Class 11

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 7 Equilibrium Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 7 Equilibrium

PSEB 11th Class Chemistry Guide Equilibrium InText Questions and Answers

Question 1.
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(a) What is the initial effect of the change on vapour pressure?
(b) How do rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Answer:
(a) If the volume of the container is suddenly increa50sed, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

Question 2.
What is K_c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂] = 0.60M,[O₂] = 0.82M and [SO₃] = 1.90 M?
2SO₂(g) + O₂(g) ↔ 2SO₃(g)
Answer:
The given reaction is
2SO₂(g) + O₂(g) ↔ 2SO₃(g)
Equilibrium constant
K_c = \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}=\frac{(1.90 \mathrm{M})^{2}}{(0.60 \mathrm{M})^{2}(0.82 \mathrm{M})}\)
= 12.238 M^-1

Question 3.
At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms I₂(g) ⇌ 2I(g)
Calculate K_p for the equilibrium.
Answer:
Given, I₂(g) ⇌ 2I(g)
I atoms in iodine vapours = 40% by volume
So, iodine vapours of I₂ molecules = 60% by volume

Question 4.
Write the expression for the equilibrium constant, Kc for each of the following reactions:
(i) 2NOCl(g) ⇌ 2NO (g) + Cl₂(g)
(ii) 2CU(NO₃)₂(S) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g)
(iii) CH₃COOC₂H₅(oq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(ag)
(iv) Fe³⁺(aq) + 3OH^–(aq) ⇌ Fe(OH)₃(s)
(v) I₂(s) + 5F₂ ⇌ 2IF₅
Answer:

Question 5.
Find out the value of K_c for each of the following equilibria from the value of Kp.
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); K_p = 1.8 x 10^-2 at 500 K
(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K
Answer:
The relation between K_p and K_c is given as
K_p = K_c(RT)^Δn
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); K_p = 1.8 x 10^-2 at 500 K.
Δn = 3 – 2 = 1
R = 0.0831 bar L mol^-1K^-1
T = 500 K
K_p =1.8 x 10^-2
K_p = K_c( RT)^Δn
1.8 x 10^-2 = K_c(0.0831 x 500)¹
K_c = \(\frac{1.8 \times 10^{-2}}{0.0831 \times 500}\) = 4.33 x 10^-4

(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K
Δn = 2 -1 = 1
R = 0.0831 bar L mol^-1K^-1
T = 1073 K
K_p =167
Now, K_p = K_c(RT)^Δn
⇒ 167 = K_c(0.0831 x 1073)¹
⇒ K_c = \(\frac{167}{0.0831 \times 1073}\) = 1.87

Question 6.
For the following equilibriuih, K_c = 6.3 x 10¹⁴ at 1000 K
NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?
Answer:
It is given that K_c for the forward reaction is 6.3 x 10¹⁴ at 1000 K.
Then, K_c for the reverse reaction will be,
NO₂(g) + O₃(g) ⇌ NO(g) + O₃(g)
K_c = \(\frac{1}{K_{c}}=\frac{1}{6.3 \times 10^{14}}\) = 1.59 x 10^-15

Question 7.
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Answer:
For a pure substance (both solids and liquids),

Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.

Question 8.
Reaction between N₂ and O₂ takes place as follows:
2N₂(g) + O₂(g) ⇌ 2N₂O(g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 x 10^-37 , determine the composition of equilibrium mixture.
Answer:
Let the concentration of N₂O at equilibrium be x.
The given reaction is :

The value of equilibrium constant i.e., K_c = 2.0 x 10^-37 is very small which means negligible amounts of N₂ and O₂ react.

Question 9.
Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:
2NO(g) + Br₂(g) ⇌ 2NOBr(g)
When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO andBr₂.
Answer:
The balanced chemical equation is

Given, 2x = 0.0518
x = 0.0259 mol
Moles of NO at equilibrium = 0.087 – 2x
= 0.087-0.0518
= 0.0352 mol
Moles of Br2 at equilibrium = 0.0437 – x
= 0.0437 – 0.0259
= 0.0178 mol

Question 10.
At 450 K, K_p = 2.0 x 10¹⁰/bar for the given reaction at equilibrium.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
What is K_c at this temperature?
Answer:
The given reaction is
2SO₂(g) + O₂(g) ⇌ 2SO₃Cg)
Δn = 2 – 3 = -1
T = 450 K
R = 0.0831 bar L K^-1 mol^-1
K_p = 2.0 x 10¹⁰ bar^-1
We know that,
K_p = K_c(RT)^Δn
=> 2.0 x 10¹⁰ bar^-1 = k_c(0.0831 L bar K^-1 mol^-1 x 450 K)^-1
\(K_{c}=\frac{2.0 \times 10^{10} \mathrm{bar}^{-1}}{\left(0.0831 \mathrm{~L} \mathrm{barK} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 450 \mathrm{~K}\right)^{-1}}\)
K_c = (2.0 x 10¹⁰ bar^-1) (0.0831 L bar K^-1mol^-1450 K)
= 74.79 x 10¹⁰ L mol^-1 = 7.48 x 10¹¹ L mol^-1

Question 11.
A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is K_p for the given equilibrium?
2HI(g) ⇌ H₂(g) + I₂(g)
Answer:
The given reaction is

∵ Decrease is pressure of HI = 0.2 – 0.04 = 0.16 atm;
So equilibrium pressure of H₂ is \(\frac{0.16}{2}\) = 0.08 atm and for I₂ is \(\frac{0.16}{2}\) = 0.08 atm
as two moles of HI on dissociation gives 1 mol of H₂ and 1 mol of I₂.
Therefore,
K_p = \(\frac{p_{\mathrm{H}_{2}} \times p_{\mathrm{I}_{2}}}{\left(p_{\mathrm{HI}}\right)^{2}}=\frac{0.08 \times 0.08}{(0.04)^{2}}=\frac{0.0064}{0.0016}\) = 4.0
Hence, the value of K_p is 4.0.

Question 12.
A mixture of 1.57 mol of N₂,1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 1. 7 x 10².
Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Answer:
The given reaction is :
N₂(g) + 3H₂(g) 2NH₃(g)
Given, [N₂] = \(\frac{1.57}{20}\) = 0.0785 M
[H₂] = \(\frac{1.92}{20}\) = 0.096 M
[NH₃] = \(\frac{8.13}{20}\) = 0.4065 M
Now, reaction quotient Q_c. is :
Q_c = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}=\frac{(0.4065)^{2}}{(0.0785)(0.096)^{3}}\) = 2.4 x 10³M^-2
Since Q_c ≠ K_c the reaction mixture is not in equilibrium.
Again Q_c > K_c. Hence, the reaction will proceed in the reverse direction.

Question 13.
The equilibrium constant expression for a gas reaction is, K_c = \(\frac{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{5}}{\left[\mathrm{NO}^{4}\left[\mathrm{H}_{2} \mathrm{O}\right]^{6}\right.}\)
Write the balanced chemical equation corresponding to this expression.
Answer:
The balanced chemical equation corresponding to the given expression can be written as :
4NO(g) + 6H₂O(l) ⇌ 4NH₃(g) + 5O₂(g)

Question 14.
One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,
H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
Calculate the equilibrium constant for the reaction.
Answer:
The given reaction is

H₂O reacted = 40% of 1 mol of H₂O = 0.4 mol
x = 0.4 mol 1 – x = 1 – 0.4 = 0.6 mol
Therefore, the equilibrium constant for the reaction,
K_c = \(\) = 0.444

Question 15.
At 700 K, equilibrium constant for the reaction
H₂(g) + I₂(g) ⇌ 2HI(g)
is 54.8. If 0.5 mol L^-1 of Hl(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
Answer:
The given reaction is
H₂(g) + I₂(g) ⇌ 2HI(g); K_c = 54.8
Or the reaction
2HI(g) ⇌ H₂(g) + I₂(g); K_c‘ = \(\)
Given, [HI] = 0.5 mol L^-1
According to equation
[H₂] = [I₂] = x mol L^-1
Therefore,
\(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}{[\mathrm{HI}]^{2}}=K_{c}^{\prime}\)
⇒ \(\frac{x \times x}{(0.5)^{2}}=\frac{1}{54.8}\)
⇒ x² = \(\frac{0.25}{54.8}\)
⇒ x = 0.06754
x = 0.068 mol L^-1
Hence, at equilibrium, [H₂] = [I₂] = x = 0.068 mol L^-1

Question 16.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of IC1 was 0.78 M?
2ICl(g) ⇌ I₂(g) + Cl₂(g); K_c = 0.14
Answer:

Question 17.
K_p = 0.04 atm at9 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
Answer:

Question 18.
Ethyl acetate is formed by the reaction between ethanol acid and acetic acid and the equilibrium is represented as :
CH₃COOH(l) + C₂H₅OH (l) ⇌ CH₃COOC₂H₅(Z) + H₂O(l)
(i) Write the concentration ratio (reaction quotient), Q_c, for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Answer:

Question 19.
A sample of pure PCl₅ was introduced into an evacuated vessel at 473K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 x 10^-1 mol L^-1. If value of K is 8.3 x 10^-3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?
PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g)
Answer:
The given reaction is

It is given that the value of equilibrium constant, K = 8.3 x 10^-3.
K_c = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
[Given, [PCl₅]_equili = 0.5 x 10^-1 mol L^-1]
\(\frac{x \times x}{0.5 \times 10^{-1}}\) = 8.3 x 10^-3
⇒ x² = 4.15 x10^-4
⇒ x = 2.04 x 10^-2 = 0.0204 mol L^-1 = 0.02 mol L^-1
Therefore, at equilibrium,
[Pcl₃] = [Cl₂] = 0.02 mol L^-1

Question 20.
One of the reaction that takes place in producing steel from iron ore is the reduction of iron (H) oxide by carbon monoxide to give iron metal and CO₂.
FeO(s) + CO(g) ⇌ Fe(s) + CO₂ (g); K_p = 0.265 atm at 1050 K. What are the equilibrium partial pressures of CO and ₂ at 1050 K if the initial partial pressures are P_Co = 1.4 atm and pCO₂ = 0.80 atm?
Answer:
(i) The given reaction is

Since Q_p > K_p, the reaction will proceed in the backward direction.
Therefore, we can say that the pressure of CO will increase while the pressure of CO₂ will decrease.
Now,let the increase in pressure of CO = decrease in pressure of CO₂ be p.
Hence pCO₂ = 0.80 – p and P_CO = 1.4 + p
and K_p = \(\frac{p_{\mathrm{CO}_{2}}}{p_{\mathrm{CO}}}\)
0.265 = \(\frac{0.80-p}{1.4+p}\)
0.371 + 0.265p = 0.80 — p= 1.265p= 0.429
p = 0.339atm
Hence, at equilibrium
P_CO2 = 0.80 – 0.339 = 0.461 atm
And, equilibrium partial pressure of
P_CO = 1.4 + 0.339 = 1.739 atm.

Question 21.
Equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K is 0.06 1.
At a particular time, the analysis shows that composition of the
reaction mixture is 3.0 mol L^-1 N₂,2.0 mol L^-1 H₂ and 0.5 mol L^-1 NH₃ Is the reaction at equilibrium? if not in which direction
does the reaction tend to proceed to reach equilibrium?
Answer:
The given reaction is
N₂(g) + 3H₂(g) ⇌ 2NH₃(g);K_c = 0.061 at 500K
Given, [N₂] = 3.0mol L^-1, [H₂] = 2.0 mol L^-1, [NH₃] = 0.5 mol L^-1
So, Q = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}=\frac{(0.5)^{2}}{(3.0)(2.0)^{3}}\) = 0.0104
It is given that K_c = 0.06 1
Since Q_c ≠ K_c, the reaction is not at equilibrium.
Since Q_c < K_c, the reaction will proceed in the forward direction to reach equilibrium.

Question 22.
Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl(g) ⇌ Br₂(g) + Cl₂(g)
for which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 x 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium?
Answer:

x = 11.312(3.30 x 10^-3 – x)
x = 0.03732 – 11.312x
x + 11.312x = 0.03732
x = \(\frac{0.03732}{12.312}\)= 3.0321 x 10^-3 mol L^-1
[BrCl]_equili = (3.30 x 10^-3 – 3.032 x 10^-3) mol L^-1
= 2.68 x 10^-4 mol L^-1

Question 23.
At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass
C(s) + CO₂(g) ⇌ 2CO(g)
Calculate K_c for this reaction at the above temperature.
Answer:
Let the total mass of the gaseous mixture be 100g.
Mass of CO = 90.55 g
and, mass of CO₂ = (100 – 90.55) = 9.45 g
Now, number of moles of CO,
n_CO = \(\frac{90.55}{28}\) = 3.234 mol
(Molar mass of CO = 28 g mol^-1 )
Now, number of moles of CO₂,
n_CO = \(\)
(Molar mass of CO₂ = 44 g mol^-1 )

For the given reaction, Δn = 2 -1 = 1
We know that,
K_p = K_c(RT)^Δn
⇒ 14.19 = K_c(0.0831 x 1127)¹
⇒ K_c = 0.154 (approximately)

Question 24.
Calculate (a) \(\Delta \boldsymbol{G}^{\ominus}\) and (b) the equilibrium constant for the formation of N02 from NO and Oa at 298K
NO(g) + \(\frac{1}{2}\)O₂(g) ⇌ NO₂(g)
where \(\Delta_{f} \boldsymbol{G}^{\ominus}\) (N02)= 52.0 kJ/mol; \(\Delta_{f} \boldsymbol{G}^{\ominus}\) (NO) = 87.0kJ/mol;
\(\Delta_{f} \boldsymbol{G}^{\ominus}\) (O2) = 0 kJ/mol
Answer:
(a) The given reaction is
N0(g) + \(\frac{1}{2}\)O₂(g) ⇌ NO₂(g)

= (52.0 – 87.0 + \(\frac{1}{2}\) x 0 )kJ mol^-1 = -35.0 kJ mol^-1

(b) \(\Delta_{r} G^{\ominus}\) = – 2.303 RT logK_c
-35.0 = – 2.303 x 0.0831 x 298 log K_c
∴ log K_c= \(\frac{35}{5.7058}\)= 6.134
∴ K_c = antilog 6.134 = 1.361 x 10⁶.

Question 25.
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a) PCl₅(g) ⇌ PCl₃(g) +Cl₂(g)
(b) CaO (s) + CO₂ (g) ⇌ CaCO₃ (s)
(c) 3Fe(s) + 4H₂O (g) ⇌ Fe₃O₄ (s) + 4H₅(g)
Answer:
(a)The number of moles of reaction products will increase. According to Le-Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.
(b) The number of moles of reaction products will decrease.
(c) The number of moles of reaction products remains the same

Question 26.
Which of the following reactions will get affected hy increasing the pressure?
Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) COCl₂(g) ⇌ CO(g) +Cl₂(g)
(ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)
(iii) CO₂(g) + C(S) ⇌ 2CO(g)
(iv) 2H₂(g) +CO(g) ⇌ CH₃OH(g)
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Answer:
In all the above reactions, the reaction no. (ii) proceeds with the same no. of moles on both sides
i.e., n_p = n_r = 3 .
∴ This reaction will not be affected by the increase in pressure i. e., the direction of equilibrium will not be affected by the increase in pressure. All other reactions will be affected by the increase in pressure.
(i) COCl₂(g) ⇌ CO(g) +Cl₂(g)
n_p > n_r , n_p = 2; nr = 1
∴ Equilibrium will shift to the left increasing pressure.
(iii) CO₂(g) + C(S) ⇌ 2CO(g)
Here, n_r – 1; n_p = 2, therefore n_p > n_r
∴ Equilibrium will go to left on increase of pressure.
(iv) 2H₂(g) +CO(g) ⇌ CH₃OH(g)
Here, n_r = 3; n_p = 1 therefore n_p < n_r
∴ Equilibrium will shift to the right on increasing pressure.
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Here n_r = 0; n_p = 1, therefore n_p > n_r
∴ Equilibrium will shift backwards (left) on increasing the pressure.
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Here n_r = 9; n_p = 10, therefore n_p > n_r
∴ Equilibrium will shift backwards on increasing the pressure.

Question 27.
The equilibrium constant for the following reaction is 1.6 x 10⁵ at 1024 K.
H₂(g) + Br₂(g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.
Answer:
Given reaction is H₂(g) + Br₂(g)⇌ 2HBr(g); K_p = 1.6 x 10⁵ at 1024 K
Therefore, for the reaction 2HBr(g) ⇌ H₂(g)+Br₂(g), the equilibrium constant will be,

p = 2.5 x 1^-2(5.0 x 10^-3)p
p+(5.0 x 10^-3)p = 2.5 x 10^-2
(1005 x 10^-3)p = 2.5 x 10^-2
p = 2.49 x 10^-2 bar = 2.5 x 10^-2 bar
rherefore, at equilibrium,
[H₂] = [Br₂] = 2.49 x 10^-2 bar
[HBr] =10 — 2 x (2.49 x 10^-2) bar
= 9.95 bar = 10 bar

Question 28.
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
(a) Write an expression for K_p for the above reaction.
(b) How will the values of k_p and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) Using a catalyst?
Answer:
(a) The given reaction is
CH₄(g) + H₄O(g) ⇌ CO(g) + 3H₂(g)
\(K_{p}=\frac{p_{\mathrm{CO}} \times p_{\mathrm{H}_{2}}^{3}}{p_{\mathrm{CH}_{4}} \times p_{\mathrm{H}_{2} \mathrm{O}}}\)
(b) (1) According to LeChatelier’s principle, the equilibrium will shift in the backward direction.
(ii) According to Le-Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.
(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

Question 29.
Describe the effect of:
(a) addition of H₂
(b) addition of CH₃OH
(e) removal of CO (d) removal of CH₃OH
on the equilibrium of the reaction:
2H₂(g) + CO (g) ⇌ CH₃OH(g)
Answer:
2H₂(g) + CO(g) ⇌ CH₃OH(g)
According to Le Chatelier’s principle,
(a) Addition of H₂ (increase in concentration of reactants) shifts the equilibrium in forward direction (more product is formed).
(b) Addition of CH₃OH (increase in concentration of product) shifts the equilibrium in backward direction.
(c) Removal of CO also shifts the equilibrium in backward direction.
(d) Removal of CH₃OH shifts the equilibrium in forward direction.

Question 30.
At 473K, equilibrium constant K_c for decomposition of phosphorus pentachloride, PCl₅ is 8.3 x 10^-3. If decomposition is depicted as,
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = 1240 kJ mol^-1
(a) Write an expression for K_c for the reaction.
(b) What is the value of K_c for the reverse reaction at the same temperature?
(c) What would be the effect on K_c if
(i) more PCl₅ is added
(ii) pressure is increased?
(iii) the temperature is increased?
Answer:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); Kc = 8.3 x 10^-3
(a) K_c = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
(b) Value of K_c for the reverse reaction at the same temperature is
K’_c = \(\frac{1}{K_{c}}=\frac{1}{8.3 \times 10^{-3}}\) = 1.2048 x 10² = 120.48
(c) (i) Addition pf PCl₅ have no effect on K_c because K_c is constant at constant temperature.
(ii) K_c does not change with pressure.
(iii) The given reaction is endothermic, hence on increasing the temperature, K_c will increase.

Question 31.
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
CO(g) +H₂O (g) ⇌ CO₂(g) + H₂(g)
If a reaction vessel at 400° C is charged with an equimolar mixture of CO and steam such that p_CO = P_H2O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium? K_P = 10.1 at 400°C
Answer:
The given reaction is
CO(g) + H20(g) ⇌ C02(g) + H2(g)

p = 12.71 – 3.17p
4.17 p = 12.71
p = \(\frac{12.71}{4.17}\) = 3.04 bar
Hence PH₂ = 3.04 bar

Question 32.
Predict which of the following reaction will have appreciable concentration of reactants and products:
(a) Cl₂(g) ⇌ 2Cl(g);K_c = 5 x 10^-39
(b) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g); Kc = 3.7 x 10⁸
(c) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl(g); K_c = 1.8
Answer:
Following conclusions can be drawn from the values of K_c:
(a) Since the value of K_c is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of K_c is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of K_c is 1.8, this means that both the products and reactants have appreciable concentration.

Question 33.
The value of K_c for the reaction
3O₂(g) ⇌ 2O₃(g)
is 2.0 x 10^-50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 x 10^-2, what is the concentration of O₃?
Answer:
The given reaction is
3O₂(g) ⇌ 2O₃(g)
Then K
It is given that Kc = 2.0 x 10^-50 and [02(g)] = 1.6 x 10^-2
Then, we have,
\(2.0 \times 10^{-50}=\frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[1.6 \times 10^{-2}\right]^{3}}\)
⇒ [O₃]² = 2.0 x 10^-50 x (1.6 x 10^-2)³
⇒ [O₃]² = 8.192 x 10^-56
⇒ [O₃] = 2.86 x 10^-28 M
Hence, the concentration of O₃ is 2.86 x 10^-28 M.

Question 34.
The reaction, CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H20 and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.
Answer:
The given equation is
CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)
Therefore,
\(\frac{\left[\mathrm{CH}_{4}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]}{[\mathrm{CO}]\left[\mathrm{H}_{2}\right]^{3}}=K_{c}\)
Given, K_c = 3.90, [CO] = 0.30 mol, [H₂] = 0.10 mol and [H₂O] \(\frac{\left[\mathrm{CH}_{4}\right] \times 0.02}{0.3 \times(0.1)^{3}}\) = 3.90
[CH₄] = \(\frac{3.90 \times 0.3 \times(0.1)^{3}}{0.02}=\frac{0.00117}{0.02}\)
= 0.0585 M= 5.85 x 10^-2M
Hence, the concentration of CH4 at equilibrium is 5.85 x 10^-2 M.

Question 35.
What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species :
\(\mathrm{HNO}_{2}, \mathrm{CN}^{-}, \mathrm{HClO}_{4}, \mathrm{~F}^{-}, \mathrm{OH}^{-}, \mathrm{CO}_{3}^{2-} \text { and } \mathrm{S}^{2-}\)
Answe:
A conjugate acid-base pair is a pair that differs only by one proton.
The conjugate acid-base for the given species is mentioned in the table below:

Question 36.
Which of the followings are Lewis acids
\(\mathbf{H}_{2} \mathbf{O}, \mathbf{B F}_{3}, \mathrm{H}^{+} \text {and } \mathrm{NH}_{4}^{+}\)
Answer:
Lewis acids are those acids which can accept a pair of electrons. For example, BF₃, H⁺ and \(\mathrm{NH}_{4}^{+}\) are Lewis acids.

Question 37.
What will be the conjugate bases for the Bronsted acids : HF, H₂SO₄ and \(\mathrm{HCO}_{3}^{-}\)?
Answer:
The table below lists the conjugate bases for the given Bronsted acids :

Question 38.
Write the conjugate acids for the following Bronsted bases: \(\mathbf{N H}_{2}^{-}\), NH₃ andHCOC^–.
Answer:
The table below lists the conjugate acids for the given Bronsted bases : Bronsted base Conjugate acid

Question 39.
The species : H₂O, HCO^–₃, HSO^–₄ and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.
Answer:
The table below lists the conjugate acids and conjugate bases for the given species :

Question 40.
Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH^– (b)F^– (c)H⁺ (d) BCl₃
OH^– and F^– are electron rich species and can donate electron pair. Hence, these act as Lewis base.

H⁺ and BCl₃ are electron deficient species and can accept electron pair. Hence, these act as Lewis acid.

Question 41.
The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10^-3 M. What is its pH?
Answer:
Given,
[H⁺] = 3.8 x 10^-3 M
∴ pH value of soft drink = – log[H⁺] = – log(3.8 x 10^-3)
= – log3.8 – log10^-3 = – log3.8 + 3 log10
= – log3.8 + 3
= -0.58 + 3
= 2.42

Question 42.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer:
Given, pH = 3.76
We know that,
pH = – log[H⁺]
⇒ log[H⁺] = -pH
⇒ [H⁺] = antilog (-pH)
= antilog (-3.76) -1 +1 = antilog \(\overline{4} .24\) = 1.74 x 10-4 M Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74 x 10^-4 M.

Question 43.
The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 x 10^-4, 1.8 x 10^-4 and 4 8 x 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.
Answer:
If K_a is the ionization constant of a weak acid and Kb is the ionization constant of its conjugate base then K_a.K_b = K_w
or K_b = \(\frac{K_{w}}{K_{a}}\)
Given, K_a of HF = 6.8 x 10^-4
Hence, K_b of its conjugate base F^–
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{6.8 \times 10^{-4}}\)= 1.5 x 10^-11
(K_w = ionic product of water =1 x 10^-14 at 298 K)
Given, K_a of HCOOH = 1.8 x 10^-4
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-4}}\) = 5.6 x 10^-11
Hence, K_b of its coagulate base CN^–
Given, K_a of HCN = 4.8 x 10^-9
Hence, K_b of its coagulate base HCOO^–
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{4.8 \times 10^{-9}}\) = 2.08 x 10^-6

Question 44.
The ionization constant of phenol is 1.0 x 10^-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?
Answer:
Ionization of phenol :
C₆H₅OH + H₂O ⇌ C₆H₅O^– + H₃O⁺

Question 45.
The first ionization constant of H2S is 9.1 x 10^-8. (i) Calculate the concentration of HS^– ion in its 0.1 M solution. (ii) How will this concentration be affected if the solution is 0.1 M in HCI also? (ifi) If the second dissociation constant of H₂S is 1.2 x 10^-13, calculate the concentration of S^2- under both conditions.
Answer:

Hence, concentration of [HS^–] is decreased in the presence of 0.1 M
HCI due to common-ion effect.
(iii) For second dissociation constant,
HS + H₂O ⇌ H₃O⁺ + S^2- (In absence of HCl)
[HS^–] = 9.54 x 10^-5 M
\(K_{a_{2}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{S}^{2-}\right]}{\left[\mathrm{HS}^{-}\right]}\)

Question 46.
The ionization constant of acetic acid is 1.74 x 10^-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
Answer:
CH₃COOH CH₃COO^– + H⁺
K_a for CH₃COOH = 1.74 x 10^-5
[CH₃COOH] = c = 0.05 M
CH₃COOH CH₃COO^– + H⁺ [where α = degree of dissociation and c = molar concentration]

[CH₃COO^–] = 0.933 x 10^-3 = 9.33 x 10^-4 M
pH = – log[H⁺] = – log (9.33 x 10^-4)
= – (-4) – log9.3 = 4 – 0.9 = 3.03

Question 47.
It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_a.
Answer:
Let the organic acid be HA.
⇒ HA ⇌ H⁺ + A^–
Concentration of HA = 0.01 M
pH = 4.15
-log[H⁺] = pH= 4.15
log[H⁺] = – 4.15
log[H⁺] = 5.85
[H⁺] = antilog \(\overline{5} .85\)
= 7.080 x 10^-5
K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
Now, [H⁺] = [A^–] = 7.08 x 10^-5 M
Then K_a = \(
K_a = 5.01 x 10^-7
PK_a = – logK_a = – log(5.01 x 10^-7)
pK_a = 7 – 0.699 = 6.301

Question 48.
Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH
Answer:
(a) HCl (aq) ⇌ H⁺ (aq) + Cl^–(aq)
[HCl]= 0.003 M
As HC1 is completely dissociated into H⁺ ions
∴ [H⁺] = [HCl] = 0.003 M
pH = – log[H⁺] = – log [3 x 10^-3]
= 3 + (-0.4771) = 2.523
(b) NaOH(aq) ⇌ Na⁺(aq) + OH^– (aq)
[NaOH] = 0.005 = 5 x 10^-3 M
[OH^–] = [NaOH] = 5 x 10^-3 M
∴ [latex]\left[\mathrm{H}^{+}\right]=\frac{K_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{5.0 \times 10^{-3}}\)
[H⁺]= 2.0 x 10^-12
∴ pH = – log(2 x 10^-12) = – (-12) – log2
= 12 – 0.30 = 11.70
[log2 = 0.30]

(c)

[HBr] = 0.002 M
[H⁺]= [HBr] = 0.002 M= 2.0 x 10^-3 M
pH = – log[H⁺] = – log[2 x 10^-3]
=- (-3) – log2
= 3 – log2
= 3 – 0.3 = 2.70

(d)

[OH^–] = 0.002 M
[H⁺] = \(\frac{K_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.002}\) = 2 x 10^-12
pH = – log[H⁺] = -(-12) – log 5 = 12 – 0.70 = 11.30

Question 49.
Calculate the pH of the following solutions:
(a) 2 g of TIOH dissolved in water to give 2 litre of solution.
(b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 niL of solution.
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
(d) 1 mL of 13.6 M HC1 is diluted with water to give 1 litre of solution.
Answer:

(d) 1 mL of 13.6 M HC1 is diluted with water to give 1 litre of solution HC1 is completely dissociated to give H+ ions
[HCl] = ?
M₁V₁ = M₂V₂
1 mL of 13.6 M HCl = 1000 mL of M₂
M₂ = \(\frac{1 \times 13.6}{1000}\) = 0.0136 M
[HC1] = [H+] = 0.0136 M pH = – log[H+] = – log(1.36 x 10-2)
= – (-2) – log 1.36 = 2 – 0.13 = 1.87

Question 50.
The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid.
Answer:
α (Degree of ionization) = 0.132
c (molar cone.) = 0.1 M

∴ H⁺ = c x α = 0.1 x 0.132 = 0.0132
pH = – logH⁺ = – log(1.32 x 10^-2)
= – (-2) – log 1.32 = 2 – 0.12 = 1.88
pK_a = -logK_a
Now, K_a = cα²
K_a = 0.1 x (0.132)² = 1.74 x 10^-3
∴ pK_a = – log (1.74 x 10^-3) = – (-3) – log1.74 = 3 – 0.24 = 2.76

Question 51.
The pH of 0.005 M codeine (C₁₈H₂₁NO₃) solution is 9.95.
Calculate its ionization constant and pK_b.
Answer:
Molar cone, of codeine, c = 0.005 = 5 x 10^-3
pH = 9.95
pOH = 14 – 9.95 = 4.05 (∵ pH + pOH = 14)
pOH = – log [OH^–]
log[OH^–] = -4.05= \(\overline{5} .95\)
[OH^–] = antilog \(\overline{5} .95\)
= 8.91 x 10-5
k_b = \(\left(\frac{8.91 \times 10^{-5}}{5 \times 10^{-3}}\right)^{2}\) = 1.588 x 10^-6
pK_b = – logK_b = – log(1.588 x 10^-6)
= 6 + (-0.2009) = 5.7991 = 5.80

Question 52.
What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from table 7.7 (427 x 10^-10). Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer:
Given, K_b = 4.27 x 10^-10, c = 0.001 M
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-}\)

[0H] = 6.534 x 10^-7
pOH = — log(6.534 x 10^-7)
= 7+ (-0.8152)= 6.18
pH + pOH =14
pH 14—6.18=7.82

Thus, the ionization constant of the conjugate acid of aniline is 2.34 x 10^-5.

Question 53.
Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74.
How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl?
Answer:
PK_a = – log K_a,
4.74 =-logK_a
log K_a = -4.74 = \(\overline{5} .26\)
Ka = antilog \(\overline{5} .26\) = 1.82 x 10^-5

[CH₃COOH is a weak acid and HC1 is a strong acid, so we can assume
that (cα + 0.01) 0.01]

In the presence of strong acid, dissociation of weak acid i.e., CH₃COOH decreases due to common ion effect.

Q.54. The Ionization constant of dimethylanilne is 54 x 10.
Calculate Its degree of ionization in its 0.02 M solution. What
percentage of dimethylamine is ionized if the solution is also
0.1MInNaOH?
Ans. Given, Kb = 5.4 x 10
c=0.02M

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

Question 55.
Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below :
(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2
(c) Human blood, 7.38 (d) Human saliva, 6.4.
Answer:
(a) pH of Human muscle fluid = 6.83
pH = – log[H+] log[H⁺] = -6.83 = \(\overline{7} .17\)
[H⁺] = antilog \(\overline{7} .17\)
[H⁺] = 1.48 x 10^-7 M

(b) pH of Human stomach fluid =1.2
log[H⁺] = -1.2 = \(\overline{2} .80\)
[H⁺] = antilog \(\overline{2} .80\)
.-. [H⁺] = 6.3 x 10^-2 M

(c) pH of Human blood = 7.38
log[H⁺] = – 7.38 = \(\overline{8} .62\)
.-. [H⁺] = antilog \(\overline{8} .62\) = 4.17 x 10^-8 M

(d) pH of Human saliva = 6.4
log[H⁺] =-6.4 = \(\overline{7} .60\)
[H⁺] = antilog \(\overline{7} .60\) = 3.98 x 10^-7 M

Question 56.
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
Answer:
The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH = – log[H⁺]
(i) pH of milk = 6.8
Since, pH = -log[H⁺]
6.8 = -log[H+] log[H⁺] = -6.8 = \(\overline{7} .20\)
[H+] = antilog (\(\overline{7} .20\)) = 1.5 x 10~7 M

(ii) pH of black coffee = 5.0
Since, pH = – log[H⁺]
5.0 = – log[H+] log[H⁺] = – 5.0
[H⁺] = antilog (-5.00) = 10-5 M

(iii) pH of tomato juice = 4.2
Since, pH = – log[H⁺]
4.2 = – log[H⁺]
log[H⁺] = – 4.2 = \(\overline{5} .80\)
[H⁺] = antilog (\(\overline{5} .80\)) = 6.31 x 10^-5M

(iv) pH of lemon juice = 2.2
Since, pH = – log[H⁺]
2.2 = – log[H+]
log[H⁺] = -2.2 = \(\overline{3} .8\)
[H⁺] – antilog (\(\overline{3} .8\)) = 6.31 x 10^-3 M

(v) pH of egg white = 7.8
Since, pH = -log[H⁺]
7.8 = – log[H⁺]
log[H⁺]= -7.8 = \(\overline{8} .20\)
[H⁺] = antilog (\(\overline{8} .20\)) = 1.58 x 10^-8 M

Question 57.
0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
Answer:
Molar cone, of KOH = \(\frac{0.561 \times 1000}{56.1 \times 200}\) = 0.05M
56.1 x 200
KOH being a strong electrolyte, is completely ionized in aqueous solution.
KOH(aq) ⇌ K⁺(aq) + OH^–(aq)
[OH^–] = 0.05 M = [K⁺]
[H⁺][OH^–] = k_w
[H⁺] = \(\) = 2 x 10^-13
pH = – log[H⁺] = – log[2 x 10^-13]
= – (-13) – log2 = 13 – 0.03
∴ pH = 12.70

Question 58.
The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Answer:
Solubility of Sr(OH)₂ = 19.23 g/L
Then, concentration of Sr(OH)₂ = \(\frac{19.23}{121.63 \times 1}\) M = 0.1581 M
Sr(OH)₂(aq) Sr²⁺(aq) + 2(OH^–)(aq)
∴ [Sr²⁺] = 0.1581 M
[OH^–] – 2 x 0.1581 M = 0.3162 M
Now
K_w = [OH^–] [H⁺]
\(\frac{1 \times 10^{-14}}{0.3162}\) = [H⁺]
[H⁺] = 3.16 x 10^-14
pH = – log[H⁺]
pH = 14 – 0.4997 = 13.5003 ≈ 13.5

Question 60.
The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Answer:
Given, pH = 2.34
Molar cone, (c) = 0.1 M
HCNO H⁺ + CNO^–
pH = – log[H⁺]
2.34 = -log[H⁺]
log[H⁺] =-2.34 = \(\overline{3} .66\)
.-. [H⁺] = antilog \(\overline{3} .66\) = 4.57 x 10^-3 M
[H⁺] = \(\sqrt{K_{a}^{c}}\)
4.57 x 10^-3 = \(\sqrt{K_{a}^{c}}\)
Ionization constant,
K_a = 2.088 x 10^-4
Degree of ionization α = \(\sqrt{\frac{K_{a}}{c}}=\sqrt{\frac{2.088 \times 10^{-14}}{0.1}}\)
α = 0.0457

Question 61.
The ionization constant of nitrous acid is 45 x 104. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Answer:
Hydrolysis constant K_h = \(\frac{K_{w}}{K_{a}}\)
where K_w = Ionic product of water, K_a = Ionisation constant of the acid

pOH = -log(9.42 x 10^-7)= 7-0.97= 6.03
∴ pH = 14 – pOH = 14 – 6.03 = 7.97

Question 62.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Answer:
Given, pH = 3.44
We know that,
PH = – log[H⁺]
.-. [H⁺]= 3.63 x 10^-4
Then K_h = \(\frac{\left(3.63 \times 10^{-4}\right)^{2}}{0.02}\) (Concentration = 0.02M)
=> K_h = 6.6 x 10^-6
Now, K_h = \(\frac{K_{w}}{K_{a}}\)
K_a = \(\frac{K_{w}}{K_{h}}=\frac{1 \times 10^{-14}}{6.6 \times 10^{6}}\)
= 1.51 x 10^-9

Question 63.
Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF
Answer:

Question 64.
The ionization constant of chloroacetic acid is 1.35 x 10^-3 . What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?
Answer:
Given that Ka = 1.35 x 10^-3.
=> Ka = cα²
α = \(\sqrt{\frac{K_{a}}{c}}=\sqrt{\frac{1.35 \times 10^{-3}}{0.1}}\)
(∵ Concentration of acid = 0.1 M)
= \(\sqrt{1.35 \times 10^{-2}}\) =0.116
.-. [H⁺]= cα = 0.1 x 0.116 = 0.0116
=> pH = – log[H⁺] = – log[0.0116] = 1.94
To find pH of 0.1 M sodium salt, we use the formula
pH = – \(\frac{1}{2}\)[log C_w + log K_a – log c]
= –\(\frac{1}{2}\)[log1 x 10^-14 + log(1.35 x 10^-3) – log(0.1)]
= –\(\frac{1}{2}\)[-14 + (-3 + 0.1303) – (-1)]
= – \(\frac{1}{2}\) [-15.8697] = 7.93485 ≈ 7.94

Question 65.
Ionic product of water at 310 K is 2.7 x 10^-14. What is the pH of neutral water at this temperature?
Answer:
Ionic product,
K_w = [H₃O⁺] [OH^–]
= 2.7 x 10^-14 at 310 K
H₂O + H₂O *=* [H₃0⁺][OH^–]
[H₃0⁺]= [OH^–]
Therefore, [H₃0⁺] = \(\sqrt{2.7 \times 10^{-14}}\)
⇒ = 1.64 x 10^-7 M
⇒ [H₃0⁺] = 1.64 x 10^-7
⇒ pH = – log[H₃0⁺] = – log[1.64 x 10^-7 = 6.78
Hence, the pH of neutral water is 6.78.

Question 66.
Calculate the pH of the resultant mixtures :
(a) 10 mL of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HC1
(b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂
(c) 10 mL of 0.1 M H₂SO₄ + 10 mL of 0.1 M KOH
Answer:

Question 67.
Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants.
[K_sp(Ag₂CrO₄) = 1.1 x 10^-12, K_sp(BaCrO₄) = 1.2 x 10^-10,
K_sp[Fe(OH)₃] = 1.0 x 10^-38, K_sp(PbCl₂) = 1.6 x 10^-5
K_sp(Hg₂I₂) = 4.5 x 10^-29]
Determine also the molarities of individual ions.
Answer:
(i) Silver chromate : Ag₂CrO₄⇌ 2Ag⁺ + CrO₄^2-, K_sp = 1.1 x 10^-12
Then, K_sp = [Ag⁺]²[CrO₄^2-]
Let the solubility of Ag₂CrO₄ be s.
⇒ [Ag⁺] = 2s and [CrO₄^2-] = s
Then, K_sp = (2s)² s – 4s³
⇒ 1.1 x 10^-12 = 4s³
0.275 x 10^-12 = s³
s = 0.65 x 10^-4 M
Molarity of Ag⁺ = 2s = 2 x 0.65 x 10^-4
= 1.30 x 10^-4 M
Molarity of CrO₄^2- = s = 0.65 x 10^-4 M

(ii) Barium chromate : BaCrO₄ ⇌ Ba²⁺ + \(\mathrm{CrO}_{4}^{2-}\); K_sp = 1.2 x 10^-10
Then, K_sp = [Ba²⁺] [latex]\mathrm{CrO}_{4}^{2-}[/latex]
Let s be the solubility of BaCrO₄.
⇒ [Ba²⁺] = s and [latex]\mathrm{CrO}_{4}^{2-}[/latex] = s
⇒ K_sp = s²
⇒ 1.2 x 10^-10 = s²
⇒ s = 1.09 x 10^-5 M
Molarity of [Ba²⁺] = Molarity of [latex]\mathrm{CrO}_{4}^{2-}[/latex] = s = 1.09 x 10^-5 M

(iii) Ferric hydroxide: Fe(OH)₃ ⇌ Fe²⁺ + 3OH^–; K_sp = 1.0.x 10^-38

K_sp = [Fe²⁺][OH^–]³
Let s be the solubility of Fe(OH)₃
⇒ [Fe³⁺] = s and [OH^–] = 3s
⇒ K_sp = s. (3s)³ = s x 27x³
K_sp = 27x⁴
1.0 x 10^-38 = 27x⁴
0.037 x 10^-38 = s⁴
0.00037 x 10^-36 = s⁴
s = 1.39 x 10^-10 M
Molarity of [Fe³⁺] = s = 1.39 x 10^-10 M
Molarity of [OH^–] = 3s = 4.17 x 10^-10 M

(iv) Lead chloride : PbCl₂ ⇌ Pb²⁺ + 2Cl^–; K_sp = 1.6 x 10^-5
K_sp = [Pb²⁺][Cl^–]²
Let s be the solubility of PbCl₂.
⇒ [Pb²⁺] = s and [Cl^–] = 2s
Thus, K_sp = s. (2s)² = 4s³
⇒ 1.6 x 10^-5 = 4s³
⇒ 0.4 x 10^-5 = s³
4 x 10^-6 = s³
⇒ s = 1.59 x 10^-2 M
Molarity of [Pb²⁺] = s = 1.59 x 10^-2 M
Molarity of [Cl^–] = 2s = 3.18 x 10^-2 M

(v) Mercurous iodide : Hg₂I₂ ⇌ \(\mathrm{Hg}_{2}^{2+}\) + = 4.5 x 10^-29
K_sp = [\(\mathrm{Hg}_{2}^{2+}\) ] []I^–]
Let s be the solubility of [Hg²I²]
⇒ [Hg₂] = s and [I^–] = 2s
K_sp = (s).(2s)² = 4s³
⇒ 4s³ = 4.5 x 10^-29
⇒ s³ = 1.125 x 10^-29
s = 2.24 x 10^-10 M
Molarity of [ \(\mathrm{Hg}_{2}^{2+}\) ] = s = 2.24 x 10^-10 M
Molarity of [I^–] = 2s = 4.48 x 10^-10 M.

Question 68.
The solubility product constant of Ag₂CrO₄ and AgBr are
1.1 x 10^-12 and 5.0 x 10^-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer:
Let s be the solubility of Ag₂CrO₄
Thus, Ag₂CrO₄ ⇌ 2Ag²⁺ + \(\mathrm{CrO}_{4}^{-}\); K_sp = 1.1 x 10^-12
K_sp = [Ag²⁺]². [latex]\mathrm{CrO}_{4}^{-}[/latex]
=> [Ag²⁺] = (2s)² and [latex]\mathrm{CrO}_{4}^{-}[/latex] = s
K_sp = (2s)². s= 4s³
1.1 x 10^-12 = 4s³
s = 6.5 x 10^-5 M

Let s be the solubility of AgBr.
AgBr(s) ⇌ Ag⁺ + Br^–; K_sp = 5.0 x 10^-13
K_sp = s² = 5.0 x 10^-13
s = \(\sqrt{5.0 \times 10^{-13}}\)
∴ s = 7.07 x 10^-7 M

Therefore, the ratio of the molarities of their saturated solution is
\(\frac{s\left(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\right)}{s(\mathrm{AgBr})}=\frac{6.5 \times 10^{-5} \mathrm{M}}{7.07 \times 10^{-7} \mathrm{M}}\) = 9.19

Question 69.
Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, K_sp = 7.4 x 10^-8).
Ans. 2NaIO₃ + Cu(ClO₃)3 → 2NaClO₃ + Cu(IO₃)₂
Molar cone, of both solutions before mixing = 0.002 M
Molar cone, of both solution after mixing
\(\left[\mathrm{IO}_{3}^{-}\right]=\left[\mathrm{Cu}^{2+}\right]=\frac{0.002}{2}\)= 0.001 M
Cu(IO₃)₂ ⇌ Cu²⁺ + \(2 \mathrm{IO}_{3}^{-}\)
[Cu²⁺] = 0.001 M
\(\) = 0.001 M
Ionic product = [Cu²⁺]\(\left[\mathrm{IO}_{3}^{-}\right]^{2}\)
– 1 x 10^-3 x [1 x 10^-3]² = 1 x 10^-9
K_sp = 7.4 x 10^-8
Cu(IO₃)₂ is precipitated if [Cu²⁺] .\(\left[\mathrm{IO}_{3}^{-}\right]^{2}\) > K_sp
Since, the ionic product is less than the solubility product. Hence there will be no precipitation.

Question 70.
The ionization constant of benzoic acid is 6.46 x 10^-5 and Ksp for silver benzoate is 2.5 x 10 . How many times is silver benzoate
more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
Answer:

where s is the solubility of C₆H₅COOAg
pH = 3.19
pH = -log[H⁺]
log[H⁺] = – pH = -3.19 = \(\overline{4} .81\)
[H⁺] = antilog \(\overline{4} .81\) = 6.46 x 10^-4

C₆H₅COOAg is 3.2 times more soluble in buffer than in pure water.

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 7 Equilibrium Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 7 Equilibrium

Very Short Answer Type Questions

Question 1.
A tank is full of water. Water is coming in as well as going out at same rate. What will happen to level of water in a tank? What is name given to such state?
Answer:
It will remain the same because rate of inflow is equal to rate of outflow. This state is called state of ‘equilibrium’.

Question 2.
The ionization of hydrogen chloride in water is given t
HCl(aq) + H₂O(l) ⇌ H₃O⁺+(aq) + Cl^–(aq)
Label two conjugate acid-base pairs in this ionization.
Answer:

Question 3.
Why solution of sugar in water does not conduct electricity whereas that of common salt in water does?
Answer:
Common salt (NaCl) is an electrolyte which gives Na⁺ and Cl^– ions in the aqueous solution. Hence, it conducts electricity. Sugar is sucrose (C₁₂H₂₂O₁₁) which is a non-electrolyte and does not give ions in the solution. Hence, it does not conduct electricity.

Question 4.
Why is ammonia termed as a base though it does not contain OH^– ions?
Answer:
Ammonia is termed as a base due to its tendency to donate electron pair. Therefore it is a Lewis base.

Question 5.
K_b for NH₄O, H is 1.8 x 10^-5 and for CH₃NH₂ is 44 x 10^-4. Which of them is strongest base and why?
Answer:
CH₃NH₂ is strongest base because it has high value of base dissociation constant.

Question 6.
pK_a value of acids A, B, C, D are 1.5, 3.5, 2.0 and 5.0. Which of them is strongest acid?
Answer:
Acid A with pK_a = 1.5 is strongest acid, lower the value of pK_a stronger will be the acid.

Question 7.
What will be the pH of 1M Na₂SO₄ solution?
Answer:
Na₂SO₄ is salt of strong acid and strong base, thus its aqueous solution will be neutral. Therefore, its pH will be 7.

Question 8.
Is it possible to get precipitate of Fe(OH)₃ at pH = 2? Give reason.
Answer:
No, because Fe(OH)₃ will dissolve in strongly acidic medium.

Question 9.
What happens to ionic product of water if some acid is added to it?
Answer:
Ionic product will remain unchanged.

Question 10.
How does common ion affect the solubility of electrolyte?
Answer:
Solubility of electrolyte decreases due to common ion effect.

Short Answer Type Questions

Question 1.
A certain buffer is made by mixing sodium form ate and formic acid in water. With the help of equations explain how this buffer neutralises addition of a small amount of an acid or a base?
Answer:
HCOONa → HCOO^– + Na⁺
HCOOH ⇌ HCOO^– + H⁺

HCOO^– is common ion in the above acidic buffer. When small amount of H⁺ ions is added, these H⁺ ions combine with HCOO^– which are in excess to form HCOOH back and [H⁺] remains practically same, so pH remains constant. When small amount of OH^– ions are added, OH^– ions will take up H⁺ and association of HCOOH will increase so as to maintain concentration of H⁺ ions. So, pH would not be affected.

Question 2.
How much volume of 0.1 M CH₃COOH should he added to 50 ml of 0.2 M CH₃COONa solution to prepare a buffer solution of pH 4.91. (pA_a of AcH is 4.76).
According to Henderson’s equation

Required volume of 0.1 M acetic acid = 70.92 mL

Question 3.
Some processes are given below. What happens to the process if it is subjected to a change given in the brackets?

(ii) Dissolution of NaOH in water (Temperature is increased)
(iii) N₂(g) + O₂(g) ⇌ 2NO(g) -180.7 kJ (Pressure is increased and temperature is decreased.)
Answer:
(i) Equilibrium will shift in the forward direction, i.e., more ice will melt.
(ii) Solubility will decrease because it is an exothermic process.
(iii) Pressure has no effect. Decrease of temperature will shift the equilibrium in the backward direction.

Question 4.
50.0 g of CaCO₃ are heated to 1073 K in a 5 L vessel. What percent of the CaCO₃ would decompose at equilibrium? K_p for the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g) is 1.15 atm at 1073 K.
Answer:
The reaction is : CaCO₃(s) ⇌ CaO(s) + CO₂(g)
K_p = P_Co2 = 1.15 atm, pV = nRT
\(\mathrm{n}_{\mathrm{CO}_{2}}=\frac{p_{\mathrm{CO}_{2}} \mathrm{~V}}{R T}=\frac{1.15 \times 5}{0.082 \times 1073}\) = 0.065 mol

1 mole of CO₂ is obtained by decomposition of 1 mole CaCO₃. Therefore, moles of CaCO₃ decomposed is equal to the moles of CO₂ = 0.065 mol.
Mole of CaCO₃ initially present = \(\frac{50}{100}\) = 0.5 mol
[Molecular mass of CaCO₃ = 100]
Per cent of CaCO₃ decomposed = \(\frac{0.065}{0.5}\) x 100 = 13%

Question 5.
Arrange the following in increasing order of pH.
KNO₃(aqr), CH₃COONa(aq), NH₄Cl(aq), C₆H₅COONH₄(aq)
Answer:
(i) KNO₃ is a salt of strong acid-strong base, hence its aqueous solution is neutral; pH = 7
(ii) CH₃COONa is a salt of weak acid and strong base, hence, its aqueous solution is basic; pH < 7.
(iii) NH₄Cl is a salt of strong acid and weak base, hence its aqueous solution is acidic; pH < 7.
(iv) C₆H₅COONH₄ is a salt of weak acid, C₆H₅COOH and weak base, NH4OH. ButNH4OH is slightly stronger than C6H5COOH. Hence, pH is slightly greater than 7.
Therefore, increasing order of pH of the given salts is,
NH₄Cl < KNO₃ < C₆H₅COONH₄ < CH₃COONa

Long Answer Type Questions

Question 1.
Calculate the pH of a buffer which is 0.1 M in acetic acid and 0.15 M in sodium acetate. Given that the ionisation constants of acetic acid is 1.75 x 10^-5. Also calculate the change in pH of the buffer if the following adds in 1 L of the buffer (i) 1 cc of 1 M NaOH. (ii) 1 cc of 1 M HC1. Assume that the charge in volume is negligible, (iii) What will be the buffer index of the above buffer?
Answer:
pH = pK_a + log\(\frac{Salt}{Acid}\) = – log(1.75 x 10^-5) + log
\(\frac{0.15}{0.10}\)
= (5 – 0.2430) + 0.1761 = 4.757 + 0.1761 = 4.933.

(i) 1 cc of 1M NaOH contains NaOH = 10^-3 mol. This will convert 10^-3 mol of acetic acid into the salt so that salt formed = 10^-3 mol.
[Acid] = 0.10 – 0.001 = 0.099 M
[Salt] = 0.15 + 0.001 = 0.151 M
pH =. 4.757 + log \(\frac{0.151}{0.099}\)
= 4.757 + 0.183 = 4.940
∴ Increase in pH = 4.940 – 4.933 = 0.007 which is negligible.

(ii) 1 cc of 1 M HC1 contains HCl = 1CF₃ mol. This will convert 10^-3 mol CH₃COONa into CH₃COOH.
Now, [Acid] = 0.10 + 0.001 = 0.101 M
[Salt] = 0.15 – 0.001 = 0.149 M 0.149
∴ pH = 4.757 + log\(\frac{0.149}{0.101}\) = 4.757 + 0.169 = 4.925
∴ Decrease in pH = 4.933 = 0.007 which is again negligible.

(iii) Calculation of buffer index No. of moles of HC1 or NaOH added = 0.001 mol
Change in pH = 0.007
Hence, buffer index = \(\frac{\Delta n}{\Delta \mathrm{pH}}=\frac{0.001}{0.007}=\frac{1}{7}\)= 0.143

Question 2.
On the basis of Le-Chatelier’s principle, explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction?
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
What will be the effect of addition of argon to the above reaction mixture at constant volume?
Answer:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92.38 kJ mol^-1

It is an exothermic process. According to Le-Chatelier’s principle, low temperature is favourable for high yield of ammonia, but practically very low temperatures slow down the reaction. So, optimum temperature, 700 K is favourable in attainment of equilibrium.

Similarly, high pressure about 200 atm is favourable for high yield of ammonia. On increasing pressure, reaction goes in the forward direction because the number of moles decreases in the forward direction.

At constant volume, addition of argon does not affect the equilibrium because it does not change the partial pressures of the reactants or products involved in the reaction and the equilibrium remains undisturbed.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 4 Chemical Bonding and Molecular Structure Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

PSEB 11th Class Chemistry Guide Chemical Bonding and Molecular Structure InText Questions and Answers

Question 1.
Explain the formation of a chemical bond.
Answer:
A chemical bond is defined as an attractive force that holds the constituents (atoms, ions etc.) together in a chemical species.
Various theories have been suggested for the formation of chemical bonds such as the electronic theory, valence shell electron pair repulsion theory, valence bond theory, and molecular orbital theory.

A chemical bond formation is attributed to the tendency of a system to attain stability. It was observed that the inertness of noble gases was because of their fully filled outermost orbitals. Hence, it was postulated that the elements having incomplete outermost shells are unstable (reactive). Atoms, therefore, combine with each other and complete their respective octets or duplets to attain the stable configuration of the nearest noble gases. This combination can occur either by sharing of electrons or by transferring one or more electrons from one atom to another. The chemical bond formed as a result of sharing of electrons between atoms is called a covalent bond. An ionic bond is formed as a result of the transference of electrons from one atom to another.

Question 2.
Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
Answer:
Mg : There are two valence electrons in Mg atom (2, 8, 2). Hence, the Lewis dot symbol of Mg is :

Na : There is only one valence electron in an atom of sodium (2, 8,1). Hence, the Lewis dot Symbol is :

B : There are three valence electrons in Boron atom (2, 3). Hence, the Lewis dot symbol is :

0 : There are six valence electrons in an atom of oxygen (2, 6). Hence, the Lewis dot symbol is :

N: There are five valence electrons in an atom of nitrogen (2,5). Hence, the Lewis dot symbol is :

Br : There are seven valence electrons in bromine (2, 8, 18, 7). Hence, the
Lewis dot symbol is :

Question 3.
Write Lewis symbols for the following atoms and ions: S and S^2-; Al and Al³⁺; H and H^–
Answer:

Question 4.
Draw the Lewis structure for the following molecules and ions :
H₂S, SiCl₄, BeF₂, \(\mathrm{CO}_{3}^{2-}\), HCOOH
Answer:

Question 5.
Define octet rule. Write its significance and limitations.
Answer:
Octet rule : Atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shell.
Significance of octet rule : It help to explain why different atom combine with each other to form ionic or covalent compounds. Limitations of the Octet Rule
Although octet rule can explain the formation of a large number of compounds but it has many exceptions also, which are discussed below :

(i) Electron deficient molecules : There are some molecules in which the central atom is surrounded by less than eight electrons, i.e., their octet is incomplete. Elements having less than four valence electrons generally form molecules of this category.
e.g., BeCl₂, BF₃, AlCl₃, LiCl, BeH₂ etc.

(ii) Odd electron molecules : Molecules like NO, NO₂, O₂ etc., are examples of such molecules in which bonded atoms have odd number of electron (usually 3) in between them. That’s why these are called odd electron molecules.
In case of these molecules, the octet rule is not satisfied for all the atoms, e.g.,

Species with one unpaired electron are called free radicals. These are paramagnetic and most of them are generally unstable.

(iii) Electron rich molecules : Elements of the third and higher periods of the periodic table, because of the availability of d orbitals can expand their covalency and -can accommodate more than eight valence electrons around the central atom. This is referred as expanded octet. Here, also the octet rule is not applicable, e.g., PF₅ (10 electrons around P atom), SF₆ (12 electrons around S atom), H₂SO₄ (12 electron around S atom).
Compounds having expanded octet are also termed as hypervalent compounds.

(iv) Other drawbacks : Other drawbacks of this theory are as follows:
1. Octet rule is based on the inertness of noble gases but some noble gases like xenon and krypton form several compounds with oxygen and fluorine like. XeF₂, XeF₄, XeF₆, XeOF₄, XeO₂F₂, KrF₂ etc.
2. It does not tell anything about the shapes of molecules and their relative stabilities.
3. It fails to explain the paramagnetic behaviour of oxygen. (Which should be diamagnetic according to this rule but it is infact paramagnetic in nature).

Question 6.
Write the favourable factors for the formation of ionic bond.
Answer:
The favourable factors for ionic bond formation are as follows :

(i) Low ionization enthalpy of element forming cation.
(ii) More negative electron gain enthalpy of element forming anion.
(iii) High lattice energy of the compound formed.

Question 7.
Discuss the shape of the following molecules using the VSEPR model:
BeCl₂, BCl₃, SiCl₄, ASF₅, H₂S, PH₃
Answer:
According to VSEPR theory, the shape of a molecule depends upon the number of valence shell electron pairs (bonded or non-bonded) around the central atom. Pairs of electrons in the valence shell repel each other. The order of their repulsion is as follows :
Ip -Ip > Ip -bp > bp – bp
(i) BeCl₂ or

The central atom Be has only 2 valence electrons which are bonded to Cl, so there are only 2 bond pairs and no lone pairs. It is of the type AB₂ and hence, the shape is linear.

(ii) BCl₃:

The central atom B has only 3 valence electrons which are bonded with three Cl atoms, so it contains only 3 bond pairs and no lone pair. It is of the type AB₃ and hence, the shape is trigonal planar.

(iii) SiCl₄ :

Similarly, the central atom Si has only 4 bond pairs and no lone pair. It is of the type AB₄ and hence, the shape is tetrahedral.

(iv) AsF₅:

The central atom As has only 5 bond pairs and no lone pair. It is of the type AB₅ and hence, the shape is trigonal bipyramidal.

(v) H₂S:

The central atom S has 6 valence electrons. Out of these only two are used in bond formation with two H-atoms while four (two pairs) remains as non-bonding electrons (i.e., lone pairs). So, it contains 2 bond pairs and 2 lone pairs. It is of the type AB₂E₂ and hence, the shape
is bent or V-shaped.

(vi) PH₃

The central atom P has 5 valence electrons. Out of which three are utilised in bonding with H atoms and one pair remains as lone pair. So, it contains 3 bond pairs and one lone pair. It is of the type AB₃E and hence the shape is pyramidal.

Question 8.
Although geometries of NH₃ and H₂O molecules are distorted
tetrahedral, bond angle in water is less than that of ammonia.
Discuss.
Answer:

In H₂O molecule there, is lone pair-lone pair repulsion due to the presence of two lone pairs of electrons while in NH₃ molecule there are only lone pair-bond pair repulsion. According to VSEPR theory the former one is more stronger and hence the bond angle in water is less than that of ammonia.

Question 9.
How do you express the bond strength in terms of bond order?
Answer:
Bond strength represents the extent of bonding between two atoms forming a molecule. The larger the bond energy, the stronger is the bond and the greater is the bond order.

Question 10.
Define the bond length.
Answer:
Bond length is defined as the equilibrium distance between the nuclei of two bonded atoms in a molecule.
Bond lengths are expressed in terms of Angstrom (10^-10 m) or picometer (10^-12 m) and are measured by spectroscopic X-ray diffractions and |
electron-diffraction techniques.

Question 11.
Explain the important aspects of resonance with reference to \(\mathrm{CO}_{3}^{2-}\) the ion.
Answer:
According to experimental findings, all carbon to oxygen bonds in \(\mathrm{CO}_{3}^{2-}\) are equivalent. Hence, it is inadequate to represent \(\mathrm{CO}_{3}^{2-}\) ion by a single Lewis structure having two single bonds and one double bond.
The \(\mathrm{CO}_{3}^{2-}\) ion is best described as a resonance hybrid of the canonical forms I, II and III.

All canonical forms have similar energy, same positions of atoms and same number of bonded and non-bonded pairs of electrons.

Question 12.
H₃PO₃ can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃? If not, give reasons for the same.

Answer:
The given structures cannot be taken as the canonical forms of the resonance hybrid of H₃PO₃, because the positions of the atoms have been changed.

Question 13.
Write the resonance structures for SO₃, NO₂ and \(\mathrm{NO}_{3}^{-}\)
Answer:
The resonance structures are :
(a) SO₃:

(b) NO₂:

(c) \(\mathrm{NO}_{3}^{-}\)

Question 14.
Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and S (b) Ca and O (c) Al and N.
Answer:
(a) K and S :
The electronic configurations of K and S are as follows :
K : 2, 8, 8, 1
S : 2, 8, 6

Sulphur (S) requires 2 more electrons to complete its octet. Potassium (K) requires one electron more than the nearest noble gas i.e., Argon. Hence, the electron transfer can be shown as:

(b) Ca and O :
The electronic configurations of Ca and O are as follows :
Ca : 2, 8, 8, 2
O : 2, 6
Oxygen requires two electrons more to complete its octet, whereas calcium has two electrons more than the nearest noble gas i.e., Argon. Hence, the electron transfer takes place as :

(c) Al and N :
The electronic configurations of Al and N are as follows :
A1: 2, 8, 3
N : 2, 5
Nitrogen is three electrons short of the nearest noble gas (Neon), whereas aluminium has three electrons more than neon. Hence, the electron transference can be shown as :

Question 15.
Although both CO₂ and H₂O are triatomic molecules, the shape of H₂O molecule is bent while that of CO₂ is linear. Explain this on the basis of dipole moment.
Answer:
According to experimental results, the dipole moment of carbon dioxide is zero. This is possible only if the molecule is linear so that the dipole moments of C—O bonds are equal and opposite to nullify each other.

Resultant µ = 0 D
On the other hand, H₂O molecule is found to have a net dipole moment value of 1.84 D (thoughit is a triatomic molecule as CO₂). The value of the dipole moment suggests that the structure of H20 molecule is bent where the dipole moment of O—H bonds are unequal.

Question 16.
Write the significance/applications of dipole moment.
Answer:
The applications of dipole moment are as follows :
1. In determining the polarity of bonds : As µ = e x d, obviously greater is the magnitude of dipole moment, higher will be the polarity of the bond. This is applicable to molecules containing only one polar bond like HC1, HBr etc. In non-polar molecules like, H₂, O₂, N₂ the dipole moment is zero. It is because there is no charge separations in these molecules [e = 0]. Thus, dipole moment can also be used to distinguish between polar and non-polar molecules.

2. In the calculation of percentage ionic character :
Take the example of HCl. Its µ = 1.03 D
If HCl is 100% ionic, each end would carry charge of one unit
i.e., 4.8 x 10^-10 e.s.u.
d (bond length) in H—Cl = 1.275Å
∴ for 100% ionic character, dipole moment will be
µ_ionic = e x d
= 4.8 x 10^-10 e.s.u x 1.275 x 10^-8 cm
= 6.12 x 10^-18 e.s.u cm = 6.12 D
∴ Percentage of ionic character =

= \(\frac{1.03}{6.12}\) x 100 = 16.83.

3. In determining the symmetry (or shape) of the molecules : Dipole moment is an important property in determining the shape of molecules containing 3 or more atoms. For instant if any molecule possesses two or more polar bonds, it will not be symmetric if it possesses some net molecular dipole moment as in case of water (\(\mu_{\mathrm{H}_{2} \mathrm{O}}\) = 1-84 D) and ammonia (\(\mu_{\mathrm{NH}_{3}}\) = 1.49 D). But if a molecule contains a number of similar atoms linked to a central atom the overall dipole moment of the molecule is found out to be zero, this will imply that the molecule is symmetrical as in the case of CO₂, BF₃, CH₄,CCl₄,etc.

Question 17.
Define electronegativity. How does it differ from electron gain enthalpy?
Answer:
Electronegativity is the ability of an atom in a chemical compound to attract a bond pair of electrons towards itself. Electronegativity of any given element is not constant. It varies according to the element to which it is bound. It is not a measurable quantity. It is only a relative number.

On the other hand, electron gain enthalpy is the enthalpy change that takes place when an electron is added to a neutral gaseous atom to form an anion. It can be negative or positive depending upon whether the electron is added or removed. An element has a constant value of the electron gain enthalpy that can be measured experimentally.

Question 18.
Explain polar covalent bond with the help of suitable example.
Answer:
When two dissimilar atoms having different electronegativities combine to form a covalent bond, the bond pair of electrons is not shared equally. The bond pair shifts towards the nucleus of the atom having greater electronegativity. As a result, electron distribution gets distorted and the electron cloud is displaced towards the electronegative atom.

As a result, the electronegative atom becomes slightly negatively charged while the other atom becomes slightly positively charged. Thus, opposite poles are developed in the molecule and this type of a bond is called a polar covalent bond.

For example-In HF, the electron pair is attracted more towards F atom due to its higher electronegativity. HF may be written as

Question 19.
Arrange the bonds in order of increasing ionic character in the molecules : LiF, K₂O, N₂, SO₂ and ClF₃.
Answer:
More the difference of electronegativity, more the ionic character of the molecules
N₂ < SO₂ < ClF₂ < K₂O < LiF.

Question 20.
The skeletal structure of CH₃COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.

Answer:
The correct Lewis structure for acetic acid is given below :

Question 21.
Apart from tetrahedral geometry, another; possible geometry for CH₄ is square planar with the four H atoms at the comers of the square and the C atom at its centre. Explain why CH₄ is not square planar?
Answer:
Electronic configuration of carbon atom :
₆C : 1s² 2s² 2p²
In the excited state, the orbital picture of carbon can be represented as:

Hence, carbon atom undergoes sp³ hybridization in CH₄ molecule and takes a tetrahedral shape.

For a square planar shape, the hybridization of the central atom has to be dsp. However, an atom of carbon does not have d-orbitals to undergo dsp² hybridization. Hence, the structure of CH₄ cannot be square planar. Moreover, with a bond angle of 90° in square planar, the stability of CH₄ will be very less because of the repulsion existing between the bond pairs. Hence, VSEPR theory also supports a tetrahedral structure for CH₄.

Question 22.
Explain why BeH₂ molecule has a zero dipole moment although the Be—H bonds are polar.
Answer:
BeH₂ molecule is linear. The two equal bond dipoles point in opposite
directions and cancel the effect of each other.
That is why its dipole moment is zero.

Question 23.
Which out of NH₃ and NF₃ has higher dipole moment and why?
Answer:
In both molecules i.e., NH₃ and NF₃, the central atom (N) has a lone pair electron and there are three bond pairs. Hence, both molecules have a pyramidal shape. Since fluorine is more electronegative than hydrogen, it is expected that the net dipole moment of NF₃ is greater than NH₃. However, the net dipole moment of NH₃ (1.46 D) is greater than that of NF₃ (0.24 D).
This can be explained on the basis of the directions of the dipole moments of each individual bond in NF₃ and NH₃. These directions can be shown as :

Thus, the resultant moment of the N—H bonds add up to the bond moment of the lone pair (the two being in the same direction), whereas that of the three N—F bonds partly cancels the moment of the lone pair.
Hence, the net dipole moment of NF₃ is less than that of NH₃.

Question 24.
What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp² , sp³ hybrid orbitals.
Answer:
Hybridisation : It is defined as the mixing of the atomic orbitals belonging to the same atom but having slightly different energies so that a redistribution of energy takes place between them resulting in the formation of new orbitals of equal energies and identical shapes. The new orbitals thus formed are known as Hybrid Orbitals. sp Hybridisation : Here one s and one p orbitals of same atom mix up to
give two sp hybrid orbitals with \(\frac{1}{2}\)s and \(\frac{1}{2}\)p character and linear shape with
bond angle of 180° between them. For example, in BeH₂, BeF₂ and C₂H₂, Be and C are sp-hybridised.

sp hybridization is also called diagonal hybridization.

sp² Hybridisation : Here one s and two p-orbitals of same atom mix up to form three sp²hybrid orbitals with \(\frac{1}{3}\)s and \(\frac{2}{3}\)p character. They form Trigonal Planar shapes with an angle of 120° with themselves. For example, in BH₃ and BF₃, boron is sp² hybridised and in C₂H₄, carbon is sp² hybridised.

sp³ Hybridisation : Here one s and three p orbitals of same atom mix up to give four sp³ hybrid orbitals with \(\frac{1}{4}\) s character and \(\frac{3}{4}\)p character. They form tetrahedral shapes with angles of 109°, 28′ with themselves. For example, in methane (CH₄), ethane (C₂H₆) and all compounds of carbon containing C—C single bonds, carbon is sp³ hybridised.

Question 25.
Describe the change in hybridisation (if any) of the A1 atom in the following reaction.
AlCl₃ + Cl^– → \(\mathbf{A l C l}_{\mathbf{4}}^{-}\)
Answer:
Electronic configuration of A1 in ground state is \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2}, 3 p^{\prime}{ }_{x}\) and it is \(1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{\prime}, 3 p_{x}^{\prime}, 3 p^{\prime} y\) in excited state.

In the formation of AlCl₃ Al undergoes sp²-hybridisation and it is trigonal planar in shape. While in the formation of \(\mathrm{AlCl}_{4}^{-} \), Al undergoes sp³-hybridisation. It means empty 3p_z-orbital also involved in hybridisation. Thus, the shape of \(\mathrm{AlCl}_{4}^{-} \) ion is tetrahedral.

Question 26.
Is there any change in the hybridisation of B and N atoms as a result of the following reaction?
BF₃ + NH₃ → F₃B . NH₃
Answer:
In BF₃, B is sp² hybridised and in NH₃, N is sp³ hybridised. After the reaction hybridisation of B changes to sp³ but that of N remains unchanged.

Question 27.
Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂ molecules.
Answer:
Formation of C₂H₄ (ethylene)

Formation of C₂H₂ (acetylene)

Question 28.
What is the total number of sigma and pi bonds in the following molecules?
(a) C₂H₂ (b) C₂H₄
Answer:
(a) The structure of C₂H₂ can be represented as:

Hence, there are three sigma and two pi-bonds in C₂H₂.

(b) The structure of C₂H₄ can be represented as:

Hence, there are five sigma bonds and one pi-bond in C2H4.

Question 29.
Considering x-axis as the intemuclear axis which out of the following will not form a sigma bond and why?
(a) 1s and 1s (b) 1s and 2p_x, (c) 2p_y and 2p_y (d) Is and 2s.
Answer:
(c) 2p_y and 2p_y orbitals will not a form a sigma bond. Taking x-axis as
the intemuclear axis, 2p_y and 2p_y orbitals will undergo lateral overlapping, thereby forming a pair bond.

Question 30.
Which hybrid orbitals are used by carbon atoms in the following molecules?
(a) CH₃— CH₃; (b) CH₃—CH = CH₂; (c) CH₃—CH₂—OH;
(d) CH₃—CHO; (e) CH₃COOH
Answer:

Question 31.
What do you understand by bond pan’s and lone pah’s of electrons? Illustrate by giving one example of each type.
Ans. When two atoms combine by sharing their one or more valence electrons, a covalent bond is formed between them. The shared pairs of electrons present between the bonded atoms are called bond pairs. All valence electrons may not participate in bonding. The electron pairs that do not participate in bonding are called lone pairs of electrons.
For example, in C₂H₆ (ethane), there are seven bond pairs but no lone pair present.

In H₂O, there are two bond pairs and two lone pairs on the central atom (oxygen).

Question 32.
Distinguish between a sigma and a pi bond.
Answer:
The following are the differences between sigma and pi-bonds :

	Sigma (σ) Bond	Pi (π) Bond
(a)	It is formed by the end to end over lapping (axial over lapping) of atomic orbitals.	It is formed by the lateral overlapping (sideway overlapping) of atomic orbitals.
(b)	The orbitals involved in the overlapping are s—s, s—p or p—p.	These bonds are formed by the overlapping of p—p orbitals only.
(c)	It is a strong bond.	It is a weak bond.
(d)	The electron cloud is symmetrical about the line joining the two nuclei.	The electron cloud is not symmetrical.
(e)	It consists of one electron cloud, which is symmetrical about the internuclear axis.	There are two electron clouds lying above and below the plane of the atomic nuclei.
(f)	Free rotation about σ bonds is possible.	Rotation is restricted in case of pi-bonds.

Question 33.
Explain the formation of H₂ molecule on the basis of valence bond theory.
Answer:
Consider two hydrogen atoms A and B are approaching each other. Their nuclei are N_A and N_B and electrons present in them are represented by e_A and e_B. When the two atoms are far apart, there is no interaction between them but as these approach each other, some new ‘ attractive and repulsive force begin to operate.
Attractive forces generated between

(i) nucleus of one atom and its own electron i.e., N_A – e_A and N_B – e_B.
(ii) nucleus of one atom and electron of other atom i.e., N_A – e_B and N_B-e_A.
Similarly, repulsive forces originated in between
1. electrons of two atoms i.e., e_A – e_B
2. nuclei of two atoms N_A – N_B.
Attractive forces tend to bring the combining atoms close to each other “ whereas repulsive forces tend to push them apart as shown in the figure

The magnitude of the attractive forces is more than that of the repulsive forces. Hence, the two atoms approach each other. As a result, the potential energy decreases. Finally, a state is reached when the attractive forces balance the repulsive forces and the system acquires minimum energy. This leads to the formation of a dihydrogen molecule.

Question 34.
Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
Answer:
Conditions required for the Combination of Atomic Orbitals
The linear combination of atomic orbitals to form molecular orbitals is possible only when they satisfied the following conditions :
(i) Similar energy of combining atomic orbitals : The combining atomic orbitals must possess the same or nearly the same energy. It means that Is orbital can combine with another Is orbital but not with 2s orbital because the energy of 2s orbitals is appreciable higher than that of Is orbital. However, it is not true in case of very different atoms.
(ii) Similar symmetry of combining atomic orbitals : The combining atomic orbitals must possess the same symmetry about the molecular axis along with the same energy. If the orbitals have same energy but their symmetry is not same, they will not combine e.g., 2p_z orbital of one atom can combine with 2p_z orbital or 2s orbital of the other atom but not with the 2p_x or 2p_y orbitals as their symmetries are different.
(iii) Maximum overlap : The combining atomic orbitals must overlap to the maximum extent. Higher the extent of overlapping, more will be the electron-density between the nuclei of a molecular orbital.

Question 35.
Use molecular orbital theory to explain why the Be₂ molecule does not exist.
Answer:
The electronic configuration of Beryllium is 1s² 2s²
The electronic configuration of Be₂ molecule (4 + 4 = 8),
σ1s², σ* 1s², σ2s²s², σ* 2s²
Hence, the bond order of Be₂ is -(N_b – N_a).
where,
N_b = Number of electrons in bonding orbitals.
N_a – Number of electrons in anti-bonding orbitals.
∴ Bond order of Be₂ = \(\frac{1}{2}\) (4 – 4) = 0
A negative or zero bond order means that the molecule is unstable. Hence, Be2 molecule does not exist.

Question 36.
Compare the relative stability of the following species and indicate their magnetic properties ; \(\mathbf{O}_{2}, \mathbf{O}_{2}^{+}, \mathbf{O}_{2}^{-}\)(superoxide), \(\mathrm{O}_{2}^{2-}\) (peroxide)
Answer:

Question 37.
Write the significance of a plus and a minus sign shown in representing the orbitals.
Answer:
Molecular orbitals are represented by wave function. A plus sign in an orbital indicates a positive wave function while a minus sign in an orbital represents a negative wave function. Combination of two wave functions having similar sign gave bonding molecular orbital while that having opposite sign gave antibonding molecular orbital.

Question 38.
Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?
Answer:
The ground state and excited state outer electronic configurations of phosphorus (Z = 15) are as follows :

Phosphorus atom is sp³ d hybridised in the excited state. These orbitals are filled by the electron pairs donated by five Cl atoms as :

The five sp³d hybrid orbitals are directed towards the five corners of the trigonal bipyramidals. Hence, the geometry of PCl₅ can be represented as :

There are five P—Cl sigma bonds in PCl₅. Three P—Cl bonds lie in one plane and make an angle of 120° with each other. These bonds are called equatorial bonds.
The remaining two P—Cl bonds lie above and below the equatorial plane and make an angle of 90° with the plane. These bonds are called axial bonds.

As the axial bond pairs suffer more repulsion from the equatorial bond pairs, axial bonds are slightly longer than equatorial bonds.

Question 39.
Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Answer:
A hydrogen bond is defined as an attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule (may be of the same kind). Due to a difference between electronegativities, the bond pair between hydrogen and the electronegative atom gets drifted far away from the hydrogen atom. As a result, a hydrogen atom becomes electropositive with respect to the other atom and acquires a positive charge.

The magnitude of H-bonding is maximum in the solid state and minimum in the gaseous state.
There are two types of H-bonds:
(i) Intermolecular H-bonds e.g., HF, H₂O etc
(ii) Intramolecular H-bonds e.g., o-nitrophenol
Hydrogen bonds are stronger than Van der Waals forces since hydrogen bonds are regarded as an extreme form of dipole-dipole interaction.

Question 40.
What is meant by the term bond order? Calculate the bond order of: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}\) and \(\mathbf{O}_{2}^{-}\).
Answer:
Bond order is defined as half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule.
Bond order = \(\frac{1}{2}\) (N_b – N_a)
If N_b > N_a, then the molecule is said be stable. However, if N_b ≤ N_a, then the molecule is considered to be unstable.
Bond order values 1, 2 or 3 correspond to single, double or triple bonds respectively.

Calculation of the bond order of \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{O}_{2}^{+}\) and \(\mathbf{O}_{2}^{-}\).
Electronic configuration of N₂

zxPunjab State Board Syllabus PSEB 11th Class Maths Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 11.

PSEB 11th Class Maths Guide | Maths Guide for Class 11 PSEB in English Medium

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Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties

Very Short Answer Type Questions

Question 1.
Which properties of the elements depend on’ the electronic configuration of the atoms and which do not?
Answer:
Chemical and many physical properties of the elements depends on the electronic configuration of the atoms, whereas the nuclear properties do not.

Question 2.
Write the number designation of a group that has 2 electrons beyond a noble gas configuration.
Answer:
The number designation of a group that has 2 electrons beyond a noble gas configuration will be 2 which means it will belong to group 2 of periodic table.

Question 3.
Why is it more logical to call the atomic radius as the effective atomic radius?
Answer:
This is because the size of atom is very small and it has no sharp boundaries.

Question 4.
A boy has reported the radii of Cu, Cu⁺ and Cu²⁺ as 0.096 nm, 0.122 nm and Question072 nm respectively. However, it has been noticed that he interchanged the values by mistake. Assign the correct values to different species.
Answer:
Cu [0.122 nm], Cu⁺ [0.096 nm], Cu²⁺ [0.072 nm].
∵ Size ∝ \(\frac{1}{\text { positive charge }}\)

Question 5.
Atomic radii of fluorine is 72 pm where as atomic radii of neon is 160 pm. Why? [NCERT Exemplar]
Answer:
Atomic radius of F is expressed in terms of covalent radius while, atomic radius of neon is usually expressed in terms of van der Waals’ radius, van der Waals’ radius of an element is always larger than its covalent radius.
Therefore, atomic radius of F is smaller than atomic radius of Ne (F = 72 pm, Ne = 160 pm)

Question 6.
Arrange the following elements in order of decreasing electron gain enthalpy : B, C, N, O.
Answer:
N has positive electron gain enthalpy while all others have negative electron gain enthalpies. Since size decreases on moving from B → C → O, therefore, electron gain enthalpies become more and more negative from B → C → O. Thus, the overall decreasing order of electron gain enthalpies is N, B, C, O.

Question 7.
Which of the following atoms would most likely form an anion (i) Be, (ii) Al, (iii) Ga, (iv) I ?
Answer:
I, because of high electron gain enthalpy, it can accept an electron readily to form an anion F < Cl < Br > I.

Question 8.
Explain why chlorine can be converted into chloride ion more easily as compared to fluoride ion from fluorine.
Answer:
Electron gain enthalpy of Cl is more negative than that of F.

Question 9.
Among alkali metals which element do you expect to be least electronegative and why? [NCERT Exemplar]
Answer:
On moving down the group, electronegativity decreases because atomic size increases. Fr has the largest size, therefore it is least electronegative.

Question 10.
Arrange the following elements in the increasing order of non-metallic character. B, C, Si, N, F
Answer:
The given non-metals are arranged in the increasing order of non-metallic character as follows:

Short Answer Type Questions

Question 1.
What would be IUPAC names and symbols for elements with atomic numbers 122, 127, 135, 149 and 150? .
Answer:
The roots 2, 7, 5, 9 and 0 are referred as bi, sept, pent, enn and nil respectively. Therefore, their names and symbol are

Z (Atomic number)	Name	Symbol
122	Unbibium	Ubb
127	Unbiseptium	Ubs
135	Untripentium	Dtp
149	Unquadennium	Uqe
150	Unpentnilium	Upn

Question2.
All transition elements are d-block elements, but all d-block elements are not transition elements. Explain.
Answer:
Elements in which the last electron enters in the d-orbitals, are called d-block elements or transition elements. These elements have the general outer electronic configuration (n – 1)d^1-10ns^0-2 Zn, Cd and Hg having the electronic configuration, (n – l1)d¹⁰ns² do not show most of the properties of transition elements. The d-orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements. Thus, on the basis of properties, all transition elements are d-block elements but on the basis of electronic configuration, all d-block elements are not transition elements.

Question 3.
Arrange the elements N, P, O and S in the order of
(i) increasing first ionisation enthalpy.
(ii) increasing non-metallic character.
Give reason for the arrangement assigned.
Answer:

	Group 15	Group 16
2nd period	N	0
3rd period	P	S

(i) Ionisation enthalpy of nitrogen (₇N = 1s², 2s², 2p³) is greater than oxygen (₈O = 1s² , 2s² , 2p⁴ ) due to extra stable half-filled 2p-orbitals. Similarly, ionisation enthalpy of phosphorus (₁₅P = 1s², 2s², 2p⁶, 3s², 3p³) is greater than sulphur (₁₆S = 1s², 2x², 2p⁶, 3s², 3p⁴).
On moving down the group, ionisation enthalpy decreases with increasing atomic size. So, the increasing order of first ionisation enthalpy is S < P < O < N.

(ii) Non-metallic character across a period (left to right) increases but on moving down the group it decreases. So, the increasing order of non-metallic character is P < S < N < 0.

Question 4.
What do you understand by exothermic reaction and endothermic reaction? Give one example of each type.
Answer:
Exothermic reactions : Reactions which are accompanied by evolution of heat are called exothermic reactions. The quantity of heat produced is shown either along with the products with a ‘+’ sign or in terms if ΔH with a sign, e.g.,

C(s) + O₂(g) → CO₂(g) + 393.5 kJ
H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l) ΔH = -285.8 kJ mol^-1

Endothermic reactions : Reactions which proceed with absorption of heat are called endothermic reactions. The quantity of heat absorbed is shown either alongwith the products with a sign or in terms of ΔH with a ‘-‘ sign, e.g.,

C(s) + H₂O(g) → CO(g) + H₂(g) -131.4 kJ
N₂(g) + 3H₂(g) → 2NH₃(g); ΔH = +92.4 kJ mol^-1

Question5.
How does the metallic and non-metallic character vary on moving from left to right in a period?
Answer:
As we move from left to right in a period, the number of valence electrons increases by one at each succeeding element but the number of shells remains same. Due to this, effective nuclear charge increases.

More is the effective nuclear charge, more is the attraction between nuclei and electron.
Hence, the tendency of the element to lose electrons decreases, this results in decrease in metallic character.
Furthermore, the tendency of an element to gain electrons increases with increase in effective nuclear charge, so non-metallic character increases on moving from left to right in a period.

Long Answer Type Questions

Question 1.
Write the drawbacks in Mendeleev’s Periodic Table that led to its modification.
Answer:
The main drawbacks of Mendeleev’s Periodic Table are:
(i) Some elements having similar properties were placed in different groups whereas some elements having dissimilar properties were placed in the same group. For example alkali metals such as Li, Na, K, etc., (I A group) are grouped together with coinage metals such as Cu, Ag, Au (I B group) though their properties are quite different. Chemically similar elements such as Cu (I B group) and Hg (II B group) have been placed in different groups.

(ii) Some elements with higher atomic weights are placed before the elements with lower atomic weights in order to maintain the similar chemical nature of elements. For example,

(iii) Isotopes did not find any place in the Periodic Table. However, according to Mendeleev’s classification, these should be placed at different places in the Periodic Table.
(All the above three defects were however removed when modern periodic law based on atomic number was given.)

(iv) Position of hydrogen in the Periodic Table is not fixed but is
controversial. ,
(v) Position of elements of group VIII could not be made clear which have been arranged in three triads without any justification.
(vi) It could not explain the even and odd series in IV, V and VI long periods.
(vii) Lanthanides and actinides which were discovered later on, have not been given proper positions in the main frame of Periodic Table.

Question 2.
p-block elements form acidic, basic and amphoteric oxides. Explain each property by giving two examples and also write the reactions of these oxides with water.
Answer:
In p-block, when we move from left to right in a period, the acidic character of the oxides increases due to increase in electronegativity, e.g.,

(i) 2nd period
B₂O₃ < CO₂ < N₂O₃ acidic character increases.

(ii) 3rd period
Al₂O₃ < SiO₂ < P₄O₁₀ < SO₃ < Cl₂O₇ acidic character increases.
On moving down the group, acidic character decreases and basic character increases, e.g.,

(a) Nature of oxides of 13 group elements

(b) Nature of oxides of 15 group elements

Among the oxides of same element, higher the oxidation state of the element, stronger is the acid. e.g., SO₃ is a stronger acid than SO₂. B₂O₃ is weakly acidic and on dissolution in water, it forms orthoboric acid. Orthoboric acid does not act as a protonic acid (it does not ionise) but acts as a weak Lewis acid.

Al₂O₃ is amphoteric in nature. It is insoluble in water but dissolves in alkalies and react with acids.

Tl₂O is as basic as NaOH due to its lower oxidation state (+1).
Tl₂O + 2HCl → 2TlCl + H₂O

P₄O₁₀ on reaction with water gives orthophosporic acid.

Cl₂O₇ is strongly acidic in nature and on dissolution in water, it gives perchloric acid.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 3 Classification of Elements and Periodicity in Properties Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 3 Classification of Elements and Periodicity in Properties

PSEB 11th Class Chemistry Guide Classification of Elements and Periodicity in Properties InText Questions and Answers

Question 1.
What is the basic theme of organisation in the periodic table?
Answer:
The basic theme of organisation of elements in the periodic table is to classify the elements in periods and groups according to their properties. This arrangement makes the study of elements and their compounds simple and systematic. In the periodic table, elements with similar properties are placed in the same group.

Question 2.
Which important property did Mendeleev use to classify the elements in his periodic table and did he stick to that? ‘
Answer:
Mendeleev arranged the elements in his periodic table ordered by atomic weight or mass. He arranged the elements in periods and groups in order of their increasing atomic weight. He placed the elements with similar properties in the same group.
However, he did not stick to this arrangement for long. He found out that if the elements were arranged strictly in order of their increasing atomic weights, then some elements did not fit within this scheme of classification. Therefore, he ignored the order of atomic weights in some cases. For example, the atomic weight of iodine is lower than that of tellurium. Still Mendeleev placed tellurium (in Group VI) before iodine (in Group VII) simply because iodine’s properties are so similar to fluorine, chlorine and bromine.

Question 3.
What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modem Periodic Law?
Answer:
Mendeleev’s periodic law : It states that the properties of the elements are a periodic function of their atomic weights.
Modern periodic law : It states that the properties of the elements are a periodic function of their atomic numbers.
Thus, change in the base of classification of elements from atomic weight to atomic number is the basic difference between Mendeleev’s periodic law and the modern periodic law.

Question 4.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
Answer:
In the periodic table of the elements, a period indicates the value of the principal quantum number (n) for the outermost shells. Each period begins with the filling of principal quantum number (n). The value of n for the sixth period is 6. For n = 6, azimuthal quantum number (0 can have values of 0, 1, 2, 3, 4, 5.
According to Aufbau’s principle, electrons are added to different orbitals in order of their increasing energies. The energy of the 6d subshell is even higher than that of the 7s subshell.

In the 6th period, electrons can be filled in only 6s, 4f, 5d and 6p subshells. Now 6s has one orbital, 4f has seven orbitals, 5d has five orbitals, and 6p has three orbitals. Therefore, there are a total of sixteen (1+ 7 + 5 + 3 = 16) orbitals available. According to Pauli’s exclusion principle, each orbital can accommodate a maximum of 2 electrons. Thus, 16 orbitals can accommodate a maximum of 32 electrons.

Hence, the sixth period of the periodic table should have 32 elements.

Question 5.
In terms of period and group where would you locate the element with Z = 114?
Answer:
₁₁₄Z = ₈₆[Rn] 7s², 5f¹⁴, 6d¹⁰, 7p²
In the periodic table the element with Z = 114 is located in
Block : p-block (as last electron enters in p-subshell).
Period : 7th (as n = 7 for valence shell).
Group : 14th (for p-block elements, group number = 10 + number of electrons in the valence shell).

Question 6.
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
Answer:
There are two elements in the 1st period and eight elements in the 2nd period. The third period starts with the element with Z = 11. Now, there are eight elements in the third period. Thus, the 3rd period ends with the element with Z = 18 i.e., the element in the 18th group of the third period has Z = 18. Hence, the element in the 17th group of the third period has atomic number Z = 17.

Question 7.
Which element do you think would have been named by
(i) Lawrence Berkeley Laboratory
(ii) Seaborg’s group.
Answer:
(i) Lawrencium (Lr) with Z = 103 and Berkelium (Bk) with Z = 97
(ii) Seaborgium (Sg) withZ = 106 ,

Question 8.
Why do elements in the same group have similar physical and chemical properties?
Answer:
Same group elements have similar electronic configuration therefore, have similar physical and chemical properties.

Question 9.
What does atomic radius and ionic radius really mean to you?
Answer:
Atomic radius is the radius of an atom. It measures the size of an atom. If the element is a metal, then the atomic radius refers to the metallic radius, and if the element is a non-metal, then it refers to the covalent radius.

Metallic radius is calculated as half the internuclear distance separating the metal cores in the metallic crystal. For example, the internuclear distance between two adjacent copper atoms in solid copper is 256 pm. Thus, the metallic radius of copper is taken as \(\frac{256}{2}\) pm = 128 pm.

Covalent radius is measured as the distance between two atoms when they are found together by a single bond in a covalent molecule. For example, the distance between two chlorine atoms in chlorine molecule is 198 pm. Thus, the covalent radius of chlorine is taken as \(\frac{198}{2}\)pm = 99 pm.

Ionic radius means the radius of an ion (cation or anion). The ionic radii can be calculated by measuring the distances between the cations and anions in ionic crystals.
Since a cation is formed by removing an electron from an atom, the cation has fewer electrons than the parent atom resulting in an increase in the effective nuclear charge. Thus, a cation is smaller than the parent atom. For example, the ionic radius of Na+ ion is 95 pm, whereas the atomic radius of Na atom is 186 pm. On the other hand, an anion is larger in size than its parent atom. This is because an anion has the same nuclear charge, but more electrons than the parent atom resulting in an increased repulsion among the electrons and a decrease in the effective nuclear charge. For example, the ionic radius of F^– ion is 136 pm, whereas the atomic radius of F atom is 64 pm.

Question10.
How do atomic radius vary in a period and in a group? How do you explain the variation?
Answer:
Atomic radius generally decreases from left to right across a period. This is because within a period, the outer electrons are present in the same valence shell and the atomic number increases from left to right across a period, resulting in an increased effective nuclear charge. As a result, the attraction of electrons to the nucleus increases.
On the other hand, the atomic radius generally increases down a group. This is because down a group, the principal quantum number (n) increases which results in an increase of the distance between the nucleus and valence electrons.

Question 11.
What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
(i) F^– (ii) Ar (iii) Mg²⁺ (iv) Rb⁺
Answer:
Atoms and ions having the same number of electrons but different nuclear charges are called isoelectronic species. In case of isoelectronic species, as the nuclear charge increases their size decreases.
(i) F^– ion has 9+1 = 10 electrons.
(ii) Ar has 18 electrons.
(iii) Mg²⁺ ion has 12 – 2 = 10 electrons.
(iv) Rb⁺ ion has 37 -1 = 36 electrons.

Question12.
Consider the following species :
N^3-, 0^2-, F^–, Na⁺, Mg²⁺ and Al³⁺
(a) What is common in them?
(b) Arrange them in the order of increasing ionic radii.
Ans. (i) All the given species have same number of electrons (10e^–). Therefore, all are isoelectronic.
(ii) The ionic radii of isoelectronic species decreases with increase in atomic number (as magnitude of the nuclear charge increases with increase in atomic number). Therefore, their ionic radii increase in the order.
Isoelectronic ions = Al³⁺ < Mg²⁺ < Na⁺ < F^– < O^2- < N^3-
Atomic number =13 12 11 9 8 7

Question 13.
Explain why cations are smaller and anions larger in radii than their parent atoms?
Answer:
A cation has a fewer number of electrons than its parent atom, while its nuclear charge remains the same. As a result, the attraction of electrons to the nucleus is more in a cation than in its parent atom. Therefore, a cation is smaller in size than its parent atom.

On the other hand, an anion has one or more electrons than its parent atom, resulting in an increased repulsion among the electrons and a decrease in the effective nuclear charge. As a result, the distance between the valence electrons and the nucleus is more in anions than in its parent atom. Hence, an anion is larger in radius than its parent atom.

Question 14.
What is the significance of the terms-‘isolated gaseous atom’ and ‘ground state’ while defining the ionization enthalpy and electron gain enthalpy?
Answer:
Ionization enthalpy : It is the minimum amount of energy required to remove an electron from an isolated gaseous atom (A) in its ground state.
X(g) → X⁺ (g) + e^–
The force by which an electron is attracted by nucleus is also affected by the presence of other atoms within its molecule or in the neighbourhood. Therefore, ionization enthalpy is determined in gaseous state because in gaseous state interatomic distances are larger and interatomic forces of attractions are minimum. Further more, ionization enthalpy is determined at a low pressure because it is not possible to isolate a single atom but interatomic attractions can be further reduced by reducing pressure. Due to these reasons, the term isolated gaseous atom in ground state has been included in definition of ionization enthalpy.

Electron gain enthalpy : It is the energy released when an isolated gaseous atom (X) in ground state gains an electron to form gaseous anion.
X(g) + e^– → X^– (g)
The most stable state of an atom is ground state. If isolated gaseous atom is in excited state, comparatively lesser energy will be released on addition of an electron. So, electron gain enthalpies of gaseous atoms must be determined in their ground states. Therefore, the terms ground state and isolated gaseous atom (explained above) has been also included in the definition of electron gain enthalpy.

Question 15.
Energy of an electron in the ground state of the hydrogen atom is -2.18 x 10^-18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol^-1.
Answer:
It is given that the energy of an electron in the ground state of the hydrogen atom is -2.18 x 10^-18 J.
Therefore, the energy required to remove that electron from the ground state of hydrogen atom is 2.18 x 10^-18 J.
∴ Ionization enthalpy of atomic hydrogen = 2.18 x 10^-18 J
Hence, ionization enthalpy per mol of hydrogen atoms
= 2.18 x 10^-18 x 6.022 x 10²³ J mol^-1
= 1.31 x 10⁶ J mol^-1

Question16.
Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < Cl < N < F < Ne. Explain why
(i) Be has higher A^H than B
(ii) O has lower AjH than N and F?
Answer:
(i) Δ_iH of Be is higher than that of B .
Electronic configuration of Be is 1s², 2s²
whereas that of B is 1s², 2s², 2p_x¹
In the case of Be, electron has to be removed from an s-orbital whereas in the case of B, it has to be removed from a p-orbital. It is difficult to remove an s-electron because it is closer to the nucleus than a p-electron hence more energy is required to remove an electron from 2s and Be, than 2p-electron in the case of B. Hence, ionization enthalpy of Be is higher than that of B.

(ii) Electronic configuration of O is
1s², 2s², \(2 p_{x}^{2} 2 p_{y}^{1} 2 p_{z}^{1}\) (neither exactly half-filled nor completely filled)
whereas N is 1s², 2s², \(2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1}\) (exactly half-filled)
It is difficult to remove an electron from the valence shell of N because its p-subshell is exactly half-filled and so has more stability whereas O has electronic configuration which is neither completely filled nor exactly half-filled. Therefore, it is easier to remove one electron from O atom. F, due to increased nuclear charge, has more ionization enthalpy than either O or N.

Question17.
How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium?
Answer:
The first electron in both the cases has to be removed from 3s orbital, but nuclear charge of Na is less than that of Mg. Hence ionization enthalpy of Na is lower than that of Mg.

After the loss of first electron the electronic configuration of Na⁺ is 1s², 2s², 2p⁶, i.e., that of noble gas which is very stable and hence the removal of second electron from Na+ is very difficult. In the case of Mg after the loss of first electron, electronic configuration of Mg⁺ ion is 1s² , 2s² 2p⁶ , 3s¹ . The second electron to be removed is from 3s orbital which is easier.
Hence, second ionization enthalpy of sodium is much larger than second . ionization enthalpy of Mg.

Question 18.
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer:
The factors responsible for the ionization enthalpy of the main group elements to decrease down a group are listed below:
(i) Atomic size : On moving down the group atomic size increases due to addition of new higher energy shell. As a result, force of attraction of nucleus for valence electrons decreases and ionization enthalpy also decreases.
(ii) Screening effect : On moving down the group, screening effect or shielding effect increases so ionization enthalpy decreases.

Question19.
The first ionization enthalpy values (in kJ mol^-1) of group 13 elements are :

How would you explain this deviation from the general trend?
Answer:
On moving down a group, ionization enthalpy generally decreases due to an increase in the atomic size and shielding. Thus, on moving down group 13, ionization enthalpy decreases from B to Al. But, Ga has higher ionization enthalpy than Al. Al follows immediately after s-block elements, whereas Ga follows after d-block elements. The shielding provided by d-electrons is not very effective. These electrons do not shield the valence electrons very effectively. As a result, the valence electrons of Ga experience a greater effective nuclear charge than those of Al. Further, moving from Ga to In, the ionization enthalpy decreases due to an increase in the atomic size and shielding. But, on moving from In to Tl, the ionization enthalpy again increases. In the periodic table, Tl follows after 4f and Ad electrons. The shielding provided by the electrons in both these orbitals is not very effective. Therefore, the valence electron is held quite strongly by the nucleus. Hence, the ionization energy of Tl is on the higher side.

Question20.
Which of the following pairs of elements would have more negative electron gain enthalpy?
(i) OorF (ii) F or Cl
Answer:
(i) O and F are present in the same period of the periodic table. An F atom has one proton and one electron more than O and as an electron is being added to the same shell, the atomic size of F is smaller than that of O. As F contains one proton more than O, its nucleus can attract the incoming electron more strongly in comparison to the nucleus of O atom. Also, F needs only one more electron to attain the stable noble gas configuration. Hence, the electron gain enthalpy of F is more negative (- 328 kJ mol^-1) than that of O (-141 kJ mol^-1).

(ii) F and Cl belong to the same group of the periodic table. The electron gain enthalpy usually becomes less negative on moving down a group.
However, in this case, the value of the electron gain enthalpy of Cl is more negative than that of F. This is because the atomic size of F is smaller than that of Cl. In F, the electron will be added to quantum level n = 2, but in Cl, the electron is added to quantum level n = 3. Therefore, there are less electron-electron repulsion in Cl and an additional electron can be accommodated easily. Hence, the electron gain enthalpy of Cl is more negative (- 349kJ mol^-1) than that of F (-328kJ mol^-1).

Question 21.
Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
Answer:
When an electron is added to O atom to form O^– ion, energy is released. Thus, the first electron gain enthalpy of O is negative.
0(g) + e^– → O^– (g); Δ_egH = -141 kJ mol^-1

On the other hand, when an electron is added to O ion to form O ion, energy has to be given out in order to overcome the strong electronic repulsion. Thus, the second electron gain enthalpy of O is positive.
O^–(g) + e^– → O^2-(g); Δ_egH = +780 kJ mol^-1

Question 22.
What is the basic difference between the terms electron gain enthalpy and electronegativity? •
Answer:
Electron gain enthalpy is the measure of the tendency of an isolated gaseous atom to accept an electron, whereas electronegativity is the measure of the tendency of an atom in a chemical compound to attract a shared pair of electrons.

Question 23.
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Answer:
On Pauling scale, the electronegativity of nitrogen, (3.0) indicates that it is sufficiently electronegative. But it is not correct to say that the electronegativity of nitrogen in all the compounds is 3. It depends upon its state of hybridisation in a particular compound, greater the percentage of s-character, more will be the electronegativity of the element. Thus, the electronegativity of nitrogen increases in moving from sp³ hybridised orbitals to sp hybridised orbitals i.e., as sp³ < sp² < sp.

Question 24.
Describe the theory associated with the radius of an atom as it
(a) gains an electron
(b) loses an electron
Answer:
(a) When an atom gains an electron, its size increases. When an electron is added, the number of electrons goes up by one. This results in an increase in repulsion among the electrons. However, the number of protons remains the same. As a result, the effective nuclear charge of the atom decreases and the radius of the atom increases. For example, ionic radius of Cl^– ion is greater than the radius of its parent atom Cl.

(b) When an atom loses an electron, the number of electrons decreases by one while the nuclear charge remains the same. Therefore, the interelectronic repulsions in the atom decrease. As a result, the effective nuclear charge increases. Hence, the radius of the atom decreases.
For example, ionic radius of Na⁺ is smaller than the radius of its parent atom Na.

Question 25.
Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer.
Answer:
The ionization enthalpy of an atom depends on the number of electrons and protons (nuclear charge) of that atom. Now, the isotopes of an element have the same number of protons and electrons. Therefore, the first ionization enthalpy for two isotopes of the same element should be the same.

Question 26.
What are the major differences between metals and non-metals?
Answer:

Metals	Non-metals
1. Metals can lose electrons easily.	Non-metals cannot lose electrons easily.
2. Metals cannot gain electrons easily.	Non-metals can gain electrons easily.
3. Metals generally form ionic compounds.	Non-metals generally form covalent compounds.
4. Metal oxides are basic in nature.	Non-metal oxides are acidic in nature.
5. Metals have low ionization enthalpies.	Non-metals have high ionization enthalpies.
6. Metals have less negative electron gain enthalpies.	Non-metals have high negative electron gain enthalpies.
7. Metals are less electronegative. They are rather electropositive elements.	Non-metals are electronegative.
8. Metals have a high reducing power.	Non-metals have a low reducing power.

Question27.
Use the periodic table to answer the following questions.
(a) Identify an element with five electrons in the outer subshell.
(b) Identify an element that would tend to lose two electrons.
(c) Identify an element that would tend to gain two electrons.
(d) Identify the group having metal, non-metal liquid as well as gas at the room temperature.
Answer:
(a) General electronic configuration of elements having five electrons in the outer sub shell is ns2 np . This configuration belongs to halogen family, i.e., F, Cl, Br, I, At.
(b) Elements of second group are known as alkaline earth metals (Mg, Ca, Sr, Ba, etc). Their general electronic configuration for valence shell is ns2. These elements form dipositive cations by the lose of two electrons easily.
(c) 16th group elements such as O, S, Se, etc., have a tendency to accept two electrons because by the gain of two electrons they attain noble gas configuration. Their general electronic configuration for valence shell is ns² np⁴.
(d) Group 1 or 17 of the periodic table contains metal, non-metal, liquid as well as gas at the room temperature, e.g., H2 is a non-metal and in gaseous state at room temperature. All other elements of this group are metals. Cs is a liquid metal. Similarly, Br₂ is a liquid non-metal while other elements of this group are gaseous non-metals. Iodine can form I⁺ so it consists some metallic properties.

Question 28.
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb < Cs whereas that among group 17 elements is F > Cl > Br > I. Explain.
Answer:
The elements present in group 1 have only 1 valency electron, which they tend to lose. Group 17 elements, on the other hand, need only one electron to attain the noble gas configuration. On moving down group 1, the ionization enthalpies decrease. This means that the energy required to lose the valence electron decreases. Thus reactivity increases on moving down a group. Thus, the increasing order of ( reactivity among group 1 elements is as follows :
Li < Na < K < Rb < Cs In group 17, as we move down the group from Cl to I, the electron gain enthalpy becomes less negative i. e., its tendency to gain electrons decreases on moving down a group. Thus, reactivity decreases down a group. The electron gain enthalpy of F is less negative than Cl. Still, it is the most reactive halogen. This is because of its low bond dissociation energy. Thus, the decreasing order of reactivity among group 17 elements is as follows : F > Cl > Br > I.

Question 29.
Write the general outer electronic configuration of s-, p-, d- and f- block elements.
Answer:

Element	General outer electronic configuration
s-block	ns1-2, where n = 2 – 7
p-block	ns2 np1-6      , where n = 2 – 6
d-block	(n – 1)d1–10ns(0 – 2), where n = 4 – 7
f-block	(n – 2)f1–14 (n – 1)d0- 1ns2, where n = 6 – 7

Question30.
Assign the position of the element having outer electronic configuration
(i) ns² np⁴ for n = 3 (ii) (n – 1 )d²ns² for n = 4, and
(iii) (n – 2)f⁷(n – 1)d¹ns² for n = 6, in the periodic table.
Answer:
(i) ns²np⁴ for n = 3; it is 3s²3p⁴
The complete electronic configuration is 1s², 2s², 2p⁶, 3s², 3p⁴
Atomic number = 2 + 2+ 6+ 2 + 4 = 16
The element is sulphur in the 3rd period and in Group 16 (p-block)

(ii) (n – 1)d²ns² for n = 4; it is 3d²4s²
The complete electronic configuration is
1s², 2s²,2p⁶, 3s², 3p⁶, 3d², 4s² Atomic number is 22, the element is Titanium.
It is a transition element present in the 4th period and in Group 4.

(iii) (n – 2)f⁷(n – l)d¹ns² for n = 6; it is 4f⁷5d¹6s²
Its complete electronic configuration is
1s², 2s², 2p⁶, 3s², 3p⁶, 3d¹⁰, 4s², 4p⁶, 4d¹⁰, 4f⁷, 5s², 5p⁶, 5d¹, 6s²
It is Gadolinium (Gd)
It is an inner transition element, belongs to Lanthanoid series or 4f series. It is an f-block element.

Question 31.
The first (Δ_i,H₁) and the second (Δ_iH₂) ionization enthalpies (in kJ mol^-1) and the (Δ_egH) electron gain enthalpy (in kJ mol^-1) of a few elements are given below:

Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX₂ (X = halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X = halogen)?
Answer:
(a) Element V is likely to be the least reactive element. This is because it has the highest first ionization enthalpy (Δ_i,H₁) and a positive electron gain enthalpy (Δ_eg,H₁).
(b) Element II is likely to be the most reactive metal as it has the lowest first ionization enthalpy (Δ_i,H₁) and the highest negative electron gain enthalpy (AegH).
(c) Element III is likely to be the most reactive non-metal as it has a high first ionization enthalpy Δ_i,H₁) and the highest negative electron gain enthalpy (Δ_eg,H).
(d) Element IV is likely to be the least reactive non-metal since it has a very high first ionization enthalpy (Δ_i,H₂) and a positive electron gain enthalpy (Δ_eg,H).
(e) Element VI has a low negative electron gain enthalpy (Δ_eg,H). Thus it is a metal. Further, it has the lowest second ionization enthalpy (Δ_i,H₂). Hence, it can form a stable binary halide of the formula MX₂ ( X = halogen).
(f) Element I has low Δ_i,H₁ but a very high Δ_i,H₂. It has less negative electron gain enthalpy. So, element (I) is alkali metal. The given values for element I match with Li. Lithium firms predominantly stable covalent halide of the formula Mx.

Question32.
Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.
(a) Lithium and oxygen
(b) Magnesium and nitrogen
(c) Aluminium and iodine
(d) Silicon and oxygen
(e) Phosphorus and fluorine
(f) Element 71 and fluorine
Answer:
(a) Li₂0
(b) Mg₃N₂
(c) AlI₃
(d) SiO₂
(e) PF₃ or PF₅
(f) The element with the atomic number 71 is Lutetium (Lu). It has valency 3. Hence, the formula of the compound is LuF₃.

Question 33.
In the modem periodic table, the period indicates the value of:
(a) atomic number
(b) atomic mass
(c) principal quantum number
(d) azimuthal quantum number.
Answer:
(c) The value of the principal quantum number (n) for the outermost shell or the valence shell indicates a period in the Modern periodic table.

Question 34.
Which of the following statements related to the modem periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The df-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
Answer:
The statement (b) is incorrect. The correct statement (b) is that the d-block has 10 columns because a maximum of 10 electrons can occupy all the orbitals in a d-subshell.
All other given statements are correct.

Question 35.
Anything that influences the valence electrons will affect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons.
Answer:
(c) Nuclear mass (protons + neutrons) does not affect the valence shell, only protons i. e., nuclear charge affects the valence shell.

Question 36.
The size of isoelectronic species F^–, Ne and Na⁺ is affected by
(a) Nuclear charge (Z)
(b) Valence principal quantum number (n)
(c) Electron-electron interaction in the outer orbitals.
(d) None of the factors because their size is same.
Answer:
(a) The size of an isoelectronic species increases with a decrease in the nuclear charge (Z).

Question 37.
Which one of the following statements is incorrect in relation to ionization enthalpy?
(a) Ionization enthalpy increases for each successive electron.
(b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
(c) End of valence electrons is marked by a big jump in ionization enthalpy.
(d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.
Answer:
(d) Electrons in orbitals bearing a lower n value are more attracted to the nucleus than electrons in orbitals bearing a higher n value. Hence, the removal of electrons from orbitals bearing a higher n value is easier than the removal of electrons from orbitals having a lower n value.

Question 38.
Considering the elements B, Al, Mg and K the correct order of their metallic character is:
(a) B > Al > Mg > K
(b) Al > Mg > B > K
(c) Mg > Al > K > B
(d) K > Mg > Al > B
Answer:
(d) In a group, metallic character increases from top to bottom as ; ionisation energy decreases and in a period metallic character decreases from left to right as tendency to lose electron decreases. Therefore, the correct order is K > Mg > Al > B.

Question 39.
Considering the elements B, C, N, F and Si the correct order of their non-metallic character is
(a) B > C > Si > N > F
(b) Si > C > B > N > F
(c) F > N > C > B > Si
(d) F > N > C > Si > B
Answer:
(c) The non-metallic character of elements increases from left to right across a period. Thus, the decreasing order of non-metallic character is F > N > c > B.
Again, the non metallic character of elements decreases down a group. ;
Thus, the decreasing order of non-metallic characters of C and Si are C > Si. However, Si is less non-metallic than B i.e., B > Si.
Hence, the correct order of non-metallic characters is F > N > C > B > Si.

Question 40.
Considering the elements F, Cl, O and N the correct order of their chemical reactivity in terms of oxidizing property is :
(a) F > Cl > O > N
(b) F > O > Cl > N
(c) Cl > F > O > N
(d) O > F > N > Cl
Answer:
(b) In a group, oxidising power decreases from top to bottom as the size increases but when we move left to right in a period it increases because size decreases.
Therefore, among F, Cl, O and N, the oxidising power decreases in the order F > O > Cl > N.

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 6 Thermodynamics Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 6 Thermodynamics

Very Short Answer Type Questions

Question 1.
Identify the state functions and path functions out of the following.
Enthalpy, entropy, heat, temperature, work, free energy.
Answer:
State function Enthalpy, entropy, temperature, free energy.
Path function Heat, work

Question 2.
At 1 atm will the Δ_fH⁰ be zero for Cl₂(g) and Br₂(g)? Explain.
Answer:
Δ_fH⁰ for Cl₂(g) will be zero but Δ_fH⁰ for Br₂(g) will not be zero because liquid bromine state is elementary state not gaseous.

Question 3.
Why for predicting the spontaneity of a reaction, free energy criteria is better than the entropy criteria?
Answer:
Criteria of free energy change is better because it requires free energy change of the system only whereas the entropy change requires the total entropy change of the system and the surroundings.

Question 4.
Water can be lifted into the water tank at the top of the house with the help of a pump. Then why is it not considered to be spontaneous?
Answer:
A spontaneous process should occur continuously by itself after initiation. But this is not so in the given case because water will go up so long as the pump is working.

Question 5.
Given that ΔH = 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
Answer:
It is a spontaneous process because although ΔH = 0, i.e., energy factor has no role to play but randomness increases, i.e., randomness factor favours the process.

Question 6.
Under what condition, the heat evolved or absorbed in a reaction is equal to its free energy change?
Answer:
As ΔG = ΔH – TΔS. Thus, ΔG = ΔH only when either the reaction is carried out at 0 K or the reaction is not accompanied by any entropy change, i.e., ΔS = 0.

Question 7.
In the equation, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) what would be the sign of work done?
Answer:
The sign of work done will be positive, i.e., work will be done on the system due to decrease in volume.

Question 8.
The molar enthalpy of vaporisation of acetone is less than that of water. Why?
Answer:
Enthalpy of vaporisation of water is more than that of acetone because there is strong hydrogen bonding in H₂O molecules.

Question 9.
One mole of acetone requires less heat to vaporise than 1 mole of water. Which of the two liquids has higher enthalpy of vaporisation?
Answer:
Less the heat required to vaporise 1 mole of a liquid, less is its enthalpy of vaporisation. Hence, water has higher enthalpy of vaporisation.

Question 10.
Which quantity out of Δ_rG and Δ_rG° will be zero at equilibrium?
Answer:
Δ_rG = Δ_rG° + RT In K
At equilibrium, 0 (zero) = Δ_rG° + RT In K
(v Δ_rG = 0)
or Δ_rG° = -RT In It;
Δ_rG° = 0 when it = 1
For all other values of K, Δ_rG° will be non-zero.

Short Answer Type Questions

Question 1.
Define the following :
(i) First law of thermodynamics.
(ii) Standard enthalpy of formation.
Answer:
(i) First law of thermodynamics : It states that energy can neither be created nor be destroyed. The energy of an isolated system is constant. ΔU = q + w
(ii) Standard Enthalpy of Formation : It is defined as the amount of heat evolved or absorbed when one mole of the compound is formed from its constituent elements in their standard states.

Question 2.
Give reason for the following:
(i) Neither q nor w is a state function but q + w is a state function.
(ii) A real crystal has more entropy than an ideal crystal.
Answer:
(a) q + w = ΔU
As ΔU is a state function hence, q + w is a state function.
(b) A real crystal has some disorder due to the presence of defects in its structural arrangement whereas ideal crystal does not have any disorder. Hence, a real crystal has more entropy than an ideal crystal.

Question 3.
Represent the potential energy/enthalpy change in the following processes graphically
(i) Throwing a stone from the ground to roof.
(ii) \(\frac{1}{2}\)H₂(g) + \(\frac{1}{2}\)Cl₂(g) ⇌ HCl(g); Δ_rH^s = – 92.32kJ mol^-1
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?
Answer:

Energy increases in (a) and it decreases in (b). Hence, in process (b), enthalpy change is the contributing factor to the spontaneity.

Question 4.
A man takes a diet equivalent to 10000 kJ per day and does work, by expending his energy in all forms equivalent to 12500 kJ per day. What is change in internal energy per day? If the energy lost was stored as sucrose (1632 kJ per 100 g), how many days should it take to lose 2 kg of his weight? (Ignore water loss)
Answer:
Energy taken by a man = 10000 kJ
Change in internal energy per day = 12500 -10000 = 2500 kJ
The energy is lost by the man as he expends more energy than he takes.
Now 100 g of sugar corresponds to energy = 1632 kJ loss in energy.
2000 g of sugar corresponds to energy = \(\frac{1632 \times 2000}{100}\) = 32640 kJ
∴ Number of days required to lose 2000 g of weight or 32640 kJ of energy = \(\frac{32640}{2500}\) = 13 days

Question 5.
Give the appropriate reason :
(i) It is preferable to determine the change in enthalpy rather than the change in internal energy.
(ii) It is necessary to define the ‘standard state’.
(iii) It is necessary to specify the phases of the reactants and products in a thermochemical equation.
Answer:
(i) Because it is easier to make measurement under constant pressure than under constant volume conditions.
(ii) Enthalpy change depends upon the conditions in which a reaction is carried out. For making the comparison of results obtained by different people meaningful, the reaction conditions must be well-defined.
(iii) Because enthalpy depends upon the phase of reactants and products.

Long Answer Type Questions

Question 1.
(i) A cylinder of gas supplied by a company is assumed to contain 14 kg of butane. If a normal family requires 20000 kJ of energy per day for cooking, how long will the cylinder last?
(ii) If the air supplied to the burner is insufficient, a portion of gas escapes without combustion. Assuming that 25% of the gas is wasted due to this inefficiency, how long will the cylinder last (Heat of combustion of butane = 2658kJ!mol.)?
Answer:
(i) Molecular formula of butane = C₄H₁₀
Molecular mass of butane = 4 x 12 +10 x 1 = 58
Heat of combustion of butane 2658 kJ mol^-1
1 mole.or 58 g of butane on complete combustion gives heat = 2658 kJ
∴ 14 x 10³ g of butane on complete combustion will give heat
= \(\frac{2658 \times 14 \times 10^{3}}{58}\) = 641586 kJ
The family needs 20000 kJ of heat per day.
∴ 20000 kJ of heat is used for cooking by a family in 1 day.
∴ 641586 kJ of heat will be used for cooking by a family in
= \(\frac{641586}{20000}\) = 32days
The cylinder will last for 32 days

(ii) 25 per cent of the gas is wasted due to inefficiency. This means that only 75% of butane gets combusted. Therefore, the energy produced by
75% combustion of butane = \(\frac{641586 \times 75}{100}\) = 481189.5 kJ
∴ The number of days the cylinder will last = \(\frac{481189.5}{20000}\) = 24 days.

Question 2.
10 moles of an ideal gas expand isothermally and reversibly from a pressure of 5 atm to 1 atm at 300 K. What is the largest mass that can be lifted through a height of 1 m by this expansion?
Answer:
W_exp = -2.303 nRT log \(\frac{p_{1}}{p_{2}}\)
= -2.303(10) x (8.314)(300) log \(\frac{5}{1}\) = – 40.15 x 10³ J
If M is the mass that can be lifted by this work through a height of 1 m, then work done = Mgh
40.15 x 10³ J = M x 9.81 ms^-1 x 1 m
or M = \(\frac{40.15 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}}{9.81 \mathrm{~m} \mathrm{~s}^{-2} \times 1 \mathrm{~m}}\) [∵ J = kg m²s^-2]
= 4092.76 kg

Punjab State Board Syllabus PSEB 11th Class Agriculture Book Solutions Guide Pdf in English Medium and Punjabi Medium are part of PSEB Solutions for Class 11.

PSEB 11th Class Agriculture Guide | Agriculture Guide for Class 11 PSEB in English Medium

Agriculture Guide for Class 11 PSEB | PSEB 11th Class Agriculture Book Solutions

PSEB 11th Class Agriculture Book Solutions in Hindi Medium

• Chapter 1 ख़रीफ़ की फसलें
• Chapter 2 ख़रीफ़ की सब्ज़ियाँ
• Chapter 3 फूलों की खेती
• Chapter 4 कृषि उत्पादों का मंडीकरण
• Chapter 5 बीज, खाद और कीड़ेमार दवाइयों का क्वालिटी कन्ट्रोल
• Chapter 6 पशु पालन
• Chapter 7 दूध और दूध से बनने वाले पदार्थ
• Chapter 8 मुर्गी पालन
• Chapter 9 सूअर, भेड़ें/बकरियां और खरगोश पालना
• Chapter 10 मछली पालन
• Chapter 11 कुछ नए कृषि विषय
• Class 11 Agriculture Objective Questions

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 5 States of Matter Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 5 States of Matter

Very Short Answer Type Questions

Question 1.
Name two intermolecular forces that exist between HF molecules in liquid state.
Answer:
HF are polar covalent molecules. In liquid state, there are dipole-dipole interactions and H-bonding.

Question 2.
Explain why Boyle’s law cannot be used to calculate the volume of a real gas when it is converted from its initial state to final state by an adiabatic expansion.
Answer:
During adiabatic expansion, temperature is lowered and therefore, Boyle’s law cannot be applied.

Question 3.
Boyle’s law states that at constant temperature, if pressure is increased on a gas, volume decreases and vice-versa. But when we fill air in a balloon, volume as well as pressure increase. Why?
Answer:
The law is applicable only for a definite mass of the gas. As we fill air into the balloon, we are introducing more and more air into the balloon.
Thus, we are increasing the mass of air inside. Hence, the law is not applicable.

Question 4.
What will be the molar volume of nitrogen and argon at 273.15 K and 1 atm?
Answer:
Every gas has 22.4 L molar volume at 273.15 K and 1 atm pressure (STP).

Question 5.
A gas that follows Boyle’s law, Charles’ law and Avogadro’s law is called an ideal gas. Under what conditions a real gas would behave ideally?
Answer:
At low pressure and high temperature, a real gas behaves as an ideal gas.

Question 6.
Explain why temperature of a boiling liquid remains constant?
Answer:
This is because at the boiling point, the heat supplied is used up in breaking off the intermolecular forces of attraction of the liquid to change it into vapour and not for raising the temperature of the liquid.

Question 7.
Assuming C02 to be van der Waals’ gas, calculate its Boyle temperature.
Given a = 3.59 L² atm mol^-2 and b = 0.0427 L mol^-1.
\(T_{b}=\frac{a}{R b}=\frac{3.59 \mathrm{~L}^{2} \mathrm{~atm} \mathrm{} \mathrm{mol}^{-2}}{\left(0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\left(0.0427 \mathrm{~L} \mathrm{~mol}^{-1}\right)}\) = 1025.3 K

Question 8.
Name two phenomena that can be explained on the basis of surface tension.
Answer:
Surface tension can explain
(i) capillary action, i.e., rise or fall of a liquid in capillary,
(ii) spherical shape of small liquid drops.

Question 9.
Why are the gases helium and hydrogen not liquefied at room temperature by applying very high pressure?
Answer:
Because their, critical temperature is lower than room temperature. (Gases cannot be liquefied above the critical temperature by applying even very high pressure).

Question 10.
What would have happened to the gas if the molecular collisions were not elastic?
Answer:
On every collision, there would have been loss of energy. As a result, the molecules would have slowed down and ultimately settle down in the vessel. Moreover, the pressure would have gradually reduced to zero.

Short Answer Type Questions

Question 1.
(i) What do you mean by ‘Surface Tension’ of a liquid?
(ii) Explain the factors which can affect the surface tension of a liquid.
Answer:
(i) Surface tension : It is defined as the force acting per unit length perpendicular to the line drawn on the surface. It’s unit is Nm-1.
(ii) Surface tension of a liquid depends upon the following factors :
(a) Temperature : Surface tension decreases with rise in temperature. As the temperature of the liquid increases, the average kinetic energy of the molecules increases. Thus, there is a decrease in intermolecular force of attraction which decrease the surface tension.
(b) Nature of the liquid : Greater the magnitude of
intermolecular forces of attraction in the liquid, greater will be the value of surface tension.

Question 2.
A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5g neon. If pressure of the mixture of gases in the cylinder is 25 bar.
What is the partial pressure of dioxygen and neon in the mixture?
Answer:

Alternatively, mole fraction of neon = 1 – 0.21 = 0.79
Partial pressure of a gas = mole fraction x total pressure
⇒ Partial pressure of oxygen = 0.21 x (25bar) = 5.25bar
Partial pressure of neon = 0.79 x (25bar) = 19.75bar

Question 3.
Give reasons for the following:
(i) The size of weather balloon becomes larger and larger as it ascends into higher altitudes.
(ii) Tyres of automobiles are inflated to lesser pressure in summer than in winter.
Answer:
(i) As we go to higher altitudes, the atmospheric pressure decreases.
Thus, the pressure outside the balloon decreases. To regain equilibrium with the external pressure, the gas inside expands to decrease its pressure, Hence, the size of the balloon increases.
(ii) In summer, due to higher temperature, the average kinetic energy of the air molecules inside the tyre increases, i.e., molecules start moving faster. Hence, the pressure on the walls of the tube increases. If pressure inside is not kept low at the time of inflation, at higher temperature, the pressure may become so high that the tyre may burst.

Question 4.
On the basis of intermolecular forces and thermal energy, explain why .
(i) a solid has rigidity but liquids do not have rigidity?
(ii) gases have high compressibility but liquids and solids have poor compressibility?
Answer:
(i) It is because in solids, the intermolecular forces are very strong and predominate over thermal energy but in liquid, these forces are no longer strong enough.
(ii) Because of very weak intermolecular forces and high thermal energy, molecules of gases are far apart. That is why gases are highly compressible.

Question 5.
A gas is enclosed in room. The temperature, pressure, density and number of moles respectively are t°C,p atm, g cm^-3 and n moles.
(i) What will be the pressure, temperature, density and number of moles in each compartment, if room is partitioned into four equal compartments?
(ii) What will be the value of pressure, temperature, density and number of moles in each compartment if the walls between the two compartments (say 1 and 2) are removed?
(iii) What will be the values of pressure, temperature, density and number of moles, if an equal volume of gas at pressure
(p) and temperature (t) is let inside the same room? .
Answer:
(i) (a) Pressure in each compartment is same, (p atm)
(b) Temperature will remain same (t°C).
(c) Density will remain same (d g cm^-3).
(d) Because of partition, volume of each compartment becomes 1/4 and the number of molecules also become 1/4. The number of moles in each compartment will be n/4.

(ii) (a) Pressure will remain same (p atm).
(b) Temperature will remain same (t°C).
(c) Density will remain same (d g cm^-3).
(d) The number of moles in each compartment will be n/2.

(iii) (a) Pressure will be doubled (2p atm).
(b) Temperature will remain same.
(c) Density will remain same (d g cm^-3) ,
(d) Number of moles will be doubled i.e., 2n.

Long Answer Type Questions

Question 1.
Explain the following:
(i) The boiling point of a liquid rises on increasing pressure. ;
(ii) Drops of liquid assume spherical space.
(iii) The boiling point of water (373 K) is abnormally high when compared to that of H₂S (211.2 K).
(iv) The level of mercury in capillary tube is lower than the s level outside when a capillary tube is inserted in the mercury.
(v) Tea or coffee is sipped from a saucer when it is quite hot.
Answer:
(i) A liquid boils when its vapour pressure becomes equal to the atmospheric pressure. An increase in pressure on liquid, therefore, causes a rise in the boiling temperature of the liquids.
(ii) Liquids have a property, called surface tension, due to which liquids tend to contract (to decrease the surface area). For a given volume of a liquid, since a sphere has the least surface area, hence the liquids tend to form spherical droplet.
(iii) The extensive hydrogen bonding in water gives a polymeric structure. This makes the escape of molecules from the liquid more difficult. Therefore, water requires higher temperature to bring its vapour pressure equal to the atmospheric pressure.
On the other hand, sulphur being less electronegative, does not form hydrogen bonds with H of H2S. As a result, H2S has low boiling point.
(iv) The cohesive forces in mercury are much stronger than the force of adhesion between glass and mercury. Therefore, mercury-glass contract angle is greater than 90°C.
As a result, the vertical component of the surface tension forces acts vertically downward, thereby lowering the level of mercury column in the capillary tube.
(v) Evaporation causes cooling and the rate of evaporation increases with an increase in the surface area. Since, saucer has a large surface area, hence tea/coffee taken in a saucer cools quickly.

Question 2.
Nitrogen molecule (N₂) has radius of about 0.2 nm. Assuming that nitrogen molecule is spherical in shape, calculate
(i) volume of a single molecule of N₂.
(ii) the percentage of empty space in one mole of N₂ gas at STP.
Answer:
(i) The volume of a sphere = \(\frac{4}{3}\)πr³ nr where Volume of a molecule of N₂
= \(\frac{4}{3} \times \frac{22}{7}\) x (2 x 10^-8)³ cm³ – 3.35 x 10^-23 cm³

(ii) To calculate the empty space, let us first find the total volume of 1 mole (6.022 x 10²³ molecules) of N₂.
Volume of 6.022 x 10²³ molecules of N₂
= 3.35 x 10^-23 x 6.022 x 10²³ = 20.17 cm³
Now, volume occupied by 1 mole of gas at STP
= 22.4 litre = 22400 cm³
Empty volume = Total volume of gas – Volume occupied by molecules
= (22400 – 20.17) cm³ – 22379.83 cm³
∴ Percentage of empty space = \(\frac{Empty space}{Total volume}\) x 100
= \(\frac{22379.83}{22400}\) x 100 = 99.9%
Thus, 99.9% of space of 1 mole of N₂ at STP is empty.