is the charge, 2d is the distance between the two charges.

2. The dipole moment is a vector quantity and the direction of the dipole moment points from the negative charge to positive charge.

3. The unit of dipole moment is Coulomb metre (C m). It is usually expressed in Debye unit (D).

4. 1 Debye = 3.336 x 10-30 Cm

Question 40.
Linear form of carbon dioxide molecule has two polar bonds. yet the molecule has zero dipole moment, why?
Answer:

  1. The linear form of carbon dioxide has zero dipole moment, even though it has two polar bonds.
  2. In CO2, there are two polar bonds [C = O], which have dipole moments that are equal in magnitude but have opposite direction.
  3. Hence the net dipole moment of the CO2 is µ = µ1 + µ2 = µ1 +( – µ1) = 0
  4. Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-28
  5. In this case Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-187

Question 41.
Draw the Lewis structures for the following species.

  1. NO3
  2. SO42-
  3. HNO3
  4. O3

Answer:
1. NO3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-29

2. SO42-
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-30

3. HNO3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-31
4. O3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-32

Question 42.
Explain the bond formation in BeCl2 and MgCl2. BeCl2 bond formation:
Answer:
1. Electronic confiuration of Be(Z = 4) is 1s2 2s2 and electronic configuration of Cl (Z = 17) is 1s2 2s2 2p6 3s2 3p5.

2. Beryllium has 2 electrons in its valence shell and chlorine atoms (2) have 7 electrons in their valence shell.

3. By losing two electrons, Beryllium attains the inert gas configuration of Helium and becomes a dipositive cation, Be2+ and each chlorine atom accepts one electron to become (Cl) uninegative anion and attains the stable electronic configuration of Argon.

4. Then Be2+ combine with 2Cl ions to form an ionic crystal in which they are held together by electrostatic attractive forces.

5. During the formation of 1 mole of BeCl2, the amount of energy released is – 468 kJ/mol. This favours the formation of BeCl, and its stabilisation.

MgCI2 bond formation:
1. Electronic configuration of Mg (z = 12) is 1s2 2s2 2p6 3s2.
Electronic configuration of Cl (z = 17) is 1s2 2p6 3p6 3p5

2. Magnesium has 2 electrons in its valence shell and chlorine has 7 electrons in its valence shell.

3. By losing two electrons, magnesium attains the inert gas configuration of Neon and becomes a dipositive cation (Mg2+) and two chlorine atoms accept these electrons to become two uninegative anions [2Cl] by attaining the stable inert gas configuration of Argon.

4. These ions, Mg2+ and 2Cl combine to form an ionic crystal in which they are held together by electrostatic attractive forces.

5. The energy released during the formation of 1 mole of MgCl2 is – 783 kJ/mole. This favours the formation of MgCI2 and its stabilisation.

Question 43.
Which bond is stronger or π? Why?
Answer:
1. Sigma bonds (σ) are stronger than Pi bonds (π). Because, sigma bonds are formed from bonding orbitals directly between the nuclei of the bonding atoms, resulting in greater overlap and a strong sigma bond (axial overlapping).

2. π bonds results from overlap of atomic orbitals that are in contact through two areas of overlap (lateral overlapping). Pi bonds are more diffused bonds than sigma bonds.

Question 44.
Define bond energy.
Answer:
Bond energy:
Bond energy (or) Bond enthalpy is defined as the minimum amount of energy required to break one mole of a bond in molecules in their gaseous state. The unit of bond energy is kJ mol-1

Question 45.
Hydrogen gas is diatomic whereas inert gases are monoatormic – explain on the basis of MO theory.
Answer:
1. Hydrogen gas is diatomic. According to MO theory. which is based on quantum mechanics H2 molecule can be represented in terms of the following diagram called M.O. diagram.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-33
H – H. i.e.. H2 molecule has two atoms which are connected by 1 σ bond. So it is diatomic.

2. But in the case of inert gases. the valence shell is fully filled i.e.. an octet (8 electrons) (or) duplet (2 electrons) in case of Helium, due to which they are in monoatomic state and remain stable. So they do not combine with any atom (neither of same or of different elements). Due to this they do no exist in diatomic state and always exist in mono – atomic state.

Question 46.
What is Polar Covalent bond? Explain with example.
Answer:
1. If a covalent bound is formed between atoms having different electronegativities. the atom with higher electronegativity will have greater tendency to attract the shared pair of electrons towards itself than the other atom. As a result, the cloud of shared electron pair gets distorted and polar covalent bond is formed.

2. Example – HF – Hydrogen fluoride:
The electronegativities of hydrogen and fluorine on Pauling’s scale are 2.1 and 4 respectively. It means that fluorine attract the shared pair of

electrons approximately twice as much as hydrogen which leads to partial negative charge on fluorine atom and partial positive charge on hydrogen atom. Hence, the H – F bond is said to be a polar covalent bond.

Question 47.
Considering x-axis as molecular axis, which out of the following will form a sigma bond.

  1. 1s and 2py
  2. 2px and 2px
  3. 2px and 2pz
  4. 1s and 2Pz

Answer:
Along X-axis as molecular axis, only 2p and 2p can form a sigma bond
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-34

Question 48
Explain resonance with reference to carbonate ion
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-35
1. For the above structure, we can draw two additional lewis structures by moving the lone pairs from the other two oxygen atoms OB and OC. and thus creating three similar structures in which the relative positive of the atoms are same.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-36

2. They only differ in the position of bonding and lone pair of electrons. Such structures are called resonance structures and this phenomenon is called resonance.

3. It is evident from the experimental results that all carbon-oxygen bonds in carbonate ion are equivalent. The actual structure of the molecule is said to be a resonance hybrid, an average of these 3 resonance forms. The following structure gives a qualitative idea about the correct structure of CO32- (carbonate) ion.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-37

Question 49.
Explain the bond formation in ethylene and acetylene.
Answer:
Bonding in Ethylene, C2H4
1. Bonding in ethylene can he explained by hybridisation concept.

2. The valency of carbon is 4. The electronic configuration of carbon is 1s2 2s2 2px1 2py1 2pz0. One electron from 2s orbital is promoted to 2pz. orbital in the excited state to satisfy the valency of carbon.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-38

3. In ethylene both the carbon atoms undergo sp2 hybridisation involving 2s, 2px and spy orbitals resulting in 3 equivalent sp2 hybridised orbitals lying in the XY plane at an angle of 1200 to each other. The unhybridised 2pz orbital lies perpendicular to the xy plane.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-39

4. One of the sp2 hybndised orbitals of each carbon atoms lying along the X – axis linearly overlaps with each other resulting in the formation of C – C sigma bond. The other two sp2 hybridised orbitals of both carbon atom linearly overlap with the four is orbitals of four hydrogen atoms leading to the formation of two C – H sigma bonds on each carbon atom.

5. The unhybridised 2pz orbital of both carbon atoms can overlap only sideways as they are not in the molecular axis. This lateral overlap results in the formation of a pi bond between the two carbon atoms.

Bonding in acetylene (C2H2):
1. The electronic configuration of valence shell of carbon atom in the ground state is [He] 2s2 2px1 2pz0. One electron from 2s orbital is promoted to 2pz orbital in the excited state to satisfy the valency of carbon.

2. In acetylene molecule, both the carbon atoms are in sp hybridised state. The 2s and 2px orbitals resulting in two equivalent sp hybridised orbitals are formed lying in a straight line along the X – axis. The unhybridised 2py, and 2pz orbitais lie perpendicular to the X-axis.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-41
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-42
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-43

3. One of the two sp hybridised orbitals of each carbon atom linearly overlaps with each other resulting in the formation of a C – C sigma bond. The other sp hybridised orbital of both carbon atoms linearly overlap with the two is orbitals of two hydrogen atoms leading to the formation of one C – H sigma bond on each carbon atom.

4. The unhybridised 2py and 2pz orbitals of each carbon atom overlap sideways. This lateral overlap results in the formation of two pi bonds. (py – py) and (pz – pz) between the two carbon atoms.

Question 50.
What type of hybridisations are possible ¡n the following geometeries?

  1. octahedral
  2. tetrahedral
  3. square planar.

Answer:

  1. Octahedral geometry is possible by sp3d2 (or) d2sp3 hybridisation.
  2. Tetrahedral geometry is possible by sp3 hybridisation.
  3. Square planar geometry is possible by dsp2 hybridisation.

Question 51.
Explain VSEPR theory. Applying this theory to predict the shapes of F7 and SF6.
Answer:
VSEPR theory:
1. The shape of the molecules depend on the number of valence shell electron pair around the central atom.

2. There are two types of electron pairs namely, bond pairs and lone pairs.

3. Each pair of valence electrons around the central atom repel each other and hence they are located as far away as possible in three dimensional space to minimise the repulsion between them.

4. The repulsive interaction between the different types of electron pairs is in the following
order:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-44

5. The lone pair of electrons are localised only on the central atom and interact with only one nucleus whereas the bond pairs are shared between two atoms and they interact with two nuclei. Because of this, the lone pairs occupy more space and have greater repulsive power than the bond pairs in a molecule.

IF7:
It is an AB7 type molecule. This molecule has 7 bond pair of electrons and no lone pair of electrons. Due to bond pair-bond pair interaction of electrons, IF7 has pentagonal bipyramidal shape.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-45

SF6:
It is an AB6 type molecule. This molecule has 6 bond pairs of electrons and no lone pair of electrons. Due to bond pair-bond pair interaction of electrons, SF6 has octahedral shape.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-46

Question 52.
CO2 and H2O both are triatomic molecules but their dipole moment values are different. Why?
Answer:
1. Linear form of carbon dioxide has zero dipole moment. In CO2 the dipole moment of two polar bonds are equal in magiitude but have opposite direction. Hence, the net dipole moment of the CO2 molecule is
µ = µ1 + µ2
µ = µ1 + (- µ1) = 0
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-47
In this caseSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-48

2. But in the case of water, net dipole moment is the vector sum µ1 + µ2 as follows:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-49
Dipole moment in water is found to be 1.85 D.

3. CO2 and H2O both are triatomic molecules but their dipole moment values are zero and 1.85 D respectively.

Question 53.
Which one of the following has highest bond order? N2, N2+ or N2 ?
Answer:
N2 (14 electrons)
Bond order = 3, Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-50
N2+ (13 electrons)
Bond order = 2.5, Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-51
N2(15 electrons)

Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-52
So N2 has the highest bond order.

Question 54.
Explain the covalent character in ionic bond.
Answer:
1. Ionic compounds like lithium chloride shows covalent character and it is soluble in organic solvents such as ethanol.

2. The partial covalent character in ionic compounds can be explained on the basis of a phenomenon called polarisation.

3. In an ionic compound, there is an electrostatic attractive force between the cation and anion. The positively charged cation attract the valence electrons of anion while repelling the nucleus.

This cause a distortion in the electron cloud of the anion and its electron density drills towards the cation, which results in some sharing of valence electrons between these ions. Thus, a partial covalent character is developed between them. This phenomenon is called polarisation.

4. Thus due to polarisation, ionic compounds shows covalent character.

Question 55.
Describe Fajan’s rule.
Answer:
1. The ability of a cation to polarise an anion is called its polarising ability and the tendency of an anion to get polarised is called its polarisibility. The extent of polarisation in an ionic compound is given by the Fajans rule.

2. To show greater covalent character, both the cation and anion should have high charge on them. Higher the positive charge on the Cation greater will be the attraction on the electron cloud of the anion.

Similarly higher the magnitude of negative charge on anion, greater is its polansability. For example, Na+ < Mg2+ < Al3+, the covalent character also follows the order – NaCI < MgCl2 < AICI3

3. The smaller cation and larger anion show greater covalent character due to the greater extent of polarisation. e.g., LiCl is more covalent than NaCI.

4. Cation having ns2np6nd10 configuration shows greater polarising power than the cations with ns2np6 configuration. e.g., CuCI is more covalent than NaCl.

Samacheer Kalvi 11th Chemistry Chemical Bonding In Text Questions Evaluate yourself

Question 1.
Draw the lewis structures for

  1. Nitrous acid (HNO2)
  2. Phosphoric acid
  3. Sulphur troxide (SO3)

Answer:
1. Nitrous acid (HNO2)
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-53

2. Phosphoric acid
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-54

3. Sulphur troxide (SO3)
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-55

Question 2.
Calculate the formal charge on each atom of carbonyl chloride (COCl2)
Answer:
Formal charge  Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-188
Carbonyl chloride COCl2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-56
Formal charge on carbon atom = 4 – \(\left[ 0+\frac { 8 }{ 2 } \right]\) = 4 – 4 = 0
Formal charge on chlorine atom = 7 – \(\left[ 6+\frac { 2 }{ 2 } \right]\) = 7 – 7 = 0
Formal charge on oxygen atom = 6 – \(\left[ 4+\frac { 4 }{ 2 } \right]\) = 6 – 6 = 0

Question 3.
Explain the ionic bond formation in MgO and CaF2
Magnesium oxide (MgO):
Answer:
Electronic configuration of Mg – 1s2 2s2 2p5 3s2
Electronic configuration of O – 1s2 2s2 2p6 3s6 3p4.
1. Magnesium has two electrons in its valence shell and oxygen has six electrons in its valence shell.

2. By losing two electrons, Mg acquires the inert gas configuration of Neon and becomes a dipositive cation Mg2+
Mg → Mg2+ + 2e

3. Oxygen accepts the two electrons to become a dinegative oxide anion, O2- thereby attaining the inert gas configuration of Neon.
O + 2e → O2-

4. These two ions, Mg2+ and O2- combine to form an ionic crystal in which they are held together by electrostatic attractive forces.

5. During the formation of magnesium oxide crystal 601.6 kJ mol-1 energy is released . This favours the formation of magnesium oxide (MgO) and its stabilisation.

CaF2, Calcium fluoride
1.  Calcium, Ca: [Ar] 4s2, Fluorine F: [He] 2s2 2p5

2. Calcium has two electrons in its valence shell and fluorine has seven electrons in its valence shell.

3. By losing two electrons, calcium attains the inert gas configuration ofArgon and becomes a dipositive cation, Ca2+.

4. Two fluorine atoms, each one accepts one electron to become two unincgative fluoride ions (F) thereby attaining the stable configuration of Neon.

5. These three ions combine to form an ionic crystal in which they are held together by electrostatic attractive force.

6. During the formation of calcium fluoride crystal 1225.91 kJ mol-1 of energy is released. This favours the formation of calcium fluoride, CaF2 and its stabilisation.

Question 4.
Write the resonance structures for

  1. Ozone molecule
  2. N2O

Answer:
1. Ozone molecule, O3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-57

2. Nitrous oxide, N2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-58

Question 5.
Of the two molecules OCS and CS2 which one has higher dipole moment value? Why?
Answer:
OCS and CS2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-59

The dipole moment µOCS = 0.7 149 ± 0.0003 Debye.
CS2:
S = C = S
In CS2, the bond dipoles of 2C = S have same values and bond dipoles cancel each other so dipole moment of CS2 is zero. Among OCS and CS2, OCS has a higher dipole moment because in OCS oxygen is more electronegative than sulphur and C = S and C = O bonds in OCS molecules do not cancel each other.

On the other hand CS2, due to linear structure, the bond dipole of two C = S bonds cancel each other. On the other hand, CS2 due to linear structure, the bond dipoles of two C = S bonds cancel each other and thc recultant dipo’c moment value is zero. So OCS has a higher dipole moment than CS2.

Question 6.
Arrange the following in the decreasing order of Bond angle

  1. CH4, H2O, NH3
  2. C2H2, BF3, CCl4

Answer:
1. CH4, H2O, NH3:
NH3 = 107°, Water= 104.5°, CH4 = 109.5°
Decreasing order of bond angle: H2O < NH3 < CH4

2. C2H2, BF3, CCl4:
C2H2 = 1800, BF3 = 120°, CCl4 = 109.5°
Decreasing order of bond angle: CCl4 < BF3 < C2H2

Question 7.
Bond angle in PH4+ is higher than in PH3. Why?
Answer:
Phosphorous in both PH3 and PH4+ is sp3 hybridised. Due to the absence of lone pair – bond pair repulsion and presence of four identical bond pair – bond pair interactions, PH4+ assumes tetrahedral geometry with a bond angle of 109° 28’.

But PH3 has three bond pairs and one lone pair around P. Due to greater lone pair-bond pair repulsion than bond pair-bond pair repulsion, the tetrahedral angle decreases from 109° 28’ to 93.6°. As a result PH3 is pyramidal.
PH3 – Pyramidal with bond angle of 93.6°. PH4+ Tetrahedral with bond angle of 109° 28’.

Question 8.
Explain the bond formation in SF4 and CCl4 using hybridisation concept.
Answer:
In SF4, the central atom is sp3d hybridised.
1. The molecule SF3 will have a total 34 valence electron 6 form sulphur, 7 each from four fluorine atoms.

2. Sulphur atom will from 4 single bonds with fluorine atoms. These bonds account for the 8 electrons out of the 34 valence electrons. Each fluorine atom will have 3 lone pair of electrons in order to have a complete octet structure.

These lone pairs will use up 24 valence electrons. So the total used valence electrons, are 32. The remaining 2 electrons will be placed on the sulphur atom as a lone pair. Sulphur atom gets a total of 10 electrons 8 from the bonds and 2 as lone pair.

3. This is quite Sulphur atom gets a total of 10 electrons 8 from the bonds and 2 as lone pair. This is quite possible for sulphur because it has easy access to its 3d orbital which means that it can expand its octet and accommodate more than 8 electrons.

4. Sulphur forms 4 single bonds and has 1 lone pair which means that its steric number is equal to 5. In this case sulphur will use five hybrid orbitais, such as one 3s orbital three 3p orbitais and one 3d orbital. So the central atom is sp3d hybridised.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-60

CCl4:
1. It is not necessary to invoke hybridisation especially in CCl4. It must be invoked for all tetrahedral bonds of carbon and other atoms.

2. The electronic configuration of an isolated carbon atom in its ground state is 1s2 s2 2p2.

3. CCl4 is a tetrahedral molecule comprising of four single bonds known as a bonds between the carbon atom and the chlorine atoms. In this type of bonding, the 2s orbital and three 2p orbitals of carbon atoms are mixed to produce four identical orbitals, a process known as sp3 hybridisation.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-61

Question 9.
The observed bond length of N2+ is larger than N2 while the bond length in NO+ is less than in NO. Why?
Answer:
(a)
(1) By molecular orbital theory, the bond order of both N2+ is 2.5 whereas N2 is 3.

2. N2 has 5e in the antibonding molecular orbital whereas N2+ has 4e in the antibonding molecular orbital. So N2+ will make a stronger and shorter bond length.

3. More the bond order and bond strength, and lesser will be the bond length.

4. So we can easily conclude N2 has more bond length than N2
Bond order in Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-62
Bond order inSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-63
So, N2 is more stable than N2+. But bond length N2+ is greater than N2.

(b) NO+ & NO
Bond order of NO = 2.5
Bond order of NO+ = 3
Due to lesser bond order in NO, the bond length is greater than NO+ So, NO+ bond length is shorter than NO bond length.

Question 10.
Draw the MO diagram for acetylide ion C22- and calculate its bond order.
Answer:
Acetylide ion, C22- in acetylene
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-64
Electronic configuration of C2 ion is:
σ1s2 σ*1s2 σ1s2 π2px2 π2py2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-65
Bond order
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-66

Samacheer Kalvi 11th Chemistry Chemical Bonding Additional Questions Solved

Samacheer Kalvi 11th Chemistry Chemical Bonding 1 Mark Questions and Answer:

I. Choose the correct answer.

Question 1.
Which is the correct Lewis structure of Helium?
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-67
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-68
Solution:
Helium has only two electrons in its valence shell which is represented as a pair of dots (duplet).

Question 2.
Which one of the following form only covalent bonds?
(a) Alkali metals
(b) Metals
(c) Non metals
(d) Metalloids
Answer:
(c) Non metals

Question 3.
In which one of the following molecule triple bond is present?
(a) O2
(b) H2
(c) CO2
(d) N2
Answer:
(d) N2
Solution:
N ≡ N

Question 4.
Which of the following is the lewis structure of water?
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-69
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-70

Question 5.
Which one of the following element forms only one bond?
(a) Carbon
(b) Oxygen
(c) Fluorine
(d) Nitrogen
Answer:
(c) Fluorine

Question 6.
Which one is the preferred structure of CO2?
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-71
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-189

Question 7.
Which is the correct lewis structure of BF3?
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-72
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-73

Question 8.
Statement I: In sulphur hexafluoride, the central atom has more than eight valence electrons.
Statement II: The central atom can accommodate additional electron pairs by using outer vacant d orbitals.
(a) Statements I and II are correct and statement II is the correct explanation of statement I.
(b) Statements I and II are correct but statement II is not the correct explanation of statement I.
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(a) Statements I and II are correct and statement II is the correct explanation of statement I.

Question 9.
Which one of the following molecule has complete octet?
(a) BF3
(b) BeCl2
(c) BCl3
(d) CCI4
Answer:
(d) CCI4

Question 10.
Which one of the following does not have electrovalent bond?
(a) KCI
(b) NaI
(c) MgO
(d) CCI4
Answer:
(d) CCI4

Question 11.
Which one of the following has an ionic bond?
(a) CO2
(b) CH4
(c) CaF2
(d) BeCI2
Answer:
(c) CaF2

Question 12.
During the formation of 1 mole of KCI crystal, the amount of energy released is ………….
(a) 418.81 kJ
(b) 348.56 kJ
(c) 718 kJ
(d) 70.25 kJ
Answer:
(c) 718 kJ

Question 13.
Which one of the following has coordinate covalent bond?
(a) CaF2
(b) MgO
(c) [Fe(CN)6]4-
(d) KCI
Answer:
(c) [Fe(CN)6]4-

Question 14.
The distance between the nuclei of the two covalently bonded atoms is called …………….
(a) bond order
(b) bond length
(c) bond angle
(d) bond enthalpy
Answer:
(b) bond length

Question 15.
The length of a bond can be determined by ……………
(a) spectroscopic method
(b) x – ray diffraction method
(c) electron-diffraction method
(d) all the above
Answer:
(d) all the above

Question 16.
The value of carbon-carbon single bond length is ………..
(a) 1.43A
(b) 1.54Å
(c) 1.33A
(d) 1.20A
Answer:
(b) 1.54Å

Question 17.
The value of carbon – carbon double bond length is …………..
(a) 1.43Å
(b) 1.20Å
(c) 1.54A
(d) 1.33A
Answer:
(d) l.33A

Question 18.
The value of carbon – carbon triple bond length is ……………..
(a) 1.33A
(b) l.20Å
(C) 1.54A
(d) 1.43A
Answer:
(b) 1.20Å

Question 19.
Among the following which one has bond order as 3?
(a) N2
(b) O2
(c) HCHO
(d) CH4
Answer:
(a) N2

Question 20.
Which one of the flowing has bond order as 2?
(a) N2
(b) C2 – H4
(c) CH4
(d) HCN
Answer:
(b) C2H4

Question 21.
Identify the molecule with bond order 1.
(a) N2
(b) O2
(c) H2
(d) C2H4
Answer:
(c) H2

Question 22.
Which one of the following has zero dipole moment?
(a) HF
(b) H2
(c) CO
(d) NO
Answer:
(b) H2

Question 23.
Which one of the following is called polar molecule?
(a) H3
(b) O2
(c) F2
(d) NO
Answer:
(d) NO

Question 24.
Statement I: CuCl is more covalent than NaCI.
Statement II: As compared to Na+. Cu+ is small and have 3s2 3p6 3d10 configuration and show greater polarisation.
(a) Statement I & II are correct and II is the correct explanation of I
(b) Statement I & II are correct but II is not the correct explanation of I
(c) Statement I & II are correct but II is wrong
(d) Statement I & II are wrong and II is the correct.
Answer:
(a) Statement I & II are correct and II is the correct explanation of I

Question 25.
Which of the following has see saw shape?
(a) PCl5
(b) IO2F2
(c) SOF4
(d) ClO33
Answer:
(b) IO2F2

Question 26.
Which one of the following has trigonal bipyramidal shape?
(a) SF6
(b) IF4+
(c) AsF5
(d) SF4
Answer:
(c) AsF5

Question 27.
Which one of the following does not have tetrahedral shape?
(a) NH4+
(b) ClO4
(c) HCHO
(d) CH4
Answer:
(c) HCHO

Question 28.
Which one of the following has linear shape?
(a) O3
(b) CO32-
(c) NO3
(d) BCl3
Answer:
(a) O3

Question 29.
Which of the following has linear shape?
(a) PCI5
(b) SnBr2
(c) BeCl2
(d) CCl2F2
Answer:
(c) BeCl2

Question 30.
Which one of the following has tetrahedral shape?
(a) HCHO
(b) BeCl2
(c) PbCl2
(d) CF2Cl2
Answer:
(d) CF2CI2

Question 31.
Which one of the following pair has T – shapcd structure?
(a) BrF3, CIF3
(b) SF4, IF4+
(c) PCl5, AsF5
(d) NH3, PF3
Answer:
(a) BrF3, CIF3

Question 32.
Which one of the following has pentagonal bipyramidal shape?
(a) XeF4
(b) XeOF4
(c) IF7
(d) IOF5
Answer:
(c) IF7

Question 33.
Which one of the following has linear shape?
(a) I3
(b) ICI4
(c) BrF5
(d) IOF5
Answer:
(a) I3

Question 34.
Which one of the following is the correct increasing order of bond angle?
(a) H2O < CH4 < BF3 < BeCI2
(b) BeCI2 < BF3 < CH4 < H2O
(c) BF3 < CH4 < BeCI2 < H2O
(d) CH4 < BeCI2 < H2O < BF3
Answer:
(a) H2O < CH4 < BF3 < BeCI2

Question 35.
Which one of the following hybridisation takes place in the formation of BeCI2?
(a) sp2
(b) sp
(c) sp3
(d) dsp2
Answer:
(b) sp

Question 36.
Which hybridisation is possible in BF3?
(a) sp2
(b) sp
(c) sp3
(d) sp3d
Answer:
(a) sp2

Question 37.
Which one of the following has bond order as 2.5?
(a) O2
(b) NO
(c) CO
(d) H2
Answer:
(b) NO

Question 38.
Which one of the following is an electron deficient compound?
(a) Al2Cl6
(b) AlBr3
(c) SF6
(d) BF3
Answer:
(d) BF3

Question 39.
Apply the VSEPR model to XeF4, which of the following molecular shape is consistent with the model?
(a) Square planar
(b) Tetrahedral
(c) Square pyramidal
(d) Octahedral
Answer:
(a) Square planar

Question 40.
On the basis of molecular orbital theory, select the most appropriate option.
(a) The bond order of O2 is 2.5 and it is paramagnetic
(b) The bond order of O2 is 1.5 and it is paramagnetic
(c) The bond order of O2 is 2 and it is diamagnetic
(d) The bond order of O2 is 2 and it is paramagnetic
Answer:
(d) The bond order of O2 is 2 and it is paramagnetic

Question 41.
Which of the following molecule does not exist due to its zero bond order?
(a) H2
(b) He2+
(c) He2
(d) H2+
Answer:
(c) He2

Question 42.
Which of the following molecules have bond order equal to 1?
(a) NO, HF, HCl, Li2, CO
(b) H2, Li2, HF, Br2, HCI
(c) Li2, B2, CO, NO, He2+
(d) B2, CO, He2+, NO, HF
Answer:
(b) H2, Li2, HF, Br2, HCI
Solution:
Bond order of He2+ = 0.5
Bond order of NO = 2.5
Bond order of CO = 3

Question 43.
Arrange the following molecules in decreasing order of bond length.
(a) O2 >O2 > O2+ > O22-
(b) O22- > O2 > O2 > O2
(c) O22- > O2 > O2+ > O2
(d) O2+ > O2+ > O22- > O2
Answer:
(b) O22- > O2 > O2 > O2
Solution: Since the bond length is inversely proportional to the bond order so option ‘b’ is correct.

Question 44.
Among the following which shows the maximum covalent character?
(a) MgCI2
(b) FeCI2
(c) SnCI2
(d) AICI3
Answer:
(d) AICI3

Question 45.
Which of the following has maximum number of lone pairs associated with Xe?
(a) XeF2
(b) XeO3
(c) XeF4
(d) XeF6
Answer:
(a) XeF2

Question 46.
During the formation of a chemical bond …………………
(a) energy decreases
(b) energy increases
(c) energy remains zero
(d) energy remains constant
Answer:
(a) energy decreases

Question 47.
Using MO theory, predict which of the following species has the shortest bond length?
(a) O2+
(b) O2
(c) C22-
(d) O22+
Answer:
(d) O22+

Question 48.
Identify the incorrect statement regarding the molecule XeO4.
(a) XeO4 molecule is tetrahedral
(b) XeO4 molecule is square planar
(c) There are four pπ – dπ bonds
(d) There are four sp3 – p, s bonds
Answer:
(b) XeO4 molecule is square planar

Question 49.
Which of the following contains maximum number of lone pairs on the central atom?
(a) ClO3
(b) XeF4
(c) SF4
(d) I3
Answer:
(d) I3

Question 50.
Which one of the following is a correct set?
(a) H2O, sp3, bent
(b) H2O, sp2, linear
(c) NH4+, dsp2, square planar
(d) CH4+, dsp2 tetrahedal
Answwer:
(a) H2O, sp3, bent

II. Match the following

Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-76
Answer:
(b) 4 3 1 2

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-77
Answer:
(a) 3 4 1 2

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-78
Answer:
(a) 3 4 2 1

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-79
Answer:
(b) 3 4 1 2

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-80
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-80.1
Answer:
(c) 2 3 4 1

Question 6.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-81
Answer:
(b) 2 4 1 3

Question 7.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-82
Answer:
(c) 4 3 1 2

Question 8.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-83
Answer:
(d) 2 3 4 1

Question 9.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-84
Answer:
(a) 2 4 1 3

Question 10.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-85
Answer:
(a) 4 3 2 1

Question 11.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-86
Answer:
(a) 4 1 2 3

III. Fill in the blanks.

Question 1.
The electrovalent bond is present in ………..
Answer:
NaCI
Solution:
Na+ cation and Cl anion are held together by electrostatic attractive forces and this is called electrovalent bond.

Question 2.
The structure (or) shape of water molecule is …………
Answer:
inverted ‘V’ shape
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-87

Question 3.
The structure of CO2 is ………..
Answer:
linear
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-88

Question 4.
In the formation of a chemical bond between Na and C[, they attain the stable configuration of …………….
Answer:
Ne, Ar
Solution:
Na+: 1s2 2s2 2p6 = [Ne]
Cl: 1s2 2s2 3s2 3p6 = [Ar]

Question 5.
The mutual sharing of one or more pair of electrons between the two combining atoms results in the formation of …………..
Answer:
Covalent bond

Question 6.
Formal charge of an atom can be calculated by the formula ………….
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-89

Question 7.
The formal charge on the carbon atom in the following structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-90 is …………………
Answer:
zero
Solution:
Formal charge on carbon atom
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-91

Question 8.
The formal charge on both oxygen atoms in the structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-92 is …………..
Answer:
0
Solution:
Formal charge on oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-93

Question 9.
The formal charge on singly bonded oxygen atom in the structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-94 is …………..
Answer:
-1
Solution:
Formal charge on singly bonded oxygen atom
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-95

Question 10.
The formal charge on the triply bonded oxygen atom in the structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-96 is …………
Answer:
+ 1
Solution:
Formal charge on triply bonded oxygen atom
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-97

Question 11.
The complete transfer of one or more valence electron from one atom to another leads to the formation of ………….
Answer:
Ionic bond

Question 12.
The shape of the molecule is determined approximately by ……………
Answer:
bond angle

Question 13.
The unit of bond enthalpy is …………..
Answer:
kJ mol-1

Question 14.
The electronegativity of hydrogen and fluorine on Pauling’s scale are …………..
Answer:
2.1 and 4

Question 15.
The unit of dipole moment is ……………
Answer:
Coulomb-1 m2

Question 16.
The dipole moment of CO2 is ……………
Answer:
0
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-98

Question 17.
The shape of sulphur hexafluoride is …………
Answer:
Octahedral

Question 18.
The type of hybridisation takes place in methane is ………………
Answer:
sp3

Question 19.
The type of hybridisation takes place in SF6 is …………..
Answer:
sp3d2

Question 20.
The number of lone pair of electrons on C – atom present in CO2 are …………
Answer:
4

Qustion 21.
In SF6, the bond angle is …………….
Answer:
900

Question 22.
The ions have noble gas electronic configuration was suggested by ……………….
Answer:
Kossel

Question 23.
Tetrachlorornethane is a molecule …………..
Answer:
non polar

Question 24.
In C2H4, type of bonds present are ……………
Answer:
Covalent bonds only

Question 25.
Molecule with bond of shape trigonal pyramid is …………..
Answer:
BF3

Question 26.
When magnesium reacts with oxygen, nature of bond formed is …………….
Answer:
ionic

Question 27.
The number of lone pair of electrons in water molecule is …………..
Answer:
2

Question 28.
Double bonds as compared to single bonds are ……………….
Answer:
Shorter

Question 29.
Number of chlorine atoms which form equatorial bonds in PCI5 molecule are/is ………..
Answer:
3

Question 30.
The hybridisation of B in BF3 is ……………
Answer:
sp2

Question 31.
Bond order of O2, F3, N2 respectively are ……………
Answer:
2, 1, 3

Question 32.
Hybridisation which takes place in acetylene is …………….
Answer:
sp

Question 33.
Bond order of O2, F2, N2 respectively are ………….
Answer:
2, 1,3

Question 34.
Hybridisation which takes place in acetylene is …………..
Answer:
sp

Question 35.
The hybndisation of orbitais of N atom in NO3, NO3+ and NH4+ are respectively ……………….
Answer:
sp2, sp, sp3

Question 36.
Malleability and ductility of metals can be accounted due to the capacity of layers of …………………. to slide over one another.
Answer:
metal ions

Question 37.
For a stable molecule, the value of bond order must be ……………
Answer:
positive

Question 38.
In acetylene molecule between the carbon atoms there are ………….. σ and ……………. bonds.
Answer:
one, two

IV. Choose the odd one out.

Question 1.
(a) Hydrogen
(b) Chlorine
(c) Neon
(d) Argon
Answer:
(c) Neon. It is mono atomic whereas others are diatomic.

Question 2.
(a) NaCl
(b) CO2
(c) LiF
(d) MgO
Answer:
(b) CO2. It contains covalent bond whereas others have ionic bond.

Question 3.
(a) Methane
(b) Ceasium chloride
(c) Ammonia
(d) Boron trifluoride
Answer:
(b) Ceasium chloride. It is an ionic compound whereas others are covalent compounds.

Question 4.
(a) H2
(b) O2
(c) Cl2
(d) F2
Answer:
(b) O2. It’s bond order is 2 whereas in others bond order is 1.

Question 5.
(a) BeCI2
(b) CS2
(c) BF3
(d) HCN
Answer:
(c) BF3. It is AB3 type whereas others are AB2 type.

Question 6.
(a) XeO2F2
(b) PCI5
(c) AsF5
(d) SOF4
Answer:
(a) XeO2F2. It is AB4L type whereas others are AB5 type.

V. Choose the correct pair.

Question 1.
(a) NaCI – ionic compound
(b) NH3 – coordinate compound
(c) BF3 – ionic compound
(d) H2 – ionic compound
Answer:
(a) NaCl – ionic compound

Question 2.
(a) O2 – Bond order 3
(b) H2 – Bond order 2
(c) N2 – Bond order 3
(d) Cl2 – Bond order 2
Answer:
(c) N2 – Bond order 3
[N = N] Bond order is 3.

Question 3.
(a) CH4 – ionic bond
(b) BF3 – dative bond
(c) NH3 – metallic bond
(d) CCI4 – covalent bond
Answer:
(d) CCI4 – covalent bond

Question 4.
(a) CH4 – 107° 18’
(b) H2O – 109°28’
(c) NH3 – 104°35’
(d) BF3 – 120°
Answer:
(d) BF3 – 120°

Question 5.
(a) AB3 – Linear
(b) AB3 – V-shape(or)bent
(c) AB4 – Trigonal planar
(d) AB5 – T-shape
Answer:
(a) AB3 – Linear

VI. Choose the incorrect pair.

Question 1.
(a) CS2 – Linear
(b) BF1 – Trigonal planar
(c) CH4 – T-shape
(d)NH3 – Pyramidal
Answer:
(c) CH4 – T-shape

Question 2.
(a) AB3 – Trigonal planar
(b) AB3L2 – T-shape
(c) AB5 – Trigonal bipyramidal
(d) AB3L – Bent
Answer:
(a) AB3L : Bent.
Actually AB3L is pyramidal.

Question 3.
(a) AB7 – IF7
(b) AB4L2 – ICI4
(c) AB6 – XeOF4
(d) AB5L – IF5
Ans.
(c) AB6 – XeOF4
Actually XeOF4 is AB5L type.

Question 4.
(a) Fluorine – Bond order 1
(b) Oxygen – Bond order 2
(c) Nitrogen – Bond order 2
(d) Cyanide – Bond order 3
Answer:
(c) Nitrogen – Bond order 2.
Actually N = N bond order is 3.

Question 5.
(a) CH4 – sp3
(b) PCI5 – sp3d
(c) BeCl2 – sp
(d) BF3 – sp3d2
Answer:
(d) BF3 – sp3d2
Actually BF4 is sp2 hybridised.

VII. Assertion and Reason.

Question 1.
Assertion (A): Diatomic molecules such as H2, O2, F2 are non-polar molecules.
Reason (R): H2, O2, F2 have zero dipole moment.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 2.
Assertion (A): HF, HCl, CO and No are polar molecules.
Reason (R): They have non zero dipole moments and so they are polar molecules.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).

Question 3.
Assertion (A): H2, Li2, C2, N2 are diamagnetic.
Reason (R): All have no unpaired electrons and so they are diamagnetic.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 4.
Assertion (A): B2, O2, NO are paramagnetic in nature.
Reason (R): They have unpaired electrons and are paramagnetic.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 5.
Assertion (A): Metals have high thermal conductivity.
Reason (R): Absence of bond gap is the main reason for high thermal conductivity.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).

Question 6.
Assertion (A): Metals have high thermal conductivity.
Reason (R): Due to thermal excitation of many electrons from the valence band to the conductance band, metals have high thermal conductivity.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is wrong but (R) is correct.
(d) (A) is correct but (R) is wrong.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).

VIII. Choose the correct statement.

Question 1.
(a) The metallic luster is due to reflection of light by the electron cloud.
(b) Metals have low inciting point and low boiling point.
(c) Metals have low thermal conductivity.
(d) Electrical conductivity of metals is low.
Answer:
(a) The metallic luster is due to reflection of light by the electron cloud.

Question 2.
(a) NO molecules is diamagnetic
(b) O2 molecules is paramagnetic
(c) N2 molecules is paramagnetic
(d) Li2 molecules is paramagnetic
Answer:
(b) O2 molecules is paramagnetic

Question 3.
(a) BeCl2 undergoes sp3 hybridisation
(b) 8F3 undergoes sp3d hybridisation
(c) CH4 undergoes sp3d2 hybridisation
(d) PCl5 undergoes sp3d hybridisation
Answer:
(d) PCl5 undergoes sp3d hybridisation

Samacheer Kalvi 11th Chemistry Chemical Bonding 2 Mark Questions and Answers

I. Write brief answer to the following questions.

Question 1.
What are chemical bonds?
Answer:
The interatomic attractive forces which holds the constituent atoms/ions together in a molecule are called chemical bonds.

Question 2.
State octet rule.
Answer:
The atoms transfer or share electrons so that all the atoms involved in chemical bonding obtain eight electrons in their outer shell (valence shell). It is called octet rule.

Question 3.
What is meant by covalent bond?
Answer:
The mutual sharing of one or more pair of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond.

Question 4.
Draw the lewis structure of

  1. H2O
  2. SO3.

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-99

Question 5.
Calculate the formal charge on the carbon atom and oxygen atom in the structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-100
Answer:
Formal charge on carbon
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-101
Formal charge on oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-102

Question 6.
Calculate the formal charge on the carbon atom and oxygen atom in the structure
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-103
Answer:
Formal charge on carbon
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-104
Formal charge on singly bonded oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-105
Formal charge on triply bonded oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-106

Question 7.
AmongSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-107 and Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-108 which is a preferable structure for CO2 molecule why?
Answer:
Structure I of Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-109
Formal charge on carbon
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-110
Formal charge on oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-111

Structure II of Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-112
Formal charge on carbon
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-113
Formal charge on singly bonded oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-114
Formal charge on triply bonded oxygen
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-115
A structure in which all formal charges are zero is preferred over the one with non – zero charges. In case of CO2 structure , structure I is preferred over the structure II as it has zero formal charge for all the atoms.

Question 8.
Draw the lewis structures of a few molecules containing odd electrons.
Answer:
Few molecules have a central atom with an odd number of valence electrons. For example, in nitrogen dioxide and nitric oxide all the atoms does not have octet configuration.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-116

Question 9.
Draw the lewis structure of PCl5 and SF6
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-117

Question 10.
Define bond length.
Answer:
The distance between the nuclei of two covalently bonded atoms is called bond length. For e.g., in a covalent molecule A – B. the bond length is equal to the sum of the radii of bonded atoms. i.e., rA + rß = bond length.

Question 11.
Prove that bond order is inversely proportional to bond length.
Answer:
1. Bond order ∝ Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-118

2. An example for illustrating the above equation is Carbon – carbon single bond length (I .54Å) is longer than the carbon-carbon double bond length (1 .34Å) and the carbon- carbon triple bond length (1 .20Å).

Question 12.
Define Bond angle.
Answer:
Covalent bonds are directional in nature and are oriented in specific direction in space. This directional nature creates a fixed angle between two covalent bonds in a molecule and this angle is termed as bond angle.

Question 13.
Define Resonance.
Answer:
The similar structures in which the relative position of the atoms are same but they differ in the position of bonding and lone pair of electrons are called resonance structures and this phenomenon is called resonance.

Question 14.
What are polar and non-polar molecules?
Answer:
1. Diatomic molecules such as H2, O2, F2 have zero dipole moment and are called non polar molecules.

2. Molecules such as HF, HCl, CO, NO have non zero dipole moment values and are called polar molecules.

Question 15.
What is meant by polarisaion?
Answer:
The ability of a cation to polarise an anion is called its polarising ability and the tendency of the anion to get polarised is called its polarisability. This phenomenon is known as polarisation.

Question 16.
Among NaCI, MgCI2 and AICI3 which shows more covalent character? Why?
Answer:
Among, the ionic compounds NaCI, MgCl2 and AICI3 the charge of the cation increases in the order Na+ < Mg2+ < Al3+, thus the covalent character also follows the same order NaCl < MgCI2 < AlCI3. So AICI3 shows more covalent character.

Question 17.
Lithium chloride is more covalent than sodium chloride. Justify this statement.
Answer:
1. The smaller cation and larger anion shows greater covalent character due to greater extent of polarisation.

2. The size of Li+ ion is smaller than Na+ ion and hence the polarising power of Li+ ion is more. So lithium chloride is more covalent than sodium chloride.

Question 18.
Lithium iodide is more covalent than Lithium chloride. Give reason.
Answer:
Lithium iodide is more covalent than Lithium chloride as the size of I ion is larger than Cl ion. Hence I ion will be more polarised than Cl ion by the cation Li+. So LiI is more covalent than LiCl.

Question 19.
Draw the structure of AB4L and AB3L2 type of molecules with example.
Answer:
AB4L:
Bond pairs = 4
Lone pair = 1
Shape = see saw
e.g., SF4 Strucutre

Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-119

AB3L2:
Bond pairs = 3
Lone pair = 2
Shape = T shaped
e.g., CIF3 Strucutre
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-120

Question 20.
Draw the structure of AB4L2 and AB7 type of molecules with example
Answer:
1. AB4L2:
Bond pairs = 4
Lone pairs = 2
Shape = Square planare.g., XeF4
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-121

2. AB7:
Bond pairs = 7
Lone pairs = Nil
Shape = pentagonal bipyramidal
e.g., IF7
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-122

Question 21.
Explain the bond formation of hydrogen molecule.
Answer:
1. Electronic configuration of hydrogen atom is 1s1.

2. During the formation of H2 molecule, the 1 s orbitais of two hydrogen atoms containing one unpaired electron with opposite spin overlaps with each other along the internuclear axis.

This overlap is called s – s overlap. Such axial overlap results in the formation of a sigma (σ) covalent bond.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-123

Question 22.
Explain the bond formation of fluorine molecule.
Answer:
1. Valence shell electronic confIguration of fluorine atom is 2s2 2px2 2py2 2pz1

2. When the half filled pz orbitais of two fluorine atoms overlap along the z – axis, a σ covalent bond is formed between them.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-124

Question 23.
How is HF molecule formed?
Answer:

  1. Electronic configuration of hydrogen atom is 1s1.
  2. Valence shell electronic configuration of fluorine atom is 2s2 2px2 2py2 2pz1.
  3. When half filled is orbital of hydrogen linearly overlaps with a halt filled 2pz orbital of fluorine, a σ covalent bond is formed between hydrogen and fluorine.

Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-125

Question 24.
What is meant by metallic bond?
Answer:
The forces that keep the atoms of the metal so closely in a metallic crystal constitute what is generally known as the metallic bond.

Question 25.
Why metallic bonding is referred as electronic bonding?
Answer:
1. Metallic crystals are an assemblage of positive ions immersed in a gas of free electrons. The free electrons are due to the ionisation of the valence electrons of the atom of the metal.

2. As the valence electrons of the atoms are freely shared by all the ions in the crystal, the metallic bonding is also referred to as electronic bonding.

Question 26.
Metals have high density. Give reason.
Answer:
The electrostatic attraction between the metal ions and the free electrons yields a three dimensional close packed crystal with a large number of nearest metal ions. So metals have high density.

Question 27.
Metals are ductile in nature. why?
Answer:
In the close packed structure of metallic crystal. it contains many slip planes along which movement can occur during mechanical loading, so the metal acquires ductility.

Question 28.
Give reason behind the lustrous nature, high melting point and boiling point of metals?
Answer:
1. The metallic lustre is due to the reflection of light by the electron cloud.

2. As the metallic bond is strong enough. the metal atoms are reluctant to break apart into a liquid or gas, so the metals have high melting and boiling points.

Question 29.
Metals are very good electrical conductors. Why?
Answer:
1. The bonding in metal is explained by molecular orbital theory. As per this theory. the atomic orbitals of large number of atoms in a crystal overlap to form numerous bonding and anti-bonding molecular orbitals without any band gap.

2. The bonding molecular orbitals are completely filled with an electron pair in each, and the anti-bonding molecular orbitals are empty.

3. Absence of band gap accounts for high electrical conductivity of metals.

Question 30.
Metals have high thermal conductivity. Give reason.
Answer:
High thermal conductivity of metals is due to thermal excitation of many electrons from the valence band to the conduction band.

Question 31.
Except Cu, Ag and Au, most metals are black. Why?
Answer:
Most metals are black except copper, silver and gold. It is due to the absorption of light of all wavelengths. Absorption of light of all wavelengths is due to the absence of band gap in metals.

Question 32.
Write the favourable factors for the formation of ionic bond.
Answer:

  1. Low ionisation enthalpy of metal atoms.
  2. High electron gain enthalpy of non-metal atoms.
  3. High lattice enthalpy of compound formed.

Question 33.
Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-126
Because of two lone pair of electrons on O – atom, repulsion on bond pairs is greater in H2O in comparison to NH3. Thus, the bond angle is less in H2O molecules.

Question 34.
Write the significance/applications of dipole moment.
Answer:

  1. In predicting the nature of the molecules: Molecules with specific dipole moments are polar in nature and those with zero dipole moments are non-polar in nature.
  2. In the determination of shapes of molecules.
  3. In calculating the percentage of ionic character.

Question 35.
Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.
Answer:
In CO2, there are two C = O bonds. Each C = O bond is a polar bond. The net dipole moment of CO2 moleculc is zero. This is possible only if CO2 is a linear molecule. (O = C = O). The bond dipoles of two C = O bonds cancels the dipole moment of each other.

Whereas, H2O molecule has a net dipole moment (1.84 D). H2O molecule has a bent structure because here the O – H bonds are oriented at an angle of 104.5° and do not cancel the bond dipole moments of each other.

Question 36.
What is the total number of sigma and pi bonds in the following molecules?

  1. C2H2
  2. C2H4

Answer:

1. H – C = C – H
Sigma bond = 3
π bond = 2

2.Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-127
Sigma bond = 5
π bond = 1

Question 37.
Use molecular orbital theory to explain why the Be2 molecule does not exist
Answer:
E.C of Be = 1 s2 2s2
M.O.E.C of Be2 = σ1s2 σ2s2 σ*2s2
Bond order = \(\frac { 1 }{ 2 }\) (4 – 4) = 0
Hence, Be2

Question 38.
Compare the relative stabililty of the following species and indicate their magnetic properties O2, O2+, O2 (superoxide), O22-(peroxide)
Answer:
O2 – Bond order = 2 paramagnetic
O2+ – Bond order = 2.5, paramagnetic
O2 – Bond order = 1.5, paramagnetic
O22- – Bond order = 1, diagmagnetic
Order of relative stability is:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-128

Question 39.
Account for the following:

  1. Water is a liquid while H2S is a gas
  2. NH3 has higher boiling point than PH3.

Answer:

  1. In case of water, hydrogen bonding causes association of the H2O molecules. There is no such hydrogen bonding in H2S, that is why it is a gas.
  2. There is hydrogen bonding in NH3 but not in PH3.

Question 40.
Why B2 is paramagnetic in nature while C2 is not?
Answer:
The molecular orbital electronic configuration of both B2 and C2 are:
B2:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-129
C2:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-130
Since, B2 has two unpaired electrons, therefore, B2 is paramagnetic C2 has no unpaired electron, therefore, C2 is diamagnetic.

Samacheer Kalvi 11th Chemistry Chemical Bonding 3 Mark Questions and Answers

Question 1.
Draw the lewis structure of

  1. Nitrogen
  2. Carbon
  3. Oxygen

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-131
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-132

Question 2.
Draw the lewis structure of

  1. Ammonia
  2. Methane
  3. Dinitrogen pentoxide.

Answer:
Lewis dot structures of Molecules
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-133

Question 3.
Calculate the bond enthalpy of OH bond in water.
Answer:
1. In the case of polyatomic molecules with two or more same bond types, the arithmetic mean of the bond energy value of the same type of bonds is considered as average bond enthalpy.

2. For e.g., in water, there are two OH bonds present and the energy needed to break them are not same.

3. H2O(g) → H(g) + OH(g)
∆H1 = 502 kJ mol-1
OH(g) → H(g) + O(g)
∆H = 427 kJ mol
The average bond enthalpy of OH bond in water = \(\frac { 502 + 427 }{ 2 }\) = 464.5 kJ mol-1

Question 4.
Explain how the ionic character in a covalent bond is related to electronegativity?
Answer:
1. The extent of ionic character in a covalent bond can be related to the electronegativity difference of the bonded atoms.

2. In a typical polar molecule Aδ- – Bδ+ the electronegativity difference (XA – XB) can be used to predict the percentage of the ionic character as follows

3. If the electronegativity difference XA – XB is equal to 1.7, then the bond A – B has 50% ionic character.

4. If it is greater than 1.7, then the bond XA – XB has more than 50% ionic character.

5. If it is greater than 1.7, then the bond A – B has more than 50% ionic character.

6. If it is lesser than 1.7, then the bond A – B has less than 50% ionic character.

Question 5.
CuCI is more covalent than NaCl. Give reason.
Answer:
1. Cations having ns2np6nd10 configuration show greater polansing power than the cations with ns2 np6 configuration. Hence they show greater covalent character.

2. CuCI is more covalent than NaCI. As compared to Na+ (1.13Å), Cu+(0.6Å)is small and has 3s23p63d10 configuration.

3. Electronic configuration of Cu+: [Ar] 3d10
Electronic configuration of Na+: [He] 2s22p6
So CuCI is more covalent than NaCI

Question 6.
Draw the structure of AB2, AB3, AB3L type of molecules with example.
Answer:
1. AB2
Number of bond pairs = 2
Shape = Linear
Example – BeCl2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-134

2. AB3
Number of bond pairs = 3
Shape = Tirgonal planar
Example – BF3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-135

3. AB2L
Number of bond pairs = 2
Number of lone pairs = 1
Shape = Bent (or) inverted V shape
Example – O3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-136

Question 7.
Give example and structure of

  1. AB3L
  2. AB5
  3. AB2L2

type of molecules with example.
Answer:
1. AB3L
Number of bond pairs = 3
Number of lone pairs = 1
Shape = Pyramidal
Example – NH3
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-137

2. AB5
Number of bond pairs = 5
Number of lone pairs = 0
Shape = Trigonal bipyramidal
Example = PCIC5
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-138

3. AB2L2
Number of bond pairs = 2
Number of lone pairs = 2
Shape = Bent
Example – H2O
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-139

Question 8.
Draw the shape of

  1. XeF2
  2. IOF5
  3. XeOF4

Answer:
1. XeF2
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-140

2. IOF5
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-141

3. XeOF4
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-142

Question 9.
Explain the bonding in oxygen molecule.
Answer:
1. Valence shell electronic configuration of oxygen atom is
2s2 2px2 2py12pz1
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-143

2. When the half filled pz orbitaIs of two oxygen atoms overlap along the z – axis a σ covalent bond is formed between them. Other two hail filled py orbitais of two oxygen atoms overlap laterally to form a π – cova1ent bond between the oxygen atoms.

3. Thus in oxygen molecule, two oxygen atoms are connected by two covalent bonds (double bond). The other two pair of electrons present in the 2s and 2px orbital do not involve in bonding and remains as lone pair on the respective oxygen atoms.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-144

Question 10.
Explain about the molecular orbital diagram of hydrogen molecule.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-145

  1. Electronic configuration of H atom 1s1
  2. Electronic configuration of H, molecule – σ1s1
    Bond order Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-146
  3. Molecule (H2) has no unpaired electrons, hence it is diamagnetic.

Question 11.
Draw and explain the M.O. diagram of lithium molecule.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-147

  1. Electronic configuration of Li atom – 1s1
  2. Electronic configuration of Li2 molecule is ais2 σ*1s2 σ*1s2 σs2
  3. Bondorder = Nb – Nb/2 = 4 – 2/2
  4. Li2 molecule has no unpaired electrons, hence it is diamagnetic.

Question 12.
Draw and explain the M.O. diagram of Boron molecule.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-148
Answer:

  1. Electronic configuration of B = 1s2 2s2 2p3
  2. Electronic configuration of B, = σ1s2 σ*1s2σ2s2 σ*2s2 π2px1 π 2pz1
  3. Bond order Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-149
  4. B2 molecule has two unpaired electrons hence it is paramagnetic.

Question 13.
Draw and explain the molecular orbital diagram of carbon molecule.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-150

  1. Electronic configuration of C atom – 1s2 2s2 2p2
  2. Electronic configuration of C2 molecule is σ1s2 σ*1s2 σ*2s2 σ*2s2 π 2px2 π 2py2
  3. Bond orderSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-151

Question 14.
Write Lewis dot symbols for atoms of the following elements: Mgq Naq B O, N, Br.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-152

Question 15.
write Lewis symbols for the following atoms and ions: S and S2-; Al and Al3+; H and OH
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-153

Question 16.
Draw the Lewis structures for the following molecules and ions H2S, SiCl4, BeF2, CO32-, HCOOH
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-154

Question 17.
Define Octet rule. Write its significance and limitations.
Answer:
Octet rule:
Atoms of elements combine with each other in order to complete their respective octet so as to acquire the stable nearest noble gas configuration.

Significance:
It helps to explain why dilfferent atoms combine with each other to form ionic compounds or covalent compounds.

Limitations of Octet rule:
1. According to octet rule, atoms take part in chemical combination to achieve the configuration of nearest noble gas elements.

However, some of noble gas elements like Xenon have formed compounds with fluorine and oxygen. For example: XeF2, XeF4, XeO3 etc. Therefore, validity of the octet rule has been challenged.

2. This theory does not account for the shapes of molecules.

Question 18.
Write the resonance structure for SO3, NO2 and NO3
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-155

Question 19.
What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
Answer:
The electron pair involved in sharing between two atoms during covalent bonding is called shared pair or bond pair. At the same time, the electron pair which is not involved in sharing is called lone pair of electrons.
For example, in Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-156
there are only 4 bond pairs, but in Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-157 there are two bond pairs and two lone pairs.

Question 20.
Distinguish between a sigma bond and a pi bond
Answer:
Sigma (σ) Bond

  1. σ – bond is formed by the axial overlap of the atomic orbitais.
  2. The bond is quite strong.
  3. Only one lobe ofthep-orbitals is involved in the overlap.
  4. Electron cloud of the molecular orbital is symmetrkal around the internuclear axis.

Pi (π) Bond

  1. π – bonnd is formed by the sidewise overlap of atomic orbitais.
  2. It is comparatively a weaker bond.
  3. Both lobes of the p-orbitais are involved in the overlap.
  4. The electron cloud is not symmetrical.

Question 21.
Write the Important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer:
1. The combining atomic orbitals should have comparable energies. For example, is orbital of one atom can combine with 1s atomic orbital of another atom, 2s orbitai can combine with 2s orbital and so on.

2. The combining atomic orbitals must have proper orientations so that they are able to overlap to a considerable extent.

3. The extent of overlapping should be large.

Question 22.
What are Lewis structures? Write the Lewis structure of H2, BeF2 and H2O.
Answer:
The outer shell electrons are shown as dots surrounding the symbol of the atom. These symbols are known as Lewis symbols or Lewis structures. The Lewis structure of
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-158

Question 23.
What are the main postulates of Valence Shell Electron Pair Repulsion (VSEPR) theory?
Answer:

  1. The shape of a molecule depends upon the no. of electron pairs around the central atom.
  2. There is a repulsive force between the electron pairs, which tend to repel one another.
  3. The electron pairs in space tend to occupy such positions that they arc at maximum distance, so that the repulsive force will be minimum.
  4. A multiple bond is treated as lilt is a single bond and the remaining electron pairs which constitute the bond may be regarded as single super pair.

Question 24.
Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with four H atoms at the corners of the square and C atom at its centre. Explain why CH4 is not square planar?
Answer:
Electronic configuration of carbon atom: C: σ1s22s22p2.
in the excited state, the orbital picture of carbon can be represented as:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-159
Hence, carbon atom undergoes sp3 hybridisation in CH4 molecule and takes tetrahedral shape.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-160
For a square planar shape, the hybridisation of the central atom has to be dsp3. However, an atom of carbon does not have d – orbitals to undergo dsp3 hybridisation. Hence, the structure of CH4 is tetrahedral.

Question 25.
Explain why BeH2 molecule has a zero dipole moment although the Be – H bonds are polar.
Answer:
The Lewis structure for BeH2 molecule is as follows:Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-161. There is no lone pair at the central atom (Be) and there are two bond pairs. Hence, BeH2, is of the type AB2. It has a linear structure.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-162
Dipole moments of each Be – H bond are equal and opposite in direction. Therefore, they nullify each other. Hence, BeH2 has a net zero dipole moment.

IV. Answer the following questions in detail:

Question 1.
Explain about Kossel-Lewis approach to chemical bonding.
Answer:
1. Kossel and Lewis approach to chemical bonding is based on the inertness of the noble gases which have little or no tendency to combine with other atoms.

2. They proposed that noble gases are stable due to their completely filled outer electronic configuration.

3. Elements other than noble gases try to attain the completely filled outer electronic configuration by losing, gaining or sharing one or more electrons from their outer shell.

4. For e.g., sodium loses one electron to form Na ion and chlorine accepts that electron to give chloride ion, Cl.These two ions are held together by electrostatic attractive forces, a bond known as an electrovalent bond.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-163

5. In diatomic molecules such as nitrogen and oxygen, they achieve the stable noble gas electronic configuration by mutual sharing of electrons.

6. Lewis introduced a scheme to represent the chemical bond and the electrons present in the outer shell of the atom called Lewis dot structure.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-164

7. For example, the electronic configuration of nitrogen is 1s22s22p3. It has 5 electrons in its outer shell. The lewis structure of nitrogen isSamacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-165

8. In N, molecule, equal sharing of 3 electrons from each nitrogen atom takes place as fol lows
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-166

Question 2.
What is meant by covalent bond?
Explain the covalent bonding in H2, O2, N2.
Answer:
1. Mutual sharing of one or more pair of electrons between two combining atoms results in the formation of a chemical bond called a covalent bond.

2. If two atoms share just one pair of electron, a single covalent bond is formed as in the çase of hydrogen molecule (H2).

3. If two or three electron pairs are shared benveen the two combining atoms, then the covalent bond is called double bond and triple bond respectively, as in the case of O2 and N2 molecules respectively.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-167

Question 3.
What is an ionic bond?
Explain about the formation of ionic bond with a suitable example.
Answer:
1. The complete transfer of electrons leads to the fomiation of a cation and an anion. Both these ions are held together by electrostatic attractive forces which is known as ionic bond.

2. KCl: Potassium chloride
Electronic configuration of K [Ar] 4s
Eleçtronic configuration of Cl = [Ne] 3s2 3p5

3. Potassium has 1 electron in its valence shell and chlorine has 7 electrons in its valence shell.

4. By losing one electron potassium attains the nearest inert gas configuration of Argon and becomes a unipositive cation (K) and chlorine accepts this electron to become uninegative chloride ion (CI) to attain the stable configuration of nearest noble gas, Argon.

5. These two ions combine to form an ionic crystal in which they are held together by electrostatic attractive forces.

6. During the formation of one mole of potassium chloride crystal from its constituent ions, 718 kJ of energy is released. This favours the formation of KCl and its stabilisation.

Question 4.
Define coordinate covalent bond. Illustrate the formation of coordinate covalent bond with a suitable example.
Answer:
1. In the bond formation, one of the combining atoms donates a pair of electrons i.e., two electrons which are necessary for the covalent bond formation and these electrons are shared by both the combining atoms, and the bond formed is called coordinate covalent bond.

2. The combining atom which donates the pair of electron is called the donor atom and the other atom is called the acceptor atom. This bond is denoted by an arrow starting from the donor atom pointing towards the acceptor atom.

3. For example, in ferricyanide ion [Fe(CN)6]4- each cyanide ion (CN) donates a pair of electrons to form a coordinate bond with iron (Fe2+) and these electrons are shared by Fe2+ and CN ions.

4. Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-168

5. Ammonia having a lone pair of electrons donates its pair to an electron deficient molecule such as BF3 to form a coordinate covalent bond
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-169

Question 5.
Explain about valence bound theory for the formation of H2 molecule.
Answer:
1. Two hydrogen atoms Ha and Hb are separated by infinite distance. At this stage, there is no interaction between these two atoms and the potential energy of this system is arbitrarly taken as zero.

2. As these two atoms approach each other, in addition to electrostatic attractive forces between the nucleus and its own electrons, the following new forces begins to operate
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-170

3. The new attractive forces Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-171 arise between:

  • nucleus of Ha and valence electron of Hb
  • nucleus of Hb and the valence electron of Ha

4. The new repulsive forces Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-172 arise between:

  • the nucleus Of Ha and Hb
  • the valence electrons of Ha and Hb

5. The attractive forces tend to bring Ha and Hb together whereas the repulsive forces tends to push them apart.

6. At the initial stage, as the two hydrogen atoms approach each other, the attractive forces are stronger than repulsive forces and the potential energy decreases.

7. A stage is reached where the net attractive forces are exactly balanced by repulsive forces and the potential energy of the system acquires a minimum energy.

8. At this stage, there is a maximum overlap between the atomic orbitals of Ha and Hb and atoms Ha and Hb are now said to be bonded together by a covalent bond.

Question 6.
What arc the salient features of Valence Bond (VB) theory?
Answer:
1. When half filled orbitals of two atoms overlap, a covalent bond will be formed between them.

2. The resultant overlapping orbitals are occupied by the two electrons with opposite spins. For example when H2 is formed, the two is electron of two hydrogen atoms get paired up and occupy the overlapped orbitals.

3. The strength of a covalent bond depends upon the extent of overlap of atomic orbitals. Greater the overlap, larger is the energy released and stronger will be the bond formed.

4. Each atomic orbital has a specific direction (excepts-orbital which is spherical) and hence orbital overlap takes place in the direction that maximises overlap.

5. Depending upon the nature of overlap, the bonds are classified as σ covalent bond and π it covalent bond.

6. When two atomic orbitals overlap linearly along the axis, the resultant bend is called a sigma (σ) bond. This overlap is also called or “axial overlap”.

7. When two atomic orbitals overlap sideways the resultant covalent bond is called a pi (π) bond.

Question 7.
Explain about sp hybridisation with suitable example.
Answer:

  1. bond rormation in Beryllium chloride takes place by sp hybridisation.
  2. The valence shell of Beryllium has the electronic configuration as follows:

Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-173

3. In BeCl2, both the Be – Cl bonds are equivalent and it was observed that the molecule is linear. VB theory explains this observed behaviour by sp hybridisation. One of the paired electrons in the 2s orbital gets excited to 2p orbital.

4. Now the 2s and 2p orbitals hybridise and produce two equivalent sp hybridised orbitals which have 50% s-character and 50% p-character. These sp hybridised orbitals are oriented in opposite direction.

5. Each of the sp hybridised orbitals linearly overlap with p orbital of the chlorine to form a covalent bond between Be and Cl atoms as follow:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-174

Question 8.
Explain the formation of methane using VB theory?
Answer:
1. Methane is formed by sp3 hybridisation. In CH4 molecule, the central carbon atom is bounded to four hydrogen atoms.

2. The ground state valence shell electronic configuration of carbon is [He] 2s22px2 2py1 2px0

3. En order to form four covalent bonds with the four hydrogen atoms, one of the paired electrons in 2s orbital of carbon is promoted to its 2Pz orbital in the excited state.

4. The one 2s orbital and three 2p orbitals of carbon atom mixes to give four equivalent sp3 hybridised orbitals. The angle between any of the two sp3 hybridised orbitals is 109°•28’
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-175

5. The Is orbital of the four hydrogen atoms overlap linearly with the four sp3 hybridised orbitais of carbon to form four C – H σ bonds in the methane molecule as follows
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-176

Question 9.
Explain sp3d hybridisation with a suitable example.
Answer:
1. In the PCl5 molecule, the central atom phosphorous is covalently bonded to five chlorine atoms. Here the atomic orbitals of phosphorous undergoes sp3d2 hybridisation which involves its one 3s orbital, three 3p orbitals and one vacant 3d orbital (dz2)

2. The ground state electronic configuration of phosphorous is [Ne] 3s23px23py13pz1
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-177

3. One of the paired electrons in the 3s orbital of phosphorous ¡s promoted to one of its vacant 3d orbital (dz2) in the excited state.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-178

4. The 3pz orbitals of the five chlorine atoms linearly overlap along the axis with the five sp3d hybridised orbitals of phosphorous to form the five P – CI bonds as follows.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-179

Question 10.
Explain about. sp3d2 hybridisation with an example.
Answer:
1. In sulphur hexafluoride SF6, the central atom sulphur extend its octet to undergo sp3d hybridisation to generate six sp3d2 hybridised orbitals which accounts for six equivalent S – F bonds.

2. The ground state electronic configuration of sulphur is [Ne] 3s23px13py13pz1
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-180

3. One electron each form 3s orbital and 3p orbital of sulphur is promoted to its two vacant 3d orbitals dz2 and dx2-y2 in the excited state.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-181

4. A total of six valence orbitals from sulphur (one 3s orbital, three 3p orbitals and two 3d orbitals) (dx2 and dx2-y2) which mixes to give six equivalent sp3d2 hybridised orbitals. The orbital geometry is octahedral.

5. The six sp3d2 hybridised orbitals of sulphur overlaps linearly with 2pz orbital of six fluorine atoms to form the six S – F bonds in sulphur hexa fluoride.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-182

Question 11.
Explain about the salient features of molecular orbital theory.
Answer:
1. When atoms combine to form molecules, their individual atomic orbitals lose their identity and form new orbitals called molecular orbitals.

2. The shape of molecular orbitals depend upon the shapes of combining atomic orbitals.

3. The number of molecular orbitals formed is the same as the number of combining atomic orbitals. half the number of molecular orbitals formed will have lower energy and are called bonding orbitals, while the remaining half molecular orbitals will have higher energy and are called anti-bonding molecular orbitals.

4. The bonding molecular orbitals are represented as σ (sigma), π (pi), δ (delta) and the corresponding anti-bonding orbitals are called σ*, π* and δ*.

5. The electrons in the molecule are accommodated in the newly formed molecular orbitals. The filling of electrons in these orbitals follow Autbau’s Principle, Pauli’s exclusion principle and Hund’s rule as in the case of filling of electrons in the atomic orbitals.

6. Bond order gives the number of covalent bonds between the two combining atoms. The bond order of a molecule can be calculated using
the foLlowing equation:
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-183
Nb = Number of electrons in bonding molecular orbitals.
Na = Number of electrons in anti-bonding molecular orbitals.

7. A bond order of zero value indicates that the molecule does not exist.

Question 12.
Explain the MO diagram for NO molecule.
Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-184
Answer:

  1. Electronic configuration of N atom is 1s2 2s2 2p3
  2. Electronic configuration of O atom is 1s2 2s2 2p4
  3. Electronic configuration of NO molecule is
    Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-185
  4. Bond order Samacheer Kalvi 11th Chemistry Solutions Chapter 10 Chemical Bonding-186
  5. NO molecule has one unpaired electron, hence it is paramagnetic.

Question 13.
Explain about metallic bonding.
Answer:
1. The forces that keep the atoms of the metal so closely in a metallic crystal constitute what is known as metallic bond.

2. According to Drude and Lorentz, metallic crystal is an assemblage of positive ions immersed in a gas of free electrons. The free electrons are due to ionisation of the valence electrons of the atoms of the metal.

3. As the valence electrons of the atoms are freely shared by all the ions in the crystal, the metallic bonding is referred to as electronic bonding.

4. The electrostatic attraction between the metal ions and the free electrons yield a three dimensional close packed crystal with a large number of nearest metal ions. So metals have high density.

5. As the close packed structure contains many slip planes along which movement can occur during mechanical loading, metal acquires ductility.

6. As metal ion is surrounded by electron cloud in all directions, the metallic bonding has no and thermal conductivity. The metallic lustre is due to the reflection of light by the electron cloud.As the metallic bond is strong enough, the metal atoms are reluctant to break apart into a liquid or gas, so the metals have high melting and boiling points.

7. High thermal conductivity of metals is due. to thermal excitation of many electrons from the valence bond to the conduction band.

Question 14.
Explain about the bonding in metals by molecular orbital theory.
Answer:
1. According to molecular orbital theory the atomic orbitals of large number of atoms in a crystal overlap to form numerous bonding and anti-bonding molecular orbitals without any band gap.

2. The bonding molecular orbitals are completely filled with an electron pair in each and the anti-bonding molecular orbitals are empty.

3. Absence of band gap accounts for high electrical conductivity of metals.

4. High thermal conductivity is due to thermal excitation of many electrons from the valence band to the conduction band.

5. With an increase in temperature, the electrical conductivity decreases due to vigorous thermal motion of lattice ions that disrupts the uniform lattice structure. that is required for free motion of electrons within the crystal.

Common Errors

  1. The number of bonds formed by elements may go wrong.
  2. When writing Lewis structure, electrons may be written in an irregular way.
  3. Coordinate covalent bond should not be written as a line

Rectifications

  1. Always hydrogen and fluorine form 1 bond Oxygen 2 bonds, Nitrogen 3 bonds, Carbon 4 bonds
  2. When writing lewis structure, each atom should be surrounded by eight electrons in such a way as 4 pairs of electrons.
  3. Coordinate covalent bond should be written from donor atom to acceptor atom as an arrow mark Donor→ Acceptor

We as a team believe the information prevailing regarding the Samacheer Kalvi Chemical Bonding for 11th Chemistry Chapter 10 Chemical Bonding has been helpful in clearing your doubts to the fullest. For any other help do leave us your suggestions and we will look into them. Stay in touch to get the latest updates on Tamilnadu State Board Chemical Bonding for different subjects in the blink of an eye.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Enhance your subject knowledge with Tamilnadu State Board for Chapter 9 Solutions and learn all the underlying concepts easily. Make sure to Download Samacheer Kalvi 11th Chemistry Book Solutions, Notes Pdf Chapter 9 Solutions Questions and Answers PDF on a day to day basis and score well in your exams. Are given after enormous research by people having high subject knowledge. You can rely on them and prepare any topic of Chemistry as per your convenience easily.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions

Students looking for Chapter 9 Solutions Concepts can find them all in one place from our Tamilnadu State Board Solutions. Simply click on the links available to prepare the corresponding topics of Chemistry easily. Samacheer Kalvi 11th Chemistry Chapter wise Questions and Answers are given to you after sample research and as per the latest edition textbooks. Clarify all your queries and solve different questions to be familiar with the kind of questions appearing in the exam. Thus, you can increase your speed and accuracy in the final exam.

Samacheer Kalvi 11th Chemistry Chapter 9 Solutions Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Solutions Multiple Choice Questions

Question 1.
The molality of a solution containing 1 .8g of glucose dissolved in 250g of water is …………
(a) 0.2 M
(b) 0.01 M
(c) 0.02 M
(d) 0.04 M
Answer:
(d) 0.04 M
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-1

Question 2.
Which of the following concentration terms is/are independent of temperature?
(a) molality
(b) molarity
(c) mole fraction
(d) (a) and (c)
Answer:
(d) (a) and (c)
Solution:
Molality and mole fraction are independent of temperature.

Question 3.
Stomach acid, a dilute solution of HCI can be neutralised by reaction with Aluminium hydroxide
Al(OH)3 + 3HCl(aq) → AlCl3 + 3H2O
How many millilitres of 0.1 M Al(OH)3 solution are needed to neutralise 21 mL of 0.1 M HCl
(a) 14 mL
(b) 7 mL
(c) 21 mL
(d) none of these
Answer:
(b) 7 mL
Solution:
M1 x V1 = M2 x V2
∵ 0.1 M Al(OH)3 gives 3 x 0.1 = 0.3 M OH ions .
0.3 x V1 = 0.1 x 21
V1= \(\frac { 0.1 x 21 }{ 0.3 }\) = 7ml

Question 4.
The partial pressure of nitrogen in air is 0.76 atm and its Henry’s law constant is 7.6 x 104 atm at 300K. What is the mole fraction of nitrogen gas in the solution obtained when air is bubbled through water at 300K?
(a) 1 x 10-4
(b) 1 x 10-6
(c) 2 x 10-5
(d) 1 x 10-5
Answer:
(d) 1 x 10-5
Solution:
PN2 = 0.76atm
KH = 7.6 x 104
x = ?
PN2 = KH . x
0.76 = 7.6 x 104x x
x = \(\frac { 0.76 }{ 7.6\times { 10 }^{ 4 } }\) = 1 x 10-5

Question 5.
The Henry’s law constant for the solubility of Nitrogen gas in water at 350K is 8 x 104 atm. The mole fraction of nitrogen in air is 0.5. The number of moles of Nitrogen from air dissolved in 10 moles of water at 350K and 4 atm pressure is ………….
(a) 4 x 10-4
(b) 4 x 104
(c) 2 x 10-2
(d) 2.5 x 10-4
Answer:
(d) 2.5 x 10-4
Solution:
KH = 8 x 104
(xN2 )in air = 0.5
Total pressure = 4 atm
Partial pressure of nitrogen = Mole fraction Total pressure
= O.5 x 4 = 2
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-3

Question 6.
Which one of the following is incorrect for ideal solution?
(a) ∆Hmix = 0
(b) ∆Umix = o
(c) ∆P = PObserved – PCalculated by raoults law = 0
(d) ∆Gmix = 0
Answer:
(d) ∆Gmix = 0
Solution:
For an ideal solution, ∆Smix \(\neq\) 0; Hence ∆Gmix \(\neq\) 0
∴ Incorrect is ∆Gmix = 0

Question 7.
Which one of the following gases has the lowest value of Henry’s law constant?
(a) N2
(b) He
(c) CO2
(d) H2
Answer:
(c) CO2
Solution:
Carbon dioxide; most stable gas and has lowest value of Henry’s Law constant.

Question 8.
P1 and P2 are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution if x1 represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be ………
(a) P1 + x1(P2 – P1)
(b) P2 – x1(P2 + P1)
(c) P1 – x2(P1 – P2)
(d) P1 + x2(P1 – P2)
Answer:
(c) P1 – x2(P1 – P2)
Solution:
Ptotal = P1 + P2
= P1 x1 + P2x2
= P1(1 – x2) + P2x2
= P1 – P1x2 + P2x2 = P1 – x2(P1 – P2)
[∵x1 + x2 = 1
x1 = 1 – x2]

Question 9.
Osomotic pressure (π) of a solution is given by the relation ……………
(a) π = nRT
(b) πV = nRT
(c) πRT = n
(d) none of these
Answer:
(b) πV = nRT
Solution:
n = CRT
n = \(\frac { n }{ V }\)
π V = nRT

Question 10.
Which one of the following binary liquid mixtures exhibits positive deviation from Raoults law?
(a) Acetone + chloroform
(b) Water + nitric acid
(c) HCI + water
(d) ethanol + water
Answer:
(d) ethanol + water

Question 11.
The Henry’s law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B is 0.2. The ratio of mole fraction of B and A dissolved in water will be …………
(a) \(\frac { 2x }{ y }\)
(b) \(\frac { y }{ 0.2x }\)
(c) \(\frac { 0.2x }{ y }\)
(d) \(\frac { 5x }{ y }\)
Answer:
(d) \(\frac { 5x }{ y }\)
Solution:
Given,
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-4

Question 12.
At 100°C the vapour pressure of a solution containing 6.5g a solute in 100g water is 732mm. If Kb = 0.52, the boiling point of this solution will be …………..
(a) 102°C
(b) 100°C
(c) 101°C
(d) 100.52°C
Answer:
(c) 101°C
Solution:
\(\frac { ΔP }{ P° }\) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } }\)
W2 = 6.5g
W1 = 100g
Kb = 0.52
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-5
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-6

Question 13.
According to Raoults law, the relative Lowering of vapour pressure for a solution is equal to….
(a) molefraction of solvent
(b) mole fraction of solute
(c) number of moles of solute
(d) number of moles of solvent
Answer:
(b) mole fraction of solute
Solution:
\(\frac { ∆P }{ P° }\) = x2 (Mole fraction of the solute)

Question 14.
At same temperature. which pair of the following solutions are isotonic?
(a) 0.2 M BaCl2 and 0.2M urea
(b) 0.1 M glucose and 0.2 M urea
(c) 0.1 MNaCl and 0.1 MK2SO4
(d) 0.1 MBa(NO3)2 and 0.1 MNa2 SO4
Answer:
(d) 0.1 M Ba (NO3)2 and 0.1 M Na2 SO4
Solution:
0.1 x 3 ion [Ba2 + 2NO3], 0.1 x 3 ion [2Na+, SO4]

Question 15.
The empirical formula of a non-electrolyte(X) is CH2O. A solution containing six gram of X exerts the same osmotic pressure as that of 0.025 M glucose solution at the same temperature. The molecular formula of X is
(a) C2H4O2
(b) C8H16O8
(c) C4H8O4
(d) CH2O
Answer:
(b) C8H16O8
Solution:
1)non electrolute = (π2)glucose
C1RT = C2RT
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-7
\(\frac { 6 }{ n(30) }\) = 0.025
n = \(\frac { 6 }{ 0.025 x 30 }\) = 30
∴ Molecular formula C8H16O8

Question 16.
The KH for the solution of oxygen dissolved in water is 4 x 104 atm at a given temperature. If the partial pressure of oxygen in air is 0.4 atm, the mole fraction of oxygen in solution is …………..
(a) 4.6 x 103
(b) 1.6 x 104
(c) 1 x 10-5
(d) 1 x 105
Answer:
(c) 1 x 10-5
Solution:
KH = 4 x 104 atm,
(PO2)air = 0.4 atm,
(xo2)in solution = ?
air – in solution
(PO2)air = KH(xo2)in solution
0.4 = 4 x 104(xo2)in solution
(xo2)in solution = \(\frac { 0.4 }{ 4\times { 10 }^{ 4 } }\) = 1 x 10-5

Question 17.
Normality of 1.25M sulphuric acid is …………
(a) 1.25 N
(b) 3.75 N
(c) 2.5 N
(d) 2.25 N
Answer:
(c) 2.5 N
Solution:
Normality of H2SO4 = (No. of replacable H+) x M = 2 x 1.25 = 2.5 N

Question 18.
Two liquids X and Y on mixing gives a warm solution. The solution is …………..
(a) ideal
(b) non-ideal and shows positive deviation from Raoults law
(c) ideal and shows negative deviation from Raoults Law
(d) non – ideal and shows negative deviation from Raoults Law
Answer:
(d) non – ideal and shows negative deviation from Raoults Law
Solution:
∆Hmix is negative and show negative deviation from Raoults law.

Question 19.
The relative lowering of vapour pressure of a sugar solution in water is 3.5 x 10-3. The mole fraction of water in that solution is …………
(a) 0.0035
(b) 0.35
(c) 0.0035/18
(d) 0.9965
Answer:
(d) 0.9965
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-9

Question 20.
The mass of a non-volatile solute (molar mass 80 g mol-1) which should be dissolved in 92g of toluene to reduce its vapour pressure to 90% ………..
(a) 10g
(b) 20g
(c) 9.2 g
(d) 8.89g
Answer:
(d) 8.89g
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-10
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-11

Question 21.
For a solution, the plot of osmotic pressure (π) verses the concentration (e in mol L-1) gives a straight line with slope 310 R where ‘R’ is the gas constant. The temperature at which osmotic pressure measured is ………..
(a) 310 x 0.082 K
(b) 3 10°C
(c) 37°C
(d) \(\frac { 310 }{ 20.082}\)
Answer:
(c) 37°C
Solution:
π = CRT
y = x(m)
m = RT
310 R = RT
T = 310 K
= 37°C

Question 22.
200 ml of an aqueous solution of a protein contains 1 .26g of protein. At 300K, the osmotic pressure of this solution is found to be 2.52 x 10-3 bar. The molar mass of protein will be (R =0.083 Lhar mol-1 K-1) ……………
(a) 62.22 Kg mol-1
(b) 12444 g mol-1
(c) 300g mol-1
(d) none of these
Answer:
(a) 62.22 Kg mol-1
Solution:
π = CRT
M = \(\frac { WRT }{ π1 }\) = \(\frac { 1.26\times 0.083\times 300 }{ 2.52\times { 10 }^{ -3 }\times 0.2 }\) = 62.22Kg mol-1

Question 23.
The Van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is ………..
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1
Solution:
Ba(OH)2 dissociates to form Ba2+ and 2OH-1 ion
α = \(\frac { (i – 1) }{ (n – 1) }\)
i = α (n – 1) + 1
n = i = 3 ( for Ba (OH)2, α = 1 )

Question 24.
What is the molality of a 10% w/w aqueous sodium hydroxide solution?
(a) 2.778
(b) 2.5
(c) 10
(a) 0.4
Answer:
(b) 2.5
Solution:
100% \(\frac { w }{ w }\) aqueous NaOH solution means that 10 g of sodium hydroxide in 100g solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-13

Question 25.
The correct equation for the degree of an associating solute, ‘n’ molecules of which undergoes association in solution, is ………
(a) α = \(\frac { n(i – 1) }{ n – 1 }\)
(b) α2 = \(\frac { n(1 – i) }{ n – 1 }\)
(c) α = \(\frac { n(i – 1) }{ 1 – n }\)
(d) α = \(\frac { n(1 – i) }{ n(1 – i) }\)
Answer:
(c) α = \(\frac { n(i – 1) }{ 1 – n }\)
Solution:
α = \(\frac { (i – 1)n }{ (n – 1) }\) (or) \(\frac { n(i – 1) }{ (1 – n) }\)

Question 26.
Which of the following aqueous solutions has the highest boiling point?
(a) 0.1 M KNO3
(b) 0.1 M Na3PO4
(c) 0.1 M BaCl2
(d) 0.1 M K2SO4
Answer:
(a) 0.1 M KNO3
Solution:
Elevation of boiling point is more in the case of Na3PO4(no. of ions 4; 3 Na+, PO43-)

Question 27.
The freezing point depression constant for water is 1.86° k kg mo1-1 . If 5g Na2SO4 is dissolved in 45g water, the depression in freezing point is 3.64°C. The van’t Hoff factor for Na2SO4 is ……..
(a) 2.50
(b) 2.63
(c) 3.64
(d) 5.50
Answer:
(a) 2.50
Solution:
Kf = 1.86
W2 = 5g
∆Tf = 3.64
M2 = 142
W1 = 45g
ΔTf = i x Kf
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-14

Question 28.
Equimolal aqueous solutions of NaCI and KCI are prepared. If the freezing point of NaCI is – 2°C, the freezing point of KCI solution is expected to be ………
(a) – 2°C
(b) – 4°C
(c) – 1°C
(d) 0°C
Answer:
(a) – 2°C
(b) – 4°C
(c) – 1°C
(d) 0°C
Solution:
Equimolal aqueous solution of KCI also shows 2° C depression in freezing point.

Question 29.
Phenol dimerises in henzene having van’t Hoff factor 0.54. What is the degree of association?
(a) 0.46
(b) 92
(c) 46
(d) 0.92
Answer:
(d) 0.92
Solution:
α = \(\frac { (1-i)n }{ (n-1) }\) = \(\frac { (1 – 0.54)2 }{ (2 – 1) }\) = 0.46 x 2 = 0.92
Question 30.
Assertion : An ideal solution obeys Raoults Law
Reason : In an ideal solution, solvent-solvent as well as solute-solute interactions are similar to solute-solvent interactions.
(a) both assertion and reason are true and reason is the correct explanation of assertion
(b) both assertion and reason are true but reason is not the correct explanation of assertion
(c) assertion is true but reason is false
(d) both assertion and reason are false
Answer:
(a) both assertion and reason are true and reason is the correct explanation of assertion

Samacheer Kalvi 11th Chemistry Solutions Short Answer Questions

Question 31.
Define

  1. Molality
  2. Normality

Answer:
1. Molality (m):
It is defined as the number of moles of the solute present in 1 kg of the solvent
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-15

2. Normality (N):
It is defined as the number of gram equivalents of solute in I litre of the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-16

Question 32.
What is a vapour pressure of liquid? What is relative lowering of vapour pressure?
Answer:
1. The pressure of the vapour in equilibrium with its liquid ¡s called vapour pressure of the liquid at the given temperature.

2. The relative lowering of vapour pressure is defined as the ratio of lowering of vapour. pressure to vapour pressure of pure solvent. Relative lowering of vapour pressure
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-17

Question 33.
State and explain Henry’s law.
Answer:
Henry’s law:
This law states “that the partial pressure of the gas in vapour phase is directly proportional to the mole fraction (x) of the gaseous solute in the solution at low concentrations.”
Psolute ∝ xsolute in solution
Psolute = KH. xsolute in solution
xsolute = mole fraction of solute in the solution
KH = empirical constant.
Psolute = Vapour pressure of the solute (or) the partial pressure of the gas in vapour state. The value of KH depends on the nature of the gaseous solute and solvent.

Question 34.
State Raoult law and obtain expression for lowering of apour pressure when nonvolatile solute is dissolved in solvent.
Answer:
Raoult’s law:
This law states that “in the case of a solution of volatile liquids the partial vapour pressure of each component (A & B) of the solution is directly proportional to its mole fraction.
PA ∝ x A
when xA = 1,
then k = P°A
(P°A = vapour pressure of pure component)
PA = P°A . xa
PB = P°B . xb
when a non volatile is dissolved in pure water, the vapour pressure of the pure solvent will decrease. In such solution, the vapours pressure of the solution will depend only on the solvent molecules as the solute is non-volatile.
Psolution ∝ xA
Psolution = k . xA
xA = 1, k = P°solvent
solution = P°solvent – Psolution
Lowering of vapour pressure = P°solvent – Psolution
Relative lowering of vapour pressure = \(\frac { P° – P }{ P° }\) = xB
where xB = Mole fraction of solute.

Question 35.
What is molal depression constant? Does it depend on nature of the solute?
Answer:
Kf = molar freezing point depression constant or cryoscopic constant.
∆Tf = Kf . m,
where
∆Tf = depression in freezing point.
m = molality of the solution
Kf = cryoscopic constant
If m = I
∆Tf = Kf
i.e., cryoscopic constant is equal to the depression in freezing point for 1 molal solution cryoscopic constant depends on the molar concentration of the solute particles. Kf is directly proportional to the molal concentration of the solute particles.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-18
WB = mass of the solute
WA = mass of solvent
MB = molecular mass of the solute.

Question 36.
What is osmosis?
Answer:
Osmosis is a spontaneous process by which the solvent molecules pass through a semipermeable membrane from a solution of lower concentration to the solution of higher concentration.

Question 37.
Define the term bisotonic
Answer:
1. Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

2. When such solutions arc separated by a semipermeable membrane, solvent flow between one to the other on either direction is same. i.e.. the net solvent flow between two isotonic solutions is zero.

Samacheer Kalvi 11th Chemistry Solutions Long Answer Questions

Question 38.
You are provided with a solid ‘A’ and three solutions of A dissolved in water – one saturated, one unsaturated, and one super saturated. How would you determine each solution?
Answer:
1. Saturated solution:
When maximum amount of solute is dissolved in a solvent, any more addition of solute will result in precipitation at a given temperature and pressure. Such a solution is called a saturated solution.

2. Unsaturated solution:
When minimum amount of solute is dissolved in a solvent at a given temperature and pressure is called an unsaturated solution.

3. Super saturated solution:
It is a solution that holds more solute than it normally could in its saturated form.

Example:

  1. A saturated solution where the addition of more compound would not dissolve in the solution. 359 g of NaCI in 1 litre of watcr at 25°C.
  2. An unsaturated solution has the capacity to dissolve more of the compound. 36 g of NaCI in 1 litre of water at 25°C.
  3. A super saturated solution is the solution in which crystals can start growing. 500 g of NaCI in 1 litre of water at 25°C.

Question 39.
Explain the effect of pressure on the solubility.
Answer:
1. The change in pressure does not have any significant effect in the solubility of solids and liquids as they are not compressible. However the solubility of gases generally increases with increase of pressure.

2. According to Le – chatlier’s principle, the increase in pressure will shift the equilibrium in the direction which will reduce the pressure. Therefore, more number of gaseous molecules dissolves in the solvent.

3. If pressure increases, solubility of gas also increases.

Question 40.
A sample of 12 M Concentrated hydrochloric acid has a density 1.2 gL-1. Calculate the molality.
Answer:
Given:
Molarity = 12 M HCI
Density of the solution = 1.2 g L-1
In 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-19
Calculate mass of water (solvent)
Mass of 1 litre HCI solution = density x volume
= 1.2gmL-1 x 1000 mL = 1200g
Mass of LICI = No. of moles of HCI x molar mass of HCI
= 12mol x 36.5 g mol-1 = 438g
Mass of waler = mass of HCI solution – mass of HCI
Mass of waler = 1200 – 438 = 762 g
Molalily =\(\frac { 12 }{ 0.762 }\) = 15.75m

Question 41.
A 0.25 M glucose solution at 370.28 K has approximately the pressure as blood. What is the osmotic pressure of blood?
Solution.
C = 0.25 M
T = 37O.28 K
(π)gIucose = CRT
(π) = 0.25 mol L-1 x 0.082 L atm K-1 morl-1 x 370.28K
= 7.59 atm

Question 42.
Calculate the molality of a solution containing 7.5g of glycine (NH2 – CH2 – COOH) dissolved in 500g of water.
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-20

Question 43.
Which solution has the lower freezing point? 10g of methanol (CH3OH) in 100g of water (or) 20g of ethanol (C2H5HO) In 200g of water.
Solution:
∆Tf = Kf . m i.e. ∆Tf ∝ m
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-21
∴ Depression in freezing point is more in methanol solution and it will have lower freezing point.

Question 44.
How many moles of solute particles are present in one litre of 10-4 M potassium sulphate?
Solution:
In 10-4M K2SO4 solution, there are 10-4 moles of potassium sulphate.
K2SO4 molecule contains 3 ions (2 K+ and 1SO42-)
1 mole of K2SO4 contains 3 x 6.023 x 1023 ions
10 mole of K2SO4 contains 3 x 6.023 x 102 x 10-4 ions = 18.069 x 1019

Question 45.
Henry’s law constant for solubility of methane in benzene is 4.2 x 10-5 mm Hg at a particular constant temperature. At this temperature, calculate the solubiiitv of methane at

  1. 75O mm Hg
  2. 84OnimHg

Solution:
(KH)Benzene = 4.2 x 10-5 mm Hg. Solubility of methane = ? P = 750 mm Hg, p = 840 mm Hg
According to Henry’s Law,
P = KH . xsolution
750 mm Hg = 4.2 x 10-5 mm Hg . xsolution
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-24

Question 46.
The observed depression in freezing point of water for a particular solution is 0.093°C. Calculate the concentration of the solution in molality. Given that molal depression constant for water is 1.86 K kg mol-1.
Solution:
T1= 0.093°C = 0.093K
m = ?
Kf = 1.86K kg mol-1
∆Tf = kf . m
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-24

Question 47.
The vapour pressure of pure henzene (C6H6) at a given temperature is 640 mm Hg. 2.2 g of non – volatile solute is added to 40 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solute?
Solution:
P0C6H6 = 640 mm Hg
W2 = 2.2 g (non volatile solute)
W1 = 40 g (benzene)
Psolution = 600 mm Hg
M2 = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-25

Samacheer Kalvi 11th Chemistry Solutions In Text Questions – Evaluate Yourself

Question 1.
If 5.6 g of KOH is present in (a) 500 mL and (b) I litre of solution, calculate the molarity of each of these solutions.
Solution.
Mass of KOH = 5.6g
No. of moles = \(\frac { 5.6 }{ 5.6 }\) = 0.1 mol
1. Volume of the solution = 500 ml = 0.5 L

2. Volume of the solution = IL
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-26

3. Volume of the solution = IL
Molarity = \(\frac { 0.1 }{ 1 }\) M

Question 2.
2.82 g of glucose is dissolved in 30 g of water. Calculate the mole fraction of glucose and water.
Solution:
Mass of glucose = 2.82 g
No. of moles of glucose = \(\frac { 2.82 }{ 180 }\) = 0.0 16
Mass of water = 30g = \(\frac { 30 }{ 18 }\) = 1.67
xH2O = \(\frac { 1.67 }{ 1.67 + 0.016 }\) = \(\frac { 1.67 }{ 1.686 }\) = 0.99
xH2O + xglucose = 1
0.99 + xglucose = 1
xglucose = 1 – 0.99 = 0.01

Question 3.
The antiseptic solution of iodopovidone for the use of external application contains 10% w/v of iodopovidone. Calculate the amount of iodopovidone present in a typical dose of 1.5 mL.
Solution:
10% \(\frac { w }{ v }\) means that 10 g of solute in 100 ml solution
∴ Amount of iodopovidone in 1.5 ml = \(\frac { 10g }{ 100ml }\) x 1.5 ml = 0.15 g

Question 4.
A litre of sea water weighing about 1.05 kg contains 5 mg of dissohed oxygen (O2). Express the concentration of dissolved oxygen in ppm.
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-28

Question 5.
Describe how would you prepare the following solution from pure solute and solvent

  1. 1 L of aqueous solution of 1.5 M COCI2.
  2. 500 mL of 6.0 % (v/v) aqueous methanol solution.

Solution:

  1. mass of 1.5 moles of COCI2 = 1.5 x 129.9 = 194.85g
  2. 194.85g anhydrous cobalt chloride is dissolved in water and the solution is make up to one litre in a standard flask.

Question 6.
How much volume of 6 M solution of NaOH is required to prepare 500 mL of 0.250 M NaOH solution.
Solution:
6% \(\frac { v }{ v }\) aqueous solution contains 6g of methanol in 100 ml solution. To prepare 500 ml of 6% v/v solution of methanol 30g methanol is taken in a 500 ml standard flask and required quantity of water is added to make up the solution to 500 ml.

Question 7.
Calculate the proportion of O2 and N2 dissolved in water at 298 K. When air containing 20% O2 and 80% N2 by volume is in equilibrium with water at 1 atm pressure. Henry’s law constants for two gases are KH(O2) = 4.6 x atm and KH (N2) 8.5 x 104 atm.
Solution:
C1V1 = C2V2
6M (V1) = 0.25M x 500 ml
V1 = \(\frac { 0.25 x 500 }{ 6 }\)
V1 = 20.3 mL

Question 8.
Explain why the aquatic species are more comfortable in cold water during winter season rather than warm water during the summer.
Solution:
Total pressure = 1 atm
PN2 = \((\frac { 80 }{ 100 })\) x Total pressure = \(\frac { 80 }{ 100 }\) x 1 atm = 0.8 atm
PO2 = \((\frac { 20 }{ 100 })\) x 1 = 0.2 atm
According to Henry’s Law
Psolute = KH x solute in solution
PN2 = (KH)Nitrogen x Mole fraction of Nitrogen in solution
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-29

Question 9.
Calculate the mole fractions of benzene and naphthalene in the vapour phase when an ideal liquid solution is formed by mixing 128 g of naphthalene with 39g of benzene. It is given that the vapour pressure of pure benzene is 50.71 mm Hg and the vapour pressure of pure naphthalene is 32.06 mm Hg at 300 K.
Solution:
P0pure benzene = 50.71 mm Hg
P0nepthalene = 32.06 mm Hg
Number of moles of benzene = \(\frac { 39 }{ 78 }\) = 0.5 mol
Number of moles of naphthalcne = \(\frac { 128 }{ 128 }\) =1 mol
Mole fraction of benzene = \(\frac { 0.5 }{ 1.5 }\) = 0.33
Mole fraction of naphthalene = 1 – 0.33 = 0.67
Partial vapour pressure of benzene =P0benzene x Mole fraction of benzene
= 50.71 x 0.33 = 16.73 mm Hg
Partial vapour pressure of naphthalene = 32.06 x 0.67 = 21.48mm Hg
Mole fraction of benzene in vapour phase = \(\frac { 16.73 }{ 16.73 + 21.48 }\) = \(\frac { 16.73 }{ 38.21 }\) = 0.44
Mole fraction of naphthalene in vapour phase = 1 – 0.44 = 0.56

Question 10.
Vapour pressure of a pure liquid A is 10.0 torr at 27°C. The vapour pressure is lowered to 9.0 torr on dissolving one grani of B in 20g of A. If the molar mass of A is 200 g mol-1 then calculate the molar mass of B.
Solution:
P0A = 10 torr
Psolution = 9 torr
WA = 20 g
WB = 1 g
MA = 200 g mol-1
MB = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-30

Question 11.
2.56g of Sulphur is dissolved in 100g of carbon disuiphide. The solution boils at 319.692K. What is the molecular formula ofSulphur in solution? The boiling pointof CS2 is 319. 450K. Given that Kb for CS2 = 2.42 K kg mol-1
Solution:
W2 = 2.56g
W1 = 100g
T = 319.692 K
Kb = 2.42 K kg mol-1
∆Tb = (319.692 – 319.450) K = 0.242 K
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-31
M2 = 256g mol-1
Molecular mass of sulphur in solulion = 256 g mol-1
Atomic mass of one mole of sulphur atom = 32
No. of atoms in a molecule of sulphur = \(\frac { 256 }{ 32 }\) = 8
Hence, molecular tòrmula of sulphur is S8.

Question 12.
2g of a non electrolyte solute dissolved in 75g of benzene lowered the freezing point of benzene by 0.20 K. The freezing point depression constant of benzene is 5.12 K Kg mol-1. Find the molar mass of the solute.
Solution:
W2 = 2g
W1 = 75g
∆Tf = 0.2 K
kf = 5.12 K kg mol-1
M2 = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-32

Question 13.
What is the mass of glucose (C6H12O6) in it one litre solution is isotonic with 6g L-1 of urea (NH2CONH2)?
Solution:
Osmotic pressure of urea solution (π1) = CRT
\(\frac { { W }_{ 2 } }{ { M }_{ 2 }V }\)RT = \(\frac { 6 }{ 60 x 1 }\) x RT
Osmotic pressure of glucose solution
2) \(\frac { { W }_{ 2 } }{ 180\times 1 }\) x RT
For isotonic solution, π1 = π2
\(\frac { 6 }{ 60 }\) = \(\frac { { W }_{ 2 } }{ 180\times 1 }\) RT ⇒ W2 = \(\frac { 6 }{ 60 }\) x 180
W2 = 18 g

Question 14.
0.2m aqueous solution of KCI freezes at – 0.68°C calculate van’t Hoff factor. Kf for water is 1.86 K kg mol-1.
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-36
Given,
∆Tf = 0.680 K
m = 0.2 m,
∆Tf (observed) = 0.680K
∆Tf(Calculated) = kf
m = 1.86 K kg mol-1 x 0.2 mol kg-1 = 0.372K
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-37

Samacheer Kalvi 11th Chemistry Solutions Example problems Solved

Question 1.
What volume of 4M HCI and 2M HCI should be mixed to get 500 mL of 2.SM HCI?
Solution:
Let the volume of 4M HCl required to prepare 500 mL of 2.5 M HCI = x mL
Therefore, the required volume of 2M HCI = (500 – x) mL
We know from the equation x = \(\frac { 250 }{ 2 }\) = 125 mL
Hence, volume of 4M HCI required = 125 mL
Volume of 2M HCl required = (500 – 125) mL = 375 mL

Question 2.
0.24g of a gas dissolves in 1 L of water at 1.5 atm pressure. Calculate the amount of dissolved gas when the pressure is raised to 6.0 atm at constant temperature.
Solution:
Psolute = KH xsolute in solution
At pressure 1.5 atm, p1 = KH x1 ………..(1)
At pressure 6.0 atm, p2 = KHx2 …………..(2)
Dividing equation (1) by (2)
Weget
\(\frac { { P }_{ 1 } }{ { P }_{ 2 } }\) = \(\frac { { x }_{ 1 } }{ { x }_{ 2 } }\)
\(\frac { 1.5 }{ 6.0 }\) = \(\frac { { 0.24 } }{ { x }_{ 2 } }\)
Therefore
x2 = \(\frac { 0.24×6.0 }{ 1.5 }\) = 0.96 g/L

Question 3.
An aqueous solution of 2% nonvolatile solute exerts a pressure of 1.004 bar at the boiling point of the solvent. What is the molar mass of the solute when PA° is 1.013 bar?
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-38
In a 2% solution weight of the solute is 2g and solvent is 98g
ΔP = PA0 – Psolution = 1.013 – 1.004 bar = 0.009 bar
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-39

Question 4.
0.75 g of an unknown substance is dissolved in 200 g solvent. If the elevation of boiling point is 0.15 K and molal elevation constant is 7.5K kg more then, calculate the molar mass of unknown substance.
Solution:
∆Tb = Kb m = Kb x W2 x 1000/M2 x W1
M2 = Kb x W2 x 1000/∆Tb x W1
= 7.5 x 0.75 x 1000/0.15 x 200 = 187.5g mol-1

Question 5.
Ethylene glycol (C2H6O2) can be used as an antifreeze in the radiator of a car. Calculate the temperature when ice will begin to separate from a mixture with 20 mass percent of glycol in water used in the car radiator. Kf for water = 1.86 K kg mol-1 and molar mass of ethylene glycol is 62g mol-1.
Solution:
Weight of solute (W2) = 20 mass percent of solution means 20g of ethylene glycol
Weight of solvent (water) W1 = 100 – 20 = 80g
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-40
The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e. 7.5 K lower than the normal freezing point of water (273 – 7.5)K = 265.5K

Question 6.
At 400K 1.5 g of an unknown substance is dissolved in solvent and the solution is made to 1.5 L. Its osmotic pressure is found to be 0.3 bar. Calculate the molar mass of the unknown substance.
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-41

Question 7.
The depression in freezing point is 0.24K obtained by dissolving 1g NaCI in 200g water. Calculate van’t – Hoff factor. The molal depression constant is 1.86 K kg mol-1.
Solution:
Sol. Molar mass of solute
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-42

Samacheer Kalvi 11th Chemistry Solutions Additional Questions Solved

Samacheer Kalvi 11th Chemistry Solutions 1 Mark Questions and Answers

I. Choose the correct answer.

Question 1.
Among the following, which one is mostly present in sea water?
(a) NaCI
(b) Nal
(c) KCI
(d) MgBr2
Answer:
(a) NaCI

Question 2.
Statement I: The most common property of sea water and air is homogeneity.
Statement II: The homogeneity implies uniform distribution of their constituents through the mixture.
(a) Statements I and II arc correct and II is the correct explanation of I.
(b) Statements I and II are correct but II is not the correct explanation of I.
(c) Statement I is correct but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) StatementI I and II are correct and II is the correct explanation I.

3. Which one of the following is a homogeneous mixture?
(a) Sea water
(b) Air
(c) Alloys
(d) All the above
Answer:
(d) All the above

Question 4.
Statement I: Salt solution is an aqueous solution.
Statement II: If water is used as the solvent, the resultant solution is called an aqueous solution.
(a) Statements I and II are correct but II is not the correct explanation of I.
(b) Statements I and II are correct and II is the correct explanation of I.
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(b) Statements I and II are correct and II is the correct explanation of I.

Question 5.
Statement I: The dissolution of ammonium nitrate increases steeply with increase in temperature.
Statement II: The dissolution process of ammonium nitrate is endothermic in nature.
(a) Statement I and II are correct and statement II is the correct explanation of statement I.
(b) Statement I and II are correct but II is not the correct explanation of I.
(c) Statement I is correct but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) Statement I and II are correct and statement II is the correct explanation of statement I.

Question 6.
In which of the following compound the solubility decreases with increase of temperature?
(a) sodium chloride
(b) ammonium nitrate
(c) cerie sulphate
(d) calcium chloride
Answer:
(c) ceric sulphate

Question 7.
Which of the following is not an ideal solution?
(a) Benzene & toluene
(b) n – Hexane & n – Heptane
(c) Ethyliodide & ethyl bromide
(d) Ethanol and water
Answer:
(d) Ethanol and water

Question 8.
Which one of the following shows positive deviation from Raoult’s law?
(a) Ethyliodide and Ethyl bromide
(b) Ethyl alcohol and cyclohexane
(c) Chioro benzene & bromo benzene
(d) Benzene & toluenc
Answer:
(b) Ethyl alcohol and cyclohexane

Question 9.
Which one of the following is not an non-ideal solution showing positive deviation?
(a) Benzene & acetone
(b) CCl4 & CHCI3
(c) Acetone & ethyl alcohol
(d) Benzene and toluene
Answer:
(d) Benzene and toluene

Question 10.
Which of the following shows negative deviation from Raoults law?
(a) Phenol and aniline
(b) Benzene and toluene
(c) Acetone and ethanol
(d) Bcnzene and acetone
Answer:
(a) Phenol and aniline

Question 11.
Which of the following is not an non-ideal solution showing negative deviation?
(a) Phenol and aniline
(b) Ethanol and water
(c) Acetone + Chlorotorm
(d) n – Heptane and n – Hexane
Answer:
(d) n – Heptanc and n – Hexane

Question 12.
Statement I: A solution of potassium chloride in water deviates from ideal behavior.
Statement II: The solute dissociates to give K and Cl ion which form strong ion dipole interaction with water molecules.
(a) Statement I & II are correct and II is the correct explanation of I
(b) Statement I & II are correct but II is not correct explanation of I
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(a) Statement I & II are correct and II is the correct explanation of I

Question 13.
Statement I: Acetic acid deviates from ideal behaviour.
Statement II: Acetic acid exists as a dimer by forming inter molecular hence deviates from Raoults law.
(a) Statement I & II are correct and II is the correct explanation of I.
(b) Statement I & II are correct but II is not the correct explanation of I.
(c) Statement I is true but II is wrong.
(d) Statement I is wrong but II is correct.
Answer:
(a) Statement I & II are correct but II is the correct explanation of I.

Question 14.
Which one of the following has found to have abnormal molar mass? hydrogen bonds and
(a) NaCl
(b) KCI
(c) Acetic acid
(d) all the above
Answer:
(d) All the above

Question 15.
What would be the value of van’t Hoff factor for a dilute solution of K2SO4 in water.
(a) 3
(b) 2
(c) 1
(d) 4
Answer:
(a) 3
Solution:
ions produced = n = 3
Since
K2SO4 → 2K+ + SO42-
K2SO4 is completely dissociated so
∝ = \(\frac { i – 1 }{ n – 1 }\) = \(\frac { i – 1 }{ 3 – 1 }\) = 1
i – 1 = 1 x 2
i – 1 = 2
i = 2+1 = 3

Question 16.
In the determination of molar mass of AB using a colligative property, what may be the value of van’t Hoff factor if the solute is 50% dissociates?
(a) 0.5
(b) 1.5
(c) 2.5
(d) 1
Answer:
(b) 1.5
Solution:
∝ = \(\frac { i – 1 }{ n – 1 }\) = 0.5
\(\frac { i – 1 }{ 2 – 1 }\) = 0.5
i – 1 = 0.5
i = 0.5 + 1 = 1.5

Question 17.
Which of the following solution has the highest boiling point?
(a) 5.85% solution of NaCI
(b) 18.0% solution of glucose
(c) 6.0% solution of urea
(d) All have same boiling point
Answer:
(a) 5.85% solution of NaCl

Question 18.
Which one of the following pair is called an ideal solution?
(a) nicotine – water
(b) water – ether
(c) water – alcohol
(d) Chiorobenzene – bromobenzene
Answer:
(d) Chiorobenzene – bromobenzene

Question 19.
Which of the following is not a colligative property?
(a) optical activity
(b) osmotic pressure
(c) elevation boiling point
(d) depression in freezing point
Answer:
(a) optical activity

Question 20.
On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?
(a) Sugar crystals in cold water
(b) Sugar crystals in hot water
(c) powdered sugar in cold water
(d) powdered sugar in hot water
Answer:
(d) powdered sugar in hot water

II. Match the following.
Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-43
Answer:
(a) 3 4 1 2

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-44
Answer:
(d) 3 4 2 1

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-45
Answer:
(c) 2 4 1 3

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-46
Answer:
(a) 4 3 1 2

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-47
Answer:
(b) 2 4 1 3

III. Fill in the blanks.

Question 1.
……… covers more than 70% of the earth’s surface.
Answer:
Seawater

Question 2.
……… is an important naturally occurring solution.
Answer:
Air

Question 3.
An example of solid homogeneous mixture is ……….
Answer:
Brass

Question 4.
A mixture of N2, O2, CO2 and other traces of gases is known as ………
Answer:
Air

Question 5
……… a non – aqueous solution.
Answer:
Br2 in CCl4

Question 6.
……… is an example for gaseous solution.
Answer:
Camphor in nitrogen gas

Question 7 .
……… is used for dental filling.
Answer:
Amalgam of potassium

Question 8.
Carbonated water is an example for ………
Answer:
Liquid solution

Question 9.
Humid oxygen is an example of ………
Answer:
Gaseous solution

Question 10.
The concentration of commercially available H2O2 is ………
Answer:
3%

Question 11.
The molality of the solution containing 45g of glucose dissolved in 2kg of water is ………
Answer:
0.125m
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-48

Question 12.
5.845 g of NaCl is dissolved in water and the solution was made up to 500 mL using a standard flask. The strength of the solution in molarity is ………
Answer:
0.2 M
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-49

Question 13.
3.15 g of oxalic acid dihydrate is dissolved in water and the solution was made up to 100 ml using a standard flask. The strength of the solution in normality is ………
Answer:
0.5N
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-50

Question 14.
5.85 g of NaCI is dissolved in water and the solution was made upto 500 ml using a standard flask. The strength of the solution in formality is ………
Answer:
0.2 F
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-51

Question 15.
Neomycin, amino glycoside antibiotic cream contains 300 mg of neomycin sulphate the active ingredient in 30 g of oinment base. The mass percentage of neomycin is ………
Answer:
1%
Solution:
The mass percentage of neomycin
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-52

Question 16.
0.5 mole of ethanol is mixed with 1.5 mole of water. Then the mole fraction of ethanol and water are ……….
Answer:
0.25, 0.75
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-53
= \(\frac { 0.5 }{ 1.5 + 0.5 }\) = \(\frac { 0.5 }{ 2.0 }\) = 0.25
Mole fraction of water = \(\frac { 1.5 }{ 2.0 }\) = 0.75

Question 17.
50 mL of tincture of benzoin, an antiseptic solution contains 10 ml of benzoin. The volume percentage of benzoin is ……….
Answer:
20%
Solution:
Volume percentage of benzoin
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-54
= \(\frac { 10 }{ 50 }\) x 100 = 20%

Question 18.
A 60 ml of paracetamol pediatric oral suspension contains 3g of paracetamol. The mass percentage of paracciamol is …………
Answer:
5%
Solution:
Mass percentage of paracetamol =
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-55
= \(\frac { 3 }{ 60 }\) x 100 = 5%

Question 19.
50 ml of tap water contains 20 mg of dissolved solids. The TDS value in ppm is ………..
Answer:
400 ppm
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-56

Question 20.
The concentration term used in the neutralisation reactions is …………
Answer:
Normality

Question 21.
The concentration term is used in the calculation of vapour pressure of solution is …………..
Answer:
Mole fraction

Question 22.
The term used to express the active ingredients present in therapeutics is ………
Answer:
Percentage units

Question 23.
When maximum amount of solute is dissolved in a solvent at a given temperature, the solution is called ………..
Answer:
Saturated solution

Question 24.
The solvent in which sodium chloride readily dissolves is …………
Answer:
Water

Question 25.
………… is used by deep-sea divers.
Answer:
Helium, nitrogen and oxygen

Question 26.
The mathematical expression of Raoult’s law is ………..
Answer:
PA = PA0 . XA

Question 27.
……….. is an ideal solution?
Answer:
Chloro benzene & bromo benzene

Question 28.
………….. is important in some vital biological systems.
Answer:
osmotic pressure

Question 29.
………. is not a colligative property.
Answer:
vapour pressure

Question 30.
According to van’t Hoff equation. the value of osmotic pressure t is equal to …………
Answer:
π = CRT

Question 31.
The osmotic pressure of the blood cells is approximately equal to at 37°C.
Answer:
7 atm.

Question 32.
Which one of the following is applied in water purification?
Answer:
reverse osmosis

Question 33.
In commercial reverse osmosis process, the semi permeable membrane used is ………..
Answer:
cellulose acetate

Question 34.
The degree of dissociation α is equal to ……….
Answer:
\(\frac { i – 1 }{ n – 1 }\)

Question 35.
The degree of association a is equal to ……….
Answer:
\(\frac { (i – 1)n }{ n – 1 }\)

Question 36.
The estimated vantt Hoff factor for acetic acid solution in benzene is ………..
Answer:
0.5

Question 37.
The estimated van’t Hoff factor for sodium chloride in water is ………..
Answer:
2

Question 38.
Number of moles of the solute dissolved per dm3 of solution is ……….
Answer:
molarity

Question 39.
Molarity of pure water is ………….
Answer:
55.55
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-57
Question 40.
18 g of glucose is dissolved in 90 g of water. The relative lowering of vapour pressure is equal to ………..
Answer:
0.1
Solution:
\(\frac { P° – P }{ P° }\) = x2
x2 = No. of moles of glucose
\(\frac { 18 }{ 180 }\) = 0.1
\(\frac { P° – P }{ P° }\) = 0.1

Question 41.
When NaCl is dissolved in water, boiling point ………..
Answer:
increases

Question 42.
Use of glycol as antifreezer in automobile is an important application of …………….
Answer:
Colligative property

Question 43.
Ethylene glycol is mixed with water and used as antifreezer in radiators because …………..
Answer:
it lowers the freezing point of water

Question 44.
Colligative properties of a solution depend on ………… present in it.
Answer:
Number of solute particles

Question 45.
Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ………….
Answer:
low atmospheric pressure

IV. Choose the odd one out.

Question 1.
(a) Air
(b) Camphor in nitrogen gas
(c) Humid oxygen
(d) Salt water
Answer:
(d) Salt water.
a, b and e are gaseous solution whereas d is a liquid solution.

Question 2.
(a) CO2 dissolve in water
(b) Salt water
(c) Solution of H2 in palladium
(d) Ethanol dissolved in water
Answer:
(c) Solution of H2 in palladium
a, b and d are liquid solutions whereas c is a solid solution.

Question 3.
(a) Amalgam of potassium
(b) Camphor in nitrogen gas
(c) Solution of H2 in palladium
(d) Gold alloy
Answer:
(b) Camphor in nitrogen gas
a, b and d arc solid solutions whereas b is gaseous solution.

Question 4.
(a) Vapour pressure
(b) Lowering ofvapour pressure
(c) Osmotic pressure
(d) Elevation of boiling point
Answer:
(a) Vapour pressure
b, e and dare colligative properties whereas a is a physical property.

Question 5.
(a) Benzene and tolucne
(b) Chlorobenzene and Bromobenzene
(c) Benzene and acetone
(d) n – hexane and n – heptane
Answer:
(a) Benzene and acetone
a, b and dare ideal solutions whereas c is a non-ideal solution.

Question 6.
(a) Ethyl alcohol and cyclohexane
(b) Ethyl bromide and ethyl iodide
(c) Acetone and ethyl alcohol
(d) Benzene and acetone
Answer:
(a) Ethyl bromide and ethyl iodide
a, e and dare non-ideal solutions whereas b is an ideal solution.

V. Choose the correct pair.

Question 1.
(a) Humid oxygen – Liquid solution
(b) Gold alloy – Solid solution
(c) Salt water – Gaseous solution
(d) Solution of H2 in palladium – Gaseous solution
Answer:
(b) Gold alloy – Solid solution

Question 2.
(a) Air – Gaseous solution
(b) Amalgam of potassium – Liquid solution
(c) Salt water – Solid solution
(d) Carbonated water – Solid solution
Answer:
(a) Air – Gaseous solution

Question 3.
(a) Benzene and toluene – Non-ideal solution
(b) Benzcnc and acetone – Non-ideal solution
(c) Chlorobenzene and bromo henzene – Non-ideal solution
(d) Carbon tetrachloride and Chloroform – ideal solution
Answer:
(b) Benzene and acetone – Non-ideal solution

Question 4.
(a) Benzene and toluene – Ideal solution
(b) n-hexane and n-heptane – Non-ideal solution
(c) Ethyl iodide and ethyl bromide – Non-ideal solution
(d) Chiorobenzene and bromo benzene – Non-ideal solution
Answer:
(a) Benzene and toluene – Ideal solution

VI. Choose the incorrect pair.
Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-59
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-60

Question 2.
(a) Benzene and acetone – Ideal solution
(b) Ethyl alcohol and cyclohexane – Non-ideal solution
(C) n-hexane and n-heptanc – Ideal solution
(d) Chioro benzene – Ideal solution
Answer:
(a) Benzene and acetone – Ideal solution

VII Assertion & Reason.

Question 1.
Assertion (A) : When NaCI is added to water, a depression in freezing point is observed.
Reason (R): The lowering of vapour pressure of a solution causes the depression in freezing poi nl.
(a) Assertion and Reason are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Assertion and Reason are correct and R is the correct explanation of A.

Question 2.
Assertion (A): Ammonia reacts with water does not obey Henry’s law.
Reason (R): The gases reacting with the solvent does not obey Henry’s law.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct hut (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 3.
Assertion (A): Acetic acid solution deviates from Raoult’s law.
Reason (R): Association of solute molecules exists as a dimer by forming intermolecular. hydrogen bonds and hence deviates from Raoult’s law.
(a) Both (A) and (R) arc wrong.
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

VIII. Choose the correct statement.

Question 1.
(a) Raoult’s law is applicable to volatile solid solute in liquid solvent
(b) Henry’s law is applicable to solution containing solid solute in liquid solvent
(c) For very dilute solutions, the solvent obeys Raoult’s law and the solute obeys Henry’s law.
(d) For saturated solution containing volatile solid solute in liquid solvent both laws are obeyed.
Answer:
(c) For very dilute solutions. the solvent obeys Raoult’s law and the solute obeys LIenrys law.

Samacheer Kalvi 11th Chemistry Solutions 2 Marks Questions and Answers

I. Write brief answer to the following questions:

Question 1.
What is the common property observed in naturally existing solution? Explain it.
Answer:

  1. Sea water, air are the naturally existing homogeneous mixture. The common property observed in these is homogeneity.
  2. The homogeneity implies uniform distribution of their constituents or components through out the mixture.

Question 2.
Define solution with an example.
Answer:
1. A solution is a homogeneous mixture of two or more substances consisting of atoms. ions or molecules.

2. For example, when a small amount of NaCl is dissolved in water, a homogeneous solution is obtained. In this solution, Na+ and C ions are uniformly distributed in water. Here NaCI is the solute and water is the solvent.

Question 3.
What are aqueous and non aqueous solution? Give example.
Answer:

  1. If the solute is dissolved in the solvent water, the resultant solution is called as an aqueous solution. e.g., salt in water.
  2. If the solute is dissolved in the solvent other than water such as benzene, ether, CCl4 etc, the resultant solution is called a non aqueous solution. e.g., Br, in CCI4.

Question 4.
Define molality.
Answer:
Molality is defined as the number of moles of solute present in 1 kg of the solvent.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-61

Question 5.
Define molaritv.
Answer:
Molarity is defined as the number of moles of solute present in 1 litre of the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-62

Question 6.
Define normality.
Answer:
Normality is deflncd as the number of gram equivalents of solute present in 1 litre of the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-63

Question 7.
Define forniality.
Answer:
Formality (F) is defined as the number of formula weight of solute present in 1 litre of the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-64

Question 8.
Define mole fraction.
Answer:
Mole fraction of a component is the ratio of number of moles of the component to the total number of moles of all components present in the solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-65

Question 9.
Show that the sum of mole fraction of a solution is equal to one.
Answer:
Consider a solution containing two components A and 13 whose mole fractions are xA and xB respectively. Let the number of moles of two components A and B are nA and nB respectively.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-66

Question 10.
Define mass percentage.
Answer:
Mass percentage is defined as the ratio of the mass of the solute in g to the mass of solution in g multiplied by 100.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-67

Question 11.
Define volume percentage.
Answer:
Volume percentage is defined as the ratio of volume of solute in mL to the volume of solution in ml multiplied by 100.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-68

Question 12.
Define mass by volume percentage.
Answer:
It is defined as the ratio of the mass of the solute in g to the volume of the solution in ml multiplied by 100.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-69

Question 13.
What is meant by ppm? Where is it used?
Answer:
1. part per million =
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-70

2. ppm is used to express the quantity of solutes present in small amounts in solutions.

Question 14.
What is meant by stock solution (or) standard solution? What is meant by working standard?
Answer:
1. A standard solution or a stock solution is a solution whose concentration is accurately known.

2. At the time of experiment, the solution with required concentration is prepared by diluting the stock solution. This diluted solution is called working standard.

Question 15.
Define solubilitv.
Answer:
The solubility of a substance is defined as the amount of the solute that can be dissolved in loo g of the solvent at a given temperature to form a saturated solution.

Question 16.
Ammonia is more soluble than oxygen in water. Why?
Answer:
Ammonia forms hydrogen bonding with water molecules, this intermolecular bonds arc very strong and thus the ammonia is more soluble in water. Ammonia is strongly interact with water to form ammonium hydroxide. But oxygen is more electronegative it is not able to interact with water more. So NH3 is more soluble than O2 in water.

Question 17.
Solubility of a solid solute in a liquid solvent increases with increase in temperature. Justify this statement.
Answer:
When the temperature is increased,the average kinetic energy of the molecules of the solute and the solvent increases. The increase in the kinetic energy facilitates the solvent molecules to break the intermolecular attractive forces that keep the solute molecules together and hence the solubility increases.

Question 18.
Dissolution of ammonium nitrate increases with increase in temperature. Why?
Answer:
The dissolution process of ammonium nitrate is endothermic. So the solubility increases with increase in temperature.

Question 19.
What is the relationship between the solubility of eerie sulphate with temperature?
Answer:
The dissolution of eerie sulphate is exothermic and the solubility decreases with the increase in temperature.

Question 20.
Why in the dissolution of CaCl2, the solubilit increases moderately with high temperature?
Answer:
Even though the dissolution of CaCI2, is cxothcrmic, the soluhility increases moderately with increase in temperature. Here the entropy factor plays a significant role in deciding the position of equilibrium.

Question 21.
Why the carbonated drinks are stored in pressurized container?
Answer:
1. The carbonated beverages contain CO2 dissolved in them. To dissolve the CO2 in these drinks, CO2 gas is bubbled through them under high pressure.

2. These containers are sealed to maintain the pressure. When we open these containers at atmospheric pressure, the pressure of the CO2 drops to the atmospheric pressure level and hence bubbles of CO2 rapidly escape from the solution and show effervescence.

Question 22.
Define

  1. Evaporation
  2. Condensation.

Answer:
1. Evaporation:
If the kinetic energy of molecules in the liquid state overcomes the intermolecular force of attraction between them, then the molecules will escape from the liquid state. This process in called evaporation.

2. Condensation:
The vapour molecules are in random motion during which they collide with each other and also with the walls of the container. As the collision is inelastic, they lose their energy and as a result the vapour returns back to liquid state. This process is called as condensation.

Question 23.
State Dalton’s law of partial pressure.
Answer:
According to Dalton’s law of partial pressure, the total pressure in a closed vessel will be equal to the sum of the partial pressure of the individual components.
Ptotal = PA + PB

Question 24.
Give the reason behind the lowering of vapour pressure in the dissolution of NaCl in water?
Answer:
NaCI is a non volatile solute. When a non volatile solute is dissolved in pure solvent, the vapour pressure of pure solvent will decrease. In such solution, vapour pressure of the solution will depend only on the solvent molecules as the solute is non-volatile.

Question 25.
What are ideal solution? Give example.
Answer:
An ideal solution is a solution in which each component i.e., the solute as well as the solvent obeys the Raoult’s law over the entire range of concentration.

Question 26.
What are non-ideal solution? Give example.
Answer:

  1. The solutions which do not obey Raoult’s law over the entire range of concentration are called non-ideal solutions.
  2. The deviation of the non-ideal solution from the Raoult’s law may be positive (or) negative.
  3. Example, Ethyl alcohol and cyclohexane.

Question 27.
What are colligative properties? Give example.
Answer:
The properties which do not depend on the chemical nature of the solute but depends only on the number of solute particles present in the solution are called colligative properties. e.g.,

  1. Relative lowering of vapour pressure – \(\frac { P° – P}{ P° }\)
  2. Osmotic pressure – π
  3. Elevation of boiling point – ∆Tb
  4. Depression in freezing point – ∆Tf

Question 28.
What is meant by elevation of boiling point?
Answer:
1. The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the atmospheric pressure.

2. When a non-volatile solute is added to pure solvent at its boiling point, the vapour pressure of the solution is lowered below 1 atm. To bring the vapour pressure again to 1 atm, the temperature of the solution has to be increased.

3. As a result, the solution boils at a higher temperature (Tb) then the boiling point of pure solvent (Tb°). This increase in the boiling point is known as elevation of boiling point.

Question 29.
Define ebullioscopic constant.
Answer:
Ebullioscopic constant kb, is equal to the elevation in boiling point for 1 molal solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-71

Question 30.
Define osmotic pressure.
Answer:
Osmotic pressure can be defined as the pressure that must be applied to the solution to stop the influx of the solvent (to stop osmosis) through the semipermeable membrane.

Question 31.
Write the Van’t Hoff equation of osmotic pressure.
Answer:
Van’t Hoff equation states that for dilute solutions, the osmotic pressure is directly proportional to the molar concentration of the solute and the temperature of the solution.
π = CRT
where
π = Osmotic pressure
C = concentration
T = Temperature
R = gas constant

Question 32.
Define Van’t Hoff factor.
Answer:
van’t Hoff factor (I) is defined as the ratin of the actual molar mass to the abnormal molar mass of the solute.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-72

Question 33.
How is degree of dissociation and degree of association are related with van’t Hoff factor?
Answer:
The degree of dissociation or association can be related to van’t Hoff factor
1. using the following relationship

  • αdissociation = \(\frac { i – 1 }{ n – 1 }\)
  • αassociation = \(\frac { (1 – i)n }{ n – 1 }\)

where n = number of solute particles

Question 34.
Give an example of a solid solution ¡n which the solute is a gas.
Answer:
Solution of hydrogen in palladium.

Question 35.
What role does the molecular interaction play in solution of alcohol and water?
Answer:
There is strong hydrogen bonding in alcohol molecules as well as water molecules. The intermolecular forces both in alcohol and water are H-bonds. When alcohol and water are mixed,

they form solution because of formation of H-bonds between alcohol and H2O molecules hut these interactions are weaker and less extensive than those in pure water. Hence, they show positive deviation from ideal behaviour.

Question 36.
Why do gases always tend to be less soluble in liquids as the temperature is raised?
Answer:
Mostly dissolution of gases in liquid is an exothermic process. it is because the fact that this process involves decrease of entropy. Thus, increase of temperature tends to push the equilibrium towards backward direction as a result of which solubility of the gas decrease with rise in temperature.
(Gas + Solvent \(\rightleftharpoons\) Solution + Heat)

Question 37.
Why is the freezing point depression of 0.1 M NaCl solution nearly twice that of 0.1M glucose solution?
Answer:
NaCl is an electrolyte and it dissociates completely whereas glucose being a non-electrolyte does not dissociate. Hence, the number of particles in 0.1 M NaCl solution is nearly double for NaCI solution than that for glucose solution of same molarity.

Therefore depression in freezing point being a colligative property ¡s nearly twice for NaCl solution than that for glucose solution of same molarity.

Question 38.
Why a person suffering from high blood pressure is advised to take minimum quantity of common salt?
Answer:
Osmotic pressure is directly proportional to the concentration of solutes. Our body fluid contains a number of solutes. On taking large amount of salt, ions entering into the body fluid thereby raises the concentration of solutes. As a result, osmotic pressure increases which may rupture the blood cells.

Samacheer Kalvi 11th Chemistry Solutions 3 Marks Questions and Answers

Question 1.
What are gaseous solution ? Give its various types with example
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-73

Question 2.
What are liquid solutions ? Explain with example
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-74

Question 3.
What are solid solution? Give example.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-75

Question 4.
How will you prepare a standard solution?
Answer:

  1. A standard solution or a stock solution is a solution whose concentration is accurately known.
  2. A standard solution of required concentration can be prepared by dissolving a required amount of a solute in a suitable amount of solvent.
  3. It is done by transforming a known amount of solute to a standard flask of definite volume. A small amount of water is added lo the flask and shaken well to dissolve the salt.
  4. Then water is added to the flask to bring the solution level lo the mark indicated at the top end of the flask.
  5. The flask is stoppered and shaken well to make concentration uniform.

Question 5.
What are the advantages of standard solution.
Answer:
1. The error due to weighing the solute can be minimised by using concentrated stock solution that requires large quantities of solute.

2. We can prepare working standards of different concentrations by diluting the stock solution which is more efficient since consistency is maintained.

3. Some of the concentrated solutions are more stable and are less likely to support microbial growth than working standards used in the experiments.

Question 6.
Explain the solubilities of ammonium nitrate, calcium chloride, ceric sulphate and sodium chloride in water at different temperature with a graph.
Answer:
1. The solubility of sodium chloride does not vary appreciably as the maximum solubility is achieved at normal temperature. In fact, there is only 10% increase in solubility between 0°C to 100°C.

2. The dissolution process of ammonium nitrate is endothermic, the solubility increases with

3. In the case of eerie sulphate. the dissolution is exothermic and the solubility decreases with increase in temperature.

4. Even though the dissolution of calcium chloride is exothermic, the solubility increases moderately with increase in temperature. Here the entropy factor also plays a significant role in deciding the position of equilibrium.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-76

Question 7.
Explain the effect of temperature gaseous solute ¡n liquid solvent.
Answer:
1. In the case of gaseous solute in liquid solvent, the solubility decreases with increase in temperature.

2. When a gaseous solute dissolves in a liquid solvent, its molecules interact with solvent molecules with weak inter molecular forces when the temperature increases, the average. kinetic energy of the molecules present in the solution also increases.

3. The increase in kinetic energy breaks (he weak inter molecular forces between the gaseous solute and liquid solvent with results in the release of the dissolved gas molecules to gaseous state.

4. The dissolution of most of the gases in Liquid solvents is an endothermic process, the increase in temperature decreases the dissolution of gaseous molecules.

Question 8.
Give reason why aquatic species are less sustained in hot water?
Answer:
There will be decrease in solubility of gases in solution with increase in temperature. During summer, in hot water rivers, due to high temperature. the availability of dissolved oxygen decreases. So the aquatic species are less sustained in hot water.

Question 9.
Deep – sea divers use air diluted with helium gas in their tanks. Why? (or) Justify this statement.
Answer:
1. Deep-sea divers carry a compressed air tank for breathing at high pressure under water. This air tank contains nitrogen and oxygen which are not very soluble in blood and other body fluids at normal pressure.

2. As the pressure at the depth is far greater than the surface atmospheric pressure, more nitrogen dissolves in the blood when the diver breathes from tank.

3. When the divers ascends to the surface, the pressure decreases, the dissolved nitrogen comes out of the blood quickly forming bubbles in the blood stream.

These bubbles restrict blood flow, affect the transmission of nerve impulses and can even burst the capillaries or block them. This condition is called “the bends” which are painful and dangerous to life.

4. To avoid such dangerous condition they use air diluted with helium gas (11.7 % helium, 56.2% nitrogen and 32.1% oxygen) of lower solubility of helium in the blood than nitrogen.

Question 10.
What are the limitations of Henry’s law?
Answer:

  1. Henry’s law is applicable at moderate temperature and pressure only.
  2. Only the less soLuble gases obey Henry’s law.
  3. The gases reacting with solvent do not obey Henry’s law.
  4. The gases obeying Henrys law should not associated or dissociated while dissolving in the solvent.

Question 11.
Explain how benzene in toluene obeys Raoult’s law.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-77
Answer:
The variation of vapour pressure of pure benzene and toluenc with its mole fraction is given in the graph.
1. The vapour pressure of pure toluene and pure benzene are 22.3 and 74.7 mm Hg respectively.

2. The graph shows the partial vapour pressure of pure components increases linearly with the increase of the mole fraction of the respective components. The total pressure at any composition of the solute and solvent is given by the straight line.

3. Psolution = P0toluene + xbenzene (P0benzene – P0toluene)

Question 12.
Derive the relationship between the relative lowering of vapour pressure and mole fraction of the solute.
Answer:
Psolution ∝ xA by Raoult’s law. where xA is the mole fraction of the solvent.
Psolution = k . xA
When
xA = 1
k = P0solvent
P0solvent = partial pressure of pure solvent
Psolution = P0solvent . xA
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-78
where
xB = mole fraction of the solute
xA + xB = 1
xB = 1 – xA
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-79

Question 13.
How would you compare Raoult’s law and Henry’s law.
Answer:
1. According to Raou It’s law, for a solution containing a non volatile solute.
Psolution = P0solvent . xsolute

2. According to henry’s law, Psolution = KH . xsolute in solution

3. The difference between the above two laws is the proportionality constant P° (Raoult’s law) and KH (Heniys law).

4. henry’s law is applicable to solution containing gaseous solute in liquid solvent, while Raoult’s law is applicable to non volatile solid solute in the liquid solvent.

5. If the solute is non volatile then the Henry’s law constant will become equal to the vapour pressure of pure solvent Po. thus Raoult’s law becomes a special case of Henry’s law.

6. For very dilute solutions, the solvent obeys Raoult’s law and the solute obeys Henry’s law.

Question 14.
What are the necessary conditions for an ideal solution? Give two example. For an ideal solution
1. There is no change in volume on mixing two components (solute and solvent)
∆Vmixing = O

2. There is no exchange of heat when the solute is dissolved in solvent (∆Hmixing = 0)

3. Escaping tendency of the solute and the solvent present in it should be same as in pure liquids.

4. Examples – For ideal solution: Benzene and toluene, n-Hexane and n-Heptane, ethyl bromide and ethyl iodide, chlorobenzene and bromo benzene.

Question 15.
Explain how non-ideal solutions shows positive deviation from Raoult’s law.
Answer:

  1. Let us consider the positive deviation shown by a solution of ethyl alcohol and water.
  2. In this solution, the hydrogen bonding interaction between ethanol and water is weaker than those hydrogen bonding interactions amongst themselves (ethyl alcohol-ethyl alcohol and water-water interaction).
  3. This results in the increased evaporation of both componeins from the aqueous solution of ethanol.
  4. Consequently, the vapour pressure of the solution is greater than the vapour pressure predicted by Raoult’s law.
  5. Here, the mixing OCCSS is endothermic i.e., (∆Hmixing > O) and there will be a slight increase in volume (∆Vmixing > O)

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-80

Question 16.
Explain with suitable example about negative deviation from law.
Answer:
1. Let us consider a solution of phenol and aniline. Both phenol and aniline form hydrogen bonding interactions amongst themselves.

2. When mixed with aniline, the phenol molecule forms hydrogen bonding interactions with aniline, which are Stronger than the hydrogen bonds formed amongst themselves.

3. Formation of new hydrogen bonds considerably reduce the escaping tendency of phenol and aniline from the solution.

4. Asa result, the vapour pressure of the solution is less and there is a slight decrease in volume (∆Vmixing < 0) on mixing.

5. During this process evolution of heat takes place i.e., ∆Vmixing < 0 (exothermic).

6. Examples – Acetone + Chloroform, Chloroform + Diethyl ether
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-81

Question 17.
The vapour pressure of a solution containing a non volatile, non-electrolyte solute is always lower than that of pure solvent. Give reason.
Answer:
1. The vapour pressure of a solution (P) containing flOfl volatile solute is lower than that of pure solvent (P°).

2. Consider a closed system is which a pure solvent is in equilibrium with its vapour. At equilibrium the molar Gibbs free energies of solvent in a liquid and gaseous phase are equal (∆G = O).

3. When a solute is added to this solvent the dissolution takes place and its free energy (G) decreases due to increase in entropy.

4. In order to maintain the equilibrium, the free energy of the vapour phase must also decrease.

5. At a given temperature, the only way to lower the free energy of the vapour is to reduce its pressure.

6. Thus the vapour pressure of the solution must decrease to maintain the equilibrium.

Question 18.
Show that relative lowering of vapour pressure is a colligative property.
Answer:
According to Raoult’s law,
Psolution xA, where xA = mole fraction of solvent.
Psolution = k . xA, where k = proportionality constant
For a pure solvent,
Vapour pressure = P°, xA = 1
solution = k x 1 = k
Substituting P°solvent in Raoult’s law
Psolution = P°solvent . xA
Relative lowenng of vapour pressure
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-82
substituting Psolution as P°xB in the above eaquation
Relative lowering of vapour pressure
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-83
xA + xB = I
where xB = mole fraction of solute. It is clear that the relative lowering of vapour pressure depends only on the mole fraction ofthe solute (xB) and is independent of its nature. Therefore relative lowering of vapour pressure is a colligative property.

Question 19.
Explain why boiling point of solution is greater than that of pure solvent?
Answer:
When a non volatile solute is added to a pure solvent at its boiling point, the vapour pressure of the solution is lowered below 1 atm. To bring the vapour pressure again to I atm the temperature of the solution has to be increased.

As a result, the solution boils at a higher temperature (Tb) than the boiling point of the pure solvent (T°b). This increase in the boiling point is known as elevation of boiling point ∆Tb = Tb – T°b.

Question 20.
Graphically prove that Tb ¡s greater than T°b.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-84
1. The vapour pressure of the solution increases with increase in temperature. The variation of vapour pressure with respect to temperature of pure water is given by the curve – A.

2. At 100°C, the vapour pressure of water is equal to I atm. Hence, the boiling point of water is 100°C (T°b).

3. When a solute is added to water, the vapour pressure of the resultant solution is lowered. The variation of vapour pressure with respect to temperature for the solution is given by curve-B.

4. From the graph, it is evident that the vapour pressure of the solution is equal to 1 atm. pressure at the temperature Tb which is greater than T°b. The difference between these two temperatures (Tb – T°b) gives the elevation of boiling point.
∆Tb = Tbb.

Question 21.
Derive the relationship between the elevation of boiling point and molar mass of non volatile solute.
Answer:
The elevation of boiling point ∆Tb = Tbb.
∆Tb is directly proportional to the concentration of the solute particles.
∆Tb ∝ m, (m = molaLiiy)
∆Tb = kb. m, where kb = ebullioscopic constant
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-85

Question 22.
Define

  1. freezing point
  2. Depression in freezing point.

Explain with graph.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-86
1. Freezing point is defined as the temperature at which the solid and the liquid states of the substances have the same vapour pressure.

2. When a non volatile solute is added to water at its freezing point, the freezing point of water is lowered from 0°C. The lowering of freezing point of the solvent when a solute is added is called depression in freezing point AT1..

3. ∆Tf = T0f – Tf

Question 23.
Define

  1. cryoscopic constant
  2. ebullioscopic constant

Answer:

1. ∆Tf = kf. m, where kf = cryoscopic constant. If m = 1. then ATf = kf
kf is defined as depression in freezing point for 1 molal solution.

2. ∆Tf = kf. m where kf ebullioscopic constant. If m = 1 then ATf = kf
kb is defined as elevation in boiling point for 1 molal solution.

Question 24.
What are the significances of osmotic pressure over other colligative properties ?
Answer:
1. Unlike elevation of boiling point and the depression in freezing point, the magnitude of osmotic pressure is large.

2. The osmotic pressure can be measured at room temperature enables to determine the molecular mass of biomolecules which are unstable at higher temperature.

3. Even for a very dilute solution, the osmotic pressure is large.

Question 25.
What is haemolysis ? intravenous fluid are isotonic to blood?
Answer:
1. The osmotic pressure of the blood cells is approximately equal to 7 atm at 37°C.

2. The intravenous injections should have saine osmotic pressure as that of the blood (isotonic vith blood).

3. If the intravenous solutions are too dilute that is hypotonie, the solvent from outside of the cells flow into the cell to normalise the osmotic pressure and this process is called haernolysis causes the cells to burst.

4. On the other hand, if the solution is too concentrated, that is hypertonic. the solvent molecules will flow out of the cells,which causes the cells to shrink and die.

5. For this reason, the intravenous fluids are prepared such that they are isotonic to blood (0.9% mass/volume sodium chloride solution).

Question 26.
Explain reverse osmosis.
Answer:
1. The pure water moves through the semipermeable membrane to the NaCl solution due to osmosis.

2. This process can be reversed by applying pressure greater than the osmotic pressure to the solution side. Now the pure water moves from the solution side to the solvent side and this process is called reverse osmosis.

3. Reverse osmosis can be defined as a process in which a solvent passes through a semipermeable membrane in the opposite direction of osmosis, when subjected to a hydrostatic pressure greater than the osmotic pressure.

Question 27.
Explain about the application of reverse osmosis in water purification.
Answer:
1. Reverse osmosis is used in the desalination of sea water and also in the purification of drinking water.

2. When a pressure higher than the osmotic pressure is applied on the solution side (sea water) the water molecules moves from solution side to the solvent side through semi permeable membrane (opposite to osmotic flow). The pure water can be collected.

3. Cellulose acetate (or) polyamide membranes are commonly used in commercial system.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-87

Question 28.
Acetic acid is found to have molar mass as 120 g mol-1. Prove it.
Answer:
1. In certain solvent, solute molecules associate to form a dimer. This reduces the total number of molecules formed in solution and as a result the calculated molar mass will be higher than the actual molar mass.

2. Acetic acid in benzene exist as a dimer
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-88

3. The molar mass of acetic acid calculate using colligative properties is found to be around 120 g mol-1 is two times of the actual molar mass 60 g mol-1.

Question 29.
Depression in freezing point of NaCI is twice that of in urea. Why?
Answer:
1. The electrolyte NaCI dissociates completely into its constituent ions in their aqueous solution. This causes an increase in the total number olparticles present in the solution.

2. When we dissolve 1 mole of NaCI in water it. dissociates and gives 1 mole of Na+ and 1 mole of Cl. Hence the solution will have 2 moles of particles.

But when we dissolve 1 mole of urea (non electrolyte) in water it appears as 1 mole only. So the colligative property value would be double in NaCl than in urea.

Question 30.
What is van’t Hoff factor? Calculate the van’t Hoff factor value for

  1. acetic acid
  2. NaCl

Answer:
1. van’t Hoff factor is defined as the ratio ofthe actual molar mass to the abnormal (calculated) molar mass of the solute.

2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-89
van’t Hoff factor (1) for acetic acid = \(\frac { 60 }{ 120 }\) = 0.5

3. van’t Hoff factor (2) for NaCl = \(\frac { 117 }{ 58.5 }\) = 2

Question 31.
Differentiate between ideal solution and non-ideal solution.
Answer:
Ideal solution
An ideal solution is a solution in which each component obeys the Raoult’s law over the entire range of concentration.
For an ideal solution,

  • ∆Hmixing = 0
  • ∆Vmixing = 0

Example: Benzenc and toluene n – Hexane and n – Heptane

Non-ideal solution
The solutions which do not obey Raoult’slaw over the entire range of concentrationsare called non-ideal solution.
For a non-ideal solution.

  • ∆Hmixing \(\quad \neq\) 0
  • ∆Vmixing \(\quad \neq\) 0

Example: Ethyl alcohol and Cyclo hexane, Benzene and acetone .

Question 32.
Explain the factors when i = 1, i < 1 and i >1 ?
Answer:
1. For a solute that does not dissociate or associate the vant’s hoff factor is equal to 1 (i = 1) and the molar mass will be close to the actual molar mass.

2. For that solute that associate to form higher oligomers in solution, the van’t Hoff factor will be less than 1 (i < 1) and the observed molar mass will be greater than the actual molar mass.

3. For solutes that dissociates into their constituent ions the van’t Hoff factor will be more than one (i > 1) and the observed molar mass will be less than the normal molar mass.

Question 33.
State Henry’s law and mention some of its important applications.
Answer:
Henry’s law: The solubility of a gas in a liquid is directly proportional to the pressure of the gas.
Application of Henry’s law:

  1. In the production of carbonated beverages
    (as solubility of CO2 increase at high pressure).
  2. In the deep sea diving.
  3. In the function of lungs.
  4. For climbers or people living at high altitudes,

Question 34.
What type of non – idealities are exhibited by cyclohexane – ethanol and acetone – chloroform mixture? Give reason for your answer.
Answer:
Ideal solutions are those which obey Raoult’s law over extreme range of concentration. Ideal solutions have another important properties:

  • ∆Hmix = 0
  • ∆Vmix = 0

Here-forces of attraction between A – A. B – B and A – B are of the same order. Non ideal solutions do not obey Raoult’s law over the entire range of concentration.
∆Hmixing\(\quad \neq\) 0 and ∆Vmixing\(\quad \neq\) 0

Cyclohexane – ethanol mixture shows positive deviation from Raoult’s law because forces of attraction between cyclohexane and ethanol are less than in between pure cyclohexane as well as pure ethanol.

Acetone-Chloroform mixture shows negative deviation from Raoults law because forces of attraction between acetone and chloroform are higher than that in between pure acetone and pure chloroform molecules.

Question 35.
Given below is the sketch of a plant for carrying out a process.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-90

  1. Name the process occurring in the above plant.
  2. To which container does the net flow of solvent take place?
  3. Name one SPM which can he used in this plant.
  4. Give one practical use of the plant.

Answer:

  1. Reverse osmosis
  2. In fresh water container from salt water container.
  3. Cellulose acetate is semipermeable membrane (SPM)
  4. Purification of water

Question 36.
Define the term osmotic pressure. Describe how the molecular mass of a substance can be determined by a method based on measurement of osmotic pressure?
Answer:
π = CRT
π = \(\frac { n }{ V }\)RT
πV = nRT
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-91
Osmotic pressure is inversely proportional to the molecular mass of the soLute.

Question 37.
1. Menthol is a crystalline substance with peppermint taste. A 6.2% solution of menthol in cyclohexane freezes at – 1.95°C.

Determine the formula mass of menthol. The freezing point and molal depression constant of cyclohexane are 6.5°C and 20.2 K m-1, respectively.

2. State Henry’s Law and mention its two important applications.

3. Which of the following has higher boiling point and why? 0.1 M NaCl or 0.1 M Glucose
Answer:
1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-92
MB = 158 g mol-1

2. Henry’s Law:
The solubility of a gas in a liquid is directly proportional to the pressure of the gas.
Applications:

  • Solubility of CO2 is increased at high pressure.
  • Mixture of He and O2 are used by deep sea divers because he is less soluble than nitrogen.

3. 0.1 M NaCI, because it dissociates in solution and furnishes greater number of particles per unit volume while glucose being a non-electrolyte does not dissociate.

Question 38.
Water is a universal solvent. But alcohol also dissolves most of the substances soluble in water and also many more. Boiling point of water is 100°C and that of alcohol is 80°C. The specific heat of water is much higher than the specific heat of alcohol.

  1. List out three possible differences if instead of water as the liquid ¡n our body we had alcohol.
  2. What value can you derive from this special property of water and its innumerable uses in sustaining life on earth?

Answer:
1.
(i) Even a small rise in temperature in the surroundings will raise the temperature of’ the body because the specific heat of alcoholis much less than the specific heat of water. So, in order to cool the body, more sweating will take place.

(ii) As there is less H bonding in alcohol, it will gel evaporated faster. The alcohol will be evaporated at such a faster rate that the liquid has to be ingested all the time.

(iii) Ice which floats on water helps aquatic life to exist even in winter as water insulates the heat from liquid below it to go back to the surroundings. Solid alcohol does not have such special properties.

2. Praise is to the almighty that has so thoughtfully given such special properties to water and made it a liquid that could sustain life on earth.

Question 39.
State Henry’s law for solubility of a gas in a liquid. Explain the significance of Henry’s law constant (KH). At the same temperature, hydrogen is more soluble in water than helium. Which of theni will have a higher value of KH and Why?
Answer:
Henry’s law states that the solubility of a gas in liquid at a given temperature is directly proportional to the partial pressure of the gas.
P = KH x
where P is the pressure of the gas, x is the mole fraction of the gas in the solution and KH is the Henry’s law constant. KH is a function of the nature oIgas.

Higher the value of KH at a given temperature. lower is the solubility of the gas in the liquid. As helium is less soluble in water, so it has a higher value of KH than hydrogen.

Henry’s Law:
As dissolution of agar in liquid is an exothermic process, therefore, the solubility should decrease with in increase in temperature.

Question 40.
What is meant by positive and negative deviations from Raoult’s law and how is the sign ∆Hmix of related to positive and negative deviations from Raoult’s law?
Answer:
Negative deviations:
In these type of deviations, the partial vapour pressure of each component A and B of solution is higher than the vapour pressure calculated from Raoult’s law. For example -Water and ethanol, chloroform and water.

Positive deviations:
In case of positive deviation A – B interactions are weaker than those between A – A or B – B. This means that in such solutions molecules or A (or B) will find it easier to positive deviation from Raoult’s law.

Samacheer Kalvi 11th Chemistry Solutions 5 Marks Questions and Answers

II. Answer the following questions in detail:

Question 1.

  1. Define solution.
  2. Explain the types of solutions with suitable example.

Answer:
1. A solution is a homogeneous mixture of two or more substances, consisting of atoms. ions or molecules.

2. Types and examples of solution.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-93

Question 2.

  1. Define Solubility
  2. Explain about the factors that influences the solubility

Answer:

1. The solubility of a substance at a given temperature is defined as the amount of the solute that can be dissolved in 100 g of the solvent at a given temperature to form a saturated solution.

2. Factors influencing solubility
(a) Nature of solute and solvent: Sodium chloride, an ionic compound readily dissolves in polar solvent such as water but it does not dissolve in non polar solvent such as benzene. Most of the organic compounds dissolve in organic solvent and do not dissolve in water.

(b) Effect of temperature: Generally, the solubility of a solid solute in a liquid solvent increases with increase in temperature. The dissolution of NaCl does not vary as the maximum solubility is achieved at normal temperature.

The dissolution of ammonium nitrate is endothermic, the solubility increases with increase in temperature. The dissolution of eerie sulphate is exothermic and the solubility decreases with increase of temperature. In the case of gaseous solute in liquid solvent the soluhility decreases with increase in temperature.

Effect of pressure:
Generally the change in pressure does not have any significant effect in the solubility of solids and l?quids as they are not compressible. However, the solubility of gases generally increases with increase of pressure.

Question 3.
Explain about the factors that are responsible for deviation from Raoult’s law.
Answer:
1. Solute-solvent interactions:
For an ideal solution, the interaction between the solvent molecules (A – A), the solute molecules (B – B) and between the solvent and solute molecules (A – B) are expected to be similar. if these interactions are dissimilar, there will be a deviation from ideal behaviour.

2. Dissolution of solute:
When a solute present in a solution dissociates to give its constituent ions, the resultant ions interact strongly with the solvent and causes deviation from Raoult’s law. e.g., KCI in water deviates from ideal behaviour due to dissociation as K+ and Cl ion which form strong ion-dipole interaction with water molecules.

3. Association of solute:
Association of solute molecules can also cause deviation from ideal behaviour. For example in solution acetic acid exists as a dimer by forming intermolecular hydrogen bonds and hence deviates from Raoult’s law.

4. Temperature:
An increase in temperature of the solution increases the average kinetic energy of the molecules present in the solution which cause decrease in the attractive force between them. As result, the solution deviates from Raoult’s law.

5. Pressure:
At high pressure, the molecules tends to stay close to each other and therefore there will be an increase in their intermolecular attraction. Thus a solution deviates from Raoult’s law at high pressure.

6. Concentration:
When the concentration is increased by adding solute, the solvent-solute interaction becomes significant. This causes deviation from Raoult’s law.

Question 4.
How would you determine molar mass from relative lowering of vapour pressure.
Answer:
1. The measurement of relative lowering of vapour pressure can be used to determine the molar mass of a non-volatile solute.

2. A known mass of the solute is dissolved in a known quantity of solvent. The relative lowering of vapour pressure is measured experimentally.

3. According to Raoult’s law, the relative lowering of vapour pressure is
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-94
WA = weight of solvent
WB = weight of solute
MA = Molar mass of solvent
MB = molar mass of solute
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-95
where nA = number of moles of solvent
nB = number of moles of solute.
For dilute solution, nA > > nB, nA + nB = nA
Then, Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-96
Number of moles of solvent and solute arc
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-97
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-98
From the above equation, molar mass of the solute MB can be calculated using the known values of WA, WB, MA and the measured relative lowering of vapour pressure.

Question 5.
How would you determine the molar mass from osmotic pressure.
Answer:
According to van’t Hoff equation
π = CRT
C = \(\frac { n }{ V }\)
Here n = number of moles of solute dissolved in ‘V’ litre of the solution.
π = \(\frac { n }{ V }\) . RT = πV = nRT
If the solution is prepared by dissolving WB of the non-volatile solute in WA g of solvent, then the number of moles of ‘n’ is
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-99
where
MB = molar mass of the solute
Substituting n value, we get
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-100
From the above equation, we can calculate the molar mass of the solute.

Question 6.
What are ideal and non-ideal solutions? Explain with suitable diagram the behaviour of ideal solutions.
Answer:
Ideal solutions:
The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. Ideal solutions are formed by mixing the two components which are identical in molecular size, in structure and have almost identical intermolecular forces.

Examples:

  1. Benzene and toluene
  2. n – Hexane and n-Heptane
  3. Chiorobenzene and bromobenzene.

Characteristics:

  1. They must obey Raoult’s law.
  2. ∆H mixing should be zero.
  3. ∆W mixing should be zero, i.e. volume change on mixing is zero.

Non – ideal solutions:
The solutions which do not obey Raoult’s law are called non-ideal solutions. In case of non – ideal solutions there is a change in volume and heat energy when the two components are mixed.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-101

Characteristics:

  1. They does not obey Raoult’s law.
  2. ∆Vmix \(\quad \neq\) 0
  3. ∆Hmix \(\quad \neq\) 0

Behaviour of Ideal Solutions:
A plot of P1 or P2 versus the mole fraction x1 and x2 for an ideal solution gives a linear plot. These Lines (I and II)pass through the points and respectively whenx1 and x2 is equal to unity.

Similarly the plot (Line III) of Ptotal versus x2 is also linear. The minimum value of is P1° and the maximum value is P2°, assuming that component 1 is less volatile than component 2, i.e. P1° < P20.

Question 7.
Explain with a suitable diagram and appropriate example, why some non-ideal solution shows positive deviation from Raou It’s law.
Answer:
Some non-ideal solutions show positive deviation from Raoult’s law. Consider a solution of two components A and B.  If A-B interactions in the solution are weaker than the A – A and B – B interactions in the two liquids forming the solution, then the escaping tendency of molecules A and B from the solution become more than in pure liquids.

The total vapour pressure will be greater than the corresponding vapour pressure as expected on the basis of Raoult’s law. This type of behaviour of solution is called positive deviation from Raoult’s law. The boiling point of such solutions are lowered. Mathematically,
PA < PA0 xA
PB < PB0 xB
The total vapour pressure is less than PA + PB
P < PA + PB
P < P°A x xA + P°B xB
Hence
P1 = PA
P2 = PB
Examples of solutions showing positive deviations

  1. Ethyl alcohol and water
  2. Benzene and acetone
  3. Ethyl alcohol and cyclohcxanc
  4. Carbon tetrachloride and chloroform.

Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-102

Question 8.

  1. What is Osmotic pressure and how is it related to the molecular mass of a non volatile substances?
  2. What advantage the osmotic pressure method has over the elevation of boiling point method for determining the molecular mass?

Answer:
1. Osmotic pressure:
It is the pressure of the solution column that can prevent the entry of solvent molecules through a semi-permeable membrane, when the solution and the solvent are separated by the same. It is denoted by π. Its unit is mm 11g or atmosphere.
We know that, π = CRT
where π is the osmotic pressure and R is the gas constant
π = \(\frac { { n }_{ 2 } }{ V }\) RT
where V is volume of solution per litre containing n2 moles of solute.
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-103
By the above relation molar mass of solute can be calculated.

2. The osmotic pressure method has the advantage over other methods as pressure measurement is around the room temperature and molarity of the solution is used instead of molality.

The technique of osmotic pressure for determination of molar mass of solutes is particularly useful for biomolecules as they are generally not stable at higher temperature and polymers have poor solubility.

III. Numerical Problems

Question 1.
Calculate the mole fraction of benzene ¡n solution containing 30% by mass in carbon tetrachioride.
Solution:
30% of benzene in carbon tetrachloride by mass means that Mass of benzene in the solution = 30g
Mass of solution = 100g
Mass of carbon tetrachloride = 100g – 30g = 70g
Molar mass of benzene (C6H6) = 78 g mol-1
Molar mass of CCl4 = 12 + (4 x 35.5) 154g mol-1
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-104

Question 2.
Calculate (he molarity of each of the following solutions:
Solution:

  1. 30 g of CO(NO3)2. 6H2O = in 4.3 L of solution
  2. 30 mL of 0.5 M H2SO4 diluted to 500 mL.

1. Molar mass of CO(NO3)2. 6H2O = 58.7 + 2(14 + 48) + (6 x 8)g mol-1
= 58.7 + 124 + 108g mol-1 = 290.7 gmol-1
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-105
Volume of solution = 4.3 L
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-106

2. 1000 mL of 0.5 M H2SO4 Contain H2SO4 = 0.5 moles
30 mL of 0.5 M H2SO4 contain H2SO4 = \(\frac { 0.5 }{ 1000 }\) x 30 mole = 0.015 mole
Volume of solution = 500 mL = 0.500 L
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-107

Question 3.
Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution.
Solution:
0.25 molal aqueous solution means that
Moles of urea = 0.25 mole
Mass of solvent (water) = 1 kg = 1000 g
Molar mass of urea = 14 + 2 + 12 + 16 + 14 + 2 = 6Og mol-1
0.25 mole of urea = 60 x 0.25 mole = 15 g
Total mass of the solution = 1000 + 15 g = 1015 g = 1.015 g
Thus, 1.015 kg of solution contain urea = 15 g
2.5 kg of solution will require urea = \(\frac { 15 }{ 1.015 }\) x 2.5 kg = 37g

Question 4.
H2S, a toxic gas with rotten egg like smell is used for the qualitative analysis. 1f the solubility of H2S in water at STP is 0.195 m. Calculate Henrvs law constant.
Solution:
Solubility of H2S gas = 0.195 m
= 0.195 mole in 1 kg of the Solvent (water)
1 kg of the solvent (water) = 1000 g Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-108
Mole fraction of H2S gas in the solution (x) = \(\frac { 0.195 }{ 0.195 + 55.55 }\) = 0.0035
Pressure at STP = 0.98 7 bar
Applying Henry’s law
PH2S = KH x xH2S
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-109

Question 5.
Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Solution:
Here
1 = 23.8 mm
W2 = 50g
M2 (urea) = 60g mol-1
W1 = 850g
M1(Water) = 18g mol-1
Here we have to calculate Ps
Applying Raoult’s law,
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-110
Thus, relative lowering of vapour pressure = 0.017
Substituting P° = 23.8 mm Hg
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-111
We get,
23.8 – Ps = 0.017 Ps
Ps = 23.4 mm Hg
Thus, vapour pressure of water in the solution = 23.4 mm Hg

Question 6.
Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500g of water such that it boils at 100°C? .
Solution:
Elevation in boiling point required, ∆Tb = 100 – 99.63° = 0.37°
Mass of solvent (water) W1 = 500g
Mass of solute, C12H22O11 = 342 g mol-1
Molar mass of solvent M1 = 18 g mol-1
Applying the formula,
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-112

Question 7.
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504g mL-1?
Solution:
68% nitric acid by mass means that
Mass of nitric acid = 68 g
Mass of solution = 100 g
Molar mass of HNO3 = 63g mol-1
68g HNO3 = mole = 1.079 mole
Density of solution = 1.504 g mL-1
Volume of solution = \(\frac { 100 }{ 1.504 }\) mL = 66.5 mL = 0.0665 L
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-113

Question 8.
A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate tile mass percentage of the resulting solution.
Solution:
300 g of 25% solution contains solute = 75g
400g of 40% solution contains solute = 160 g
Total mass of solute = 160 + 75 = 235 g
Total mass oÍ’ solution = 300 + 400 = 700 g
% of solute in the final solution = \(\frac { 235 }{ 700 }\) x 1oo = 33.5%
% of water in the finaI solution = 100 – 33.5 = 66.5%

Question 9.
A sample of drinking water was found to be severely contaminated with chloroform (CHCI3) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass)

  1. express this in percentage by mass
  2. determine the molality of chloroform in the water sample.

Solution:
1. 15 ppm means 15 parts in million (106) parts by mass in the solution
% of mass = \(\frac { 15 }{ { 10 }^{ 6 } }\) x 100 = 1.5 x 10-4

2. Taking 15g chloroform in 106g of the solution
Mass of the solvent = 106 g
Molar mass of CHCl3 = 12 + 1 + (3 x 35.5) = 119.5 g mol-1
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-114

Question 10.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at normal boiling point of the solvent. What is the molar mass of the solute?
Solution:
Vapour pressure of pure water at the boiling point
(P°) = 1 atm 1.013 bar
Vapour pressure of solution Ps = 1.004 bar
M1 = 18 g mol-1
M2 = ?
Mass of solute = W2 = 2g
Mass of solution = 100g
Mass of solvent W1 = 98 g
Applying Raoult’s law for dilute solution
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-115

Question 11.
A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15K.
Solution:
Molar mass of cane sugar
C12H22O11 = 342 g mol-1
Molality of sugar = \(\frac { 5 x 1000 }{ 342 x 100 }\) = 0.146
∆T2 for sugar solution = 273.15 – 271 = 2.15°
∆Tf = Kf x m
Kf = \(\frac { 2.15 }{ 0.146 }\)
Molality of glucose solution = \(\frac { 5 }{ 180 }\) x \(\frac { 1000 }{ 100 }\) = 0.278 m
∆Tf (Glucose) = \(\frac { 2.15 }{ 0.146 }\) x 0.278 = 4.09°K
Freezing point of glucose solution = 273.15 – 4.09 = 269.06 K

Question 12.
Calculate the amount of benzoic acid (C6HCOOH) required for preparing 250 mL of 0.15 M solution in methanol.
Solution:
0.15 M solution means that 0.15 moles of C6H5COOH is present in IL
= 1000 mL of the solution
Molar mass of C6H5COOH = 72 + 5 + 12 + 32 + 1 = 122 g mol-1
Thus, 1000 mL of solution contains benzoic acid = 18.3g
250 mL of solution will contain benzoic acid
= \(\frac { 18.3 }{ 1000 }\) x 250 = 4.575 g

Question 13.
A solution containing 8 g of a substance in 100 g of diethyl ether boils at 36.86°C, whereas pure ether boils at 35.60 °C. Determine the molecular mass of the solute (For ether Kb 2.02 K kg mol-1)
Solution:
We have, mass of solute, W2 = 8 g
Mass of solvent, W1 = 100 g
Elevation of boiling point
∆Tb = 36.86 – 35.60 = 1.260C
Kb = 2.02
Molecular mass of the solute
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-116
= 128.25g mol-1

Question 14.
Ethylene glycol (molar mass = 62 g mol-1) is a common automobile antifreeze. Calculate the freezing point of a solution containing 12.4 of this substance in 100 g of water. Would it be advisable to keep this substance ¡n the car radiator during summer? (Kf for water = 1.86 k kg/mol-1) (Kb for water = 0.512 K kg/mol-1)
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-117
Since water boils at 100°C, so a solution containing ethylene glycol will boil at 101.024 °C, SO it is advisable to keep this substance in car radiator during summer.

Question 15.
15.0 g of an unknown molecular material is dissolved in 450g of water. The resulting solution freezes at – 0.34°C. What is the molar mass of the material? Kf for water = 1.86 K kg mol-1.
Solution:
W(solute) = 1 5.0 g
W(solvent) = 450 g
∆Tf = T°f – Tf = 0 – ( – 0.34) = 0.34 °C
∆Tf = kf m
0.34 = 1.86 x \(\frac { 15 }{ M }\) x \(\frac { 1000 }{ 450 }\)
M = \(\frac { 1.86 x 15 x 1000 }{ 0.34 x 450 }\) = 182.35 g mol-1

Common Errors
Common Errors:

  1. Students are writing solute, solvent, students get confused to write A (or) B, 1 (or)2.
  2. Mole, mole fraction may be confused.
  3. In writing osmosis definition Students get confused in mentioning concentration terms.
  4. van’t Hoff factor I formula may be con fused.
  5. Students may get contused rhen they write solute and solvent.
  6. Mole and mole fraction may be confused by students.
  7. Standard solutions must be known.
  8. When students write Raoult’s law, they get confused with solute and solvent.
  9. When they write the definition of osmosis, the conc. term may be little con fused.
  10. van’t Hoff equation may be written wrongly.

Rectifications:

  1. Always solvent is first so it is denoted as A (or) 1 solute is second so, it is denoted as B (or)2.
  2. Mole = n ; Mole fraction x
  3. Osmosis-movenient of solvent from low concentration to high concentration through a semipermeable membrane.
  4. Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-118
  5. For e.g.. solid in liquid means solid is the solute and liquid is the solvent.
  6. Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-119
  7. 1 N = 1 normal solution, 0.1 N Decinormal solution, 0.01 N = Centinormal solution
  8. In Raoult’s law, “A” is always solvent and “B” is always solute.
  9. In osmosis, always solvent rnoes through semi permeable membrane from low concentration to high concentration.
  10. van’t Hoff factory = i
    Samacheer Kalvi 11th Chemistry Solutions Chapter 9 Solutions-120

We as a team believe the information prevailing regarding the Samacheer Kalvi Solutions for 11th Chemistry Chapter 9 Solutions has been helpful in clearing your doubts to the fullest. For any other help do leave us your suggestions and we will look into them. Stay in touch to get the latest updates on Tamilnadu State Board Solutions for different subjects in the blink of an eye.

Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium

Enhance your subject knowledge with Tamilnadu State Board for Chapter 8 Physical and Chemical Equilibrium and learn all the underlying concepts easily. Make sure to Download Samacheer Kalvi 11th Chemistry Book Solutions, Notes Pdf Chapter 8 Physical and Chemical Equilibrium Questions and Answers PDF on a day to day basis and score well in your exams. Are given after enormous research by people having high subject knowledge. You can rely on them and prepare any topic of Chemistry as per your convenience easily.

Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium

Students looking for Chapter 8 Physical and Chemical Equilibrium Concepts can find them all in one place from our Tamilnadu State Board Solutions. Simply click on the links available to prepare the corresponding topics of Chemistry easily. Samacheer Kalvi 11th Chemistry Chapter wise Questions and Answers are given to you after sample research and as per the latest edition textbooks. Clarify all your queries and solve different questions to be familiar with the kind of questions appearing in the exam. Thus, you can increase your speed and accuracy in the final exam.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Multiple Choice Questions

Question 1.
If Kb and Kf for a reversible reactions are 0.8 x 10-5 and 1.6 x 10-4 respectively, the value of the equilibrium constant is
(a) 20
(b) 0.2 x 10-1
(c) 0.05
(d) None of these
Answer:
(a) 20
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-110

Question 2.
At a given temperature and pressure, the equilibrium constant values for the equilibria
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-111
The relation between K1 and K2 is
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-112
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-113
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-114

Question 3.
The equilibrium constant for a reaction at room temperature is K1 and that at 700 K is K2. If K1 > K2 then
(a) The forward reaction is exothermic
(b) The forward reaction is endothermic
(c) The reaction does not attain equilibrium
(d) The reverse reaction is exothermic
Answer:
(a) The forward reaction is exothermic
Solution:
T1 = 25 + 273 = 298 K, T2 = 700 K
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-115
In this case, T2 > T1 and K1 > K2
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-116
ΔH° is – ve i.e., forward reaction is exothermic.

Question 4.
The formation of ammonia from N2(g) and H2(g) is a reversible reaction
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) + Heat
What is the effect of increase of temperature on this equilibrium reaction
(a) equilibrium is unaltered
(b) formation of ammonia is favoured
(c) equilibrium is shifted to the left
(d) reaction rate does not change
Answer:
(c) equilibrium is shifted to the left
Solution:
Increase in temperature, favours the endothermic reaction, given that formation of NH3 is exothermic i.e., the reverse reaction is endothermic.
∴ Increase in temperature, shift the equilibrium to left

Question 5.
Solubility of carbon dioxide gas in cold water can be increased by ………….
(a) increase in pressure
(b) decrease in pressure
(c) increase in volume
(d) none of these
Answer:
(a) increase in pressure
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-117
increase in pressure, favours the forward reaction.

Question 6.
Which one of the following is incorrect statement?
(a) for a system at equilibrium, Q is always less than the equilibrium constant
(b) equilibrium can be attained from either side of the reaction
(c) presence of catalyst affects both the forward reaction and reverse reaction to the same extent
(d) Equilibrium constant varied with temperature
Answer:
(a) for a system at equilibrium, Q is always less than the equilibrium constant
Solution:
Correct statement is, for a system at equilibrium, Q = Keq

Question 7.
K1 and K2 are the equilibrium constants for the reactions respectively.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-118
What is the equilibrium constant for the reaction NO2(g) \(\rightleftharpoons\) \(\frac { 1 }{ 2 }\)N2(g) + O2(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-119
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-120
Solution:
Let equilibrium constant for the required reaction be K. Then,
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-121

Question 8.
In the equilibrium, 2A(g) \(\rightleftharpoons\) 2B(g) + C2(g) the equilibrium concentrations of A, B and C, at 400 K are 1 x 104 M, 2.0 x 10-3 M, 1.5 x 10-4 M respectively. The value of Kc. for the equilibrium at 400 K is
(a) 0.06
(b) 0.09
(c) 0.62
(d) 3 x 10-2
Answer:
(a) 0.06
Solution:
[A] = 1 x 10-4 M
[B] = 2 x 10-3 M
[C] = 1.5 x 10-4 M
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-122

Question 9.
An equilibrium constant of 3.2 x 10-6 for a reaction means, the equilibrium is ………
(a) largely towards forward direction
(b) largely towards reverse direction
(c) never established
(d) none of these
Answer:
(b) largely towards reverse direction
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-123

Question 10.
\(\frac { { K }_{ c } }{ { K }_{ p } }\) for the reaction, N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) is ………
(a) \(\frac { 1 }{ RT }\)
(b) \(g\sqrt { RT } \)
(c) RT
(d) (RT)2
Answer:
(d) (RT)2
Solution:
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
∆ng = 2 – 4 = – 2
Kp = Kc (RT)-2
\(\frac { { K }_{ c } }{ { K }_{ p } }\) = (RT)2

Question 11.
For the reaction, AB(g) \(\rightleftharpoons\) A(g) + B(g), at equilibrium, AB is 20% dissociated at a total pressure of R The equilibrium constant K is related to the total pressure by the expression ………..
(a) P = 24 Kp
(b) P = 8 Kp
(c) 24 p = Kp
(d) none of these
Answer:
(a) P =24 Kp
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-125

Question 12.
In which of the following equilibrium, K and K are not equal?
(a) 2NO(g) \(\rightleftharpoons\) N2(g) + O2(g)
(b) SO2(g) + NO2(g) \(\rightleftharpoons\) SO3(g) + NO(g)
(c) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
(d) PCI5(g) \(\rightleftharpoons\) PCl3(g) + Cl2(g)
Answer:
(d) PCI5(g) \(\rightleftharpoons\) PCl3(g) + Cl2(g)
Solution:
For reactions given in options (a), (b) and (c) ∆ng = 0
For option (d) ∆ng = 2 – 1 = 1
∴ Kp = Kc (RT)

Question 13.
If x is the fraction of PCI5 dissociated at equilibrium in the reaction,
PCl5 \(\rightleftharpoons\) PCl3 + Cl2
then starting with 0.5 moIe of PCI5 the total number of moles of reactants and products at equilibrium is
(a) 0.5 – x
(b) x + 0.5
(e) 2x + 0.5
(d) x ± 1
Answer:
(b) x + 0.5
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-126
Total no. of moles at equilibrium = 0.5 – x + x + x = 0.5 + x

Question 14.
The values of Kp1 and Kp2 for the reactions X \(\rightleftharpoons\) Y + Z and A \(\rightleftharpoons\) 2B are in the ratio 9 : 1 if degree of dissociation and initial concentration of X and A be equal then total pressure at equilibrium P1, and P2 are in the ratio
(a) 36 : 1
(b) 1: 1
(c) 3 : 1
(d) 1: 9
Answer:
(a) 36: 1
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-127
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-128

Question 15.
In the reaction, Fe(OH)3(s) \(\rightleftharpoons\) Fe3+(aq) + 3OH(aq), if the concentration of OH ions is decreased by ¼ times, then the equilibrium concentration of Fe3+ will …………
(a) not changed
(b) also decreased by ¼ times
(c) increase by 4 times
(d) increase by 64 times
Answer:
(d) increase by 64 times
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-129
[∵ Concentration of solids is constant)
When concentration of OH ions decreased by \(\frac { 1 }{ 4 }\) times, then
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-130
To maintain KC as constant, concentration of Fe3+ will increase by 64 times.

Question 16.
Consider the reaction where Kp = 0.5 at a particular temperature
PCI5(g) \(\rightleftharpoons\) PCI3(g) + Cl2(g) if the three gases are mixed in a container so that the partial pressure of each gas is initially 1 atm, then which one of the following is true?
(a) more PCI3 will be produced
(b) more Cl2 will be produced
(c) more PCI5 will be produced
(d) none of these
Answer:
(c) more PCI5 will be produced .
Solution:
Kp = 0.5
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-131
Q = \(\frac { 1 x 1 }{ 1 }\)
Q > Kp
∴Reverse reaction is favoured; i.e., more PCI5 will be produced.

Question 17.
Equimolar concentrations of H2 and I2 are heated to equilibrium in a 1 litre flask. What percentage of initial concentration of H2 has reacted at equilibrium if rate constant for both forward and reverse reactions are equal
(a) 33%
(b) 66%
(c)(33)2 0/0
(d) 16.5%
Answer:
(a) 33%
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-132
As degree of dissociation cannot be negative. Therefore, degree of dissociation
= \(\frac { a }{ 3 }\) x 100 = 33.33%

Question 18.
In a chemical equilibrium, the rate constant for the forward reaction is 2.5 x 1 and the equilibrium constant is 50. The rate constant for the reverse reaction is ……….
(a) 11.5
(b) 5
(c) 2 x 102
(d) 2 x 103
Answer:
(b) 5
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-133

Question 19.
Which of the following is not a general characteristic of equilibrium involving physical process
(a) Equilibrium is possible only in a closed system at a given temperature
(b) The opposing processes occur at the same rate and there is a dynamic but stable condition
(c) All the physical processes stop at equilibrium
(d) All measurable properties of the system remains constant
Answer:
(c) All the physical processes stop at equilibrium
Solution:
Correct statement – Physical processes occurs at the same rate at equilibrium.

Question 20.
For the fórmation of two moles of SO3(g) from SO2 and O2, the equilibrium constant is K1. The equilibrium constant for the dissociation of one mole of SO3 into SO2 and O2 is …………
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-134
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-135
Solution:
2SO2 +O2 \(\rightleftharpoons\) 2SO3
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-136
Dissociation of 1 mole of 2SO3
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-137

Question 21.
Match the equilibria with the corresponding conditions …………
(i) Liquid \(\rightleftharpoons\) Vapour – 1. Melting point
(ii) Solid \(\rightleftharpoons\) Liquid – 2. Saturated solution
(iii) Solid \(\rightleftharpoons\) Vapour – 3. Boiling point
(iv) Solute (s) \(\rightleftharpoons\) Solute (Solution) – 4. Sublimation point
5. Unsaturated solution
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-138
Answer:
(b) 3, 1, 4, 2

Question 22.
Consider the following reversible reaction at equilibrium, A + B \(\rightleftharpoons\) C, if the concentration of the reactants A and B are doubled, then the equilibrium constant will ………
(a) be doubled
(b) become one fourth
(c) be halved
(d) remain the same
Answer:
(d) remain the same
Solution:
A + B \(\rightleftharpoons\) C
KC = \(\frac { [C] }{ [A] [B] }\)
If [A] and [B] are doubled, [C] increases 4 times to maintain KC as constant.
∴ Equilibrium constant will remain the same.

Question 23.
[CO(H2O)6]2+ (aq) (pink) + 4C (aq) \(\rightleftharpoons\) [COCI4]2- (aq) (blue) + 6H2O (1) In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in colour. On the basis of this information, which one
of the following is true?
(a) ∆H > 0 for the forward reaction
(b) ∆H = 0 for the reverse reaction
(c) ∆H < 0 for the forward reaction
(d) Sign of the ∆H cannot be predicted based on this information.
Answer: (a) ∆H > 0 for the forward reaction
Solution:
On cooling, reverse reaction predominates and the solution is pink in colour. Decrease in temperature, favours the reverse reaction i.e. reverse reaction is exothermic (∆H < 0)and for the forward reaction is endothermic (∆H > 0).

Question 24.
The equilibrium constants of the following reactions are:
N2 + 3H2 \(\rightleftharpoons\) 2NH3 ; K1
N2 + O2 \(\rightleftharpoons\) 2NO ; K2
H2+ \(\frac { 1 }{ 2 }\)O2 \(\rightleftharpoons\) H2O ; K3
The equilibrium constant (K) for the reaction; Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-140 will be
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-141
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-167
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-143

Question 25.
A 20 litre container at 400 K contains CO2(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO2 attains its maximum value will be.
Given that: SrCO3(s) \(\rightleftharpoons\) SrO(s) + CO2(g)
(a) 2 litre
(b) 5 litre
(c) 10 Litre
(d) 4 litre
Answer:
(b) 5 litre
Solution:
Given that Kp = 1.6 atm
V1 = 20 L
V2 = ?
T1 = 400 K
T2 = 400 K
Kp = Pco2
Pco2 = 1.6 atm
P1 = 0.4 atm
P2 = 1.6 atm
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-144

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Short Answer Questions

Question 26.
If there is no change in concentration, why is the equilibrium state considered dynamic?
Answer:
At chemical equilibrium the rate of two opposing reactions are equal and the concentration of. reactants and products do not change with time. This condition is not static and is dynamic, because both the forward and reverse reactions are still occurring with the same rate and no macroscopic change is observed. So chemical equilibrium is in a state of dynamic equilibrium.

Question 27.
For a given reaction at a particular temperature, the equilibrium constant has constant value. Is the value of Q also constant? Explain.
Answer:
In the chemical reaction, as the reaction proceeds, there is a continuous change in the concentration of reactants and products and also the Q value until the reaction reaches the equilibrium. So even at particular temperature, Q is not constant. Even once the equilibrium is achieved then change in concentration of reactants or products, pressure, volume will change the value of Q.

Question 28.
What is the relation between Kp and KC. Give one example for which Kp is equal to KC.
Answer:
The relation between Kp and Kc is Kp = KC (RT)∆ng
Kp = equilibrium constant is terms of partial pressure.
Kc = equilibrium constant is terms of concentration.
R = gas constant
T = Temperature.
∆ng = Difference between the sum of the number of moles of products and the sum of number of moles of reactants in the gas phase. When ∆ng = 0
Kp = KC(RT)0 = KC i.e., Kp = KC
Example: H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
∆ng = 2 – 2 = 0
∴ Kp = KC for the synthesis of HI.

Question 29.
For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is KC is larger or smaller than Kp ?
Answer:
For a homogeneous reaction at equilibrium, number of moles of products (np) are greater than the number of moles of reactants (nR) then ∆ng = + ve
np > nR
∆ng = + ve
If ∆ng is ve, Kp value is greater than KC
Kp = KC. (RT)+ve
Kp > KC
Example: PCl5 \(\rightleftharpoons\) PCl3(g) + Cl2(g)
2 – 1 = 1
Kp = KC (RT)1
Kp > KC

Question 30.
When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant (K), in which direction does the reaction proceed to reach equilibrium?
Answer:
When Q > KC the reaction will proceed in the reverse direction, i.e, formation of reactants.

Question 31.
For the reaction: A2(g) + B2(g) \(\rightleftharpoons\) 2AB(g); ∆H is – ve.
Answer:
The following molecular scenes represent different reaction mixture
(A – dark grey, B – light grey)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-145

  1. Calculate the equilibrium constant Kp and KC
  2. For the reaction mixture represented by scene (x), (y) the reaction proceed in which directions?
  3. What is the effect of increase in pressure for the mixture at equilibrium?

Answer:
1. Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-146
At equilibrium
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-147

2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-148

3.  Since ∆ng = 2 – 2 = 0, thus, pressure has no effect. So by increasing the pressure, equilibrium will not be affected.

Question 32.
State Le – Chatelier principle.
Answer:
It states that If a system at equilibrium is disturbed, then the system shifts itself in a direction that nullifies the effect of that disturbance.

Question 33.
Consider the following reactions

  1. H2(g) + l2(g) \(\rightleftharpoons\) 2HI(g)
  2. CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g)
  3. S(s) + 3F2(g) \(\rightleftharpoons\) SF6(g)

In each of the above reaction find out whether you have to increase (or) decrease the volume to increase the yield of the product.
Answer:
1.  H2(g) + l2(g) \(\rightleftharpoons\) 2HI(g)
In the above equilibrium reaction, volume of gaseous molecules is equal on both sides. So increase or decrease the volume will not affect the equilibrium and there will be no change in the yield of product.

2. Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-149
Volume is greater in product side. By decreasing the pressure, volume will increase thus, to get more of product CO2, the pressure should be decreased or volume should be increased.

3. Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-150
Volume is lesser in product side. So by increasing the pressure, equilibrium shifts to the product side.

Question 34.
State law of mass action.
Answer:
The law states that “At any instant, the rate of chemical reaction at a given temperature is directly proportional to the product of the active masses of the reactants at that instant.”

Question 35.
Explain how will you predict the direction of an equilibrium reaction.
Answer:

  1. A large value of KC indicates that the reaction reaches equilibrium with high product yield.
  2. A low value of KC indicate that the reaction reaches equilibrium with low product formed.
  3. In general, if the K is greater than I the reaction proceeds nearly to completion. If is less than 10-3, the reaction rarely proceeds.
  4. If K < 10-3, reverse reaction is favoured. If Kc > 103, forward reaction is favoured.

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Long Answer Questions

Question 36.
Derive a general expression for the equilibrium constant Kp and KC for the reaction.
3H2(g) + N2(g) \(\rightleftharpoons\) 2NH3(g)
Answer:
In the formation of ammonia, ‘a’ moles of Nitrogen and ‘b’ moles of hydrogen gas are allowed to react in a container of volume of ‘V’. Let ‘x’ moles of nitrogen react with 3x moles of hydrogen to give 2x moles of ammonia.
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-151
Applying law of mass action
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-152
Total number of moles at equilibrium n = a – x + b -3x + 2x = a + b – 2x
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-153
Total number of moles at equilibrium
n = a – x + b – 3x + 2x = a + b – 2x
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-154

Question 37.
Write a balanced chemical equation for an equilibrium reaction for which the equilibrium constant is given by expression?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-155

Question 38.
What is the effect of added inert gas on the reaction at equilibrium at constant volume?
Answer:
When an inert gas (i.e., a gas which does not react with any other species involved in equilibrium) is added to an equilibrium system at constant volume, the total number of moles. of gases present in the container increases, that is, the total pressure of gases increases, the partial pressure of the reactants and the products are unchanged. Hence at constant volume, addition of inert gas has no effect on equilibrium.

Question 39.
Derive the relation between Kp and KC.
Answer:
Consider a general reaction in which all reactants and products are ideal gases.
x A + y B \(\rightleftharpoons\)  lC+mD
The equilibrium constant KC is
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-156
The ideal gas equation is
PV = nRT or P = \(\frac { n }{ V }\) RT
Since,
Active mass = molar concentration = \(\frac { n }{ V }\)
P = Active mass x RT
Based on the above expression, the partial pressure of the reacants and products can be expressed as
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-157
On substitution in equation (2),
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-158
By comparing equation (1) and (4), we get
Kp = KC (RT) ∆ng
where ∆ng is the difference between the sum of number of moles of products and the sum of number of moles of reactants in the gas phase.
(i) If ∆ng = 0, Kb = KC(RT)0
Kp = KC
Example: H2(g) + I2 \(\rightleftharpoons\) 2HI(g)

(ii) where
∆ng = +Ve
Kp = KC (RT)+ve
Kp = KC
Example: 2NH3(g) N2(g) + 3H2(g)

(iii) When
∆ng = -Ve
Kp = KC (RT)-ve
Kp < KC
Example: 2SO2(g) +O2(g) \(\rightleftharpoons\) 2SO3(g)

Question 40.
One mole of PCl5 is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant.
Answer:
Given that
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-159

Question 41.
For the reaction: SrCO2(s) \(\rightleftharpoons\) SrO (s) + CO2(g), the value of equilibrium constant Kp = 2.2 x 10-4 at 1002K. Calculate KC for the reaction.
Answer:
For the reaction, SrCO2(s) \(\rightleftharpoons\) SrO (s) + CO2(g)
∆ng = 1 – 0 = 1
Kp = Kp (RT)
2.2 x 10-4 = KC (0.0821)(1002)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-160

Question 42.
To study the decomposition of hydrogen iodide, a student fills an evacuated 3 litre flask with 0.3 mol of HI gas and allows the reaction to proceed at 500°C. At equilibrium he found the concentration of HI which is equal to 0.05 M. Calculate Kp and KC for this reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-161
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-162
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-168

Question 43.
Oxidation of nitrogen monoxide was studied at 200°C with initial pressures of 1 atm NO and 1 atm of O2. At equilibrium partial pressure of oxygen is found to be 0.52 atm. Calculate Kp value.
Answer:
2NO(g) + O2(g) \(\rightleftharpoons\) 2NO2(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-164
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-165

Question 44.
1 mol of CH4, 1 mole of CS2 and 2 mol of H2S are 2 moI of H2 are mixed in a 500 mL flask. The equilibrium constant for the reaction KC = 4 x 10-2 mol2 lit-2. In which direction will the reaction proceed to reach equilibrium?
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-166
∴ The reaction will proceed in the reverse direction to reach the equilibrium.

Question 45.
At particular temperature KC = 4 x 10-2 for the reaction
H2S(g) \(\rightleftharpoons\) 2H2(g) + \(\frac { 1 }{ 2 }\) S2(g). Calculate KC for each of the following reaction
(i). H2S(g) \(\rightleftharpoons\) 2H2(g) + S2(g)
(ii). 3H2S(g) \(\rightleftharpoons\) 3H2(g) + \(\frac { 3 }{ 2 }\) S2(g)
Answer:
KC = 4 x 10-2 for the reaction
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-167
For the reaction 3H2S(g) \(\rightleftharpoons\) 3H2(g) + \(\frac { 3 }{ 2 }\) S2(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-168

Question 46.
28 g of nitrogen and 6 g of hydrogen were mixed in a 1 litre closed container. At equilibrium 17g NH3 was produced. Calculate the weight of nitrogen, hydrogen at equilibrium.
Answer:
Given
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-169
[NH3] = \((\frac { 17 }{ 17 })\) = 1 mol = 2x
x = 0.5 mol
At equilibrium, [N2] = 1 – x = 0.5 mol
[H2] = 3 – 3x = 3 – 3 (0.5) = 1.5 mol
Weight of N2 (no. of moles of N2) x molar mass of N2 = 0.5 x 28 = 14g
Weight of H2 = (no. of moles of H2) x molar mass of H2= 1.5 x 2 = 3g

Question 47.
The equilibrium for the dissociation of XY2 is given as,
2XY2 (g) \(\rightleftharpoons\) 2XY (g) + Y2(g).
If the degree of dissociation x is so small compared to one. Show that 2 Kp = Px3 where P is the total pressure and K is the dissociation equilibrium constant of XY2.
Answer:
2XY2 (g) \(\rightleftharpoons\) 2XY (g) + Y2(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-170
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-171
Question 48.
A sealed container was filled with I mol of A, (g), I mol B, (g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 1 for the reaction:
A2(g) + B2(g) \(\rightleftharpoons\) 2AB(g)
Answer:
A2(g) + B2(g) \(\rightleftharpoons\) 2AB(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-172
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-173
[B2]eq = 1 – x = 1 – 0.33 = 0.667
[AB]2 = 2x = 2 x 0.33 = 0.66
Question 49.
Deduce the Vant Hoff equation.
Answer:
This equation gives the quantitative temperature dependance of equilibrium constant K. The relation between standard free energy change ∆G° and equilibrium constant is
∆G° = – RT ln K  …………(1)
We know that,
∆G° = ∆H° – T∆S°        ………….(2)
Substituting (1) in equation (2)
– RT lnK = ∆H° – T∆S°

Rearranging, ln K = \(\frac { – ∆H° }{ RT }\) + \(\frac { ∆S° }{ R }\)   ………….(4)
Differentiating equation (3) with respect to temperature
d(lnK) ∆H° \(\frac { d(lnK) }{ dT }\) = \(\frac { { \triangle H }^{ 0 } }{ { RT }^{ 2 } }\)
Equation (4) is known as differential form of van’t Hoff equation. On integrating the equation 4, between T1 and T2 with their respective equilibrium constants and K2
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-174
Equation is known as integrated form of van’t Hoff equation.

Question 50.
The equilibrium constant K for the reaction
N2(g) + 3H2(g) = 2NH3(g) is 8.19 x 102
at 298 K and 4.6 x 10-1 at 498 K. Calculate ∆H° for the reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-175
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-176

Question 51.
The partial pressure of carbon dioxide in the reaction CaCO3(s) \(\rightleftharpoons\) CaO (s) + CO2(g) is 1.017 x 10-3 atm at 500° C. Calculate Kp, at 600° C for the reaction. ∆H for the reaction is 181 kJ mol-1 and does not change in the given range of temperature.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-177

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium In – Text problems Solved

Question 1.
Consider the following reaction
Fe3+(aq) + SCN(aq) [Fe(SCN)]2+(aq)
A solution is made with initial Fe3+, SCN concentration of 1 x 10-3 M and 8 x 10-4M respectively. At equilibrium [Fe(SCN)]2+ concentration is 2 x 10-4M. Calculate the value of equilibrium constant.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-1

Question 2.
The atmospheric oxidation of NO
2NO(g) + O2(g) \(\rightleftharpoons\) 2NO2(g)
was studied with initial pressure of 1 atm of NO and I atm of O2. At equilibrium, partial pressure of oxygen is 0.52 atm. Calculate Kp of the reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-2
As,
1 – x = 0.52
x = 0. 48
= At equilibrium,
PNO = 1 – 2x = 1 – 2(0.48) = 0.04
PNO2 = 2x = 2(0.48) = 0.96
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-3

Question 3.
The following water gas shift reaction is an important industrial process for the production of hydrogen gas.
CO(g) + H2O(g) \(\rightleftharpoons\) CO2(g) + H2(g) At a given temperature Kp = 2.7.  If 0.13 moI of CO, 0.56 moI of water, 0.78 mol of CO2 and 0.28 mol of H2 are introduced into a 2L flask, find out in which direction must the reaction proceed to reach equilibrium.
Answer:
CO(g) + H2O(g) \(\rightleftharpoons\) CO2(g) + H2(g)
Given
Kp = 2.7
[CO] = 0.13 mol, [H2O] = 0.56 mol
[CO2] = 0.78 mol; [H2] = 0.28 mol
V = 2L
Kp = KC (RT)Δng
2.7 = KC(RT)°
KC = 2.7
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-1.2
Q = 3
Q > KC Hence, the reaction proceed in the reverse direction.

Question 4.
1 moI of PCl5, kept in a closed container of volume 1 dm3 and was allowed to attain equilibrium at 423 K. Calculate the equilibrium composition of reaction mixture. (The KC value for PCl5 dissociation at 423 K is 2)
Answer:
PCl5 \(\rightleftharpoons\) PCL3 + Cl2
Given that [PCl5]initial = 1 mol
V = 1 dm3
KC = 2
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-4
∴ Equilibrium concentration of
[PCl5]eq = \(\frac { 1-x }{ 1 }\) = 1 – 0.732 = 0.268 M
[PCl3]eq = \(\frac { x }{ 1 }\) = \(\frac { 0.732 }{ 1 }\) = 0.732
[Cl2]eq = \(\frac { x }{ 1 }\) = \(\frac { 0.732 }{ 1 }\) = 0.732

Question 5.
The equilibrium constant for the following reaction is 0.15 at 298 K and 1 atm pressure.
Answer:
N2O4(g) \(\rightleftharpoons\) 2NO2(g)
T1 = 298 K
Kp1 = 0.15
T2 = 1000C = 100 + 273 = 373 K
Kp2 = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-5

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium In – Text problems Solved

Question 1.
One mole of H2 and one mole of I2 are allowed to attain equilibrium in 1 lit container. If the equilibrium mixture contains 0.4 mole of HI. Calculate the equilibrium constant.
Answer:
Given data: [H2] = I mole, [I2] = 1 mole
At equilibrium, [HI] = 0.4 mole, KC = ?
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-6

Question 2.
The equilibrium concentrations of NH3, N2 and H2 are 1.8 x 102 M, 1.2 x 10-2 M and 3 x 10-2 M respectively. Calculate the equilibrium constant for the formation of NH3 from N2 and H2. [Hint: M = mol lit-1]
Answer:
Given data:
[NH3] 1.8 x 10-2 M, [N2] = 1.2 x 10-2M, [H2] 3 x 10-2 M, KC = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-7

Question 3.
The equilibrium constant at 298 K for a reaction is 100.
A+ B \(\rightleftharpoons\) C + D
If the initial concentration of all the four species is 1 M, the equilibrium concentration of D (in mol lit-1) will be
Answer:
Given data: [A] = [B] = [C] = [D] = 1 M, KC = 100, [D]eq = ?
Let x be the number of moles reactants reacted
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-8

Question 4.
For an equilibrium reaction Kp = 0.0260 at 25° C and ?H = 32.4 kJ mor-1. Calculate Kp at 37° C.
Answer:
T1 = 25 + 273 = 298 K
T2 = 37 + 273 = 310 K
ΔH = 32.4 kJ mor-1 = 32400 J mol-1
R = 8.314 JK-1 mol-1
Kp1 = 0.0260
Kp2 = ?
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-9
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-10

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium Additional Questions 

I. Choose the correct answer.

Question 1.
Which of the following represents physical equilibrium?
(a) PCl5(g) \(\rightleftharpoons\) PCl3(g) + Cl2(g)
(b) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
(c) H2O(l) \(\rightleftharpoons\) H2O(g)
(d) N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
Answer:
(c) H2O(l) \(\rightleftharpoons\) H2O(g)
Solution:
Physical states arc in equilibrium i.e., liquid – vapour equilibrium.

Question 2.
Which one of the following is an example of chemical equilibrium?
(a) 2NO(g) + O2(g) \(\rightleftharpoons\) 2NO2(g)
(b) I2(s) \(\rightleftharpoons\) I2(g)
(c) H2O(s) \(\rightleftharpoons\) H2O(1)
(d) NH2CI(s) \(\rightleftharpoons\) NH4CI(g)
Answer:
(a) 2NO(g) + O2(g) \(\rightleftharpoons\) 2NO2(g)
Solution:
All the other three are physical equilibrium. Only (a) is chemical equilibrium.

Question 3.
Which one of the following does not undergo sublimation?
(a) Iodine
(b) water
(c) Camphor
(d) Ammonium chloride
Answer:
(b) Water

Question 4.
At chemical equilibrium,
(a) rate of forward reaction = rate of backward reaction
(b) rate of forward reaction > rate of backward reaction
(c) rate of forward reaction < rate of backward reaction
(d) rate of forward reaction = rate of backward reaction
Answer:
(a) rate of forward reaction = rate of backward reaction

Question 5.
Which of the following is an example of homogeneous equilibrium?
(a) H2O(1) \(\rightleftharpoons\) H2O(g)
(b) CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO(g)
(c) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
(d) 2CO(g) \(\rightleftharpoons\) CO2(g) + C(s)
Answer:
(c) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Solution:
Here all reactants and products are in same phase i.e., gaseous phase.

Question 6.
Which of the following is an example of heterogeneous equilibrium?
(a) Synthesis of HI
(b) Dissociation of PCI5
(c) Acid hydrolysis of ester
(d) Decomposition of limestone
Answer:
(d) Decomposition of limestone
Solution:
CaCO3(s) → CaO(s) + CO2(g)
Here CO2 is in gaseous state while CaCO3 and CaO are in solid state.

Question 7.
Statement I: In dissociation of PCI5 to PCI3 and CI2, Kp > KC
Statement II: In dissociation of PCI5, Δng = – ve and so Kp > KC.
(a) Statement I & II are correct and statement II is the correct explanation of statement I.
(b) Statement I & II are correct but statement II is not the correct explanation of statement I.
(c) Statement I is correct but statement II is wrong.
(d) Statement I is wrong but statement II is correct.
Answer:
(c) Statement I is correct but statement II is wrong.
Solution:
Δng = 2 – 1 = 1 = +ve

Question 8.
In the reaction, 2NH3(g) \(\rightleftharpoons\) N2(g) + 3H2(g)
(a) Kp = KC
(b) Kp < KC
(c) Kp > KC
(d) Kp = \(\frac { 1 }{ { K }_{ C } }\)
Answer:
(c) Kp > KC
Solution:
Kp = KC (RT)Δng and Δng = 4 – 2 = 2
∴ Kp = KC (RT)2 = Kp > KC

Question 9.
In which of the following reaction, Kp is equal to KC ?
(a) N2(g) ± O2(g) \(\rightleftharpoons\) 2NO(g)
(b) 2NH3(g) \(\rightleftharpoons\) N2(g) + 3H2(g)
(c) 2H2(g) + O2(g) \(\rightleftharpoons\) 2H2O(g)
(d) PCI5(g) \(\rightleftharpoons\) PCI3(g) + CI2
Answer:
(a) N2(g) ± O2(g) \(\rightleftharpoons\) 2NO(g)
Solution:
KC (RT)Δng when Δng = 0 then K = K for option (a), Δng = 2 – 2 = 0

Question 10.
In the equilibrium reaction CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g) whose concentration remains constant at a given temperature?
(a) CaO
(b) CO2
(e) CaCO3
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)
Solution:
Concentration of solids remains constant at a particular temperature.

Question 11.
Consider the following equilibrium reaction and relate their equilibrium constants
(i) N2 + O2 \(\rightleftharpoons\) 2NO; K1
(ii) 2NO + O2 \(\rightleftharpoons\) 2NO2; K2
(iii) N2 + 2O2 \(\rightleftharpoons\) 2NO2; K3
(a) K3 = K2 = K1
(b) K1 x K3 = K2
(c) K1 x K2 = K3
(d) \(\frac { { K }_{ 1 } }{ { K }_{ 2 } }\) = K3
Answer:
(c) K1 x K2 = K3
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-11

Question 12.
Statement I: A pure solid in an equilibrium reaction has the same concentration at a given temperature.
Statement II: The solid does not expand to fill its container and it has same number of moles of its volume.
(a) Statement I and II are correct and statement II is the correct explanation of statement of I.
(b) Statement I and II are correct but II is not the correct explanation of!.
(c) Statement I and II are not correct.
(d) Statement I is wrong but 11 is correct.
Answer:
(a) Statement I and II are correct and statement II is the correct explanation of statement of I.

Question 13.
Find the Q value of the reaction H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g) at an instant where concentration of H2, I2 and HI are found to be 0.2 moI L-1, 0.2 mol L-1, and 0.6 moI L-1 respectively.
(a) 48
(b) 9
(c) 0.9
(d) 90
Answer:
(b) 9
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-12

Question 14.
For the reaction N2O4(g) \(\rightleftharpoons\) 2NO2(g) KC = 0.21 at 373 K. The concentrations of N2O4 and NO2 are found to be 0.125 mol dm-3 and 0.5 mol dm-3 respectively at a given temperature. Predict the direction of the reaction.
(a) At equilibrium
(b) reverse direction
(c) forward direction
(d) Both reverse and forward direction
Answer:
(b) reverse direction
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-13
KC = 0.21
Q = 2
Q > KC
Hence the reaction will proced in the reverse direction

Question 15.
Which of the following does not alter the equilibrium?
(a) catalyst
(b) concentration
(c) temperature
(d) pressure
Answer:
(a) catalyst

Question 16.
Statement I. In Haber’s process, NH3 is liquefied and removed.
Statement II. In manufacture of NH3, liquefied and removal of NH3, keeps the reaction moving in forward direction.
(a) Statement I and II are correct and II is the correct explanation of I.
(b) Statement I and II are correct but II is not the correct explanation of I.
(c) Statement I is wrong but statement II is correct.
(d) Statement I is correct but statement II is wrong.
Answer:
(a) Statement I and II are correct and II is the correct explanation of I.
Solution:
Removal of NH3 will decrease its concentration which favours the production of NH3 according to the Le – Chatelier’s principle.

Question 17.
In which of the following reaction, pressure has no effect?
(a) N2 + 3N2 \(\rightleftharpoons\) 2NH3(g)
(b) 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g)
(c) N2O4(g) \(\rightleftharpoons\) 2NO2(g)
(d) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Answer:
(d) H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Solution:
In the reaction H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g) the volumes are equal on both sides and so pressure has no effect.

Question 18.
Among the following reactions which one has Kp = KC
(a) N2O4 \(\rightleftharpoons\) 2NO(g)
(b) 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g)
(c) N2(g) + O2(g) \(\rightleftharpoons\) 2NO(g)
(d) N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
Answer:
(c) N2(g) + O2(g) \(\rightleftharpoons\) 2NO(g)
Solution:
Kp = KC. RTΔng
In equation N2(g) + O2(g) \(\rightleftharpoons\) 2NO(g)
Δng = 0
Kp = KC. RT°
Kp = KC

Question 19.
Statement I. Addition of an inert gas at constant volume has no effect on equilibrium.
Statement II. When an inert gas is added, the total number of moles of gases present in the container increases and total pressure also increases, the partial pressure of the products and reactants are unchanged.
(a) Statement I and II are correct but statement II is not the correct explanation of I.
(b) Statement I and II are correct and statement II is the correct explanation of 1.
(c) Statement I is correct but statement II is not correct.
(d) Statement I is wrong but statement II is correct.
Answer:
(b) Statement I and II are correct and statement II is the correct explanation of I.

Question 20.
Which one of the following equation is not correct?
(a) ΔG° = – RTInK
(b) ΔG° = ΔH° – TΔS°
(c) – RTInK = ΔH° – TΔS°
(d) In k = \(\frac { ΔH° }{ T }\) – \(\frac { ΔS° }{ R }\)
Answer:
(d) In k = \(\frac { ΔH° }{ T }\) – \(\frac { ΔS° }{ R }\)

Question 21.
The equilibrium expression, KC = [CO2] represents the reaction.
(a) C(s) + O2(g) \(\rightleftharpoons\) CO2(g)
(b) CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g)
(c) 2CO(g) + O2(g) \(\rightleftharpoons\) 2CO2(g)
(d) CaO(s) + CO2(g) \(\rightleftharpoons\) CaCO3(s)
Answer:
(b) CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g)

Question 22.
Hydrogen molecule (H2) can be dissociated into hydrogen atoms (H). Which one of the following changes will not increase the number of atoms present at equilibrium?
(a) adding H atoms
(b) increasing the temperature
(c) increasing the total pressure
(d) increasing the volume of the container
Answer:
(d) increasing the total pressure container
Solution:
It favours backward reaction i.e., formation of H2 molecule.

Question 23.
What is the expression for Keq for the reaction, 2N2O(g) + O2(g) \(\rightleftharpoons\) 4NO(g)?
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-14
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-15

Question 24.
What is the correct expression for the representation of the solubility product constant of Ag2CrO4 ?
(a) [Ag+]2 [CrO42-]
(b) [2Ag+] [CrO42-]
(c) [Ag+] [CrO42-]
(d) [2Ag+]2 [CrO42-]
Answer:
(a) [Ag+]2 [CrO42-]

Question 25.
H2 + S \(\rightleftharpoons\) H2S + energy. In this reversible reaction, select the factor which will shift the equilibrium to the right.
(a) adding heat
(b) adding H2S
(c) blocking hydrogen gas reaction
(d) removing hydrogen suiphide gas
Answer:
(a) removing hydrogen suiphide gas

Question 26.
What effect does a catalyst have on the equilibrium position of a reaction?
(a) a catalyst favours the formation of products
(b) a catalyst favours the formation of reactants
(c) a catalyst does not change the equilibrium position of a reaction
(d) a catalyst may favour reactants or product formation, depending upon the direction in which the reaction is written.
Answer:
(c) a catalyst does not change the equilibrium position of a reaction

Question 27.
A chemist dissolves an excess of BaSO4 in pure water at 25°C if its Ksp = 1 x 10-10. What is the concentration of barium in the water?
(a) 10-4 M
(b) 10-5 M
(c) 10-15 M
(a) 10-6 M
Answer:
(c) 10-15 M
Solution:
Ksp = [Ba2+] [SO42-]
1 x 10-10 = (x) (x)
10-5 = x

Question 28.
If in a mixture where Q = K, then what happens?
(a) the reaction shift towards products
(b) the reaction shift towards reactants
(c) nothing appears to happen, but forward and reverse reactions are continuing at the same rate
(d) nothing happens
Answer:
(c) nothing appears to happen, but forward and reverse reactions are continuing at the same rate

Question 29.
If dissociation for reaction PCI5 \(\rightleftharpoons\) PCI3 + Cl2 is 20% at 1 atm pressure. Calculate the value of KC.
(a) 0.04
(b) 0.05
(c) 0.07
(d) 0.06
Answer:
(d) 0.05
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-18

Question 30.
What would be the value of Δng for the reaction NH4CI(s) \(\rightleftharpoons\) NH2(g) + HCI(g)?
(a) 1
(b) 0.5
(c) 2
(d) 1.5
Answer:
(c) 2
Solution:
Δng = np – nr = 2 – 0 = 2

Question 31.
Which of the following is not a general characteristic of equilibria involving physical processes?
(a) Equilibrium is possible only in a close system at a given temperature
(b) All measurable properties of the system remain constant
(c) All the physical processes stop at equilibrium
(d) The opposing processes occur at the same rate and there is dynamic but stable condition
Answer:
(c) All the physical processes stop at equilibrium

Question 32.
At 500K, equilibrium constant KC for the following reaction is 5, \(\frac { 1 }{ 2 }\)H2(g) + \(\frac { 1 }{ 2 }\)I2(g) \(\rightleftharpoons\) HI(g) what would be the equilibrium constant KC. for the reaction 2HI(g) \(\rightleftharpoons\) H2(g) + I2(g)
(a) 0.44
(b) 0.04
(c) 25
(d) 2.5
Answer:
(b) 0.04
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-19

Question 33.
For the reaction 2NO2(g) \(\rightleftharpoons\) 2NO(g) + O2 (g), KC = 1.8 x 10-6 at 1850C. At the same temperature the value of KC for the reaction. NO(g) + \(\frac { 1 }{ 2 }\)O2 \(\rightleftharpoons\) NO2(g) is ……….
(a) 0.9 x 106
(b) 7.5 x 102
(c) 1.95 x 10-3
(d) 1.95 x 103
Answer:
(b) 7.5 x 102
Solution:
The reaction is reversed and halved.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-20

Question 34.
Which of the following reaction will be favoured at low pressure?
(a) N2 + O2 \(\rightleftharpoons\) 2NO
(b) H2 ± I2 \(\rightleftharpoons\) 2HI
(c) PCI5 \(\rightleftharpoons\) PCI3 + Cl2
(d) N2 + 3H2 \(\rightleftharpoons\) 2NH3
Answer:
(c) PCI5 \(\rightleftharpoons\) PCI3 + Cl2

Question 35.
Consider the reaction CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g) is a closed container at equilibrium. What would be the effect of addition of CaCO3 on the equilibrium?
(a) increases
(b) remains unaffected
(c) decreases
(d) unpredictable
Answer:
(b) remains unaffected

Question 36.
For the reaction PCI5(g) \(\rightleftharpoons\) PCI3(g) + Cl2(g) the forward reaction at constant temperature is favoured by …………………
(a) introducing an inert gas at constant volume
(b) introducing PCl3(g) at constant volume
(c) introducing PCl5(g) at constant volume
(d) introducing CI2(g) at constant volume
Answer:
(c) introducing PCI5(g) at constant volume

Question 37.
The equilibrium of the reaction N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) will shift to product side when….
(a) Kp > 1
(b) Q < Kp
(c) Q = Kp
(d) Q = 2Kp
Answer:
(b) Q < Kp

Question 38.
NO2 is involved in the formation of smog and acid rain. A reaction that is important in the formation of NO2 is O3(g) + NO(g) \(\rightleftharpoons\) O2(g) + NO2(g) KC = 6.0 x 1034. If the air over a section of New Delhi contained 1.0 x 10-6 M of O3, 1.0 x 10-5 M of NO, 2.5 x 10-4 M of NO2 and 8.2 x 10-3 of O2, what can we conclude?
(a) there will be a tendency to form more NO and O2
(b) there will be a tendency to form more NO2 and O2
(c) there will be a tendency to form more NO2 and O3
(d) there will no tendency for change because the reaction is at equilibrium
Answer:
(b) there will be a tendency to form more NO2 and O2
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-21
As Q < KC the reaction will hive a tendency to move forward.

Question 39.
Haemoglobin (Hb) forms bond with oxygen and given oxyhaemoglobin (HbO2). This process is partially regulated by the concentration of H3O+ and dissolved CO2 in blood as HbO2 +H3O+ + CO2 \(\rightleftharpoons\) H+ – Hb – CO2 + O2 + H2O. If there is production of lactic acid and CO2 during a muscular exercise, then
(a) more HbO2 is formed
(b) more O2 is released
(c) CO2 is released
(d) both (b) and (c)
Answer:
(b) more O2 is released

Question 40.
In the reaction N2 + 3H2 \(\rightleftharpoons\) 2NH3 + x k Cal, one mole of N2 reacts with. 3 moles of H2 at equilibrium. Then the value of a (degree of dissociation) is approximately ……………… P is the pressure at equilibrium
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-22
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-23
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-24

II. Match the following.

Question 1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-25
Answer:
(a) 2, 1, 4, 3

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-26
Answer:
(b) 2, 3, 4, 1

Question 3.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-27
Answer:
(a) 4, 1, 2, 3

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-28
Answer:
(a) 3, 4, 1, 2

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-29
Answer:
(b) 2, 3, 4, 1

III. Fill in the blanks.

Question 1.
Transport of oxygen by Hemoglobin in our body is ………… a reaction.
Answer:
reversible

Question 2.
The temperature at which the solid and liquid phases of a substance are at equilibrium is called …………
Answer:
Freezing point

Question 3.
Thejemperature at which the liquid and vapour phases are at equilibrium is called …………
Answer:
Condensation point

Question 4.
………… law is used to explain gas-solution equilibrium processes.
Answer:
Henry’ law

Question 5.
In the reaction 2H2(g) + O2(g) \(\rightleftharpoons\) 2H2O(g), the Kp value is equal to ………
Answer:
< KC
Solution:
Δng = 2 – 3 = -1
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-30

Qustion 6.
The expression of K for the reaction CO2(g) + H2O(l) \(\rightleftharpoons\) H+(aq) + HCO3(aq) is equal to …………..
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-31

Question 7.
The expression of K for the reversible reaction 2CO(g) \(\rightleftharpoons\) CO2(g) + C(s)
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-32

Question 8.
The Ang value for the reaction 2NO(g) + O2(g) \(\rightleftharpoons\) 2NO2(g) is …………….
Answer:
– 1
Solution:
Δng = np – nr = 2 – 3 = -1

Question 9.
The correct differential form of van’t Hoff equation is ………….
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-33

Question 10.
For the reaction H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g), the equilibrium constant K is
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-34

Question 11.
PCI5 is kept in a closed container at a temperature of 250K the equilibrium concentrations of PCI5, PCl3 and Cl2 are 0.045 moles L-1, 0.096 moles L-1, 0.096 moles L-1 respectively. The value of equilibrium constant for the reaction PCI5(g) \(\rightleftharpoons\) PCl3(g) + Cl2(g) will be ……………………
Answer:
0.205
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-35

Question 12.
Equilibrium constant changes with ………………..
Answer:
Both temperature and pressure

Question 13.
For the reaction 2HI(g) \(\rightleftharpoons\) H2(g) + I2(g) at 720 K, the equilibrium constant value is 50. The equilibrium constant for the reaction H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g) at the same temperature will be ………..
Answer:
0.02
Solution:
Forward reaction equilibrium constant K1 = 50
Reverse reaction equilibrium constant K2 = ?
K2 = \(\frac { 1 }{ { K }_{ 1 } }\) = \(\frac { 1 }{ 50 }\) = 0.02

Question 14.
If equilibrium constant for the reaction N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) at 298 K is 2.54, the value of equilibrium constant for the reaction \(\frac { 1 }{ 2 }\)N2 + \(\frac { 3 }{ 2 }\)H2 \(\rightleftharpoons\) NH3 will be ………….
Answer:
1.59
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-37

Question 15.
The chemical system at equilibrium is not affected by addition of ……………..
Answer:
catalyst

Question 16.
A catalyst will increase the rate of a chemical reaction by lowering the ……………
Answer:
activation energy

Question 17.
In a closed system, A(s) \(\rightleftharpoons\) 3B(g) + 3C(g) If partial pressure of C is doubled, then partial pressure of B will be ………….. times the original value.
Answer:
\(\frac { 1 }{ 2\sqrt { 2 } }\)

Question 18.
Consider the following gaseous equilibria with equilibrium constants K1 and K2 respectively
SO2 + \(\frac { 1 }{ 2 }\)O2(g) \(\rightleftharpoons\) SO3(g) – K1
2SO3(g) \(\rightleftharpoons\) 2SO2(g) + O2(g) – K2
The equilibrium constants are related as ………….
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-39.1
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-39

Question 19.
K2 for the following reaction at 700 K is 1.3 x 10-3 atm-1. The K at same temperature for the reaction 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g) will be ………….
Answer:
7.4 x 10-2
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-40

Question 20.
For the reaction PCI23(g) + CI2(g) \(\rightleftharpoons\) PCI5(g) at 250°C, the value of KC is 26 then the value of Kp on the same temperature will be ……………
Answer:
0.61
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-41

Question 21.
In the reaction N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g), the value of the equilibrium constant depends on ………………
Answer:
the temperature

Question 22.
K1 and K2 are velocity constant o forward and backward reactions. The equilibrium constant KC of the reaction is …………
Answer:
\(\frac { { K }_{ 1 } }{ { K }_{ 2 } }\)

Question 23.
The equilibrium constant of the reaction 3C2H2 \(\rightleftharpoons\) C6H6 is 4.0 at a temperature of T K. If the equilibrium concentration of C2H2 is 0.5 molesL-1, the concentration of C6H6 is …………..
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-42

Question 24.
In an equilibrium reaction for which ΔG0 = 0, the equilibrium constant K should be = ……………
Answer:
1

Question 25.
The equilibrium constant for the reaction 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g) is 5. If the equilibrium constant mixture contains equal moles of SO3 and SO2, the equilibrium partial pressure of O2,
gas is ……………..
Answer:
0.2 atm
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-43

Question 26.
In the reaction NH4CI(s) \(\rightleftharpoons\) NH3(g) + HCI(g) the value Offlg is Δng ………………….
Answer:
2
Solution:
Δng = np(g) – nr(g) = 2 – 0 = 2

IV. Choose the odd one out.

Question 1.
(a) see – saw
(b) tug – of – war
(c) sublimation of camphor
(d) Acid hydrolysis of an ester
Answer:
(d) Acid hydrolysis of an ester Acid hydrolysis of an ester is a chemical equilibrium.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-44
Whereas a, b, c are examples of physical equilibrium.

Question 2.
(a) Synthesis of hydrogen iodide
(b) Decomposition of calcium carbonate
(c) Sublimation of iodine
(d) Dissociation of PCl5
Answer:
(c) Sublimation of iodine Sublimation of iodine I2(s) \(\rightleftharpoons\) I2(g) is an example of physical equilibrium, whereas a, b and dare examples of chemical equilibrium.

Question 3.
(a) Synthesis of HI
(b) Dissociation of PCl5
(e) Synthesis of NH3
(d) Decomposition of CaCO3
Answer:
(a) Decomposition of CaCO3
Decomposition of CaCO3 is an example of heterogeneous equilibrium.
CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g)
Where as a, b, c are examples of homogeneous equilibrium.

Question 4.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-46
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-47
Synthesis of SO3 is an example of homogeneous equilibrium whereas the others a, b and d are heterogeneous equilibrium.

Question 5.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-48
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-49
(b) is an example of heterogeneous equilibrium whereas the others a, c and d are homogeneous equilibrium.

V. Choose the correct pair.

Question 1.
(a) Q = KC : Reaction is in equilibrium state
(b) Q < KC : Reaction proceed in reverse direction
(c) Q > KC : Reaction proceed in ftrward direction
(d) Q = KC : Reaction proceed in both directions
Answer:
(a) Q = KC : Reaction is in equilibrium state

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-50
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-51

Question 3.
(a) Kp = KC : Synthesis of HI
(b) Kp > KC : Dissociation of PCl5
(c) Kp < KC : Synthesis of SO3
(d) Kp = KC : Synthesis of HI
Answer:
(a) Kp = KC : Synthesis of HI
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g) : Δng = 2 – 2 = 0
ΔKp = KC

VI. Choose the incorrect pair.

Question 1.
(a) Acid hydrolysis of an ester – Homogeneous equilibrium
(b) Synthesis of Ammonia – Homogeneous equilibrium
(c) Decomposition of CaCO3 – Homogeneous equilibrium
(d) Synthesis of HI – Homogeneous equilibrium
Ans.
(c) Decomposition of CaCO3: Homogeneous equilibrium It is a heterogeneous equilibrium.

Question 2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-52
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-53

VII. Assertion and Reason.

Question 1.
Assertion (A): Chemical equilibrium is in a state of dynamic equilibrium.
Reason (R): At equilibrium the forward and backward reactions are proceeding at the same rate and no macroscopic change is observed.
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of(A).

Question 2.
Assertion (A): In Haber’s process, NH3 is liquefied and removed.
Reason (R): Because of the reaction keeps moving in the backward direction.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of(A).

Question 3.
Assertion (A): In the dissociation of PCI5 at constant pressure and temperature addition of helium at equilibrium increases the dissociation of PCl5.
Reason (R): Helium removes CI, from the field of action.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is the not the correct explanation of A.
(c) A is true but R is false
(d) Both A and R are false
Answer:
(d) Both A and R are false

VIII. Choose the incorrect statement.

Question 1.
(a) In equilibrium mixture of ice and water kept in perfectly insulted flask, mass of ice and water does not change with time.
(b) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate
(c) On addition of catalyst the equilibrium constant value is not affected
(d) Equilibrium constant for a reaction with negative .H value decreases as the temperature increases.
Answer:
(b) The intensity of red colour increases when oxalic acid is added to a solution containing iron (III) nitrate and potassium thiocyanate

Solution:
Oxalate ions of oxalic acid form complex with ferric ions thus decrease its concentration thus, concentration of red complex in product decreases.

Samacheer Kalvi 11th Chemistry Physical and Chemical Equilibrium 2 Mark Questions and Answers.

I. Write brief answer to the following questions:

Question 1.
Define the state of equilibrium.
Answer:
At a particular stage, the rate of the reverse reaction is equal to that of the forward reaction indicating a state of equilibrium.

Question 2.
What are the different types of equilibrium? Explain with example?
Answer:
1. Physical equilibrium:
A system in which the amount of matter constituting different phases does not change with time is said to be in physical equilibrium.
H2O(s) \(\rightleftharpoons\) H2O(1). Solid-liquid equilibrium.

2. Chemical equilibrium:
Chemical reactions in which the forward and backward reactions are proceeding at the same rate and no macroscopic change is observed is said to be in chemical equilibrium. H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)

Question 3.
Explain about the equilibrium involving dissolution of solid in liquid with suitable example.
Answer:
When sugar is added to water at a particular temperature. it dissolves to form sugar solution. When more sugar is added to that solution, a particular stage sugar remains as solid and results in the formation of saturated solution. Here a dynamic equilibrium is established between the solute molecules ii the solid phase and in the solution phase.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-54

Question 4.
How is a gas – solution equilibrium exist?
Answer:
When a gas dissolves in a liquid under a given pressure, there will be an equilibrium between gas molecuLes in the gaseous state and those dissolved in the liquid. Example – In carbonate beverages the following equilibrium exists.
CO2(g) \(\rightleftharpoons\)  CO2 (Solution)

Question 5.
What is meant by active mass? Give its unit.
Answer:
The active mass represents the molar concentration of the reactants (or) products.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-55

Question 6.
Show that Kp = KC with two example
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-56

Question 7.
Give two examples of equilibrium reactions where Kp > KC.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-57

Question 8.
When will be Kp < KC? Give two example.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-58

Question 9.
Write the KC for the reaction CO2(g) + H2O(l) \(\rightleftharpoons\) H+(aq) + HCO3(aq)
Answer:
H2O(l) is a pure liquid and its concentration remains constant.
Solution:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-59

Question 10.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-60 What is the value of K4 in A \(\rightleftharpoons\) D
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-61

Question 11.
Write the Kp and KC for the following reactions
(i) 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g)
(ii)2CO(g) \(\rightleftharpoons\) CO2(g) + C(s)
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-62

Question 12.
Explain how the equilibrium constant K. predict the extent of a reaction.
Answer:
1. The value of equilibrium constant KC tells us the extent of the reaction i.e., it indicates how far the reaction has proceeded towards product formation at a given temperature.

2. A large value of KC indicates that the reaction reaches equilibrium with high product yield on the other hand, lower value of KC indicates that the reaction reaches equilibrium with low product yield.

3. If KC > 103 the reaction proceeds nearly to completion.

4. If KC < 10-3 the reaction rarely proceeds.

5. If the KC is in the range 10-3 to 103 significant amount of both reactants and products are present at equilibrium.

Question 13.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-63
Predict the extent of the above two reactions.
Answer:
In the reactions, decomposition of water at 500 K and oxidation 01 nitrogen at 1000 K, the value of KC is very less KC < 10-3. So reverse reaction is favoured.
∴ Products << reactants

Question 14.
Explain about the extent of reaction of dissociation of bromine mono chloride at 1000 K.
Answer:
2BrCl(g) \(\rightleftharpoons\) Br2(g) + Cl2(g) KC = 5
10-3 <  KC <  103
both forward and backward reaction make significant progress. Neither forward nor reverse reaction predominates.

Question 15.
What is the KC value for formation of HCl at 700 K? Predict the extent of the reaction?
Answer:
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g) at 700 K
KC = 57.0
10-3 <KC <103
So both forward and backward reaction make significant progress. Neither forward nor reverse reaction predominates.

Question 16.
What is the KC value of formation of HCI at 300 K? Explain it.
Answer:
H2(g) – Cl2(g) \(\rightleftharpoons\) 2HCI(g) at 300 K
K2 = 4 x 1031
KC > 103 So[products] >> [Reactant]
Reaction nearly goes to completion. So forward reaction is favoured.

Question 17.
2CO(g) + O2(g) \(\rightleftharpoons\) 2CO2(g) at 1000 K. What is the Kc this reaction? Predict the extent of this reaction.
Answer:
2CO(g) + O2(g) \(\rightleftharpoons\) 2CO2(g) at 1000 K
KC = 2.2 x 1022
KC > 103.
So [Products] >> [Reactants]
Reaction nearly goes to completion and forward reaction is favoured.

Question 18.
Define Q value for a chemical equilibrium reaction.
Answer:
Consider a homogeneous reversible reaction xA + yB \(\rightleftharpoons\) lC + mD For the above reaction under non-equilibrium conditions, reaction quotient Q is defined as the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants. Under non equilibrium conditions,
Q = \(\frac { d(Ink) }{ dt }\) ΔH° = \(\frac { { \triangle H }^{ 0 } }{ { RT }^{ 2 } }\)

Question 19.
Explain the diagrammatic expression about the direction of reaction.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-65
In (i) Q < KC, the reaction will proceed in forward direction.
In (ii) Q = KC, the reaction is in equilibrium state.
In (iii) Q> KC, the reaction will proceed in the reverse direction.

Question 20.
Explain about the effect of catalyst in an equilibrium reaction?
Answer:
Addition of catalyst does not affect the state of equilibrium. The catalyst increases the rate of both the forward and reverse reactions to the same extent. Hence it does not change the equilibrium composition of the reaction mixture.

Question 21.
For the following equilibrium. KC = 6.3 x 1014 at 1000K
NO(g) + O3(g) \(\rightleftharpoons\) NO2(g) + O2(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions what is KC for the reverse reaction?
Answer:
For the reverse reaction Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-66

Question 22.
Explain why pure liquids and solids can be ignored while writing the value of equilibrium constants.
Answer:
This is because molar concentration of a pure solid or liquid is independent of the amount present.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-67
Since density of pure liquid or solid is fixed and molar mass is also fixed, therefore molar concentration are constant.

Question 23.
A sample of Hl(g) is placed in a flask at a pressure of 0.2 atm. At equilibrium partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium?
2HI(g) \(\rightleftharpoons\) H2 (g) + I2(g)
Answer:
pHI = 0.04 atm. pH2 = 0.08 atm; pl2 = 0.08 atm
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-68

Question 24.
The equilibrium constant expression for a gas reaction is.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-69
Write the balanced chemical equation corresponding to this expression.
Answer:
Balanced chemical equation for the reaction is
4 NO(g) + 6 H2O(g) \(\rightleftharpoons\) 4NH3(g) + 5O2(g)

Question 25.
Predict which of the following will have appreciable concentration of reactions and products:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-70
Answer:
Following conclusion can be drawn from the values of
(a) Since the value of K is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of K is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of K is 1.8, this means that both the products and reactants have appreciable concentrations.

Question 26.
Write the equilibrium constant (KC) expression for the following reactions.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-71
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-72

Question 27.
For the equilibrium 2NOCI(g) \(\rightleftharpoons\) 2NO(g) + CI2(g) the value of the equilibrium constant KC is 3.75 x at 1069 K. Calculate the Kp for the reaction at this temperature?
Answer:
We know that Kp = Kc(RT)∆ng
For the above reaction, ∆ng = (2 + 1) – 2 = 1
Kp = 3.75 x 10-6 (0.0831 x 1069) = 3.3 x 10-4

Question 28.
The value of Kc for the reaction 2A \(\rightleftharpoons\) B + C is 2 x 10-3. At a given time, the composition of reaction mixture is [A] [B] [C] = 3 x 10-4 M. In which direction the reaction will proceed?
Answer:
For the reaction, the reaction quotient Q is given by QC = [B] [C]/[A]2
as [A] = [B] = [C] = 3 x 10-4 M
Qc (3 x 10-4)(3 x 10-4)/(3 x 10-4)2 = l
as QC > KC so, the reaction will proceed in the reverse direction.

Question 29.
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH4(g) + H2O(g) \(\rightleftharpoons\) CO(g) + 3H2(g)
(a) Write an expression for Kp for the above reaction.
(b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-73
(b) (I) value of Kp will not change, equilibrium will shift in backward direction.
(ii) value of Kp will increase and reaction will proceed in forward direction.
(iii) no effect.

II. Answer the following questions:

Question 1.
Explain about the formation of solid-liquid equilibrium with suitable example.
Answer:
1. Consider melting of ice in a closed container at 273 K. This system reach a state of physical equilibrium in which the amount of water in the solid phase and liquid phase does not change with time.

2. In this process. the total number of water molecules leaving from and returning to the solid phase at any instant are equal.

3. If some ice cubes and water are placed in a thermos flask (at 273K and I atm) then there will be no change in the mass of ice and water.

4. At equilibrium: Rate of melting of ice Rate of freezing of water
H2O(l) \(\rightleftharpoons\) H2O(g)

Question 2.
How is liquid – vapour equilibrium exist?
Answer:
1. Liquid water is in equilibrium with vapour at 373 K and I atm pressure in a closed vessel

2. Rate of evaporation = Rate of condensation

Question 3.
What is meant by boiling point and condensation point of the liquid?
Answer:
The temperature at which the liquid and vapour phases are at equilibrium is called the boiling point and condensation point of the liquid.

Question 4.
Define melting point (or) freezing point of the substance.
Answer:
The temperature at which the solid and liquid phases of a substance are at equilibrium is called the melting point or freezing point of substance.

Question 5.
Illustrate the formation of solid – vapour equilibrium with suitable example.
Answer:
1. Consider a system in which the solid sublimes to vapour. e.g., I, (or) camphor.
2. When solid iodine is placed in a closed transparent vessel, after sometime, the vessel gets filled up with violet vapour due to sublimation of iodine.
3. Initially the intensity of the violet colour increases, after some time it decreases and finally it becomes constant as the following equilibrium is attained.
I2(s) \(\rightleftharpoons\) I2 (g)

Question 6.
Give three examples for solid vapour equilibrium.
Answer:
I2(s) \(\rightleftharpoons\) I2 (g)
Camphor (s) \(\rightleftharpoons\) Camphor (g)
NH4CI(s) \(\rightleftharpoons\) NH4CI(g)

Question 7.
Explain the following diagrams.
Diagram – 1
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-74
Answer:

  1. As the concentration of the products increases, more products collide and react in the backward direction.
  2. As the rate of the reverse reaction increases, the rate of the forward reaction decreases.
  3. Eventually the rate of both reactions becomes equal.

Diagram-II
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-75

  1. Concentration of reactants decreases with time initially and concentration of products increases with time.
  2. After sometime, equilibrium is reached i.e., concentration of reactants and products remains constant.

Question 8.
What are the types of chemical equilibrium? Explain with suitable example.
Answer:

  1. Chemical equilibrium is of two types:
    • Homogeneous equilibrium
    • Heterogeneous equilibrium.
  2. In a homogeneous equilibrium, all the reactants and products are in the same phase
    H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
    In the above equilibrium H2, I2 and HI are in the gaseous state.
  3. If the reactants and products of a reaction in equilibrium are in different phases, then it is calLed heterogeneous equilibrium.
    e.g., CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO3(g)

Question 9.
Write the value of K and K equation for CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g)
Answer:
A pure solid always has the same concentration at a given temperature, as it does not expand to fill its container i.e., it has the same no. of moles of its volume. Therefore the concentration of pure solid is a constant. So the expression if K and K is K [CO2], K = PCO2

Question 10.
Consider the following equilibrium reaction and relate their equilibrium constants

  1. N2 + O2 \(\rightleftharpoons\) 2NO, K1
  2. 2NO + O2 \(\rightleftharpoons\) 2NO2, K1
  3. N2 + 2O2 \(\rightleftharpoons\) 2NO2, K3

Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-77

Question 11.
Explain the effect of concentration in an equilibrium state?
Answer:
At equilibrium, the concentration of the reactants and products does not change. The addition of more reactants or products to the reacting system at equilibrium causes an increase in their respective concentration.

According to Le – Chatelier’s principle, the effect of increase in concentration of a substance is to shift the equilibrium in a direction that consumes the added substance. For example,
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)

The addition of H2 or I2 to the equilibrium mixture, disturbs the equilibrium. In order to minimize the stress, the system shifts the reaction in a direction where H2 and I2 are consumed i.e., formation of additional HI would balance the effect of added reactant.

Hence the equilibrium shifts to the right (forward direction). i.e., the equilibrium is re – established. Similarly, removal of HI (Product) also favours forward reaction. If HI is added to the equilibrium mixture, the concentration of HI is increased and system proceeds in the reverse direction to nullify the effect of increase in concentration of HI.

Question 12.
Consider the reaction, N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g). Explain the effect of pressure on this equilibrium reaction.
Answer:
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
In the above equilibrium, if the pressure is increased, the volume wiLl decreases. The system responds to this effect by reducing the number of gas molecules. i.e., it favours the formation of ammonia. If the pressure is reduced, the volume will increases. It favours the decomposition of ammonia.

Question 13.
Why pressure has no effect on the synthesis of HI?
Answer:
(i) When the total number of moles of gaseous reactants and gaseous products are equal, the change in pressure has no effect on system at equilibrium.
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Here the number of moles of reactants and products are equal. So the pressure has no effect on such equilibrium with ∆ng = O.

Question 14.
Explain the effect of temperature on the following equilibrium reaction.
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) ∆H = – 92.2 kJ.
Answer:
In this equilibrium, N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) ∆H = – 92.2 kJ.

Forward reaction is exothermic while the reverse reaction is endothermic. If the temperature of the system increased, the system responds by decomposing some of ammonia molecules to nitrogen and hydrogen by absorbing the supplied heat energy.

Similarly, the system responds to a drop in the temperature by forming more ammonia molecule from nitrogen and hydrogen which release heat energy.

Question 15.
How does oxygen exchanges between maternal and fetal blood in a pregnant woman?
Answer:
1. In a pregnant women, the oxygen supply for the fetus is provided by the maternal blood in the placenta where the blood vessels of both mother and fetus arc in close proximity. Both fetal and maternal hemoglobin binds tO oxygen reversibly as follows.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-78

2. In the above two equilibrium, the equilibrium constant value for the oxygenation of fetal hemoglobin is higher which is due to its higher affinity for oxygen compared to adult hemoglobin. Hence in placenta, the oxygen from the mother’s blood s effectively transferred to fetal hemoglobin.

Question 16.
What is K for the following reaction in state of equilibrium?
2SO2(g) + O2 \(\rightleftharpoons\) 2SO3(g)
(Given:[SO2] = 0.6 M; [O2] = 0.82 M; and [SO3] = 1.90 M
Answer:
2SO2(g) + O2 \(\rightleftharpoons\) 2SO3(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-79

Question 17.
At a certain temperature and total pressure of 105Pa, iodine vapours contain 40% by volume of iodine atoms in the equilibrium 12(g) \(\rightleftharpoons\) 2I(g). Calculate Kp for the equilibrium.
Answer:
According to available data:
Total pressure of equilibrium mixture = 105 Pa
Partial pressure of iodine atoms (1) = \(\frac { 40 }{ 100 }\) x (105 Pa) = 0.4 x 105 Pa
Partial pressure of iodine molecules (I2) = \(\frac { 60 }{ 100 }\) x (105 Pa) = 0.6 X 105 Pa
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-80

Question 18.
A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant KC for the reaction
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) is 1.7 x 10-2
Is this reaction at equilibrium? if not. what is the direction of net rection?
Answer:
The reaction is: N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
Concentration quotient (Q ) =
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-81
The equilibrium constant (Kr) for the reaction = 1.7 x 10-2
As QC \(\neq\) KC; this means that the reaction is not in a state of equilibrium.

Question 19.
What is the effect of:

  1. addition of H2
  2. addition of CH3OH
  3. removal of CO
  4. removal of CH3OH

On the equilibrium 2H2(g) + CO(g) \(\rightleftharpoons\) CH3OH(g)
Answer:

  1. Equilibrium will be shifted in the forward direction.
  2. Equilibrium will be shifted in the backward direction.
  3. Equilibrium will be shifted in the backward direction.
  4. Equilibrium will be shifted in the forward direction.

Question 20.
At 473 K, the equilibrium constant Kc for the decomposition of phosphorus pentachioride (PCl5) is 8.3 x 10-3 . If decomposition proceeds as:
Answer:
PCI5(g) \(\rightleftharpoons\) PCI3(g) + Cl2(g);
∆H = + 124.0 kJ mol-1

  1. Write an expression for K. for the reaction.
  2. What is the value of K for the reverse reaction at the same temperature.
  3. What would be the effect on KC if
    • More of PCI5 is added
    • Temperature is increased.

Answer:
1. The expression for Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-82
2.  For reverse reaction Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-83
3.
(I) By adding more of PCI5, value of Kc will remain constant because there is no change in temperature.
(ii) By increasing the temperature the forward reaction will be favoured since it is endotherniic in nature. Therefore, the value of equilibrium constant will increase.

Question 21.
Dihydrogen gas used in Haher’s process is produced by reacting methane from natural gas with high temperature stam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction.
Answer:
CO(g) + H2O(g) \(\rightleftharpoons\) CO2(g) + H2(g)
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam so that PCO = PH2O = 4.0 bar., what will be the partial pressure of 2 at equilibrium? Kp = 0.1 at 400°C.
Answer:
Let the partial pressure of hydrogen (H2) at equilibrium point p bar
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-84

Question 22.
The value of KC for the reaction 3O2(g) \(\rightleftharpoons\) 2O3(g) is 2.0 x 10-50 So at 25°C. If equilibrium concentration of 0, in 25°C is 1.6 x 102 what is the concentration of O3?
Answer:
3O2(g) \(\rightleftharpoons\) 2O3(g)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-85

Question 23.
The reaction CO(g) + 3H2(g) \(\rightleftharpoons\) CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1K flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, KC for the reaction at the given temperature is 3.90.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-86

Question 24.
The following concentration were obtained for the formation of NH3 from N2 and H2 at equilibrium at 500 K.

  • [N2(g)] 1.5 x 10-2 M
  • [H2(g)] = 3.0 x 10-2 M
  • [NH3]= 1.2 x 10-2M.

Calculate equilibrium constant.
Answer:
N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
Calculate equilibrium constant
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-87

Question 25.

  1. In the reaction A+ B → C + D, what will happen to the equilibrium if concentration of A is increased?
  2. The equilibrium constant for a reaction is 2 x 10-23 at 25°C and 2 x 10-2 at 50°C. Is the reaction endothermic or exothermic?
  3. Mention at least three ways by which the concentration of SO3 can be increased in the following reaction in a state of equilibrium. 2SO2(g) + O2(g) \(\rightleftharpoons\) 2SO3(g)

Answer:

  1. The reaction will shift in the forward direction.
  2. Endothermic
  3. following reaction
    • increasing concentration of SO2
    • increasing pressure.
    • increasing concentration of oxygen.

Samacheer Kalvi 11th Chemistry Chapter 8 Physical and Chemical Equilibrium

Question 26.
PCI5, PCI3 and CI2 are at equilibrium at 500 K and having concentration I.59M PCl3, l.59M
CI2 and 1.41M PCI5. Calculate K. for the reaction PCl5 \(\rightleftharpoons\) PCl3 + Cl2
Answer:
The equilibrium constant K. for the above reaction can be written as:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-88

Question 27.
Given the equilibrium
N2O4(g) \(\rightleftharpoons\) 2NO2(g) with Kp = 0.15 atm at 298 K
(a) What is Kp using pressure in torr?
(b) What is KC using units of moles per litre.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-89

III. Answer the following questions in detail:

1. Derive the values of equilibrium constants Kp and KC for a general reaction
x A + y B \(\rightleftharpoons\) lC + mD
Answer:
Let us consider a reversible reaction x A + y B \(\rightleftharpoons\) lC + mD
where A, B are the reactants C and D are the product and x, y. l and m are the stoichiometric coefficients of A. B, C and D respectively. Applying the law of mass action the rate of forward reaction.
rf ∝ [A]x [B]y or rf Kf [A]x [B]y
Similarly the rate of backward reaction
rb ∝ [C]l [D]m or rb = Kb [C]l [D]m
where Kf and Kb are proportionality constants.
At equilibrium, Rate of forward reaction (rf) = Rate of backward reaction (rb)
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-90
where KC is the equilibrium constant in terms of concentration. At a given temperature, the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants is a constant known as equilibrium constant.

If the reactants and products of the above reaction are in gas phase, then the equilibrium constant can be written in terms of partial pressures
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-91
where PA, PB, PC and PD are the partial pressure of gases A, B, C and D respectively.

Question 2.
Derive the values of K and K for the synthesis of HI.
Answer:
H2(g) + I2(g) \(\rightleftharpoons\) 2HI(g)
Let us consider the formation of HI in which ‘a’ moles of hydrogen, ‘b’ moles of iodine gas are allowed to react in an container of volume ‘V’. Let ‘x’ moles of each of H2, and I2, react together to form 2x moles of HI.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-92

Question 3.
Derive the values of K and K for dissociation of PCI5.
Answer:
Consider that ‘a’ moles of PCl5 is taken in container of volume ‘V’ Let x moles of PCI5 be dissociated into x moles of PCI3 and x moles of Cl2.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-93
Applying law of mass action
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-94
Where ‘n’ is the total number of moles at equilibrium
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-95

Question 4.
At certain temperature and under a pressure of 4 atm, PCl5 is 10% dissociated. Calculate the pressure at which PCI5 will be 20% dissociated at temperature remaining constant.
Calculation of Kp
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-96
Total no. of moles in the equilibrium mixture = 1 – α + α + α = (1 + α) mol.
Let the total pressure of equilibrium mixture = ρ atm
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-97
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-98
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-99
calculation of P under new condition
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-100

IV. Numerical Problems

Question 1.
Find the value of K for each of the following equlibria from the value of K
(a) 2NOCI(g) \(\rightleftharpoons\) 2NO(g) + Cl2(g); Kp= 1.8 x 102 atm at 500 K
(b) CaCO3(s) \(\rightleftharpoons\) CaO(s) + CO2(g): Kp = 167 atm at 1073 K.
Answer:
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-102

Question 2.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICI was 0.78 M?
2ICI(g) \(\rightleftharpoons\) I2(g) + CI2(g); KC = 0.14
Answer:
Suppose at equilibrium, the molar concentration of both I2(g) and Cl2(g) is x mol L-1.
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-103

Question 3.
Equilibrium constant K for the reaction. N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g) at 500 K is 0.061. At particular time, the analysis shows that the composition of the reaction mixture is 3.0 mol L-1 of N2; 2.0 mol L-1 of H2; 0.50 moI L-1 of NH3. Is the reaction at equilibrium?
Answer:
The given reaction is: N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g)
According to available data
Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-104

Common Errors

  1. In writing Kp, KC values from the equations, students may confuse to write whether products or reactants in the numerator or in the denominator
  2. When Kp = KC.(RT)0 students may confuse.
  3. In writing An values, students will consider all reactants and products.
  4. Students are confused to understand the concept of Q value and KC value.
  5. In writing chemical equilibrium reaction, you may miss the physical states of reactants and products.
  6. ∆ngvalue calculation will go wrong if you consider all the reactants and products.
  7. Equilibrium constant value is calculated under equilibrium condition and reaction quotient value defined at sometimes at equilibrium condition wrongly by the students.
  8. Chemical equilibrium condition must be known. Students may wrongly write in an open vessel, the equilibrium take place.
  9. Equilibrium symbol, students may wrongly write as =

Rectifications

  1. Always we have to write Kp & KC as Samacheer Kalvi 11th Chemistry Solutions Chapter 8 Physical and Chemical Equilibrium-105
  2. (Anything)° = 1. So Kp = KC.
  3. Only they have to consider gaseous reactants and gaseous products. Since solid and liquid are constant in their concentration.
  4. Q value is calculated under non equilibrium conditions. KC value is calculated under equilibrium condition.
  5. Physical states of reactants and products must be written as a subscript by the words s, l, g. (solid, liquid, gas)
  6. ∆ng = ngp – ngr You should consider only gaseous products and gaseous reactants.
  7. Reaction quotient Q value is calculated only under non-equilibrium conditions.
  8. Chemical equilibrium reactions are always take place at closed vessel only.
  9. Equilibrium reaction symbol is ⇔.

We as a team believe the information prevailing regarding the Samacheer Kalvi Solutions for 11th Chemistry Chapter 8 Physical and Chemical Equilibrium has been helpful in clearing your doubts to the fullest. For any other help do leave us your suggestions and we will look into them. Stay in touch to get the latest updates on Tamilnadu State Board Solutions for different subjects in the blink of an eye.