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Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 7 Chemical Kinetics

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Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Text Book Evalution

I. Choose the correct answer.

12th Chemistry Chapter 7 Book Back Answers Question 1.
For a first order reaction A → B the rate constant is x min^-1. If the initial concentration of A is 0.01 M, the concentration of A after one hour is given by the expression.
(a) 0.01 e^-x
(b) 1 x 10^-2 (1 – e^-60x)
(c) (1 x 10^-2) e^-60x
(d) none of these
Answer:
(c) (1 x 10^-2) e^-60x
Answer:
Solutions:

In this case
k = x min^-1 and [A₀] = 0.01 M
= 1 x 10^-2 M
t = 1 hour = 60 min
[A] = 1 x 10^-2(e^-60x)

12th Chemistry 7th Lesson Book Back Answers Question 2.
A zero order reaction X → Product, with an initial concentration 0.02M has a half life of 10 min. If one starts with concentration 0.04M, then the half life is …………….
(a) 10 s
(b) 5 min
(c) 20 min
(d) cannot be predicted using the given information
Answer:
(c) 20 min
Solutions:

Given,
[A₀] = 0.02 M ; t_1/2 = 10 min
[A₀] = 0.04 M ; t_1/2 = ?
Substitute in (1)
10 min ? 0.02 M ……………………..(2)
t_1/2 ∝ 0.04 M ……………………..(3)
Dividing Eq.(3) by Eq. (2) we get,
\(\frac { { t }^{ 1/2 } }{ 10min }\) = \(\frac { 0.04M }{ 0.02M }\)
t_1/2 = 2 x 10 min = 20 min

Chemical Kinetics Book Back Answers Question 3.
Among the following graphs showing variation of rate constant with temperature (T) for a reaction, the one that exhibits Arrhenius behavior over the entire temperature range is ……………

Answer:

Solution:

In k = In A – \(\left( \frac { { E }_{ a } }{ R } \right)\) \((\frac { 1 }{ T })\)
this equation is in the form of a straight
line equation y = c + m x
a plot of ink vs \((\frac { 1 }{ T })\) is a straight line with negative slope.

Samacheer Kalvi Guru 12th Chemistry Question 4.
For a first order react ion A → product with initial concentration x mol L^-1, has a half life period of 2.5 hours. For the same reaction with initial concentration mol L^-1 the half life is
(a) (2.5 x 2) hours
(b) \((\frac { 2.5 }{ 2 })\) hours
(c) 2.5 hours
(d) Without knowing the rate constant, t_1/2 cannot be determined from the given data
Answer:
(d) Without knowing the rate constant, t_1/2 cannot be determined from the given data.
Solutions:
For a first order reaction
t_1/2 = \(\frac { 0.693 }{ k }\) t_1/2 does not depend on the initial concentration and it remains constant (whatever may be the initial concentration)
t_1/2 = 2.5 hrs .

Samacheer 12 Chemistry Solutions Question 5.
For the reaction, 2NH₃ → N₂ + 3H₂, if

then the relation between
k₁, k₂ and k₃ is
(a) k₁ = k₂ = k₃
(b) k₁ = 3 k₂ = 2 k₃
(c) 1.5k₁ = 3 k₂ = k₃
(d) 2k₁ = k₂ = 3 k₃
Answer:
(c) 1.5k₁ = 3 k₂ = k₃
Solution:

Samacheer Kalvi 12th Chemistry Question 6.
The decomposition of phosphine (PH₃) on tungsten at low pressure is a first order reaction. It is because the …………….
(a) rate is proportional to the surface coverage
(b) rate is inversely proportional to the surface coverage
(c) rate is independent of the surface coverage
(d) rate of decomposition is slow
Answer:
(c) rate is independent of the surface coverage
Solution:
Given:
At low pressure the reaction follows first order, therefore Rate ∝ [reactant]¹ Rate ∝ (surface area) At high pressure due to the complete coverage of surface area, the reaction follows zero order. Rate ∝ [reactant]°. Therefore the rate is independent of surface area.

Chemical Kinetics Solutions Question 7.
For a reaction Rate = k [acetone]^3/2 then unit of rate constant and rate of reaction respectively is …………..
(a) (mol L^-1 s^-1), (mol^-1/2 L^1/2 s^-1)
(b) (mol^-1/2 L^1/2 s^-1), (mol L^-1 s^-1)
(c) (mol^1/2 L^1/2 s^-1), (mol L^-1 s^-1)
(d) (mol L s^-1), (mol^1/2 L^1/2 s)
Answer:
(b) (mol^1/2 L^1/2 s^-1), (mol L^-1 s^-1)
Solution:
Rate = k [A]ⁿ
Rate = \(\frac { -d[A] }{ dt } \)
unit of rate = \(\frac { mol{ L }^{ -1 } }{ s }\) = mol L^-1 s^-1
unit of rate constant = \(\frac { (mol{ L }^{ -1 }{ S }^{ -1 }) }{ { (mol{ L }^{ -1 }) }^{ n } }\)
= mol^1-n L^n-1 S^-1
in this case, rate k [Acetone]^3/2
n = 3/2
mol^1-(3/2) L^(3/2)-1 s^-1
mol^-(1/2) L^(1/2) s^-1

Samacheer Kalvi Class 12 Chemistry Solutions Question 8.
The addition of a catalyst during a chemical reaction alters which of the following quantities?
(a) Enthalpy
(b) Activation energy
(c) Entropy
(d) Internal energy
Answer:
(b) Activation energy
Solution:
A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.

Chemical Kinetics In Tamil Question 9.
Consider the following statements:
(i) increase in concentration of the reactant increases the rate of a zero order reaction.
(ii) rate constant k is equal to collision frequency A if E_a = o
(iii) rate constant k is equal to collision frequency A if E_a = o
(iv) a plot of ln (k) vs T is a straight line.
(v) a plot of In (k) vs \((\frac { 1 }{ T })\) is a straight line with a positive slope.

Correct statements are
(a) (ii) only
(b) (ii) and (iv)
(c) (ii) and (v)
(d) (i), (ii) and (v)
Answer:
(a) (ii) only
Solutions:
In zero order reactions, increase in the concentration of reactant does not alter the rate, So statement (i) is wrong.

if E_a = O (so, statement (ii) is correct, and statement (iii) is wrong)
k = A e°
k = A
in k = A – \(\left( \frac { { E }_{ a } }{ R } \right)\) \(\frac { 1 }{ T }\)
this equation is in the form of a straight line equation yc + m x. a plot of Ink vs \(\frac { 1 }{ T }\) is a straight line with negative slope so statements (iv) and (v) are wrong.

Samacheer Kalvi Guru Class 12 Chemistry Question 10.
In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively – x kJ mol^-1 and y kJ mol^-1. Therefore, the energy of activation in the backward direction is ………..
(a) (v – x)kJ mol^-1
(b) (x + y) J mol^-1
(c) (x – y) kJ mol^-1
(d) (x + y) x 10³ J mol^-1
Answer:
(d) (x + y) x 10³ J mol^-1
Solution:

12th Chemistry Samacheer Kalvi Question 11.
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200K to 400K? (R 8.314 JK^-1 mol^-1)
(a) 234.65 kJ mol^-1 K^-1
(b) 434.65 kJ mol^-1 K^-1
(c) 434.65 J mol^-1 K^-1
(d) 334.65 J mol^-1 K^-1
Answer:
(c)434.65 J mol^-1 K^-1
Solutions:

Question 12.

This reaction follows first order kinetics. The rate constant at particular temperature is 2.303 x 10² hourd. The initial concentration of cyclopropane is 0.25 M. What will be the concentration of cyclopropane after 1806 minutes? (Log 2 = 0.30 10)
(a) 0.125 M
(b) 0.215 M
(c) 0.25 x 2.303 M
(d) 0.05 M
Answer:
(b) 0.2 15 M
Solution:

Question 13.
For a first order reaction, the rate constant is 6.909 min^-1.The time taken for 75% conversion in minutes is …………
(a) \((\frac { 3 }{ 2 })\) log 2
(b) \((\frac { 3 }{ 2 })\) log 2
(c) \((\frac { 3 }{ 2 })\) log \((\frac { 3 }{ 4 })\)
(d) \((\frac { 2 }{ 3 })\) log \((\frac { 4 }{ 3 })\)
Answer:
(b) \((\frac { 3 }{ 2 })\) log 2
Solution:
k = \((\frac { 2.303 }{ t })\) log \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
[A₀] = 100
[A] = 25
[A₀]= 100; [A]=25
6.909 = \((\frac { 2.303 }{ t })\) log \((\frac { 100 }{ 25 })\)
t = \((\frac { 2.303 }{ 6.909 })\) log (4) ⇒ t = \((\frac { 1 }{ 3 })\) log 2²
t = \((\frac { 2 }{ 3 })\) log 2

Question 14.
In a first order reaction x → y; if k is the rate constant and the initial concentration of the reactant x is 0.1 M, then, the half life is ……..
(a) \((\frac { log2 }{ k })\)
(b) \((\frac { 0.693 }{ (0.1)k })\)
(c) \((\frac { In2 }{ k })\)
(d) none of these
Answer:
(c) \((\frac { In2 }{ k })\)
Solution:
k = \((\frac { 1 }{ t })\) In \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
[A₀] = 0.1
[A] = 0.05
k = \(\left( \frac { 1 }{ { t }_{ 1/2 } } \right)\) In \((\frac { 0.1 }{ 0.05 })\)
k = \(\left( \frac { 1 }{ { t }_{ 1/2 } } \right)\) In (2) ⇒ t_1/2 = \((\frac { In(2) }{ k })\)

Question 15.
Predict the rate law of the following reaction based on the data given below:
2A + B → C + 3D

(a) rate = k [A]² [B]
(b) rate = k [A][B]²
(c) rate = k [A][B]
(d) rate = k [A]^1/2 [B]^3/2
Answer:
(b) rate = k [A][B]²
Solution:
rate₁ = k [0.1]ⁿ [0.1]^m ……………(1)
rate₂ = k [0.2]ⁿ [0.1]^m …………(2)
Dividing Eq.(2) by Eq.(1)

\(\frac { 2x }{ x }\) = 2ⁿ
∴ n = 1
rate₃ = k [0.1]ⁿ [0.2]^m …………..(3)
rate₄ = k [0.2]ⁿ [0.2]^m …………..(4)
Dividing Eq.(4) by Eq.(2)

\(\frac { 8 }{ 2 } \) = 2^m
∴m = 1
∴ rate = k [A]¹ [B]²

Question 16.
Assertion: rate of reaction doubles when the concentration of the reactant is doubles if it is a first order reaction.
Reason: rate constant also doubles
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(c) Assertion is true but reason is false.
Solution:
For a first reaction, when the concentration of reactant is doubled, then the rate of reaction also doubled. Rate constant is independent of concentration and is a constant at a constant temperature, i.e., it depends on the temperature and hence, it will not be doubled and when the concentration of the reactant is doubled.

Question 17.
The rate constant of a reaction is 5.8 x 102 s¹. The order of the reaction is ………….
(a) First order
(b) zero order
(c) Second order
(a) Third order
Answer:
(a) First order
Solution:
The unit of rate constant is s^-1 and it indicates that the reaction is first order.

Question 18.
For the reaction N₂ O_5(g) → 2NO_2(g) +\(\frac { 1 }{ 2 }\) – O_2(g) the value of rate of disappearance of N₂O₅ is given as 6.5 x 10^-2 mol L^-1s^-1 The rate of formation of NO₂ and O₂ is given respectively as …………….
(a) (3.25 x 10^-2 mol L^-1s^-1) and (1.3 x 10^-2 mol L^-1s^-1)
(b) (1.3 x 10^-2 mol L^-1s^-1) and (3.25 x 10² mol L^-1s^-1)
(c) (1.3 x 10^-1 mol L^-1s^-1) and (3.25 x 10^-2 mol L^-1s^-1)
(d) None of these
Answer:
(c) (1.3 x 10^-1 mol L^-1s^-1) and (3.25 x 10^-2 mol L^-1s^-1)

Question 19.
During the decomposition of H₂O₂ to give dioxygen, 48g O₂ is formed per minute at certain point of time. The rate of formation of water at this point is …………….
(a) 0.75 mol min^-1
(b) 1.5 mol min^-1
(c) 2.25 mol min^-1
(d) 3.0 mol min^-1
Answer:
(d) 3.0 mol min^-1
Solution:
H₂O₂ → H₂O + \(\frac { 1 }{ 2 }\)O₂
Rate =
No. of moles of oxygen = \((\frac { 48 }{ 32 })\) = 1.5 mol
Rate of formation of oxygen = 2 x 1.5
= 3 mol min^-1

Question 20.
If the initial concentration of the reactant is doubled, the time for half reaction is also doubled. Then the order of the reaction is …………
(a) Zero
(b) one
(c) Fraction
(d) none
Answer:
(a) Zero
Solution:
For a first order reaction t^1/2 is independent of initial concentration .i.e., n \(\neq\) 1 for such cases

Question 21.
In a homogeneous reaction A ? B + C + D, the initial pressure was P₀ and after time t it was P. Expression for rate constant in terms of P₀, P and t will be ……….

Answer:

Solution:

Question 22.
If 75% of a first order reaction was completed in 60 minutes, 50% of the same reaction under the same conditions would be completed in ………
(a) 20 minutes
(b) 30 minutes
(c) 35 minutes
(d) 75 minutes
Answer:
(b) 30 minutes
Solution:

Question 23.
The half life period of a radioactive element is 140 days. After 560 days, 1 g of element will be reduced to
(a) \(\frac { 1 }{ 2 }\) g
(b) \(\frac { 1 }{ 4 }\) g
(c) \(\frac { 1 }{ 8 }\) g
(d) \(\frac { 1 }{ 16 }\) g
Answer:
(d) \(\frac { 1 }{ 16 }\) g
Solution:
in 140 days ⇒ initial concentration reduced to \(\frac { 1 }{ 2 }\) g
in 280 days ⇒ initial concentration reduced to \(\frac { 1 }{ 4 }\) g
in 420 days ⇒ initial concentration reduced to \(\frac { 1 }{ 8 }\) g
in 560 days ⇒ initial concentration reduced to \(\frac { 1 }{ 8 }\) g

Question 24.
The correct difference between first and second order reactions is that …………
(a) A first order reaction can be catalysed a second order reaction cannot be catalysed.
(b) The half life of a first order reaction does not depend on [A₀] the half life of a second order reaction does depend on [A₀].
(c) The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.
(d) The rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations,
Answer:
(b) The half life of a first order reaction does not depend on [A₀]; the half life of a second order reaction does depend on [A₀].
Solution:
For a first order reaction
t_1/2 = \(\frac { 0.6932 }{ k }\)
For a second order reaction

Question 25.
After 2 hours, a radioactive substance becomes \((\frac { 1 }{ 16 })\)^th of original amount. Then the half life (in mm) is ………………
(a) 60 minutes
(b) 120 minutes
(c) 30 minutes
(d) 15 minutes
Answer:
(c) 30 minutes
Solution:

II . Answer the following questions:

Question 1.
Define average rate and instantaneous rate.
Answer:
1. Average rate:
The average rate of a reaction is defined as the rate of change of concentration of a reactant (or of a product) over a specified measurable period of time.

2. insantaneous rate:
Instantaneous rate of reaction gives the tendency of the reaction at a particular point of time during its course (or) The time derivative of the concentration of a reactant (or product) converted to a positive number is called the instantaneous rate of reaction.

Question 2.
Define rate law and rate constant.
1. Rate law:
The expression in which reaction rate is given in terms of molar concentration of the reactants with each term raised to some power, which may or may not be same as the Stoichiometric coefficient of the reacting species in a balanced chemical equation.
x A + y B → products
Rate = k [A]^m [B]^m
k = Rate constant

2. Rate constant:
For a reaction involving the reactants A and B, Reaction rate = k [A]^m [B]^m The constant k is called rate constant of the reaction. If [A] = 1 M and [B] = 1 M; Reaction rate = k Thus, the rate constant (k) of a reaction is equal to the rate of reaction when the concentration of each reactant is equal to 1 mol L^-1. The change in the concentration of reactant or product per unit time under the condition of unit concentration of all the reactant.

Question 3.
Derive integrated rate law for a zero order reaction A product. A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentrations is called as zero order reactions. Such reactions are rare. Let us consider the following hypothetical zero order reaction.
A → Product
The rate law can be written
Rate = k [A]°
(∴[A]° = 1)
– d [A] k (l)
\(\frac { -d[A] }{ dt }\) = k(1)
-d[A] = k dt
Integrate the above equation between the limits of [A₀] at zero time and [A] at some later time ‘t’,

Question 4.
Define half life of a reaction. Show that for a first order reaction half life is independent of Initial concentration.
Answer:
Half life of a reaction is defined as the time required for the reactant concentration to reach one half of its initial value. For a first order reaction, the half life is a constant i.e., it does not depend on the initial concentration. The rate constant for a first order reaction is given by,

Question 5.
What is an elementary reaction? Give the differences between order and molecularity of a reaction.
Answer:
Elementary reaction – Each and every single step in a reaction mechanism is called an elementary reaction. Differences between order and molecularity:
Order of a reaction:

1. It is the sum of the powers of concentration terms involved in the experimentally determined rate law.
2. It can be zero (or) fractional (or) integer.
3. It is assigned for a overall reaction.

Molecularity of a reaction:

1. It is the total number of reactant species that are involved in an elementary step.
2. It is always a whole number, cannot be zero or a fractional number.
3. It is assigned for each elementary step of mechanism.

Question 6.
Explain the rate determining step with an example.
Answer:
1. Most of the chemical reactions occur by multistep reactions. In the sequence of steps it is found that one of the steps is considerably slower than the others. The overall rate of the reaction cannot be lower in value than the rate of the slowest step.

2. Thus in a multistep reaction the experimentally determined rate corresponds to the rate of the slowest step. The step which has the lowest rate value among the other steps of the reaction is called as the rate determining step (or) rate limiting step.

3. Consider the reaction,
2A + B → C + D
going by two steps as follows,

Here the overall rate of the reaction corresponds to the rate of the first step which is the slow step and thus the first step is called as the rate determining step of the reaction. In the above equation, the rate of the reaction depends upon the rate constant k ( only. The rate of second step dosn’t contribute experimentally determined overall rate of the reaction.
For example,
NO_2(g) + CO_2(g) → NO_(g) + CO_2(g)
Which occurs in two elementary steps:

• NO₂ + NO₂ → NO + NO₃ (Slow)
• NO₃+ CO → NO₂ + CO₂ (fast)

Because the first step is the lowest step, the overall reaction cannot proceed any faster than the rate of the first elementary step. The first elementary step in this example is therefore the rate determining step.

The rate equation for this reaction is equal to the rate is constant of step-1 multiplied by the reactants of that first step. If the rate constant of step-1 is denoted as k₁ then the rate of the first step in the reaction (and the total reaction) will be,
Rate = k, [NO₂] [NO₂]
= k₁ [NO₂]₂

Question 7.
Describe the graphical representation of first order reaction.
Answer:
Rate constant for first order reaction is,
kt = ln\(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
kt = In [A₀] – In [A]
In[A] = In [A₀] – kty = c + mx
If we follow the reaction by measuring the concentration of the reactants at regular time interval ‘t’, a plot of ln[A] against ‘t’ yields a straight line with a negative slope. From this, the rate constant is calculated.

Question 8.
Write the rate law for the following reactions.

1. A reaction that is 3/2 order in x and zero order in y.
2. A reaction that is second order in NO and first order in Br₂.

Answer:
1. \(\frac { 3 }{ 2 }\) x + y (excess) → products
– \(\frac { 3 }{ 2 }\) \(\frac { d[x] }{ dt }\) = k [x]^3/2

2. 2NO + Br₂ → products
– \(\frac { 1 }{ 2 }\) \(\frac { d[NO] }{ dt }\) = k [NO]² [Br₂]

Question 9.
Explain the effect of catalyst on reaction rate with an example.
Answer:

1. Significant changes in the reaction can be brought out by the addition of a substance called catalyst.
2. A catalyst is substance which alters the rate of a reaction without itself undergoing any permanent chemical change.
3. They may participate in the reaction, but again regenerated and the end of the reaction.
4. In the presence of a catalyst, the energy of activation is lowered and hence, greater number of molecules can cross the energy barrier and change over to products, thereby increasing the rate of the reaction.
5. For example, decomposition of potassium chlorate is enhanced by addition of MnO₂.

2KClO₃ \(\frac{\mathrm{MnO}_{4}}{\Delta} 2 \mathrm{KCl}\) + 3O₂ (MnO₂ – Catalyst)

Question 10.
The rate law for a reaction of A, B and C has been found to be rate = k[A]² [B][L]^3/2. How would the rate of reaction change when

1. Concentration of [L] is quadrupled
2. Concentration of both [A] and [B] are doubled
3. Concentration of [A] is halved
4. Concentration of [A] is reduced to(1/3) and concentration of [L] is quadrupled.

Solution:
Rate = k [A]² [B] [L]^3/2 ………….(1)
1. when [L] = [4L]
Rate = k [A]² [B] [4L]^3/2
Rate = 8 (k[A]² [B] [L]^3/2) …………………..(2)
Comparing (1) and (3) rate is increased by 8 times.

2. when [A] = [2A] and [B] = [2B]
Rate = k[2A]² [2B ] [L]^3/2
Rate = 8 (k[A]² [B] [L]^3/2 …………….(3)
Comparing (1) and (3); rate is increased by 8 times.

3. when [A] = \([\frac { A }{ 2 }]\)
Rate = k \([\frac { A }{ 2 }]\)² [L]\(\frac { 3 }{ 2 }\)
Rate = \(\frac { 1 }{ 4 }\) (k[A]² [B] [L]^3/2) ……………..(4)
Comparing (1) and ( 4); rate is reduced to \(\frac { 1 }{ 4 }\) times.

4. when [A] = \([\frac { A }{ 3 }]\) and [L] = [4L]
Rate k\(\frac { A }{ 3 }\)² [B] [4L]^3/2
Rate = \([\frac { 8 }{ 9 }]\) (k[A]² [B] [L]^3/2) ……………….(5)
Comparing (1) and (5); rate is reduced to 8/9 times.

Question 11.
The rate of formation of a dimer in a second order reaction is 7.5 x 10^-3 mol L^-1s^-1 at 0.05 mol L^-1 monomer concentration. Calculate the rate constant.
Solution:
Let us consider the dimensation of a monomer M
2M → (M)₂
Rate = k [M]ⁿ
Given that n =2 and [M] = 0.05 mol L^-1
Rate = 7.5 x 10^-3 mol L^-1s^-1
Rate 7.5 x 10³ mol L^-1 s^-1
k = \(\frac { Rate }{ { \left[ M \right] }^{ n } }\)
k= =\(\frac { 7.5\times { 10 }^{ -3 } }{ { \left( 0.05 \right) }^{ 2 } }\) = 3 mol^-1 Ls^-1

Question 12.
For a reaction x +y + z → products, the rate law is given by rate = k [x]^3/2 [y]^1/2 what is the overall order of the reaction and what is the order of the reaction with respect to z.
Solution:
Rate = k [x]^3/2 [y]^1/2
overall order = \(\left( \frac { 3 }{ 2 } +\frac { 1 }{ 2 } \right)\) = 2
i.e., second order reaction.
Since the rate expression does not contain the concentration of Z , the reaction is zero order with respect to Z.

Question 13.
Explain briefly the collision theory of bimolecular reactions.
Answer:
Collision theory is based on the kinetic theory of gases. According to this theory, chemical reactions occur as a result of collisions between the reacting molecules. Let us understand this theory by considering the following reaction.
A_2(g) + B_2(g) → 2AB_(g)

If we consider that, the reaction between A₂ and B₂ molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second. Rate ix Number of molecules colliding per litre per second (or) Rate ∝ Collision rate. The number of collisions is directly proportional to the concentration of both A₂ and B₂.
Collision rate ∝ [A₂] [B₂] …………………(1)
Collision rate = Z [A₂] [B₂] ……………………..(2)
Where, Z is a constant.

The collision rate ¡n gases can be calculated from kinetic theory of gases. For a gas at room temperature (298K) and 1 atm pressure, each molecule undergoes approximately 10⁹ collisions per second, i.e., I collision in 10⁹ second. Thus, if every collision resulted in reaction, the reaction would be complete in 10⁹ second.

In actual practice this does not happen. It implies that all collisions are not effective to lead to the reaction. In order to react, the colliding molecules must possess a minimum energy called activation energy. The molecules that collide with less energy than activation energy will remain intact and no reaction occurs.

Fraction of effective collisions (f) is given by the following expression, \({ e }^{ \frac { { -E }_{ a } }{ RT } }\)
Fraction of collisions is further reduced due to orientation factor i.e., even if the reactant collide with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state. The fraction of effective collisions (f) having proper orientation is given by the steric factor P.

Rate = P x f x collision rate
Rate= P x \({ e }^{ \frac { { -E }_{ a } }{ RT } }\) x Z [A₂] [B₂] ……..(1)
As per the rate law, Rate = k [A₂] [B₂] ………………….(2)
Where k is the rate constant
On comparing equation (1) and (2), the rate constant k is,
k = p Z \({ e }^{ \frac { { -E }_{ a } }{ RT } }\)

Question 14.
Write Arrhenius equation and explains the terms involved.
Answer:
Arrhenius equation:
k = A\({ e }^{ \frac { { -E }_{ a } }{ RT } }\)
A = Arrhenius factor (frequency factor)
R = Gas constant
k = Rate constant
E_a = Activation energy
T = Absolute temperature (in K)

Question.15.
The decomposition of Cl₂O₇ at 500K in the gas phase to Cl₂ and O₂ is a first order reaction. After 1 minute at 500K, the pressure of Cl₂O₇ falls from 0.08 to 0.04 atm. Calculate the rate constant in s^-1.
Answer:
Solution:
k = \(\frac { 2.303 }{ t }\) log \(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)
k = \(\frac { 2.303 }{ 1 min }\) log \(\frac { [0.08] }{ [0.04] }\)
k = 2.303 log 2
k = 2.303 x 0.3010
k = 0.693 2 min^-1
k = \((\frac { 0.6932 }{ 60 })\) s-1
k = 1.153 x 10^-2 s^-1

Question 16.
Give the examples for a zero order reaction.
Answer:
Examples for a zero order reaction:
1. Photochemical reaction between H₂ and Cl
H_2(g) + Cl_2(g) \(\underrightarrow { h\nu }\) 2HCI_(g)

2. Decomposition of N₂O on hot platinum surface
N₂ O_(g) \(\rightleftharpoons\) N_2(g) + \(\frac { 1 }{ 2 }\) O_2(g)

3. lodination of acetone in acid medium is zero order with respect to iodine.
CH₃COCH₃ + I₂ \(\underrightarrow { { H }^{ + } } \) ICH₂COCH₃ + HI
Rate k [CH₃COCI₃] [H⁺]

Question 17.
Explain pseudo first order reaction with an example.
Answer:
A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction. Let us consider the acid hydrolysis of an ester,
CH₃COOCH_3(aq) +H₂ O₍₁₎ \(\underrightarrow { { H }^{ + } } \) CH₃COOH_(aq) + CH₃OH_(aq)
Rate = k [CH₃COOCH₃] [H₂O]

If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis. i.e., concentration of water remains almost a constant. Now we can define k [H₂O] = k
∴ The above rate equation becomes
Rate k [CHCOOCH] Thus it follows first order kinetics.

Question 18.
Identify the order for the following reactions

1. Rusting of Iron
2. Radioactive disintegration of ₉₂U²³
3. 2 A+ B → products; rate = k [A]^1/2 [B]²

Answer:
1.
Theoritically order value may be more than one but practically one.

2. All radioactive disintegrations are first order reactions

3. 2A + 3B → products:
rate = k[A]^1/2 [B]²
Order = \(\frac { 1 }{ 2 }\) + 2 = \(\frac { 5 }{ 2 }\) = 2.5

Question 19.
A gas phase reaction has energy of activation 200 kJ mol^-1. If the frequency factor of the reaction is 1.6 x 10¹³ s^-1. Calculate the rate constant at 600 K. (e^-40.09 = 3.8 x I0^-18 )
Solution:

Question 20.
For the reaction 2x +y → L find the rate law from the following data.

Answer:
Rate = k [x]ⁿ [y]^m
0.15 = k [0.2]ⁿ [0.02]^m ……………..(1)
0.30 = k [0.4]ⁿ [0.02]^m ……………… (2)
1.20 = k [0.4]ⁿ [0.08]^m ……………… (3)
Dividing Eq(3) by Eq (2) we get

Question 21.
How do concentrations of the reactant influence the rate of reaction?
Answer:
The rate of a reaction increases with the increase in the concentration of the reactants. The effect of concentration is explained on the basis of collision theory of reaction rates.

According to this theory, the rate of a reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate.

Question 22.
How do nature of the reactant influence rate of reaction?
Answer:
Nature and state of the reactant:
We know that a chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product.

The net energy involved in this process is dependent on the nature of the reactant and hence the rates arc different for different reactants. Let us compare the following two reactions that we carried out in volumetric analysis.

1. Redox reaction between ferrous ammonium sulphate (FAS) and KMnO₄
2. Redox reaction between oxalic acid and KMnO₄

The oxidation of oxalate ion by KMnO₄ is relatively slow compared to the reaction between KMnO₄ and Fe . In fact heating is required for the reaction between KMnO₄ and Oxalate ion and is carried out at around 60°C. The physical state of the reactant also plays an important role to influence the rate of reactions. Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants.

For example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine. Let us consider another example that we carried out in inorganic qualitative analysis of lead salts.

If we mix the aqueous solution of colorless potassium iodide with the colorless solution of lead nitrate, precipitation of yellow lead iodide take place instantancously, whereas if we mix the solid lead nitrate with solid potassium iodide, yellow coloration will appear slowly.

Question 23.
The rate constant for a first order reaction is 1.54 x 10 s^-1. Calculate its half life time.
Answer:
We know that, t, 0.693 k
t_1/2 = 0.693/1.54 x 1o^-3 = 450 s

Question 24.
The half life of the homogeneous gaseous reaction SO₂CI₂ → SO₂ + Cl₂ which obeys first order kinetics Is 8.0 minutes. How long will it take for the concentration of SO₂Cl₂ to be reduced to 1% of the initial value?
Answer:
We know that, k = 0.693/ t_1/2
k = 0.693/8.0 minutes = 0.087 minutes ^-1
For a first order reaction,
k = \(\frac { 2.303 }{ k }\) log \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
t = \(\frac { 2.303 }{ 0.087{ min }^{ -1 } }\) log\(\frac { 100 }{ 1 }\)
t = 52.93 mm

Question 25.
The time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?
Answer:
1. Order of a reaction = 1
t_1/2 = 60
seconds, k = ?
k = \(\frac { 2.303 }{ 60 }\)
We know that, k = \(\frac { 2.303 }{ { t }_{ 1/2 } }\)
k = \(\frac { 2.303 }{ 60 }\) = 0.01155 s^-1

2. [A₀] = 100%
t = 180 s
k = 0.01155 seconds^-1
[A] = ?
For the first order reaction k = \(\frac { 2.303 }{ 60 }\) log \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
0.9207 = log 100 – log [A]
log [A] = log 100 – 0.9207
log [A] = 2 – 0.9207
log[A] = 1.0973
[A] = antilog of (1.0973)
[A] = l2.5%

Question 26.
A zero order reaction is 20% complete in 20 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?
Answer:
1. A = 100%, x = 20%, Therefore, a – x =100 – 20 = 80
For the zero order reaction k= \((\frac { x }{ t })\) ⇒
k = \((\frac { 20 }{ 20 })\) = 1
Rate constant for a reaction = 1

2. To calculate the time for 80% of completion
k = 1, a = l00, x = 80%, t = ?
Therefore, t = \((\frac { x }{ k })\) = \((\frac { 80 }{ 1 })\) = 80 min

Question 27.
The activation energy of a reaction is 225 k cal mol^-1 and the value of rate constant at 40°C is 1.8 x 10^-5 s^-1. Calculate the frequency factor, A. Here, we arc given that
Answer:
E_a = 22.5 kcal mol^-1 = 22500 cal mol^-1
T = 40°C = 40 + 273 = 313 K
k = 1.8 x 10^-5 sec^-1
Substituting the values in the equation
log A = log k + \(\left( \frac { { E }_{ a } }{ 2.303RT } \right)\)
log A = log (l .8 x 10^-5) + \(\left( \frac { 22500 }{ 2.303\times 1.987\times 313 } \right)\)
log A = log (l.8) – 5 + (15.7089)
log A = (10.9642)
A = antilog ( 10.9642)
A = 9.208 x 10¹⁰ collisions s^-1

Question 28.
Benzene diazonium chloride in aqueous solution decomposes according to the equation
C₆H₅N₂CI C₆H₅CI + N₂. Starting with an initial concentration of 10 g L^-1 volume of N₂. gas obtained at 50°C at different intervals of time was found to be as under:

Show that the above reaction follows the first order kinetics. What is the value of the rate constant ?
Solution:
For a first order reaction
k = \(\frac { 2.303 }{ t }\) log \(\frac { a }{ (a-x) }\)
k = \(\frac { 2.303 }{ t }\) log \(\frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ t } }\)
In this case, V_∞ = 58.3 ml
The value of k at different time can be calculated as follows:

Since the value of k comes out to be nearly constant, the given reaction is of the first order. The mean value of k = 0.0674 min^-1

Question 29.
From the following data, show that the decomposition of hydrogen peroxide is a reaction of the first order:

Where t is the volume of standard KMnO₄ solution required for titrating the same volume of the reaction mixture.
Solution:
Volume of KMnO₄ solution used Amount of H₂O₂ present. Hence if the given reaction is of the first order, it must obey the equation
k = \(\frac { 2.303 }{ t }\) log \(\frac { a }{ (a-x) }\)
k = \(\frac { 2.303 }{ t }\) log \(\frac { { V }_{ 0 } }{ { V }_{ t } }\)
In this case,V₀ = 46.1 ml
The value of k at each instant can be calculated as follows:

Thus, the value of k comes out to be nearly constant. Hence it is a reaction of the first order.

Question 30.
A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. in what time will the reaction be 80% complete?
Answer:
1. For the first order reaction k = \(\frac { 2.303 }{ t }\) log \(\frac { a }{ (a-x) }\)
Assume, a = 100 %, x = 40%, t = 50 minutes
Therefore, a – x = 100 – 40 = 60
k = (2.303/50) log (100/60)
k = 0.010216 min^-1
Hence the value of the rate constant is 0.010216 min^-1

2. t = ?, when x = 8O%
Therefore, a – x = 100 – 80 = 20
From above, k = 0.0102 16 min^-1
t = (2.303 / 0.010216) log (100 / 20)
t = 157.58 min
The time at which the reaction will be 80% complete is 157.58 min.

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Evaluate yourself

Question 1.
Write the rate expression for the following reactions, assuming them as elementary reactions.

1. 3A + 5B₂ → 4CD
2. X₂ + Y₂ → 2XY

Answer:
1. 3A + 5B₂ → 4CD
Rate = – \(\frac { 1 }{ 3 }\) \(\frac { \triangle [A] }{ dt }\)
= – \(\frac { 1 }{ 5 }\) \(\frac { \triangle [{ B }_{ 2 }] }{ dt }\)
= + \(\frac { 1 }{ 4 }\) \(\frac { \triangle [CD] }{ dt }\)

2. X₂ + Y₂ → 2XY
Rate = – \(\frac { \triangle [{ X }_{ 2 }] }{ dt }\)
= + \(\frac { 1 }{ 2 }\) \( [latex]\frac { \triangle [{ XY }_{ 2 }] }{ dt }\)

Question 2.
Consider the decomposition of N₂O_5(g) to form NO_2(g) and O_2(g). At a particular instant N₂O₅ disappears at a rate of 2.5 x 10^-2 mol dm^-3 s^-1. At what rates are NO₂ and O₂ formed? What is the rate of the reaction?
Solution:
2N₂O_5(g) → 4NO_2(g) + O_2(g)
from the stoichiometry of the reaction.
– \(\frac { 1 }{ 2 }\) \(\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\)
= \(\frac { 1 }{ 4 }\) \(\frac { d[{ N }{ O }_{ 2 }] }{ dt }\)
= –\(\frac { d[{ N }{ O }_{ 2 }] }{ dt }\)
= 2 –\(\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\)
Rate of disappearance of N₂O₅ is 2.5 x 10^-2 mol dm^-3 s^-1
∴ The rate of formation of NO2 at this temperature is 2 x 2.5 x 10^-2 = 5 x 10^-2 mol dm^-3 s^-1.
– \(\frac { 1 }{ 2 }\) \(\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\)
= – \(\frac { d[{ O }_{ 2 }] }{ dt }\)
∴ \(\frac { d[{ O }_{ 2 }] }{ dt }\) = \(\frac { 1 }{ 2 }\) x 2.5 x 10^-2 mol dm^-3 s^-1
= 1.25 x 10^-2 mol dm^-3 s^-1

Question 3.
For a reaction, X + Y → Product quadrupling [x], increases the rate by a factor of 8. Quailrupling both [x] and [y] increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall order of the reaction?
Solution:

z = k[x]^m [y]ⁿ …………(1)
8z = k[x]^m [y]ⁿ …………(2)
16z = k[x]^m [y]ⁿ ……………(3)
Dividing Eq (2) by Eq (1) we get,

8 = 4^m ⇒2³ = (2²)m ⇒ 2 = 2^2m
2m = 3
m = 3/2
1.5 order with respect to x.
Dividing Eq (3) by Eq (1) we get,

16 = 4^m. 4ⁿ
16 = 4². 4ⁿ
\(\frac { 16 }{ 16 }\) = 4ⁿ
1 = 4n
∴ n = 0 [Zero order with respect to y]
Overall order of the reaction.
k [x]^m [y]ⁿ
k [x]^1.5 [y]⁰
Order (1.5+0) = 1.5

Question 4.
Find the individual and overall order of the following reaction using the given data.
2NO(g) + Cl₂ (g) → 2NOCI(g)

Solution:
Rate = k [NO]^m [CI₂]
For experiment 1, the rate law is,
Rate₁ = k [NO]^m [CI₂]ⁿ
7.8 x 10^-5 k[0.1]^m [0.1]ⁿ ………………(1)

For experiment 2, the rate law is.
Rate₂ = k [NO]^m [CI₂]ⁿ
3.12 x 10^-4 = k[O.2]^m [0.1]ⁿ ……………….(2)

For experiment 3, the rate law is,
Rate₃ = k [NO]^m [CI₂]ⁿ
9.36 x 10^-4 = k [O.2]^m [0.3]^m ……………(3)
Dividing Eq (2) by Eq (l) we get,

4 = \(\frac { 0.2 }{ 0.1 }\)^m
⇒ 2² = 2^m
∴m = 2
Therefore the reaction is secondary order with respect to NO.
Dividing Eq (3) by Eq (2) we get,

Therefore the reaction is first order with respect to Cl₂
The rate law is, Rate = k [NO]₂ [Cl₂]¹
The overall order of the reaction (2 +1) = 3.

Question 5.
In a first order reaction A → products, 60% of the given sample of A decomposes in 40 min. what is the half life of the reaction?
Solution:
k = \(\frac { 2.303 }{ t }\) log \(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)
k = \(\frac { 2.303 }{ 40min }\) log \(\frac { 100 }{ (100-60) }\)
k = 0.0575 (0.3979) ⇒ k = 0.02287 min^-1
t_1/2 = \(\frac { 0.6932 }{ k }\) log \(\frac { 0.6932 }{ 0.02287 }\)
t_1/2 = 30.31 min.

Question 6.
The rate constant for a first order reaction is 2.3 x 10^-4 s^-1. If the initial concentration of the reactant is 0.01 M. what concentration will remain after 1 hour?
Solution:
Rate constant of a first order reaction k = 2.3 x 10^-4 s^-1
Initial concentration of the reactant [A₀] = 0.01 M
Initial concentration ot the reactant [A₀] = 0.01 M
Concentration will remain after 1 hour [A] =7
k = \(\frac { 2.303 }{ t }\) log \(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)
2.3 x 10^-4 = \(\frac { 2.303 }{ 1 hour }\) log \(\frac { [0.01] }{ [A] }\)
\(\frac { 2.3\times { 10 }^{ -4 }\times 1 }{ 2.303 }\) = log [0.01] – log[A]
9.986 x 10^-5 = – 2 – log [A]
11.986 x 10^-5 = – log [A]
[A] = Antilog (-11.986 x 10^-5)
[A] = 0.997 M

Question 7.
Hydrolysis of an ester in an aqueous solution was studied by titrating the liberated carboxylic acid against sodium hydroxide solution. The concentrations of the ester at different time intervals are given below.

Show that, the reaction follows first order kinetics.
Solution:
The value of k at different time can be calculated as follows:

This value shows that reaction follows first order kinetics.

Question 8.
For a first order reaction the rate constant at 500K is 8 x 10^-4 s^-1. Calculate the frequency factor, if the energy of activation for the reaction is 190 kJ mol^-1.
k = 8 x 10^-4s
T = 500K
E_a = 190 kJ mol^-1 A = ?
According to Arrhenius equation,

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Text book Example problems

Question 1.
The oxidation of nitric oxide (NO)
2NO_(g) + O_2(g) → NO_2(g)
Series of experiments are conducted by keeping the concentration of one of the reactants constant and the changing the concentration of the others.

Find out the individual overall order of the reaction.
Solution:
Rate = k [NO]^m [O₂]ⁿ
For experiment 1, the rate law is
Rate₂ = k [NO]^m [O₂]ⁿ
19.26 x 10^-2 = k[1.3]^m [1.1]ⁿ ………………(1)
Similarly for experiment 2
Rate₂ = k [No]^m [O₂]ⁿ
38.40 x 10^-2 = k [1.3]^m [2.2]ⁿ …………………..(2)
For experiment 3
Rate₃ = k [NO]^m [O₂]ⁿ
76.8 x 10^-2 = k [2.6]^m [1.1]ⁿ ……….(3)
Dividing Eq (2) by Eq (1)
\(\frac { 38.40\times { 10 }^{ -2 } }{ 19.26\times { 10 }^{ -2 } }\) = \(\frac { k{ \left[ 1.3 \right] }^{ m }{ \left[ 2.2 \right] }^{ n } }{ k{ \left[ 1.3 \right] }^{ m }{ \left[ 1.1 \right] }^{ n } }\)
2 = \({ \left( \frac { 2.2 }{ 1.1 } \right) }^{ n }\)
2 = 2ⁿ
i.e., n= 1
Therefore the reaction is first order with respect to O₂
Dividing Eq (3) byEq (1)
\(\frac { 76.8\times { 10 }^{ -2 } }{ 19.26\times { 10 }^{ -2 } }\) = \(\frac { k{ \left[ 2.6 \right] }^{ m }{ \left[ 1.1 \right] }^{ n } }{ k{ \left[ 1.3 \right] }^{ m }{ \left[ 1.1 \right] }^{ n } }\)
4 = \({ \left( \frac { 2.6 }{ 1.3 } \right) }^{ m }\)
4 = 2^m
i.e., m = 2
Therefore the reaction is second order with respect to NO
The rate law is Rate₁ = k [NO]² [O₂]¹
The overall order of the reaction = (2 + 1) = 3

Question 2.
Consider the oxidation of nitric oxide to form NO₂
2NO_(g) + O_2(g) → 2NO_2(g)

• Express the rate of the reaction in terms of changes in the concentration of NO, O₂ and NO₂.
• At a particular instant, when [O₂] is decreasing at 0.2 mol L^-1 s^-1 at what rate is
  [NO₂] increasing at that instant?

Solution:

Question 3.
What is the order with respect to each of the reactant and overall order of the following reactions?
1.  5Br^–_(aq) + BrO₃^– _(aq)+ 6H⁺_(aq)  → 3Br₂₍₁₎ +3H₂O₍₁₎
The experimental rate law is
Rate = k [Br^–] [BrO₃^–][H⁺]²

2.  CH₃ CHO_(g) \(\underrightarrow { \triangle }\) CH_4(g) + CO_(g)
the experimental rate law is
Rate = k [CH₃CHO]^3/2
Solution:
1. First order with respect to Br^–, first order with respect to BrO₃^– and second order with respect to H⁺. Hence the overall order of the reaction is equal to 1+1+2=4

2. Order of the reaction with respect to acetaldehyde is \(\frac { 3 }{ 2 }\)and overall order is also \(\frac { 3 }{ 2 }\)

Question 4.
The rate of the reaction x + 2y → product is 4 x 10^-3 mol L^-1 s^-1, if [x] = [y] = 0.2 M and rate constant at 400K is 2 x 10^-2 s^-1. what is the overall order of the reaction.
Solution:

Comparing the powers on both sides, the overall order of the reaction n + m = 1

Question 5.
A first order reaction takes 8 hours for 90% completion. Calculate the time required for 80% completion. (log 5 = 0.6989; log10 = 1)
Solution:
For a first order reaction,

Let [A₀] = 100M
When
t = t_90%; [A] = 10M (given that t_90% = 8 hours)
t = t_80%; [A] = 20M

Find the value of k using the given data
k = \(\frac{2.303}{t_{90 \%}}\) log \((\frac { 100 }{ 10 })\)
k = \(\frac { 2.303 }{ 8 hours }\) log 10
k = \(\frac { 2.303 }{ 8 hours }\) (1)
Substitute the value of k in equation (2)
t_80% = \(\frac { 2.303 }{ 2.303/8 hours }\) log (5)
t_80% = 8 hours x 0.6989
t_80% = 5.59 hours

Question 6.
The half life of a first order reaction x → products is 6.932 x 10⁴ s at 500K . What percentage of x would be decomposed on heating at 500K for 100 min. (e^0.06 = 1.06)
Solution:
Given t_1/2 = 0.6932 x 10⁴ s
To solve: when t = 100 min

We know that for a first order reaction,t_1/2 = \(\frac { 0.6932 }{ k }\)

Question 7.
Show that in case of first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction.
Solution:
Let [A₀] = 100
When t = t_99.9%; [A] = (100 – 99.9) = 0.1

Question 8.
The rate constant of a reaction at 400 and 200K are 0.04 and 0.02 s^-1 respectively. Calculate the value of activation energy.
Answer:
According to Arrhenius equation.

Question 9.
Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation. log k = log A \(\frac { { E }_{ a } }{ 2.303R }\) \(\frac { 1 }{ T }\).

Where E is the activation energy. When a graph is plotted for log k Vs \(\frac { 1 }{ T }\) a straight line with a slope of 4000K is obtained. Calculate the activation energy.
Solution:

E_a = – 2.303 Rm
E_a = – 2.303 x 8.314 JK^-1 mol^-1 x (-4000k)
E_a = 76,589 J mol^-1
E_a = 76,589 kJ mol^-1

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Additional Questions

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 1 Mark Questions and Answers

I. Choose the correct answer.

Question 1.
Which one of the following is a slow reaction?
(a) Rusting of iron
(b) Combustion of carbon
(c) Reaction between BaCl₂ and dil. H₂SO₄
(d) Reaction between acidified K₂Cr₂O₇ with NaCl.
Answer:
(a) Rusting of iron

Question 2.
Which one of the following is the unit of rate of reaction?
(a) s^-1
(b) mol s^-1
(c) mol L^-1 s^-1
(d) mol L s
Answer:
(c) mol L^-1 s^-1

Question 3.
For a gas phase reaction, the unit of reaction rate is ……………
(a) s^-1
(b) atm s^-1
(c) mol L^-1 s^-1
(d) mol^-1 L^-1 s^-1
Answer:
(b) atm s^-1

Question 4.
For the reaction A → 2B, the rate of the reaction is ………….
(a) +\(\frac { d[B] }{ dt }\) = 2 – \(\frac { d[A] }{ dt }\)
(b) +\(\frac { d[A] }{ dt }\) = \(\frac { 1 }{ 2 }\) \(\frac { d[B] }{ dt }\)
(c) Rate = \(\frac { 1 }{ 2 }\) = \(\frac { d[A] }{ dt }\)
(d) Rate = 2\(\frac { d[B] }{ dt }\)
Answer:
(a) +\(\frac { d[B] }{ dt }\) = 2 – \(\frac { d[A] }{ dt }\)

Question 5.
Consider the following statement.
(i) In ionisation of cyclopropane, if the concentration of cyclopropane is reduced half, the rate increases twice.
(ii) The rate of the reaction depends upon the concentration of the reactant.
(iii) Order values must be determined experimentally.

Which of the above statement (s) is / are not correct?
(a) (i) only
(b) (ii) and (iii)
(c) (iii) only
(d) (ii) only
Answer:
(a) (i) only

Question 6.
In the reaction 2NO_(g) + O_2(g) → 2NO_2(g) the order of the reaction with respect to NO is…………
(a) first order
(b) second order
(c) third order
(d) zero order
Answer:
(b) second order

Question 7.
In the reaction 2NO_(g) + O_2(g) → 2NO_2(g). the order of the reaction with respect to O₂is …….
(a) zero order
(b) first order
(c) second order
(d) third order
Answer:
(b) first order

Question 8.
The overall order of the reaction 2NO_(g) + O_2(g) → 2NO_2(g) is …………….
(a) 2
(b) 1
(c) 3
(d) 0
Answer:
(c) 3

Question 9.
Consider the following statements.
(i) Rate of the reaction does not depend on the initial concentration of the reactants.
(ii) Rate constant of the reaction depends on the initial concentration of reactants.
(iii) Rate constant of the reaction is equal to the rate of the reaction, when the concentration of each of the reactants is unity.
Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) only
(c) (i) and (ii)
(d) (iii) only
Answer:
(a) (iii) only

Question 10.
The overall molecularity of the reaction 2H₂ O_2(aq) \(\underrightarrow { { I }^{ – } }\) 2H₂O₁ + O_2(g) is …………
(a) unimolecular
(b) bimolecular
(c) termolecular
(d) pentamolecular
Answer:
(b) bimolecular

Question 11.
Which of the following is the order of decomposition of hydrogen peroxide catalysed by I^– ………….
(a) First order
(b) Second order
(c) Zero order
(a) Third order
Answer:
(a) First order

Question 12.
Consider the following statements.
(i) order cannot be zero.
(ii) Molecularity can be zero (or) fractional (or) integer.
(iii) order can be determined only by experiment.

Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i) and (ii)
Answer:
(c) (iii) only

Question 13.
The overall order of the reaction 5Br^– + BrO₃^– + 6H⁺ is ……..
(a) 4
(b) 3/2
(c) 12
(d) 1
Answer:
(a) 4

Question 14.
Which one of the following reaction is a fractional order reaction?
(a) 2NO +O₂ → 2NO₂
(b) CH₃CHO_(g) → CH_4(g) + CO_(g)
(c) 2H₂O₂ → 2H₂)O₍₁₎ + O_2(g)
(d) H₂ + Br₂ → 2HBr
Answer:
(b) CH₃CHO_(g) → CH_4(g) + CO_(g)

Question 15.
The order of decomposition of acetaldehyde is …………….
(a) 1
(b) 1.5
(c) 2
(d) 5/2
Answer:
(b) 1.5

Question 16.
Which one of the following is the unit of rate constant for a first order reaction?
(a) mol^-1 L s^-1
(b) mol L^-1 s^-1
(c) s^-1
(d) mol L S
Answer:
(c) s^-1

Question 17.
Which one of the following is an example for first order reaction?
(a) 2NO_(g)+ O_2(g) → 2NO_2(g)
(b) CH₃CHO_(g) → CH_4(g) + CO_(g)
(c) SO₂ Cl₂₍₁₎ → SO_2(g) + Cl_2(g)
(d) 2HBr → H₂ + Br₂
Answer:
(c) SO₂ Cl₂₍₁₎ → SO_2(g) + Cl_2(g)

Question 18.
Which one of the following is not an example for first order reaction?
(a) N₂O_5(g) → 2NO_2(g) \(\frac { 1 }{ 2 }\) O_2(g)
(b) SO₂Cl₂₍₁₎ → SO_2(g) + Cl_2(g)
(e) H₂O_2(aq) → H₂O₁\(\frac { 1 }{ 2 }\)O_2(g)
(d) CH₃CHO_(g) → CH_4(g) + CO_(g)
Answer:
(d) CH₃CHO_(g) → CH_4(g) + CO_(g)

Question 19.
What is the order of isomerisation of cyclopropane to propene?
(a) 1.5
(b) 3/2
(c) 5/2
(d) 1
Answer:
(d) 1

Question 20.
Which one of the following is an example of pseudo first order reaction?
(a) CH₃CHO_4(g) → CH_4(g) + CO_(g)
(b) 2H₂O_2(aq) → H₂O₍₁₎ +O_2(g)
(c) CH₃COOCH_3(aq) + H₂O₍₁₎ \(\underrightarrow { { H }^{ + } }\) CH₃COOH_(aq) + CH₃OH_(aq)
(d) Isomerisation of cyclo propane to propene
Answer:
(c) CH₃COOCH_3(aq) + H₂O₍₁₎ \(\underrightarrow { { H }^{ + } }\) CH₃COOH_(aq) + CH₃OH_(aq)

Question 21.
Which one of the following is called pseudo first order reaction?
(a) Decomposition of acetaldehyde
(b) Acid hydrolysis of an ester
(c) Isomerisation of cyclopropane to propene
(d) Decomposition of hydrogen peroxide
Answer:
(b) Acid hydrolysis of an ester

Question 22.
Which of the following is an example of zero order reaction?
(a) lodination of acetone in acid medium
(b) Hydrolysis of an ester in acid medium
(c) Decomposition of acetaldehyde
(d) Isomerisation of cyclopropane to propene
Answer:
(a) lodination of acetone in acid medium

Question 23.
Which one of the follow is not zero order reaction?
(a) H_2(g) + Cl_2(g) \(\underrightarrow { h\nu }\)  2HCI_(g)
(b) N₂O_(g) \(\rightleftharpoons\) N_2(g) + \(\frac { 1 }{ 2 }\) O_2(g)
(c) CH₃CHO_(g) → CH_4(g)+ CO_(g)
(d) CH₃COCH₃ + I₂ \(\underrightarrow { { H }^{ + } }\) CH₂COCH₃ + HI
Answer:
(c) CH₃CHO_(g) → CH_4(g)+ CO_(g)

Question 24.
Consider the following statements.
(i) For a first order reaction, half life period is independent of initial concentration.
(ii) Photo chemical reaction between H₂ and Cl₂ is a zero order reaction
(iii) Acid hydrolysis of an ester is a second order reaction

Which of the above statement is/are correct?
(a) (i) only
(b) (iii) only
(c) (i) & (ii)
(d) (ii) & (iii)
Answer:
(c) (i) & (ii)

Question 25.
The formula of half life for an nth order reaction involving reactant A and n \(\neq\) 1 is

Answer:

Question 26.
The half life period of first order reaction is 10 seconds. What is the time required for 99.9% completion of that reaction?
(a) 20 seconds
(b) 1000 seconds
(c) 100 seconds
(d) 999 seconds
Answer:
(c) loo seconds
Hint:
10 x t_1/2 = t_99.9%
∴ t_99.9% = 10 x 10 sec = 100 sec

Question 27.
Which one of the following is known as arrhenius equation?

Answer:

Question 28.
Which one of the following does not affect the rate of the reaction?
(a) Nature of the reactant
(b) Concentration of the reactants
(c) Surface area and temperature
(d) pressure
Answer:
(d) pressure

Question 29.
Consider the following statements.
(i) Higher the concentration, slower is the possibility for collision and rate also slower
(ii) Increase in surface area of reactant leads to more collisions per litre per second
(iii) Gas phas reactions are slower as compared to solid or liquid reactants

Which of the above statement is/are not correct?
(a) (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i) & (ii)
Answer:
(b) (i) & (iii)

Question 30.
Which of the following reaction take place at a faster rate?

Answer:

Question 31.
Which one of the following graph is not correct ………..

Answer:

Question 32.
The half life of paracetamol with in the body is ………
(a) 2 hours
(b) 2.5 hours
(c) 6 hours
(d) 10 hours
Answer:
(b) 2.5 hours

Question 33.
What is the order of radioactive decay?
(a) first order
(b) zero order
(c) second order
(d) third order
Answer:
(a) first order

Question 34.
t_1/2 of the reaction increases with increase in initial concentration of the reaction means the order of the reaction will be …………
(a) first order
(b) zero order
(c) second order
(d) third order
Answer:
(b) zero order

Question 35.
The reaction rate that does not decrease with time is …………
(a) pseudo first order reaction
(b) first order reaction
(c) zero order reaction
(d) second order reaction
Answer:
(c) zero order reaction

Question 36.
The rate of the reaction X → Y becomes 8 times when the concentration of the reactant ‘X’ is doubled. The rate law of the reaction is ………
(a) – \(\frac { d[x] }{ dt }\) = k[X]²
(b) – \(\frac { d[x] }{ dt }\) = k[X]³
(c) – \(\frac { d[x] }{ dt }\) = k[X]⁴
(d) – \(\frac { d[x] }{ dt }\) = k[X]⁸
Answer:
(b) – \(\frac { d[x] }{ dt }\) = k[X]³

Question 37.
The decomposition of ammonia gas on platinum surface has a rate constant k = 2.5 x 10^-4 mol L^-1 s^-1 What is the order of the reaction?
(a) first order
(b) second order
(c) third order
(d) zero order
Answer:
(d) zero order

Question 38.
A reaction is 50% completed in 2 hours and 75% completed in 4 hours. Then the order of the reaction is ………….
(a) first order
(b) zero order
(c) second order
(d) third order
Answer:
(a) first order
Answer:
(a) first order

Question 39.
What is the rate equation for the reaction A + B → C has zero order?
(a) Rate = k
(b) Rate = k [A]
(c) Rate = k [A]. [B]
(a) Rate = k. \(\frac { 1 }{ [c] }\)
Answer:
(c) Rate = k [A]. [B]

Question 40.
How does the value of rate constant vary with reactant concentration?

Answer:

Question 41.
Identify the reaction order if the unit of rate constant is s^-1 ……….
(a) zero order reaction
(b) second order reaction
(c) first order reaction
(d) third order reaction
Answer:
(c) first order reaction

Question 42.
What is unit of zero order reaction?
(a) s^-1
(b) mol^-1 L^-1 s^-1
(c) mol L^-1 s^-1
(d) mol L s^-1
Answer:
(c) mol L^-1 s^-1

Question 43.
Which of the following factor affect the rate of the reaction’?
(a) volume
(b) pressure
(c) cone
(d) all the above
Answer:
(c) cone

Question 44.
Acid hydrolysis of an ester is an example of ………
(a) zero order reaction
(b) Pseudo first order reaction
(c) second order reaction
(d) first order reaction
Answer:
(b) Pseudo first order reaction

Question 45.
Polymerisation reactions follows ………………. order kinetics.
(a) fractional
(b) first
(c) zero
(d) Pseudo first
Answer:
(a) fractional

Question 46.
Activation energy of a chemical reaction can be determined by ……….
(a) changing concentration of the reactants
(b) Evaluating rate constants at standard temperature
(c) Evaluating rate constants at two different temperature
(d) Evaluating reIocities of reaction at two different temperature
Answer:
(c) Evaluating rate constants at two different temperature

Question 47.
Which of the following is the fastest reaction?

Answer:

Question 48.
Half life period of a reaction is found to be inversely proportional to the cube of its initial concentration. The order of the reaction is ………
(a) 2
(b) 5
(c) 3
(d) 4
Answer:
(d) 4

Question 49.
A large increase in the rate of a reaction for a rise in temperature is due to ………
(a) the decrease in the number of collisions
(b) increase in the number of activated molecules
(c) the shortening of mean free path
(d) the lowering of activation energy
Answer:
(b) increase in the number of activated molecules

Question 50.
The minimum energy of a molecule would possess in order to enter into a fruitful collision is known as …………..
(a) Reaction energy
(b) collision energy
(c) Activation energy
(d) Threshold energy
Answer:
(a) Threshold energy

II. Fill in the blanks.

1. The unit of the rate of a reaction is ………..
2. For a ……….. reaction, the unit of the reaction rate is atm s
3. An elementary step is characterised by its ………..
4. The total number of reactant species involved in an elementary step is called ………..
5. The sum of powers of concentration terms involved in the experimentally determined rate law is called ………..
6. The overall order of decomposition of acetaldehyde to methane and carbon monoxide is ………..
7. A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, it is called ………..
8. A reaction in which rate is independent of the concentration of the reactant over a wide range of concentration is called ………..
9. All radioactive disentegration reactions follow ……….. kinetics.
10. For a first order reaction, half life does not depend on ………..
11. Half life period of zero order reaction is ……….. proportional to initial concentration of the reactant.
12. Half life period ……….. reaction is directly proportional to initial concentration of the reactant.
13.  ……….. was proposed by Max Trautz and William lewis.
14. Collision theory was proposed by ……….. in 1916 and in ……….. 1918.
15. For a gas at room temperature (298 K) and I atm, each molecule undergoes approximately ……….. per second.
16. In order to react, the collidng molecules must possess a minimum energy called ………..
17. Generally the reaction rate tends to double when the temperature is increased by ………..
18. The number of collisions of reactant molecules per second is known as ………..
19. Heating is required for the reaction between KMnO₄ and oxalate ion and is carried out at around ………..
20. ……….. reactions are faster as compared to reactions involving solid or liquid reactants.
21. The rate of the reaction ……….. with the increase in the concentration of the reactants.
22. Higher the concentration of reactants greater is the possibility of and hence the ………..
23. In the presence of catalyst the energy of activation is ……….. and hence greater number of molecules change over to products there by increasing the rate of the reaction.
24. Bio availability of drugs within the body and this branch of study is called ………..
25. ……….. has a half life of 2.5 hours within the body.
26. The change in concentration of species per unit time gives the ……….. of the reaction.
27. The rate constant is equal to the rate of the reaction when concentration of reactants is ………..
28. Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of the reaction is ………..
29. Acid hydroJysis of an ester is an example of ………..
30. Molecularity of a chemical reaction will never be equal to ………..

Answer:

1. moI L^-1 s^-1
2. gas phase
3. molecularity
4. Molecularity
5. order
6. 3/2 or 1.5
7. Pseudo first order reaction
8. Zero order reaction
9. First order
10. initial concentration
11. directly
12. zero order
13. Collision theory
14. Max Trautz, William lewis
15. 10 collisions
16. Activation energy
17. 10°C
18. Frequency factor (A)
19. 60°C
20. Gas phase
21. increases
22. collisions, rate
23. lowered
24. Pharmaco kinetics
25. Paracetamol
26. rate
27. unity
28. increased
29. Pseudo first order reaction
30. zero

III. Match the following

Match the list I and II using the code given below the list.

Quetion 1.

Answer:
(a) 3, 4, 1, 2

Question 2.

Answer:
(a) 3, 4, 1, 2

Question 3.

Answer:
(a) 3, 1, 4, 2

Question 4.

Answer:
(a) 3, 4, 2, 1

IV. Assertion and Reason

Question 1.
Assertion (A): Decomposition of hydrogen peroxide catalysed by I^– is a bimolecular first order reaction.
Reason (R): The above reaction take place in two steps, step 1 involves both H₂O₂ and I and so it is bimolecular but order is determined experimentally as 1.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R arc correct and R is the correct explanation of A

Question 2.
Assertion (A):  The overall order of the reaction is equal to 4.
Reason (R): The experimental rate law is.
Rate = K [Br^–] [BrO₃^–] [H⁺]².
So 1 + 1 + 2 = 4 order value is 4.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation olA.

Question 3.
Assertion (A): The rate of a reaction increases with the increase in the concentration of the reactants.
Reason (R): The rate of the reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate.
(a) Both A and R are correct and R is the correct explanation of A,
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 4.
Assertion (A): Powdered calcium carbonate reacts much faster with dilute
HCL than with the same mass of CaCO₃ as marble.
Reason (R): For a given mass of a reactant, when the particle size decreases, surface area increases. Increase in surface area of the reactant leads to more collisions per litre per second and hence the rate of the reaction also increases.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R arc correct but R is not correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 5.
Assertion (A): Catalyst presence increases the rate of the reaction
Reason (R): In the presence of a catalyst, energy of activation is lowered and hence greater number of molecules can across the energy harrier and change over to products thereby increasing the rate of the reaction.
(a) Both A and R are correct but R is not correct explanation of A
(b) Both A and R are correct and R is the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(b) Both A and R are correct and R is the correct explanation of A

Question 6.
Assertion (A): Doctors adviced to take paracetamol once in 6 hours during fever and body pain
Reason (R): Paracetarnol has a half life of 2.5 hours within the body. After 10 hours (4 half lives) only 6.25% of drug remains. Based on this, doctors adviced to take it once in 6 hours.
(a) Both A and R are wrong
(b) A is correct but R is wrong
(c) A and R are correct and R is the correct explanation of A
(d) A and R are correct but R is not correct explanation of A
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 7.
Assertion (A): Order of the reaction can be zero or fractional
Reason (R): We cannot determine order from balanced chemical equation
(a) Both A and R are correct but R is not correct explanation of A.
(b) Both A and R are correct and R is the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is not correct explanation of A

Question 8.
Assertion (A): If the activation enery of a reaction is zero, temperature will have no effect on the rate constant
Reason (R): Lower the activation energy, faster is the reaction.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(b) Both A and R are correct but R is not correct explanation of A

V. Find the odd one out

Question 1.

Answer:

Hint: It is a fractional order reaction with order value 3/2 where as others are first order reaction.

Question 2.

Answer:

Hint: It is first order reaction whereas others are zero order reaction.

Question 3.
(a) Decomposition of dinitrogen pentoxide
(b) Iodination of acetone in acid medium
(c) Decomposition of N20 on hot Pt surface
(d) photochemical reaction between H2 and CI2
Answer:
(a) Decomposition of dinitrogen pentoxide
Hint: It is a first order reaction whereas others are zero order reactions

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 2 Mark Questions and Answers

Question 1.
Define rate of the reaction. Give the unit of rate.
Answer:
1. In a chemical reaction, the change in the concentration of the species involved in a chemical reaction per unit time gives the rate of a reaction.
A → B
Rate = –\(\frac { \triangle [A] }{ dt }\) (or) \(\frac { \triangle [B] }{ dt }\)

2.

= mol L^-1 s^-1

Question 2.
Define molecularity of a reaction.
Answer:
Molecularity of a reaction is the total number of reactant species that are involved in an elementary step.

Question 3.
Define order of a chemical reaction.
Answer:
Order of a chemical reaction is the sum of powers of concentration terms involved in the experimentally determined rate law.

Question 4.
Define Half life period.
Answer:
The half life ola reaction is defined as the time required for the reactant concentration to reach one half its initial value.

Question 5.
Mention the factors affecting the reaction rate.
Answer:
The rate of the reaction is affected by the following factors.

1. Nature and state of the reactant
2. Concentration of the reactant
3. Surface area of the reactant
4. Temperature of the reaction
5. Presence of a catalyst

Question 6.
How is surface area of the reactant affect the rate of the reaction?
Answer:

1. In heterogeneous reactions, the surface area of the solid reactants play an important role in deciding the rate.
2. For a given mass of a reactant, when the particle size decreases surface area increases. Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of reaction is increased.
3. For example, powdered calcium carbonate reacts much faster with dilute HCl than with the same mass of CaCO₃ as marble.

Question 7.
Paracetamol is prescribed to take once in 6 hours. Justify this statement.
Answer:
1. Paracetamol is a well known antipyretic and analgesic that is prescribed in cases of fever and body pain.

2. Paracetamol has a half life of 2.5 hours within the body. (Le) the plasma concentration of the drug is halved after 2.5 hours. So after 10 hours (4 half lives), only 6.25% of drug remains. Based on this, the dosage and frequency will be decided.

3. In the case of paracetamol, it is usually prescribed to take once in 6 hours.

Question 8.
For a reaction, A + B → product; the rate law is given by r = k[A]^1/2 [B]². What is the order of the reaction?
Answer:
Order of the reaction = \(\frac { 1 }{ 2 }\) + 2 = 2 \(\frac { 1 }{ 2 }\) or 0.5

Question 9.
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will ¡t affect the rate of formation of Y?
Answer:
For the reaction, X → Y as it follows second order kinetics, therefore the rate law equation will be
Rate = k[X]² = ka²
if [X] = a mol^-1
if the concentration of X is increased three times, then
[X] = 3a mol L^-1
∴ Rate = k (3a)² = 9 ka²
Thus, the rate of the reaction will become 9 times. Hence the rate of formation of Y will increase by 9 times.

Question 10.
Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Answer:
For a first order reaction,

Question 11.
What will be the effect of temperature on rate constant?
Answer:
Rate constant of a reaction is nearly doubled with rise in temperature by 10°C. The exact dependence of the rate constant on temperature is given by Arrhenius equation:
Rate constant,

Question 12.
A reaction is first order in A and second order in B.

1. Write the differential rate equation.
2. How is the rate affected on increasing the concentration of B three times?
3. How is the rate affected when the concentrations of both A and B arc doubled?

Answer:

1. \(\frac { dx }{ dt }\) = k [A]¹ [B]²
2. If concentration of ‘B’ is tripled, then the rate will become 9 times.
3. When concentration of both A and B are doubled, then the rate will become 8 times.

Question 13.
Define zero order reaction. Give the unit for its rate constant(k).
Answer:
Zero Order Reaction. The reaction in which the rate of reaction is independent of the concentration of the reactants is called zero order reaction.
Rate = k [A]⁰ ⇒ k
Where k is the rate constant. Its unit is mol L^-1 s^-1

Question 14.
Write units of rate constant k for zero order, first order, second order and n order reaction.
Answer:
Order of Reaction

1. Zero order reaction
2. First order reaction
3. Second order reaction
4. nth order reaction

Unit of k:

1. mol L^-1 s^-1
2. s^-1
3. mol L s^-1
4. (mol/ L)^1-n s^-1

Question 15.
What is the effect of catalyst on the activation energy? Why?
Answer:
A Catalyst lower down the activation energy. It provides an alternate path to the reaction. It forms an unstable intermediate which readily changes into products.

Question 16.
Give two differences between zero order and first order reaction.
Answer:
Zero Order:

1. Its ‘k’ has unit = mol L^-1 s^-1
2. Its t 1/2 is directly proportional to initial conc. of reactant

First order:

1. Its ‘k’ has unit = time^-1 = sec^-1
2. Its half life is independent of the initial conc. of the reactant.

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 3 Mark Questions and Answers

Question 1.
Write the differences between the rate and rate constant of the reaction.
Answer:
Rate of a reaction:

1. It represents the speed at which the reactants are converted into products at any instant
2. It is measured as decrease in the concentration of the reactants (or) increase in the concentration of products
3. It depends on the initial concentration of reactants

Rate constant of a reaction:

1. It is a proportionality constant
2. It is equal to the rate of the reaction, when the concentration of each of the reactants is unity
3. It does not depend on the initial concentration of the reactants

Question 2.
What are the examples of first order reaction?
Answer:

1. Decompostion of dinitrogen pentoxide
   N₂O_2(g) → 2NO_2(g) + \(\frac { 1 }{ 2 }\) O_2(g)
2. Decomposition of thionylchloride
   SO₂Cl_2(g) → SO_2(g) + CI_2(g)
3. Decomposition of H₂O₂ in aqueous solution
   H₂O_2(aq) → H₂O₍₁₎ + \(\frac { 1 }{ 2 }\) O_2(g)
4. Isomerisation of cyclopropane to propene

Question 3.
For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer:

Question 4.
In a reaction, 2 A → products. the concentration of A decreases from 0.5 moI L^-1 to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this interval.
Answer:
Average rate

Question 5.
A first order reaction has a rate constant, 1.15 x 10^-3 s^-1. How long will 5 g of this reactant take to reduce to 3 g?
Answer:
Here,
[R]₀ = 5g
[R] = 3 g
k = 1.15 x 10^-3 s^-1 As the reaction is of first order,

Question 6.
Time required to decompose SO₂CI₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. For a first order reaction,
Answer:

Question 7.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is-

1. doubled
2. reduced to half.

Answer:
1. Reaction is second order with respect to the reactant
∴ Rate = k[A]² = ka².
when [A] = 2a
Rate k (2a)²
= 4ka²
= 4 times
Therefore, when concentration of the reactant is doubled the rate will become 4 times

2. when [A] = \(\frac { 1 }{ 2 }\) a
Rate = k \({ \left( \frac { 1 }{ 2 } a \right) }^{ 2 }\) = \(\frac { 1 }{ 4 }\) ka² = \(\frac { 1 }{ 4 }\) k
Therefore, rate will be reduced to one-fourth of the initial rate.

Question 8.
The rate constant for a first order reaction is 60 s¹. How much time will It take to reduce the initial concentration of the reactant to its 1/6^th value?
Answer:

Question 9.
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer:
For a first order reaction, t = k = \(\frac { 2.303 }{ t }\) log \(\frac { a }{ a-x }\)
99% completion means that
x = 99% of a =0.99 a
t_99% = \(\frac { 2303 }{ k }\) log \(\frac { a }{ a-0.99a }\) = \(\frac { 2.303 }{ k }\) log 10² = 2 x \(\frac { 2303 }{ k }\)
90% completion means that
x = 90% of a = 0.90 a
t_99% = \(\frac { 2303 }{ k }\) log \(\frac { a }{ a-0.99a }\) = \(\frac { 2.303 }{ k }\) log 10 = \(\frac { 2.303 }{ k }\)
\(\frac{t_{99 \%}}{t_{90 \%}}=\left(\frac{2 \times 2303}{k}\right) / \frac{2.303}{k}=2\)
or
t_99% = 2 x t_90%

Question 10.
Calculate the half life of a first order reaction whose rate constant is 200 s^-1
Answer:
Here rate constant
k= 200 s^-1
∴ Half – life of a first order reaction is
t_1/2 = \(\frac { 0.693 }{ k }\) = \(\frac { 0.693 }{ 200 }\) = 3.46 x 10^-3 sec

Question 11.
The decomposition of dinitrogen pentoxide (N₂O) follows the first order rate law. Calculate the rate constant from the given data.
t = 800 sec, [N₂O₅] = 1.45 moI L^-1 = [A₂]
t = 1600 sec
[N₂O₃] = 0.88 moI L^-1 = [A₂]
Answer:
Applying the formula,
k = \(\frac { 2.303 }{ \left( { t }_{ 2 }-{ t }_{ 1 } \right) }\) log 10 \(\frac { \left[ { A }_{ 1 } \right] }{ \left[ { A }_{ 2 } \right] }\)
= \(\frac { 2.303 }{ (1600 – 800) }\) log 10 \(\frac { 1.45 }{ 0.88 }\) = \(\frac { 2.303 }{ 800 }\) x 0.2169
= 6.24 x 10^-4 sec^-1

Question 12.
A second order reaction in which both the reactants have same concentration, is 20% completed in 500 seconds. How much time it will take for 60% completion?
Answer:
The second order equation when both the reactants have same concentration is
k = \(\frac { 1 }{ t }\). \(\frac { x }{ a(a – x) }\)
If a = 100, x = 20, 1= 500 seconds
Then k = \(\frac { 1 }{ 500 }\) x \(\frac { 20 }{ 100  x  (100-20) }\)
When
a = 100
x = 60
t = ?
t = \(\frac { 1 }{ k }\) \(\frac { 60 }{ 100  x  40 }\)
Substituting the value of k,
t = \(\frac { 500  x  100  x  80 }{ 20 }\) x \(\frac { 60 }{ 100  x  40 }\)
or
t = 3000 seconds

Question 13.
A first order reaction is 20% completed in 10 minutes. Calculate the time taken for the reaction to go to 80% completion.
Answer:
Applying the first order equation,
k = \(\frac { 2303 }{ t }\) l0g \(\frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] }\)
At t = 10 min
R = 100 – 20
k = \(\frac { 2303 }{ t }\) log 10 \(\frac { 100 }{ (100 – 20) }\)
t = \(\frac { 2303 }{ 10 }\) log 10 \(\frac { 100 }{ 80 }\)
= 0.0223 min^-1

Question 14.
For a reaction: 2NH_3(g)  \(\underrightarrow { Pt }\)  N_2(g)+ 3H_2(g) Rate = K

1. Write the order and molecularity of this reaction.
2. Write the unit of K.

Answer:

1. Order of reaction Zero order. Molecularity = 2
2. Unit of K = mol L^-1 sec^-1

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 5 Marks Questions and Answers

Question 1.
How would you calculate the order of the reaction 2NO + O_2(g) → 2NO_2(g) by an experiment?
(or)
prove that 2NO + O₂ → 2 NO₂ is a third order reaction.
Answer:
2NO_(g) + O_2(g) → 2NO_2(g)
Series of experiments are conducted by keeping the concentration of one of the reactants as constant and changing the concentration of the others.

Rate = k [NO]^m[O₂]ⁿ
For experiment 1, the rate law is
Rate₁ = k [NO]^m [O₂]ⁿ
19.26 x 10^-2 = k [1.3]^m [1.1]ⁿ ………………….(1)
For experiment 2
Rate₂ = k [NO]^m [O₂]ⁿ
38.40 x 10^-2 = k [1.3]^m [2.2]ⁿ ……………….(2)
For experiment 3
Rate₃ = k [NO]^m [O₂]ⁿ
76.8 x 10^-2 = k [2.6]^m [1.1]ⁿ ………………(3)

2 = \({ \left( \frac { 2.2 }{ 1.1 } \right) }^{ n }\)
4 = 2^m
⇒ n = 1
Therefore the reaction is first order with respect to O₂

2 = \({ \left( \frac { 2.6 }{ 1.3 } \right) }^{ m }\)
4 = 2^m
⇒ m = 1
Therefore the reaction is second order with respect to NO
The rate law is Rate_– = k [NO]² [O₂]¹
The overall order of the reaction = 2 + 1 = 3

Question 2.
Derive the integrated rate law for a first order reaction?
Answer:
A reaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction. First order reaction is A → product. Rate law can be expressed as, Rate = k [A]¹. Where, k is the first order rate constant
\(\frac { -d[A] }{ dt }\) = k[A]¹
\(\frac { -d[A] }{ [A] }\) = k.dt ……………………(1)
Integrate the above equation (I) between the limits of time t = 0 and time equal to t, while the concentration varies from initial concentration [A₀] to [A] at the later time.

This equation (2) is in natural logarithm. To convert it into usual logarithm with base 10, we have to multiply the term by 2.303

Question 3.
Explain the effect of temperature on reaction rate based on Arrhenius theory.
Answer:
1. Generally, the rate of a reaction increases with increasing temperature. However, there are very few exceptions.

2. As a rough rule, for many reactions near room temperature, reaction rate tends to double when the temperature is increased by 10°C.

3. A large number of reactions are known which do not take place at room temperature but occur readily at higher temperature. Example – Reaction between H₂ and O₂ to form H₂O takes place only when an electric spark is passed.

4. Arrhenius suggested that the rates of most reactions vary with temperature in such a way that the rate constant is directly proportional to
and he proposed a relation between the rate constant and temperature.

where
k = frequency factor
R = gas constant ,
E_a = Activation energy
T = Absolute temperature (in kelvin)
The factor A does not vary significantly with temperature and hence it may be taken as a constant.

5. Taking logarithm on both side of the equation (1)

6. The plot of Ink vs \((\frac { 1 }{ T })\) is a straight line with negative slope\(\frac { { E }_{ a } }{ RT }\). If the rate constant for a reaction at two different temperatures is known, we can calculate the activation energy.

This equation can be used to calculate E from rate E_a constants k₁ & k₂ at temperature T₁ and T₂

Question 4.
Explain about the factors that affecting the reaction rate.
Answer:
The rate of a reaction is affected by the following factors.
1. Nature and state of the reactant
(a) A chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product. The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.

(b) Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants. For example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine.

2. Concentration of the reactant
The rate of the reaction increases with the increase in the concentration of the reactants. According to collision theory, the rate of the reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate.

3. Effect of surface area of the reactant:
In heterogeneous reactions, the surface areas of the solid reactants plays an important role in deciding the rate. For a given mass of a reactant, when the particle size decreases surface area increases.

Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of reaction is increased. For example, powdered calcium carbonate reacts much faster with dilute HCI than with the same mass of CaCOl as marble

4. Temperature:
For many reactions near room temperature, the reaction rate tends to double when the temperature is increased by 10°C . For eg, Reaction between H₂ and O₂ to form H₂O take place only when an electric spark is passed. So when the temperature increases, the rate of the reaction also increases.

5. Effect of presence of catalyst
(a) A catalyst is substance which alters the rate of a reaction without itself undergoing any permanent chemical change. They may participate in the reaction, but again regenerated and the end of the reaction.

(b) In the presence of a catalyst, the energy of activation is lowered and hence greater number of molecules can cross the energy barrier and change over to products,thereby increasing the rate of the reaction.

Question 5.
The decomposition of A into product has value of k as 4.5 x 10³ s^-1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 x 10⁴ s^-1 ?
Answer:
k₁ = 4.5 x 10³ s^-1
T₁ = 10 + 273 K = 283 K
k₂ = 1.5 x 10⁴ s^-1
T₂ = ?
E_a = 6o kJ mol^-1
According to Arrhenius equation

Question 6.
For a decomposition reaction the values of rate constant k at two different temperatures are given below:
k₁ = 2.15 x 10 L mol^-1 s^-1 at 650 K
k₂ = 2.39 x 10 L mol^-1 s^-1 at 700 K
Calculate the value of activation energy for this reaction. (R = 8.314 J K^-1 mol^-1 )
Answer:
Here
k₁ = 2.15 x 10 L mol^-1 s^-1 at 650 K
T₁ = 650 K
T₂ = 700K and
k₂ = 2.39 x 10 L mol^-1 s^-1 at 700 K
R = 8.314 J K^-1 mol^-1
Using the formula

Question 7.
For a certain chemical reaction variation In concentration, In IRI Vs time (s) plot is given below.
For this reaction write/draw:

1. What is the order of the reaction?
2. What is the units of rate constant (k)?
3. Give the relationship between k and t_1/2 (half-life period).
4. What does the slope of above line indicate?
5. Draw the plot of log [R₀]/[R] vs time (s)

Answer:
1. First order
2. time^-1 (s^-1)
3.
4.  Rate constat (k) of reaction
5.

Question 8.
A substance reacts according to the first order rate law and the specific reaction rate for the reaction ¡s 1 x 10^-2 s^-1. If the initial concentration is 1.0 M.

1. What is the initial rate?
2. What ¡s the reaction rate after 1 minute?

Answer:

1. Initial rate of a first order reaction
= k C
= l x 10^-2 x 1.0
= l x 10^-2 mol L^-1 s^-1

2. Concentration after 60 seconds is calculated by applying the first order kinetic equation,

Rate of reaction after 1 minute
= k x C
= l x 10^-2 x 0.5489
5.489 x 10^-3 mol L^-1 s^-1

Question 9.
A first order reaction is 50% completed in 30 minutes at 27°C and in 10 minutes at 47°C. Calculate the reaction rate constant at 27°C and the energy of activation of the reaction in kJ mol^-1
Answer:
For a first order reaction

Now applying the following equation:

Question 10.
In Arrhenius equation for a certain reaction, the values of A and E (activation energy) are 4 x 10¹³ sec^-1 and 98.6 KJ mol^-1 respectively. If the reaction is of first order, at what temperature will its half life period be 10 minutes?
Answer:
According to the Arrhenius equation.

For a first order reaction
t_1/2 = \(\frac { 0.693 }{ k }\) \(\frac { 0.693 }{ 600 }\)
So,
k = \(\frac { 0.693 }{ 600 }\) sec^-1 (t_1/2 = 10 min = 600 sec)
= 1.1 x 10^-3 sec^-1
Hence, log (1.1 x 10^-3)

Common Errors

Common Errors:

1. Order and molecularity may get confused
2. Unit of first order Rate constant and zero order may get con fused.
3. t_1/2 – Half liefe period may be difficult to remember

Rectifications:

1. Order and molecularity are same for the single step process. But for reactions of more than one step, they may be different.
2. First order sec^-1, Zero order – mol litre^-1 sec^-1
3. t_1/2 = 0.693 / k₁ for first order reaction.

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Tamilnadu Samacheer Kalvi 6th Social Science Civics Solutions Term 2 Chapter 1 National Symbols

Samacheer Kalvi 6th Social Science National Symbols Textual Evaluation

I. Choose the correct answer:

Samacheer Kalvi Guru 6th Social Science Term 2 Question 1.
The National Song Vande Mataram was composed by _______
(a) Pingali Venkayya
(b) Rabindra Nath Tagore
(c) Bankim Chandra Chatterjee
(d) Gandhiji
Answer:
(c) Bankim Chandra Chatterjee

National Symbols In Tamil Question 2.
Which is the National Anthem of India?
(a) Jana Gana Mana
(b) Vande Mataram
(c) Amar Sonar Bangla
(d) Neerarum Kaduluduththa
Answer:
(a) Jana Gana Mana

Questions On National Symbols Of India Question 3.
Who wrote the most ramous novel Anand Math?
(a) Akbar
(b) Rabindra Nath Tagore
(c) Bankim Chandra Chatterjee
(d) Jawaharlal Nehru
(c) Bankim Chandra Chatterjee
Answer:
(c) Bankim Chandra Chatterjee

Samacheer Kalvi 6th Social Term 2 Question 4.
________ birthday is celebrated as the International Day of non-violence?
(a) Mahatma Gandhi
(b) Subash Chandra Bose
(c) Sardar Vallabhai Patel
(d) Jawaharlal Nehru
Answer:
(a) Mahatma Gandhi

6th Standard Symbol Question 5.
The colour of the Asoka chakra found in our National flag is _______
(a) sky blue
(c) blue
(b) navy blue
(d) green
Answer:
(b) navy bluel

Question 6.
The first flag ever flown after the Independence is stored in ……………..
(a) Chennai fort Museum
(b) Delhi Museum
(c) Saranath Museum
(d) Kolkata Museum
Answer:
(a) Chennai fort Museum

Question 7.
The National Anthem was written by _______
(a) Devandranath Tagpre
(c) Rabindranath Tagore
(b) Bharathiyar
(d) Balagangadhar Tilak
Answer:
(c) Rabindranath Tagore

Question 8.
The time taken to play the National Anthem is ……………..
(a) 50 seconds
(b) 52 minutes
(c) 52 seconds
(d) 20 seconds
Answer:
(c) 52 seconds

Question 9.
“Vande Mataram” was first sung by _______ at the 1896 session of the National Congress
(a) Bankim Chandra Chatterjee
(c) Mahatma Gandhi
(b) Rabindranath Tagore
(d) Sarojini Naidu
Answer:
(b) Rabindranath Tagore

Question 10
________ hoists the flag on Independence day in Delhi.
(a) The Prime Minister
(b) The President
(c) Vice President
(d) Any Political leader
Answer:
(a) The Prime Minister

II. Fill in the blanks :

1. The National emblem was adapted from the Ashoka pillar of ______
2. The National fruit of India is______
3. The National Bird of India is______
4. Our National tree is the______
5. The Flag which was flown in 1947 Independence day was weaved in______
6. The Indian National Flag was designed by ______
7. ______ started the Saka Era
8. The longest river in India is______
9. The Indian Rupee symbol was designed by ______
10. The Chakra of the National Flag has______ spokes

Answer:

1. Sarnath
2. Mango
3. peacock
4. Banyan tree
5. Gudiyatham
6. Pingali Venkayya
7. Kanishka
8. The Ganges
9. D. Udhaya
10. 24

III. Choose the correct answer

1. The Lion Capital is now in the ______ museum (Kolkata/Samath)
2. The National Anthem was adopted in ______
3. ______ is declared as our National Micro organism (Lacto bacillus / Rhizobium)

Answer:

1. Sarnath
2. 1950
3. Lacto bacillus

IV. Fill in the blanks:

1. Saffron — Courage ; White – _______
2. Horse — Energy; Bull – _______
3. 1947 — Independence day; 1950 – _______

Answer:

1. Honesty, peace, purity
2. Hardwork and dedication
3. Republic day

V. Chooose the Correct Option:

1) Rabindranath Tagorea – a.National Song
2) Bankim Chandra Chatteijee – b. National Flag
3) Pingali Venkayyac. – c. Astro Physicist
4) Meghnad Saha – d. National Anthem
a) a d b c
b) d a c b
c) d a b c
Answer:
c) d a b c

VI. Match and choose the wrong answer:

1. National Reptile – Tiger
2. National Aquatic Animal – Lacto bacillus
3. National Heritage Animal – King Cobra
4. National Micro organism – Dolphin

Answer:

1. National Reptile – King Cobra
2. National Aquatic Animal – Dolphin
3. National Heritage Animal – Tiger
4. National Micro organism – Lacto bacillus

VII. Choose the Wrong sentence:

Question 1.
a) The ratio of our National Flag’s length and breadth is 3:2
b) The Chakra has 24 spokes
c) The Chakra is Sky Blue in colour
Answer:
c) The Chakra is Blue in colour [Clue : Navy Blue is correct]

Question 2.
a) The National Flag was designed by Pingali Venkayya
b) The First ever flown Flag after the Independence is stored in Kolkata Museum
c) The First National Flag was weaved in Gudiyattam
Answer:
b) The First ever flown Flag after the Independence is stored in Kolkata Museum,

VIII. Choose the correct sentence:

a) August 15 is celebrated as the Independence day
b) November 26 is celebrated as the Republic day
c) October 12 is celebrated as Gandhi Jayanti
Answer:
a) August 15 is celebrated as the Independence day

IX. Answer the following:

Question 1.
What do the colours in our National Flag represent?
Answer:

1. The saffron at the top represents valour and sacrifice.
2. The green at the bottom represents fertility and prosperity.
3. The white band in between represents honesty, peace and purity.
4. The Ashoka chakra or the wheel in navy blue represents truth and peace.

Question 2.
What are the parts of our National emblem?
Answer:

1. Our National emblem consists of two parts the upper and the lower parts.
2. The upper part has four lions facing the North, South, East and West on a circular Pedestal.
3. The lower part has an elephant, a horse, a bull and a lion. The wheel of right sourness is placed between them.

Question 3.
What are the salient features of the National anthem?
Answer:

1. ‘Jana Gana Mana’ is our National anthem.
2. It symbolises the sovereignty and intergrity of our Nation.
3. This anthem was written by Rabindranath Tagore in Bengali.

The rules to be observed while singing the Anthem

1. This anthem has to be sung at a duration of 52 seconds.
2. Everyone should stand erect while singing the national anthem,
3. One should understand the meaning while singing.

Question 4.
Draw and define the Indian Rupee Symbol.
Answer:

1. The Indian Rupee sign is the official currency of India. (Designed by D. Udhaya Kumar)
2. It was presented to the public by the Government of India on 15 July 2010.
3. Shershah Sur’s currency was Rupiya.
4. This Rupiya has been transformed into Rupees the Indian Currency.
5. The symbol of Rupees in.
6. This was designed by D. Udhayakumar from Tamil Nadu in 2010.

Question 5.
Where do we use our National emblem?
Answer:
The National emblem is found at the top of the Government Communication, Indian Currency and passport.

Question 6.
Who wrote the National Pledge?
Answer:
Pydimarri Venkata Subha Rao wrote the National Pledge.

Question 7.
What are the animals found in the bottom of the emblem?
Answer:
In the bottom of the emblem, we see an elephant, a horse, a bull and a lion.

Question 8.
What are the natural national Symbols?
Answer:

1. Banyan tree
2. Peacock
3. River Ganges
4. Dolphin
5. King Cobra
6. Lotus
7. Tiger
8. Elephant
9. Lactobacillus
10. Mango

Question 9.
Where is the peacock sanctuary located in Tamil Nadu?
Answer:
There is a peacock sanctuary at Viralimalai in the district of Pudukottai (Tamil Nadu).

X. Activities:

Question 1.
What should we do to protect the endangered plants and animals – Discuss.
Answer:

1. Grow native plants.
2. Reduce your water consumption.
3. Recycle and buy sustainable products.
4. Do not buy plastic products.
5. Volunteer your time to protect the wildlife in your area.
6. Do not purchase products from companies
7. Avoid supporting the market in illegal wildlife including tortoise-shell, ivory, and coral.
8. Herbicides and pesticides are hazardous pollutants that affect wildlife at many levels.
9. Educate friends and relatives about endangered species in your area.

Question 2.
Celebrate the national eyents in your school and prepare a news item for a local
newspaper.
Answer:
Republic Day Celebrations in our XYZ Hr. Sec. School
A Report
On the morning of 26th Jan, our school, XYZ Hr. Sec. School, celebrated Republic Day in the school premises. It was a matter of great pride for all of us. Our Headmaster, teachers and the students didn’t want to leave any stone unturned in making this monumental day a memorable one.

The DIG of the City Police was the Chief Guest. He came on time. He was warmly welcomed by the whole staff led by our Headmaster. Then our Headmaster took him to the Function venue. There our hon’ble guest unfurled the national flag. With that everybody stood erected for National Anthem sung by the students. Then there were parades by NCC cadets of the school. After this, students presented cultural items.

These cultural items included patriotic songs, folk dances and speeches. Some students also presented skit based on fight against terrorist. Then there were speakers who through their speeches invoked the precious contribution of great freedom fighters.

With a couple more cultural items, the function ended with national anthem. Towards the end, all the people there flew colored balloons symbolizing prosperity and peace. Despite weather being inclement, we enjoyed Republic Day with great enthusiasm. Along with other students, I will never forget this wonderful day.

XI. Life skill activity:

Question 1.
Why are certain organisms adopted as natural National symbols? Analyse.
Answer:
Along with various animals and birds, which form part of the national symbols of India, the Indian government adopted a microorganism called lactobacillus delbrueckii in the year 2012 as our national microorganism.

The reason for doing so, is primarily educational and to create awareness among people about the importance of microorganisms. Lactobacillus is an important bacteria used in food production.

Lactobacillus species are used for the production of yogurt, cheese, sauerkraut, pickles, beer, wine, cider, kimchi, cocoa, and other fermented foods, as well as animal feeds, such as silage.

Samacheer Kalvi 6th Social Science National Symbols InText Questions

HOTS

Question 1.
Who has been given the right to manufacture the National flag?
Answer:

1. By law the flag is to be made of Khadi.
2. The right to manufacture the flag is held by the Khadi Development and Village Industries Commission who allocates it to regional groups.
3. As of 2009, the Karnataka Khadi Gramodyoga Samyukta Sangha has been the sole manufacturer of the flag.

Samacheer Kalvi 6th Social Science National Symbols Additional Questions

I. Choose the Correct answer

Question 1.
Wildlife Protection Act was passed in the year.
(a) 1970
(b) 1971
(c) 1972
(d) 1973
Answer:
(c) 1972

Question 2.
The National flower of India is
(a)’ Lotus
(b) Jasmine
(c) Rose
(d) Lily
Answer:
(a) Lotus

Question 3.
The Government of India adopted Elephant as a natural National symbol in the year.
(a) 1950
(b) 1963
(c) 1973
(d) 2010
Answer:
(d) 2010

Question 4.
Congress committee in 1911 Was held at
(a) Chennai
(b) Delhi
(c) Kolkata
(d) Bombay
Answer:
(c) Kolkata

Question 5.
The symbol Indian National Currency was designed by
(a) Ravikumar
(b) Udhyakumar
(c) Krishnakumar
(d) Manojkumar
Answer:
(b) Udhyakumar

II. Fill in the blanks:

1. Tirupur Kumaran was bom in _______
2. Tirupur Kumaran honoured with the titile _______
3. Satyameva Jayate means _______
4. Lactobacillus was accepted as our National Micro organism in the year _______
5. SherShah sur introduced rupee currency Rupiya in the _______
6. During a leap year the spring equinox begins on _______
7. World’s biggest democracy is _______
8. Aaduvome Pallu Paduvome” was sung by _______ Over the AIR
9. The first citizen of India is the _______
10. On Republic Day, the _______ of India the National Flag at the red fort,New Delhi
11. The International Day of Non-violence is observed on _______

Answer:

1. Chennimalai
2. Kodi Kaatha Kumaran
3. Truth alone Triumphs
4. 2012
5. Sixteenth Century
6. March 21th
7. India
8. T.K. pattammal
9. President
10. President
11. 2nd October

III. Match the following:

IV. Answer the following questions :

Question 1.
Explain the ‘Beating retreat’ Ceremony.
Answer:
On January 29, the third day of the Republic day, the celebrations are brought to an end with the “ Beating Reteat” ceremony. This is performed by the bands of Indian Army, Navy and Airforce. The President of India is the chief guest of this day. Rashtrapati Bhavan will be illuminated at 6pm as a part of the celebration.

Question 2.
Write a note on Kodi Katha Kumaran.
Answer:

1. Tirupur Kumaran was bom in Chennimalai of Erode district.
2. As a youth, he actively participated in the freedom struggle of India,
3. in 1932, when Gandhiji was arrested, there were protests all over India.
4. Tirupur Kumaran took active part in these protests.
5. He lost his life when the police attacked violently.
6. He held on to the tricolour flag even when he died.
7. He was honoured with the title, ‘Kodi Kaatha Kumaran’.

Question 3.
Write about Tamil Nadu’s National Symbols.
Answer:

Question 4.
Explain the natural national symbols.
Answer:

1. Banyan tree (1950). It is a symbol of pride and has many medicinal values.
2. Peacock (1963). It is native to Asia and the only bird which has a tail.
3. River Ganges (2008) is a perennial river and many royal capitals flourished on its banks.
4. River Dolphin (2010). It is the reliable indicator of the health of the entire river eco-system. It is in the endangered list.
5. King Cobra – It is the World’s longest venomous snake and lives in the rain forests and plains of India.
6. Lotus (1950) Though it grows in muddy water, it blooms with beauty.
7. Tiger (1973) It is the largest cat species. India has 70% of tigers population in the world.
8. Elephant (2010) It is native to mainland Asia and plays a critical role in maintaining the regions forests.

You can Download Samacheer Kalvi 6th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Science Solutions Term 3 Chapter 1 Magnetism

Samacheer Kalvi 6th Science Magnetism Textual Evaluation

I. Choose the appropriate answer:

Samacheer Kalvi Guru 6th Science Question 1.
An object that is attracted by magnet.
(a) wooden piece
(b) plain pins
(c) eraser
(d) a piece of paper
Answer:
(b) plain pins

Samacheer Kalvi 6th Science Term 3 Question 2.
People who made mariner’s compass for the first time.
a. Indians
b. Europeans
c. Chinese
d. Egyptians
Answer:
c. Chinese

Magnetism Lesson 6th Grade Question 3.
A freely suspended magnet always comes to rest in the
(a) North – East
(b) South – West
(c) East – West
(d) North – South
Answer:
(d) North-South

Samacheer Kalvi 6th Science Book Back Answers Question 4.
Magnets lose their properties when they are
a. used
b. stored
c. hit with a hammer
d. cleaned
Answer:
c. hit with a hammer

Samacheer Kalvi 6th Science Question 5.
Mariner’s compass is used to find the
(a) speed
(b) displacement
(c) direction
(d) motion
Answer:
(c) direction

II. Fill in the blanks:

1. Artificial magnets are made in different shapes such as _______ , _______ and _______
2. The materials which are attracted towards the magnet are called _______
3. Paper is not a _______ material.
4. In olden days, sailors used to find direction by suspending a piece of _______
5. A magnet always has _______ poles.

Answers:

1. Bar-magnet, Horseshoe magnet, Ring magnet
2. magnetic substances
3. magnetic
4. lodestones
5. two

III. True or False. If False, give the correct statement:

Samacheer Kalvi 6th Science Guide Question 1.
A cylindrical magnet has only one pole.
Answer:
False. A cylindrical magnet has two poles.

6th Standard Samacheer Kalvi Science 3rd Term Question 2.
Similar poles of a magnet repel each other.
Answer:
True.

Samacheer Kalvi 6th Standard Third Term Question 3.
Maximum iron filings stick in the middle of a bar magnet when it is brought near them.
Answer:
False. Maximum iron filings stick in the poles of a bar magnet when it is brought near them.

Science Term 3 Question 4.
A compass can be used to find East-West direction at any place.
Answer:
True. A magnetic compass always points towards the North-South direction. If the North-South direction is known, then the East-West direction can also be determined. This direction is perpendicular to the North-South direction, ie., perpendicular to the compass needle in the same plane.

6th Science Samacheer Kalvi Question 5.
Rubber is a magnetic material.
Answer:
False. Rubber is anon-magnetic material.

IV. Match the following:

1. Compass – Maximum magnetic strength
2. Attraction – Like poles
3. Repulsion – Opposite poles
4. Magnetic poles – Magnetic needle

Answer:

1. Compass – Magnetic needle
2. Attraction – Opposite poles
3. Repulsion – Like poles
4. Magnetic poles – Maximum magnetic strength

V. Circle the odd ones and give reasons:

6th Science Term 3 Question 1.
Iron nail, pins, rubber tube, needle.
Answer:
Rubber tube.
Rubber tube is a non-magnetic substance, others are magnetic substances.

Samacheer Kalvi Guide 6th Science Question 2.
Lift, escalator, electromagnetic train, electric bulb.
Answer:
Electric bulb.
Electric bulb does not have magnets others have electromagnets.

Samacheer Kalvi 6th Science Book Solutions Question 3.
Attraction, repulsion, pointing direction, illumination.
Answer:
Illumination
Illumination is not a property of magnet, others are magnetic properties.

VI. The following diagrams show two magnets near one another. Use the words, ‘Attract, Repel, Turn around’ to describe what happens in each case.

Answer:
a. Unlike poles attract one another.
b. Like poles repel each other.
c. Unlike poles attract one another.
d. Perpendicular poles turn around and attract one another.
e. Like poles repel each other.
f. Perpendicular poles turn around and attract one another.

VII. Write down the names of substances :

Answer:

VIII. Give Short Answer:

6th Samacheer Kalvi Science Question 1.
Explain the attraction and repulsion between magnetic poles.
Answer:

1. Like poles (N – N, S – S) repel each other.
2. Unlike poles (N – S, S – N) attract each other.

Samacheer Kalvi 6th Science Book Question 2.
A student who checked some magnets in the school laboratory found out that their magnetic force is worn out. Give three reasons for that?
Answer:
Magnets lose their properties if they are:

1. heated
2. dropped from a height
3. hit with a hammer

These are the reasons for that their magnetic force is worn out.

IX. Answer in detail:

Samacheer Kalvi 6 Science Question 1.
You are provided with an iron needle. How will you magnetize it?
Answer:

1. Place the iron needle on the table.
2. Take a bar magnet and place one of its poles near one edge of the iron needle.
3. Rub from one end to another without changing the direction of the pole of the magnet.
4. Repeat the process for 30 to 40 times. The needle will be magnetized.
5. If it will not attract pin or iron fillings continue the same process for some more time.

Samacheer Kalvi 6th Standard Science Third Term Question 2.
How does an electromagnetic train work?
Answer:

1. Electromagnets are used in Electromagnetic train.
2. Electromagnets are magnetised only when current flows through them.
3. When the direction of current is changed, the poles of the electromagnets are also changed.
4. Like poles of the magnets which are attached at the bottom of the train and rai l track repel each other.
5. So, the train is lifted from the track up to a height of 10 cm.
6. We know that we can move any magnetic object with the force of attraction or repulsion properties of magnets.
7. This train also moves with the help of the magnets attached on the sides of track and the magnets fitted at the bottom sideway of the train.
8. By controlling the current, we can control the magnets and movement of the train.

X. Questions based on Higher Order Thinking Skills:

Question 1.
You are provided with iron filings and a bar magnet without labelling the poles of the magnet. Using this.

1. How will you identify the poles of the magnet?
2. Which part of the bar magnet attracts more iron filings? Why?

Answer:

1. When we place the bar magnet in iron fillings large amount of iron fillings stick on the two ends of the bar magnet. These ends are poles of the magnet.
2. Poles will attract more iron filings. Because poles have high magnetic strength.

Question 2.
Two bar magnets are given in the figure A and B. By the property of attraction, identify the North pole and the South pole in the bar magnet (B)

Answer:
The Fig -A has S and N poles.
In the Fig -B magnet, nearer to the North pole of Fig-A is South pole and the next pole is North pole.

Question 3.
Take a glass of water with a few pins inside. How will you take out the pins without dipping your hands into the water?
Answer:
If we keep a strong bar magnet above the glass of water, all pins inside the water come up and attract the magnet.

Samacheer Kalvi 6th Science Magnetism Intext Activities

Activity 1

Question 1.
Take a magnet. Take the magnet Closer to the objects surrounding you.
What happens? Observe and note.
i. The objects attracted by the magnet: ______________
ii. The objects, not attracted by the magnet: ______________
iii. Which substance is used to make the objects attracted by the magnet? ______________
Answer:
i. The objects attracted by the magnet: irons nail, iron rod, pins, needle, key, iron filings.
ii. The objects, not attracted by the magnets: rubber, paper, plastic pen, scale, pencil, water bottle.
iii. Which substance is used to make the objects attracted by the magnet?
Iron is used to make the objects attracted by the magnet.

Activity 2

LET US MAKE MAGNETS

Take a nail / a piece of Iron and place it on a table. Now take a bar magnet and place one of its poles near one edge of the nail/piece of Iron and rub from one end to another end without changing the direction of the pole of the magnet. Repeat the process for 30 to 40 times.

Bring a pin or some iron filings near the nail/piece of Iron to check whether it has become a magnet. Does the nail/piece of iron attract the pin/iron filings? If not, continue the same process for some more time.
Answer:
Yes, The nail / a piece of Iron acquires the same ability to attract other pieces of pin/ iron filings.

Activity 3

Make your own magnetic compass

Insert the magnetized needle, that you made in the activity 2, in to two styrofoam balls and place the needle in bowl of water. Test whether the floating needle is always turned in rest on north-south direction.

Note : If you don’t have styrofoam balls you can use dry leaf or a cork piece.

Inference: The floating needle is always turned in nest on North-South direction.

Samacheer Kalvi 6th Science Magnetism Additional Questions

I. Choose the appropriate answer:

Question 1.
_______ made objects are attracted by magnets.
(a) Plastic
(b) Iron
(c) Glass
(d) Wax
Answer:
(b) Iron

Question 2.
Magnets lose their property if they
(a) heated
(b) dropped from a height
(c) hit with hammer
(d) all of these
Answer:
(d) all of these

Question 3.
_______ poles repel to each other.
i. N – N
ii. N – S
iii. S – N
iv. S – S
(a) i and ii
(b) ii and iii
(c) iii and iv
(d) i and iv
Answer:
(d) i and iv

Question 4.
When N pole of bar magnet is moved closer to the north pole of a magnet it will
(a) attract
(b) repel
(c) rotate
(d) none
Answer:
(b) repel

Question 5.
Electro magnetic trains can easily attain a speed of _______ km per hour.
(a) 1200
(b) 600
(c) 100
(d) 800
Answer:
(b) 600

II. Fill up the blanks:

1. The magnetic Ore is called as _______
2. _______ magnets do not have a definite shape.
3. Man-made magnets are called _______ magnets.
4. A _______ is an instrument which is used to find directions.
5. _______ will also get affected by magnetic field.
6. For a _______ magnet a single piece of soft iron can be used as a magnetic keeper across the pole.

Answer:

1. magnetite
2. Natural
3. Artificial
4. compass
5. Cell phone
6. horse-shoe

III. True or False. If False, give the correct statement:

Question 1.
Magnetites are artificial magnets.
Answer:
False. Magnetites are natural magnets.

Question 2.
Cube shaped magnets are also available.
Answer:
False. Oval shaped, Disc shaped, Cylindrical shaped magnets are also available.

Question 3.
Substancess which are attracted by magnet are called non-magnetic substance.
Answer:
False. Substances which are attracted by magnets are called magnetic substance.

Question 4.
The end of the magnet that points to the north is called south pole.
Answer:
False. The end point of the magnet that points to the south is called south pole.

Question 5.
The compass has a magnetic needle that can rotate easily.
Answer:
True.

Question 6.
Magnets lose their properties if they are dropped from a height.
Ans:
True.

Question 7.
Proper storage can also cause magnets to lose their properties.
Answer:
False. Improper storage can also cause magnets to lose their properties.

Question 8.
Electro magnetic train is also called as flying train.
Answer:
True.

IV. Match the following :

Answer:
I – f
II – d
III – e
IV – b
V – c
VI – a.

V. Analogy:

Question 1.
Natural magnet: Magnetic stones.
Artificial magnet: _________
Answer:
Bar magnet.

Question 2.
Magnetic substance : Attracted by magnets
Non magnetic substance : _________
Answer:
Not attracted by magnets.

Question 3.
Repel to each other: Like poles.
Attract to each other : _________
Answer:
Unlike poles.

Question 4.
Demagnetisation : _________
Making magnets: Rubbing with one end to another end without changing direction.
Answer:
Hit with hammer

Question 5.
Electromagnet: Magnetic crane.
Ordinary magnet: _________
Answer:
Mobile phone covers.

VI. Give Short Answer:

Question 1.
Give the different shapes of Artificial magnets.
Answer:

1. Bar magnet
2. Horseshoe magnet
3. Ring magnet
4. Needle magnet
5. Oval shape magnet
6. Disc shape magnet
7. Cylindrical shape magnet.

Question 2.
Differentiate magnetic and non-magnetic substance.
Answer:
Magnetic substance:

1. It is attracted by magnets
2. Ex: Iron, Cobalt, Nickel

Non-magnetic substance:

1. It is not attracted by magnets
2. Ex: Paper, Plastic, Glass

Question 3.
Define – Poles of a magnet.
Answer:
The attractive force of the magnet is very large near the two ends. These two ends are called poles of a magnet.

Question 4.
Give the properties of magnets.
Answer:

1. Like poles repel each other.
2. Unlike poles attract each other.

Question 5.
What are the objects affected by magnetic field?
Answer:
Cellphone, Computer and DVD are the objects affected by magnetic field.

Question 6.
Give any two tips to store bar magnets.
Answer:

1. Bar magnet should be kept in pairs with their unlike poles on the same side.
2. They must be seperated by a piece of wood and two pieces of soft iron should be placed across their ends.

VII. Answer in detail :

Question 1.
Give the uses of magnets.
Answer:

1. Magnets are used in speakers to power up.
2. Magnets are used in Ladies hand bags, Pencil boxes etc., to make a tight seal.
3. Magnets are also used in some door locks.
4. In home, magnets are used as magnetic toys, magnetic smiley stickers, magnetic Alphabet stickers to stick on the refrigerators.
5. Magnets are used to store data in computers.

Question 2.
How will you make levitating propeller?
Answer:
1. Make a propeller from a 500-ml plastic bottle. Make a hole in the bottle lid.

2. Screw the lid with the hole on a bottle half filled with sand. Press fit a stiff straw in the lid. Embed the straw in the sand to make it stand erect. Place a few ring magnets in the straw. Similar poles will repel each other.

3. Place two magnets each inside and outside the propellor lid. These magnets will automatically stick to each other.

4. Like poles repel and this levitates the bottle fan. The ceiling fan makes it spin.

5. Place the propeller on the stiff straw. The magnets in the straw and the propeller should repel each other. This will make the propeller levitate. On placing it under a ceiling fan the propeller will spin very fast.

You can Download Samacheer Kalvi 6th Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Science Solutions Term 1 Chapter 6 Health and Hygiene

Samacheer Kalvi 6th Science Health and Hygiene Textual Evaluation

I. Choose the appropriate answer

Samacheer Kalvi Guru 6th Science Question 1.
Our body needs _______ for muscle-building.
(a) Carbohydrate
(b) Fat
(c) Protein
(d) Water
Answer:
(c) Protein

Health And Hygiene Class 6 Questions And Answers Samacheer Question 2.
Scurvy is caused due to the deficiency of ………..
(a) Vitamin A
(b) Vitamin B
(c) Vitamin C
(d) Vitamin D
Answer:
(c) Vitamin C

Samacheer Kalvi 6th Science Question 3.
Calcium is an example of a _______
(a) Carbohydrate
(b) Fat
(c) Protein
(d) Minerals
Answer:
(d) Minerals

Samacheer Kalvi 6th Science Guide Question 4.
We should include fruits and vegetables in our diet, because ……….
(a) They are the best source of Carbohydrates.
(b) They are the best source of Proteins.
(c) They are rich in minerals and vitamins.
(d) They have high water content.
Answer:
(c) They are rich in minerals and vitamins.

Samacheer Kalvi 6th Science Book Back Answers Question 5.
Bacteria are very small _______ microorganism.
(a) Prokaryotic
(b) Eukaryotic
(c) Protozoa
(d) Acellular
Answer:
(a) Prokaryotic

II. True or false

6th Science Samacheer Kalvi Question 1.
There are three main nutrients present in food.
Answer:
False. There are six main nutrients present in food.

Samacheer Kalvi.Guru 6th Science Question 2.
Fats are used as an energy store by our body.
Answer:
True.

Samacheer Kalvi Science 6th Standard Question 3.
All bacteria have flagella.
Answer:
False. Not all bacteria have flagella only, some bacteria have flagella.

Samacheer Kalvi 6 Science Question 4.
Iron helps in the formation of haemoglobin.
Answer:
True

Samacheer Kalvi 6th Science Solution Question 5.
Virus can grow and multiply outside host.
Answer:
False. Virus can grow and multiply inside host.

III. Fill in the Blanks.

1. Malnutrition leads to _______
2. Iodine deficiency leads to _______ in adults.
3. Vitamin D deficiency causes _______.
4. Typhoid is transmitted due to contamination of _______ and water.
5. Influenza is a _______ disease.

Answers:

1. deficiency disease
2. goitre
3. Rickets
4. food
5. viral (virus)

IV. Complete the Analogy.

Samacheer Kalvi 6th Standard Question 1.
Rice: Carbohydrate :: Pulses: _______
Answer:
Protein.

Samacheer Kalvi 6th Science Solutions Question 2.
Vitamin D : Rickets :: Vitamin C: _______
Answer:
Scurvy.

Samacheer Kalvi 6th Science Book Question 3.
Iodine: Goitre :: Iron: _______
Answer:
Anaemia.

Samacheer Kalvi Class 6 Science Solutions Question 4.
Cholera: Bacteria :: Smallpox: _______
Answer:
Virus.

V. Match the following

1. Vitamin A – Rickets
2. Vitamin B – Night blindness
3. Vitamin C – Sterility
4. Vitamin D – Beriberi
5. Vitamin E – Scurvy

Answer:

1. Vitamin A – Night blindness
2. Vitamin B – Beri beri
3. Vitamin C – Scurvy
4. Vitamin D – Rickets
5. Vitamin E – Sterility

VI. Complete the diagram

Question:

Answer:

VII. Write a Short Answer.

Samacheer Kalvi Science 6th Question 1.
Write two examples for each of the following.
Answer:
(a) Food items were rich in fat – Egg yolk, meat.
(b) Vitamin deficiency – diseases.
(c) Vitamin C – Scurvy
(d) Vitamin D – Rickets.

6th Samacheer Kalvi Science Question 2.
Differentiate between carbohydrate and protein.
Answer:
Carbohydrate:

1. Energy giving component of the food.
2. The sources of carbohydrate are nuts, fruits, rice and maize.

Protein:

1. It is body building foods.
2. The sources of proteins are pulses, soyabean, nuts, egg, and fish.

6th Standard Science Question 3.
Define the term ‘balanced diet’?
Answer:
A balanced diet is a diet which contains an adequate amount of all the necessary nutrients like carbohydrate fat protein minerals and vitamins required for healthy growth and activity.

Samacheer Kalvi Guru 6th Question 4.
Why should the fruits and vegetables not to be washed after cutting?
Answer:
We should not wash the fruits and vegetables after cutting, because the minerals and protein in the fruits and vegetables will also be washed away.

Samacheer Kalvi 6th Solutions Question 5.
Write any two viral diseases.
Answer:

1. AIDS
2. Hepatitis.

Samacheerkalvi.Guru 6th Science Question 6.
What is the main feature of a microorganism?
Answer:
Microorganism will be seen with the help of microscope. They are very small in size.

VIII. Long Answer.

Samacheer Kalvi Guru 6th Standard Science Question 1.
Tabulate the vitamins and their corresponding deficiency diseases.
Answer:

Samacheer Kalvi 6th Science Health and Hygiene Intext Activities

Samacheer Kalvi Guru 6th Science Guide Activity 1

Identify the following food items and complete the table given below.

Answer:

Samacheer Kalvi Guru 6 Science Question 1.
Do your favorite foods make you healthy?
Answer:
Some of my favorite foods make me healthy.

Samacheer Kalvi 6th Question 2.
Do you choose your food by taste or by its nutritive value?
Answer:
I choose my food by taste and also by its nutritive value.

Samacheer Kalvi 6th Guide Activity 2
Collect as many food items as you can and classify them according to the major nutrient content in it.
Answer:

Activity 3
Aim: To test the presence of Carbohydrate as Starch in the given food item.

Question 1.
What do you need?
Answer:
Boiled potato, dropper and dilute Iodine solution.

Question 2.
How to do?
Answer:
Smash the boiled potato. Add two or three drops of dilute Iodine solution on the Sample.

Question 3.
What do you observe?
Answer:
The potato turns blue-black in colour.

Question 4.
What do you infer?
Answer:
Iodine reacts with Starch to form Starch-Iodine complex which is blueblack in colour. Thus, the appearance of blue-black colour confirms the presence of Starch in the food item.

Activity 4
Aim: To test the presence of Fat in the given food item.

Question 1.
What do you need?
Answer:
Coconut Oil, groundnut oil, and any Paper.

Question 2.
How to do?
Answer:
Pour few drops of oil onto the paper and rub it gently with your finger. In case of ground nut, crush the groundnut and place it on a paper. Now rub the groundnut on the paper.

Question 3.
What do you see?
Answer:
The paper turns translucent and becomes greasy.

Question 4.
What do you learn?
Answer:
The given food sample contains fat.

Activity 5
Aim: To test the presence of Protein in the given food item.

Question 1.
What do you need?
Answer:
Egg white, Copper sulphate solution, Sodium hydroxide, Test tube and Bunsen burner.

Question 2.
How to do?
Answer:
Take a small amount of the food sample (egg white) and put in the test tube. Add some water to the test tube and shake it. Next, heat the test tube for about one minute. After the test tube has cooled down, add two drops each of Copper sulphate solution and Sodium hydroxide to it.

Question 3.
What do you see?
Answer:
The food sample turns purple or violet.

Question 4.
What do you learn?
Answer:
Change in colour of the given food sample turns purple or violet confirms the presence of Protein.

Just Think

A medical camp was conducted in School. Most of the children were healthy. Some of the students had some health issues
Priya had bleeding gums.
Raja could not see clearly in dim light.
Arun had bent legs .
Can you guess what could be the reasons?

• Priya was affected by Vitamin ‘C’ deficiency. So she had bleeding gums.
• Raja was affected by vitamin ‘A’ deficiency. So he could not see clearly in dim light.
• Arun was affected by vitamin ‘D’ deficiency. So he had bent legs.

Activity 6

Question 1.
Make your food little healthier. What do you need?
Answer:
A small cup of green gram seeds, Water and thin cloth.

Question 2.
How to do?
Answer:
Soak the green gram seeds in water over night. Take out the seeds and strain the water. Wrap the seeds in wet thin cloth. Keep it for a day or two. Sprinkle some water whenever it is dry.

Question 3.
What do you see?
Answer:
You can see white sprouts coming out of the seeds.

Question 4.
What do you learn?
Answer:
Green gram sprouts are low in calories, have fibre and Vitamin B. It has comparatively high amount of vitamin C and vitamin K.

Complete the following Table – 4

Answer:

Activity 7

Question:
Why do we need a balanced diet?
Prepare a diet chart to provide balanced diet to a 12 year old child. The diet chart should include food item which are not expensive and are commonly available in your area.
Answer:
We need a balanced diet because,
a balanced diet contains sufficient amount of various nutrients to ensure good health.
Balance diet for 12 year old boy / girl.

Activity 8

Question:
Visit a nearby Anganwadi centre and find the steps taken by the government to overcome malnutrition and ensure health in the age group 0-5 years.
Answer:
Activity to be done by the students themselves

Activity 9

One day Rahim, a class six boy vomited three times. He was looking tired and dehydrated. His mother who was a nurse prepared a solution and gave it to him drink. He felt better after sometime and asked his mother what the solution was. His mother said it was Oral Rehydration Solution – ORS. Shall we see what an ORS is? Vomiting or loose motions result in loss of water and cause salt imbalance in the body. Loss of water (dehydration) can lead to serious problems. This can be prevented by consuming ORS at short intervals.

Follow the steps to make ORS at home:

• Take a litre of boiled water. Cool it.
• Add half a teaspoon of salt and six teaspoons of sugar to it.
• You can also add a few drops of lemon juice to it. Stir it and give it to the person suffering from vomiting, loose motions or dehydration.

Answer:
Activity to be done by the students themselves

Discuss in your class room
“Is virus a living thing or non living thing?”
Answer:
Viruse is a non-living thing.

Reasons:

1. Virus is an a cellular, meaning it lacks a cell membrane.
2. It is an obligate parasite. It requires a host cell to reproduce.
3. It lacks an energy-generating system. It cannot produce its own energy.
4. It cannot perform metabolic activities.
5. Viruses are not composed of cells and do not reproduce themselves from other pre-existing viruses.
6. The genetic material (DNA or RNA) inside the virus invades the living host cell, and causes the cell to replicate virus parts and assemble them to make new viruses, which are released to infect more cells.

Samacheer Kalvi 6th Science Health and Hygiene Additional Questions

I. Choose the correct answer

Question 1.
_______ provides more energy than Carbohydrates.
(a) Fat
(b) Vitamin
(c) Protein
(d) Water
Answer:
(a) Fat

Question 2.
Name the vitamin which is rich in Gooseberry.
(a) C
(b) E
(c) B
(d) D
Answer:
(c) B

Question 3.
Among the Vitamins, which one is the water soluble Vitamin?
(a) Vitamin A
(b) Vitamin B
(e) Vitamin D
(d) Vitamin E
Answer:
(b) Vitamin B

Question 4.
Name the vitamin present in Muringa leaves
(a) A and B
(b) C and D
(c) K and A
(d) A and C
Answer:
(d) A and C

Question 5.
_______ is made in our skin using sunlight.
(a) Vitamin D
(b) Vitamin C
(c) Vitamin A
(d) Vitamin B
Answer:
(a) Vitamin D

Question 6.
_______ are required for growth as well as for the regulation of normal body function.
(a) Fats
(b) Proteins
(c) Carbohydrates
(d) Minerals
Answer:
(d) Minerals

Question 7.
80% of the world production of Moringa leaves is in _______
(a) China
(b) Germany
(c) India
(d) Canada
Answer:
(c) India

Question 8.
Any human being should take minimum _______ of water every day.
(a) 2 litres
(b) 3 litres
(c) 8 litres
(d) 6 litres
Answer:
(a) 2 litres

Question 9.
_______ is the bacterial disease and it transmits through contamination of wounds, with the bacteria.
(a) Cholera
(b) Tetanus
(c) Typhoid
(d) Pneumonia
Answer:
(b) Tetanus

Question 10.
_______ is a cellular agent that replicates only inside the cells of the other living organism.
(a) Bacteria
(b) Protozoa
(c) Fungi
(d) Virus
Answer:
(d) Virus

Il. Fill in the blanks.

Question 1.
We can obtain Carbohydrates in the form of _______
Answer:
Sugar, Starch, Dietary Fibers

Question 2.
Vitamins are called as _______ food.
Answer:
Protective

Question 3.
The vitamins A, D, E, K are _______ soluble vitamins.
Answer:
Fat

Question 4.
_______ is a disease, due to the deficiency of Vitamin E.
Answer:
Nervous Weakness

Question 5.
Moringa leaves contains powerful anti _______
Answer:
Oxidantsl

Question 6.
Skinny appearance and slow body growth are the symptoms of _______ disease.
Answer:
Marasmusl

Question 7.
_______ is strengthening muscles and the cardiovascular system.
Answer:
Physical exercisel

Question 8.
_______ can kill damage or change the cells and make you sick.
Answer:
Virus

Question 9.
Sun screen lotion reduces your skin’s ability to produce _______ by up to 95%.
Answer:
Vitamin D

Question 10.
Goose bernes contain nearly _______ the vitamin C than orange.
Answer:
20 times

Question 11.
India has the _______ highest number of obese children ¡n the world.
Answer:
Second

III. True of false. If false give the correct answer.

Question 1.
Minerals are required for carrying out various biochemical reactions in our body.
Answer:
False. Vitamins are required for carrying out various biochemical reactions in our body.

Question 2.
Night blindness is a disease due to deficiency of Vitamin A.
Answer:
True.

Question 3.
Vitamin D abundantly found in orange and gooseberry.
Answer:
False. Vitamin C abundantly found in orange and gooseberry.

Question 4.
Sun screen lotion reduces ability to produce Vitamin D. It leads to Vitamin D deficiency diseases.
Answer:
True.

Question 5.
Iodine maintains strong bones, teeth and helps in clotting of blood.
Answer:
False. Calcium maintains strong bones, teeth and helps in clotting of blood.

Question 6.
Moringa leaves are rich in the minerals potassium, calcium and iron.
Answer:
True.

IV. Complete the Analogy

Question 1.
Polio : Virus :: Tetanus : _______
Answer:
Bacteria

Question 2.
_______ : Diarrhoea:: Marasmus: Slow body growth.
Answer:
Kwashiorkar

Question 3.
Synthesis of thyroid harmone: Iodine:: Formation of haeniogiobin: _______
Answer:
Iron

Question 4.
Fish oil : Vitamin D :: Vegetable oil : _______
Answer:
Vitamin II

Question 5.
Vitamin K : Clotting of blood :: Vitamin E : _______
Answer:
Fertility

Question 6.
Protein: Soya bean:: Fat: _______
Answer:
Meat

V. Match the following.

Question 1.

1. Carbohydrate – (a) Carrying out various biochemical reactions
2. Proteins – (b) Regulation of normal body function
3. Vitamins – (c) energy giving component
4. Minerals – (d) Body building food

Answer:

1. – c
2. – d
3. – a
4. – b

Question 2.

Answer:

1. – d
2. – e
3. – b
4. – a
5. – c

VI. Complete the diagram

Answer:
A – Exercise
B – Yoga.

VII. Short Answers.

Question 1.
Define homeostasis.
Answer:
It is to maintain a stable equilibrium of body in accordance with the pressures and changes of body environment.

Question 2.
What are the nutrients obtained from food?
Answer:

1. Carbohydrate
2. Proteins
3. Fats
4. Vitamins
5. Minerals
6. Water.

Question 3.
What is meant by deficiency diseases?
Answer:
The diseases that are caused due to lack of nutrients in the diet are called deficiency diseases.

Question 4.
Sun Screen lotion is not good for our health. Justify.
Answer:
Sun Screen lotion reduces our skin’s ability to produce vitamin D upto 95%. It may lead to Vitamin D deficiency. So it not for good our health.

Question 5.
Fill in the blanks.
A – Calcium – Rickets
Phosphorus – …………
Answer:
Osteomalacia

B – Iodine – cretinism
Iron – ………….
Answer:
anaemia

Question 6.
What are the nutrients present in Moringa leaves?
Answer:
Moringa leaves are rich in Vitamin A, Vitamin C, Potassium, Calcium, Iron and Protein. It also contains powerful anti-oxidants.

Question 7.
How does Balanced diet help in our body?
Answer:
Balanced diet helps in our body, in the following ways :

1. An increased capacity to work
2. Good physical and mental health
3. Increased capacity to resist diseases.
4. Help in proper growth of the body.

Question 8.
Give some mineral deficiency diseases.
Answer:
Rickets, Osteomalatia, Cretinism in Child, Goitre in adult, Anaemia are the mineral deficiency diseases.

Question 9.
How does Physical exercise help us?
Answer:
Physical exercise helps us in the following ways:

1. increase in growth and development
2. strengthening muscles and the cardiovascular system
3. developing athletic skills
4. weight loss or maintenance, and enjoyment.

Question 10.
How are the following bacterial diseases transmitted?
(i) Cholera
(ii) Tetanus
(iii) Typhoid
(iv) Tuberculosis
Answer:

Question 11.
Name the countries which import Moringa leaves.
Answer:
China, United States, German, Canada, South Korea and European Countries import Moringa leaves.

Question 12.
Define health.
Answer:
Health is a state of complete physical, mental and social well-being and not merely absence of diseases. Eating a healthy diet keeps us physically and mentally fit.

Question 13.
Name the four major groups of Microbes.
Answer:
Microbes divided into four major groups. They are

• Bacteria
• Virus
• Protozoa
• Fungi

Question 14.
Define Disease.
Answer:
Disease is a definite pathological process having a characteristic set of signs and symptoms.

Question 15.
What is a Retrovirus?
Answer:
A Virus that contains R.N.A. instead of D.N.A. is called a Retrovirus.

VIII. Long Answer.

Question 1.
Fill in the Table :

Answer:

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Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 6 Solid State

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Samacheer Kalvi 12th Chemistry Solid State Text Book Evalution

I. Choose the correct answer.

12th Chemistry Chapter 6 Book Back Answers Question 1.
Graphite and diamond are ………..
(a) Covalent and molecular crystals
(b) ionic and covalent
(c) both covalent crystals
(d) both molecular crystals
Answer:
(c) both covalent crystals

12th Chemistry Solid State Book Back Answers Question 2.
An ionic compound A_x B_y crystallizes in fee type crystal structure with B ions at the centre of each face and A ion occupying centre of the cube, the correct formula of A B is ………..
(a) AB
(b) AB₃
(c) A₃B
(d) A₈B₆
Answer:
(b) AB₃
Hint: Number of A ions = \(\left( \frac { { N }_{ c } }{ 8 } \right)\) = \(\left( \frac { 8 }{ 8 } \right)\) = 1
Number of B ions = \(\left( \frac { { N }_{ f } }{ 2 } \right)\) = \(\left( \frac { 6 }{ 2 } \right)\) = 3
Simplest formula AB₃

12th Chemistry Unit 6 Book Back Answers Question 3.
The rano of close packed atoms to tetrahedral hole in cubic packing is ………..
(a) 1:1
(b) 1:2
(c) 2:1
(d) 1:4
Answer:
(b) 1:2
Hint: If number of close packed atoms =N; then, The number of Tetrahedral holes formed = 2N, Number of Octahedral holes formed = N. Therefore N : 2N = 1 : 2

Solid State Questions Pdf Question 4.
Solid CO₂ is an example of ………..
(a) Covalent solid
(b) metallic solid
(c) molecular solid
(d) ionic solid
Answer:
(c) molecular solid
Hint: Lattice points are occupied by CO₂ molecules

Solid State Numericals Class 12 Question 5.
Assertion: monoclinic sulphur is an example of monoclinic crystal system.
Reason: for a monoclinic system, a \(\neq\) b \(\neq\) c and α = γ = 90° , β \(\neq\) 90°.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion

Chemistry Class 12 Samacheer Kalvi Question 6.
In calcium fluoride, having the flurite structure the coordination number of Ca²⁺ ion and F Ion are ………..
(a) 4 and 2
(b) 6 and 6
(c) 8 and 4
(d) 4 and 8
Answer:
(c) 8 and 4
Hint: CaF₂ has cubical close packed arrangement Ca²⁺ ions are in face centered cubic arrangement, each Ca²⁺ ions is surrounded by 8 F^– ions and each F^– ion is surrounded by 4 Ca²⁺ ions.Therefore coordination number of Ca²⁺ is 8 and of F^– is 4 pattern is (NA is the Avogadro number)

Samacheer Kalvi 12th Chemistry Question 7.
The number of unit cells in 8gm of an element X (atomic mass 40) which crystallizes in bcc pattern is (N_A is the Avogadro number)
(a) 6.023 x 10²³
(b) 6.023 x 10²²
(c) 60.23 x 10²³
(d) \(\left( \frac { 6.023\times { 10 }^{ 23 } }{ 8\times 40 } \right)\)
Answer:
(b) 6.023 x 10²²
Hint: In bcc unit cell, 2 atoms = 1 unit cell
Number of atoms in 8g of element is, Number of moles = \(\left( \frac { 8g }{ 40g,{ mol }^{ -1 } } \right)\) = 0.2 mol
1 mole contains 6.023 x 10²³ atoms 0.2 mole contains 0.2 x 6.023 x 10²³ atoms
\((\frac{1unit cell}{2 atoms})\) x 0.2 x 6.023 x 10²³ = 6.023 x 10²² unit cells

Samacheer Kalvi Guru 12th Chemistry Question 8.
The number of carbon atoms per unit cell of diamond is ………..
(a) 8
(b) 6
(c) 1
(d) 4
Answer:
(a) 8
Hint: In diamond carbon forming fee. Carbon occupies comers and face centres and also occupying half of the tetrahedral voids.
\(\left( \frac { { N }_{ c } }{ 8 } \right)\) + \(\left( \frac { { N }_{ f } }{ 2 } \right)\) + 4C atoms in Td voids
\(\left( \frac { 8 }{ 8 } \right)\) + \(\left( \frac { 6 }{ 2 } \right)\) + 4 = 8.

Samacheerkalvi.Guru 12th Chemistry Question 9.
In a solid atom M occupies ccp lattice and \(\left( \frac { 1 }{ 3 } \right)\) of tetrahedral voids are occupied by atom N. Find the formula of solid formed by M and N.
(a) MN
(b) M₃N
(C) MN₃
(d) M₃N₂
Answer:
(d) M₃N₂
Hint: If the total number of M atoms is n, then the number of tetrahedral voids = 2 n. Given that \(\left( \frac { 1 }{ 3 } \right)\)^rd of tetrahedral voids are occupied i.e., \(\left( \frac { 1 }{ 3 } \right)\) x 2 n are occupied by N atoms.
∴ M : N ⇒ n : \(\left( \frac { 2 }{ 3 } \right)\)
1 : \(\left( \frac { 1 }{ 3 } \right)\) 3 : 2 ⇒ M₃N₂

12th Chemistry Samacheer Kalvi Question 10.
The Composition of a sample of wurizite is Fe_0.93 O_1.00 what % of Iron present in the form of Fe³⁺
(a) 16.05%
(b) 15.05%
(c) 18.05%
(d) 17.05%
Answer:
(b) 15.05%
Hint: Let the number of Fe²⁺ ions in the crystal be x. The number of Fe³⁺ ions in the crystal be y. Total number of Fe²⁺ and Fe³⁺ ions is x + y. Given that x + y = 0.93. The total charge= 0 x (2+) + (0.93 – x) (+ 3) – 2 = 0 ⇒ 2x + 2.97 – 3x – 2 = 0 x 0.79 Percentage of Fe³⁺ = \(\left( \frac { (0.93-0.79) }{ 0.93 } \right)\)100 = 15.05%

Samacheer Kalvi Guru Class 12 Chemistry Question 11.
The ionic radii of A⁺ and B^– are 0.98 x 10^-10 m and 1.81 x 10^-10 m , the coordination number of each ion in AB is ………..
(a) 8
(b) 2
(c) 6
(d) 4
Answer:
(c) 6
Hint: \(\frac { { r }_{ { c }^{ 1 } } }{ { r }_{ { A }^{ – } } }\) = \(\left( \frac { 0.98\times { 10 }^{ -10 } }{ 1.81\times { 10 }^{ -10 } } \right)\) = 0.54. It is in the range of 0.414, 0.732,hence the coordination number of each ion is 6.

Class 12 Chemistry Samacheer Kalvi Question 12.
CsCl has bcc arrangement, its unit cell edge length is 400pm, its inter atomic distance is ………..
(a) 400pm
(b) 800pm
(c) \(g\sqrt { 3 } \) x 100pm
(d) \(\left( \frac { \sqrt { 3 } }{ 2 } \right)\) x 400 pm
Answer:
(d) \(\left( \frac { \sqrt { 3 } }{ 2 } \right)\) x 400 pm
Hint:
\(g\sqrt { 3 } \)a = r_Cs⁺ + 2r_Cs^– + r_Cs⁺
\(\left( \frac { \sqrt { 3 } }{ 2 } \right)\)a = (r_Cs⁺ + r_Cs^–)
\(\left( \frac { \sqrt { 3 } }{ 2 } \right)\)400 = inter ionic distance

12 Chemistry Samacheer Kalvi Question 13.
A solid compound XY has NaCl structure, if the radius of the cation is 100pm , the radius of the anion will be ………..
(a) \(\left( \frac { 100 }{ 0.414 } \right)\)
(b) \(\left( \frac { 0.732 }{ 100 } \right)\)
(c) 100 x 0.414
(d) \(\left( \frac { 0.414 }{ 100 } \right)\)
Answer:
(a) \(\left( \frac { 100 }{ 0.414 } \right)\)
Hint: For an fcc structure

12th Chemistry Solutions Samacheer Kalvi Question 14.
The vacant space in bcc lattice unit cell is ………..
(a) 48%
(b) 23%
(c) 32%
(d) 26%
Answer:
(c) 32%
Hint: Packing efficiency = 68%. Therefore empty space percentage = (100-68) = 32%

Solid State Book Back Answers Question 15.
The radius of an atom is 300pm, if it crystallizes in a face centered cubic lattice, the length of the edge of the unit cell is ………..
(a) 488.5pm
(b) 848.5pm
(c) 884.5pm
(d) 484.5pm
Answer:
(b) 848.5pm
Hint: Let edge length = a \(g\sqrt { 2 } \)a = 4r
a = \(\left( \frac { 4\times 300 }{ \sqrt { 2 } } \right)\)
a = 600 x 1.414
a = 848.4 pm

Samacheer Kalvi Guru Chemistry Question 16.
The fraction of total volume occupied by the atoms in a simple cubic is ………..
(a) \(\left( \frac { \pi }{ 4\sqrt { 2 } } \right)\)
(b) \(\left( \frac { \pi }{ 6 } \right)\)
(c) \(\left( \frac { \pi }{ 4 } \right)\)
(d) \(\left( \frac { \pi }{ 3\sqrt { 2 } } \right)\)
Answer:
(b) \(\left( \frac { \pi }{ 6 } \right)\)
Hint:

12th Chemistry Solid State Pdf Question 17.
The yellow colour in NaCl crystal is due to ………..
(a) excitation of electrons in F centers
(b) reflection of light from Cl^– ion on the surface
(c) refraction of light from Na⁺ ion
(d) all of the above
Answer:
(a) excitation of electrons in F centers

Solid State Chemistry Class 12 Pdf Question 18.
If ’a’ stands for the edge length of the cubic system; sc ,bcc, and fcc. Then the ratio of radii of spheres in these systems wilL be respectively.

Answer:

Hing:

Solid State Chemistry Class 12 Question 19.
If a is the length of the side of the cube, the distance between the body centered atom and one comer atom in the cube will be ………..
(a) \(\left( \frac { 2 }{ \sqrt { 3 } } \right) a\)
(b) \(\left( \frac { 4 }{ \sqrt { 3 } } \right) a\)
(c) \(\left( \frac { \sqrt { 3 } }{ 4 } \right) a\)
(d) \(\left( \frac { \sqrt { 3 } }{ 2 } \right) a\)
Answer:
(d) \(\left( \frac { \sqrt { 3 } }{ 2 } \right) a\)
Hint: If a is the length of the side, then the length of the leading diagonal passing through the body centered atom is \(g\sqrt { 3 } \)a. Required distance = \(\left( \frac { \sqrt { 3 } }{ 2 } \right) a\)

Question 20.
Potassium has a bcc structure with nearest neighbor distance 4.52 A. its atomic weight is 39. Its density will be ………..
(a) 915 kg m^-3
(b) 2142 kg m^-3
(c) 452 kg m^-3
(d) 390 kg m^-3
Answer:
(a) 915 kg m^-3
Hint:
(ρ) = \(\frac { nM }{ { a }^{ 3 }{ N }_{ A } }\)For bcc n = 2 M 39. Nearest distance 2r = 4.52

ρ = 915 kg m^-3 N_A

Question 21.
Schottky defect in a crystal is observed when ………..
(a) unequal number of anions and anions are missing from the lattice
(b) equal number of anions and anions are missing from the lattice
(e) an ion leaves its normal site and occupies an interstitial site
(d) no ion is missing from its lattice.
Answer:
(b) equal number of anions and anions are missing from the lattice

Question 22.
The cation leaves its normal position in the crystal and moves to some interstitial position, the defect in the crystal is known as ………..
(a) Schottky defect
(b) F center
(c) Frenkel defect
(d) non-stoichiometric defect
Answer:
(c) Frenkel defect

Question 23.
Assertion – due to Frenkel defect, density of the crystalline solid decreases.
Reason – in Frenkel defect cation and anion leaves the crystal.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false
Answer:
(d) Both assertion and reason are false

Question 24.
The crystal with a metal deficiency defect is ………..
(a) NaCI
(b) FeO
(c) ZnO
(a) KCI
Answer:
(b) FeO

Question 25.
A two dimensional solid pattern formed by two different atoms X and Y is shown below. The black and white squares represent atoms X and Y respectively. The simplest formula for the compound based on the unit cell from the pattern is ………..

(a) XY₈
(b) X₄Y₉
(c) XY₂
(d) XY₄
Answer:
(a) XY₈

II. Answer the following questions:

Question 1.
Define unit cell.
Answer:
A basic repeating structural unit of a crystalline solid is called a unit ccli.

Question 2.
Cive any three characteristics of ionic crystals.
Answer:

1. Ionic solids have high melting points.
2. These solids do not conduct electricity, because the ions are fixed in their lattice positions.
3. They do conduct electricity in molten state (or) when dissolved in water because, the ions are free to move in the molten state or solution.

Question 3.
Differentiate crystalline solids and amorphous solids.
Answer:
Crystalline solids:

1. Long range orderly arrangement of constituents
2. Definite shape
3. Generally crystalline solids are anisotropic in nature
4. They are true solids
5. Definite Heat of fusion
6. They have sharp melting points.
7. Examples: NaCl, diamond etc.,

Amorphous solids:

1. Short range, random arrangement of constituents
2. Irregular shape
3. They are isotropic like liquids
4. They are considered as pseudo solids (or) super cooled liquids
5. Heat of fusion is not definite
6. Gradually soften over a range of temperature and so can be moulded.
7. Examples: Rubber , plastics, glass etc

Question 4.
Classify the following solids.

1. P₄
2. Brass
3. Diamond
4. NaCI
5. iodine

Answer:

1. P₄ – Molecular solid
2. Brass – Metallic solid
3. Diamond –
4. NaCl – Ionic solid
5. Iodine – Molecular solid

Question 5.
Explain briefly seven types of unit cell.
Answer:
Seven types of unit cell:

1. Cubic – NaCl
2.  Tetragonal – TiO₂
3. Orthorhombic – BaSO₄
4. Hexagonal – ZnO
5. Monoclinic – PbCrO₄
6. Triclinic – H₃BO₃
7. Rhombohedral – Cinnabar Cubic

They differ in the arrangements of their crystallographic axes and angles. Corresponding to the above seven, Bravis defined 14 possible crystal system as shown in the figure.

Question 6.
Distinguish between hexagonal close packing and cubic close packing.
Answer:
Hexagonal close packing

1. aba arrangement
2. In this case, the spheres of the third layer are exactly aligned with those of the first layer
3. In HCP, tetrahedral voids of the second layer may be covered by the spheres of the third layer

Cubic close packing:

1. abc arrangement
2. In this case, the spheres of the third layer are not aligned with those of the first layer or second layer. Only when fourth layer is placed, its spheres are aligned with the first layer
3. In cep third layer may be placed above the second layer in a manner such that its sphere cover the octahedral voids

Question 7.
Distinguish tetrahedral and octahedral voids.
Answer:
Tetrahedral void

1. A single triangular void in a crystal is surrounded by four (4) spheres and is called a tetrahedral void
2. A sphere of second layer is above the void of the first layer, a tetrahedral void is formed
3. This constitutes four spheres, three in the lower and one in upper layer. When the centres of these four spheres are joined a tetrahedron is formed
4. The radius of the sphere which can be accommodated in an octahedral hole without disturbing the structure should not exceed 0.414 times that of the structure forming sphere
5. Radius of an tetrahedral void \(\frac { r }{ R }\) = 0.225

Octahedral void

1. A double triangular void like c is surrounded by six(6) spheres and is called an octahedral void
2. The voids in the first layer are partially covered by the spheres of layer now such a void is called a octahedral void
3. This constitutes six spheres, three in the lower layer and three in the upper layer. When the centers of these six spheres are joined an octahedron is formed
4. The sphere which can be placed in a tetrahedral hole without disturbing the close packed structure should not have a radius larger than 0.225 times the radius of the sphere forming the structure
5. Radius of a octahedral void \(\frac { r }{ R } \) = 0.414

Question 8.
What are point defects?
Answer:
If the deviation occurs due to missing atoms, displaced atoms or extra atoms the imperfection is named as a point defect. Such defects arise due to imperfect packing during the original crystallisation or they may arise from thermal vibrations of atoms at elevated temperatures.

Question 9.
Explain Schottky defect.
Answer:
Schottky defect arises due to the missing of equal number of cations and anions from the crystal lattice. This effect does not change the stoichiometry of the crystal.Ionic solids in which the cation and anion are of almost of similar size show schottky defect.
Example: NaCl.

Presence of large number of schottky defects in a crystal, lowers its density. For example, the theoretical density of vanadium monoxide (VO) calculated using the edge length of the unit cell is 6.5 g cm^-3 but the actual experimental density is 5.6 gcm³. It indicates that there is approximately 14% Schottky defect in VO crystal. Presence of Schottky defect in the crystal provides a simple way by which atoms or ions can move within the crystal lattice.

Question 10.
Write short note on metal excess and metal deficiency defect with an example. Metal excess defect.
Answer:
Metal excess defect arises due to the presence of more number of metal ions as compared to anions.Alkali metal halides NaCl, KCl show this type of defect.The electrical neutrality of the crystal can be maintained by the presence of anionic vacancies equal to the excess metal ions (or) by the presence of extra cation and electron present in interstitial position.

For example, when NaCl crystals are heated in the presence of sodium vapour, Na⁺ ions are formed and are deposited on the surface of the crystal. Chloride ions (Cl^–) diffuse to the surface from the lattice point and combines with Na⁺ ion.

The electron lost by the sodium vapour diffuse into the crystal lattice and occupies the vacancy created by the Cl^– ions. Such anionic vacancies which are occupied by unpaired electrons are called F centers. Hence, the formula of NaCl which contains excess Na⁺ ions can be written as Na_1+xCl.

Metal deficiency defect:
Metal deficiency defect arises due to the presence of less number of cations than the anions. This defect is observed in a crystal in which, the cations have variable oxidation states. For example, in FeO crystal, some of the Fe²⁺ ions are missing from the crystal lattice.

To maintain the electrical neutrality, twice the number of other Fe²⁺ ions in the crystal is oxidized to Fe³⁺ ions. In such cases, overall number of Fe²⁺ and Fe³⁺ ions is less than the O^2- ions. It was experimentally found that the general formula of ferrous oxide is Fe_xO, where x ranges from 0.93 to 0.98.

Question 11.
Calculate the number of atoms in a fee unit cell.
Answer:
Number of atoms in a fee unit cell,
= \(\frac { { N }_{ c } }{ 8 }\) + \(\frac { { N }_{ f } }{ 2 }\)
= \(\frac { 8 }{ 8 } \) + \(\frac { 6 }{ 2 } \) = 1 + 3 = 4

Question 12.
Explain AAAA and ABABA and ABCABC type of three dimensional packing with the help of neat diagram.
Answer:
1. AAAA type of three dimensional packing
This type of three dimensional packing arrangements can be obtained by repeating the AAAA type two dimensional arrangements in three dimensions, i.e., spheres in one layer sitting directly on the top of those in the previous layer so that all layers are identical.

All spheres of different layers of crystal are perfectly aligned horizontally and also vertically, so that any unit cell of such arrangement as simple cubic structure as shown in fig.

In simple cubic packing, each sphere is in contact with 6 neighbouring spheres – Four in its own layer, one above and one below and hence the coordination number of the sphere in simple cubic arrangement is 6.

2. ABABA type of three dimensional packing:
In this arrangement, the spheres in the first layer (A type) are slightly separated and the second layer is formed by arranging the spheres in the depressions between the spheres in layer A as shown in figure.

The third layer is a repeat of the first. This pattern ABABAB is repeated throughout the crystal. In this arrangement, each sphere has a coordination number of 8, four neighbors in the layer above and four in the layer below.

3. ABCABC type of three dimensional packing:
In this arrangement (FCC) second layer spheres are arranged at the dips of first layer. Third layer spheres are arranged in a manner such that it cover the octahedral void. Then no longer third layer is similar to first or second layer.

Third layer gives different arrangement. Fourth layer spheres are similar to first layer. If the first, second and third layer are represented as A,B,C then this type of packing gives the arrangement of layers as ABCABC… (i.e.,), the first three layers do not resemble first, second and third layers respectively and the sequence is repeated.

with the addition of more layers. In this arrangement atoms occupy 74% of the available space and thus has 26% vacant space. The coordination number is 12. Voids – The empty spaces between the three dimensional layers are known as voids. There are two types of common voids possible. They are tetrahedral and octahedral voids.

Tetrahedral void – A void formed by three spheres of a layer in contact with each other and also with a sphere on the top or bottom layer is a hole between four spheres. The spheres are arranged at the vertices of a regular tetrahedron such a hole or void is called tetrahedral void.

Octahedral void:
A hole or void formed by three spheres of a hexagonal layer and another three spheres of the adjacent layer is a hole between six spheres. The spheres are arranged at the vertices of a regular octahedron. Such a hole or void is abc arrangement – ccp structure called octahedral void.

Question 13.
Why ionic crystals are hard and brittle?
Answer:
The ionic compounds are very hard and brittle. In ionic compounds the ions are rigidly held in a lattice because the positive and negative ions are strongly attracted to each other and difficult to separate. But the brittleness of a compound is now easy to shift the position of atoms or ions in a lattice.

If we apply a pressure on the ionic compounds the layers shifts slightly. The same charged ions in the lattice comes closer. A repulsive forces arises between ‘ same charged ions, due to this repulsions the lattice structure breaks down chemical bonding.

Question 14.
Calculate the percentage efficiency of packing in case of body centered cubic crystal. Packing efficiency.
Answer:
In body centered cubic arrangement the spheres are touching along the leading diagonal of the cube as shown in the In ∆ABC,

AC² = AB² + BC²
AC = \(g\sqrt { { AB }^{ 2 }+{ BC }^{ 2 } }\)
AC = \(g\sqrt { { a }^{ 2 }+{ a }^{ 2 } }\) = a²
In ∆ACG,
AG² = AC² + CG²
AG =\(g\sqrt { { AC }^{ 2 }+{ CG }^{ 2 } }\)
AG = \(\sqrt { { \left( \sqrt { 2 } a \right) }^{ 2 }+{ a }^{ 2 } }\)
AG = \(g\sqrt { { 2a }^{ 2 }+a^{ 2 } }\) = \(g\sqrt { 3a^{ 2 } } \) = \(\sqrt { 3 } a\)
i.e., \(g\sqrt { 3 } \)a = 4r
r = \(\frac { \sqrt { 3 } }{ 4 } a\)
∴ Volume of the sphere with radius ‘r’
= \(\frac { 4 }{ 3 } \) πr³
= \(\frac { 4 }{ 3 } \) π \(\left( \frac { \sqrt { 3 } }{ 4 } a \right)\)
= \(\frac { \sqrt { 3 } }{ 6 }\) πa³
Number of spheres belong to a unit cell in bee arrangement is equal to two and hence the total volume of all spheres

i.e., 68% of the available volume is occupied.The available space is used more efficiently than in simple cubic packing

Question 15.
What is the two dimensional coordination number of a molecule in square close packed layer?
Answer:
Square close packing – When the spheres of the second row are placed exactly above those of the first row. This way the spheres are aligned horizontally as well as vertically. The arrangement is AAA type. Coordination number is 4.

Question 16.
Experiment shows that Nickel oxide has the formula Ni_0.96 O_1.00. What fraction of Nickel exists as of Ni²⁺ and Ni³⁺ ions ?
Answer:
Let the number of Ni²⁺. Then the number of Ni³⁺ ion will be = (0.96 – x). Total number of cation, = 2 x + 3 (0.96 – x)
= 2x + 2.88 – 3x
= ( – x) + 2.88
Number of anions O2- ( – 2) x 1 = -2. Number of cations = Number of anions
– x + 2.88 2
– x = – 2.88 + 2
x =0.88
%of Ni as Ni² =\(\frac { 0.08 }{ 0.96 } \times 100\) = 91.66%
Number of Ni³⁺ ion will be = 0.96 – x
= 0.96 – 0.88 = 0.08
% of Ni as Ni³⁺ = \(\frac { 0.08 }{ 0.96 } \times 100\) = 8.3 %

Question 17.
What is meant by the term “coordination number”? What is the coordination number of atoms in a bcc structure?
Answer:

1. Coordination number – The number of nearest neighbours that surrounding a particle in a crystal is called the coordination number of that particle.
2. Coordination number of atoms in a bcc structure is 8

Question 18.
An element has bcc structure with a cell edge of 288 pm. the density of the element is 7.2 gcm^-3. How many atoms are present ¡n 208g of the element.
Answer:
An elemeñt has bec structure with a cell edge of 288 pm. The density of the element is 7.2 gcm^-3. For the Bec structure, n = 2

By mole concept, 51.42 g of the element contains 6.023 x 10²³ atom 208 g of the element will contain
= \(\frac { 6.023\times { 10 }^{ 23 }\times 208 }{ 51.42 }\) atoms
= 24.17 x 10²³ atoms (or) 2.417 x 10²⁴ atoms

Question 19.
Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125pm. Calculate the edge length of unit cell.
Answer:
Given, Radius (r) = 125 pm
Edge length of unit cell (a) = ?
Since aluminium crystallizes in Face centered cubic
r= \(\frac { a\sqrt { 2 } }{ 4 } \) (or) r=\(\frac { a }{ 2\sqrt { 2 } } \)
a = r x 2 x \(g\sqrt { 2 } \)
= 125 x 2 x 1.414
a = 353.5 pm

Question 20.
If NaCI is doped with 10² mol percentage of strontium chloride, what is the concentration of cation vacancy?
Answer:
We know that two Na⁺ ions are replaced by each of the Sr²⁺ ions while SrCl², is doped with NaCI. But in this case, only one lattice point is occupied by each of the Sr²⁺ ions and produce one cation vacancy.

Here 10^-2 mole of SrCI₂, is doped with 100 moIes of NaCI. Thus, cation vacancies produced by NaCi = 10^-2 mol. Since, 100 moles of NaCl produces cation vacancies after doping = 10^-2 mol. Therefore, I mole of NaCl will produce cation vacancies after doping

= \(\frac { { 10 }^{ -2 } }{ 100 }\)= 10^-4mol
:. Total cationic vacancies,
= 10^-4 x Avogadro’s number
= 10 x 6.023 x 10²³
= 6.023 x 10¹⁹ vacancies

Question 21.
KF crystallizes in fcc structure like sodium chloride, calculate the distance between K⁺ and F^– in KF. (given : density of KF is 2.48 g cm^-3)
Answer:
Density of KF 2.48 g cm^-3

Question 22.
An atom crystallizes ¡n fcc crystal lattice and has a density of 10 gcm3 with unit cell edge length of 100pm. calculate the number of atoms present in 1 g of crystal.
Answer:
Given, Density = 10 g cm^-3
mass = 1 g
Edge length of unit cell = 100 pm

= 0.1 cm³
Volume of unit cell = a³
= (100 x 10^-10cm)³
= 1 x 10^-24 cm³
Number of unit cell in 1 g of crystal,

The given unit cell is of FCC type. Therefore. it contains 4 atoms.
0.1 x 10²⁴ unit cells will contain 4 x 0.1 x 10²⁴ = 4 x 10²³atoms

Question 23.
Atoms X and Y form bcc crystalline structure. Atom X is present at the corners of the cube and Y is at the centre of the cube. What is the formula of the compound?
Answer:
Atoms X and Y form bcc crystalline structure. Atom X is present at the corners of the cube Atom Y is present at the centre of the cube.
No of atoms of X in the unit cell = \(\frac { { N }_{ c } }{ 8 }\) = \(\frac { 8 }{ 8 } \) = 1
No of atoms of Y in the unit cell = \(\frac { { N }_{ b } }{ 1 } \) = \(\frac { 1 }{ 1 } \) = 1
Ratio of atoms X : Y = 1 : 1
Hence formula of the compound = XY.

Question 24.
Sodium metal crystallizes in bcc structure with the edge length of the unit cell 4.3 x 10^-8 cm. Calculate the radius of sodium atom.
Answer:
Edge length of the unit cell (a) = 4.3 x 10^-8cm
Radius of sodium atom (r) = ?
For bcc structure, r = \(\frac { \sqrt { 3 } }{ 4 } a\)
= \(\frac { \sqrt { 3 } }{ 4 } a\) (4.3 x 10^-8cm)
\(\frac { 1.732\times 4.3\times { 10 }^{ -8 } }{ 4 }\)
= \(\frac { 1.732 }{ 4 } \times { 10 }^{ -8 }cm\)
= 1.86 x 10^-8 cm = l.86Å

Question 25.
Write a note on Frenkel defect.
Answer:
Frenkel defect arises due to the dislocation of ions from its crystal lattice. The ion which is missing from the lattice point occupies an interstitial position. This defect is shown by ionic solids in which cation and anion differ in size.

Unlike Schottky defect, this defect does not affect the density of the crystal. For example AgBr, in this casc, small Ag ion leaves its normal site and occupies an interstitial position as shown in the figure.

Samacheer Kalvi 12th Chemistry Solid State Evaluate Yourself

Question 1.
An element has a face centered cubic unit cell with a length of 352.4 pm along an edge. The density of the element is 8.9 gcm^-3. How many atoms are present in loo g of an element?
Answer:
Mass = 100g
Density = 8.9 g cm^-3
Edge length = 352.4 pm
(a) 352.4 x 10^-10cm
Volume of the unit cell,
a³ = (352.4 x 10¹⁰cm) 34.37 x 10²³ cm³
Volume of 100 g of an element.

Therefore number of unit cells,
\(\frac { 11.23 }{ 4.37\times { 10 }^{ -23 } }\) = 11.23 = 2.56 x 10²³
Since each Fcc cube contains 4 atoms, therefore total number of atoms in 100 g.
= 4 x (2.56 x 10²³) = 10.24 x 10²³ atoms

Question 2.
Determine the density of CsCl which crystallizes in a bcc type structure with an edge length 412.1 pm.
Answer:
Molar mass of cscl = 168.5 g /mol
Number atoms present in per unit cell for bcc (cscl)
n = 1
Edge length (a) = 412.1 pm
Density (ρ) = ?
ρ = \(\frac { nM }{ { a }^{ 3 }{ N }_{ A } }\)

Question 3.
A face centered cubic solid of an element (atomic mass 60) has a cube edge of 4 Å. Calculate its density.
For FCC unit cell n = 4
Edge length (a) = 4Å = 4 x 10^-8cm
Mass (M) = 60 g mol^-1
Density (p) = ?
ρ = \(\frac { nM }{ { a }^{ 3 }{ N }_{ A } }\)

Samacheer Kalvi 12th Chemistry Solid State Additional Questions

Samacheer Kalvi 12th Chemistry Solid State 1 Mark Questions and Answers

I. Choose the best answer.

Question 1.
Consider the following statements
(i) Solids have definite volume and shape
(ii) Solids are rigid and compressible
(iii) Solids have weak cohesive forces

Which of the above statemetns is / are not correct?
(a) (i) only
(b) (ii) & (iii) only
(c) (iii) only
(d) (i) & (ii) only
Answer:
(b) (ii) & (iii) only

Question 2.
Which one of the following is an ionic crystal?
(a) Glass
(b) Rubber
(c) NaCl
(d) SiO₂
Answer:
(c) NaCl

Question 3.
Which one of the following is an amorphous solid?
(a) Glass
(b) SiO₂
(c) NaCl
(d) Na
Answer:
(a) Glass

Question 4.
Which one of the following is an example for molecular crystals?
(a) Diamond
(b) Silica
(c) Glass
(d) Naphthalene
Answer:
(d) Naphthalene

Question 5.
Which one of the following is an example for atomic solids?
(a) Frozen elements of group 18
(b) Group 17 elements
(c) Inner – transition elements
(d) chalcogens
Answer:
(a) Frozen elements of group 18

Question 6.
Which one of the following is a covalent crystal?
(a) Glass
(b) Diamond
(c) Anthracene
(d) Glucose
Answer:
(b) Diamond

Questions 7.
Consider the following statements.
(i) Crystalline solids have irregular shape
(ii) Generally crystalline solids are anisotropic in nature
(iii) Heat of fusion of crystalline solids are not definite

Which of the above statements is / are correct?
(a) (i) & (iii)
(b) (i) only
(c) (iii) only
(d) (ii) only
Answer:
(d) (ii) only

Question 8.
Consider the following statements.
(i) Amorphous solids are isotropic like liquids
(ii) Amorphous solids are considered as pseudo solids
(iii) Amorphous solids have sharp melting points

Which of the above statemetns is / are correct?
(a) (i) only
(b) (ii) only
(c) (i) & (ii)
(d) (i) & (iii)
Answer:
(c) (i) & (ii)

Question 9.
In an ionic crystal, both cations and anions are bound together by ………..
(a) Strong electrostatic attractive forces
(b) Weak electrostatic attractive forces
(c) Vanderwaals forces of attraction
(d) Weak cohesive forces
Answer:
(a) Strong electrostatic attractive forces

Question 10.
Molecular solids contains neutral molecules held together by ………..
(a) strong cohesive forces
(b) weak vanderwaals forces
(c) weak ionic forces
(d) strong electrostatic forces
Answer:
(b) weak vanderwaals forces

Question 11.
Which is used inside pencils and in many lubricants?
(a) Lead nitrate
(b) charcoal
(c) graphite
(d) coke
Answer:
(c) graphite

Question 12.
In non polar molecular solids, molecules are held together by ………..
(a) London forces
(b) weak vanderwaals forces
(c) Strong electrostatic forces
(d) strong cohesive forces
Answer:
(a) London forces

Question 13.
Which one of the following is non-polar molecular solids?
(a) Diamond
(b) SiC
(c) Anthracene
(d) Glass
Answer:
(c) Anthracene

Question 14.
Silicon carbide is an example of ………..
(a) Ionic solid
(b) Covalent solid
(c) Polar molecular solid
(d) Non – polar molecular solid
Answer:
(b) Covalent solid

Question 15.
Naphthalene is an example of ………..
(a) ionic solid
(b) covalent solid
(c) non polar molecular solid
(d) polar molecular solid
Answer:
(c) non polar molecular solid

Question 16.
Solid NH₃ solid CO₂ are examples of ………..
(a) Covalent solids
(b) polar molecular solids
(c) molecular solids
(d) ionic solids
Answer:
(b) polar molecular solids

Question 17.
Solids ice, glucose are examples of ………..
(a) metallic solids
(b) ionic solids
(c) hydrogen bonded molecular solids
(d) non polar molecular solids
Answer:
(c) hydrogen bonded molecular solids

Question 18.
Consider the following statements.
(i) metallic solids possess high electrical and thermal conductivity
(ii) solid ice are soft solids under room temperature
(iii) In non polar molecular solids constituent molecules are held together by strong electrostatic forces of attraction

Which of the above statements is / are not correct?
(a) (i) & (ii)only
(b) (iii) only
(c) (ii) only
(d) (i) only
Answer:
(b) (iii) only

Question 19.
Each atom in the comer of the cubic unit cell is shared by how many unit cells?
(a) 8
(b) 6
(c) 1
(d) 12
Answer:
(a) 8

Question 20.
Which is the coordination number of each atom in a simple cubic unit cell?
(a) 8
(b) 6
(c) 12
(d) 4
Answer:
(b) 6

Question 21.
The number of atoms belongs to fcc unit cell is ………..
(a) 2
(b) 4
(c) 6
(d) 12
Answer:
(a) 2

Question 22.
The number of atoms in fee unit cell is ………..
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(b) 4

Question 23.
The atoms the face centre is being shared by ………..
(a) 4
(b) 8
(c) 2
(d) 6
Answer:
(c) 2

Question 24.
An atom present at the body centre be longs to only unit cell ………..
(a) 1
(b) 2
(c) 4
(d) 8
Answer:
(a) 1

Question 25.
Which one of the following is known as Bragg’s equation’?
(a) d = \(\frac { 2sinθ }{ nλ } \)
(b) d = \(\frac { nλ }{ 2sinθ } \)
(c) d = \(\frac { d }{ sinθ } \)
(d) d = \(\frac { 2sinθ }{ nλ } \)
Answer:
(b) d = \(\frac { nλ }{ 2sinθ } \)

Question 26.
Which one of the following formula is used to calculate the density of the unit cell ?
(a) ρ = \(\frac { nM }{ { a }^{ 3 }{ N }_{ A } } \)
(b) ρ = \(\frac { { a }^{ 3 }{ N }_{ A } }{ nM } \)
(c) ρ = \(\frac { { N }_{ A } }{ { a }^{ 3 }{ N }M } \)
(d) ρ = \(\frac { { a }^{ 3 }{ N }_{ A } }{ n } \)
Answer:
(a) ρ = \(\frac { nM }{ { a }^{ 3 }{ N }_{ A } } \)

Question 27.
Which is the packing fraction in simple cubic unit cell?
(a) 52.31%
(b) 100%
(c) 68%
(d) 75%
Answer:
(a) 52.3 1%

Question 28.
The packing fraction in bcc arrangement is ………..
(a) 52.3 1%
(b) 68%
(c) 100%
(d) 80%
Answer:
(b) 68%

Question 29.
Which is the coordination number in both hep and ccp arrangements?
(a) 12
(b) 6
(c) 4
(d) 8
Answer:
(a) 12

Question 30.
What is the coordination number of B₂O₃?
(a) 4
(b) 6
(c) 8
(d) 3
Answer:
(d) 3

Question 31.
Which one of the following is the structure of B₂O₃ ?
(a) Tetra hedral
(b) Octahedral
(c) Trigonal planar
(d) Cubic
Answer:
(c) Trigonal planar

Question 32.
The coordination number of zinc sulphide is ………..
(a) 3
(b) 4
(c) 6
(d) 8
Answer:
(b) 4

Question 33.
The coordination number of CSCI ¡s ………..
(a) 3
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

Question 34.
Which one of the following is the coordination number of NaCl?
(a) 3
(b) 4
(c) 6
(d) 8
Answer:
(c) 6

Question 35.
Which one of the following is the packing efficiency in fcc unit cell?
(a) 74%
(b) 52.6 1%
(c) 100%
(d) 68%
Answer:
(a) 74%

Question 36.
Which one of the following is an example for schoriky defect?
(a) NaCI
(b) AgBr
(c) KCI
(d) FeS
Answer:
(a) NaCl

Question 37.
Which one of the following is an example for Frenkel defect?
(a) NaCl
(b) AgCI
(c) AgBr
(d) AgNO₃
Answer:
(c) AgBr

Question 38.
Metal excess defect is possible in ………..
(a) AgCI
(b) AgBr
(c) KCl
(d) Fes
Answer:
(c) KCl

Question 39.
Which one of the following is the metal deficiency defect?
(a) FeO
(b) ZnO
(c) KCl
(d) NaCl
Answer:
(a) FeO

Question 40.
Which one of the following shows non- stoichiometric defect?
(a) FeO
(b) AgBr
(c) ZnO
(d) Both a and c
Answer:
(d) Both a and c

II. Fill in the blanks

1. Naphthalene,Anthracene and glucose are examples of ……………
2. The best examples of covalent crystals are …………… and ……………
3. Frozen elements of group 18 are called ……………
4. Glass, Rubber, plastics are the examples of …………… solids.
5. …………… means uniformity in all directions.
6. Crystalline solids are …………… and they show different values of physical properties when measured along different directions.
7. NaCl and KCl are the examples of …………… crystals.
8. Diamond and silicon carbide are the examples of …………… solids.
9. In molecular solids, the neutral molecules are held together by weak ……………
10. …………… is a component of many lubricants for example cycle chain oil.
11. In non polar molecular solids constituent molecules are held together by ……………
12. In solids CO₂ solid NH₃ the molecules are held together by strong ……………
13. Glucose and urea are generally …………… under room temperature.
14. …………… solids possess excellent electrical and thermal conductivity.
15. The regular arrangement of the ions throughout the crystal is called a ……………
16. The basic repeating structural unit of a crystalline solid is called a ……………
17. The number of the nearest neighbours that surrounding a particle in a crystal is called the ……………
18. A unit cell that contains only one lattice point is called a ……………
19. There are …………… primitive crystal systems.
20. The coordination number of bcc is ……………
21. Each atom in the comer of cubic unit cell is shared by …………… neighbouring unit cells.
22. The number of atoms in bcc unit cell is ……………
23. The number of atoms in a fee unit cell is ……………
24. …………… is the most powerful tool for the determinaiton of crystal structure.
25. Only …………… of the available volume is occupied by the spheres in simple cubic packing
26. Of all the metals in the periodic table, only …………… crystallizes in simple cubic pattern.
27. In bcc cubic pattern …………… of the available volume is occupied.
28. If the third layer arrangement is aba arrangement, it is called …………… arrangement.
29. If third layer arrangement is abc arrangment, it is known as …………… arrangement.
30. In both …………… and …………… arrangements, the coordination number of each sphere is 12.
31. The packing efficiency of fee unit cell is ……………
32. …………… defect arises due to the missing of equal number of cations and anions from the crystal lattice.
33. Presence of large number of schottky defects in a crystal, lowers its ……………
34. Vanadium monoxide shows …………… defect.
35. …………… arises due to the dislocation of ions from its crystal lattice.
36. …………… arises due to the presence of more number of metalions as compared to anions.
37. Zno is …………… at room temperature but when it is heated it becomes in colour.
38. …………… arises due to the presence of less number of cations than the anions.
39. …………… is the appearance of an electrical potential across the sides of the crystal, when it is subjected to mechanical stress.
40. Stoichiometric defects in an ionic solid is also called …………… or …………… defect.

Answer:

1. Molecular crystals
2. Diamond, SiO₂
3. Atomic solids
4. Amorphous
5. Isotropy
6. Anisotropic
7. Ioni
8. Covalent
9. Vander waals forces
10. Graphite
11. London forces
12. dipole-dipole interactions
13. Soft solids
14. Metallic
15. Crystal lattice
16. Unit cell
17. Coordination number
18. Primitive unit cell
19. Seven
20. Eight
21. Eight
22. Two
23. Four
24. X-ray diffraction analysis
25. 52.31%
26. Polonium
27. 68%
28. hep or hexagonal close packed
29. ccp or cubic close packed
30. hep, ccp
31. 74%
32. Schottky
33. density
34. Schottky defect
35. Frenkel defect
36. metal excess defect
37. Colourless, yellow
38. Metal deficiency
39. Piezoelectricity
40. intrinsic or thermodynamic

III. Match the following:

Question 1.

Answer:
(a) 3, 4, 1, 2

Question 2.

Answer:
Answer:
(b) 3, 1, 4, 2

Question 3.

Answer:
(a) 2, 4, 1, 3

Question 4.

Answer:
(b) 3, 1, 4, 2

Question 5.

Answer:
(c) 4, 1, 2, 3

IV. Assertion and Reason

Question 1.
Assertion (A): Amorphous solids are isotropic in nature.
Reason (R): In amorphous solids, they have identical values of physical properties such as refractive index, electrical conductance in all directions which is called isotropy.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a)Both A and R are correct and R is the correct explanation of A.

Question 2.
Assertion (A): Crystalline solids are anisotropic in nature.
Reason (R): Anisotropy is the property which depends on the direction of measurement. Crystalline solids are anisotropic and they show different values of physical properties when measured along different directions.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a)Both A and R are correct and R is the correct explanation of A.

Question 3.
Assertion (A): Ionic solids do not conduct electricity in solid state but in molten state they conduct electricity.
Reason (R): In solid state, the ions are fixed in their lattice positions but in molten state, the ions are free to move and conduct electricity.
(a) Both A and R are correct but R is not correct explanation of A.
(b) Both A and R are correct and R is the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(A)Both A and R are correct and R is the correct explanation of A

Question 4.
Assertion (A): Diamond and Silicon carbide are very hard and have high melting point.
Reason (R): In covalent solids, the atoms are bound together in a three dimensional network entirely by covalent bonds.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a)Both A and R are correct and R is the correct explanation of A.

Question 5.
Assertion (A): Solid CO₂ Solid NH₃ have higher melting points.
Reason (R): The constituents are molecules formed by polar covalent bonds. They are held together by relatively strong dipole- dipole interactions.
(a) A is correct but R is wrong.
(b) A is wrong but R is correct
(c) A and R are correct and R is the correct explanation of A
(d) A and R are correct but R is not the correct explanation of A
Answer:
(c) A and R are correct and R is the correct explanation of A

Question 6.
Assertion (A): Solid ice, Glucose are generally soft solids under room temperature.
Reason (R): The constituents are held together by strong electrostatic forces of attraction.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct and R is not the correct explanation of A
(c) A is correct but R is wrong.
(d) A is wrong but R is correct
Answer:
(c) A is correct but R is wrong.

Question 7.
Assertion (A): In bcc, the available volume is more efficiently used than in simple cubic packing.
Reason (R): In simple cubic arrangement, the number of spheres belongs to a unit cell is equal to one whereas in bcc, it is equal to 2.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct R is not the correct explanation of A
(c) A is correct but R is wrong.
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 8.
Assertion (A): B₂ O₃ has trigonal planar structure.
Reason (R): The ratio of radius of cation and anion \(\frac { { r }_{ { c }^{ + } } }{ { r }_{ { A }^{ – } } } \) = 0.155 – 0.225 plays an important role in determining the structure . and B₂ O₃ has coordination number as 3 and has trigonal planar structure.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong.
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 9.
Assertion (A): Schottky defect does not change the stoichiometry of the crystal.
Reason (R): This defect arises due to the missing of equal number of cations and anions from the crystal lattice.
(a) Both A and R are correct but R is not the correct explanation of A
(b) Both A and R are correct and R is the correct explanation of A
(c) A is correct but R is wrong.
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

Question 10.
Assertion (A): Zinc oxide is colourless at room temperature but on heating it becomes yellow in colour.
Reason (R): On heating Zinc loses oxygen and thereby forming free zn²⁺ ions. The excess zn²⁺ ions move to interstitial sites and the electrons also occupy interstitial positions.
(a) Both A and R are correct and R is the correct explanation of A
(b) Both A and R are correct R is not the correct explanation of A
(c) A is correct but R is wrong.
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A

V. Find the odd one out

Question 1.
(a) Plastic
(b) Rubber
(c) Glucose
(d) Glass
Answer:
(c) Glucose It is a molecular solid where as others are amorphous solids.

Question 2.
(a) Anthracene
(b) Naphthalene
(c) Glucose
(d) Sodium chloride
Answer:
(d) Sodium chloride It is ionic crystall where as others are molecular crystals.

Question 3.
(a) Sodium
(b) Pottasium
(c) Frozen elements of group 18
(d) Gold
Answer:
(c) Frozen elements of group 18 It is atomic solid where as others are metallic solids.

Question 4.
(a) Solid CO₂
(b) Solid ice
(c) Glucose
(d) Urea
Answer:
(a) Solid CO₂ It is a polar molecular solid where as others are hydrogen bonded molecular solids.

Question 5
(a) Cubic
(b) Rhombohedral
(c) Hexagonal
(d) Cyclic
Answer:
(d) Cyclic It is a ring structure where as other primitive crystal systems.

VI. Find the odd correct pair

Question 1.
(a) Glass, plastic
(b) Rubber, ice
(c) Nacl, Glucose
(d) Urea, solid NH₃
Answer:
(a) Glass, plastic (Amorphous solids)

Question 2.
(a) NaCI, KCI
(b) FeO, ZnO
(e) AgBr, AgNO₃
(d) VO, ZnO
Answer:
(a) Naci, Kcl (schottky defect)

Question 3.
(a) Solid CO₂ Solid ice
(b) Solid CO₂ Solid NH₃
(c) Graphite, Silicon carbide
(d) Naphthalene, Phenol
Answer:
(b) Solid CO₂ Solid NH₃ Polar molecular solids

Question 4.
(a) NaCI, SiC
(b) Naphthalene, anthracene
(c) Solid ice, graphite
(d) Copper, KO
Answer:
(b) Naphthalcne, anthracenc – It is non-polar molecular solid examples.

VII. Find out the odd incorrect pair

Question 1.
(a) NaCl, KCl
(b) Naphthalenc, anthracene
(c) Solid CO₂, Solid NH₃
(d) Diamond, solid ice
Answer:
(d)Diamond, solid ice

Question 2.
(a) Cu, Fc
(b) Glucose, Urca
(c) Diamond, SiC
(d) Benzene, glucose
Answer:
(d) Benzene, glucose

Samacheer Kalvi 12th Chemistry Solid State 2 Mark Question and Answers

Question 1.
What are crystalline solid? Give example.
Answer:
A crystalline solid is one in which its constituents (atoms, ions or molecules) have an orderly arrangement extending over a long range which has three dimensional pattern. Example – NaCl.

Question 2.
What are amorphous solid? Give example.
Answer:
In an amorphous solid, the constituents are randomly arranged. It is a short range arrangement of constituents. Eg – Glass.

Question 3.
What are covalent solids? Give example.
Answer:
In covalent solids, the constituents (atoms) are bound together in a three dimensional network entirely by covalent bonds. Examples. Diamond, Silicon carbide.

Question 4.
Silicon carbide is very hard. Justify this statement.
Answer:
Silicon carbide is very hard. It is a covalent solid contains the atoms which are bound together in a three dimensional network entirely by covalent bonds. So the covalent network crystal SiC is very hard and have high melting point.

Question 5.
Write a note about molecular solids.
Answer:

1. In molecular solids, the constituents are neutral molecules. They are held together by weak vander waals forces.
2. Molecular solids are soft and they do not conduct electricity. Eg – Solid CO₂

Question 6.
What are non-polar molecular solids? Give example.
Answer:

1. In non polar molecular solids, constituent molecules are held together by weak dispersion forces or London forces.
2. They have low melting points and are usually in liquids or gaseous state at room temperature. Eg., Naphthalene, anthrancene.

Question 7.
What are hydrogen bonded molecular solids? Give example.
Answer:

1. Molecular solids in which the constituents are held together by hydrogen bonds.
2. They are generally soft solids under room temperature. Examples., Solid ice, Glucose, Urea.

Question 8.
Define crystal lattice.
Answer:
The regular arrangement of these species throughout the crystal is called a crystal lattice.

Question 9.
Define coordination number.
Answer:

1. A crystal may be considered to consist of large number of unit cells, each one in direct contact with its nearer neighbour and all similarly oriented in space.
2. The number of the nearest neighbours that surrounding a particle in a crystal is called coordination number of that particle.

Question 10.
Draw the tetragonal crystal systems.
Answer:

Question 11.
Draw the hexagonal primitive crystal structure.
Answer:

Question 12.
Draw the typec of monoclinic primitive cubic crystals.
Answer:

Question 13.
Draw the primitive cubes.

1. trigonal
2. triclinic

Answer:

Question 14.
What is Bragg’s equation?
Answer:
1. X-ray diffraction analysis is the most powerful tool fiw the determination of crystal structure.

2. The interplanar distance (d) between two successive planes of atoms can be calcuLated using the following equation form the X-ray diffraction data 2d sin θ = nλ. The equation is known as Bragg’s equation.

Where λ = wavelength of X-ray – d = Interplanar distance, θ The angle of diffraction – n = order of reflection. By knowing the values of θ, λ and n. we can calculate the value of d. d = \(\frac { nλ }{ 2Sinθ } \) Using these values, the edge of the unit cell can be calculated.

Question 15.
What is meant by linear arrangement of spheres in one direction?
Answer:
1. In a specific direction, there is only one possibility to arrange the spheres in one direction.

2. In this arrangement, each sphere is in contact with two neighbouring spheres on either.

Question 16.
What is meant by piezo electricity?
Answer:
Piezo electricity is the appearance of an electrical potential across the sides of a crystal. When you subject it to mechanical stress. The word piezo electricity means electricity resulting from pressure and latent heat. Even the inverse is possible which is known as inverse piezo electric effect.

Question 17.
Why are solids rigid?
Answer:
In a solid, the constituent particles are very closely – packed. Hence, the forces of attraction among these particles are very strong, that is why they are rigid.

Question 18.
Why do solids have a definite volume?
Answer:
The constituent particles of a solid have fixed positions and are not free to move i.e., they possess rigidity. That is why they have a definite volume.

Question 19.
Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.
Answer:
Amorphous solids: Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass. Crystalline solids: Benzoic acid, naphthalene, potassium nitrate, copper.

Question 20.
Why is glass considered as super cooled liquid?
Answer:
Glass is an amorphous solid. Like liquids it has a tendency to flow, though very slowly. The proof of this fact is that glass panes in the windows or doors of old buildings are invariably found to be slightly thicker at the bottom than at the top.

Question 21.
Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?
Answer:
As the solid has same value of refractive index along all directions, this means that it is isotropic and hence amorphous. Being an amorphous solid, it would not show a clean cleavage when cut with a knife. Instead it would break into pieces with irregular surfaces.

Question 22.
Classify the following solids in different categories based on the nature of intermolecular forces operating in them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.
Answer:

1. Potassium sulphate – Ionic
2. Tin – Metallic
3. Benzene – Molecular (non polar)
4. Urea – Molecular (polar)
5. Ammonia – Molecular (hydrogen bonded)
6. Water – Molecular (hydrogen bonded)
7. Zinc Suiphide – Ionic
8. Graphite – Covalent or Network
9. Rubidium – Metallic
10. Argon – Molecular (non polar)
11. Silicon carbide – Covalent or Network

Question 23.
Solid A is a very hard electrical insulator in solid as well as in molten state, and melts at extremely high temperature. What type of solid is it?
Answer:
Covalent or Network solid like SiO₂ (quartz) or SIC or C (diamond).

Question 24.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer:
In the molten state, ionic solids dissociate to give free ions and hence can conduct electricity. However in the solid state, as the ions are not free to move but remain held together by strong electrostatic forces of attraction, so they cannot conduct electricity in the solid state.

Question 25.
An element with molar mass 2.7 x 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If the density is 2.7 x 10³ kg m^-3, what ¡s the nature of the cubic unit cell?
Answer:
Density

Thus, there are 4 atoms of elements present per unit cell, hence, the cubic unit cell must be facecentred or cubic close – packed (cep).

Question 26.
Ionic solids, which have anionic vacancies due to metal excess defect develop colour. Explain with the help of suitable example.
Answer:
Taking the example of NaCl, when its crystals arc heated in presence of sodium vapour some chloride ion leave their lattice sites to combine with sodium to form NaCI. For this released diffuses into the crystal to occupy the anion vacancies created by Cl ions.

The crystal now has excess of sodium. The sites occupied by unpaired electrons are called F – centres. They impart yellow colour to the crystal because they absorb energy from the visible light and get excited.

Question 27.
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

1. Tetra phosphorous decoxide (P₄ O₁₀)
2. Ammonium Phosphate (NH₄)₃PO₄
3. SIC
4. I₂
5. P₄
6. Plastic
7. Graphite
8. Brass
9. Rb
10. LiBr
11. Si

Answer:

1. Tetra phosphorous decoxide (P₄ O₁₀) = Molecular solid
2. Ammonium Phosphate (NH₄)₃PO₄ = Ionic solid
3. SiC = Covalent solid
4. I₂ = Molecular solid
5. P₄ = Molecular solid
6. Plastic Amorphous solid
7. Graphite = Covalent solid
8. Brass = Metallic solid
9. Rb = Metallic solid
10. LiBr = Ionic solid
11. Si = Covalent solid

Question 28.
What ¡s the formula of a compound in which the element Y forms ccp Lattice and atoms of X occupy 2/3rd of tetrahedral voids?
Answer:
Number of element Y = n, Number of element X = 2 n x \(\frac { 2 }{ 3 } \). As number of tetrahedral void = 2n
X : Y= \(\frac { 4n }{ 3 } \) : n
Formula = X₄ Y₃

Question 29.
The energy required to vapourise one mole of copper is smaller than that of energy required to vapourise I mol of diamond. Why?
Answer:
Copper is a metallic solid having metal-metal bonds while diamond is a covalent solid having covalent bonds.Metallic bonds are weaker than covalent bonds and thus less amount of energy is required to break metallic bonds than covalent bonds.

Question 30.
Assign reason for the following:

1. phosphorous doped silicon is a semiconductor.
2. Schottky defect lowers the density of a solid.

Answer:

1. It is because its conductance is intermediate between conductor and insulator.
2. In Schottky defect, both cations and anions are missing which leads to lowering the density of a solid.

Samacheer Kalvi 12th Chemistry Solid State 3 Mark Questions and Answers

Question 1.
Distinguish between isotropy and anisotropy?
Answer:
Isotropy:

1. Isotropy means uniformity in all directions.
2. Isotropy means having identical values of physical properties such as refractive index, electrical conductance in all directions.
3. Isotropy is the property of amorphous solids.

Anisotropy:

1. Anisotropy means non-uniformity in all directions.
2. Anisotropy is the property which depends on the direction of measurement. They show different values of physical properties when measured along different directions.
3. Anisotropy is the properly of crystalline solids.

Question 2.
What are polar molecular solids? Give example.
Answer:

1. The constituents are molecules formed by polar covalent bonds.
2. They are held together by relatively strong dipole-dipole interactions.
3. They have higher melting points than the non-polar molecular solids. Eg., Solid CO₂, Solid NH₃.

Question 3.
Write a note about metallic solids.
Answer:

1. In metallic solids, the lattice points are occupied by positive metal ions and a cloud of electrons pervades the space.
2. They are hard and have high melting point.
3. Metallic solids possess excellent electrical and thermal conductivity. They possess bright lustre.
4. Examples – Metals and metal alloys Cu, Fe, Zn, Ag, Ay, Cu – Zn etc.

Question 4.
What are characteristic parameters of a unit cell?
Answer:
1. A basic repeating structural unit of a crystalline solid is called a unit cell.

2. A unit cell is characterised by the three edge lengths or lattice constants a,b and c and the angle between the edges a,P and y.

3.

Question 5.
What are type of unit cells? Give their names.
Answer:

1. There are two types of unit cells, a) primitive b) Non – primitive.
2. A unit cell that contains only one lattice point is called a primitive unit cell, which is made up from the lattice points at each of the comers.
3. In the case of non-primitive unit cells, there are additional lattice points, either on a face of the unit cell or with in the unit cell.

Question 6.
Draw the seven types of primitive crystal systems.
Answer:

Question 7.
Draw the types of cubic crystal systems.
Answer:

Question 8.
Draw the types of orthorhombic cubic crystal systems.
Answer:

Question 9.
Calculate the number of atoms belong to one unit cell of simple cubic unit cell(sc).
Answer:
1. In the simple cubic unit cell, each corner is occupied by an identical atoms (or) ions (or) molecules and they touch along the edges of cube,do not touch diagonally. The coordination number of each atom is 6.

2. Each atom in the comer of the cubic unit cell is shared by 8 neighbouring unit cells and therefore atoms per unit cell is equal to \(\frac { { { N }_{ c } } }{ 8 }\). where Nc is the number of atoms at the corners.

3. no of atoms in a Sc unit cell = \(\left( \frac { { N }_{ c } }{ 8 } \right)\) = \(\left( \frac { 8 }{ 8 } \right) \) = 1

Question 10.
Calculate the number of atoms per unit cell of bec type.
Answer:
1. In a body centered cubic unit cell, each corner is occupied by an identical particle and in addition to that one atom occupied the body centre.

2. Those atoms which occupy the corners do not touch each other, however they all touch the one that occupies the body centre.

3. Hence each atom is surrounded by eight nearest neighbours and coordination number is 8. An atom present at the body centre belongs to only a particular unit cell i.e., unshared by other cell.
∴number of atoms in a bcc unit cell = \(\frac { { N }_{ c } }{ 8 }\) + \(\frac { { N }_{ b } }{ 1 }\) + \(\frac { 8 }{ 8 } \) = \(\frac { 1 }{ 1 } \) = 1 + 1 = 2

Question 11.
How will derive the formula of density of a unit cell?
Answer:
1. Using the edge length of a unit cell, we can calculate the density (p )of the crystal by considering a cubic unit cell as follows.

Substitute the value (3) in (2)
Mass of the unit cell = n x \(\frac { M }{ { N }_{ A } }\)
For a cubic unit cell, all the edge lengths are equal. i.e., a = b = c
Volume of the unit cell = a x a x a = a³
∴Density of the unit cell = ρ = \(\frac { nM }{ { a }^{ 3 }{ N }_{ A } }\)

Question 12.
Calculate the packing fraction of siniple cubic arrangement
Answer:
In a simple cubic arrangement

Consider the cube with an edge length ‘a’
Volume of the cube with edge length a is = a x a x a = a³ …………….(l)
Let ‘r’ ¡s the radius of the sphere
From the figure a = 2 r ⇒ r = \(\frac { a }{ 2 } \)
:. volume of the sphere with radius r = \(\frac { 4 }{ 3 } \) πr³
= \(\frac { 4 }{ 3 } \) π \({ \left( \frac { a }{ 2 } \right) }^{ 3 }\)
= \(\frac { 4 }{ 3 } \) π \(\left( \frac { { a }^{ 3 } }{ 8 } \right) \)
= \(\frac { { πa }^{ 3 } }{ 6 }\) ……………………….(2)

In a simple cubic arrangement. number of spheres belongs to a unit cell equal to one.
∴Total volume occupied by the spheres in sc unit cell = 1 x \(\left( \frac { { πa }^{ 3 } }{ 6 } \right) \) ……(3)
Dividing 3 by 1

Packing fraction = \(\left( \frac { \frac { \pi { a }^{ 3 } }{ 6 } }{ { a }^{ 3 } } \right) \times 100\) = \(\frac { 100π }{ 6 }\) = 52.3 1% Only 52.31% of the available volume is occupied by the spheres in simple cubic packing, making in efficient use of available space and hence minimizing the attractive forces.

Question 13.
What is meant by packing efficiency? How is ¡t measured?
Answer:
1. There is some free space between the spheres of a single layer and the spheres of successive layers.

2. The percentage of total volume occupied by these constituent spheres gives the packing efficiency of an arrangement. For eg., in simple cubic arrangement.

Question 14.
Calculate the packing efficiency in fcc unit cell?
Answer:
Total number of spheres belongs to a single fcc unit cell is 4. Volume of the sphere with radius r is = \(\frac { n }{ 2 }\) π \({ \left( \frac { \sqrt { 2 } a }{ 4 } \right) }^{ 3 }\)

Question 15.
How is radius ratio ¡s useful in determination of structure of an Ionic compound?
Answer:

1. The structure of an ionic compound depends upon the stoichiometry and the size of the ions.
2. Generally in ionic crystals, the bigger anions are present in the close packed arrangements the cations occupy the voids.
3. The ratio of radius of cation and anion plays an important role in determining the structure.
4. For Eg;

Question 16.
What Is meant by impurity defect? Explain with example?
Answer:

1. A method of introducing defects in ionic solids by adding impurity ions.
2. If the impurity ions are in different valence state from that of host, vaccancies are creited in the crystal lattice of the host.
3. For cg., addition of CdCl² to silver chloride yields solid solutions where the divalent cation Cd²⁺ occupies the position of Ag²⁺
4. This will disturb the electrical neutrality of the crystal. In order to maintain the same, proportional number of Ag ions leave the lattice. This produces a cation vaccancy in the lattice, such kind of crystal defects are called impurity defects.

Question 17.
A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3 rd of the tetrahedral voids. What is the formula of the compound?
Answer:
Suppose the atoms N in the ccp = n
∴ No. of tetrahedral voids = 2n
As 1/3 rd of the tetrahedral voids are occupied by atoms M, therefore,
No. of atoms M = \(\frac { 2n }{ 3 } \)
∴ Ratio of M : N = \(\frac { 2n }{ 3 } \) : n = 2 : 3
Hence, the formula is M₂N₃

Question 18.
How many lattice points are there in one unit cell of each of the following lattice?

1. Face – centred cubic
2. Face – centred tetragonal
3. Body – centered

Answer:

1. Lattice points in face-centred cubic lattice = 8 (at corners) + 6 (at the face centre) = 14
2. Face centred tetragonal = 8 (at corners) + 6 (at the face centre) = 14
3. Lattice points in body-centred cube = 8 (at corners) +1 (at the body centre) = 9

Question 19.
Explain:

1. The basis of similarities and differences between metallic and ionic crystals.
2. Ionic solids are hard and brittle.

Answer:
1. Similarities:

• Both ionic and metallic crystals have electrostatic forces of attractiàn.
• In ionic crystals these forces are between oppositely charged ions. In metals,these forces are among the valence electrons and posityely charged kernels.
• Both have high melting point.

Differences:

• Ionic bond is strong due to electrostatic forces of attraction whereas metallic bond may be weak or strong depending upon the number of valence electrons and the size of kernels.
• In ionic bond, ions are not free to move. Hence, they cannot conduct electricity in solid state. They can do so only in molten state or in aqueous solution. ¡n metals, electrons are free to move. Hence, they conduct electricity in solid state.

2. Ionic crystals are hard due to strong electrostatic forces between them. They are brittle because ionic bond is non – directional.

Question 20.
ZnO is colourless at room temperature, while yellow when hot, why?
Answer:
ZnO is colourless at room temperature. When it is heated, it becomes yellow in colour. On heating, it loses oxygen and thereby forming free Zn²⁺ ions. The excess Zn²⁺ ions move to interstitial sites and the electrons also occupy the interstitial positions.

Samacheer Kalvi 12th Chemistry Solid State 5 Mark Questions and Answers

Question 1.
What are general characteristics of solids?
Answer:

1. Solids have definite volume and shape.
2. Solids are rigid and incompressible.
3. Solids have strong cohesive forces.
4. Solids have short interatomic, ionic (or) molecular distances.
5. Their constituents (atoms, ions or molecules) have fixed positions and can only oscillate about their mean positions.
6. Unlike gases, in solids, the atoms, ions (or) molecules are held together by strong force of attraction.

Question 2.
Write a note about classification of solids with suitable examples.
Answer:

Question 3.
What are ionic solids? Give their characteristics.
Answer:

1. The structural units of an ionic crystals are cations and anions. They are bound together by strong Na electrostatic attractive forces.
2. To maximize the attractive force, cations are surrounded by as many anions as possible and vice versa.
3. Ionic crystals possess definite crystals structure.
4. Many solids are cubic close packed.

Characteristics

1. Ionic solids have high melting points.
2. These solids do not conduct electricity, because the ions are fixed in their lattice positions.
3. They do not conduct electricity in molten state (or) when dissolved in water, because the ions are free to move in the molten state or solution.
4. They are hard as only strong external force can change the relative positions of ions.
5. Example – The arrangement of Na and Cl ions in NaCI crystal.

Question 4.
What are molecular solids?Explain their classification with suitable examples.
Answer:
In molecular solids, the constituents are neutral molecules. They are held together by weak vander Waals forces. Generally molecular solids are soft and they do not conduct electricity. These molecular solids are further classified into three types.

1. Non polar molecular solids.
2. Polar molecular solids.
3. Hydrogen bonded molecular solids.

1. Non polar molecular solids

• In this type, molecules are held together by weak dispersion forces or London forces.
• They have low melting points and are usually in liquids or gaseous state at room temperature. Examples – Naphthalene, Anthracene etc.

2. Polar molecular solids

• In this type, molecules formed by polar covalent bonds.
• They are held together by strong dipole-dipole interactions.
• They have higher melting points than the non – polar molecular solids. Examples., solid CO₂, solid NH₃

3. Hydrogen bonded molecular solids

• The constituents are held together by hydrogen bonds.
• The constituents are held together by hydrogen bonds.
• They are soft solids under room temperature Examples: Solid ice, glucose, urea.

Question 5.
An element crystallizes in a fcc lattice with cell edge of 400 pm. The density of the element is 7 g/cm³. How many atoms one present in 280 g of the element?
Answer:
Volume of unit cell a³ (a = edge length)
= 400 pm
= (400 x 10^-12 m)³
= (400 x 100^-10 cm)³ = 64 x 10^-24 cm³

Since each f.c.c. unit cell contains 4 atoms therefore,
Total number =4 x 0.46 x 1024,
= 1.84 x 1024 atoms.

Common Errors

Common Errors

1. sc, bcc, fee structures may get confused if they ask in different order.
2. Calculation of atoms at each cube may get confused.
3. Sharing of atoms may get con¬fused.

Rectifications

1. In sc, only simple cube can be drawn, bcc structure is with one dot at centre, fee structure is with sc with six dots at each face.
   sc – only one atom, bcc – two atoms, fee – four atoms (we can remember as f and f).
2. Atom at comer is shard by 8 unit cells. Atom at centre is not shared. Atom at face is shared by 2 unit cells. Atom at edge is shared by 4 unit cells.

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Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 11 Hydroxy Compounds and Ethers

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Samacheer Kalvi 12th Chemistry Chapter 11 Hydroxy Compounds and Ethers Textual Evaluation Solved

Samacheer Kalvi 12th Chemistry Hydroxy Compounds and Ethers Multiple Choice Questions

12th Chemistry Chapter 11 Book Back Answers Question 1.
An alcohol (x) gives blue colour in victormayer’s test and 3.7g of X when treated with metallic sodium liberates 560 mL of hydrogen at 273 K and 1 atm pressure what will be the possible structure of X?
(a) CH₃ CH (OH) CH₂CH₃
(b) CH₃ – CH(OH) – CH₃
(c) CH₃ – C (OH) (CH₃)₂
(d) CH₃ – CH₂ – CH (OH) – CH₂ – CH₃
Answer:
(a) CH₃ CH (OH) CH₂CH₃
Hint:
2R – OH + Na → 2RONa + 2H₂ ↑ 2 moles of alcohol gives 1 mole of H₂ which occupies
22.4L at 273K and 1 atm
number of moles of alcohol = \(\frac{2 \text { moles of } \mathrm{R}-\mathrm{OH}}{22.4 \mathrm{L} \text { of } \mathrm{H}_{2}}\) x 560 mL = 0.05 moles
number of moles = \(\frac{\text { mass }}{\text { molar mass }}\)
= molar mass = \(\frac{3.7}{0.05}\) = 74 g mol^-1
General formula for
R – OH C_n H_2n+1 – OH
n(12) + (2n+1) (1) + 16 +1 = 74
14n = 74 – 18
14n = 56
n = \(\frac { 56 }{ 4 }\) = 4
The 2° alcohol which contains 4 carbon is CH_n CH(OH)CH₂ CH₃

Hydroxy Compounds And Ethers Question 2.
Which of the following compounds on reaction with methyl magnesium bromide will give tertiary alcohol.
(a) benzaldehyde
(b) propanoic acid
(c) methyl propanoate
(d) acetaldehyde
Answer:
(c) methyl propanoate
Hint:

12th Chemistry Evaluate Yourself Answers Question 3.

This ‘X’ is …………..

Answer:

Hint:
hydro boration – Anti markownikoff product
i.e CH₃ – CH₂ – CH – CH₂ – CH₂ – OH

Hydroxy Compounds And Ethers Pdf Question 4.
In the reaction sequence, Ethane

Ethan – 1, 2 – diol. A and X respectively are ………….
(a) Chioroethane and NaOH
(b) ethanol and H₂SO₄
(c) 2 – chloroethan – 1 – ol and NaHCO₃
(d) ethanol and H₂O
Answer:
(c) 2 – chloroethan – 1 – ol and NaHCO₃
Hint:

Hydroxy Compounds And Ethers Class 12 Question 5.
Which one of the following is the strongest acid ………..
(a) 2 – nitrophenol
(b) 4 – chlorophenol
(c) 4 – nitrophenol
(d) 3 – nitrophenol
Answer:
(c) 4 – nitrophenol

Hydroxy Compounds And Ethers Important Questions Question 6.

on treatment with Con. H₂SO₄, predominately gives ……………..

Answer:

Hint:

12th Chemistry Evaluate Yourself Answers Samacheer Question 7.
Carbolic acid is …………..
(a) Phenol
(b) Picric acid
(c) benzoic acid
(d) phenylacetic acid
Answer:
(a) Phenol

12th Chemistry Book Inside Evaluate Yourself Answers Question 8.
Which one of the following will react with phenol to give salicyladehyde after hydrolysis …………..
(a) Dichioro methane
(b) trichioroethane
(c) trichloro methane
(d) CO₂
Answer:
(c) trichloro methane (Riemer Tiemann reaction)
Hint:

12th Chemistry Evaluate Yourself Answers Pdf Question 9.

(a) (CH₃)₃ CCH = CH₂
(b) (CH₃)₂ C = C (CH₃)₂
(c) CH₂ = C(CH₃)CH₂ – CH₂ – CH₃
(d) CH₂ = C (CH₃) – CH₂ – CH₂ – CH₃
Answer:
(b) (CH₃)₂ C = C (CH₃)₂
Hint:

12th Chemistry Evaluate Yourself Answers Pdf Download Question 10.
The correct IUPAC name of the compound,

(a) 4 – chloro – 2, 3 – dimethyl pentan – 1 – ol
(b) 2, 3 – dimethyl – 4 – chloropentan – 1 – ol
(c) 2, 3, 4 – trimethyl – 4 – chiorobutan – 1 – ol
(d) 4 – chioro – 2, 3, 4 – trimethyl pentan – 1 – ol
Answer:
(a) 4 – chloro – 2, 3 – dimethyl pentan – 1 – ol

12 Chemistry Evaluate Yourself Answers Question 11.
Assertion: Phenol is more acidic than ethanol
Reason: Phenoxide ion is resonance stabilized
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false.
Answer:
if both assertion and reason are true and reason is the correct explanation of assertion.

12th Std Chemistry Evaluate Yourself Answers Question 12.
In the reaction Ethanol
is ………………
(a) ethane
(b) ethoxyethane
(c) ethylbisuiphite
(d) ethanol
Answer:
(d) ethanol
Hint:

12th Chemistry Evaluate Yourself Answers Chapter 11 Question 13.
The reaction

can be classified as
(a) dehydration
(b) Williams on alcoholsynthesis
(c) Williamson ether synthesis
(d) dehydrogenation of alcohol
Answer:
(c) Williamson ether synthesis
Hint: Cyclic alcohol → sodium cyclic alkoxide → williamson ether synthesis

Samacheer Kalvi Guru 12th Chemistry Question 14.
Isoprophylbcnzene on air oxidation in the presence of dilute acid gives …………
(a) C₆H₅COOH
(b) C₆H₅COCH₃
(c) C₆H₅COC₆H₅
(d) C₆H₅ – OH
Answer:
(a) C₆H₅ – OH (phenol)

11th Chemistry Evaluate Yourself Answers Chapter 11 Question 15.
Assertion: Phenol is more reactive than benzene towards electrophilic substitution reaction
Reason: In the case of phenol. the intermediate arenium ion is more stabilized by resonance.
(a) if both assertion and reason are true and reason is the correct explanation of assertion.
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false,.
Answer:
(a) if both assertion and reason are true and reason is the correct explanation of assertion.

Samacheer Kalvi 12th Chemistry Solutions Question 16.
HO CH₂ CH₂ – OH on heating with periodic acid gives ………..
(a) methanoic acid
(b) Glyoxal
(c) methanol
(d) CO₂
Answer:
(c) methanol

Question 17.
Which of the following compound can be used as artireeze in automobile radiators?
(a) methanol
(b) ethanol
(c) Neopentyl alcohol
(d) ethan -1, 2-diol
Answer:
(d) ethan -1, 2-diol

Question 18.

is an example of …………..
(a) Wurtz reaction
(b) cyclic reaction
(c) Williamson reaction
(d) Kolbe reactions
Answer:
(c) Kolbe reactions

Question 19.
One mole of an organic compound (A) with the formula C₃H₈O reacts completely with two moles of HI to form X and Y. When Y is boiled with aqueous alkali it forms Z. Z answers the iodoform test. The compound (A) is ……………
(a) propan – 2 – ol
(b) propan- 1- ol
(c) ethoxy ethane
(d) methoxy ehane
Answer:
(d) methoxy ehane
Hint:

Question 20.
Among the following ethers which one will produce methyl alcohol on treatment with hot HI?

Answer:

Hint:

Question 21.
Williamson synthesis of preparing dimethyl ether is a / an /
(a) SN¹ reactions
(b) SN² reaction
(c) electrophilic addition
(d) electrophilic substitution
Answer:
(b) SN² reaction

Question 22.
On reacting with neutral ferric chloride, phenol gives
(a) red colour
(b) violet colour
(c) dark green colour
(d) no colouration
Answer:
(b) violet colour

II. Answer the following questions

Question 1.
IdentIfy the product (s) is / are formed when 1 – methoxy propane is heated with excess HI. Name the mechanism involved in the reaction
Answer:
1-methoxy propane is heated with excess HI, yields two products named as Methyl iodide and 1- iodo propane.
Step 1:

Step 2.

Step 3:

Step 4:

Overall reaction:

This reaction ivvolves nucleophilic substitution reaction mechanism. (SN¹)

Question 2.
Draw the major product formed when 1 – ethoxyprop – 1 – ene is heated with one equivalent of HI
Answer:

This reaction follows SN¹ mechanism because in this reaction the more stable carbocation is formed that is double bonded carbocation. Therefore, the given molecule reacts with HI to form ethanol and 1- iodo prop – 1 – ene.

Question 3.
Suggest a suitable reagent to prepare secondary alcohol with identical group using Grignard reagent.
Answer:
Acetaldehyde reacts with Grignard reagent to give addition product, which on further undergoes acid hydrolysis to yield secondary alcohol, that is isopropyl alcohol.

Question 4.
What is the major product obtained when two moles of ethyl magnesium bromide is treated with methyl benzoate followed by acid hydrolysis
Answer:

Question 5.
Predict the major product, when 2-methyl but – 2 – ene is converted into an alcohol in each of the following methods.

1. Acid catalysed hydration
2. Hydroboration
3. Hydroylation using bayers reagent

Answer:
1. Acid catalysed hydration:

2. Hydroboration.

3. Hydroxylation using bayers reagent:

Question 6.
Arrange the following in the increasing order of their boiling point and give a reason for your ordering

1. Butan – 2 – ol, Butan – 1 – SI, 2 – methylpropan – 2 – ol
2. Propan – 1 – ol, propan – 1, 2, 3 – triol, propan – 1, 3 – diol, propan – 2 – ol

Answer:
1. Boiling points increases regularly as the molecular mass increases due to a corresponding increase in their Van der waal’s force of attraction. Among isomeric alcohols 2° – alcohols have lower boiling points than 1° – alcohols due to a corresponding decreases in the extent

of H-bonding because of steric hindrance. Thus the boiling point of Butan – 2 – ol is lower than that of Butan – 1 – ol. Overall increasing order of boiling points is, 2 – methylpropan – 2 – ol < Butan – 2 – ol < Butan – 1 – ol

2. 2°-alcohols have lower boiling points than 1° – alcohols due to a corresponding decrease in the extent of H – bonding because of steric hindrance. Therefore Propan – 1 – ol has higher boiling point than Propan – 2 – ol. Hydrogen group increases, boiling point also increases. Overall increasing order of boiling points is, propan – 2 – ol < Propan – 1 – ol < propan – 1, 3 – diol < propan -1, 2, 3 – triol

Question 7.
Can we use nucelophiles such as NH₃,CH₃O for the Nucleophilic substitution of alcohols
Answer:
1. Increasing order of nucelophilicity,
NH₃ < – OH^⊕ < CH₃O^⊖-

2. Higher electron density will increase the nucelophilicity.

3. Negatively charged species are almost always more nucelophiles than neutral species.

4. RO^⊖ has an alkyl group attached, allowing a greater amount of polarizability. This means oxygen’s lone pairs will be more readily available to reach in RO^⊖ than in OH^⊖. Hence CH₃O – is the better nucelophile for the nucleophilic substitution of alcohols. NH₃ cannot act as nucelophile for the nucleophilic substitution of alcohols.

Question 8.
Is it possible to oxidise t – butyl alcohol using acidified dichromate to form a carbonyic compound.
Answer:
3° – alcohols do not undergo oxidation reaction under normal condition, but at elevated temperature, under strong oxidising agent cleavage of C – C bond takes place to give a mixture of carboxylic acid.

Yes, it is possible. t – butyl alcohol is readily oxidsing in acidic solution (K₂Cr₂O₇ / H₂SO₄) to a mixture of a ketone and an acid each containing lesser number of carbon atoms than the original alcohol. The oxidation presumably occur via alkenes formed through dehydration of alcohols under acidic conditions.

Question 9.
What happens when 1 – phenyl ethanol is treated with acidified KMnO₄.
Answer:
1 – phenyl ethanol reacts with acidified KMnO₄ to give Acetophenone.

Question 10.
Write the mechanism of acid catalysed dehydration of ethanol to give ethene.
Mechanism of acid catlaysed dehydration of ethanol:
Step1:
Protonation of ethanol.

Step 2:
Elimination of water molecule.

Question 11.
How is phenol prepared form

1. chloro benzene
2. isopropyl benzene

Answer:
1. From Chioro benzene:
According to Dow’s process, when Chiorobenzene is hydrolysed with 6 – 8% NaOHat 300 bar and 633K in a closed vessel, sodium phcnoxidc is formed which on treatment with dilute HCl gives phenol.

2. For isopropyl benzene:
On passing air to a mixture of cumene (isoprophy benzene) and 5% aqueous sodium carbonate solution, cumene hydro peroxide is formed by oxidation. It is treated with dilute acid to get phenol and acetone.

Question 12.
Explain Kolbe’s reaction
Answer:
Kolbe’s (or) Kolbe’s Schmitt reaction:
In this reaction, phenol is first converted into sodium phenoxide which is more reactive than phenol towards electrophilic substitution reaction with CO₂. Treatment of sodium phenoxide with CO₂ at 400K, 4 -7 bar pressure followed by acid hydrolysis gives salicylic acid.
Answer:

Question 13.
Writes the chemical equation for Williamson synthesis of 2 – ethoxy – 2 – methyl pentane starting from ethanol and 2 – methyl pentan – 2 – ol
Answer:
Step 1:

Step 2:

Step 3:

Williamsons synthesis occurs by SN² – mechanism and primary alkyl halides are more reactive in SN² reactions. Therefore ethanol is converted into ethyl bromide.

Question 14.
Write the structure of the aldehyde, carboxylic acid and ester that yield 4 – methylpent – 2 – en – 1 – ol.
Answer:
1. Aldehyde yield 4 – methylpent – 2 – 3n – ol is

2. Acid yield 4 – methylpent – 2 – en – 1 – ol is

3. Ester yield 4 – methylpent – 2 0 en – l – ol is

The above shown compounds undergo reduction reaction to yield 4 – methylpent – 2 – en – l – ol.

Question 15.
What is meta merism? Give the structure and IUPAC name of metamers of 2 – methoxy propane
Answer:
Metamerism:
It is a special type of isomerism in which molecules with same formula, same functional group, but different only in the nature of the alkyl group attached to oxygen.

Ethoxy ethane and 1-methoxy propane are metamers of 2-methoxy propane.

Question 16.
How are the following conversions effected

1. benzylchlorjde to benzvlalcohol
2. benzvl alcohol to benzoic acid

Answer:
1. Conversion of benzyl chloride into benzyl alcohol:

2. Conversion of benzyl alcohol into benzoic acid:

Question 17.
Complete the following reactions

Answer:

Question 18.
O.44g of a monohydric alcohol when added to methyl magnesium iodide in ether liberates at STP 112 cm³ of methane with PCC the same alcohol form a carbonyl compound that answers silver mirror test. Identify the compound.
Answer:
0.44g of a monohydric alcohol liberates 112 cm³ of methane.

Mass of monohydric alcohol which gives 22400 cm3 of methane = \(\frac{22400 \times 0.44}{112}\) = 88
C₅H₁₂O molecular fórmula has mass number 88 and it shows eight possible isomers. But neopentyl alcohol reacts with PCC to form neopentyl aldehyde, which shows positive silver mirror test. Therefore, compound is. neopentyl alcohol (or) 2, 2 – dimethyl propan – l – ol.

Question 19.
Complete the following reactions

Answer:

Question 20.
Phenol is distilled with Zn dust gives (A) followed by friedel – crafts alkylation with propyl chloride to give a compound B, B on oxidation gives (C). Identify A,B and C.
Answer:

Question 21.

Identify A, B, C, D and write the complete equation.
Answer:

Question 22.
What will be the product for the following reaction

Answer:

Question 23.
How will you convert acetylen in to n – butyl alcohol.
Answer:

Question 24.
Predict the product A, B, X and Y in the following sequence of reaction

Answer:

Question 25.
3,3 – dimethylbutan – 2 – ol on treatment with conc. H₂SO₄ to give tetramethyl ethylene as a major product. Suggest a suitable mechanisms
Answer:

According to Saytzeff’s rule the dehydration of 3,3 – dimethylbutan – 2 – ol gives a mixture of alkenes. But the secondary carbocation formed in this reaction undergoes rearrangement to form a more stable tertiary

carbocation which further, undergoes to 13 – elimination leads more stable product, that is 2,3 – dimethyl but – 2 – ene (more yield). According to Saytzeff’s nile, 2, 3 – dimethyl pent- 2 – ene is the major product.

Samacheer Kalvi 12th Chemistry Hydroxy Compounds and Ethers Evaluate yourself

Question 1.
Classify the following alcohols as 1⁰, 2⁰, and 3⁰ and give their IUPAC Names.

Answer:

Question 2.
Write all the possible isomers of an alacohol having the molecular formula C₁₅H₁₂O and their IUPAC names.
Answer:
Eight isomers are possibe for C₁₅H₁₂O. They are,

Question 3.
Suggest a suitable carbonyl compound for the preparation of pent – 2 – en – 1 ol using LiAlH₄.
Answer:

Question 4.
2 – methylpropan – 2 – ene
Answer:

Question 5.
How will you prepare the following using Grignard reagent.

1. t – butyl alcohol
2. allyl alcohol

Answer:

Question 6.
Identify the products in the following reactions. Write their IUPAC names and mention the mechanism involved in the reactions.

Answer:
1.
This reaction involves E₁ – mechanism.

2.
This reaction ivolves SN² – mechanims.

3. Step 1:

Step 2:

This reaction involves SN² Mechanims.

Question 7.
What is the major product obtained when 2,3 – dimethyl pentan – 3 – ol is heated in the presence of H₂SO₄
Answer:

According to Saytzeff’s rule, 2, 3 – dimethyl pent – 2 – ene is the major product.

Question 8.
Which of the following set of reactants will give 1 – methoxy – 4 – nitrobenzene.

Answer:
Step 1:

step 2:

Chemically both sets equally possible. In Set – 1, the Br-atom is activated by electron withdrawing effect of – NO₂ group. Therefore nucleophilic attack by CH₃ONa followed by elimination of NaBr gives the desired ether.

Step 1:

Step 2:

In set – 2, nucleophilic attack by 4 – nitrosodium phenoxide ion on methyl bromide gives the desired ether.

Since alkyl halides (CH₃Br) are more reactive than aryl halides in nucleophilic substitution reactions, therefore set – 2 reactants are preferred.

Question 9.
What happens when m – cresol is treated with acidic solution of sodium dichromate?
Answer:

When m – cresol is treated with acidic solution of sodium dichromate it gives 4 – hydroxy beazoic acid.

Question 10.
When phenol is treated with propan – 2 – ol in the presence of HF, Friedel – Craft reaction takes place. Identify the products.
Answer:

Question 11.
Given the IUPAC name for the following ethers and classify them as simple or mixed.
Answer:

Question 12.
1. Which of the following reaction will give 1 – methoxy – 4 – nitrobenzene.

1. 4 – nitro – 1 – bromobenzene + sodium methoxide.
2. 4 – nitrosodium phenoxide + bromomethane

Answer:
1. Set – 1:

2. set – 2:

Chemically both sets equally possible. In Set – 1, the Br – atom is activated by electron withdrawing effect of – NO₂ group. Therefore nucleophilic attack by CH₃ONa followed by elimination of NaBr gives the desired ether.
Step 1:

Step 2:

In set – 2, nucleophilic attack by 4 – nitrosodium phenoxide ion on methyl bromide gives the desired ether.

Since alkyl halides (CH₃Br) are more reactive than aryl halides in nucleophilic substitution reactions, therefore set – 2 reactants are preferred.

Question 13.
Arrange the following compounds in the increasing order of their acid strength. propan – 1 – ol, 2, 4, 6 – trinitroptienol, 3 – nitrophenol, 3,5 – dinitrophenol, phenol, 4 – methyiphenol.
Answer:
Phenols are stronger acids than alcohols because the phenoxide ion left after the removal of proton is stabilized by resonance while the alkoxide ion left after the removal of a proton from alcohol is not stabilized. Thus propan – 1 – ol is much weaker acid than any phenol.

Thus propan- 1 – ol is a much weaker acid than any phenol. We know that electron donating groups decrease the acidic character and stronger is the electron donating group, weaker is the phenol.

Compare to propan – 1 – ol, 4 – methyl phenol is stronger acidic character. But comparing phenol and 4-methyl phenol, phenol is stronger acidic. Since electron withdrawing groups increase the acidic character of phenols and the effect is more pronounced at the para position than at the meta position.

Therefore 4 – nitro phenol is a stronger acid than 3 – nitro phenol. Further as the number of electron withdrawing groups increases the acidic strength further increases. Therefore 2, 4, 6 – trinitro phenol is a stronger acid than 3, 5 – dintiro phenol. It may be noted here that although the two nitro groups in 3, 5 – dinitro phenol are at m – position with respect to OH group,

their combined effect is however greater than one nitro group at p – position. Therefore 3, 5 – dinitro phenol is a stronger acid than 4-nitro phenol. Thus, the overall increasing order of acid strength is. Propan – 1 – 01 < 4 – methyl phenol < phenol < 3 – nitrophenol < 3, 5 – dinitro phenol < 2, 4, 6 – trinitro phenol.

Question 14.
1 mole of HI is allowed to react with t – butyl methylether. Identify the product and write down the mechanism of the reaction.
Answer:
Step 1:

Step 2

Step 3:

Overall reaction:

Samacheer Kalvi 12th Chemistry Hydroxy Compounds and Ethers Additional Questions

Samacheer Kalvi 12th Chemistry Hydroxy Compounds and Ethers 1 Mark Questions and Answers

I. Choose the best answer.

Question 1.
Which one of the following is a trihydric alcohol?
(a) Glycol
(b) Ethanol
(c) Glycerol
(d) Sorbitol
Answer:
(c) Glycerol

Question 2.
Identify the monohydric unsaturated alcohol.

Answer:

Question 3.
Which one of the following is named as sorbital?

Answer:

Question 4.
Which one of the following is a primary alcohol?

Answer:

Question 5.
Which of the following is a dihydric alcohol?
(a) Ethenol
(b) Ethanol ]
(c) Ethane – 1, 2 – diol
(d) Propan – 2 – ol
Answer:
(c) Ethane – 1, 2 – diol

Question 6.
Which one of the following is an example of secondary (2°) alcohol?
(a) Propan – 2 – ol
(b) Phenyl methanol
(c) Ethenol
(d) 2 – methyl – propan – 2 – ol
Answer:
(a) Propan – 2 – ol

Question 7.
Which one of the following is a tertiary alcohol?

Answer:

Question 8.
Which of the following is a primary alcohol?

Answer:

Question 9.
Which one of the following find application in proper functioning of our eyes?
(a) Cholesterol
(b) Retinol
(c) Phenol
(d) Ethanol
Answer:
(b) Retinol

Question 10.
Which is the storage of vitamin – A?
(a) Retinol
(b) Benzyl alcohol
(c) Phenol
(d) Ascorbic acid
Answer:
(a) Retinol

Question 11.
The important component in our cell membrane is ………….
(a) Retinol
(b) Phenol
(c) Cholesterol
(d) Methanol
Answer:
(c) Cholesterol

Question 12.
Which acts as an additive to petrol?
(a) Glycerol
(b) Ethanol
(c) Phenol
(d) Methanol
Answer:
(b) Ethanol

Question 13.
Which one of the following vitamin is stored in Retinol?
(a) Vitamin – B₁₂
(b) Vitamin – A
(c) Vitamin – C
(a) Vitamin – D
Answer:
(b) Vitamin – A

Question 14.
Which alcohol is used as skin cleanser for injection?
(a) Methanol
(b) Ethanol
(c) 1 -propanol
(d) 2-propanol
Answer:
(d) 2-propanol

Question 15.
Which one of the following is used as an industrial solvent?
(a) Methanol
(b) Benzyl alcohol
(c) Phenol
(d) Cholesterol
Answer:
(a) Methanol

Question 16.
2 – methyl but – 3 – en – 2 – ol belongs to which type of alcohol?
(a) 3° alcohol
(b) 2° alcohol
(c) 1° alcohol
(d) Aromatic alcohol
Answer:
(a) 3° alcohol

Question 17.
The IUPAC name of

(a) 1 – methyl – 2 – propanol
(b) 2 – methyl – p ropan – 2 – ol
(c) Tertibutyl alcohol
(d) 2 – propanol
Answer:
(b) 2 – methyl – p ropan – 2 – ol

Question 18.
The TUPAC name of CH₂ = CH – CH₂OH is …………
(a) Allyl alcohol
(b) Propenc – 2 – ol
(c) Prop – 2 – en – 1 – oI
(d) Isopropyl alcohol
Answer:
(c) Prop – 2 – en – 1 – oI

Question 19.
In methanol, – OH group attached to carbon is ………..
(a) sp hybridised atom
(b) sp³ hybridised atom
(c) sp² hybridised atom
(d) dsp² hybridised atom
Answer:
(c) sp² hybridised atom

Question 20.
Which one of the following is C – O – H bond angle in methanol?
(a) 109.5°
(b) 104°
(c) 90°
(d) 108.9°
Answer:
(d) 108.9°

Question 21.
Primaiy alkyl halides undergoes substitution by ……….
(a) SN¹ reaction
(b) SN_i reaction
(c) SN² reaction
(d) SN reaction
Answer:
(c) SN² reaction

Question 22.
What is the product formed when propene is hydrolysed in the presence of mineral acid?
(a) Propan – 1 – ol
(b) Propan – 2 – ol
(c) Iso butyl alcohol
(d) 2 – mcthyl – propan – 2 – ol
Answer:
(b) Propan – 2 – ol

Question 23.
The product formed when phenyl magnesium bromide treated with methanal and hydrolysed is ………..
(a) Phenyl methanal
(b) Phenol
(c) Phenyl methanol
(d) Benzyl benzoate
Answer:
(c) Phenyl methanol

Question 24.
To get Butan – 2 – ol, Ethyl magnesium bromide is treated with followed by hydrolysis.
(a) HCHO
(b) CH₃COCH₃
(c) CO₂
(d) CH₃CHO
Answer:
(d) CH₃CHO

Question 25.
Which one of the following is formed when Butyl magnesium bromide is treated with propanone followed by hydrolysis?
(a) Tertiary butyl alcohol
(b) Isopropyl alcohol
(c) 2 – methyl hexan – 2 – ol
(d) Propan – 1 -ol
Answer:
(c) 2 – methyl hexan – 2 – ol

Question 26.
Which one of the tbllowing is used to get propan – 2 – ol by the reaction with CH₃MgBr?
(a) Ethanol
(b) Ethanal
(c) Ethyl inethanoate
(d) Propanone
Answer:
(c) Ethyl inethanoate

Question 27.
Crotanaldehyde on reaction with LiAlH₄ and water produces
(a) Ethanol
(b) Propan – 2 – ol
(c) Methanol
(d) But – 2 – en – 1 – ol
Answer:
(d) But – 2 – en – 1 – ol

Question 28.
Which one of the following is used as a catalyst in the conversion of Bcnzoic acid to Benzyl alcohol?
(a) Ni
(b) LiAIH₄ / H₂O
(c) Sn / HCI
(d) Zn / NaOH
Answer:
(b) LiAIH₄ / H₂O

Question 29.
What is the product formed when acetone is treated with LiA1H4 and 1120?
(a) Isobutyl alcohol
(b) n – butyl alcohol
(c) Propan – 2 – ol
(d) Propan – 1 – ol
Answer:
(c) Propan – 2 – ol

Question 30.
Which one of the following is formed when ethene reacts with Baeyer’s reagent?
(a) Ethane
(b) Ethylene glycol
(c) Propane – 1, 2 – diol
(d) Glycerol
Answer:
(b) Ethylene glycol

Question 31.
Which one of the following is named as Baeyer’s reagent?
(a) acidified K₂Cr₂O₇
(b) acidified KMnO₄
(c) Cold dilute alkaline KMnO₄
(d) LiAlH₄
Answer:
(c) Cold dilute alkaline KMnO₄

Question 32.
The alkaline hydrolysis of fats to give glycerol is known as …………
(a) Esterification
(b) Hydroboration
(c) Hydration
(d) Saponification
Answer:
(d) Saponification

Question 33.
Which one of the following alcohol reacts immediately with Lucas reagent?
(a) Primaiy alcohol
(b) Tertiary alcohol
(c) Phenol
(d) Secondary alcohol
Answer:
(b) Tertiary alcohol

Question 34.
Which one of the following is called Lucas reagent?
(a) Conc. HCl + Anhydrous ZnCl₂
(b) Conc. HCl + Anhydrous A1CI₃
(c) LiAIH₄ + H₂O
(d) Cold dilute alkaline KMnO₄
Answer:
(a) Conc. HCl + Anhydrous ZnCl₂

Question 35.
Which alcohol gives red colour in Victor Meyer’s test?
(a) 2° alcohol
(b) 3° alcohol
(c) Phenol
(d) 1° alcohol
Answer:
(d) 1° alcohol

Question 36.
Which colour is given by secondary alcohol in Victor Meyer’s test?
(a) Red
(b) Green
(c) Blue
(d) Yellow
Answer:
(c) Blue

Question 37.
Which mechanism is followed in the reaction of 2 – methyl – 2 – propanol with HBr?
(a) E₁ mechanism
(b) E₂ mechanism
(c) SN² mechanism
(d) SN¹ mechanism
Answer:
(d) SN¹ mechanism

Question 38.
Which mechanism is followed in the conversion of ethanol to bromoethane by HBr?
(a) SN¹ mechanism
(b) SN² mechanism
(c) E₁ mechanism
(d) E₂ mechanism
Answer:
(c) E₁ mechanism

Question 39.
Which one of the following is used as a catalyst in the reaction of methanol with thionyl chloride?
(a) Pyridine
(b) pyrrole
(c) THF
(d) Nickel
Answer:
(a) Pyridine

Question 40.
The mechanism of the reaction of ethanol with PCl₃ is ……………..
(a) SN¹
(b) SN²
(c) E₂
(d) E₁
Answer:
(b) SN²

Question 41.
Which one of the following reagent is used in the conversion of Ethanol to ethene?
(a) Zn + Hg / H₂O
(b) LiAlH₄
(c) acidified K₂Cr₂O₇
(d) Conc. H₂SO₄
Answer:
(d) Conc. H₂SO₄

Question 42.
Primary alcohol undergo dehydration by ………
(a) E₁ mechanism
(b) E₂ mechanism
(c) SN¹ mechanism
(d) SN² mechanism
Answer:
(b) E₂ mechanism

Question 43.
Tertiary alcohols undergo dehydration by ……….
(a) SN¹ mechanism
(b) E₂ mechanism
(c) E₁ mechanism
(d) SN² mechanism
Answer:
(c) E₁ mechanism

Question 44.
Which one of the following is the correct order of relative reactivities of alcohols in the dehydration reaction?
(a) 1° < 2° < 3°
(b) 2° < 1° < 3°
(c) 3° < 2° < 1°
(d) 3° < 1° < 2°
Answer:
(a) 1° < 2° < 3°

Question 45.
Which of the following is the product formed when 3,3 – dimethyl – 2 – butanol reacts with conc.H₂SO₄?
(a) 2, 3 – dirnethyl but – 1 – ene
(b) 2,3 – dimethyl but – 2 – ene
(c) 3, 3 – dimethyl but – 1 – ene
(d) all the above
Answer:
(d) all the above

Question 46.
The oxidising agent used to prepare aldehyde (or) ketone from alcohol, the reagent used is …………..
(a) acidified Na₂Cr₂O₇
(b) alkaline KMnO₄
(c) Pyridinium chlorochromate
(d) conc. H₂SO₄
Answer:
(c) Pyridinium chlorochromate

Question 47.
The product formed when propan – 2 – ol is treated with dimethyl sulfoxide (DMSO) and oxalyl chloride followed by the addition of Et₃N is ……….
(a) Oxalyl chloride
(b) Propanal
(c) Ethanoic aicd
(d) Propanone
Answer:
(d) Propanone

Question 48.
Which reaction is used to convert alcohol to ketone / aldehyde in the presence of DMSO?
(a) Lucas test
(b) Swern oxidation
(c) Biological oxidation
(d) Kolbe’s reaction
Answer:
(b) Swern oxidation

Question 49.
Which product is formed when propan- 1 – ol is oxidised by pyridinium chlorochromate (PCC)?
(a) Propanal
(b) Propanone
(c) Propane
(d) Propene
Answer:
(a) Propanal

Question 50.
Which one of the enzyme is produced in liver to detoxify the alcohol?
(a) Diastase
(b) Zymase
(c) Invertase
(d) Dehydrogenase alcohol
Answer:
(d) Dehydrogenase alcohol

Question 51.
What isADH and NAD?
(a) Alcohol dehydrogenase and nicotinamide adenine dinucleotide
(b) Acid dehydration and Nitrogen addition
(c) Alcohol dehydration and Nicotine addition
(d) Adeninc hydrogenase and Nicotinamide adenine dinucleotide
Answer:
(a) Alcohol dehydrogenase and nicotinamide adenine dinucleotide

Question 52.
What is the main reaction take place when 2 – methyl propan – 2 – ol reacts with Cu at 573 K?
(a) Dehydrogenation
(b) Oxidation
(c) Dehydration
(d) Hydrogenation
Answer:
(c) Dehydration

Question 53.
Name the product formed when tertiary butyl alcohol is treated with Cu at 573 K?
(a) 2 – methyl prop – 1 – ene
(b) 2 – methyl prop – 2 – ene
(c) propene
(d) 1 – butene
Answer:
(a) 2 – methyl prop – 1 – ene

Question 54.
Which one of the following product is formed when propan – 2 – ol is treated with Cu at 573 K?
(a) Propanal
(b) Propanone
(c) Propan – 1 – ol
(d) Propane
Answer:
(b) Propanone

Question 55.
What is the name of the reaction between ethanol and ethanoic acid?
(a) Esterification
(b) Saponification
(c) Ethenfication
(d) Hydroxylation
Answer:
(a) Esterification

Question 56.
Which one of the following is formed when ethan – 1, 2 – diol is treated with PI₃?
(a) Ethane
(b) Ethyne
(c) Ethene
(d) Ethanol
Answer:
(c) Ethene

Question 57.
Which reagent is used to convert ethylene glycol to ethylene?
(a) HI
(b) I₂
(c) PI₃
(d) Conc. H₂ SO₄
Answer:
(c) PI₃

Question 58.
What is the product formed when ethylene glycol is heated at 773 K?
(a) Ethanal
(b) Ethene
(c) Ethane
(d) Oxirane
Answer:
(d) Oxirane

Question 59.
Which reagent is used to convert ethan – 1, 2 – diol into Ethanal?
(a) Anhydrous ZnCI₂
(b) Dilute. H₂SO₄
(c) Either (a) or (b)
(d) Conc. H₂SO₄
Answer:
(c) Either (a) or (b)

Question 60.
Name the product formed when ethan- 1, 2-diol is treated with anhydrous ZnCl₂.
(a) Ethanol
(b) Ethene
(c) Ethane
(d) Ethanal
Answer:
(d) Ethanal

Question 61.
Which one of the following is formed when ethane – 1, 2 – diol is treated with Conc. H₂ SO₄?
(a) 1, 4 – dioxane
(b) Ethanal
(c) Ethanoic acid
(d) Ethene
Answer:
(a) 1, 4 – dioxane

Question 62.
Which one of the following is formed when ethylene glycol is treated with periodic acid?
(a) Methanal
(b) Methanol
(c) Ethanol
(d) Ethanal
Answer:
(a) Methanal

Question 63.
Identify the product formed when glycerol is treated with nitric acid and conc. H₂SO₄?
(a) Nitroglycerine
(b) Glyceryl triacetate
(c) Prop – 2 – enal
(d) Glyceric acid
Answer:
(a) Nitroglycerine

Question 64.
What will be the product formed when propan – 1, 2, 3 – triol is treated with KHSO₄?
(a) Nitroglycerine
(b) TNG
(c) Prop – 2 – enal
(d) Allyl alcohol
Answer:
(a) Nitroglycerine

Question 65.
Oxidation of glycerol with dil.HNO₃ gives ………
(a) Meso oxalic acid
(b) Glyceric acid and tartronic acid
(c) Glycerose
(d) Glyceraldehyde and dihydroxy acetone
Answer:
(b) Glyceric acid and tartronic acid

Question 66.
Oxidation of glycerol with Fenton reagent gives ………..
(a) Glyceraldehyde + Dihydroxy acetone
(b) Glyceric acid + Tartronic acid
(c) Meso oxalic acid
(d) Oxalic acid
Answer:
(a) Glyceraldehyde + Dihydroxy acetone

Question 67.
Which one of the following product is formed when glycerol is oxidised with acidified KMnO₄?
(a) Meso oxalic acid
(b) Oxalic acid
(c) Formic acid
(d) Glyceric acid
Answer:
(b) Oxalic acid

Question 68.
Which one of the following is used as a solvent for paints, varnishes and gum?
(a) Ethanol
(b) Methanol
(c) Methanal
(d) Ethanal
Answer:
(b) Methanol

Question 69.
Which one of the following is used as fuel for aeroplane?
(a) Methanol + Ethanol
(b) Ethanol + Petrol
(c) Ethanol + Propanol
(d) Butanol + Methanol
Answer:
(b) Ethanol + Petrol

Question 70.
Which one of the following is used as beverage as well as preservative for biological specimens?
(a) Ethanol
(b) Methanol
(c) Phenol
(d) Benzyl alcohol
Answer:
(a) Ethanol

Question 71.
Which one of the following is used as an anti-freezer in automobile radiator?
(a) Glycerol
(b) Phenol
(c) Benzyl alcohol
(d) Ethylene glycol
Answer:
(d) Ethylene glycol

Question 72.
Which one of the following is used as a sweetening agent in confectionery and beverages?
(a) Glycerol
(b) Phenol
(c) Benzyl alcohol
(d) Ethylene glycol
Answer:
(a) Glycerol

Question 73.
Which one of the following is used in the manufacture of cosmetics and transparent soaps?
(a) Methanol
(b) Ethanol
(c) Glycerol
(d) Phenol
Answer:
(c) Glycerol

Question 74.
Which one of the following is used in the manufacture of explosive dynamite and cordite by mixing it with clay?
(a) Glycol
(b) Glycerol
(c) Ethanol
(d) Benzaldehyde
Answer:
(b) Glycerol

Question 75.
Which alcohols is used in making printing inks and stamp pad ink?
(a) Glycol
(b) Ethanol
(c) Glycerol
(d) Phenol
Answer:
(c) Glycerol

Question 76.
Except which alcohol, other alcohols are weaker acid than water?
(a) Ethanol
(b) Phenol
(c) Methanol
(d) Propanol
Answer:
(c) Methanol

Question 77.
Which one of the following is the correct decreasing order of acidity in alcohol?
(a) 1° alcohol > 2° alcohol > 3° alcohol
(b) 3° alcohol > 2° alcohol> 1° alcohol
(c) 2° alcohol> 1° alcohol > 3° alcohol
(d) 3° alcohol > 1° alcohol > 2° alcohol
Answer:
(a) 1° alcohol > 2° alcohol > 3° alcohol

Question 78.
Which one of the following is more acidic?
(a) Benzyl alcohol
(b) Phenol
(c) Ethanol
(d) Methanol
Answer:
(b) Phenol

Question 79.
The JUPAC name of Phioroglucinol is ………….
(a) 4 – methyl phenol
(b) 1, 4 – dihydroxy benzene
(c) 1, 3, 5 – trihydroxy benzene
(d) 1, 2, 3 – trihydroxy benzene
Answer:
(c) 1, 3, 5 – trihydroxy benzene

Question 80.
The other name of 1 , 2, 3 – trihydroxy benzene is called ……….
(a) Pholoroglucinol
(b) Quinol
(c) Pyrogallol
(d) Hydroxy quinol
Answer:
(c) Pyrogallol

Question 81.
The other name of 3, 5 – dihydroxy toluene is known as …………
(a) Orcinol
(b) Quinol
(c) Pyrogallol
(d) Resorcinol
Answer:
(a) Orcinol

Question 82.
The IUPAC name of Catechol is known as ………..
(a) 1 , 3 – dihydroxy benzene
(b) 1, 2 – dihydroxy benzene
(c) 1, 4 – dihydroxy benzene
(d) 1, 3, 5 – trihydroxy benzene
Answer:
(b) 1, 2 – dihydroxy benzene

Question 83.
The name of

is …………
(a) Phloroglucinol
(b) pyrogallol
(c) Quinol
(d) Resorcinol
Answer:
(a) Phloroglucinol

Question 84.
The name of

(a) Pyrogallol
(b) Hydroxy cresol
(c) Orcinol
(d) Phloroglucinol
Answer:
(c) Orcinol