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Students can Download Tamil Nadu 11th Chemistry Model Question Paper 4 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Chemistry Model Question Paper 4 English Medium

Time: 2½ Hours
Total Score: 70 Marks

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I. II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail. Draw diagrams wherever necessary.

PART -1

Answer all the questions. [15 x 1 = 15]
Choose the most suitable answer from the given four alternatives.

Question 1.
Carbon forms two oxides, namely carbon monoxide and carbon dioxide. The equivalent mass of which element remains constant?
(a) Carbon
(b) Oxygen
(c) Both carbon and oxygen
(d) Neither carbon nor oxygen
Answer:
(b) Oxygen

Reaction 1 : 2 C + O₂ → 2 CO
2 x 12 g carbon combines with 32 g ofoxygen.
Hence, Equivalent mass of carbon = \(\frac{2 \times 12}{32}\) x 8 = 6

Reaction 2 : C + O₂ → CO₂
12 g carbon combines with 32 g of oxygen.
Hence, Equivalent mass of carbon = \(\frac{12}{32}\) x 8 = 3

Question 2.
Electronic configuration of species M²⁺ is Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ and its atomic weight is 56.
The number of neutrons in the nucleus of species M is ………………… .
(i) 26
(b) 22
(c) 30
(iv) 24
Answer:
(c) 30

Solution:
M²⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ ; M : Is² 2s² 2p⁶ 3s² 3p⁶ 3d⁸
Atomic number = 26
Mass number = 56
No. of neutrons = 56 – 26 = 30

Question 3.
Assertion: Helium has the highest value of ionization energy among all the elements known Reason: Helium has the highest value of electron affinity among all the elements known
(a) Both assertion and reason are true and reason is correct explanation for the assertion
(b) Both assertion and reason are true but the reason is not the correct explanation for the assertion
(c) Assertion is true and the reason is false
(d) Both assertion and the reason are false
Answer:
(c) Assertion is true and the reason is false

Question 4.
At room temperature normal hydrogen consists of
(a) 25% ortho form + 75% para form
(b) 50% ortho form + 50% para form
(c) 75% ortho form + 25% para form
(d) 60% ortho form + 40% para form
Answer:
(c) 75% ortho form + 25% para form

Question 5.
Match the following.
List-I List-II
A. Beryllium – 1. Sacrificial anode
B. Calcium – 2. X-ray tube radiation window
C. Magnesium – 3. Scavenger to remove oxygen in TV
D. Barium – 4. Getter in vacuum tubes
Code: A B C D
(a) 4 2 3 1
(b) 2 4 1 3
(c) 3 1 4 .2
(d) 1 3 2 4
Answer:
(b) 2 4 1 3

Question 6.
Which of the following pair will diffuse at the same rate?
(a) CO₂ and N₂O
(b) CO₂ and NO
(c) CO₂ and CO
(d) N₂O and NO
Answer:
(a) CO₂ and N₂O

Question 7.
An ideal gas expands from the volume of 1 x 10^-3 m³ to 1 x 10^-2 m³ at 300K against a constant pressure at 1 x 10³ Nm^-2. The work done is ………………… .
(a) -900 J
(b) 900 kJ
(c) 270 kJ
(d) -900 kJ
Answer:
(a) -900 J

w = PΔV
w = – (1 x 105 Nm^-2) (1 x 10^-2 m³ – 1 x 10^-3 m³)
w = 10⁵ (10^-2 – 10^-3) Nm
w = 10⁵(10 – 1) 10^-3) J
w = 10⁵ (9 x 10^-3) J
w = 9 x 10² J
w = -900 J

Question 8.
At a given temperature and pressure, the equilibrium constant values for the equilibria

The reaction between K₁ and K₂ is …………………. .

Answer:
b

Solution:

Question 9.
Consider the following statements.
(i) Henry’s law is applicable at moderate temperature and pressure only.
(ii) Highly soluble gases obey’s Henry’s law.
(iii) The gases react with the solvent do not obey Henry’s law.
Which of the above statements is/are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i) and (ii)
Answer:
(b) (ii) only

Question 10.
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°

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 11.
The method used to estimate nitrogen in foods and fertilisers is
(a) Dumas method
(b) Kjeldahl’s method
(c) Carius method
(d) Oxide method
Answer:
(b) Kjeldahl’s method

Question 12.
Which of the following carbocation will be most stable?

Answer:
d

Question 13.
Which one of the following is used as a soil sterilizing agent?

Answer:

Question 14.
Which of the following pair of oxides is responsible for acid rain?
(a) Chloroform
(b) Chloral
(c) Iodoform
(d) Chloropicrin
Answer:
(d) Chloropicrin

Question 15.
Which of the following pair of oxides is responsible for acid rain?
(a) SO₃ + NO₂
(b) CO₂ + CO
(c) N₂O + CH₄
(d) O₂ + H₂
Answer:
(a) SO₃ + NO₂

PART- II

Answer any six questions.in which question No. 20 is compulsory.

Question 16.
Calculate the number of moles present in 9g of ethane?
Answer:
Mass of ethane = 9 g
Molar mass of ethane C₂H₆ = 30 g mol^-1.
\(\text { No. of moles }=\frac{\text { Mass }}{\text { Molar mass }}=\frac{9}{30}=0.3 \mathrm{mol}\)

Question 17.
Why did halogens act as oxidizing agents?
Answer:
Halogens act as oxidizing agents. Their electronic configuration is ns2 np5. So all the halogens are ready to gain one electron to attain the nearest inert gas configuration. An oxidizing agent is the one which is ready to gain an electron. So all the halogens act as oxidizing agents. Also halogens are highly electronegative with low dissociation energy and high negative electron gain enthalpies. Therefore, the halogens have a high tendency to gain an electron. Hence they act as oxidizing agents.

Question 18.
Would it be easier to drink water with a straw on the top of Mount Everest?
Answer:
It is difficult to drink water with a straw on the top of Mount Everest. This is because the reduced atmospheric pressure is less effective in pushing water into the straw at the top of the mountain because gravity falls off gradually with height. The air pressure falls off, there isn’t enough atmospheric pressure to push the water up in the straw all the way to the mouth.

Question 19.
Be(OH)₂ is amphoteric in nature. Prove it.
Answer:
Be(OH)₂ is amphoteric in nature as it reacts with both acid and alkali.

• Be(OH)₂ + 2NaOH → Na₂BeO₂ + 2H₂O
  (Acidic)
• Be(OH)₂ + 2HC1 → BeCl₂ + 2H₂O
  (Basic)

Question 20.
An engine operating between 127°C and 47°C takes some specified amount of heat from a high temperature reservoir. Calculate the percentage efficiency of an engine.
Answer:
Given: T₁ = 127°C = 127 + 273 = 400K
T₂ = 47°C = 47 + 273 = 320 K
% efficiency q?

Question 21.
Calculate the formal charge on each atom of carbonyl chloride (COCl₂)?
Answer:

Question 22.
Why we need to purify the organic compounds?
Answer:
In order to study the structure, physical properties, chemical properties and biological properties of organic compounds, they must be in the pure state. So organic compounds must be purified.

Question 23.
What are electrophiles? Give an example.
Answer:
Electrophiles: Electrophiles are reagents that are attracted towards negative charge or electron rich center. They are either positively charged ions or electron deficient neutral molecules.

Example:
CO₂, A1Cl₃, BF₃, FeCl₃, NO⁺, NO_2⁺, etc.

Question 24.
Complete the reaction, C₂H₅NH₂

Answer:

Part – III

Answer any six questions in which question No. 30 is compulsory. [6 x 3 = 18]

Question 25.
On the formation of SF₆ by the direct combination of S and F₂, which is the limiting reagent? Prove it.
Answer:
SF₆ is formed by burning Sulphur in an atmosphere of Fluorine. Suppose 3 moles of S is allowed to react with 12 moles of Fluorine.
\(\mathrm{S}_{(\mathrm{o}}+3 \mathrm{F}_{2_{(\mathrm{g})}} \rightarrow \mathrm{SF}_{6(\mathrm{g})}\)

As per the stoichiometric reaction, one mole of S reacts with 3 moles of fluorine to complete the reaction. Similarly, 3 moles of S requires only 9 moles of fluorine.

∴ It is understood that the limiting reagent is Sulphur and the excess reagent is Fluorine.

Question 26.
Explain about the significance of de-Broglie equation.
Answer:

• \(\lambda=\frac{h}{m v}\) This equation implies that a moving particle can be considered as a wave and a wave
  can exhibit the properties of a particle.
• For a particle with high linear momentum (mv) the wavelength will be so small and cannot be observed.
• For a microscopic particle such as an electron, the mass is of the order of 10 31 kg, hence the wavelength is much larger than the size of atom and it becomes significant.
• For the electron, the de Broglie wavelength is significant and measurable while for the iron ball it is too small to measure, hence it becomes insignificant.

Question 27.
What are all the factors that influences electron affinity?
Answer:
Factors that affects electron affinities are size of the atom, effective nuclear charge, screening effect.

• \(Electron affinity \propto \frac{1}{\text { Size of the atom }}\)
• Electron affinity QC Effective nuclear charge
• \(Electron affinity \propto \frac{1}{\text { Screening effect }}\)

Question 28.
Describe about the biological importance of sodium and potassium.
Answer:

• Monovalent sodium and potassium ions are found in large proportions in biological fluids.
• These ions perform important biological functions such as maintenance of ion balance and nerve impulse conduction.
• Sodium – Potassium play an important role in transmitting nerve signals.
• A typical 70 kg man has 90 g of Na and 170 g of K.
• Sodium ions are found on the outside of cells, being located in blood plasma and in the interstitial fluid which surrounds the cells. These ions participate in the transmission of nerve signals, in regulating the flow of water across cell membranes and in the transport of sugars and amino acids into cells.
• Potassium ions are the most abundant cations within cell fluids, where they activate many enzymes, participate in the oxidation of glucose to produce ATP and with sodium, are responsible for the transmission of nerve signals.

Question 29.
What are the conventions adopted in writing the thermochemical equation?
Answer:
A thermochemical equation is a balanced stoichiometric chemical equation that includes the enthalpy change (ΔH).

Conventions adopted in thermochemical equations:

• The coefficients in a balanced thermochemical equation refer to number of moles of reactants and products involved in the reaction.
• The enthalpy change of the reaction ΔH has unit kJ.
• When the chemical reaction is reversed, the value of ΔH is reversed in sign with the same magnitude.
• Physical states (gas, liquid, aqueous and solid) of all species is important and must be specified in a thermochemical reaction since ΔH depends on the phases of reactants and products.
• If the thermochemical equation is multiplied throughout by a number, the enthalpy change is also be multiplied by the same number value.
• The negative sign of \(\Delta \mathrm{H}_{r}^{o}\) indicates the reaction to be an exothermic and the positive sign of \(\Delta \mathrm{H}_{r}^{o}\) indicates an endothermic type of reaction.

Question 30.
Calculate the work done when 2-moles of an ideal gas expands reversibly and isothermally from a volume of 500 mL to a volume 2 L at 25°C and normal pressure.
Answer:
Given

Question 31.
For a gaseous homogeneous reaction at equilibrium number of moles of products are greater than the number of moles of reactants. Is K is larger or smaller than K_r.
Answer:
For a homogeneous reaction at equilibrium, number of moles of products (n_p) are greater than the number of moles of reactants (n_R), then Δn_g = +ve

Question 32.
Explain – Resonance.
Answer:
(i) Certain organic compounds can be represented by more than one structure and they differ only in the position of bonding and lone pair of electrons. Such structures are called resonance structure and this phenomenon is called as resonance.

This phenomenon is also called as mesomerism or mesomeric effect.
(ii) For example, the structure of aromatic compounds such as benzene and conjugated system like 1,3 butadiene cannot be represent by a single structure and their observed properties can be explained on the base of a resonance hybrid.
(iii) Resonance structure of benzene.

(I) and (II) are called, as resonance hybrids of benzene.
(iv) For 1,3 butadiene:

(I), (II) and (III) are called as resonance hybrids of 1,3 butadiene.

Question 33.
Toluene undergoes nitration easily than benzene. Why?
Answer:

• Toluene has a methyl group on the benzene ring which is electron releasing group and hence activate the benzene ring by pushing the electrons on the benzene ring.
• CH₃ group is ortho – para director and ring activator. Therefore in toluene, ortho and para positions are the most reactive towards an electrophile, thus promoting electrophilic substitution reaction.
• The methyl group hence makes it around 25 times more reactive than benzene. Therefore it undergoes nitration easily than benzene.
  

Part – IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) (i) Explain about the classification of matter. (3)
(ii) What is a combination reaction? Give an example. (2)
[OR]
(b) (i) Define – electronegativity. (2)
(ii) How Moseley determined the atomic number of an element using X-rays. (3)
Answer:
(a) (i) Classification of matter:

(ii) When two or more substances combine to form a single substance, the reactions are called combination reactions.
A + B → C
Example: 2 Mg + O₂ → 2MgO
[OR]
(b) (i) Electronegativity is the relative tendency of an element present in a covalently bonded molecule, to attract the shared pair of electrons towards itself.

(ii)

• Henry Moseley studied the X-ray spectra of several elements and determined their atomic numbers (Z).
• He discovered a correlation between atomic number and the frequency of X-rays generated by bombarding an element with high energy of electrons.
• Moseley correlated the frequency of the X-ray emitted by an equation as, \(\sqrt{\mathfrak{v}}=a(\mathrm{Z}-b)\)
  Where u = Frequency of the X-rays emitted by the elements. a and. b = Constants.
• From the square root of the measured frequency of the X-rays emitted, he determined the atomic number of the element.

Question 35.
(a) Explain the following observations,
(i) Aerated water bottles are kept under water during summer. (3)
(ii) Liquid ammonia bottle is cooled before opening the seal. (2)
[OR]
(b) Derive the relation between ΔH and ΔU for an ideal gas. (5)
Answer:
(a) (i) In aerated water bottles, CO₂ gas is passed through the aqueous solution under pressure because the solubility of the gas in water is not very high. In summer, the solubility of the gas in water is likely to decrease because of the rise in temperature. Thus, in summer, more ’ of gas will be present above the liquid surface in the glass bottle. In case, the pressure of the gas becomes too high, the glass will not be able to withstand the pressure and the bottle may explode. To avoid this, the bottles are kept under water. As a result, the temperature is likely to decrease and the solubility of CO₂ is likely to increase in aqueous solution resulting in decreased pressure.

(ii) Liquid ammonia bottle contains the gas under very high pressure. If the bottle is opened as such, then the sudden decrease in pressure will lead to a large increase in volume of the gas. As a result, the gas will come out of the bottle all of a sudden with force. This will lead to the breakage of the bottle and also causes accident However, if the bottle is cooled under tap water for sometime, there will be a decrease in the volume of a gas to a large extent. If the seal is opened now, the gas will come out of the bottle at a slower rate, reduces the chances of accident.
[OR]
(b) Relation between ΔH and ΔU for an ideal gas.
1. When the system at constant pressure undergoes changes from an initial state with Hp U₁, V₁ and P parameters to a final state with H₂, U₂, V₂ and P parameters, the change in enthalpy AH, is given by
∆H = U + PV
2. At initial state H₁ = U₁ + PV₁ ……………….. (1)
At final state H₂ = U₂ + PV₂ ……………….. (2)
(2) – (1) ⇒ (H₂ – H₁) = (U₂ – U₁) + P (V₂ – V₁)
∆H = ∆U + P∆V …………………… (3)
Considering ∆U = q + w ; w = – P∆V
∆H = q + w + PAV
∆H = q_p – PAV+ PAV
∆H = q_p ………………… (4)

q_p is the heat absorbed at constant pressure and is considered as heat content.

3. Consider a closed system of gases which are chemically reacting to produce product gases at constant temperature and pressure with V. and Vf as the total volumes of the reactant and product gases respectively, and n. and nf are the number of moles of gaseous reactants and products. Then,
For reactants: PV_i = n_iRT
For products : P V_f = n_f RT

Then considering reactants as initial state and products as final state,
P (V_f – V_i) = (n_f – n_i) RT
P∆V = ∆n_g RT

We know
∆H = ∆U + P∆V
∆H = ∆U + ∆n_g RT …………………. (5)

Question 36.
(a) (i) What are heterogenous equilibrium? Give an example. (2)
(ii) The atmospheric oxidation of NO 2NO_(g) + O_2(g) = 2NO_2(g) was studied with initial pressure of 1-atm of NO and 1-atm of O₂. At equilibrium partial pressure of oxygen is 0.52 atm. Calculate K_p of the reaction. (3)
[OR]
(b) (i) Explain the factors influencing the solubility of the solutes. (3)
(ii) Why the carbonated drinks are stored in a pressurised container? (2)
Answer:
(a) (i) Heterogeneous equilibrium: If the reactants and products of a reaction in equilibrium are in different phases, then it is called as heterogeneous equilibrium.

(b) (i) Factors influencing solubility:
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.

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 ceric sulphate is exothermic and the solubility decreases with increase of temperature. In the case of gaseous solute in liquid solvent the solubility 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 liquids as they are not compressible. However, the solubility of gases generally increases with increase of pressure.

(ii)

• The carbonated beverages contain CO₂ dissolved in them. To dissolve the CO₂ in these drinks, CO₂ gas is bubbled through them under high pressure.
• 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 37.
(a) (i) Explain – paper chromatography. (3)
(ii) What are stereo-isomerism? (2)
[OR]
(b) (i) Explain the homolytic fission of a covalent bond? (3)
(ii) Why chloroacetic acid is more acidic than acetic acid? (2)
Answer:
(a) (i) 1. It is an example of partition chromatography. A strip of paper acts as an adsorbent.
This method involves continues differential pardoning of components of a mixture between stationary and mobile phase. In paper chromatography, a special quality paper known as chromatographic paper is used. This paper act as a stationary phase.
2. A strip of chromatographic paper spotted at the base with the solution of the mixture is suspended in a suitable solvent which acts as the mobile phase. The solvent rises up and flows over the spot. The paper selectivity retains different components according to their different partition in the two phases where a chromatogram is developed.
3. The spots of the separated coloured components are visible at different heights from the position of initial spots on the chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent.

(ii) Stereo-isomerism: The isomers which have same bond connectivity but different arrangement of groups or atoms in space are known as stereoisomers. This phenomenon is known as stereoisomerism.

[OR]

(b) (i)

• Homolytic cleavage is the process in which a covalent bond breaks symmetrically in such way that fcach of the bonded atoms retains one electron.
• This type of cleavage occurs under high temperature or in the presence of UV light.
• In a compound containing non-‘polar covalent bond formed between atoms of similar electronegativity, in such molecules the cleavage of bonds results into free radicals.
• For example, ethane undergo homolytic fission to produce, two methyl free radicals.
  

(ii) Chloro acetic acid:

Chloro acetic acid has Cl-group and it has high electronegativity and shows -I effect. Therefore Cl-atom to facilitate the dissociation of O-H bond very fastly. Whereas in the – case of acetic acid, has CH₃ group and it shows +1 effect, therefore dissociation of O-H bond will be more difficult. Thus chloro acetic acid is stronger acid than acetic acid.

Question 38.
(a)
Identify A and B (major products).
(ii) Describe the mechanism of sulphonation of benzene. (3)
[OR]
(b) An organic compounds A of a molecular formula C₆H₆ which is simple aromatic hydrocarbon. A reacts with Cl₂ in presence of FeCl₃ to give B. B reacts with NaOH at 350°C and 300 atm pressure to give C. B again reacts with ammonia at 250°C and 50 atm pressure to give D. Identify A, B, C and D explain the reaction. (5)
Answer:

(ii) Step 1: Generation of SO3 electrophile: 2H₂SO₄ → H₃O + SO₃ + HSO₄
Step 2: Attack of the electrophilic on benzene ring to form Armenian ion:

Students can Download Computer Science Chapter 13 Introduction to Object-Oriented Programming Techniques Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 13 Introduction to Object-Oriented Programming Techniques

Samacheer Kalvi 11th Computer Science Introduction to Object-Oriented Programming Techniques Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

Question 1.
The term is used to describe a programming approach based on classes and objects is ……………….
(a) OOP
(b) POP
(c) ADT
(d) SOP
Answer:
(a) OOP

Question 2.
The paradigm which aims more at procedures ……………….
(a) Object Oriented Programming
(b) Procedural programming
(c) Modular programming
(d) Structural programming
Answer:
(b) Procedural programming

Question 3.
Which of the following is a user defined data type?
(a) class
(b) float
(c) int
(d) object
Answer:
(a) class

Question 4.
The identifiable entity with some characteristics and behaviour is ……………….
(a) class
(b) object
(c) structure
(d) member
Answer:
(b) object

Question 5.
The mechanism by which the data and functions are bound together into a single unit is known as ……………….
(a) Inheritance
(b) Encapsulation
(c) Polymorphism
(d) Abstraction
Answer:
(b) Encapsulation

Question 6.
Insulation of the data from direct access by the program is called as ……………….
(a) Data hiding
(b) Encapsulation
(c) Polymorphism
(d) Abstraction
Answer:
(a) Data hiding

Question 7.
Which of the following concept encapsulate all the essential properties of the object that are to be created?
(a) Class
(b) Encapsulation
(c) Polymorphism
(d) Abstraction
Answer:
(d) Abstraction

Question 8.
Which of the following is the most important advantage of inheritance?
(a) data hiding
(b) code reusability
(c) code modification
(d) accessibility
Answer:
(b) code reusability

Question 9.
“Write once and use it multiple time” can be achieved by ……………….
(a) redundancy
(b) reusability
(c) modification
(d) composition
Answer:
(b) reusability

Question 10.
Which of the following supports the transitive nature of data?
(a) Inheritance
(b) Encapsulation
(c) Polymorphism
(d) Abstraction
Answer:
(a) Inheritance

PART – 2
II. Answers to all the questions

Question 1.
How is modular programming different from procedural programming paradigm?
Answer:
Modular programming:

• Emphasis on algorithm rather than data.
• Programs are divided into individual modules.
• Each modules are independent of each other and have their own local data.
• Modules can work with its own data as well as with the data passed to it.

Procedural programming:

• Programs are organized in the form of subroutines or sub programs.
• All data items are global.
• Suitable for small sized software application.
• Difficult to maintain and enhance the program code as any change in data type needs to be propagated to all subroutines that use the same data type.

Question 2.
Differentiate classes and objects.
Answer:
Class:

• Class is a blue print or template from which objects are created.
• Class doesn’t allocate memory when it is created.
• Class is a logical entity.

Object:

• Object is an instance of a class.
• Objects allocate memory when it is created.
• Object is a physical entity.

Question 3.
What is polymorphism?
Answer:
Polymorphism is the ability of a message or function to be displayed in more than one form.

Question 4.
How is encapsulation and abstraction are interrelated?
Answer:
Abstraction means giving only essential things and hiding unnecessary details. Encapsulation is the binding of data members and methods together in a capsule to avoid accidental changes to data from external users, i.e., encapsulation is the bundling of related algorithms and data.

Question 5.
Write the disadvantages of OOP.
Answer:

1. Size: Object Oriented Programs are much larger than other programs.
2. Effort: Object Oriented Programs require a lot of work to create.
3. Speed: Object Oriented Programs are slower than other programs, because of their size.

PART – 3
III. Answers to all the questions

Question 1.
What is paradigm? Mention the different types of paradigm.
Answer:
Paradigm means organizing principle of a program. It is an approach to programming. There are different approaches available for problem solving using computer. They are Procedural programming, Modular Programming and Object Oriented Programming.

Question 2.
Write a note on the features of procedural programming.
Answer:
Important features of procedural programming

1. Programs are organized in the form of subroutines or sub programs
2. All data items are global
3. Suitable for small sized software application
4. Difficult to maintain and enhance the program code as any change in data type needs to be propagated to all subroutines that use the same data type. This is time consuming.
5. Example: FORTRAN and COBOL.

Question 3.
List some of the features of modular programming.
Answer:
Important features of Modular programming:

1. Emphasis on algorithm rather than data
2. Programs are divided into individual modules
3. Each modules are independent of each other and have their own local data
4. Modules can work with its own data as well as with the data passed to it.
5. Example: Pascal and C.

Question 4.
What do you mean by modularization and software reuse?
Answer:

1. Modularization : where the program can be decomposed into modules.
2. Software re – use : where a program can be composed from existing and new modules.

Question 5.
Define information hiding.
Answer:
Encapsulation is the most striking feature of a class. The data is not accessible to the outside world, and only those functions which are wrapped in the class can access it. This encapsulation of data from direct access by the program is called data hiding or information hiding.

PART – 4
IV. Answers to all the questions

Question 1.
Write the differences between Object Oriented Programming and Procedural Programming
Answer:
Object Oriented Programming:

• Emphasizes on data rather than algorithm.
• Data abstraction is introduced in addition to procedural abstraction.
• Data and its associated operations are grouped in to single unit.
• Programs are designed around the data being operated.
• Example: C++, Java, VB.Net, Python

Procedural Programming:

• Programs are organized in the form of subroutines or sub programs.
• All data items are global.
• Suitable for small sized software application.
• Difficult to maintain and enhance the program code as any change in data type needs to be propagated to all subroutines that use the same data type.
• Example: FORTRAN and COBOL

Question 2.
What are the advantages of OOPs?
Answer:
Re – usability : “Write once and use it multiple times” you can achieve this by using class. Redundancy: Inheritance is the good feature for data redundancy. If you need a same functionality in multiple class you can write a common class for the same functionality and inherit that class to sub class.

Easy Maintenance : It is easy to maintain and modify existing code as new objects can be created with small differences to existing ones.

Security : Using data hiding and abstraction only necessary data will be provided thus maintains the security of data.

Question 3.
Write a note on the basic concepts that supports OOPs?
Answer:
The Object Oriented Programming has been developed to overcome the drawbacks of procedural and modular programming. It is widely accepted that object – oriented programming is the most important and powerful way of creating software.

The Object – Oriented Programming approach mainly encourages:

1. Modularization: where the program can be decomposed into modules.
2. Software re – use: where a program can be composed from existing and new modules.

Main Features of Object Oriented Programming:

1. Data Abstraction.
2. Encapsulation.
3. Modularity.
4. Inheritance.
5. Polymorphism.

Encapsulation:
The mechanism by which the data and functions are bound together into a single unit is known as Encapsulation. It implements abstraction. Encapsulation is about binding the data variables and functions together in class. It can also be called data binding. Encapsulation is the most striking feature of a class.

The data is not accessible to the outside world, and only those functions which are wrapped in the class can access it. These functions provide the interface between the object’s data and the program. This encapsulation of data from direct access by the program is called data hiding or information hiding.

Data Abstraction:
Abstraction refers to showing only the essential features without revealing background details. Classes use the concept of abstraction to define a list of abstract attributes and function which operate on these attributes. They encapsulate all the essential properties of the object that are to be created. The attributes are called data members because they hold information. The functions that operate on these data are called methods or member function.

Modularity:
Modularity is designing a system that is divided into a set of functional units (named modules) that can be composed into a larger application.

Inheritance:
Inheritance is the technique of building new classes (derived class) from an existing Class (base class). The most important advantage of inheritance is code reusability.

Polymorphism:
Polymorphism is the ability of a message or function to be displayed in more than one form.

Samacheer kalvi 11th Computer Science Introduction to Object-Oriented Programming Techniques Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
In procedural programming all data items are ……………….
(a) Cobol
(b) global
(c) fortan
(d) class
Answer:
(b) global

Question 2.
Class represents a group of similar ……………….
(a) objects
(b) modules
(c) arrays
(d) data
Answer:
(a) objects

Question 3.
………………. is an example of object oriented programming.
(a) Python
(b) Java
(c) VB.Net
(d) All the above
Answer:
(d) All the above

Question 4.
………………. refers to showing only the essential features without revealing background details.
(a) Redundancy
(b) Encapsulation
(c) Abstraction
(d) Inheritance
Answer:
(c) Abstraction

Question 5.
………………. is about binding the data variables and functions together in class.
(a) Data abstraction
(b) Modularization
(c) Redundancy
(d) Encapsulation
Answer:
(d) Encapsulation

PART – 2
II. Very Short Answers

Question 1.
What is procedural programming?
Answer:
Procedural means a list of instructions were given to the computer to do something. Procedural programming aims more at procedures. This emphasis on doing things.

Question 2.
What is a class?
Answer:
A Class is a construct in C++ which is used to bind data and its associated function together into a single unit using the encapsulation concept. Class is a user defined data type. Class represents a group of similar objects.

Question 3.
What is modularity?
Answer:
Modularity is designing a system that is divided into a set of functional units (named modules) that can be composed into a larger application.

Question 4.
What are the main features of OOP?
Answer:
Main Features of Object Oriented Programming:

1. Data Abstraction
2. Encapsulation
3. Modularity
4. Inheritance
5. Polymorphism

Question 5.
What is redundancy?
Answer:
Inheritance is the good feature for data redundancy. If you need a same functionality in multiple class you can write a common class for the same functionality and inherit that class to sub class.

PART – 3
III. Short Answers

Question 1.
Write about objects.
Answer:
Objects: Represents data and its associated function together into a single unit. Objects are the basic unit of OOP. Basically an object is created from a class. They are instances of class also called as class variables. An identifiable entity with some characteristics and behaviour is called object.

Question 2.
What is Encapsulation and data binding?
Answer:
The mechanism by which the data and functions are bound together into a single unit is known as Encapsulation. Encapsulation is the most striking feature of a class. The data is not accessible to the outside world, and only those functions which are wrapped in the class can access it. These functions provide the interface between the object’s data and the program. This encapsulation of data from direct access by the program is called data hiding or information hiding.

PART – 4
IV. Explain in Detail

Question 1.
Explain about features of object oriented programming.
Answer:
Main Features of Object Oriented Programming

1. Data Abstraction
2. Encapsulation
3. Modularity
4. Inheritance
5. Polymorphism

Encapsulation:
1. The mechanism by which the data and functions are bound together into a single unit is known as Encapsulation. It implements abstraction.

2. Encapsulation is about binding the data variables and functions together in class. It can also be called data binding.

3.. Encapsulation is the most striking feature of a class. The data is not accessible to the outside world, and only those functions which are wrapped in the class can access it. These functions provide the interface between the object’s data and the program. This encapsulation of data from direct access by the program is called data hiding or information hiding.

Data Abstraction:
Abstraction refers to showing only the essential features without revealing background details. Classes use the concept of abstraction to define a list of abstract attributes and function which operate on these attributes.

They encapsulate all the essential properties of the object that are to be created. The attributes are called data members because they hold information. The functions that operate on these data are called methods or member function.

Modularity:
Modularity is designing a system that is divided into a set of functional units (named modules) that can be composed into a larger application.

Inheritance:
Inheritance is the technique of building new classes (derived class) from an existing Class (base class). The most important advantage of inheritance is code reusability.

Polymorphism:
Polymorphism is the ability of a message or function to be displayed in more than one form.

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Samacheer Kalvi 11th Computer Science Arrays and Structures Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

Question 1.
Which of the following is the collection of variables of the same type that an referenced by a common name?
(a) int
(b) float
(c) Array
(d) class
Answer:
(c) Array

Question 2.
Array subscripts always starts with which number?
(a) – 1
(b) 0
(c) 2
(d) 3
Answer:
(b) 0

Question 3.
int age[ ]={6, 90, 20, 18, 2}; How many elements are there in this array?
(a) 2
(b) 5
(c) 6
(d) 4
Answer:
(b) 5

Question 4.
cin >> n[3]; To which element does this statement accepts the value?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(c) 4

Question 5.
By default, a string ends with which character?
(a) \o
(b) \t
(c) \n
(d) \b
Answer:
(a) \o

PART – 2
II. Answers to all the questions

Question 1.
What is Traversal in an Array?
Answer:
Accessing each element of an array at least once to perform any operation is known as “Traversal”. Displaying all the elements in an array is an example of “traversal”.

Question 2.
What is Strings?
Answer:
A string is defined as a sequence of characters where each character may be a letter, number or a symbol. Each element occupies one byte of memory. Every string is terminated by a null (‘\0’, ASCII code 0) character which must be appended at the end of the string.

Question 3.
What is the syntax to declare two – dimensional array.
Answer:
The declaration of a 2 – D array is
data – type array_name[row – size] [col – size];
In the above declaration, data-type refers to any valid C++ data – type, array _ name refers to the name of the 2 – D array, row – size refers to the number of rows and col-size refers to the number of columns in the 2 – D array.

PART – 3
III. Answers to all the questions

Question 1.
Define an Array. What are the types?
Answer:
“An array is a collection of variables of the same type that are referenced by a common name”. An array is also a derived datatype in C++.
There are different types of arrays used in C++. They are:

1. One – dimensional arrays
2. Two – dimensional arrays
3. Multi – dimensional arrays

Question 2.
Write a note on Array of strings.
Answer:
An array of strings is a two – dimensional character array. The size of the first Index (rows) denotes the number of strings and the size of the second index (columns) denotes the maximum length of each string. Usually, array of strings are declared in such a way to accommodate the null character at the end of each string. For example, the 2 – D array has the declaration:
char name [7][10];
In the above declaration,
No. of rows = 7;
No. of columns =10;
We can store 7 strings each of maximum length 10 characters.

Question 3.
Write a C++ program to accept and print your name.
Answer:
#include
using namespace std;
int main()
{
charname[5];
cout<< “Enter your name:”; cin >>name;
cout<< “My name is”<< name;
}
Output:
Enter your name: PRIYA
My name is PRIYA

PART – 4
IV. Answers to all the questions 

Question 1.
Write a C++ program to find the difference between two matrix.
Answer:

Output
Enter 3*3 Array 1 Elements :
10 11 12
13 14 15
16 17 18

Enter 3*3 Array 2 Elements :
123
456
789

Subtracting array (array1 – array2)
Result of Array 1 – Array2 is :
9 9 9
9 9 9
9 9 9

Question 2.
How will you pass two dimensional array to a function explain with example.
Answer:
Passing 2”D array to a function

Output:
Displaying Values
3 4
9 5
7 1

PART – 1
I. Choose the correct answer

Question 1.
The data elements in the structure are also known as ………………..
(a) objects
(b) members
(c) data
(d) records
Answer:
(a) objects

Question 2.
Structure definition is terminated by
(a) :
(b) }
(c) ;
(d) ::
Answer:
(c) ;

Question 3.
What will happen when the structure is declared?
(a) it will not allocate any memory
(b) it will allocate the memory
(c) it will be declared and initialized
(d) it will be only declared
Answer:
(b) it will allocate the memory

Question 4.
What is the output of this program?
Answer:
#include
#include
using namespace std;
int main()
{

struct student
{
int n;
char name[10];
};
student s;
s.n = 123;
strcpy(s.name, “Balu”);
cout <<s.n;
cout<< s.name << endl;

return 0; }
(a) 123Balu
(b) BaluBalu
(c) Balul23
(d) 123 Balu
Answer:
(d) 123 Balu

Question 5.
A structure declaration is given below,
struct Time
{
int hours;
int minutes;
int seconds;
} t;
Using above declaration which of the following refers to seconds.
(a) Time.seconds
(b) Time::seconds
(c) seconds
(d) t. seconds
Answer:
(d) t. seconds

Question 6.
What will be the output of this program?
#include
using namespace std;
struct Shoe Type
{
string name;
double price;
};
int main()
{
ShoeType shoel, shoe2;
shoe 1.name = “Adidas”;
shoel.price = 9.99;
cout << shoel.name << “#” << shoel.price<<endl;
shoe2 – shoel;
shoe2.price = shoe2.price / 9;
cout << shoe2.name << “#”<< shoe2.price;
return 0;
(a) Adidas # 9.99; Adidas # 1.11
(b) Adidas # 9.99; Adidas # 9.11
(c) Adidas # 9.99; Adidas # 11.11
(d) Adidas # 9.11; Adidas # 11.11
Answer:
(a) Adidas # 9.99; Adidas # 1.11

Question 7.
Which of the following is a properly defined structure?
(a) struct {int num;}
(b) struct sum {int num;}
(c) struct sum int sum;
(d) struct sum {int num;};
Answer:
(d) struct sum {int num;};

Question 8.
A structure declaration is given below,
struct employee
{
int empno;
char ename[10];
} e[5];
Using above declaration which of the following statement is correct?
(a) cout << e[0].empno << e[0].ename;
(b) cout << e[0].empno << ename;
(c) cout << e[0]->empno << e[0] → ename;
(d) cout << e.empno << e.ename;
Answer:
(a) cout << e[0].empno << e[0].ename;

Question 9.
Which of the following cannot be a structure member?
(a) Another structure
(b) Function
(c) Array
(d) variable of double datatype
Answer:
(b) Function

Question 10.
When accessing a structure member, the identifier to the left of the dot operator is the name of …………………
(a) structure variable
(b) structure tag
(c) structure member
(d) structure function
Answer:
(c) structure member

PART – 2
II. Answer to all the questions

Question 1.
Define structure. What is its use?
Answer:
Structure is a user – defined which has the combination of data items with different data types. This allows to group of variables of mixed data types together into a single unit. The structure provides a facility to store different data types as a part of the same logical element in one memory chunk adjacent to each other.

Question 2.
To store 100 integer number which of the following is good to use?
Array or Structure. State the reason.
Answer:
Array is good to use.
Reasons:

1. All 100 numbers are integer type.
2. Array index helps to access the numbers quickly.

Question 3.
What is the error in the following structure definition?
struct employee {inteno ;chamame [20] ;char dept;}
Employee e1,e2;
Answer:
Errors:

1. ‘i’ is missing.
2. Spaces are missing at two places
3. Structure name given wrongly.

Corrected structure:
struct employee {int eno; char ename [20];

char dept;}
employee e, e2;

Question 4.
Write a structure definition for the structure student containing examno, name and an array for storing five subject marks.
Answer:
struct student
{
long examno;
char name [50];
int marks [5];
};

Question 5.
Why for pacing a structure to a function call by reference is advisable to us?
Answer:
In this method of passing the structures to functions, the address of a structure variable /object is passed to the function using address of(&) operator. So any change made to the contents of structure variable inside the function are reflected back to the calling function.

PART – 3
III. Answer to all the questions

Question 1.
How will you pass a structure to a function?
Answer:
A structure variable can be passed to a function in a similar way of passing any argument that is of built – in data type.
If the structure itself is an argument, then it is called “call by value”. If the reference of the structure is passed as an argument then it is called, “call by reference”.

Question 2.
The following code sums up the total of all students name starting with ‘S’ and display it. Fill in the blanks with required statements.
Answer:
struct student {int exam no, lang, eng, phy, che, mat, csc, total; char name[15];};
int main()
{
student s[20];
for(int i = 0;i<20;i++)
{
………………….. //accept student details
}
for(int i=0;i<20;i++)
{
………………….. //check for name starts with letter “S”
………………….. // display the detail of the checked name
}
return 0;
}
Answer:

Question 3.
How to access members of a structure? Give example.
Answer:
Once objects of a structure type are declared, their members can be accessed directly. The syntax for that is using a dot(.) between the object name and the member name.
Example: student.name;
if the members are a pointer type, then is used to access the members.
Let name be a character pointer in student like char*name
It can be accessed student → name,
struct
{

long rollno;
int age;
float weight;
student;

}
The student can be referred as reference name to the above structure and the elements can be accessed like student.roll no, student.age, and student.weight.

Question 4.
Write the syntax and an example for structure.
Answer:
Structure is declared using the keyword ‘struct’. The syntax of creating a structure is given below.
struct structure_name {

type member_name 1;
type member_name2;

} reference_name;
An optional field reference_name can be used to declare objects of the structure type directly.
Example:
struct Student
{

long rollno;
int age;
float weight;

};

Question 5.
What is called anonymous structure? Give an example.
Answer:
A structure without a name/tag is called anonymous structure.
Example:
struct
{
long rollno;
int age;
float weight;
};
The student can be referred as reference name to the above structure and the elements can be accessed like student.rollno, student.age and student.weight.

PART – 4
IV. Answer to all the questions

Question 1.
Explain array of structures with example.
Answer:
A class may contain many students. So, the definition of structure for one student can also be extended to all the students. If the class has 20 students, then 20 individual structures are required. For this purpose, an array of structures can be used. An array of structures is declared in the same way as declaring an array with built – in data types like int or char.
Program that reads the details of 20 students and prints the same.

Output:
Enter the details of 20 students
Enter the details for student 1

Enter the age:
18
Enter the height:
160.5
Enter the weight:
46.5
Enter the details for student2
Enter the age:
18

Question 2.
Explain call by value with respect to structure.
Answer:
When a structure is passed as argument to a function using call by value method, any change made to the contents of the structure variable inside the function to which it is passed do not affect the structure variable used as an argument.
#include
using namespace std;
struct Employee
{
char name[50];
int age;
float salary;
};
void printData(Employee); // Function declaration
int main()
{

Employee p;
cout << “Enter Full name:”; cin >> p.name;
cout << “Enter age:”; cin >> p.age;
cout << “Enter salary:”; cin >> p.salary;
// Function call with structure variable as an argument
printData(p);
return 0;

}
void printData(Employee q)
{

cout << “\nDisplaying Information.” << endl;
cout << “Name:” << q.name << endl;
cout <<”Age:” << q.age << endl;
cout << “Salary:” << q.salary;

}

Output:
Enter Full name: Kumar
Enter age : 55
Enter salary : 34233.4
Displaying Information.
Name : Kumar
Age : 55
Salary : 34233.4
In the above example, a structure named Employee is declared and used. The values that are entered into the structure are name, age and salary of a Employee are displayed using a function named printData(). The argument for the above function is the structure Employee. The input can be received through a function named readData().

Question 3.
Write a C++ program to add two distances using the following structure definition, struct Distance!
int feet;
float inch;
}d1, d2, sum;
Answer:
int main()
{

cout << “Enter 1st distance:” << end1;
cout << “Enter feet:”; cin >> d1.feet;
cout << “Enter inch:”; cin >> d1.inch;
cout << “\n information for 2nd distance:” << end1;
cout << “Enter feet:”; cin >> d2.feet;
cout << “Enter inch:”; cin >> d2.inch;
sum.feet = d1 . feet + d2.feet;
sum.inch = d1.inch + d2.inch;
if (sum.inch > 12)
{
++ sum.feet;
sum.inch = 12;
}
cout << end1; “Sum of distance =” << sum.feet << “feet” << sum.inch << “inches”;
return 0;

}

Output:
Enter 1 st distance
Enter feet: 6
Enter inch: 3.4
Enter 2nd distance
Enter feet: 5
Enter inch: 10.2
Sum of distances = 12 feet 1.6 inches

Question 4.
Write a C++ program to declare and accept an array of professors. Display the details of the department= “COMP.SCI” and the name of the professors start with ‘A’. The structure “college” should contain the following members.
prof id as integer
name and Department as character array
Answer:

Output
Professor 1
Enter id of professor: 100
Enter name of the professor: John
Enter name of the department: CS

Professor 2
Enter id of professor: 101
Enter name of the professor: Janardhan
Enter name of the department: ECE

Professor 3
Enter id of professor: 102
Enter name of the professor: Albert
Enter name of the department: CS

Professor 4
Enter id of professor: 103
Enter name of the professor: Asha
Enter name of the department: CS

Professor 5
Enter id of professor: 104
Enter name of the professor: Han
Enter name of the department: EEE

Name of the professor starting with A
Professor 1
Enter id of professor: 102
Enter name of the professor: Albert
Enter name of the department: CS

Professor 2.
Enter id of professor: 103
Enter name of the professor: Asha
Enter name of the department: CS

Question 5.
Write the output of the following C++ program
Answer:
#include
#include
#include
#include
using namespace std;
struct books {
char name[20], author[20];
}a[50];
int main()

Output:
Details of Book No. 1
Book name : Programming
Book author : Dromy
Details of Book No. 2
Book Name : C++ Programming
Book Author : Bjame Stroustrup
1 Iprogrammingl Dromy
2| C++ Programming | Bjame Stroustmp

Question 6.
Write the output of the following C++ program
Answer:

Output
First student
roll no : 1
name : Brown
Phone no. : 123443
Second Student
roll no : 2
name : Sam
Phone no. : 1234567
Third Student
roll no : 3
name : Addy
Phone no. : 1234597844

Question 7.
Debug the error in the following program
Answer:

Corrected Program:

Samacheer kalvi 11th Computer Science Arrays and Structures Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
The size of the array is referred to as its ………………..
(a) dimension
(b) direction
(c) location
(d) space
Answer:
(a) dimension

Question 2.
The subscript in bracket can be a variable, a constant or an expression to ………………..
(a) character
(b) integer
(c) long double
(d) float
Answer:
(b) integer

Question 3.
Displaying all the elements in an array is an example of ………………..
(a) memory allocation
(b) call by reference
(c) traversal
(d) none of these
Answer:
(c) traversal

Question 4.
Syntax of character array declaration is ………………..
(a) int array char name [size];
(b) char array [ ];
(c) char_name[size];
(d) char array – name[size];
Answer:
(d) char array – name[size];

Question 5.
During ……………….. the array of elements cannot be initialized more than its size.
(a) declaration
(b) initialization
(c) assigning
(d) execution
Answer:
(b) initialization

Question 6.
2 – D array memory representation have ……………….. types.
(a) 2
(b) 3
(c) 4
(d) only 1
Answer:
(a) 2

Question 7.
Pass an array to a function in C++, the function needs the array name as ………………..
(a) a function
(b) an argument
(c) global object
(d) string
Answer:
(b) an argument

Question 8.
Objects declared along with structure definition are called ………………..
(a) structure
(b) nested structure
(c) global objects
(d) memory
Answer:
(c) global objects

Question 9.
A structure without a name tag is called ………………..
(a) homogenous structure
(b) anonymous structure
(c) array of structure
(d) dynamic memory
Answer:
(b) anonymous structure

Question 10.
Array of structure is declared in the same way as declaring an array with ………………..
(a) built – in data type
(b) data type
(c) undefined
(d) none of these
Answer:
(a) built – in data type

PART – 2
II. Very Short Answers

Question 1.
What is the formula to calculate memory space allocated for an array?
Answer:
Number of bytes allocated for type of array x Number of elements.

Question 2.
Write a C++ program to check palindrome or not using array.
Answer:
Program to check palindrome or not

Output:
Enter a string : madam The String is palindrome

Question 3.
Write about returning structures from functions.
Answer:
A structure can be passed to a function through its object. Therefore, passing a structure to a function or passing a structure object to a function is the same because structure object represents the structure. Like a normal variable, structure variable(structure object) can be passed by value or by references / addresses. Similar to built-in data types, structures also can be returned from a function.

Question 4.
What is the output of the following program?
Answer:

Output:
Enter a string: welcome to C++ programming
You entered char array: welcome to C++ programming

Question 5.
What is global object?
Answer:
Objects declared along with structure definition are called global objects.

Question 6.
What is the condition of structure assignments?
Answer:
Structure assignment is possible only if both structure variables/objects are same type.

Question 7.
Why structures are usually passed by reference method?
Answer:
Structures are usually passed by reference method because it saves the memory space and executes faster.

Question 8.
What is the size of the following highlighted variable in terms of byte if it is compiled in dev C++?
Answer:
struct A{ float f[3]; char ch[5];long double d;};
struct B{Aa; int arr[2][3];}b[3]

Question 9.
Is the following snippet is fully correct. If not identify the error.
Answer:
struct suml{ int n1,n2;}s1;
struct sum2{int n1,n2}s2;
cin >> s1.n1 >> s1.n2; s2=s1;

Question 10.
Differentiate array and structure.
Answer:
Array:

• An array is a collection of variables of same data type.
• Array data are accessed using index.
• Array allocates static memory
• Array element access takes lesser time.

Structure:

• A structure is a collection of variables of different data type.
• Structure elements are accessed using operator.
• Structures allocate dynamic memory.
• Structure elements takes more time.

Question 11.
What are the different ways to initialize the structure members?
Answer:
Values can be assigned to structure elements similar to assigning values to variables.
Example
balu.rollno= “702016”;
balu.age= 18;
balu.weight= 48.5;
Also, values can be assigned directly as similar to assigning values to Arrays.
balu={702016, 18, 48.5};

Question 12.
What is wrong with the following C++ declarations?
Answer:
A. struct point (double x, y )
B. struct point { double x, double y };
C. struct point { double x; double y }
D. struct point { double x; double y;};
E. struct point { double x; double y;}

PART – 3
III. Short Answers

Question 1.
Write about initialization of 2 – D array.
Answer:
The array can be initialized in more than one way at the time of 2-D array declaration.
For example
int matrix[4][3] = {
{10,20,30},// Initializes row 0
{40,50,60},// Initializes row 1
{70,80,90},// Initializes row 2
{100,110,120}// Initializes row 3
}; .
int matrix[4][3] = {10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120};
Array’s row size is optional but column size is compulsory.

Question 2.
What is row major order and column major order?
Answer:
In row – major order, all the elements are stored row by row in continuous memory locations, that is, all the elements in first row, then in the second row and so on. The memory representation of row major order is as shown below;
Row major order:

Column major order:

Question 3.
What is array of strings?
Answer:
An array of strings is a two – dimensional character array. The size of the first index (rows) denotes the number of strings and the size of the second index (columns) denotes the maximum length of each string. Usually, array of strings are declared in such a way to accommodate the null character at the end of each string. For example, the 2 – D array has the declaration: char Name[6][10];
In the above declaration, the 2 – D array has two indices which refer to the row size and column size, that is 6 refers to the number of rows and 10 refers to the number of columns.

Question 4.
Write a C++ program to access array elements using pointer with sample output.
Answer:
#include
using namespace std;
int main()
{

int data [5];
cout << “Enter elements:”;
for(int i = 0; i <5; ++i) cin >> data [i];
cout << “You entered:”;
for (int i = 0; i < 5;++i)
cout << endl << *(data +i);

}
return 0;
Output:
Enter elements:
1
2
3
5
4
You entered:
1
2
3
5
3

Question 5.
Write a program to assign data to members of a structure variable and display the contents.
Answer:
C++ Program that assigns data to members of a structure variable and displays the contents

Output:
Enter Full name:
Ezhil Enter age:
27
Enter salary:
40000.00
Displaying Information.
Name: Ezhil
Age: 27
Salary: 40000.00

Question 6.
What is called nested structure? Give example.
Answer:
The structure declared within another structure is called a nested structure. Nested structures act as members of another structure and the members of the child structure can be accessed as parent structure name. Child structure name. Member name, struct dob
{
int date;
char month[3];
int year;
} ;
Values can be assigned to this structure as follows.
dob = {25, “DEC”, 2017}

Question 7.
Rewrite the following program after removing the syntactical error(s),if any. Underline each, correction.
Answer:

Question 8.
What is the difference among the following two programs?
Answer:
(a) #include
struct point { double x; double y;};
int main() {
struct point test;
test.x = .25; testy = .75;
cout << test.x << test.y;
return 0;
}

(b) #include
struct { double x; double y;} Point;
int main(void) {
Point test={.25,.75};
return 0;
}

(a) Output

1. 0.250.75
2. Named structure point is created.

(b)

1. No output: Error
2. Anonymous structure point is created.

Question 9.
For the following structure definition write the user defined function to accept data through keyboard.
struct date{ int dd,mm,yy};
struct item {int item id;char name[10];float price;date datemanif;}
Answer:
void accept (item & i)
{

cout << “\n Enter the Item id”; cin >> i.id;
cout << “\n Enter the item name:”; cin >> i.name;
cout << “\n Enter the item price:”; cin >> i.price;
cout << “\n Enter the item manufacturing day:”; cin >> i.date.dd;
cout << “\n Enter the item manufacturing month:”; cin >> i.date.mm;
cout << “\n Enter the item manufacturing year:”; cin >> i.date.yy;

}

Question 10.
Write a user defined function to return the structure after accepting value through keyboard. The structure definition is as follows: struct Item{int item no;float price;};
Answer:
item accept (item i)
{

cout << “\n Enter the Item No:”; cin >> i.no;
cout << “\n Enter the Item Price:”; cin >> i.price;
return i;

}

PART – 4
IV. Explain in Detail

Question 1.
Write C++ program to find transpose of a matrix.
Answer:
C++ Program to find transpose of a matrix

Output:
Enter rows and columns of matrix:
2 3
Enter elements of matrix
Enter elements of a1 1 = 1
Enter elements of a1 2 = 2
Enter elements of a1 3 = 9
Enter elements of a2 1 = 0
Enter elements of a2 2 = 4
Enter elements of a2 3 = 7
Entered matrix:
1 2 9 0 4 7
Transpose of matrix:
10 2 4 9 7

Question 2.
Write C++ program to sort words in dictionary order.
C++ Program to find transpose of a matrix
Answer:

Output:
Enter 10 words:
Kalai
Artqueen
Visalatchi
Jaya
Dhanush
In lexicographical order:
Artqueen
Dhanush
Jaya
Kalai
Visalatchi

Question 3.
Explain memory representation of 2 – D array.
Answer:
Normally, the two – dimensional array can be viewed as a matrix. The conceptual view of a 2 – D array is shown below:
int A[4][3];

In the above example, the 2 – D array name A has 4 rows and 3 columns.
Like one – dimensional, the 2 – D array elements are stored in continuous memory. There are two types of 2 – D array memory representations. They are:

1. Row – Major order
2. Column – Major order

For example
intA[4][3] = {

{8,6,5},
{2,1,9},
{3,6,4},
{4,3,2},

Row major:

Column major order

Question 4.
Let an organisation have three employees. If we want to read and print all their details, write a C++program for the following structure definition
struct Employee
{

char name [50];
int age;
float salary;

};
Answer:

Output:
Enter the details of 3 employees:
Enter the details of Employee 1
Enter name:
Lilly

Enter age:
42

Enter salary:
40000.00

Enter the details of Employee 2
Enter name:
Aster

Enter age:
38

Enter salary:
60000.00

Enter the details of Employee 3
Enter name:
Jasmine

Enter age:
45

Enter salary:
80000.00

Displaying Information:
The details of Employee 1
Name : Lilly
Age : 42
Salary : 40000.00

The details of Employee 2
Name : Aster
Age : 38
Salary : 60000.00

The details of Employee 3
Name : Jasmine
Age : 45
Salary : 80000.00

Question 5.
Explain returning structures from functions with an example.
Answer:
A structure can be passed to a function through its object. Therefore, passing a structure to a function or passing a structure object to a function is the same because structure object represents the structure. Like a normal variable, structure variable (structure object) can be passed by value or by references / addresses. Similar to built – in data types, structures also can be returned from a function.

Output:
Enter Employee Id : 10
Enter Employee Name : Ajay
Enter Employee Age : 25
Enter Employee Salary : 15000
Employee Id : 10
Employee Name : Ajay
Employee Age : 25
Employee Salary : 15000

Question 6.
Write output of the following program.
Answer:
Output:
Enter the age:
18
Enter the height:
160.5
Enter the weight:
46.5
The Date of birth Enter the day:
25
Enter the month:
NOV
Enter the year:
2017
The values entered for Age, height and weight are
18 160.5 46.5
His date of Birth is:
25 – NOV – 2017

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TN State Board 11th Chemistry Previous Year Question Paper March 2019 English Medium

Time: 2½ Hours
Total Score: 90 Marks

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I. II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail. Draw diagrams wherever necessary.

Part- I

Answer all the Questions. [Answers are in Bold] [15 x 1 = 15]

Choose the most appropriate answer from the given four alternatives and write the option code and the corresponding answer.

Question 1.
Many of the organic compounds are inflammable because of its
(a) Vander Waal’s force
(b) Co-ordinate nature
(c) Covalent nature
(d) Ionic nature
Answer:
(c) Covalent nature

Question 2.
When Ang is negative in chemical equilibrium reaction, then ..
(a) K_p < K_c
(b) K_p = 1 /K_c
(c) K_p = K_c (RT)^-vc
(d) K_p > K_c
Answer:
(c) K_p = K_c (RT)^-vc

Question 3.
Find A in the following equation.
\(\mathrm{CaO}+3 \mathrm{C} \stackrel{3273 \mathrm{K}}{\longrightarrow} \mathrm{A}+\mathrm{CO}\)
(a) CaC₂
(b) CO₂
(c) Ca
(d) Ca₂O
Answer:
(d) Ca₂O

Question 4.
Splitting of spectral lines in an electric field is called
(a) Compton effect
(b) Stark effect
(c) Zeeman effect
(d) Shielding effect
Answer:
(a) Compton effect

Question 5.
Which of the following species does not exert a resonance effect?
(a) C₆H₅NH₂
(b) C₆H₅NH₃
(c) C₆H₅OH
(d) C₆H₅CI
Answer:
(b) C₆H₅NH₃

Question 6.
Match the following:
Compound – Uses
1. Chloro picrin (i) Detection of primary amine
2. Methyl Isocyanide (ii) DDT
3. Chloro benzene (iii) Paint remover
4. Methylene chloride (iv) Soil sterilizer
(a) 1- (iv), 2 – (iii), 3 – (ii), 4 – (i)
(b) 1- (Hi), 2 – (iv), 3 – (ii), 4 – (i)
(c) 1- (i), 2 – (ii), 3 – (iv), 4 – (iii)
(d) 1- (iv), 2 – (i), 3 – (ii), 4 – (iii)
Answer:
(d) 1 – (iv), 2 – (i), 3 – (ii), 4 – (iii)

Question 7.
Use of hot air balloon in the meteorological observatory is an application of
(a) Kelvin’s Law
(b) Brown’s Law
(c) Boyle’s Law
(d) Newton’s Law
Answer:
(b) Brown’s Law

Question 8.
What is the pH of rain water?
(a) 5.6
(b) 4.6
(c) 6.5
(d) 7.5
Answer:
(a) 5.6

Question 9.
Which compound is named as “Blue John” among the following compounds?
(a) Ca₃(PO₄)₂
(b) CaO
(c) CaH₂
(d) CaF₂
Answer:
(d) CaF₂

Question 10.
The element with positive electron gain enthalpy is ………………………… .
(a) Argon
(b) Fluorine
(c) Hydrogen
(d) Sodium
Answer:
(d) Sodium

Question 11.
Which of the following mMecule does not contain n bond?
(a) CO₂
(b) H₂O
(c) SO₂
(d) NO₂
Answer:
(c) SO₂

Question 12.
Which of the following compound has same percentage of Carbon as that of ethylene (C₂H₄)?
(a) benzene
(b) ethane
(c) propene
(d) ethyne
Answer:
(a) benzene

Question 13.
The SI unit of Molar heat capacity is
(a) JK^-1 mol^-1
(b) kJ mol⁺¹
(c) kJ mol^-1
(d) cm
Answer:
(c) kJ mol^-1

Question 14.
What percentage of solution of H202 is called as “100-Volume” H202?
(a) 15%
(b) 50%
(c) 20%
(d) 30%
Answer:
(a) 15%

Question 15.
Osmotic pressure (n) of a solution is given by the relation
(a) πRT = n
(b) V = πnRT
(c) π = nRT
(d) πV = nRT
Answer:
(b) V = πnRT

Part-II

Answer any six of the following questions. Question no. 24 is compulsory. [6 x 2 = 12]

Question 16.
State and explain Pauli’s Exclusion Principle.
Answer:
Pauli’s exclusion principle states that “No two electrons in an atom can have the same set of values of all four quantum numbers”.
Illustration: H(Z = 1) Is¹.

One electron is present in hydrogen atom, the four quantum numbers are n = 1, l = 0, m = 0 and s = + ½. For helium Z = 2. He: Is². In this one electron has the quantum number same as that of hydrogen, n = 1, l = 0 m = 0 and s = + ½. For other electron, fourth quantum number is different, i.e. n = 1, l = 0, m = 0 and s = – ½.

Question 17.
Define – Valency.
Answer:
The valency of an element may be defined as the combining capacities of elements. The electrons present in the outermost shell are called valence electrons and these electrons determine the valency of the atom.

Question 18.
What are ideal gases?
Answer:

• A gas that follows Boyle’s law, Charles’ law and Avogadro law strictly is called an ideal gas. It is assumed that intermolecular forces are not present between the molecules of an ideal gas.
• Ideal gases are the gases that obey gas laws or gas equation PV = nRT.
• Real gases do not obey gas equation. PV = nRT.
• The deviation of real gases from ideal behaviour is measure in terms of a ratio of PV to PV nRT. This is termed as compression factor (Z). Z = \(\frac{P V}{n R T}\)

Question 19.
State the third law of Thermodynamics.
Answer:
It states that the entropy of pure crystalline substance at absolute zero is zero, (or)

It can be stated as “it is impossible to lower the temperature of an object to absolute zero in a finite number of steps”. Mathematically \(\lim _{T \rightarrow 0} S=0\)

Question 20.
What is called Bond Length? Name the techniques through which the length of a bond can be determined.
Answer:
The distance between thS 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., γ_A + γ_B = bond length.

The length of a bond can be determined by spectroscopic, x-ray diffraction and electron- diffraction techniques.

Question 21.
Describe the reaction involved in the detection of Nitrogen in an organic compound by Lassaigne Method.
Answer:
Detection of Nitrogen: The following reactions are involved in the detection of nitrogen with formation of prussian blue precipitate conforming the presence of nitrogen in an organic compound.

Question 22.
How is Alkane prepared from Grignard reagent?
Answer:
CH₃MgX + HOH → CH₄ + Mg(OH)X.
Water would protonate the grignard reagent and destroy the grignard reagent, because the grignard carbon atom is highly nucleophilic. This would form a hydrocarbon. Therefore to make a grignard solution, only ether is the best solvent and water or alcohol are not used for that purpose.

Question 23.
Define – Acid rain.
Answer:
Rain water has a pH of 5.6 due to the dissolution of CO₂ into it. Oxides of sulphur and nitrogen in the atmosphere may be absorbed by droplets of water that make up the clouds and get chemically converted into sulphuric acid and nitric acid. Due to this the pH of rain water drops below the level of 5.6. Hence it is called acid rain.

Question 24.
Which is the suitable method for detection of Nitrogen present in food and fertilizers?
Answer:
Kjeldahl’s method is the suitable for detection of nitrogen present in food and fertilizers.

Part-III

Answer any six of the following questions. Question No. 33 is compulsory. [6 x 3 = 18]

Question 25.
Calculate the equivalent mass of H₂SO₄.
Sulphuric acid = H₂SO₄
Molar mass of Sulphuric acid = 2 + 32 + 64 = 98
Basicity of Sulphuric acid = 2
Equivalent mass of acid =
\(=\frac{\text { Molar mass of an acid }}{\text { Basicity }}\)
\(=\frac{98}{2}=49 \mathrm{geq}^{-1}\)

Question 26.
Explain diagonal relationship.
Answer:

• On moving diagonally across the periodic table, the second and the third period elements show certain similarities.
• Even though the similarity is not same as we see in a group, it is quite pronounced in the following pair of elements.
• 
• The similarity in properties existing between the diagonally placed elements is called “diagonal relationship”.

Question 27.
How is Tritium prepared?
Answer:
It occurs naturally as a result of nuclear reactions induced by cosmic rays in the upper atmosphere.

Question 28.
Define – 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 29.
State the term “Isotonic solution”.
Answer:

1. Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.
2. When such solutions are 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.

Question 30.
Both C₂H₂ and CO₂ have the same structure. Explain why.
Answer:
H – C = C – H and O = C = O
Both has,
Linear- Molecular geometry
Zero – Dipole moment
sp – Hybridization of C in C₂H₂ and CO₂. Hence C₂H₂ and CO₂ have the same structure.

Question 31.
Write note on Williamson’s Synthesis.
Answer:
Williamson’s synthesis: Haloalkanes, when boiled with sodium alkoxide, gives the corresponding ether.

Question 32.
Explain why Ca(OH)2 is used in white washing.
Answer:
(i) Ca(OH)₂ used in white washing, this is due to disinfectant nature.
(if) Calcium hydroxide reacts with CO₂ in air to form thin layer of calcium carbonate on the walls. It gives a shiny finish to the walls.

Question 33.
Give the structural formula for the following compounds.
(a) m – dinitrobenzene
(b) p – dichlorobenzene
(c) 1,3, 5, Tri-methyl Benzene
Answer:

Part – IV

Answer all the questions.

Question 34.
(a) (i) Calculate oxidation number of oxygen in H₂O₂.
(ii) Write the de-Broglie equation.
[OR]
(b) (i) State and explain Dobereiner’s “Triad”.
(ii) Complete the following equation.
(a)

Therefore oxidation number of ‘O’ in H₂O₂ is -1.

(ii) 1. Louis de Broglie extended the concept of dual nature of light to all forms of matter. To quantify this relation, he derived an equation for the wavelength of a matter wave.
2. He combined the following two equations of energy of which one represents wave character (ho) and the other represents the particle nature (mc²).
Planck’s quantum hypothesis:
E = hv …………………………. (1)
Einsteins mass-energy relationship:
E = mc² ………………………….. (2)
From (1) and (2)
hv = mc²
h/λ = mc²
∴ λ = \(\frac{h}{m c}\) ……………………………….. (3)

The equation (3) represents the wavelength of photons whose momentum is given by. me. (Photons have zero rest mass).

3. For a particle of matter with mass m and moving with a velocity v, the equation (3) can be written as λ = \(\frac{h}{m c}\) ……………………………….. (3)

[OR]

(b) (i) Johann Dobereiner noted that elements with similar properties occur in groups of three which he called triads. It was seen that invariably, the atomic weight of the middle number of the triad was nearly equal to the arithmetic mean of the weights of the other two numbers of the triad. For e.g.,

Question 35.
(a) (i) Among the alkaline earth metals BeO is insoluble in water but other oxides are soluble. Why?
(ii) State Diffusion Law.
[OR]
(b) (i) Calculate the entropy change during the melting of one mole of ice into water at 0°C. Enthalpy of fusion of ice is 6008 J mol^-1.
(ii) Write the balanced chemical equation for Kc = \(\frac{\left[\mathbf{C a O}_{(\mathbf{s})}\right]\left[\mathbf{C O}_{2(\mathbf{g})}\right]}{\left[\mathbf{C a C O}_{3(\mathbf{(})}\right]}\)
(a) (i) Lattice energy of BeO is comparatively higher than the hydrogen energy. Therefore it is almost insoluble in water. Whereas other alkaline earth metals oxides have high hydration energy. Hence they are soluble in water.

(ii) Diffusion is the spreading of molecules of a substance throughout a space or a second substance. Diffusion refers to the ability of the gases to mix with each other. E.g., Spreading of something such as brown tea liquid spreading through the water in a tea cup.

[OR]

(b)

Question 36.
(a) (i) NH₃ and HC1 do not obey Henry’s law. Why?
(ii) Write the structure of the following compunds. (A) NH₃ (B) BF₃

[OR]

(b) (i) Identify the cis and trans isomers for the following compounds.

(ii) Explain with example the Positive Mesomeric Effect.
Answer:
(a) (i) Only the less soluble gases obeys Henry’s law. NH₃ and HC1 react with (or soluble) water and hence does not obey Henry’s law.

(b)

(ii)

1. Positive mesomeric effect occurs, when the electrons move away from substituent attached to the conjugated system.
2. It occurs, if the electron releasing substituents are attached to the conjugated system.
3. The attached group has a tendency to release electrons through resonance, these electron releasing groups are usually denoted as +R or +M groups
4. Examples: —OH, —SH, —OR, —SR, —NH₂ etc.

Question 37.
(a) (i) Write the IUPAC name for the following compound.

(ii) What are Nucleophiles and Electrophiles? Give one example each.

[OR]

(b) (i) How will you get the following products with the given reactants?
(A) Acetylene → Benzene
(B) Phenol → Benzene
(C) Benzene → Toluene
(ii) Write any two different components you get during fractional distillation of Coal Tar at any two different temperatures.
Answer:
(a)

(ii) Electrophiles: Electrophiles are reagents that are attracted towards negative charge or electron rich center. They are either positively charged ions or electron deficient neutral molecules.
Example, CO₂, AlCl₃, BF₃, FeCl₃, N0⁺, NO₂⁺, etc.

Nucleophiles: Nucleophiles are reagents that has high affinity for electropositive centers. They possess an atom has a unshared pair of electrons. They are usually negatively charged ions or electron rich neutral molecules.
Example: NH₃, R-NH₂, R-SH, H₂O, R-OH, CN^–, OH^– etc.

[OR]

(b)

(ii) Coal tar is a viscous liquid obtained by the pyrolysis of coal. During fractional distillation, coal tar is heated and distills away its volatile compounds, namely, benzene, toluene, xylene in the temperature range of 350 K to 443 K. These vapours are. collected at the upper part of the fractionating column.

Question 38.
(a) (i) A compound having the empirical formula C₆H₆O has the vapour density 47. Find its Molecular formula.
(ii) The Simple Aromatic Hydrocarbon compound (A) reacts with Bromine to give (B). Compound (A) reacts with Raney Ni and gives (C). Identify (A), (B) and (C).

[OR]

(b) (i) C_(s) + O_2(g) → CO_2(g)
Calculate the standard entropy change for the above reaction. Given the standard entropies of CO_(s) + C_(s), O_2(g) are 213.6, 5.740 and 205 JK^-1 respectively.
(ii) Identify the compound (A) and (B).

(a) (i) Empirical formula = C₆H₆O
Empirical formula mass
= (6 x 12) + (6 x 1) + (16 x 1)
= 72 + 6+16
= 94
Vapour density =47
Molecular mass = 2 x Vapour density
= 2 x 47
= 94
Molecular formula = n x Empirical formula
\($\mathrm{n}=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}$\)
n = 94/94 ⇒ n = 1
Molecular formula = 1 x C₆H₆O
= c₆H₆O

[OR]

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Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 10 Flow of Control

Samacheer Kalvi 11th Computer Science Flow of Control Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

Question 1.
What is the alternate name of null statement?
(a) No statement
(b) Empty statement
(c) Void statement
(d) Zero statement
Answer:
(b) Empty statement

Question 2.
In C++, the group of statements should enclosed within:
(a) { }
(b) []
(c) ()
(d) <>
Answer:
(a) { }

Question 3.
The set of statements that are executed again and again in iteration is called as:
(a) condition
(b) loop
(c) statement
(d) body of loop
Answer:
(d) body of loop

Question 4.
The multi way branching statement:
(a) if
(b) if … else
(c) switch
(d) for
Answer:
(c) switch

Question 5.
How many types of iteration statements exist?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 6.
How many times the following loop will execute? for (int i = 0; i < 10; i++)
(a) 0
(b) 10
(c) 9
(d) 11
Answer:
(b) 10

Question 7.
Which of the following is the exit control loop?
(a) for
(b) while
(c) do … while
(d) if … else
Answer:
(c) do … while

Question 8.
Identify the odd one from the keywords of jump statements:
(a) break
(b) switch
(c) goto
(d) continue
Answer:
(a) break

Question 9.
A loop that contains another loop inside its body:
(a) Nested loop
(b) Inner loop
(c) Inline loop
(d) Nesting of loop
Answer:
(a) Nested loop

PART – 2
II. Answers to all the questions

Question 1.
What is a null statement and compound statement?
Answer:
Null statement:
The “null or empty statement” is a statement containing only a semi colon. It takes the following form:
;// it is a null statement.
Null statement are commonly used as place holders in iteration statements or as statements on which to place labels at the end of compound statements or functions.

Compound statement:
C++ allows a group of statements enclosed by pair of braces { }. This group of statements is called as a compound statement or a block.
The general format of compound statement is:

Question 2.
What is selection statement? Write it’s types.
Answer:
The selection statement means the statements are executed depends – upon a condition. If a condition is true, a true block (a set of statements) is executed, otherwise a false block is executed. This statement is also called decision statement or selection statement because it helps in making decision about which set of statements are to be executed.

Types:

1. Two way branching
2. Multiway branching

Question 3.
Correct the following code segment:
Answer:

Question 4.
What will be the output of the following code:

If the input given is

1. 2000
2. 2003
3. 2010?

Answer:
1. Output
(Input = 2000)
Leap

2. Output
(Input = 2003)
Not leap Year

3. Output (Input = 2010)
Not leap Year

Question 5.
What is the output of the following code?
for (int i = 2; i < = 10 ; i + = 2)
cout<<i;
Answer:
Output:
2 4 6 8 10

Question 6.
Write a for loop that displays the number from 21 to 30.
Answer:
# include
using namespace std;
int main()
{

for (int = 21; i < 31; i++)
cout << “value of i:” << i << endl;
return 0;
}
Output:
Value of i: 21
Value of i: 22
Value of i: 23
Value of i: 24
Value of i: 25
Value of i: 26
Value of i: 27
Value of i: 28
Value of i: 29
Value of i: 30

Question 7.
Write a while loop that displays numbers 2,4, 6, 8 ……………… 20.
Answer:

Output:
2 4 6 8 10 12 13 14 16 18 20

Question 8.
Compare an if and a? : operator.
Answer:

PART – 3 III.
III. Answers to all the questions Question

Question 1.
Convert the following if – else to a single conditional statement:

Answer:
a = (x> =10)? m + 5:m;

Question 2.
Rewrite the following code so that it is functional:
Answer:

Question 3.
Write a C++ program to print multiplication table of a given number.
Answer:
# include
using namespace std;
int main ()
{

int num;
cout << “Enter Number to find its multiplication table”; cin >> num;
for (int a = 1; a < = 10; a++)
{
cout << num << “*” << a << “=” << num*a << endl;
}
return( );

}

Question 4.
Write the syntax and purpose of switch statement.
Answer:
The switch statement is a multi – way branch statement. It provides an easy way to dispatch execution to different parts of code based on the value of the expression. The switch statement replaces multiple if – else sequence.
The syntax of the switch statement is;

Question 5.
Write a short program to print following series:
(a) 1 4 7 10 …… 40
Answer:
# include
using namespace std;
int main ()
{

cout << “\n PRINT SERIES” ;
for (int i = 1; i< 40; i = i + 3)
cout << i << “1+”;
cin.get();
return ();
}

PART – 4
IV. Answers to all the questions

Question 1.
Explain control statement with suitable example.
Answer:
Control statements are statements that alter the sequence of flow of instructions.
Types of control statements:
1. Selection statement:
The selection statement means the statement (s) executed, depends upon a condition. If a condition is true, a true block (a set of statements) is executed otherwise a false block is executed. This statement is also called decision statement or selection statement because it helps in making decision about which set of statements are to be executed.
Example:

1. If
2. Switch

2. Iteration (or looping) statement:
An iteration (or loop) is a sequence of one or more statements that are repeatedly executed until a condition is satisfied. These statements are also called as control flow statements. It is used to reduce the length of code to reduce the time to execute program and takes less memory space.

Example:

1. While
2. do ….. while
3. for

Question 2.
What is entry control loop? Explain any one of the entry control loop with suitable example.
Answer:
An entry control loop checks the condition at the time of entry and if condition or expression becomes true then control transfers into the body of the loop, for loop and while loop are the examples of Entry Controlled Loop. A while loop is a control flow statement that allows the loop statements to be executed as long as the condition is true.
Type: Entry control loop
Syntax:

Control flow:
Step 1 : Test – expression is evaluated to either True or False;
Step 2 : If test – expression is true;
(a) The body of the loop is executed.
(b) Control is transferred to step 1.
Step 3: If test – expression is false, the control exits the while loop.
Example:
int a = 1;
while (a<=10)
{

cout << a << ‘\t+’;
a+=2;

}

Output:
1 3 5 7 9

Question 3.
Write a program to find the LCM and GCD of two numbers.
Answer:

Question 4.
Write programs to find the sum of the following series:
(a) x – \(\frac{x^{2}}{2 !}\) + \(\frac{x^{3}}{3 !}\) + \(\frac{x^{4}}{4 !}\) + \(\frac{x^{5}}{5 !}\) – \(\frac{x^{6}}{6 !}\)
(b) x + \(\frac{x^{2}}{2}\) + \(\frac{x^{3}}{3}\) + ……… + \(\frac{x^{n}}{n}\)
Answer:
(a)

(b) # include
using namespace std;
# include int main()
{

int M, L;
float sum = 0;
cout << “Enter the value for M:”; cin >> M;
cout << “Enter the number of terms”; cin >> L;
for (int i = 1; i<=m; i++)
sum = Sum + Pow(m,i)/i;
cout << “Sum=” << Sum << endl;
cin.get();
return 0;

}

Question 5.
Write a program to find sum of the series
S = 1 + x + x² + ………….. + xⁿ
Answer:
# include
using namespace std;
# include int main()
{

int x, n;
float sum = 0;
cout << “Enter the value for x:”; cin >> x;
cout << “Enter the number of terms”; cin >> n;
for (int i = 0; i<=n; i++)
sum = Sum + Pow(x,i);
cout << “Sum=” << Sum << endl;
cin.get( );
return 0;

}

Samacheer kalvi 11th Computer Science Flow of Control Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
The empty statement is otherwise called as ………………..
(a) Control statement
(b) Zero statement
(c) Null statement
(d) Block statement
Answer:
(c) Null statement

Question 2.
Selection statement is also called as ………………..
(a) Decision statement
(b) Sequence statement
(c) Null statement
(d) Compound statement
Answer:
(a) Decision statement

Question 3.
Iteration statement is called as ………………..
(a) Null statement
(b) Block statement
(c) Selection statement
(d) Looping statement
Answer:
(d) Looping statement

Question 4.
In C++ any non – zero is iterated as true ……………….. and zero is treated as false.
(a) positive numbers
(b) negative numbers
(c) prime numbers
(d) none of these
Answer:
(b) negative numbers

Question 5.
……………….. is a multi – path decision making statement.
(a) if
(b) if – else
(c) else – if
(d) if – else ladder
Answer:
(d) if – else ladder

Question 6.
Syntax of the conditional operator is ………………..
(a) expression 1? expression 2: expression 3
(b) expression 1: expression 2
(c) expression 1! expression 2: expression 3
(d) expression 1: expression 2: expression 3
Answer:
(a) expression 1? expression 2: expression 3

Question 7.
……………….. is more efficient than if-else statement.
(a) Control statement
(b) Switch statement
(c) Empty statement
(d) Null statement
Answer:
(b) Switch statement

Question 8.
When a switch is a part of the statement sequence of another switch, then it is called as ………………..
(a) if – else ladder
(b) Switch statement
(c) Nested switch statement
(d) Empty statement
Answer:
(c) Nested switch statement

Question 9.
C++ supports types of iteration statements.
(a) 3
(b) 2
(c) 4
(d) 5
Answer:
(a) 3

Question 10.
Every loop has ……………….. elements that are used for different purposes.
(a) 3
(b) 4
(c) 5
(d) 2
Answer:
(b) 4

Question 11.
……………….. is used to transfer the control from one place to another place without any condition in a program.
(a) Break statement
(b) Continue statement
(c) goto statement
(d) All the above
Answer:
(c) goto statement

PART – 2
II. Short Answers

Question 1.
What is if statement? Write it’s syntax.
Answer:
The if statement evaluates a condition, if the condition is true then a true – block (a statement or set of statements) is executed, otherwise the true – block is skipped. The general syntax of the if statement is:
if (expression)
true – block;
statement – x;

Question 2.
Write a program to check whether a person is eligible to vote using if statement.
C++ program to check whether a person is eligible to vote using if statement
Answer:

Output:
Enter your age: 23
You are eligible for voting ….
This statement is always executed.

Question 3.
What is nested if? Mention it’s types.
Answer:
An if statement contains another if statement is called nested if. The nested can have, one of the following three forms.

1. If nested inside if part
2. If nested inside else part
3. If nested inside both if part and else part

Question 4.
Write the syntax for if – else ladder.
Answer:
The syntax of if – else ladder:

Question 5.
What is nested switch statement?
Answer:
When a switch is a part of the statement sequence of another switch, then it is called as nested switch statement. The inner switch and the outer switch constant may or may not be the same.

Question 6.
What are loop elements?
Answer:
Every loop has four elements that are used for different purposes. These elements are

1. Initialization expression
2. Test expression
3. Update expression
4. The body of the loop

Question 7.
Write C++ program to sum the numbers from 1 to 10.
Answer:
C++ program to sum the numbers from 1 to 10 using for loop

Output:
The sum of 1 to 10 is 55

Question 8.
Rewrite the given program.
Answer:
C++ program to sum the numbers from 1 to 10 using for loop

C++ program to sum the numbers from 1 to 10 using for loop

Output:
The sum of 1 to 10 is 55

Question 9.
Write a program to display numbers from 1 to 10 except 6 using continue statement.
Answer:
C++ program to display numbers from 1 to 10 except 6 using continue statement
#include
using namespace std;
int main()
{
if (i = 6)
continue;
else
cout << i << ” ”
}
return 0;
}
Output:
1 2 3 4 5 7 8 9 10

Question 10.
What will be the output of the following code.
Answer:
int a, b, largest;
cout << “\n Enter any two numbers:”; cin >> a >> b;
largest = (a > b)? a:b;
cout << “\n Largest number:” << largest;
return 0;
Output:
Enter any two numbers:
5 10
Largest number: 10

PART – 3
III. Explain in Brief

Question 1.
What is sequence statements?
Answer:
The sequential statement are the statements that are executed one after another only once from top to bottom. These statements do not alter the flow of execution. These statements are called as sequential flow statements. They always end with a semicolon (;).

Question 2.
Write the syntax of nested switch statement.
Answer:
The syntax of the nested switch statement is; switch (expression)

Question 3.
Write short notes on Test expression.
Answer:
The test expression is an expression or condition whose value decides whether the loop – body will be executed or not. If the expression evaluates to true (i.e., 1), the body of the loop is executed, otherwise the loop is terminated. In an entry – controlled loop, the test – expression is evaluated before the entering into a loop whereas in an exit – controlled loop, the test – expression is evaluated before exit from the loop.

Question 4.
What is do – while loop? Write it’s syntax.
Answer:
The do – while loop is an exit-controlled loop. In do-while loop, the condition is evaluated at the bottom of the loop after executing the body of the loop. This means that the body of the loop is executed at least once, even when the condition evaluates false during the first iteration.
The do – while loop syntax is:
do {
Body of the loop;
} while(condition);

Question 5.
What are the differences between break and continue statement.
Answer:
Break:

• Break is used to terminate the execution of the loop.
• It breaks the iteration.
• When this statement is executed, control will come out from the loop and executes the statement immediate after loop.
• Break is used with loops as well as switch case.

Continue:

• Continue is not used to terminate the execution of loop.
• It skips the iteration.
• When this statement is executed, it will not come out of the loop but moves/jumps to the next iteration of loop.
• Continue is only used in loops, it is not used in switch case.

Question 6.
What are the important things to know about switch statement?
Answer:
There are some important things to know about switch statement. They are

1. A switch statement can only work for quality of comparisons.
2. No two case labels in the same switch can have identical values.
3. If character constants are used in the switch statement, they are automatically converted to their equivalent ASCII codes.
4. The switch statement is more efficient choice than if in a situation that supports the nature of the switch operation.

PART – 4
IV. Explain in Detail

Question 1.
What are the key differences between if else and switch statement?
Answer:

Question 2.
Explain about parts of a loop.
Answer:
Every loop has four elements that are used for different purposes. These elements are:

1. Initialization expression
2. Test expression
3. Update expression
4. The body of the loop

1. Initialization expression(s) : The control variable(s) must be initialized before the control enters into loop. The initialization of the control variable takes place under the initialization expressions. The initialization expression is executed only once in the beginning of the loop.

2. Test Expression : The test expression is an expression or condition whose value decides whether the loop-body will be executed or not. If the expression evaluates to true (i.e., 1), the body of the loop is executed, otherwise the loop is terminated.

In an entry – controlled loop, the test – expression is evaluated before the entering into a loop whereas in an exit-controlled loop, the test – expression is evaluated before exit from the loop.

3. Update expression : It is used to change the value of the loop variable. This statement is executed at the end of the loop after the body of the loop is executed.

4. The body of the loop : A statement or set of statements forms a body of the loop that are executed repetitively. In an entry – controlled loop, first the test-expression is evaluated and if it is nonzero, the body of the loop is executed otherwise the loop is terminated. In an exit – controlled loop, the body of the loop is executed first then the test – expression is evaluated. If the test – expression is true the body of the loop is repeated otherwise loop is terminated.

Question 3.
Explain about jump statements.
Answer:
Jump statements are used to interrupt the normal flow of program. Types of Jump Statements are:

1. goto statement
2. break statement
3. continue statement

The goto statement is a control statement which is used to transfer the control from one place to another place without any condition in a program.

A break statement is a jump statement which terminates the execution of loop and the control is transferred to resume normal execution after the body of the loop.

The continue statement works quite similar to the break statement. Instead of terminating the loop (break statement), continue statement forces the loop to continue or execute the next iteration. When the continue statement is executed in the loop, the code inside the loop following the continue statement will be skipped and next iteration of the loop will begin.

Question 4.
Programs to produce the following design using nested loops.
Answer:
Output
Enter number of rows: 5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

Students can Download Computer Science Chapter 11 Functions Questions and Answers, Notes Pdf, Samacheer Kalvi 11th Computer Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Computer Science Solutions Chapter 11 Functions

Samacheer Kalvi 11th Computer Science Functions Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

Question 1.
Which of the following header file defines the standard I/O predefined functions?
(a) stdio.h
(b) math.h
(c) string.h
(d) ctype.h
Answer:
(a) stdio.h

Question 2.
Which function is used to check whether a character is alphanumeric or not?
(a) isalpha()
(b) isdigit()
(c) isalnum()
(d) islower()
Answer:
(c) isalnum()

Question 3.
Which function begins the program execution?
(a) isalpha()
(b) isdigit()
(c) main()
(d) islower()
Answer:
(c) main()

Question 4.
Which of the following function is with a return value and without any argument?
(a) x = display(int, int)
(b) x = display()
(c) y = display(float)
(d) display(int)
Answer:
(b) x = display()

Question 5.
Which is return data type of the function prototype of add(int, int);?
(a) int
(b) float
(c) char
(d) double
Answer:
(a) int

Question 6.
Which of the following is the scope operator?
(a) >
(b) &
(c) %
(d) ::
Answer:
(d) ::

PART – 2
II. Answer to all the questions

Question 1.
Define Functions.
Answer:
A large program can typically be split into small sub – programs (blocks) called as functions where each sub-program can perform some specific functionality. Functions reduce the size and complexity of a program, makes it easier to understand, test and check for errors.

Question 2.
Write about strlen() function.
Answer:
The strlen() takes a null terminated byte string source as its argument and returns its length. The length does not include the null(\0) character.

Question 3.
What are importance of void data type? void type has two important purposes:
Answer:

1. To indicate the function does not return a value.
2. To declare a generic pointer.

Question 4.
What is Parameter and list its types?
Answer:
Arguments or parameters are the means to pass values from the calling function to the called function. The variables used in the function definition as parameters are known as formal parameters. The constants, variables or expressions used in the function call are known as actual parameters.

Types : Default arguments and Constant Arguments.

Question 5.
Write a note on Local Scope.
Answer:

1. A local variable is defined within a block. A block of code begins and ends with curly braces { }.
2. The scope of a local variable is the block in which it is defined.
3. A local variable cannot be accessed from outside the block of its declaration.
4. A local variable is created upon entry into its block and destroyed upon exit.

PART – 3
III. Answer to all the questions

Question 1.
What is Built – in functions?
Answer:
C++ provides a rich collection of functions ready to be used for various tasks. The tasks to be performed by each of these are already written, debugged and compiled, their definitions alone are grouped and stored in files called header files. Such ready – to – use sub programs are called pre – defined functions or built – in functions.

Question 2.
What is the difference between isupper() and toupper() functions?
Answer:
isupper():

• This function is used to check the given character is uppercase.
• This function will return 1 if true otherwise 0.

toupper():

• This function is used to convert the given character into its uppercase.
• This function will return the upper case equivalent of the given character. If the given character itself is in upper case, the output will be the same.

Question 3.
Write about strcmp() function.
Answer:
The strcmp() function takes two arguments: string1 and string2. It compares the contents of string1 and string2 lexicographically.
The strcmp() function returns a:

1. Positive value if the first differing character in string1 is greater than the corresponding character in string2. (ASCII values are compared)
2. Negative value if the first differing character in string1 is less than the corresponding character in string2.
3. 0 if string1 and string2 are equal.

Question 4.
Write short note on pow() function in C++.
Answer:
The pow() function returns base raised to the power of exponent. If any argument passed to pow() is long double, the return type is promoted to long double. If not, the return type is double. The pow() function takes two arguments:

1. base – the base value
2. exponent – exponent of the base

Example:
cout << pow(5, 2);

Output:
25

Question 5.
What are the information the prototype provides to the compiler?
Answer:
The prototype above provides the following information to the compiler:

1. The return value of the function is of type long.
2. Fact is the name of the function.
3. The function is called with two arguments:
   • The first argument is of int data type.
   • The second argument is of double data type, int display(int, int) // function prototype//.

The above function prototype provides details about the return data type, name of the function and a list of formal parameters or arguments.

Question 6.
What is default arguments? Give example.
Answer:
In C++, one can assign default values to the formal parameters of a function prototype. The Default arguments allows to omit some arguments when calling the function.
1. For any missing arguments, complier uses the values in default arguments for the called function.

2. The default value is given in the form of variable initialization.
Example : void defaultvalue(int n1 = 10, n2 = 100);

3. The default arguments facilitate the function call statement with partial or no arguments.
Example :

1. defaultvalue (x, y);
2. defaultvalue (200, 150);
3. defaultvalue (150);
4. defaultvalue (x, 150);

4. The default values can be included in the function prototype from right to left, i.e., we cannot have a default value for an argument in between the argument list.
Example:

1. void defaultvalue (int n1=10, n2);//invalid prototype.
2. void defaultvalue (int n1, n2 = 10);//valid prototype.

PART – 4
IV. Answers to all the questions

Question 1.
Explain Call by value method with suitable example.
Answer:
This method copies the value of an actual parameter into the formal parameter of the function. In this case, changes made to formal parameter within the function will have no effect on the actual parameter.

Output:
Example : Function call by value
Enter the Value for A : 5
The Value inside display function (a * a) : 25
The Value inside main function 5

Question 2.
What is Recursion? Write a program to find GCD using recursion.
Answer:
A function that calls itself is known as recursive function. And, this technique is known as recursion.

Output:
Enter two numbers: 81 153
gcd : 9

Question 3.
What are the different forms of function return? Explain with example.
Answer:
The return statement:
Returning from the function is done by using the return statement. The return statement stops execution and returns to the calling function. When a return statement is executed, the function is terminated immediately at that point. The return statement is used to return from a function. It is categorized as a jump statement because it terminates the execution of the function and transfer the control to the called statement.
Syntax:
return expression/variable;

Example : retum(a + b); retum(a);
return; // to terminate the function

The Returning values:
The functions that return no value is declared as void. The data type of a function is treated as int, if no data type is explicitly mentioned. For example,
For Example:
int add (int, int);
add (int, int);
In both prototypes, the return value is int, because by default the return value of a function in C++ is of type int.

Returning Non – integer values:
A string can also be returned to a calling statement.

Output:
Example: Function with Non Integer Return Chennai

The Returning by reference:
#include
using namespace std;
int main( )
{

int n 1 = 150;
int &n 1 ref = n1;
cout << “\nThe Value of N1 = “<< n1 << “and n 1 Reference =”<< n 1 ref;
n 1 ref++;
cout << “\n After nl increased the Value of N1 =”<< n1;
cout << “and n 1 Reference = ”<< n 1 ref;
retum(0);

}
Output:
The Value of N1 = 150 and nl Reference =150
After n1 increased the Value of N1 = 151 and n1 Reference =151

Question 4.
Explain scope of variable with example.
Answer:
Scope refers to the accessibility of a variable.
There are four types of scopes in C++

1. Local Scope
2. Function Scope
3. File Scope
4. Class Scope

1. Local Scope:

• A local variable is defined within a block. A block of code begins and ends with curly braces {}.
• The scope of a local variable is the block in which it is defined.
• A local variable cannot be accessed from outside the block of its declaration.
• A local variable is created upon entry into its block and destroyed upon exit;
  Example:
  int main( )
  {
  int a,b;   //Local variable
  }

2. Function Scope:

• The scope of variable within a function is extended to the function block and all sub-blocks therein.
• The lifetime of a function scope variable is the lifetime of the function block.
  Example:
  int. sum(intx, int y);  //x and y has function scope.

3. File Scope:

• A variable declared above all blocks and functions (including main()) has the scope of a file.
• The lifetime of a file scope variable is the lifetime of a program.
• The file scope variable is also called as global variable.
  Example:
  #include
  using namespace std;
  int x,y; //x and y are global variable
  void main()
  {
  ……..
  }

4. Class Scope:

• Data members declared in a class has the class scope.
• Data members declared in a class can be accessed by all member functions of the class.
  Example:
  Class example
  {

int x,y; //x and y can be accessed by print() and void():
void print();
Void total();

  };

Question 5.
Write a program to accept any integer number and reverse it.
Answer:

Output:
Enter number : 1 2 3
Reverse number: 3 2 1

Samacheer kalvi 11th Computer Science Functions Additional Questions and Answers

PART – 1
I. Choose the correct answer

Question 1.
………………. is the name of the function.
(a) Pre – defined
(b) Built – in
(c) Library
(d) All the above
Answer:
(d) All the above

Question 2.
………………. is used to check whether the given character is an alphabet or not.
(a) isalnum()
(b) isalpha()
(c) isalph()
(d) isal()
Answer:
(b) isalpha()

Question 3.
The strcpy() function takes two arguments of ……………….
(a) target and source
(b) upper and lower
(c) base and exponent
(d) none of these
Answer:
(a) target and source

Question 4.
………………. takes a null terminated byte string source as its argument and returns its length.
(a) strcpy()
(b) strlen()
(c) strcmp()
(d) strcat()
Answer:
(b) strlen()

Question 5.
The pow() function takes the two arguments of ……………….
(a) target and source
(b) upper and lower
(c) base and exponent
(d) source and exponent
Answer:
(c) base and exponent

Question 6.
………………. is the name of the function.
(a) fact
(b) task
(c) arguments
(d) none of these
Answer:
(d) none of these

Question 7.
The C++ program always have main() function to begin the program execution.
(a) 1
(b) 2
(c) 3
(d) null
Answer:
(a) 1

Question 8.
Arguments are also called as ……………….
(a) variable
(b) constant
(c) function
(d) parameters
Answer:
(d) parameters

Question 9.
In C++ the arguments can be passed to a function in ………………. ways.
(a) 2
(b) 1
(c) 3
(d) 7
Answer:
(a) 2

Question 10.
Inline functions execute faster but requires more ……………….
(a) variables
(b) pointers
(c) memory
(d) functions
Answer:
(c) memory

PART – 2
II. Very Short Answers

Question 1.
Write about reusability.
Answer:

1. Few lines of code may be repeatedly used in different contexts. Duplication of the same code can be eliminated by using functions which improves the maintenance and reduce program size.
2. Some functions can be called multiple times with different inputs.

Question 2.
What is user – defined functions?
Answer:
C++ also provides the facility to create new functions for specific task as per user requirement. The name of the task and data required (arguments) are decided by the user and hence they are known as User-defined functions.

Question 3.
What is constant arguments and write its syntax?
Answer:
The constant variable can be declared using const keyword. The const keyword makes variable , value stable. The constant variable should be initialized while declaring. The const modifier enables to assign an initial value to a variable that cannot be changed later inside the body of the function.
Syntax:
(const )

Question 4.
What are the advantages of inline functions?
Answer:
Advantages of inline functions:

1. Inline functions execute faster but requires more memory space.
2. Reduce the complexity of using STACKS.

Question 5.
What is function scope?
Answer:
Function Scope:

1. The scope of variables declared within a function is extended to the function block, and all sub – blocks therein.
2. The life time of a function scope variable, is the life time of the function block. The scope of.

PART – 3
III. Short Answers

Question 1.
What is divide and conquer?
Answer:
Divide and Conquer

1. Complicated programs can be divided into manageable sub programs called functions.
2. A programmer can focus on developing, debugging and testing individual functions.
3. Many programmers can work on different functions simultaneously.

Question 2.
Define library functions.
Answer:
C++ provides a rich collection of functions ready to be used for various tasks. The tasks to be performed by each of these are already written, debugged and compiled, their definitions alone are grouped and stored in files called header files. Such ready – to – use sub programs are called pre – defined functions or built – in functions or Library Functions.

Question 3.
What is isdigit()? Give example.
Answer:
This function is used to check whether a given character is a digit or not. This function will return 1 if the given character is a digit, and 0 otherwise.

Example:
using namespace std;
#include
#include int main( )
{

char ch;
cout << “\n Enter a Character:”; cin >> ch;
cout << “\n The Return Value of isdigit(ch) is << isdigit(ch);

}
Output – 1
Enter a Character: 3
The Return Value of isdigit(ch) is : 1

Output – 2
Enter a Character: A
The Return Value of isdigit(ch) is :0

Question 4.
Write a program using pow() and sin() function.
Answer:
The pow() function returns base raised to the power of exponent. The sin() function takes a single argument in radians.
#include
#include using namespace std;
int main ()
{

double base, exponent, result;
base = 5;
exponent = 4;
result = pow(base, exponent);
cout << “pow(“ << base << “A” << exponent <<”) =” << result;
double x = 25;
result = sin(x);
cout << “\nsin(“<< x <<”)=”<<result;
return 0;

}
Output:
pow (5^4) = 625
sin (25) = – 0.132352

Question 5.
What is return statement with example?
Answer:
The return statement stops execution and returns to the calling function. When a return statement is executed, the function is terminated immediately at that point. The return statement is used to return from a function. It is categorized as a jump statement because it terminates the execution of the function and transfer the control to the called statement.

Example:
return(a + b); return(a);
return; // to terminate the function

Question 6.
What is scope resolution operation?
Answer:

1. The scope operator reveals the hidden scope of a variable. The scope resolution operator (::) is used for the following purposes.
2. To access a Global variable when there is a Local variable with same name. An example using Scope Resolution Operator.

PART – 4
IV. Explain in Detail

Question 1.
Explain about generating random numbers with suitable program.
Answer:
The srand() function in C++ seeds the pseudo random number generator used by the rand() function. The seed for rand() function is 1 by default. It means that if no srand() is called before rand(), the rand() function behaves as if it was seeded with srand( 1). The srand() function takes an unsigned integer as its parameter which is used as seed by the rand() function. It is defined inor header file.
#include
#include using namespace std; int main()
{

int random = rand(); /* No srand() calls before rand(), so seed = 1*/
cout << “\nSeed = 1, Random number =” << random;
srand(10);
/* Seed= 10 */
random = rand();
cout << “\n\n Seed =10, Random number =” << random;
return 0;

}
Output:
Seed = 1, Random number = 41
Seed =10, Random number 71

Question 2.
Explain about Inline functions with a suitable program.
Answer:
An inline function looks like normal function in the source file but inserts the function’s code directly into the calling program. To make a function inline, one has to insert the keyword inline in the function header.

Output:
Enter the Principle Amount Rs. :60000
Enter the Number of Years : 10
Enter the Rate of Interest :5
The Simple Interest = Rs. 30000

Question 3.
Write about function prototype.
Answer:
C++ program can contain any number of functions. But, it must always have only one main() function to begin the program execution. We can write the definitions of functions in any order as we wish. We can define the main() function first and all other functions after that or we can define all the needed functions prior to main(). Like a variable declaration, a function must be declared before it is used in the program. The declaration statement may be given outside the main() function

The prototype above provides the following information to the compiler:

1. The return value of the function is of type long.
2. Fact is the name of the function.
3. The function is called with two arguments:
   • The first argument is of int data type.
   • The second argument is of double data type, int display(int, int)//function prototype//

The above function prototype provides details about the return data type, name of the function and a list of formal parameters or arguments.

Question 4.
Explain about address method.
Answer:
This method copies the address of the actual argument into the formal parameter. Since the address of the argument is passed, any change made in the formal parameter will be reflected back in the actual parameter.
#include
using namespace std;
void display(int & x) //passing address of a//
{

x = x*x;
cout << “\n\n The Value inside display function (n1 x n1) :”<< x ;

}
int main()
{
intn 1;
cout << “\n Enter the Value for N1 cin >> n1;
cout << “\n The Value of N1 is inside main function Before passing:” << n1;
display(n1);
cout << “\n The Value of N1 is inside main function After passing (n1 x n1):”<< n1; retum(O);
}

Output:
Enter the Value for N1 : 45
The Value of N1 is inside main function Before passing : 45
The Value inside display function (n1 x n1) : 2025
The Value ofNl is inside main function After passing (n1 x n1): 2025

Question 5.
Given data: Principal amount Rs. 50000

Number of years 5
Rate of interest 7

To find the simple interest of the above mentioned given data. Write a C++ program using inline functions.
Answer:
#include
using namespace std;
inline float simple interest(float p1, float n 1, float r 1)
{

float sil=(pl*nl*rl)/100;
retum(sil);

}
int main ()
{

float si,p,n,r;
cout << “\n Enter the Principle Amount Rs. :”; cin >> p;
cout << “\n Enter the Number of Years :”; cin >> n;
cout << “\n Enter the Rate of Interest :”; cin >> r;
si = simple interest(p, n, r);
cout << “\n The Simple Interest = Rs.” << si;
return 0;

}

Output:
Enter the Principle Amount Rs. : 50000
Enter the Number of Years : 5
Enter the Rate of Interest : 7
The Simple Interest = Rs. 17500

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 3 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Chemistry Model Question Paper 3 English Medium

Time: 2½ Hours
Total Score: 70 Marks

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I. II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail. Draw diagrams wherever necessary.

Part – I

Answer all the questions.
Choose the most suitable answer from the given four alternatives.

Question 1.
Choose the disproportionation reaction among the following redox reactions.

Answer:

Solution:
\(P_{4}(s)+3 N a O H+3 H_{2} \rightarrow P H_{3}(g)+3 N a H_{2} P O_{2}(a q)\)

Question 2.
Consider the following sets of quantum numbers:

Which of the following sets of quantum number is not possible?
(a) (i), (ii), (iii) and (iv)
(b) (ii), (iv) and (v)
(c) (i) and (ill) (d) (ii), (iii) and (iv)
Answer:
(b) (ii), (iv) and (v)

Solution:
(ii) l can have the values from O to n-1
n = 2; possible ‘l’ values are 0, 1 hence l = 2 is not possible.
(iv) for l = O; m = – 1 not possible
(v) for n = 3 l = 4 and m = 3 not possible

Question 3.
Which of the following is arranged in order of increasing radius?
\((a) \mathrm{K}_{(\mathrm{aq})}^{+}<\mathrm{Na}_{(\mathrm{aq})}^{+}<\mathrm{Li}^{+}_{(\mathrm{eq})}\)
\((b) \mathrm{K}_{(\mathrm{aq})}^{+}>\mathrm{Na}_{(\mathrm{aq})}^{+}>\mathrm{Zn}_{(\mathrm{eq})}^{2+}\)
\((c) \mathrm{K}_{(\mathrm{aq})}^{+}>\mathrm{Li}_{(\mathrm{aq})}^{+}>\mathrm{Na}_{(\mathrm{aq})}^{+}\)
\((d) \mathbf{L}_{(\mathrm{aq})}^{+}<\mathbf{N} \mathbf{a}_{(\mathbf{a} \mathbf{q})}^{+}<\mathbf{K}_{(\mathbf{a} \mathbf{q})}^{+}\)
Answer:
\(\text { (d) } \mathbf{L} \mathbf{i}_{(\mathbf{z} q)}^{+}<\mathbf{N} \mathbf{a}_{(\mathbf{a} \mathbf{q})}^{+}<\mathbf{K}_{(\mathbf{a} \mathbf{q})}^{+}\)

Question 4.
Statement – I: The magnetic moment of para-hydrogen is zero.
Statement – II: The spins of two hydrogen atoms in para H₂ molecule neutralise each other.
(a) Statements – I and lIare correct and Statement-Il is the correct explanation of statement – I.
(b) Statements – I and H are correct but Statement-Il is not the correct explanation of statement – I.
(e) Statement – I is correct lut Statement-Il is wrong.
(d) Statement – I is wrong but Statement-Il is correct.
Answer:
(a) Statements – I and lIare correct and Statement-Il is the correct explanation of statement – I.

Question 5.
Which of the following has the highest tendency to give the reaction,
\(\mathbf{M}^{+}(\mathrm{g}) \frac{\text { Aqueous }}{\text { Medium }} \rightarrow \mathrm{M}^{+}(\mathrm{aq})\)
(a) Na
(b) Li
(c) Rb
(d) K
Answer:
(b) Li

Question 6.
The value of universal gas constant depends upon
(a) Temperature of the gas
(b) Volume of the gas
(c) Number of moles of the gas
(d) units of Pressure and volume
Answer:
(d) units of Pressure and volume

Question 7.
The heat of formation of CO and CO, are —26.4 kcal and —94 kcal, respectively. Heat of combustion of carbon monoxide will be …………………… .
(a) +26.4 keal
(b) -67.6 kcal
(c) -120.6 kcal
(d) +52.8 kcal
Answer:
(b) -67.6 kcal

Solution:

Question 8.
In a chemical equilibrium, the rate constant for the forward reaction is 2.5 x 10² 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 10^-3
Answer:
(b) 5

Solution:

Question 9.
The Vant Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is ………………………………. . [NEET]
(a) O
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Solution: Ba(OH)₂ dissociates to form Ba2 and 20H- ion
\(\alpha=\frac{(i-1)}{(n-1)}\)
i = α (n – 1) + 1
∴ n = i = 3 (for Ba (OH)₂, α = 1)

Question 10.
Match the list I and list II using the correct code given below in the list:

Code: A B C D
(a) 4 3 2 1
(b) 1 2 3 4
(c) 3 1 4 2
(d) 2 4 1 3
Answer:
(a) 4 3 2 1

Question 11.
Nitrogen detection in an organic compound is carried out by Lassaigne’s test. The blue colour formed is due to the formation of
(a) Fe₃ [Fe(CN)₆]₂
(b) Fe₄ [Fe(CN)₆]₃
(c) Fe₄ [Fe(CN)₆]₂
(d) Fe₃ [Fe(CN)₆]₃
Answer:
(b) Fe₄ [Fe(CN)₆]₃

Question 12.
What is the hybridisation state of benzyl carbonium ion?
(a) sp²
(b) spd²
(c) sp²
(d) sp²d
Answer:
(a) sp²

Question 13.
Statement – I: Boiling point of methane is lower than that of butane.
Statement – II: The boiling point of continuous chain alkanes increases with increase in length of carbon chain.
(a) Statement -I and II are correct and statement – II is correct explanation of statement – I.
(b) Statement – I and II are correct but statement – II is not correct explanation of statement – I.
(c) Statement – I is correct but statement – II is wrong.
(d) Statement – II is wrong but statement – 1 is correct.
Answer:
(a) Statement -I and II are correct and statement – II is correct explanation of statement – I.

Question 14.
In the reaction


Answer:

Question 15.
Which of the following binds with haemoglobin and reduce the oxygen-carrying capacity of blood?
(a) CO₂
(b) NO₂
(c) SO₃
(d) CO
Answer:
(d) CO

Part – II
Answer any six questions in which question No. 21 is compulsory. [6 x 2 = 12]

Question 16.
State and explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle states that “No two electrons in an atom can have the same set of values of all four quantum numbers”.
Illustration: H(Z = 1) Is¹.

One electron is present in hydrogen atom, the four quantum numbers are n – 1, / = 0, m = 0 and s = + ½. For helium Z = 2. He: Is². In this one electron has the quantum number same as that of hydrogen. n = l, l = 0m = 0 and s = + ½. For other electron, fourth quantum number is different. i.e. n = 1, l = 0, m = 0 and s = – ½.

Question 17.
Give the general electronic configuration of lanthanides and actinides.
Answer:

• The electronic configuration of lanthanides is 4f^1-14 5d^10-1 6s².
• The electronic configuration of actinides is 5f^1-14 6d^10-1 7s².

Question 18.
Why alkaline earth metals are harder than alkali metals?
Answer:

1. The strength of metallic bond in alkaline earth metals is higher than alkali metals due to the presence of 2 electrons in its outermost shell as compared to alkali metals, which have only 1 electron in valence shell. Therefore, alkaline earth metals are harder than alkali metals.
2. The alkaline earth metals have greater nuclear charge and more valence electrons, thus metallic bonding is more effective. Due to this they are harder than alkali metals.

Question 19.
What are the limitations of the thermodynamics?
Answer:

1. Thermodynamics suggests feasibility of reaction but fails to suggest rate of reaction. It is concerned only with the initial and the final states of the system. It is not concerned with the path by which the change occurs.
2. It does not reveal the mechanism of a process.

Question 20.
Write the application of equilibrium constant?
Answer:

• predict the direction in which the net reaction will take place
• predict the extent of the reaction and
• calculate the equilibrium concentrations of the reactants and products.

Question 21.
How much volume of 6 M solution of NaOH is required to prepare 500 ml of 0.250 M NaOH solution?
Answer:
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 22.
Define – Inductive effect.
Answer:

• It is defined as the change in the polarization of a covalent bond due to the presence of adjacent . bonded atoms or groups in the molecule. It is denoted as I-effect.
• Atoms or groups which lose electron towards a carbon atom are said to have a +1 effect.
  Example, CH₃ -, (CH₃)₂ CH (CH₃)₂ C- etc.
• Atoms or groups which draw electrons away from a carbon atom are said to have a -I effect.
  Example, -NO₂ -I, -Br, -OH, C₆H₅ etc.

Question 23.
How will you convert n-hexane into benzene? ,
Answer:
Alkanes with six to ten carbon atoms are converted into homologous of benzene at higher temperature and in the presence of a catalyst. This process is known as aromatisation.

For example, n -Hexane passed over Cr₂O₃ supported on alumina at 873 K gives benzene.

Question 24.
Why haloalkanes have higher boiling point than parent alkanes having the same number of carbons?
Answer:
Haloalkanes have higher boiling point than the parent alkane having the same number of carbon atoms because the intermolecular forces of attraction and dipole-dipole interactions are comparatively stronger in haloalkanes.

Part-III

Answer any six questions in which question No. 25 is compulsory. [6 x 3 = 18]

Question 25.
Balance the following equations by using oxidation number method.
K₂Cr₂O₇ + KI + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + I₂ + H₂O
Answer:

Question 26.
Explain the periodic trend of ionization potential.
Answer:
(a) The energy required to remove the most loosely held electron from an isolated gaseous atom is called as ionization energy.
(b) Variation in a period.

Ionization energy is a periodic property.
On moving across a period from left to right, the ionization enthalpy value increases. This is due to the following reasons.

• Increase of nuclear charge in a period
• Decrease of atomic size in a period

Because of these reasons, the valence electrons are held more tightly by the nucleus. Therefore, ionization enthalpy increases.

(c) Variation in a group
As we move from top to bottom along a group, the ionization enthalpy decreases. This is due to the following reasons.

• A gradual increase in atomic size
• Increase of screening effect on the outermost electrons due to the increase of number of inner electrons.

Hence, ionization enthalpy is a periodic property.

Question 27.
Distinguish between hard water and soft water.
Answer:

Question 28.
Explain how the equilibrium constant K_c predict the extent of a reaction.
Answer:
(i) The value of equilibrium constant K_c tells us the extent of the reaction i.e., it indicates how far the reaction has proceeded towards product formation at a given temperature.
(ii) A large value of K_c indicates that the reaction reaches equilibrium with high product yield on the other hand, lower value of K_c indicates that the reaction reaches equilibrium with low product yield.
(iii) If K_c > 10³, the reaction proceeds nearly to completion.
(iv) If K_c < 10^-3 the reaction rarely proceeds.
(v) It the K_c is in the range 10^-3 to 10³, a significant amount of both reactants and products are present at equilibrium.

Question 29.
How many moles of solute particles are present in one litre of 10^-4 M K₂SO₄?
Answer:
In 10^-4 M K₂ SC₄), solution, there are 10^-4 moles of potassium sulphate.
K₂SO₄ molecule contains 3 ions (2K⁺ and 1 SO_4^2-)
1 m;ole of I^SO,, contains 3 x 6.023 x 10²³ ions
10″4 mole of K₂SO₄ contains 3 x 6.023 x 10²³ x 10^-4 ions = 18.069 x 10¹⁹

Question 30.
Draw the Lewis structures for the following.
(i) SO_4^2-
(ii) O₃
Answer:

Question 31.
Write a short notes on hyperconjugation.
Answer:

• The delocalization of electrons of a bond is called as hyperconjugation. It is a special stabilizing effect that results due to the interaction of electrons of a cr bond with the adjacent empty non-bonding P-orbitals resulting in an extended molecular orbital.
• Hyperconjugation is a permanent effect.
• For example, in propene, the a- electrons of C—H bond of methyl group can be delocalised into the n- orbital of doubly bonded carbon as represented below.
  

Question 32.
How will you prepare propyne using alkyene dihalide?
Answer:

Question 33.
Explain how the oxides of sulphur pollute the atomospheric air? Give its harmful effects.
Answer:

• Sulphur dioxide and sulphur trioxide are produced by burning sulphur-containing fossil fuels and by roasting of sulphide ores.
• SO₂ is a poisonous gas for both animals and plants. SO₂ causes eye irritation, coughing and respiratory diseases like asthma, bronchitis.
• SO₂ is oxidised to more harmful SO₃ gas in the presence of particulate matter present in the polluted air:
  SO₃ combines with atmospheric water vapour to form H₂SO₄ which comes down along with rain in the form of acid rain:
• Acid rain causes stone leprosy, affect aquatic ecosystem, corrode water pipes and causes respiratory ailment in humans and animals.

Part – IV

Answer all the questions.

Question 34.
(a) Urea is prepared by the reaction between ammonia and arbondioxide.
2NH_3(g) + CO_2(g) → (NH₂)₂ CO_(aq) + H₂O_(l)
In one process, 637.2 g of NH₃ are allowed to react with 1142 g of CO₂.
(i) Which of the two reactants is the limiting reagent? (2)
(ii) Calculate the mass of (NH₂)₂ CO formed. (2)
(iii) How much Of the excess reagent in grams is left at the end of the reaction? (1)
[OR]
(b) (i) A student reported the ionic radii of isoelectronic species X³⁺, Y²⁺ and Z^– as 136 pm, 64 pm and 49 pm respectively. Is that order correct? Comment. (2)
(ii) Explain any three factors which influence the ionization energy. (3)
Answer:
(a) (i) 2NH_3(g) + CO_2(g) → (NH₄)₂ CO_(aq) + H₂O_(l)
2 moles 1 mole 1 mole
No. of moles of ammonia =\(\frac{637.2}{17}\) = 37.45 mole
No. of moles of CO₂ = \(\frac{1142}{44}\) = 25.95 mole

As per the balanced equation, one mole of C02 requires 2 moles of ammonia.
∴ No. of moles of NH3 required to react with 25.95 moles of CO₂ is
= \(\frac{2}{1}\) x 25.95 = 51.90 moles.
∴ 37.45 moles of NH₃ is not enough to completely react with CO₂ (25.95 moles).
Hence, NH₃ must be the limiting reagent, and CO₂ is excess reagent.
(ii) 2 moles of ammonia produce 1 mole of urea.
∴ Limiting reagent 37.45 moles of NH₃ can produce \(\frac{1}{2}\) x 37.45 moles of urea.
= 18.725 moles of urea.
∴ The mass of 18.725 moles of urea = No. of moles x Molar mass
= 18.725 x 60
= 1123.5 g of urea.

(iii) 2 moles of ammonia requires 1 mole of CO₂.
∴ Limiting reagent 37.45 moles of NH₃ will require \(\frac{1}{2}\) x 37.45 moles of CO₂.
= 18.725 moles of CO₂.
∴ No. of moles of the excess reagent (CO₂) left = 25.95 – 18.725 = 7.225
The mass of the excess reagent (C02) left = 7.225 x 44
= 317.9 g of CO₂.
[OR]

(b) (i) X³⁺, Y²⁺, Z^– are isoelectronic.
∴ Effective nuclear charge is in the order
\(\left(Z_{\text {eff }}\right)_{Z}<\left(Z_{\text {eff }}\right)_{Y^{2+}}<\left(Z_{\text {eff }}\right)_{X^{3+}}\) and hence, ionic radii should be in the order \(r_{z}>r_{y^{2}}>r_{x^{3+}}\)
∴ The correct values are:

(ii) 1. Size of the atom: If the size of an atom is larger, the outermost electron shell from the nucleus is also larger and hence the outermost electrons experience lesser force of attraction. Hence it would be more easy to remove an electron from the outermost shell. Thus, ionization energy decreases with increasing atomic sizes.
\(Ionization enthalpy \propto \frac{1}{\text { Atomic size }}\)

2. Magnitude of nuclear charge: As the nuclear charge increases, the force of attraction . between the nucleus and valence electrons also increases. So, more energy is required to remove a valence electron. Hence I.E increases with increase in nuclear charge.
Ionization enthalpy α nuclear charge

3. Screening or shielding effect of the inner electrons: The electrons of inner shells form a cloud of negative charge and this shields the outer electron from the nucleus. This screen reduces the coulombic attraction between the positive nucleus and the negative outer electrons. If screening effect increases, ionization energy decreases.
\(Ionization enthalpy \propto \frac{1}{\text { Screening effects }}\)

4. Penetrating power of subshells s, p, d and f: The s-orbital penetrate more closely to the nucleus as compared to p-orbitals. Thus, electrons in s-orbitals are more tightly held by the nucleus than electrons in p-orbitals. Due to this, more energy is required to remove a electron from an s-orbital as compared to a p-orbital. For the same value of ‘n’, the penetration power decreases in a given shell in the order, s > p > d > f.

5. Electronic configuration: If the atoms of elements have either completely filled or exactly half filled electronic configuration, then the ionization energy increases.

Question 35.
(a) (i) Write the important common features of group 2 elements? (3)
(ii) What is meant by retrograde solubility? (2)
[OR]
(b) (i) Derive a general expression for the equilibrium constant K_p and K_c for the reaction. (3)
(ii) Write the K_p and K_c for NH₃ formation reaction. (2)
Answer:
(a) (i)

• Group 2 elements except beryllium are commonly known as alkaline earth metals because their oxides and hydroxides are alkaline in nature and these metal oxides are found in the Earth’s crust.
• Many alkaline earth metals are used in creating colours and used in fireworks.
• Their general electronic configuration is ns².
• Atomic and ionic radii of alkaline earth metals are smaller than alkali metals, on moving down the group, the radii increases.
• These elements exhibit +2 oxidation state in their compounds.
• Alkaline earth metals have higher ionizatoin enthalpy values than alkali metals and they are less electropositive than alkali metals.
• Hydration enthalpies of alkaline earth metals decreases as we go down the group.
• Electronegativity values of alkaline earth metals decrease down the group.
• Alkaline earth metal salts moistened with concentrated hydrochloric acid gave a characteristic coloured flame, when heated on a platinum wire in a flame.

(ii) Gypsum is a soft mineral and it is less soluble in water as the temperature increases. This is known as retrograde solubility, which is a distinguishing characteristic of gypsum.

[OR]

(b) (i) Consider a general reaction in which all reactants and products are ideal gases.

The equilibrium constant k_c is

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,

By comparing equation (1) and (4), we get K_p = K_c (RT)^(Δng) ………………………. (5)

where Δn_g 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.

Question 36.
(a) (i) 0.24 g 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. (3) (ii) What is a vapour pressure of liquid? (2)

[OR]

(b) (i) Explain the ionic bond formation in MgO and CaF₂. (3)
(ii) What is bond angle?
Answer:
(a) (i) \(P_{\text {solute }}=\mathrm{K}_{\mathrm{H}} x_{\text {solute in solution }}\)
At pressure 1.5 atm, \(p_{1}=\mathbf{K}_{\mathrm{H}} x_{1}\)  ……………. (1)
At pressure 6.0 atm, \(p_{2}=\mathbf{K}_{\mathbf{H}} \boldsymbol{x}_{\mathbf{2}}\) ………………… (2)
Dividing equation (1) by (2)
We get \(\frac{p_{1}}{p_{2}}=\frac{x_{1}}{x_{2}} \Rightarrow \frac{1.5}{6.0}=\frac{0.24}{x_{2}}\)
Therefore \(x_{2}=\frac{0.24 \times 6.0}{1.5}=0.96 \mathrm{g} / \mathrm{L}\)

(ii) 1. The pressure of the vapour in equilibrium with its liquid is 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 = \(\frac{p_{\text {solvent }}^{\circ}-p_{\text {solution }}}{p_{\text {solvent }}^{\circ}}\)

[OR]

(b) (i) Magnesium oxide (MgO) :
Electronic configuration of Mg – ls² 2s² 2p⁵ 3s²
Electronic configuration of O – ls² 2s² 2p⁶ 3s² 3p⁴

• Magnesium has two electrons in its valence shell and oxygen has six electrons in its valence shell.
• By losing two electrons, Mg acquires the inert gas configuration of Neon and becomes a dipositive cation, Mg²⁺:
  \(\mathrm{Mg} \longrightarrow \mathrm{Mg}^{2+}+2 e^{-}\)
• Oxygen accepts the two electrons to become a dinegative oxide anion, O^2- thereby attaining the inert gas configuration of Neon:
  \(0+2 e^{-} \longrightarrow 0^{2-}\)
• These two ions, Mg²⁺ and O^2- combine to form an ionic crystal in which they are held together by electrostatic attractive forces.
• 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:

• Calcium, Ca : [Ar] 4s² Fluorine, F : [He] 2s² 2p⁵
• Calcium has two electrons in its valence shell and fluorine has seven electrons in its valence shell.
• By losing two electrons, calcium attains the inert gas configuration of Argon and becomes a dipositive cation, Ca²⁺.
• Two fluorine atoms, each one accepts one electron to become two uninegative fluoride ions (F-) thereby attaining the stable configuration of Neon.
• These three ions combine to form an ionic crystal in which they are held together by electrostatic attractive force.
• 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.

(ii) 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 37.
(a) (i) 0.33 g of an organic compound containing phosphorous gave 0.397 g of Mg₂P₂O₇ by the analysis. Calculate the percentage of P in the compound. (3)
(ii) Give the IUPAC names of the following compounds. (2)

[OR]

(b) (i) Describe the mechanism of nitration of benzene. (3)
(ii) How will you prepare m-dinitro benzene. (2)
Answer:
(a) (i) Weight of organic substance (w) = 0.33g
Weight of Mg₂P₂O₇ (x) = 0.397g
Percentage of phosphorous = \(\frac{62}{222} \times \frac{x}{w} \times 100=\frac{62}{222} \times \frac{0.397}{0.33} \times 100=33.59 \%\)
(ii) (a) 4-chloropent-2-ype .
(b) Acetophenone

(b) (i) Step-1: Generation of NO₂ electrophile.
HNO₃ + H₂SO₄ → NO₂ + HSO₄ + H₂O

Step-2: Attack of the electrophile on benzene ring to form arenium ion.

Question 38.
(a) Complete the following reaction, identify the products.

[OR]
(b) (i) What is green chemistry?
(ii) Differentiate – BOD and COD.
Answer:

[OR]
(b) (i)

• Green chemistry is a chemical philosophy encouraging the design of products and processes that reduces or eliminates the use and generation of hazardous substances.
• Efforts to control environmental pollution resulted in the development of science for the synthesis of chemicals favourable to the environment.
• Green chemistry means the science of environmentally favourable chemical synthesis.

(ii)

Students can Download Tamil Nadu 11th Chemistry Previous Year Question Paper June 2019 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Chemistry Previous Year Question Paper June 2019 English Medium

Time: 2½ Hours
Total Score: 90 Marks

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I. II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail. Draw diagrams wherever necessary.

Part-I

Answer all the Questions. [15 x 1 = 15]
Choose the most appropriate answers from the given alternatives and write the option code and corresponding answer.

Question 1.
The oxidation number of Carbon in CH₂F₂ is …………………………. .
(a) + 4
(b) – 4
(c) 0
(d) + 2
Answer:
(c) 0

Question 2.
The energy of an electron in the third orbit of hydrogen atom is – E. The energy of an electron in the first orbit will be.
(a) – 3E
(b) \(-\frac{E}{3}\)
(c) \(-\frac{E}{9}\)
(d) – 9E
Answer:
(d) – 9E

Question 3.
The effective nuclear charge experienced by d1 electron in the given electronic configuration, (Is)² (2s, 2p)⁸ (3s, 3p)⁸ (3d)¹ (4s)² is ……………………. .
(a) 4
(b) 3
(c) 2.1
(d) 6.9
Answer:
(b) 3

Question 4.
The type of H-bonding present in orthonitro phenol and p-nitro phenol are respectively.
(a) Inter molecular H-bonding and intra molecular H-bonding
(b) Intra molecular H-bonding and inter molecular H-bonding
(c) Intra molecular H-bonding and no H-bonding
(d) Intra molecular H-bonding and intra molecular H-bonding
Answer:
(b) Intra molecular H-bonding and inter molecular H-bonding

Question 5.
When CaC₂ is heated in atmospheric nitrogen in an electric furnace, the compound formed is
(a) Ca(CN)₂
(b) CaNCN
(c) CaC₂N₂
(d) CaNC₂
Answer:

Question 6.
When an ideal gas undergoes unrestrained expansion, on cooling occurs because the molecules …………………………. .
(a) are above the inversion temperature
(b) exert on attractive force on each other
(c) do work equal to the loss in kinetic energy
(d) collide with loss of energy
Answer:
(b) exert on attractive force on each other

Question 7.
Among the following statements, which on is/are correct?
(i) During cyclic process the amount of heat absorbed by the surrounding is equal to work done on the surrounding.
(ii) Refractive index is an example for intensive property.
(iii) If the enthalpy change of a process is positive then the process is spontaneous.
(iv) The entropy of an isolated system increases during spontaneous process.
(a) (i), (ii), (iii)
(b) (i), (iv)
(c) (ii), (iv)
(d) (ii) only
Answer:
(d) (ii) only

Question 8.
If Kb and Kf for reversible reaction are 0.8 x 10 5 and 1.6 x 10^ respectively, the value of equilibrium constant is.
(a) 20
(b) 0.2 x 10^-1
(c) 0.05
(d) 0.2
Answer:
(a) 20

Question 9.
Assertion: Mixture of carbon tetrachloride and chloroform show positive deviation from Raoult’s law.
Reason: In the mixture, the inter molecular force of attraction between chloroform and carbon tetrachloride is weaker than those between molecules of carbon tetrachloride and chloroform molecules.
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.
(b) Both assertion and reason are correct and reason is not the correct explanation for assertion.
(c) Both assertion and reason are false.
(d) Assertion is true, but reason is false.
Answer:
(a) Both assertion and reason are correct and reason is the correct explanation of assertion.

Question 10.
Shape and hybridisation of IF₅ are
(a) Trigonal bipyramidal, sp³d²
(b) Trigonal bipyramidal, sp³d
(c) Square pyramidal, sp³d²
(d) Octahedral, sp³d²
Answer:
(c) Square pyramidal, sp³d²

Question 11.
Which of the following is optically active?
(a) 3-chloro pentane
(b) 2-chloro propane
(c) meso-tartaric acid
(d) glucose
Answer:
(d) glucose

Question 12.
Which of the following species is not electrophile in nature? ‘
(a) Cl⁺
(b) BH₃
(c) H₃O⁺
(d) ⁺NO₂
Answer:
(c) H₃O⁺

Question 13.
group is ortho para directing and deactivating group.
(a) amino
(b) methyl
(c) halogen
(d) aldehyde
Answer:
(c) halogen

Question 14.
The raw material for Rasching process is
(a) chloro benzene
(b) phenol
(c) benzene
(d) anisole
Answer:
(c) benzene

Question 15.
cause kidney damage.
(a) Cadmium, Mercury
(b) Lead, Cadmium
(c) Freon, Fluoride
(d) Copper, Cadmium
Answer:
(a) Cadmium, Mercury

Part-II

Answer any six from the following questions. Question No. 24 is compulsory. [6 x 2 = 12]

Question 16.
What is syngas? How it is prepared?
Answer:
The mixture of CO and H₂ gases are called syngas.
Preparation: C + H₂O → CO + H₂

Question 17.
Write any two similarities between beryllium and aluminum.
(i) Beryllium and aluminium have same electronegativity values.
(ii) BeCl₂ and A1C1₃ forms dimeric structure. Both are soluble in organic solvents and are strong Lewis acids.

Question 18.
What is inversion temperature?
Answer:
The temperature below which a gas obey Joule-Thomson effect is called inversion temperature
\(\text { (T) } . \quad \mathrm{Ti}=\frac{2 \mathrm{a}}{\mathrm{Rb}}\)

Question 19.
What is the effect of added inert gas on the reaction of equilibrium?
Answer:
When an inert gas (i.e., a gas which does not react with any 6ther 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 20.
Linear form of carbon dioxide molecule has two polar bonds. Yet the molecule has zero dipole moment. Why?
Answer:

• The linear form of carbon dioxide has zero dipole moment, even though it has two polar bonds.
• In C02, there are two polar bonds [C = O], which have dipole moments that are equal in magnitude but have opposite direction.
• Hence the net dipole moment of the CO₂ is μ = μ₁ + μ₂ = μ₁ + (- μ₁) = 0
  • 
• In this case, \(\mu=\vec{\mu}_{1}+\vec{\mu}_{2}=\vec{\mu}_{1}+\left(-\vec{\mu}_{1}\right)=0\)

Question 21.
How do you detect the presence of nitrogen and sulphur together in an organic compound?
Answer:
If both N and S are present, a blood-red colour is obtained due to the following reactions

Question 22.
What happens when acetylene undergoes ozonolysis?
Answer:
When acetylene undergoes ozonolysis,

Question 23.
What is Green chemistry?
Answer:

• Green chemistry is a chemical philosophy encouraging the design of products and processes that reduces or eliminates the use and generation of hazardous substances.
• Efforts to control environmental pollution resulted in development of science for the synthesis of chemicals favourable to environment.
• Green chemistry means science of environmentally favourable chemical synthesis.

Question 24.
Calculate the orbital angular momentum for d and f orbital.
Answer:
Orbital angular momentum = √1(1 (1+1) h/2π.
For d orbital = √(2 x 3) h/2π = √6 h/2π
For f orbital =  √(3(3 + 1)) h/2π = √12 h/2π

Part-III

Answer any six froita the following questions. Q. No. 33 is compulsory. [6 x 3 = 18]

Question 25.
What do you understand by the term mole?
Answer:
The mole is defined as the amount of a substance which contains 6.023 x 10²³ particles such as atoms, molecules or ions. It is represented by the symbol “n”.

Question 26.
Ionisation potential of nitrogen is greater than that of oxygen. Explain by giving appropriate reason.
Answer:
N (Z = 7) Is² 2s² 2p_x¹ 2p_y¹ 2p_z¹. It has exactly half filled electronic configuration and it is more stable. Due to stability, ionization energy of nitrogen is high.

O (Z = 8) Is² 2s² 2p_x¹ 2p_y¹ 2p_z¹ It has incomplete electronic configuration and it requires less ionization energy. I.E₁ N > I.E₁ O

Question 27.
Among the alkali metal halides, which is covalent? Explain with reason.
Answer:
All alkali metal halides are ionic crystals. However lithium halides shows covalent character, as it is the smallest cation that exerts high polarising power on the halides.

Question 28.
Derive ideal gas equation.
Answer:
The gaseous state is described completely using the following four variables T, P, V and n. Each gas law relates one variable of a gaseous sample to another while the other two variables are held constant.

Therefore, combining all equations into a single equation will enable to account for the change in any or all of the variables.

• Boyle’s law: V α \(\frac{1}{P}\)
• Charles’law: V α T
• Avogadro’s law: V α n

We can combine these equations into the following general equation that describes the physical behaviour of all gases.
\(\mathrm{V} \propto \frac{\mathrm{nT}}{\mathrm{P}}\)
\(\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}\), where R = Proportionately constant.

The above equation can be rearranged to give PV = nRT – Ideal gas equation. Where, R is also known as Universal gas constant.

Question 29.
Define molar heat capacity. Give its unit.
Answer:
Molar heat capacity is defined as “the amount of heat absorbed by one mole of a substance in raising the temperature by 1 Kelvin”. It is denoted as C_m.

Unit of Molar heat capacity: SI unit of C_m is JK^-1 mol^-1.

Question 30.
What is vapour pressure of a liquid? What is relative lowering of vapour pressure?

1. The pressure of the vapour in equilibrium with its liquid is 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.
3. Relative lowering of vapour pressure = \(\frac{p_{\text {solvent }}-p_{\text {solution }}}{p_{\text {solvent }}}^{\circ}\)

Question 31.
Explain a suitable method for purifying and separating liquids present in a mixture having very close boiling point.
Answer:
Fractional distillation: This method is used to purify and separate liquids present in the mixture having their boiling-point close to each other. The process of separation of the components in liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. This method is applied in distillation of petroleum, coal tar and crude oil.

Question 32.
What is polymerisation? Explain the two types of polymerisation reaction of acetylene.
ANswer:
A polymer is a large molecule formed by the combination of large number of small molecules (monomers). This process is known as polymerisation.

Acetylene undergoes two types of polymerisation reactions, they are:

• Linear polymerisation
• Cyclic polymerisation

(i) Linear polymerisation: Acetylene forms linear polymer, when passed into a solution of cuprous chloride and ammonium chloride.

(ii) Cyclic polymerisation: Acetylene undergoes cyclic polymerisation on passing through red hot iron tube. Three molecules of acetylene polymerises to form benzene.

Question 33.
The bond length between all the four carbon atoms is same in 1,3 – butadine. Explain with reason.
Answer:

1,3-butadiene is a conjugated molecule with four overlapping p-orbital on adjacent atoms. And π-electrons are delocalised over four atoms. This shortens the bond length of central C – C bond. Thus, the bond length between all the four C-atoms are same in 1,3-butadiene.

Part – IV

Answer all the following questions. [5 x 5 = 25]

Question 34.
(a) (i) What are auto redox reactions? Give example.
(ii) Define orbital. What are the n and 1 values of 3p_x and 4d_x²-y² electron?
[OR]
(b) (i) Why hydrogen peroxide is stored in plastic containers, not in glass container?
(ii) Give the general electronic configuration of lanthanides and actinides.
Answer:
(a) (i) Auto redox reactions: It is a redox reaction in which a substance act as both the oxidizing agent and the reducing agent. (Reaction occur in the same substance)

(ii) 1. Orbital is a three dimensional space which the probability of finding the electron is maximum.
2. For 3p_x electron – n value =3, l value =1
3. For 4d_x²-y² electron n value = 4, l value = 2

[OR]

(b) (i) The aqueous solution of hydrogen peroxide is spontaneously disproportionate to give oxygen. The reaction is slow but it is explosive when it is catalyzed by metal or alkali dissolved from glass. For this reason, its solution are stored in plastic bottles.
\(\mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})} \rightarrow \mathrm{H}_{2} \mathrm{O}_{(\mathrm{i})}+1 / 2 \mathrm{O}_{2(\mathrm{g})}\)

(ii) 1. The electronic configuration of lanthanides is 4f^1-14 5d^0-1 6s².
2. The electronic configuration of actinides is 5f^1-14 6d^0-1 7s².

Question 35.
(a) (i) Why blue colour appears during the dissolution of alkali metals in liquid ammonia?
(ii) What is Boyle’s temperature? What happens to real gases above and below the Boyle’s temperature?

[OR]

(b) (i) Derive the relation between KP and Kc for general homogeneous gaseous reaction.
(ii) How do you measure heat changes of a constant pressure?
Answer:
(a) (i) M + (x + y)NH₃ → [M(NH₃)_x]⁺ + [e(NH₃)_y]^–
The blue colour of the solution is due to the ammoniated electron which absorbs energy in the visible region of light and thus imparts blue colour to the solution. The solutions are paramagnetic and on standing slowly liberate hydrogen resulting in the formation of amide. In concentrated solution, the blue colour changes to bronze colour and become diamagnetic.

(ii)

1. Over a range of low pressures, the real gases can behave ideally at a particular temperature called as Boyle temperature or Boyle point.
2. The Boyle point varies with the nature of the gas.
3. Above the Boyle point, the compression point Z > 1 for real gases i.e. real gases show positive deviation.
4. Below the Boyle point, the real gases first show a-decrease for Z, reaches a minimum and then increase with increase in pressure. So, it is clear that at low pressure and at high temperature, the real gases behave as ideal gases.
5. 

[OR]

(b) (i) Consider a generaL reaction in which all reactants and products arc ideal gases.

P = Active mass x RT
Based on the above expression, the partial pressure of the reactants and products can be expressed as.
\(\begin{array}{ll}
p_{A}^{x}=[\mathrm{A}]^{x} \cdot[\mathrm{RT}]^{x} & p_{\mathrm{B}}^{y}=[\mathrm{B}]^{y} \cdot[\mathrm{RT}]^{y} \\
p_{\mathrm{C}}^{\prime}=[\mathrm{C}]^{\prime} \cdot[\mathrm{RT}]^{\prime} & p_{\mathrm{D}}^{m}=[\mathrm{D}]^{m} \cdot[\mathrm{RT}]^{m}
\end{array}\)

On substitution in equation (2),

where Δn_g 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.

(ii) 1. Measurement of heat change at constant pressure can be done in a coffee cup calorimeter.
2. We know that ΔH = q_p (at constant P) and therefore, heat absorbed or evolved, q_p at constant pressure is also called the heat of reaction or enthalpy of reaction, ΔH_r.
3. In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, q_p will be negative and Reaction AH. will also be negative. Mixture
4. Similarly in an endothermic reaction, heat is absorbed, q_p is positive and ΔH_r will also be positive.

Question 36.
(a) (i) Draw the M.O. diagram of oxygen molecule. Calculate its bond order and magnetic character.
(ii) Draw and explain the graph obtained by plotting solubility versus temperature for calcium chloride.

[OR]

(b) (i) Write the IUPAC names of the following compunds:

(ii) Calculate the formal charge on carbon and oxygen for the following structure

(a)

1. Electronic configuration of O atom is Is² 2s² 2p⁴
2. Electronic configuration of O₂ molecule is

3. Bond order \(\frac{N_{b}-N_{a}}{2}=\frac{10-6}{2}=2\)
4. Molecule has two unpaired electrons, hence it is paramagnetic.

(ii) 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 the equilibrium.

[OR]

(b) (i) (x) 5-oxohexanoic acid (y) 4-chloropent-2-yne (z) 2-ethoxypropane
(ii)

Question 37.
(a) (i) Explain about the inductive effect.
(ii) What do you mean by confirmation? Explain about staggered conformation in ethane.

[OR]

(b) (i) Among the following compounds, o-dichloro benzene and p-dichloro benzene, which has higher melting point? Explain with reason.
(ii) Write notes on the adverse effect caused by ozone depletion.
Answer:
(a) (i) Inductive effect:
1. It is defined as the change in the polarization of a covalent bond due to the presence of adjacent bonded atoms or groups in the molecule. It is denoted as I-effect.
2. Atoms or groups which lose electron towards a carbon atom are said to have a +1 effect.
Example, CH₃ —, (CH₃)₂ CH —, (CH₃)₂ C— etc.
3. Atoms or groups which draw electrons away from a carbon atom are said to have a -I effect.
Example, —NO₂. -I, Br, -OH, C₆H₅ etc.
4. For example, consider ethane and ethyl chloride. The C—C bond in ethane is non¬polar while the C-C bond in ethyl chloride is polar. We know that chlorine is more electronegative than carbon and hence it attracts the shared pair of electrons between C—Cl in ethyl chloride towards itself.

This develops a slight negative charge on chlorine and a slight positive charge on carbon to which chlorine is attached. To compensate it, the C₁ draws the shared pair of electron between itself and C₂. This polarization effect is called inductive effect.

(ii) The rotation about C—C single bond axis yielding several arrangements of a hydrocarbon called conformers.

Staggered conformation: In this conformation, the hydrogens of both the carbon atoms are far apart from each other. The repulsion between the atoms is minimum and it is the most stable conformer.

[OR]

(b) (i) The melting point of p-dichloro benzene is higher than the melting point of the corresponding ortho and meta isomers. The higher melting which leads to more close packing of its moleules in the crystal lattice and consequently strong intermolecular attractive forces which requires more energy for melting.
p-dichlorobenzene > o-dichlorobenzene > m-dichlorobenZene

(ii) 1. The formation and destruction of ozone is a regular natural process, which never disturbs the equilibrium level of ozone in the stratosphere. Any change in the equilibrium level of ozone in the atmosphere will adversely affect the life in biosphere in the following ways.
2. Depletion of ozone layer will allow more UV rays to reach the earth surface and would cause skin cancer and also decreases the immunity level in human beings.
3. UV radiations affects plant proteins which lead to harmful mutation in plant cells.
4. UV radiations affect the growth of phytoplankton and as a result ocean food chain is disturbed and it even damages the fish productivity.

Question 38.
(a) (i) Calculate the uncertainty in the position of an electron, if the uncertainty in its velocity is 5.7 x 10⁵ ms^-1.
(ii) What is the mass of glucose (C₆ H₁₂ O₆) in it one litre solution which is isotonic with 6 gl^-1 of urea (NH₂ CONH₂)?

[OR]

(b) (i) An organic compound (A) of molecular formula C₂H₆O, on heating with conc.H₂SO₄ gives compound (B). (B) on treating with cold dilute alkaline KMnO₄ gives compound (C). Identify (A), (B) and (C) and explain the reactions.
(ii) A simple aromatic hydrocarbon (A) reacts with chlorine to give compound (B). Compound (B) reacts with ammonia to give compound (C) which undergoes carbylamine reaction. Identify (A), (B) and (C) arid explain the reactions.

(a) (i) Uncertainty in velocity = Δv = 5.7 x 10⁵ ms^-1.
Mass of an electron = m = 9.1 x 10^-31 kg.
Uncertainty in position = Δx = ?

(ii) Osmotic pressure of urea solution (π₁) = CRT

(b)

Students can Download Tamil Nadu 11th Chemistry Model Question Paper 2 English Medium Pdf, Tamil Nadu 11th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 11th Chemistry Model Question Paper 2 English Medium

Time: 2½ Hours
Total Score: 70 Marks

General Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I. II, III and IV are to be attempted separately.
4. Question numbers 1 to 15 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered in detail. Draw diagrams wherever necessary.

PART-I

Answer all the Questions. [15 x 1 = 1]
Choose the most suitable answer from the given four alternatives.

Question 1.
Match the List-I with List-Il using the correct code given below the list.

Code :
A B C D
(a) 4 3 1 2
(b) 3 4 1 2
(c) 2 1 4 3
(d) 3 1 2 4
Answer:
(b) 3 4 1 2

Question 2.
The energies E₁ and E₂ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e. Xl and X2 will be ……………… .
(a) \(\frac{\lambda_{1}}{\lambda_{2}}=1\)
(b) λ₁ = 2λ₂
(c) \(\lambda_{1}=\sqrt{25 \times 50} \lambda_{2}\)
(d) 2λ₁ = λ₂
Answer:
(b) λ₁ = 2λ₂

Solution:

Question 3.
In which of the following options the order of arrangement does not agree with the variation of property indicated against it? (NEET 2016 Phase-l)
(a) 1 < Br < Cl < F (increasing electron gain enthalpy)
(b) Li < Na < K < Rb (increasing metallic radius)
(c) AI³⁺ < Mg²⁺ < Na⁺ < F- (increasing ionic size)
(d) B < C < O < N (increasing first ionization enthalpy)
Answer:
(a) 1 < Br < Cl < F (increasing electron gain enthalpy)

Question 4.
For decolourisation of I mole of acidified KMnO4, the moles of H202 required is
(a) 1/2
(b) 3/2
(c) 5/2
(d) 7/2
Answer:
(c) 5/2

Question 5.
The compound (X) on heating gives a colourless gas and a residue that is dissolved in water to obtain (B). Excess of CO₂ is bubbled through aqueous solution of B, C is formed. Solid (C) on heating gives back X. (B) is ………………………. .
(a) CaCO₃
(b) Ca(OH)₂
(c) Na₂CO₂
(d) NaHCO₃
Answer:
(b) Ca(OH)₂

Solution:

Question 6.
Which law is used in the process of enriching the isotope of U²³⁵ from other isotopes?
(a) Boyle’s law
(b) Dalton’s law of partial pressure
(c) Graham’s law of diffusion
(d) Charles’ law
Answer:
(c) Graham’s law of diffusion

Question 7.
The enthalpies of formation of Al₂O₃ and Cr₂O₃ are -1596 kJ and -1134 kJ, respectively. ΔH for reaction 2A1 + Cr₂O₃ → 2Cr + Al₂O₃ is …………………… .
(a) -1365 kJ
(b) 2730 kJ
(c) -2730 kJ
(d) -462 kJ
Answer:
(d) -462 kJ

Solution:

Question 8.
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:

increase in pressure, favours the forward reaction.

Question 9.
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: ΔH_mix is negative and show negative deviation from Raoults law.

Question 10.
Which of the following molecules have bond order equal to 1?
(a) NO, HF, HC1, Li₂, CO
(b) H₂, Li₂, HF, Br₂, HC1
(c) Li₂, B₂, CO, NO, He₂⁺
(d) B₂, CO, He₂⁺, NO, HF
Answer:
(b) H₂, Li₂, HF, Br₂, HC₁

Question 11.
How many cyclic and acyclic isomers are possible for the molecular formula C3H60?
(a) 4
(b) 5
(c) 9
(d) 10
Answer:
(c) 9

Question 12.
Statement-I: Fluoro acetic acid is stronger acid than acetic acid
Statement-II: Fluorine has high electronegativity and it is facilitate to dissociate the O-H bond easily.
(а) Statement-I and II are correct and statement-II is correct explanation of statement-I.
(b) Statement-I and II are correct but statement-II is not 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:
(а) Statement-I and II are correct and statement-II is correct explanation of statement-I.

Question 13.
Consider the following statements.
(i) Alkanes are saturated hydrocarbons in which all the bonds between the carbon atoms are single bond.
(ii) Alkenes are saturated hydrocarbons in which atleast one carbon – carbon double bond is present.
(iii) Alkynes are unsaturated hydrocarbons in which atleast one carbon – carbon triple bond is present.
Which of the above statement is/are correct?
(a) (i) and (ii)
(b) (i) and (iii)
(c) (ii) and (iii)
(d) only (ii)
Answer:
(b) (i) and (iii)

Question 14.
Product X is ……………….. .
(a) CH₄
(b) C₂H₆
(c) HCHO
(d) CH₃OH
Answer:
(a) CH₄

Question 15.
Eutrophication causes reduction in
(a) dissolved oxygen
(b) dissolved nitrogen
(c) dissolved salts
(d) all of the above
Answer:
(a) dissolved oxygen

Part-II

Answer any six questions in which question No. 22 is compulsory. [6 x 2 = 12]

Question 16.
Explain about the factors that affect electronegativity?
Answer:

• Effective nuclear charge: As the nuclear charge increases, electronegativity also increases along the periods.
• Atomic radius: The atoms in smaller size will have larger electronegativity.

Question 17.
How do you convert para-hydrogen into ortho-hydrogen?
Answer:
Para hydrogen can be converted into ortho hydrogen by the following ways:

• By treating with catalysts platinum or iron.
• By passing an electric discharge.
• By heating > 800°C.
• By mixing with paramagnetic molecules such as O₂, NO, NO₂.
• By treating with nascent/atomic hydrogen.

Question 18.
Gases don’t settle at the bottom of a container?
Answer:
Gases by definition are the least dense state of matter. They have negligible intermolecular forces of attraction. So they are all free to roam separately. So the least dense gas particles will not sink at the bottom of a container.

Question 19.
Define – Normality.
Answer:
Normality (N): It is defined as the number of gram equivalents of solute in 1 litre of the solution.
\(\mathrm{N}=\frac{\text { Number of gram equivalent of solute }}{ \text { Volume of solution (in }\mathrm{L})}\)

Question 20.
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 21.
In CH4, NH3 and H20 the central atom undergoes sp3 hybridization – yet their bond angles are different. Why?
Answer:

1. In CH₄, NH₂ 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 are 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 Ip-lp. Again repulsion between Ip-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 Ip-bp. So 3 bonds are more restricted to form pyramidal shape with bond angle equal to 107° 18’.

Question 22.
Identify the compound in the following reaction,

Answer:

Question 23.
ClassIfy the following compounds in the form of alkyl, allylic, vinyl, benzyllc halides,
(a) CH₂ = CH – CH₂ CI
(b) C₆H₅CH₂I
(c) CH₃ – CH – CH₃
(d) CH₂ = CH – Cl
Answer:

Question 24.
Define — Smog.
Answer:

1. Smog is a combination of’ smoke and fog which form droplets that remains suspended in the air.
2. Smog is a chemical misture of gases that forms a brownish yellow haze. It mainly consists of ground level ozone, oxideš of nitrogen, volatile organic compounds, SO2, acidic aerosols and some other gases.

Part – III

Answer any six questions in which question No. 30 is compulsory. [6 x 3 = 18]

Question 25.
Explain how effective nuclear charge is related with stability of the orbital?
Answer:

1. In a multi-electron atom, in addition to the electrostatic attractive force between the electron and nucleus, there exists a repulsive force among the electrons.
2. These two forces are operating in the opposite direction. This results in the decrease in the nuclear force of attraction on electron.
3. The net charge experienced by the electron is called effective nuclear charge.
4. The effective nuclear charge depends on the shape of the orbitals and it decreases with increase in azimuthal quantum number.
5. The order of the effective nuclear charge felt by a electron in an orbital within the given shell is s > p > d > f.
6. Greater the effective nuclear charge, greater is the stability of the orbital. Hence, within a given energy level, the energy of the orbitals are in the following order s < p < d < f.

Question 26.
Write a notes on hydrogen sponge.
Answer:
(i) Hydrogen sponge (or) Metal hydride e.g., palladium-hydrogen system is a binary hydride (Pdll).
(ii) Upon heating, H atoms diffuse through the metal to the surface and recombine to form molecular hydrogen. Since no other gas behaves this way with palladium, this process has been used to separate hydrogen gas from other gases:
\(2 \mathrm{Pd}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{g})} \rightleftharpoons 2 \mathrm{PdH}_{(\mathrm{s})}\)
(iii) The hydrogen molecule readily adsorb on the palladium surface, where it dissociates into atomic hydrogen. The dissociated atoms, dissolve into the interstices or voids (octahedral or tetrahedral) of the crystal lattice.

(iv) Technically, the formation of metal hydride is by chemical reaction but it behaves like a physical storage method, i. e., it is absorbed and released like a water sponge. Such a reversible uptake of hydrogen in metals and alloys is also attractive for hydrogen storage and for rechargeable metal hydride battery applications.

Question 27.
In what way lithium differs from other metals of the same group?
Answer:

Lithium	Other elements of the family
Very hard.	Very Soft.
High melting and boiling point.	Low melting and boiling point.
Least reactive.	More reactive.
Reacts with nitrogen to get Li3N.	No reaction.
Reacts with bromine slowly.	Reacts violently.
Burnt in air gives monoxide only.	Burnt in air gives peroxides also, apart from monoxides. K, Rb and Cs gave superoxides.
Compounds are partially soluble in water.	Highly soluble in water.
Lithium nitrate decomposes to form an oxide.	Other metals on heating gives nitrite.
Extremely small in size.	Comparatively large in size.
Li+ has greater polarizing power.	Other M+ ions have comparatively larger polarizing power.

Question 28.
A 0.25 M glucose solution at 370.28 K has approximately the pressure as blood does what is the osmotic pressure of blood?
Sol.
C = 0.25 M
T = 370.28 K
(π)_glucose = CRT
(π) = 0.25 mol L^-1 x 0.082 L atm K^-1 mol^-1 x 370.28K
= 7.59 atm

Question 29.
Write the important principles of VSEPR theory to predict the shape of molecules.
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 are at maximum distance, so that the repulsive force will be minimum.
4. A multiple bond is treated as if it is a single bond and the remaining electron pairs which constitute the bond may be regarded as single super pair.

Question 30.
0.26 g of an organic compound gave 0.039 g of water and 0.245 g of CO₂ on combustion. Calculate the percentage of carbon and hydrogen.
Answer:
Weight of organic compound = 0.26 g
Weight of water = 0.039 g
Weight of CO₂ = 0.245 g

Percentage of hydrogen
18 g of water contains 2 g of hydrogen
0.039 g of water contains = \(\frac{2}{18} \times \frac{0.039}{0.26} \text { of } \mathrm{H}\)
∴ % of hydrogen = \(\frac{0.039}{0.26} \times \frac{2}{18}\) x 100 = 1.66%

Percentage of carbon
44 g of CO₂ contains 12 g of C
0. 245 g of CO₂ contains = \(\frac{12}{44} \times \frac{0.245}{0.26}\) g of C
% of Carbon = \(\frac{12}{44} \times \frac{0.245}{0.26}\) x 100 = 25.69%

Question 31.
How will you distinguish l-butyne and 2-butyne?

Answer:
In l-butyne, terminal carbon atom contains atom one acidic hydrogen, therefore it will react with silver nitrate in the presence of ammonium hydroxide to give silver butynide. Whereas 2-butyne does not undergo such type of the reaction, because of the absence of acidic hydrogen.

Question 32.
Write the structure of the following compounds,
(a) l-Bromo-4 -ethyl cyclohexane
(b) 1, 4-Dichlorobut-2-ene
(c) 2-Chloro-3-methyl pentane
Answer:
(a) l-bromo-4 ethylcyclo hexane
(b) 1,4-dichloro but-2-ene
(c) 2-chloro-3-methyl pentane Br

Question 33.
Explain how does green house effect cause global warming.
Answer:

1. The earth’s atmosphere allows most of the visible light from the sun to pass through and reach the earth’s surface. As earth’s surface is heated by sunlight, it radiates a part of this energy back towards the space as longer IR wavelengths.
2. Some of the heat is trapped by CH₄, CO₂, CFCs and water vapour present in the atmosphere. They absorb IR radiations and block a large portion of earth’s emitted radiations.
3. The radiations thus absorbed is partly remitted to the earth’s surface. Therefore the earth’s surface gets heated up by a phenomenon called greenhouse effect.
4. Thus greenhouse effect is defined as the heating up of the earth surface due to trapping of infrared radiations reflected by earth’s surface by CO₂ layer in the atmosphere. The heating up of the earth through the greenhouse effect is called global warming.

Part-IV

Answer all the questions. [5 x 5 = 25]

Question 34.
(a) (i) Write the steps to be followed for writing empirical formula. (3)
(ii) What do you understand by the terms empirical formula and molecular formula? (2)
[OR]
(b) (i) Explain the shape of s and p-orbital. (2)
(ii) The mass of an electron is 9.1 x 10^-31 kg. If its kinetic energy is 3.0 x 10^-25 J. Calculate its wavelength. (3)
Answer:
(a) (i) Empirical formula shows the ratio of numher of atoms of different elements in one molecule of the compound.

Steps for finding the Empirical formula:
The percentage of the elements in the compound is determined by suitable methods and from the data collected; the empirical formula is determined by the following steps.

1. Divide the percentage of each element by its atomic mass. This will give the relative number of atoms of various elements present in the compound.
2. Divide the atom value obtained in the above step by the smallest of them so as to get a simple ratio of atoms of various elements.
3. Multiply the figures so obtained, by a suitable integer if necessary in order to obtain whole number ratio.
4. Finally write down the symbols of the various elements side by side and put the above numbers as the subscripts to the lower right hand of each symbol. This will represent the empirical formula of the compound.
5. Percentage of Oxygen = 100 – Sum of the percentage masses of all the given elements.

(ii)

(b) (i) s-orbital: For is orbital, l = 0, m = 0, f(θ) = 1/√2 and g((φ) = 1√2π. Therefore, the angular distribution function is equal to 1/2√π. i.e. it is independent of the angle θ and φ. Hence, the probability of finding the electron is independent of the direction from the nucleus. So, the shape of the s orbital is spherical.

p-orbital: For p orbitals l = 1 and the corresponding m values are -1, 0 and +1. The three different m values indicates that there are three different orientations possible for p orbitals. These orbitals are designated as p_x, p_y and p_z. The shape of p orbitals are dumb bell shape.

(ii) Step I. Calculation of the velocity of electron
Kinetic energy =1/2 mu² = 3.0 x 10^-25 J = 3.0 x 10^-25 kg m² s^-2

Step II. Calculation of wavelength of the electron According to de Broglie’s equation,

Question 35.
(a) (i) Explain the exchange reactions of deuterium, (2)
(ii) Explain the action of soap with hard water. (3)
[OR]
(b) (i) Derive ideal gas equation. (3)
(ii) CO₂ gas cannot be liquified at room temperature. Give the reason. (2)
Answer:
(a) (i) Deuterium can replace reversibly hydrogen in compounds either partially or completely depending upon the reaction conditions. These reactions occur in the presence of deuterium.

(ii) Action of soap with hard water:

• The cleaning capacity of soap is reduced when used in hard water.
• Soaps are sodium or potassium salts of long chain fatty acids.
• When soap is added to hard water, the divalent magnesium and calcium cations present in hard water relict with soap.
• The sodium salts present in soaps are converted to their corresponding magnesium and calcium.salts which are precipitated as scum or precipitate.
  

(b) (i) Ideal gas equation:
The gaseous state is described completely using the following four variables T, P, V and n. Each gas law relates one variable of a gaseous sample to another while the other two variables are held constant.

Therefore, combining all equations into a single equation will enable to account for the change in any or all of the variables.

Boyle’s law: V α \(\frac{1}{P}\)
Charles’ law: V α T
Avogadro’s law: V α n

We can combine these equations into the following general equation that describes the physical behaviour of all gases.
\(\mathrm{V} \propto \frac{\mathrm{nT}}{\mathrm{P}}\)

The above equation can be rearranged to give PV = nRT – Ideal gas equation.

Where, R is also known as Universal gas constant.
(ii) Only below the critical temperature, by the application of pressure, a gas can be liquefied. CO₂ has critical temperature as 303.98 K. Room temperature means (30 + 273 K) 303 K. At room temperature, (critical temperature) even by applying large amount of pressure CO₂ cannot be liquefied. Only below the critical temperature, it can be liquefied. At room temperature, CO₂ remains as gas.

Question 36.
(a) (i) Why deep-sea divers use air diluted with helium gas in their air tanks? (3)
(ii) What is molal depression constant? Does it depend on nature of the solute. (2)

[OR]

(b) (i) Write the resonance structures for ozone molecule and N20? (2)
(ii) Draw MO diagram of CO and calculate its bond order. (3)
Answer:
(a) (i) 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.

(ii) K_f = molar freezing point depression constant (or) cryoscdpic constant. ΔT_f = K_f. m, where ΔT_f = depression in freezing point, m = molality of the solution K_f= cryoscopic constant If m = 1, ΔT_f = k_f

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. K_f is directly proportional to the molal concentration of the solute particles.
\(\Delta \mathrm{T}_{f}=\mathrm{K}_{f} \times \frac{\mathrm{W}_{\mathrm{B}} \times 1000}{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}\)

W_B = mass of the solute, W_A = mass of solvent, M_B = molecular mass of the solute.

[OR]

(b)

• Electronic configuration of C atom: Is², 2s² 2p²
  Electronic configuration of O atom: ls² 2s² 2p⁴
• Electronic configuration of CO molecule is: \(\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{y}^{2} \pi 2 p_{z}^{2} \sigma 2 p_{x}^{2}\)
• Bond order = \(\frac{N_{b}-N_{a}}{2}=\frac{10-4}{2}=3\)
• Molecule has no unpaired electron, hence it is diamagnetic.

Question 37.
(a) (i) What are cis-trans isomerism? Explain with example.
(ii) Give the IUPAC names of the following compounds,

[OR]

(b) (i) What is enzymatic browning?
(ii) How will you convert nitrile into primary amine?
Answer:
(a) (i)

• Geometrical isomers are the stereoisomers which have different arrangement of groups or atoms around a rigid framework of double bonds. This type of isomerism occurs due to restricted rotation of double bonds or about single bonds in cyclic compounds.
• In 2-butene, the carbon-carbon double bond is sp² hybridised. The carbon-carbon double bond consists of a a bond and a n bond. The presence of π bond lock the molecule in one position. Hence, rotation around C = C bond is not possible.
• 
• These two compounds are termed as geometrical isomers and are termed as cis and trans form.
• The cis isomer is the one in which two similar groups are on the same side of the double bond. The_trans isomer is that in which two similar groups are on the opposite side of the double bond. Hence, this type of isomerism is called cis-trans isomerism.

(ii)

[OR]

(b) (i) 1. Apples contains an enzyme called polyphenol oxidase (PPO) also known as tyrosinase.
2. Cutting an apple exposes its cells to the atmospheric oxygen and oxidizes the phenolic compounds present in apples. This is called the “enzymatic browning” that turns a cut apple brown.
3. In addition to apples, enzymatic browning is also evident in bananas, pears, avocados and even potatoes.

(ii) Converting nitrile into primary amine:

Question 38.
(a) An organic compound (A) of a molecular formula C₂H₄, decolourises bromine water. (A) on reacts with Cl₂ gives (B). (A) reacts with HBr to give (C). (A) reacts with hydrogen in the presence of Ni to give D. Identify A, B, C and D. Explain the reactions. (5)

[OR]

(b) (i) Explain the harmful effects of acid rain. (3)
(ii) What is Eutrophication? (2)
Answer:
(a) 1. C₂H₄ (A), decolorises the bromine water. Therefore it contains double bond. Hence (A) is ethylene.

[OR]

(b) (i)

1. Acid rain causes damage to buildings made us of marbles. This attack on marble is termed as stone leprosy.
   CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂↑
2. Acid rain affects plant and animal life in aquatic ecosystem.
3. It is harmful for agriculture, as it dissolves in the earth and removes the nutrients needed for the growth of plants.
4. It corrodes water pipes resulting in the leaching of heavy metals such as iron, lead and copper into drinking water which have toxic effects.
5. It causes respiratory ailment in humans and animals.

(ii) Eutrophication: When the growth of algae increases in the surface of water, dissolved oxygen in water is greatly reduced. This phenomenon is known as eutrophication. Due to this growth of fishes gets inhibited.