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Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Probability Distributions Miscellaneous Exercise 7

(I) Choose the correct option from the given alternatives:

Question 1.
P.d.f. of a c.r.v. X is f(x) = 6x(1 – x), for 0 ≤ x ≤ 1 and = 0, otherwise (elsewhere) If P(X < a) = P(X > a), then a =
(a) 1
(b) \(\frac{1}{2}\)
(c) \(\frac{1}{3}\)
(d) \(\frac{1}{4}\)
Answer:
(b) \(\frac{1}{2}\)

Question 2.
If the p.d.f. of a c.r.v. X is f(x) = 3(1 – 2x²), for 0 < x < 1 and = 0, otherwise (elsewhere), then the c.d.f. of X is F(x) =
(a) 2x – 3x²
(b) 3x – 4x³
(c) 3x – 2x³
(d) 2x³ – 3x
Answer:
(c) 3x – 2x³

Question 3.
If the p.d.f. of a c.r.v. X is f(x) = \(\frac{x^{2}}{18}\), for -3 < x < 3 and = 0, otherwise, then P(|X| < 1) =
(a) \(\frac{1}{27}\)
(b) \(\frac{1}{28}\)
(c) \(\frac{1}{29}\)
(d) \(\frac{1}{26}\)
Answer:
(a) \(\frac{1}{27}\)

Question 4.
If p.m.f. of a d.r.v. X takes values 0, 1, 2, 3, … which probability P(X = x) = k(x +1) . 5-x, where k is a constant, then P(X = 0) =
(a) \(\frac{7}{25}\)
(b) \(\frac{16}{25}\)
(c) \(\frac{18}{25}\)
(d) \(\frac{19}{25}\)
Answer:
(b) \(\frac{16}{25}\)

Question 5.
If p.m.f. of a d.r.v. X is P(X = x) = \(\frac{\left({ }^{5} \mathrm{C}_{x}\right)}{2^{5}}\), for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. If a = P(X ≤ 2) and b = P(X ≥ 3), then
(a) a < b
(b) a > b
(c) a = b
(d) a + b
Answer:
(c) a = b

Question 6.
If p.m.f. of a d.r.v. X is P(X = x) = \(\frac{x}{n(n+1)}\), for x = 1, 2, 3, ……, n and = 0, otherwise, then E(X) =
(a) \(\frac{n}{1}+\frac{1}{2}\)
(b) \(\frac{n}{3}+\frac{1}{6}\)
(c) \(\frac{n}{2}+\frac{1}{5}\)
(d) \(\frac{n}{1}+\frac{1}{3}\)
Answer:
(b) \(\frac{n}{3}+\frac{1}{6}\)

Question 7.
If p.m.f. of a d.r.v. X is P(x) = \(\frac{c}{x^{3}}\), for x = 1, 2, 3 and = 0, otherwise (elsewhere), then E(X) =
(a) \(\frac{343}{297}\)
(b) \(\frac{294}{251}\)
(c) \(\frac{297}{294}\)
(d) \(\frac{294}{297}\)
Answer:
(b) \(\frac{294}{251}\)

Question 8.
If the d.r.v. X has the following probability distribution:

then P(X = -1) =
(a) \(\frac{1}{10}\)
(b) \(\frac{2}{10}\)
(c) \(\frac{3}{10}\)
(d) \(\frac{4}{10}\)
Answer:
(a) \(\frac{1}{10}\)

Question 9.
If the d.r.v. X has the following probability distribution:

then k =
(a) \(\frac{1}{7}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{1}{9}\)
(d) \(\frac{1}{10}\)
Answer:
(d) \(\frac{1}{10}\)

Question 10.
Find the expected value of X for the following p.m.f.

(a) 0.85
(b) -0.35
(c) 0.15
(d) -0.15
Answer:
(b) -0.35

(II) Solve the following:

Question 1.
Identify the random variable as either discrete or continuous in each of the following. If the random variable is discrete, list its possible values:
(i) An economist is interested in the number of unemployed graduates in the town of population 1 lakh.
(ii) Amount of syrup prescribed by a physician.
(iii) The person on a high protein diet is interesting to gain weight in a week.
(iv) 20 white rats are available for an experiment. Twelve rats are males. A scientist randomly selects 5 rats, the number of female rats selected on a specific day.
(v) A highway-safety group is interested in studying the speed (in km/hr) of a car at a checkpoint.
Solution:
(i) Let X = number of unemployed graduates in a town.
Since the population of the town is 1 lakh, X takes the finite values.
∴ random variable X is discrete.
Range = {0, 1, 2, …, 99999, 100000}.

(ii) Let X = amount of syrup prescribed by a physician.
Then X takes uncountable infinite values.
∴ random variable X is continuous.

(iii) Let X = gain of weight in a week
Then X takes uncountable infinite values
∴ random variable X is continuous.

(iv) Let X = number of female rats selected on a specific day.
Since the total number of rats is 20 which includes 12 males and 8 females, X takes the finite values.
∴ random variable X is discrete.
Range = {0, 1, 2, 3, 4, 5}

(v) Let X = speed of .the car in km/hr.
Then X takes uncountable infinite values
∴ random variable X is continuous.

Question 2.
The probability distribution of discrete r.v. X is as follows:

(i) Determine the value of k.
(ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).
Solution:

Question 3.
The following is the probability distribution of X:

Find the probability that
(i) X is positive
(ii) X is non-negative
(iii) X is odd
(iv) X is even.
Solution:
(i) P(X is positive) = P(X = 1) + P(X = 2) + P(X = 3)
= 0.25 + 0.15 + 0.1
= 0.50

(ii) P(X is non-negative)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.20 + 0.25 + 0.15 + 0.1
= 0.70

(iii) P(X is odd)
= P(X = -3) + P(X = -1) + P(X = 1) + P(X = 3)
= 0.05 + 0.15 + 0.25 + 0.1
= 0.55

(iv) P(X is even)
= P(X = -2) + P(X = 0) + P(X = 2)
= 0.10 + 0.20 + 0.15
= 0.45.

Question 4.
The p.m.f. of a r.v. X is given by P(X = x) = x = \(\frac{{ }^{5} \mathrm{C}_{\mathrm{x}}}{2^{5}}\), for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. Then show that P(X ≤ 2) = P(X ≥ 3).
Solution:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= \(\frac{{ }^{5} \mathrm{C}_{0}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{1}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{2}}{2^{5}}\)
= \(\frac{{ }^{5} \mathrm{C}_{5}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{4}}{2^{5}}+\frac{{ }^{5} \mathrm{C}_{3}}{2^{5}}\) ………[latex]{ }^{n} \mathrm{C}_{r}={ }^{n} \mathrm{C}_{n-r}[/latex]
= P(X = 5) + P(X = 4) + P(X = 3)
= P(X ≥ 3)
∴ P(X ≤ 2) = P(X ≥ 3).

Question 5.
In the p.m.f. of r.v. X

Find a and obtain c.d.f. of X.
Solution:
For p.m.f. of a r.v. X
\(\sum_{i=1}^{5} P(X=x)=1\)
∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1

Let F(x) be the c.d.f. of X.
Then F(x) = P(X ≤ x)
∴ F(1) = P(X ≤ 1) = P(X = 1) = \(\frac{1}{20}\)
F(2) = P(X ≤ 2) = P(X = 1) + P (X = 2)
\(=\frac{1}{20}+\frac{3}{20}=\frac{4}{20}=\frac{1}{5}\)
P(3) = P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}=\frac{9}{20}\)
F(4) = P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}+\frac{10}{20}=\frac{19}{20}\)
F(5) = P(X ≤ 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\(=\frac{1}{20}+\frac{3}{20}+\frac{5}{20}+\frac{10}{20}+\frac{1}{20}=\frac{20}{20}=1\)
Hence, the c.d.f. of the random variable X is as follows:

Question 6.
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Write down the probability distribution of X. Also, find the formula for p.m.f. of X.
Solution:
When a fair coin is tossed 4 times then the sample space is
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT}
∴ n(S) = 16
X denotes the number of heads.
∴ X can take the value 0, 1, 2, 3, 4
When X = 0, then X = {TTTT}
∴ n (X) = 1
∴ P(X = 0) = \(\frac{n(X)}{n(S)}=\frac{1}{16}=\frac{{ }^{4} \mathrm{C}_{0}}{16}\)
When X = 1, then
X = {HTTT, THTT, TTHT, TTTH}
∴ n(X) = 4
∴ P(X = 1) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{{ }^{4} C_{1}}{16}\)
When X = 2, then
X = {HHTT, HTHT, HTTH, THHT, THTH, TTHH}
∴ n(X) = 6
∴ P(X = 2) = \(\frac{n(X)}{n(S)}=\frac{6}{16}=\frac{{ }^{4} \mathrm{C}_{2}}{16}\)
When X = 3, then
X = {HHHT, HHTH, HTHH, THHH}
∴ n(X) = 4
∴ P(X = 3) = \(\frac{n(X)}{n(S)}=\frac{4}{16}=\frac{{ }^{4} C_{3}}{16}\)
When X = 4, then X = {HHHH}
∴ n(X) = 1
∴ P(X = 4) = \(\frac{n(X)}{n(S)}=\frac{1}{16}=\frac{{ }^{4} \mathrm{C}_{4}}{16}\)
∴ the probability distribution of X is as follows:

Also, the formula for p.m.f. of X is
P(x) = \(\frac{{ }^{4} \mathrm{C}_{x}}{16}\), x = 0, 1, 2, 3, 4 and = 0, otherwise.

Question 7.
Find the probability distribution of the number of successes in two tosses of a die, where success is defined as
(i) number greater than 4
(ii) six appear on at least one die.
Solution:
When a die is tossed two times, we obtain (6 × 6) = 36 number of observations.
Let X be the random variable, which represents the number of successes.
Here, success refers to the number greater than 4.
P(X = 0) = P(number less than or equal to 4 on both the tosses)
= \(\frac{4}{6} \times \frac{4}{6}=\frac{16}{36}=\frac{4}{9}\)
P(X = 1) = P(number less than or equal to 4 on first toss and greater than 4 on second toss) + P(number greater than 4 on first toss and less than or equal to 4 on second toss)
= \(\frac{4}{6} \times \frac{2}{6}+\frac{4}{6} \times \frac{2}{6}\)
= \(\frac{8}{36}+\frac{8}{36}\)
= \(\frac{16}{36}\)
= \(\frac{4}{9}\)
P(X = 2) = P(number greater than 4 on both the tosses)
= \(\frac{2}{6} \times \frac{2}{6}=\frac{4}{36}=\frac{1}{9}\)
Thus, the probability distribution is as follows:

(ii) Here, success means six appears on at least one die.
P(Y = 0) = P(six appears on none of the dice) = \(\frac{5}{6} \times \frac{5}{6}=\frac{25}{36}\)
P(Y = 1) = P(six appears on none of the dice x six appears on at least one of the dice ) + P(six appears on none of the dice x six appears on at least one of the dice)
= \(\frac{1}{6} \times \frac{5}{6}+\frac{1}{6} \times \frac{5}{6}=\frac{5}{36}+\frac{5}{36}=\frac{10}{36}\)
P(Y = 2) = P(six appears on at least one of the dice) = \(\frac{1}{6} \times \frac{1}{6}=\frac{1}{36}\)
Thus, the required probability distribution is as follows:

Question 8.
A random variable X has the following probability distribution:

Determine:
(i) k
(ii) P(X > 6)
(iii) P(0 < X < 3).

Question 9.
The following is the c.d.f. of a r.v. X:

Find
(i) p.m.f. of X
(ii) P( -1 ≤ X ≤ 2)
(iii) P(X ≤ X > 0).
Solution:
(i) From the given table
F(-3) = 0.1, F(-2) = 0.3, F(-1) = 0.5
F(0) = 0.65, f(1) = 0.75, F(2) = 0.85
F(3) = 0.9, F(4) = 1
P(X = -3) = F(-3) = 0.1
P(X = -2) = F(-2) – F(-3) = 0.3 – 0.1 = 0.2
P(X = -1) = F(-1) – F(-2) = 0.5 – 0.3 = 0.2
P(X = 0) = F(0) – F(-1) = 0.65 – 0.5 = 0.15
P(X = 1) = F(1) – F(0) = 0.75 – 0.65 = 0.1
P(X = 2) = F(2) – F(1) = 0.85 – 0.75 = 0.1
P(X = 3) = F(3) – F(2) = 0.9 – 0.85 = 0.1
P(X = 4) = F(4) – F(3) = 1 – 0.9 = 0.1
∴ the p.m.f of X is as follows:

(ii) P(-1 ≤ X ≤ 2) = P(X = -1) + P(X = 0) + P(X = 1) + P(X = 2)
= 0.2 + 0.15 + 0.1 + 0.1
= 0.55

(iii) (X ≤ 3) ∩ (X > 0)
= { -3, -2, -1, 0, 1, 2, 3} n {1, 2, 3, 4}
= {1, 2, 3}

Question 10.
Find the expected value, variance, and standard deviation of the random variable whose p.m.f’s are given below:

Solution:
(i) We construct the following table to find the expected value, variance, and standard deviation:

(ii) We construct the following table to find the expected value, variance, and standard deviation:



(iii) We construct the following table to find the expected value, variance, and standard deviation:



(iv) We construct the following table to find the expected value, variance, and standard deviation:

Question 11.
A player tosses two wins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears and ₹ 2 if no head appears. Find the expected winning amount and variance of the winning amount.
Solution:
When a coin is tossed twice, the sample space is
S = {HH, HT, TH, HH}
Let X denote the amount he wins.
Then X takes values 10, 5, 2.
P(X = 10) = P(2 heads appear) = \(\frac{1}{4}\)
P(X = 5) = P(1 head appears) = \(\frac{2}{4}\) = \(\frac{1}{2}\)
P(X = 2) = P(no head appears) = \(\frac{1}{4}\)
We construct the following table to calculate the mean and the variance of X:

From the table Σx_i . P(x_i) = 5.5, \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) = 38.5
E(X) = Σx_i . P(x_i) = 5.5
Var(X) = \(\Sigma x_{i}^{2} \cdot P\left(x_{i}\right)\) – [E(X)]²
= 38.5 – (5.5)²
= 38.5 – 30.25
= 8.25
∴ Hence, expected winning amount = ₹ 5.5 and variance of winning amount = ₹ 8.25.

Question 12.
Let the p.m.f. of r.v. X be P(x) = \(\frac{3-x}{10}\), for x = -1, 0, 1, 2 and = 0, otherwise.
Calculate E(X) and Var(X).
Solution:
P(X) = \(\frac{3-x}{10}\)
X takes values -1, 0, 1, 2
P(X = -1) = P(-1) = \(\frac{3+1}{10}=\frac{4}{10}\)
P(X = 0) = P(0) = \(\frac{3-0}{10}=\frac{3}{10}\)
P(X = 1) = P(1) = \(\frac{3-1}{10}=\frac{2}{10}\)
P(X = 2) = P(2) = \(\frac{3-2}{10}=\frac{1}{10}\)
We construct the following table to calculate the mean and variance of X:

From the table
Σx_iP(x_i) = 0 and \(\Sigma x_{i}{ }^{2} \cdot P\left(x_{i}\right)\) = 1
E(X) = Σx_iP(x_i) = 0
Var(X) = \(\Sigma x_{i}{ }^{2} \cdot P\left(x_{i}\right)\) – [E(X)]²
= 1 – 0
= 1
Hence, E(X) = 0, Var (X) = 1.

Question 13.
Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.
f(x) = k(4 – x²), -2 ≤ x ≤ 2 and = 0 otherwise.
Compute
(i) P(X > 0)
(ii) P(-1 < X < 1)
(iii) P(X < -0.5 or X > 0.5).
Solution:
(i) P(X > 0)


(ii) P(-1 < X < 1)


(iii) P(X < -0.5 or X > 0.5)

Question 14.
The p.d.f. of a continuous r.v. X is given by f(x) = \(\frac{1}{2 a}\), for 0 < x < 2a and = 0, otherwise. Show that P( X < \(\frac{a}{2}\)) = P(X > \(\frac{3a}{2}\))
Solution:

Question 15.
The p.d.f. of r.v. X is given by f(x) = \(\frac{k}{\sqrt{x}}\), for 0 < x < 4 and = 0, otherwise. Determine k. Determine c.d.f. of X and hence find P(X ≤ 2) and P(X ≤ 1).
Solution:
Since f is p.d.f. of the r.v. X,

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 7 Probability Distributions Ex 7.2 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 7 Probability Distributions Ex 7.2

Question 1.
Verify which of the following is p.d.f. of r.v. X:
(i) f(x) = sin x, for 0 ≤ x ≤ \(\frac{\pi}{2}\)
(ii) f(x) = x, for 0 ≤ x ≤ 1 and -2 – x for 1 < x < 2
(iii) fix) = 2, for 0 ≤ x ≤ 1.
Solution:
f(x) is the p.d.f. of r.v. X if
(a) f(x) ≥ 0 for all x ∈ R and

Hence, f(x) is the p.d.f. of X.

(ii) f(x) = x ≥ 0 if 0 ≤ x ≤ 1
For 1 < x < 2, -2 < -x < -1
-2 – 2 < -2 – x < -2 – 1
i.e. -4 < f(x) < -3 if 1 < x < 2
Hence, f(x) is not p.d.f. of X.

(iii) (a) f(x) = 2 ≥ 0 for 0 ≤ x ≤ 1

Hence, f(x) is not p.d.f. of X.

Question 2.
The following is the p.d.f. of r.v. X:
f(x) = \(\frac{x}{8}\), for 0 < x < 4 and = 0 otherwise.
Find
(a) P(x < 1.5)
(b) P(1 < x < 2) (c) P(x > 2).
Solution:

Question 3.
It is known that error in measurement of reaction temperature (in 0°C) in a certain experiment is continuous r.v. given by
f(x) = \(\frac{x^{2}}{3}\) for -1 < x < 2
= 0. otherwise.
(i) Verify whether f(x) is p.d.f. of r.v. X
(ii) Find P(0 < x ≤ 1)
(iii) Find the probability that X is negative.
Solution:

Question 4.
Find k if the following function represents p.d.f. of r.v. X
(i) f(x) = kx. for 0 < x < 2 and = 0 otherwise.
Also find P(\(\frac{1}{4}\) < x < \(\frac{3}{2}\)).
(ii) f(x) = kx(1 – x), for 0 < x < 1 and = 0 otherwise.
Also find P(\(\frac{1}{4}\) < x < \(\frac{1}{2}\)), P(x < \(\frac{1}{2}\)).
Solution:
(i) Since, the function f is p.d.f. of X

(ii) Since, the function f is the p.d.f. of X,

Question 5.
Let X be the amount of time for which a book is taken out of the library by a randomly selected students and suppose X has p.d.f.
f(x) = 0.5x, for 0 ≤ x ≤ 2 and = 0 otherwise.
Calculate:
(i) P(x ≤ 1)
(ii) P(0.5 ≤ x ≤ 1.5)
(iii) P(x ≥ 1.5).
Solution:
(i) P(x ≤ 1)

(ii) P(0.5 ≤ x ≤ 1.5)

(iii) P(x ≥ 1.5)

Question 6.
Suppose that X is waiting time in minutes for a bus and its p.d.f. is given by f(x) = \(\frac{1}{5}\), for 0 ≤ x ≤ 5 and = 0 otherwise. Find the probability that
(i) waiting time is between 1 and 3
(ii) waiting time is more than 4 minutes.
Solution:
(i) Required probability = P(1 < X < 3)

(ii) Required probability = P(X > 4)

Question 7.
Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f.
f(x) = k(4 – x²), -2 ≤ x ≤ 2 and 0 otherwise.
Compute:
(i) P(X > 0)
(ii) P(-1 < X < 1)
(iii) P(-0.5 < X or X > 0.5).
Solution:
Since, f is the p.d.f. of X,

Question 8.
The following is the p.d.f. of continuous r.v. X
f(x) = \(\frac{x}{8}\), for 0 < x < 4 and = 0 otherwise.
(i) Find expression for c.d.f. of X.
(ii) Find F(x) at x = 0.5, 1.7 and 5.
Solution:
(i) Let F(x) be the c.d.f. of X

Question 9.
Given the p.d.f. of a continuous random r.v. X, f(x) = \(\frac{x^{2}}{3}\), for -1 < x < 2 and = 0 otherwise. Determine c.d.f. of X and hence find P(X < 1); P(X < -2), P(X > 0), P(1 < X < 2).
Solution:

Question 10.
If a r.v. X has p.d.f.
f(x) = \(\frac{c}{x}\) for 1 < x < 3, c > 0. Find c, E(X), Var (X).
Solution:
Since f(x) is p.d.f of r.v. X

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 10 States of Matter Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 10 States of Matter

1. Select and write the most appropriate alternatives from the given choices.

Question A.
The unit of viscosity is
a. dynes
b. newton
c. gram
d. poise
Answer:
d. poise

Question B.
Which of the following is true for 2 moles of an ideal gas?
a. PV = nRT
b. PV = RT
c. PV = 2RT
d. PV = T
Answer:
c. PV = 2RT

Question C.
Intermolecular forces in liquid are
a. greater than gases
b. less than solids
c. both a and b
d. greater than solids
Answer:
c. both a and b

Question D.
Interactive forces are ………. in ideal gas.
a. nil
b. small
c. large
d. same as that of real gases
Answer:
a. nil

Question E.
At constant temperature the pressure of 22.4 dm³ volume of an ideal gas was increased from 10⁵ kPa to 210 kPa, New volume could be-
a. 44.8 dm³
b. 11.2 dm³
c. 22.4 dm³
d. 5.6 dm³
Answer:
b. 11.2 dm³

2. Answer in one sentence.

Question A.
Name the term used for mixing of different gases by random molecular motion and ferquent collision.
Answer:
The mixing of different gases by random molecular motion and frequent collision is called diffusion.

Question B.
The pressure that each individual gas would exert if it were alone in the container, what do we call it as ?
Answer:
The pressure that each individual gas would exert if it were alone in the container is called as partial pressure.

Question C.
When a gas is heated the particles move more quickly. What is the change in volume of a heated gas if the pressure is kept constant ?
Answer:
The volume of the gas increases on heating if pressure is kept constant.

Question D.
A bubble of methane gas rises from the bottom of the North sea. What will happen to the size of the bubble as it rises to the surface ?
Answer:
According to Boyle’s law, the size of the bubble of methane gas increases as it rises to the surface.

Question E.
Convert the following temperatures from degree celcius to kelvin.
a. -15° C
b. 25° C
c. -197° C
d. 273° C
Answer:
a. T(K) = t°C +273.15
∴ T(K) = -15 °C + 273.15 = 258.15 K
b. T(K) = t°C +273.15
∴ T(K) = 25 °C + 273.15 = 298.15 K
c. T(K) = t°C + 273.15
∴ T(K) = -197 °C + 273.15 = 76.15 K
d. T(K) = t°C + 273.15
∴ T(K) = 273 °C + 273.15 = 546.15 K

Question F.
Convert the following pressure values into Pascals.
a. 10 atmosphere
b. 1 kPa.
c. 107000 Nm^-2
d. 1 atmosphere
Answer:
a. 10 atmosphere:
1 atm = 101325 Pa
∴ 10 atm = 1013250 Pa
= 1.01325 × 10⁶ Pa

b. 1 kPa:
1 kPa = 1000 Pa

c. 107000 N m^-2:
1 N m^-2 = 1 Pa
∴ 107000 Nm^-2 = 107000 Pa
= 1.07 × 10⁵ Pa

d. 1 atmosphere:
1 atm = 101325 Pa
= 1.01325 × 10⁵ Pa

Question G.
Convert:
a. Exactly 1.5 atm to pascals
b. 89 kPa to newton per square metre (N m^-2)
c. 101.325 kPa to bar
d. -100 °C to Kelvin
e. 0.124 torr to standard atmosphere
Answer:
a. Exactly 1.5 atm to pascals:
1 atm = 101325 Pa
∴ 1.5 atm = 1.5 × 101325
= 151987.5 Pa

b. 89 kPa to newton per square metre (N m^-2):
1 Pa = 1 N m^-2 and 1 Pa = 10^-3 kPa
∴ 10^-3 kPa = 1 N m^-2
∴ 89 kPa = \(\frac{1 \times 89}{10^{-3}}\) N m^-2 = 89000 N m^-2

c. 101.325 kPa to bar:
1 bar = 1.0 × 10⁵ Pa
= 1.0 × 10² k Pa
∴ 100 kPa = 1 bar
∴ 101.325 kPa = \(\frac{1 \times 101.325}{100}\)
= 1.01325 bar

d. -100 °C to Kelvin:
T(K) = t °C + 273.15
∴ T(K) = (- 100 °C) + 273.15 = 173.15 K

e. 0.124 torr to standard atmosphere:
1 atm = 760 torr
∴ 1 torr = \(\frac {1}{760}\)atm
∴ 0.124 torr = 0.124 × \(\frac {1}{760}\)
= 1.632 × 10^-4 atm

Question H.
If density of a gas is measured at constant temperature and pressure then which of the following statement is correct ?
a. Density is directly proportional to molar mass of the gas.
b. Greater the density greater is the molar mass of the gas.
c. If density, temperature and pressure is given ideal gas equation can be used to find molar mass.
d. All the above statements are correct.
Answer:
d. All the above statements are correct.

Question I.
Observe the following conversions.

Which of the above reactions is in accordance with the priciple of stoichiometry ?
Answer:
Both the reactions are in accordance with the principle of stoichiometry.
In the first reaction, both the reactants are completely consumed to form product according to reaction stoichiometry.
1 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride
In the second reaction, chlorine is the limiting reagent and it is completely consumed to form hydrogen chloride. Excess hydrogen remains unreacted at the end of the reaction. This reaction also follows principle of stoichiometry.
2 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride + 1 mol hydrogen

Question J.
Hot air balloons float in air because of the low density of the air inside the balloon. Explain this with the help of an appropriate gas law.

Answer:
The working of hot air balloon can be explained with the help of Charles’ law. According to Charles’ law, at constant pressure, the volume of a fixed amount of a gas varies directly with the temperature. This means that as the temperature increases, the air inside the balloon expands and occupies more volume. Thus, hot air inside the balloon is less dense than the surrounding cold air. This causes the hot air balloon to float in air.

3. Answer the following questions.

Question A.
Identify the gas laws from the following diagrams.

Answer:
a. Boyle’s law
b. Charles’ law
c. Avogadro’s law [Note: Assuming, T constant]

Question B.
Consider a sample of a gas in a cylinder with a movable piston.

Show digramatically the changes in the position of piston, if
a. Pressure is increased from 1.0 bar to 2.0 bar at constant temperature.
b. Temperature is decreased from 300 K to 150 K at constant pressure
c. Temperature is decreased from 400 K to 300 K and pressure is decreased from 4 bar to 3 bar.
Answer:

Thus, the volume of the gas remains the same.
Hence, there will be no change in the position of the piston.

Question D.
List the characteristic physical properties of the gases.
Answer:
Characteristic physical properties of the gases:

• Gases are lighter than solids and liquids (i.e., possess lower density).
• Gases do not possess a fixed volume and shape. They occupy entire space available and take the shape of the container.
• Gas molecules are widely separated and are in continuous, random motion. Therefore, gases exert pressure equally in all directions due to collision of gas molecules, on the walls of the container.
• In case of gases, intermolecular forces are weakest.
• Gases possess the property of diffusion, which is a spontaneous homogeneous inter mixing of two or more gases.
• Gases are highly compressible.

Question E.
Define the terms:
a. Polarizability
b. Hydrogen bond
c. Aqueous tension
d. Dipole moment
Answer:
a. Polarizability is defined as the ability of an atom or a molecule to form momentary dipoles, that means, the ability of the atom or molecule to become polar by redistributing its electrons.

b. The electrostatic force of attraction between positively polarised hydrogen atom of one molecule and a highly electronegative atom (which may be negatively charged) of other molecule is called as hydrogen bond.

c. The pressure exerted by saturated water vapour is called aqueous tension.

d. Dipole moment (p) is the product of the magnitude of the charge (Q) and the distance between the centres of positive and negative charge (r). It is designated by a Greek Letter (p) and its unit is Debye (D).

Question F.
Would it be easier to drink water with a straw on the top of the Mount Everest or at the base ? Explain.
Answer:
When you drink through a straw, the pressure inside the straw reduces (as the air is withdraw by mouth) and the liquid is pushed up to your mouth by atmospheric pressure. Thus, drinking with a straw makes use of pressure difference to force the liquid into your mouth. So, if the pressure difference is less it will be difficult to drink through a straw. On the top of the Mount Everest, atmospheric pressure is very low. Hence, it will be difficult to drink water with a straw on the top of Mount Everest as compared to at the base.

Question G.
Identify type of the intermolecular forces in the following compounds.
a. CH₃ – OH
b. CH₂ = CH₂
c. CHCl₃
d. CH₂Cl₂
Answer:
a. Hydrogen bonding (dipole-dipole attraction) and London dispersion forces
b. London dispersion forces
c. Dipole-dipole interactions and London dispersion forces
d. Dipole-dipole interactions and London dispersion forces

Question H.
Name the types of intermolecular forces present in Ar, Cl₂, CCl₄ and HNO₃.
Answer:
a. Ar: London dispersion forces
b. Cl₂: London dispersion forces
c. CCl₄: London dispersion forces
d. HNO₃: Flydrogen bonding (dipole-dipole attraction) and London dispersion forces.

Question I.
Match the pairs of the following :

	A		B
a.	Boyle’s law	i.	At constant pressure and volume
b.	Charles’ law	ii.	At constant temperature
		iii.	At constant pressure

Answer:
a – ii,
b – iii

Question J.
Write the statement for :
(a) Boyle’s law
(b) Charles’ law
Answer:
a. Statement of Boyle’s law: For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.
OR
At constant temperature, the pressure of fixed amount (number of moles) of a gas varies inversely with its volume.

b. Statement for Charles’ law:
‘At constant pressure, the volume of a fixed mass of a gas is directly proportional to its temperature in Kelvin.

Question K.
Differentiate between Real gas and Ideal gas.
Answer:
Ideal gas:

1. Strictly obeys Boyle’s and Charles’ law.
   \(\frac{\mathrm{PV}}{\mathrm{nRT}}\) = 1
2. Molecules are perfectly elastic.
3. No attraction or repulsion between the gas molecules i.e. collision without loss of kinetic energy (K.E.)
4. Actual volume of the gas molecules is negligible as compared to total volume of the gas.
5. Ideal gases cannot be liquified even at low temperature but continues to obey Charles’ law and finally occupies zero volume at 0 K.
6. Practically, ideal gas does not exist.

Real gas:

1. Shows deviation from Boyle’s and Charles’ law at high pressure and temperature, i.e. obeys Boyle’s law and Charles’ law at low pressure and high temperature. \(\frac{\mathrm{PV}}{\mathrm{nRT}}\) ≠ 1
2. Molecules are not perfectly elastic.
3. Intermolecular attraction is present, hence collision takes place with loss of kinetic energy.
4. Actual volume of individual gas molecule is significant at high pressure and low- temperature.
5. Real gases undergo liquefaction at low’ temperature when cooled and compressed.
6. Gases that exist in nature like H₂, O₂, CO₂, N₂, He, etc. are real gases.

4. Answer the following questions

Question A.
State and write mathematical expression for Dalton’s law of partial pressure and explain it with suitable example.
Answer:
i. Statement: The total pressure of a mixture of two or more non-reactive gases is the sum of the partial pressures of the individual gases in the mixture.
ii. Explanation:
Dalton’s law can be mathematically expressed as:
P_Total = P₁ + P₂ + P₃ …(at constant T and V)
where, P_Total is the total pressure of the mixture and P₁, P₂, P₃, … are the partial pressures of individual gases 1, 2, 3, … in the mixture.
For example, consider two non-reactive gases A and B. On mixing the two gases, pressure exerted by individual gas A in the mixture of both the gases is called partial pressure of gas A (say P₁). Likewise, partial pressure of gas B is P₂. According to Dalton’s law, total pressure of the mixture of gas A and B at constant T and V will be given as:
P_Total = P₁ + P₂

iii. Schematic illustration of Dalton’s law of partial pressures:

Question B.
Derive an Ideal gas equation. Mention the terms involved in it. Also write how it is utilised to obtain combined gas law.
Answer:
According to Boyle’s law,
V ∝ \(\frac{1}{\mathrm{P}}\) (at constant T and n) ……….(1)
According to Charles’ law,
V ∝ T (at constant P and n) ……(2)
According to Avogadro’s law,
V ∝ n (at constant P and T) ……(3)
Combining relations (1), (2) and (3), we get
V ∝ \(\frac{\mathrm{nT}}{\mathrm{P}}\)
Converting this proportionality into an equation by introducing a constant of proportionality (‘R’ known as gas constant), we get
∴ V = \(\frac{\mathrm{nRT}}{\mathrm{P}}\)
On rearranging the above equation, we get
PV = nRT
where,
P = Pressure of gas,
V = Volume of gas,
n = number of moles of gas,
R = Gas constant,
T = Absolute temperature of gas.
This is the ideal gas equation or equation of state.
[Note: In the ideal gas equation, R is called gas constant or universal gas constant, whose value is same for all the gases. In this equation, if three variables are known, fourth can be calculated. The equation describes the state of an ideal gas. Hence, it is also called as equation of state.]

The ideal gas equation is written as PV = nRT …(1)
On rearranging equation (1), we get,

The ideal gas equation used in this form is called combined gas law.

Question C.
With the help of graph answer the following –

At constant temperature,
a. Graph shows relation between pressure and volume. Represent the relation mathematically.
b. Identify the law.
c. Write the statement of law.
Answer:
a. P ∝ \(\frac{1}{\mathrm{~V}}\)
b. The graph represents Boyle’s law as it gives relation between pressure and volume at constant temperature.
c. Statement of Boyle’s law: For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.
OR
At constant temperature, the pressure of fixed amount (number of moles) of a gas varies inversely with its volume.

Question D.
Write Postulates of kinetic theory of gases.
Answer:
Postulates of kinetic theory of gases:

• Gases consist of tiny particles (molecules or atoms).
• On an average, gas molecules remain far apart from each other. Therefore, the actual volume of the gas molecules is negligible as compared to the volume of the container. Hence, gases are highly compressible.
• The attractive forces between the gas molecules are negligible at ordinary temperature and pressure. As a result, gas expands to occupy entire volume of the container.
• Gas molecules are in constant random motion and move in all possible directions in straight lines. They collide with each other and with the walls of the container.
• Pressure of the gas is due to the collision of gas molecules with the walls of the container.
• The collisions of the gas molecules are perfectly elastic in nature, which means that the total energy of the gaseous particle remains unchanged after collision.
• The different gas molecules move with different velocities at any instant and hence have different kinetic energies. However, the average kinetic energy of the gas molecules is directly proportional to the absolute temperature.

Question E.
Write a short note on
a. Vapour pressure.
b. Surface tension
c. Viscosity.
Answer:
a. Vapour pressure:

• Molecules of liquid have tendency to escape from its surface to form vapour above it. This called evaporation.
• When a liquid is placed in a closed container, the liquid undergoes evaporation and vapours formed undergo condensation.
• At equilibrium, the rate of evaporation and rate of condensation are equal.
• The pressure exerted by the vapour in equilibrium with the liquid is known as saturated vapour pressure or simply vapour pressure.
• Vapour pressure is measured by means of a manometer.
• The most common unit for vapour pressure is torr. 1 torr = 1 mm Hg.

[Note: i. The vapour pressure of water is also called aqueous tension.
ii. Water has a vapour pressure of approximately 20 torr at room temperature.]

b. Surface tension:

• The particles in the bulk of liquid are uniformly attracted in all directions and the net force acting on the molecules present inside the bulk is zero.
• But the molecules at the surface experience a net attractive force towards the interior of the liquid, or the forces acting on the molecules on the surface are imbalanced.
• Therefore, liquids have tendency to minimize their surface area and the surface acts as a stretched membrane.
• The force acting per unit length perpendicular to the line drawn on the surface of liquid is called surface tension.
• Unit: Surface tension is measured in SI unit, N m^-1 and is denoted by Greek letter ‘γ’

c. Viscosity:
i. Liquids (fluids) have tendency to flow.
ii. Viscosity measures the magnitude of internal friction in a liquid or fluid to flow as measured by the force per unit area resisting uniform flow.
iii. Different layers of a liquid flow with different velocity. This called laminar flow. Here, the layers of molecules in the immediate contact of the fixed surface remains stationary. The subsequent layers slip over one another. Strong intermolecular forces obstruct the layers from slipping over one another, resulting in a friction between the layers.
iv. Viscosity is defined as the force of friction between the successive layers of a flowing liquid. It is also the resistance to the flow of a liquid.
v. When a liquid flow through a tube, the central layer has the highest velocity, whereas the layer along the inner wall in the tube remains stationary. This is a result of the viscosity of a liquid. Hence, a velocity gradient exists across the cross-section of the tube.

vi. Viscosity is expressed in terms of coefficient of viscosity, ‘η’ (Eta). The SI unit of viscosity coefficient is N s m^-2 (newton second per square meter). In CGS system, the unit (η) is measured in poise.
1 poise = 1 g cm^-1 s^-1 = 10^-1 kg m^-1 s^-1

5. Solve the following

Question A.
A balloon is inflated with helium gas at room temperature of 25 °C and at 1 bar pressure when its initial volume is 2.27L and allowed to rise in air. As it rises in the air external pressure decreases and the volume of the gas increases till finally it bursts when external pressure is 0.3bar. What is the limit at which volume of the balloon can stay inflated ?
Answer:
Given: P₁ = Initial pressure = 1 bar
V₁ = Initial volume = 2.27 L
P₂ = Final pressure = 0.3 bar
To find: V₂ = Final volume
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
∴ V₂ = \(\frac{P_{1} V_{1}}{P_{2}}=\frac{1 \times 2.27}{0.3}\) = 7.566667 L ≈ 7.567 L
Ans: The balloon can stay inflated below the volume of 7.567 L.

Question B.
A syringe has a volume of 10.0 cm³ at pressure 1 atm. If you plug the end so that no gas can escape and push the plunger down, what must be the final volume to change the pressure to 3.5 atm?

Answer:
Given: P₁ = Initial pressure = 1 atm
V₁ = Initial volume = 10.0 cm³
P₂ = Final pressure = 3.5 atm
To find: V₂ = Final volume
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
∴ V₂ = \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2}}=\frac{1 \times 10.0}{3.5}\)
= 2.857 L
Ans: The final volume of the gas in the syringe is 2.857 L.

Question C.
The volume of a given mass of a gas at 0°C is 2 dm³. Calculate the new volume of the gas at constant pressure when
a. The temperature is increased by 10°C.
b.The temperature is decreased by 10°C.
Answer:
Given: T₁ = Initial temperature = 0 °C = 0 + 273.15 = 273.15 K,
V₁ = Initial volume = 2 dm³
a. T₂ = Final temperature = 273.15 K + 10 = 283.15 K
b. T₂ = Final temperature = 273.15 K – 10 = 263.15 K
To find: V₂ = Final volume in both the cases
Formula: \(\frac{\mathrm{V}_{\mathrm{l}}}{\mathrm{T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,

Ans: The new volume of a given mass of gas is:
a. 2.073 dm³
b. 1.927 dm³

Question D.
A hot air balloon has a volume of 2800 m³ at 99 °C. What is the volume if the air cools to 80 °C?

Answer:
Given: V₁ = Initial volume = 2800 m³, T₁ = Initial temperature = 99 °C = 99 + 273.15 = 372.15 K,
T₂ = Final temperature = 80 °C = 80 + 273.15 K = 353.15 K
To find: V₂ = Final volume
Formula: = \(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,
\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (at constant n and P)
∴ \(\mathrm{V}_{2}=\frac{\mathrm{V}_{1} \mathrm{~T}_{2}}{\mathrm{~T}_{1}}=\frac{2800 \times 353.15}{372.15}=\mathbf{2 6 5 7 \mathrm { m } ^ { 3 }}\)
Ans: The volume of the balloon when the air cools to 80 °C is 2657 m³.

Question E.
At 0 °C, a gas occupies 22.4 liters. How nuch hot must be the gas in celsius and in kelvin to reach volume of 25.0 literes?
Answer:
V₁ = Initial volume of the gas = 22.4 L,
T₁ = Initial temperature = 0 + 273.15 = 273.15 K,
V₂ = Final volume = 25.0 L
To find: T₂ = Final temperature in Celsius and in Kelvin

Ans: The temperature of the gas must be 31.7 °C or 304.9 K.

Question F.
A 20 L container holds 0.650 mol of He gas at 37 °C at a pressure of 628.3 bar. What will be new pressure inside the container if the volume is reduced to 12 L. The temperature is increased to 177 °C and 1.25 mol of additional He gas was added to it?
Answer:
Given: V₁ = Initial volume = 20 L, n₁ = Initial number of moles = 0.650 mol
P₁ = Initial pressure = 628.3 bar
T₁ = Initial temperature = 37 °C = 37 + 273.15 K = 310.15 K
n₂ = Final number of moles = 0.650 + 1.25 = 1.90 mol, V₂ = Final volume = 12 L
T₂ = Final temperature = 177 °C = 177 + 273.15 K = 450.15 K, R = 0.0821 L atm K^-1 mol^-1
To find: P₂ = Final pressure
Formula: PV = nRT
Calculation: According to ideal gas equation,
P₂V₂ = n₂RT₂.
∴ \(\mathrm{P}_{2}=\frac{\mathrm{n}_{2} \mathrm{RT}_{2}}{\mathrm{~V}_{2}}=\frac{1.90 \times 0.0821 \times 450.15}{12}=\mathbf{5 . 8 5 2} \mathrm{atm}\)
Ans: The final pressure of the gas is 5.852 atm.
[Note: In the above numerical, converting the pressure value to different units, we get: 5.852 atm = 4447.52 torr = 5.928 bar]

Question G.
Nitrogen gas is filled in a container of volume 2.32 L at 32 °C and 4.7 atm pressure. Calculate the number of moles of the gas.
Answer:
Given: V = 2.32 L, P = 4.7 atm, T = 32 °C = 32 + 273.15 K = 305.15 K
R = 0.0821 L atm K^-1 mol^-1
To find: n = number of moles of gas
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT
∴ \(\mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{4.7 \times 2.32}{0.0821 \times 305.15}=\mathbf{0 . 4 3 5} \mathrm{moles}\)
Ans: Number of moles of N₂ gas in the given volume is 0.435 moles.

Question H.
At 25 °C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at the height where temperature is 10 °C and volume of the gas 640 mL ?
Answer:
Given: V₁ = Initial volume = 600 mL, V₂ = Final volume = 640 mL
P₁ = Initial pressure = 760 mm Hg
T₁ = Initial temperature = 25 °C = 25 + 273.15 K = 298.15 K
T₂ = Final temperature = 10 °C = 10 + 273.15 K = 283.15 K
P₂ = Final pressure
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: According to combined gas law.

Ans: The final pressure of a gas is 676.654 mm Hg.

Question I.
A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5g neon. If pressure of the mixture of the gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture?
Answer:
Given: m_O2 = 70.6 g, m_Ne = 167.5 g,
P_Total = 25 bar
To find: Partial pressure of each gas
Formula: P₁ = x₁ × P_Total
Calculation: Determine number of moles (n) of each gas using formula: n = \(\frac{\mathrm{m}}{\mathrm{M}}\)

Ans: The partial pressure of dioxygen and neon are 5.2 bar and 19.8 bar respectively.

Question J.
Calculate the pressure in atm of 1.0 mole of helium in a 2.0 dm³ container at 20.0 °C.
Answer:
Given: n = number of moles = 1.0 mol, V = volume = 2.0 dm³
T = Temperature = 20.0 °C = 20.0 + 273.15 K = 293.15 K
R = 0.0821 L atm K^-1 mol^-1
To find: Pressure (P)
Formula: PV = nRT
Calculation: According to ideal gas equation,

Ans: The pressure of the given helium gas is 12.03 atm.

Question K.
Calculate the volume of 1 mole of a gas at exactly 20 °C at a pressure of 101.35 kPa.
Answer:
Given: n = number of moles = 1 mol, P = pressure = 101.35 kPa = 1.00025 atm ≈ 1 atm
T = Temperature = 20 °C = 20 + 273.15 K = 293.15 K
R = 0.0821 dm³ atm K^-1 mol^-1
To find: Volume (V)
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT

Ans: The volume of the given gas is 24.07 dm³.

Question L.
Calculate the number of molecules of methane in 0.50 m³ of the gas at a pressure of 2.0 × 10² kPa and a temperature of exactly 300 K.
Answer:
V = 0.5 m³, P = 2.0 × 10² kPa = 2.0 × 10⁵ Pa
T = 300 K, R = 8.314 J K^-1 mol^-1
To find: Number of molecules of methane gas
Formula: PV = nRT
Calculation: According to ideal gas equation,
n = \(\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{2.0 \times 10^{5} \times 0.5}{8.314 \times 300}=40 \mathrm{~mol}\)
Number of molecules = n × NA = 40 × 6.022 × 10²³ = 2.4088 × 10²³ ≈ 2.409 × 10²⁵
Ans: The number of molecules of methane gas present is 2.409 × 10²⁵ molecules.

11th Chemistry Digest Chapter 10 States of Matter Intext Questions and Answers

Do you know? (Textbook Page No. 140)

Question 1.
Consider three compounds: H₂S, H₂Se and H₂O. Identify which has the highest boiling point. Justify.
Answer:
Among the three compounds H₂O, H₂S and H₂Se, the first one, H₂O has the smallest molecular mass. But it has the highest B.P. of 100 °C. B.P. of H₂S is -60 °C and of H₂Se is -41.25 °C. The extraordinary high B.P. of H₂O is due to very strong hydrogen bonding even though it has the lowest molecular mass.

Can you tell? (Textbook Page No. 140)

Question i.
What are the various components present in the atmosphere?
Answer:
Various components present in the atmosphere are as follows:
a. Nitrogen (78%)
b. Oxygen (21%)
c. Carbon dioxide and other gases (0.03%)
d. Inert gases (mainly argon) (0.97%)
e. Traces of water vapour

Question ii.
Name five elements and five compounds those exist as gases at room temperature.
Answer:
Five elements and five compounds that exist as gases at room temperature are as follows:

No.	Elements
a.	Nitrogen
b.	Oxygen
c.	Hydrogen
d.	Chlorine
e.	Argon

No.	Compounds
a.	Carbon dioxide
b.	Carbon monoxide
c.	Nitrogen dioxide
d.	Sulphur dioxide
e.	Methane

Just think. (Textbook Page No. 140)

Question 1.
What is air?
Answer:

• Air is a mixture of various gases.
• One cannot see air but can feel the cool breeze.
• The composition of air by volume is around 78 percent N₂, 21 percent O₂ and 1 percent other gases including CO₂.

Use your brainpower. (Textbook Page No. 141)

Question 1.
Find the unit in which car-tyre pressure is measured.
Answer:
Car-tyre pressure is measured in the units of pounds per square inch (psi) or Newton per metre square (N m^-2).

Do you know? (Textbook Page No. 142)

Question 1.
How does a bicycle pump work?
Answer:
A bicycle pump works on Boyle’s law. Pushing a bicycle pump squashes the same number of particles into a smaller volume. This squashing means particles hit the walls of the pump more often, increasing the pressure. The increased pressure of a gas can be felt on palm by pushing in the piston of a bicycle pump.

Internet my friend (Textbook Page No. 143)

Question 1.
i. Watch Boyle’s law experiment.
ii. Find applications of Boyle’s law.
iii. Try to study how Boyle’s law helps in ‘scuba-diving’ i.e., importance of Boyle’s law in scuba diving an exhilarating sport.
Answer:
i. Students can refer to ‘Boyle’s law experiment’ on YouTube channel of ‘Socratica’.
ii. a. Syringes: When the plunger of a syringe is pulled back out, it causes the volume of the gas inside it to increase due to the reduction of pressure. This creates a vacuum in the syringe, which is constantly trying to adjust the pressure back to normal. However, since the only substance available, such as the blood or medication, is on the other side of the needle, this liquid is sucked into the vacuum, increasing the pressure and decreasing the volume of the gas. When we push the plunger back down, the pressure again increases, lowering the volume inside the syringe, and pushing the fluid out.

b. Respiration: Boyle’s law is essential for the human breathing process. When person breathes in, his/her lung volume increases and the pressure within decreases. Since air always moves from areas of high pressure to areas of low pressure, air is drawn into the lungs. The opposite happens when person exhales. Since the lung volume decreases, the pressure within increases, forcing the air out of the lungs

c. Storage of Gases: Many industries store gases under high pressure. This allows the gas to be stored at a low volume, saving plenty of storage space.
[Note: Students are expected to search more on the internet about various other applications of Boyle’s law on their own.]

iii. Importance of Boyle’s law in scuba diving:
a. Boyle’s law affects scuba diving in many ways.
b. It explains the role of pressure in the aquatic environment.
c. As divers descend, the water pressure surrounding them increases, causing air in their body and equipment to have a smaller volume. As the divers ascend, water pressure decreases, causing their body and equipment to expand to acquire a greater volume.
d. Furthermore, it is crucial that scuba divers never attempt to hold their breath when immersed in water.
e. According to Boyle’s law, if divers attempt this when they ascend to a body of water of less pressure, then the air that is trapped in their lungs will over-expand and rupture. This is known as Pulmonary Barotrauma. Thus, it is important for scuba divers to exhale as they ascend because the external pressure increases.
f. Also, if a diver returns to the surface too quickly, dissolved gases in the blood expand and form bubbles, which can get stuck in capillaries and organs (causing the ‘bends’).
[Note: Students are expected to collect additional information their own.]

Just think. (Textbook Page No. 144)

Question i.
Why does bicycle tyre burst during summer?
Answer:

• According to Charles’ law, at constant pressure, the volume of a fixed amount of a gas varies directly with the temperature. This means that as the temperature increases, the volume also increases.
• During summer, the temperature of the surrounding air is high. Due to the high temperature, the air inside the tyre gets heated. This will increase the volume of the tyres and it will burst.

Question ii.
Why do the hot air balloons fly high?
Answer:

• According to Charles’ law, at constant pressure, gases expand on heating and become less dense. Thus, hot air is less dense than cold air.
• In a hot air balloon, the air inside it is heated by a burner. Upon heating, the air inside the balloon expands and becomes lighter (less dense) than the cooler air on the outside. This causes the hot air balloon to fly high in air.

Just think. (Textbook Page No. 145)

Question 1.
i. List out various real-life examples of Charles’ law.
ii. Refer and watch Charles’ law experiments.
Answer:
i. Few real-life examples of Charles’ law:
a. Helium balloon: If we fill a helium balloon in a warm or hot room, and then take it into a cold room, it shrinks up and will look like it has lost some of the air inside it. This shows that gases expand on heating and contract on cooling.
b. A bottle of deodorant: If we expose a bottle of deodorant to sunlight and high temperatures, the air molecules inside the bottle will expand which can lead to the bursting of the deodorant bottle. This is another example of Charles’ law.
c. Basketball: You may have noticed that a basketball has less responsive bounce during winter than in summer. This yet another example of Charles’ law. When a basketball is inflated, the air pressure inside it is set to a fixed value. As the temperature falls, the volume of the gas inside the ball also decreases proportionally.
[Note: Students are expected to collect additional real-life examples on their own,]

ii. pi [Note: Students can scan the adjacent QR code to visualize Charles’ law with the aid of a relevant video.]

Use your brainpower. (Textbook Page No. 146)

Question 1.
Why does the pressure in the automobile tyres change during hot summer or winter season?
Answer:

• According to Gay-Lussac’s law, at constant volume, pressure of a fixed amount of a gas is directly proportional to its absolute temperature.
• During hot summer, the temperature of automobile tyre increases faster. Consequently, the air inside the tyre gets heated and the gas molecules starts moving faster.
• As the volume of the tyre remains constant, the pressure inside it increases.
• During winter, the temperature of automobile tyre decreases. Consequently, the air inside the tyre gets cooled and the gas molecules starts moving much slower and the pressure inside the tyre decreases.

Just think. (Textbook Page No. 149)

Question 1.
Do all pure gases and mixtures of gases obey the gas laws?
Answer:
Yes, the gas laws are also applicable to the mixtures of gases. The measurable properties of a mixture of the gases such as pressure, temperature, volume, and amount of gaseous mixture are all related by an ideal gas law.

Just think. (Textbook Page No. 150)

Question 1.
Where is Dalton’s law applicable?
Answer:
Air is a gaseous mixture of different gases. Dalton’s law is useful for the study of various phenomena in air, for example, air pollution.

Just think. (Textbook Page No. 155)

Question 1.
What makes the oil rise through the wick in an oil lamp?
Answer:
In an oil lamp, oil rises through the wick due to the capillary action. Such a capillary rise of oil is due to the surface tension of oil. The wick acts as a capillary tube. When the wick is placed in oil, the attractive forces between the oil and the inner wall of the capillary (wick) pull the oil up through the wick.

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 9 Elements of Group 13, 14 and 15 Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 9 Elements of Group 13, 14 and 15

1. Choose the correct option.

Question A.
Which of the following is not an allotrope of carbon?
a. buckyball
b. diamond
c. graphite
d. emerald
Answer:
d. emerald

Question B.
………… is inorganic graphite.
a. borax
b. diborane
c. boron nitride
d. colemanite
Answer:
c. boron nitride

Question C.
Haber’s process is used for the preparation of ………….
a. HNO3
b. NH3
c. NH2CONH2
d. NH4OH
Answer:
b. NH3

Question D.
Thallium shows a different oxidation state because ……………
a. of inert pair effect
b. it is an inner transition element
c. it is metal
d. of its high electronegativity
Answer:
a. of inert pair effect

Question E.
Which of the following shows the most prominent inert pair effect?
a. C
b. Si
c. Ge
d. Pb
Answer:
d. Pb

2. Identify the group 14 element that best fits each of the following description.

A. Non-metallic element
B. Form the most acidic oxide
C. They prefer +2 oxidation state.
D. Forms strong π bonds.
Answer:
i. Carbon (C)
ii. Carbon
iii. Tin (Sn) and lead (Pb)
iv. Carbon

3. Give reasons.

A. Ga³⁺ salts are better reducing agent while Tl³⁺ salts are better oxidising agent.
B. PbCl₄ is less stable than PbCl₂
Answer:
A. i. Both gallium (Ga) and thallium (Tl) belong to group 13.
ii. Ga is lighter element compared to thallium Tl. Therefore, its +3 oxidation state is stable. Thus, Ga+ loses two electrons and get oxidized to Ga³⁺. Hence, Ga+ salts are better reducing agent.
iii. Thallium is a heavy element. Therefore, due to the inert pair effect, Tl forms stable compounds in +1 oxidation state. Thus, Tl³⁺ salts get easily reduced to Tl¹⁺ by accepting two electrons. Hence, Tl³⁺ salts are better oxidizing agent.
[Note: This question is modified so as to apply the appropriate textual concept.]

B. i. Pb has electronic configuration [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p².
ii. Due to poor shielding of 6s² electrons by inner d and f electrons, it is difficult to remove 6s² electrons (inert pair).
iii. Thus, due to inert pair effect, +2 oxidation state is more stable than +4 oxidation state.
Hence, PbCl₄ is less stable than PbCl₂.

4. Give the formula of a compound in which carbon exhibit an oxidation state of

A. +4
B. +2
C. -4
Answer:
A. CCl₄
B. CO
C. CH₄

5. Explain the trend of the following in group 13 elements :

A. atomic radii
B. ionization enthalpy
C. electron affinity
Answer:
A. Atomic radii:

• In group 13, on moving down the group, the atomic radii increases from B to Al.
• However, there is an anomaly observed in the atomic radius of gallium due to the presence of 3d electrons. These inner 3d electrons offer poor shielding effect and thus, valence shell electrons of Ga experience greater nuclear attraction. As a result, atomic radius of gallium is less than that of aluminium.
• However, the atomic radii again increases from Ga to Tl.
• Therefore, the atomic radii of the group 13 elements varies in the following order:
  B < Al > Ga < In < Tl

B. Ionization enthalpy:

• Ionization enthalpies show irregular trend in the group 13 elements.
• As we move down the group, effective nuclear charge decreases due to addition of new shells in the atom of the elements which leads to increased screening effect. Thus, it becomes easier to remove valence shell electrons and hence, ionization enthalpy decreases from B to Al as expected.
• However, there is a marginal difference in the ionization enthalpy from Al to Tl.
• The ionization enthalpy increases slightly for Ga but decreases from Ga to In.
  In case of Ga, there are 10 d-electrons in its inner electronic configuration which shield the nuclear charge less effectively than the s and p-electrons and therefore, the outer electron is held fairly strongly by the nucleus. As a result, the ionization enthalpy increases slightly.
• Number of d electrons and extent of screening effect in indium is same as that in gallium. However, the atomic size increases from Ga to In. Due to this, the first ionization enthalpy of In decreases.
• The last element Tl has 10 d-electrons and 14 f-electrons in its inner electronic configuration which exert still smaller shielding effect on the outer electrons. Consequently, its first ionization enthalpy increases considerably.

C. Electron affinity:
a. Electron affinity shows irregular trend. It first increases from B to A1 and then decreases. The less electron affinity of boron is due to its smaller size. Adding an electron to the 2p orbital in boron leads to a greater repulsion than adding an electron to the larger 3p orbital of aluminium.

b. From Al to Tl, electron affinity decreases. This is because, nuclear charge increases but simultaneously the number of shells in the atoms also increases. As a result, the effective nuclear charge decreases down the group resulting in increased atomic size and thus, it becomes difficult to add an electron to a larger atom. The electron affinity of Ga and In is same.
Note: Electron affinity of group 13 elements:

6. Answer the following

Question A.
What is hybridization of Al in AlCl₃?
Answer:
Al is sp² hybridized in AlCl₃.

Question B.
Name a molecule having banana bond.
Answer:
Diborane (B₂H₆)

7. Draw the structure of the following

Question A.
Orthophosphoric acid
Answer:

Question B.
Resonance structure of nitric acid
Answer:

8. Find out the difference between

Question A.
Diamond and Graphite
Answer:
Diamond:

1. It has a three-dimensional network structure.
2. In diamond, each carbon atom is sp³ hybridized.
3. Each carbon atom in diamond is linked to four other carbon atoms.
4. Diamond is poor conductor of electricity due to absence of free electrons.
5. Diamond is the hardest known natural substance.

Graphite:

1. It has a two-dimensional hexagonal layered structure.
2. In graphite, each carbon atom is sp² hybridized.
3. Each carbon atom in graphite is linked to three other carbon atoms.
4. Graphite is good conductor of electricity due to presence of free electrons in its structure.
5. Graphite is soft and slippery.

Question B.
White phosphorus and Red phosphorus
Answer:
White phosphorus:

1. It consists of discrete tetrahedral P₄ molecules.
2. It is less stable and more reactive.
3. It exhibits chemiluminescence.
4. It is poisonous.

Red phosphorus:

1. It consists chains of P₄ molecules linked together by covalent bonds.
2. It is stable and less reactive.
3. It does not exhibit chemiluminescence.
4. It is nonpoisonous.

9. What are silicones? Where are they used?
Answer:
i. a. Silicones are organosilicon polymers having R₂SiO (where, R = CH₃ or C₆H₅ group) as a repeating unit held together by

b. Since the empirical formula R₂SiO (where R = CH₃ or C₆H₅ group) is similar to that of ketones (R₂CO), these compounds are named as silicones.

ii. Applications: They are used as

• insulating material for electrical appliances.
• water proofing of fabrics.
• sealant.
• high temperature lubricants.
• for mixing in paints and enamels to make them resistant to high temperature, sunlight and chemicals.

10. Explain the trend in oxidation state of elements from nitrogen to bismuth.
Answer:

• Group 15 elements have five valence electrons (ns² np³). Common oxidation states are -3, +3 and +5. The range of oxidation state is from -3 to +5.
• Group 15 elements exhibit positive oxidation states such as +3 and +5. Due to inert pair effect, the stability of +5 oxidation state decreases and +3 oxidation state increases on moving down the group.
• Group 15 elements show tendency to donate electron pairs in -3 oxidation state. This tendency is maximum for nitrogen.
• The group 15 elements achieve +5 oxidation state only through covalent bonding.
  e. g. NH₃, PH₃, ASH₃, SbH₃, and BiH₃ contain 3 covalent bonds. PCl₅ and PF₅ contain 5 covalent bonds.

11. Give the test that is used to detect borate radical is qualitative analysis.
Answer:
i. Borax when heated with ethyl alcohol and concentrated H₂SO₄, produces volatile vapours of triethyl borate, which bum with green edged flame.

ii. The above reaction is Used as a test for the detection and removal of borate radical \(\left(\mathrm{BO}_{3}^{3-}\right)\) in qualitative analysis.

12. Explain structure and bonding of diborane.
Answer:

• Electronic configuration of boron is 1s² 2s² 2p¹. Thus, it has only three valence electrons.
• In diborane, each boron atom is sp³ hybridized. Three of such hybrid orbitals are half filled while the fourth sp3 hybrid orbital remains vacant.
• The two half-filled sp³ hybrid orbitals of each B atom overlap with 1s orbitals of two terminal H atoms and form four B – H covalent bonds. These bonds are also known as two-centred-two-electron (2c-2e) bonds.
• When ‘1s’ orbital of each of the remaining two H atoms simultaneously overlap with half-filled hybrid orbital of one B atom and the vacant hybrid orbital of the other B atom, it produces two three-centred-two- electron bonds (3c-2e) or banana bonds.
• Hydrogen atoms involved in (3c-2e) bonds are the bridging H atoms i.e., H atoms in two B – H – B bonds.
• In diborane, two B atoms and four terminal H atoms lie in one plane, while the two bridging H atoms lie symmetrically above and below this plane.

13. A compound is prepared from the mineral colemanite by boiling it with a solution of sodium carbonate. It is white crystalline solid and used for inorganic qualitative analysis.

a. Name the compound produced.
b. Write the reaction that explains its formation.
Answer:
a. Borax
b. Borax is obtained from its mineral colemanite by boiling it with a solution of sodium carbonate.

14. Ammonia is a good complexing agent. Explain.
Answer:
i. The lone pair of electrons on nitrogen atom facilitates complexation of ammonia with transition metal ions. Thus, ammonia is a good complexing agent as it forms complex by donating its lone pair of electrons.

ii. This reaction is used for the detection of metal ions such as Cu²⁺ and Ag⁺.

15. State true or false. Correct the false statement.

A. The acidic nature of oxides of group 13 increases down the graph.
B. The tendency for catenation is much higher for C than for Si.
Answer:
A. False
The acidic nature of oxides of group 13 decreases down the group. It changes from acidic through amphoteric to basic.
B. True

16. Match the pairs from column A and B.

	Column A		Column B
i.	BCl3	a.	Angular molecule
ii.	SiO2	b.	Linear covalent molecule
iii.	CO2	c.	Tetrahedral molecule
		d.	Planar trigonal molecule

Answer:
i – d,
ii – c,
iii – b

17. Give the reactions supporting basic nature of ammonia.
Answer:
In the following reactions ammonia reacts with acids to form the corresponding ammonium salts which indicates basic nature of ammonia.

18. Shravani was performing inorganic qualitative analysis of a salt. To an aqueous solution of that salt, she added silver nitrate. When a white precipitate was formed. On adding ammonium hydroxide to this, she obtained a clear solution. Comment on her observations and write the chemical reactions involved.
Answer:
i. When silver nitrate (AgNO₃) is added to an aqueous solution of salt sodium chloride (NaCl), a white precipitate of silver chloride (AgCl) is formed.

ii. On adding ammonium hydroxide (NH₄OH) to this, the white precipitate of silver chloride gets dissolved and thus, a clear solution is obtained.

11th Chemistry Digest Chapter 9 Elements of Group 13, 14 and 15 Intext Questions and Answers

Can you recall? (Textbook Page No. 123)

Question 1.
If the valence shell electronic configuration of an element is 3s² 3p¹, in which block of the periodic table is it placed?
Answer:
The element having valence shell electronic configuration 3s² 3p¹ must be placed in the p-block of the periodic table as its last electron enters in p-subshell (3p).

Can you recall? (Textbook Page No. 127)

Question 1.
What is common between diamond and graphite?
Answer:
Both diamond and graphite are made up of carbon atoms as they are two allotropes of carbon.

Can you recall? (Textbook Page No. 129)

Question i.
Which element from the following pairs has higher ionization enthalpy?
B and TI, N and Bi
Answer:
Among B and Tl, boron has higher ionization enthalpy while, among N and Bi, nitrogen has higher ionization enthalpy.

Question ii.
Does boron form covalent compound or ionic?
Answer:
Yes, boron forms covalent compound.

Try this. (Textbook Page No. 131)

Question 1.
Find out the structural formulae of various oxyacids of phosphorus.
Answer:

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 8 Elements of Group 1 and 2 Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 8 Elements of Group 1 and 2

1. Explain the following

Question A.
Hydrogen shows similarity with alkali metals as well as halogens.
Answer:

• The electronic configuration of hydrogen is 1s¹ which is similar to the outer electronic configuration of alkali metals of group 1 i.e., ns¹.
• However, 1s¹ also resembles the outer electronic configuration of group 17 elements i.e., ns² np⁵.
• By adding one electron to H, it will attain the electronic configuration of the inert gas He which is 1s^2, and by adding one electron to ns² np⁵ we get ns² np⁶ which is the outer electronic configuration of the remaining inert gases.
• Therefore, some chemical properties of hydrogen are similar to those of alkali metals while some resemble halogens.

Hence, hydrogen shows similarity with alkali metals as well as halogens.

Question B.
Standard reduction potential of alkali metals have high negative values.
Answer:

• The general outer electronic configuration of alkali metals is ns¹.
• They readily lose one valence shell electron to achieve stable noble gas configuration and hence, they are highly electropositive and are good reducing agents.

Hence, standard reduction potentials of alkali metals have high negative values.

Question C.
Alkaline earth metals have low values of electronegativity; which decrease down the group.
Answer:

• Electronegativity represents attractive force exerted by the nucleus on shared electrons.
• The general outer electronic configuration of alkaline earth metals is ns². They readily lose their two valence shell electrons to achieve stable noble gas configuration. They are electropositive and hence, they have low values of electronegativity.

Question D.
Sodium dissolves in liquid ammonia to form a solution which shows electrical conductivity.
Answer:
i. Sodium dissolves in liquid ammonia giving deep blue coloured solutions which is electrically conducting in nature.
Na + (x + y) NH₃ → [Na(NH₃)_x]+ + [e(NH₃)_y]^–
ii. Due to formation of ions, the solution shows electrical conductivity.

Question E.
BeCl₂ is covalent while MgCl₂ is ionic.
Answer:

• Be²⁺ ion has very small ionic size and therefore, it has very high charge density.
• Due to this, it has high tendency to distort the electron cloud around the negatively charged chloride ion (Cl^–) which is larger in size.
• This results in partial covalent character of the bond in BeCl₂.
• Mg²⁺ ion has very less tendency to distort the electron cloud of Cl^– due to the bigger size of Mg²⁺ as compared to Be²⁺.

Hence, BeCl₂ is covalent while MgCl₂ is ionic.

Question F.
Lithium floats an water while sodium floats and catches fire when put in water.
Answer:

• When lithium and sodium react with water, hydrogen gas is released. Due to these hydrogen gas bubbles, lithium and sodium floats on water.
  eg. 2Na + 2H₂O → 2Na⁺ + 2OH^– + H₂↑
• The reactivity of group 1 metals increases with increasing atomic radius and lowering of ionization enthalpy down the group.
• Thus, sodium having lower ionization enthalpy, is more reactive than lithium.
• Hence, lithium reacts slowly while sodium reacts vigorously with water.
• Since the reaction of sodium with water is highly exothermic, it catches fire when put in water.

2. Write balanced chemical equations for the following.

Question A.
CO₂ is passed into concentrated solution of NaCl, which is saturated with NH₃.
Answer:

Question B.
A 50% solution of sulphuric acid is subjected to electrolyte oxidation and the product is hydrolysed.
Answer:

Question C.
Magnesium is heated in air.
Answer:

Question D.
Beryllium oxide is treated separately with aqueous HCl and aqueous NaOH solutions.
Answer:
Beryllium oxide (BeO) is an amphoteric oxide and thus, it reacts with both acid (HCl) as well as base (NaOH) to give the corresponding products.
i. \(\mathrm{BeO}+\underset{(\text { Acid })}{2 \mathrm{HCl}} \longrightarrow \mathrm{BeCl}_{2}+\mathrm{H}_{2} \mathrm{O}\)
ii. \(\mathrm{BeO}+\underset{(\text { Base })}{2 \mathrm{NaOH}} \longrightarrow \mathrm{Na}_{2} \mathrm{BeO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

3. Answer the following questions

Question A.
Describe the diagonal relationship between Li and Mg with the help of two illustrative properties.
Answer:
a. The relative placement of these elements with similar properties in the periodic table is across a diagonal and is called diagonal relationship.
b. Lithium is placed in the group 1 and period 2 of the modem periodic table. It resembles with magnesium which is placed in the group 2 and period 3.

ii. Li and Mg show similarities in many of their properties.
e. g.
a. Reaction with oxygen:
1. Group 1 elements except lithium, react with oxygen present in the air to form oxides (M₂O) as well as peroxides (M₂O₂) and superoxides (MO₂) on further reaction with excess of oxygen.
2. This anomalous behaviour of lithium is due to its resemblance with magnesium as a result of diagonal relationship.
3. As group 2 elements form monoxides i.e., oxides, lithium also form monoxides.

b. Reaction with nitrogen:
1. All the group 1 elements react only with oxygen present in the air to form oxides while group 2 elements react with both nitrogen and oxygen present in the air forming corresponding oxides and nitrides.
2. However, lithium reacts with oxygen as well as nitrogen present in the air due to its resemblance with magnesium.

Question B.
Describe the industrial production of dihydrogen from steam. Also write the chemical reaction involved.
Answer:
Three stages are involved in the industrial production of dihydrogen from steam.
i. Stage 1:
a. Reaction of steam on hydrocarbon or coke (C) at 1270 K temperature in presence of nickel catalyst gives water-gas which is a mixture of carbon monoxide and hydrogen.
1. Reaction of steam with hydrocarbon:

2. Reaction of steam with coke or carbon (C):

b. Sawdust, scrapwood, etc. can also be used in place of carbon.

ii. Stage 2:
Water-gas shift reaction: When carbon monoxide in the water-gas reacts with steam in the presence of iron chromate (FeCrO₄) as catalyst, it gets transformed into carbon dioxide. This is called water-gas shift reaction.

iii. Stage 3: In the last stage, carbon dioxide is removed by scrubbing with sodium arsenite solution.

Question C.
A water sample, which did not give lather with soap, was found to contain Ca(HCO₃)₂ and Mg(HCO₃)₂. Which chemical will make this water give lather with soap? Explain with the help of chemical reactions.
Answer:

• Soap does not lather in hard water due to presence of the soluble salts of calcium and magnesium in it. So, the given water sample is hard water.
• Hardness of hard water can be removed by removal of these calcium and magnesium salts.
• Sodium carbonate is used to make hard water soft as it precipitates out the soluble calcium and magnesium salts in hard water as carbonates. Thus, it will make water give lather with soap.
  e.g. Ca(HCO₃)_2(aq) + Na₂CO_3(aq) → CaCO_3(s) + 2NaHCO_3(aq)

Question D.
Name the isotopes of hydrogen. Write their atomic composition schematically and explain which of these is radioactive ?
Answer:
i. Hydrogen has three isotopes i.e., hydrogen \(\left({ }_{1}^{1} \mathrm{H}\right)\), deuterium \(\left({ }_{1}^{2} \mathrm{H}\right)\) and tritium \(\left({ }_{1}^{3} \mathrm{H}\right)\) with mass numbers 1, 2 and 3 respectively.
ii. They all contain one proton and one electron but different number of neutrons in the nucleus.
iii. Atomic composition of isotopes of hydrogen:

iv. Tritium is a radioactive nuclide with half-life period 12.4 years and emits low energy β^– particles.
v. Schematic representation of isotopes of hydrogen is as follows:

4. Name the following

Question A.
Alkali metal with smallest atom.
Answer:
Lithium (Li)

Question B.
The most abundant element in the universe.
Answer:
Hydrogen (H)

Question C.
Radioactive alkali metal.
Answer:
Francium (Fr)

Question D.
Ions having high concentration in cell sap.
Answer:
Potassium ions (K⁺)

Question E.
A compound having hydrogen, aluminium and lithium as its constituent elements.
Answer:
Lithium aluminium hydride (LiAlH₄)

5. Choose the correct option.

Question A.
The unstable isotope of hydrogen is …..
a. H-1
b. H-2
c. H-3
d. H-4
Answer:
c. H-3

Question B.
Identify the odd one.
a. Rb
b. Ra
c. Sr
d. Be
Answer:
a. Rb

Question C.
Which of the following is Lewis acid ?
a. BaCl₂
b. KCl
c. BeCl₂
d. LiCl
Answer:
c. BeCl₂

Question D.
What happens when crystalline Na₂CO₃ is heated ?
a. releases CO₂
b. loses H₂O
c. decomposes into NaHCO₃
d. colour changes.
Answer:
b. loses H₂O

Activity :

1. Collect the information of preparation of dihydrogen and make a chart.
2. Find out the s block elements compounds importance/uses.
Answer:
1.

2. Uses of s-block elements:
Group 1 elements (alkali metals):
a. Lithium: Lithium is widely used in batteries.
b. Sodium:

• Liquid sodium metal is used as a coolant in fast breeder nuclear reactors.
• Sodium is also used as an important reagent in the Wurtz reaction.
• It is used in the manufacture of sodium vapour lamp.

c. Potassium:

• Potassium has a vital role in biological system.
• Potassium chloride (KCl) is used as a fertilizer.
• Potassium hydroxide (KOH) is used in the manufacture of soft soaps and also as an excellent absorbent of carbon dioxide.
• Potassium superoxide (KO₂) is used as a source of oxygen.

d. Caesium: Caesium is used in devising photoelectric cells.

Group 2 elements (alkaline earth metals):
a. Magnesium: Magnesium hydroxide [Mg(OH)₂] in its suspension form is used as an antacid.
b. Calcium: Compounds of calcium such as limestone and gypsum are used as constituents of cement and mortar.
c. Barium: BaSO₄ being insoluble in H₂O and opaque to X-rays is used as ‘barium meal’ to scan the X-ray of human digestive system.
[Note: Students are expected to collect additional information about preparation of dihydrogen and uses of s-block elements on their own.]

11th Chemistry Digest Chapter 8 Elements of Group 1 and 2 Intext Questions and Answers

Can you recall? (Textbook Page No. 110)

Question 1.
Which is the first element in the periodic table?
Answer:
Hydrogen is the first element in the periodic table.

Question 2.
What are isotopes?
Answer:
Many elements exist naturally as a mixture of two or more types of atoms or nuclides. These individual nuclides are called isotopes of that element. Isotopes of an element have the same atomic number (number of protons) but different atomic mass numbers due to different number of neutrons in their nuclei.

Question 3.
Write the formulae of the compounds of hydrogen formed with sodium and chlorine.
Answer:
Hydrogen combines with sodium to form sodium hydride (NaH) while it reacts with chlorine to form hydrogen chloride (HCl).

Can you tell? (Textbook Page No. 110)

Question 1.
In which group should hydrogen be placed? In group 1 or group 17? Why?
Answer:

• Hydrogen contains one valence electron in its valence shell and thus, its valency is one. Therefore, hydrogen resembles alkali metals (group 1 elements) as they also contain one electron in their valence shell (alkali metals tend to lose their valence electron).
• However, hydrogen also shows similarity with halogens (group 17 elements) as their valency is also one because halogens tend to accept one electron in their valence shell.
• Due to this unique behaviour, it is difficult to assign any definite position to hydrogen in the modem periodic table.

Just think! (Textbook Page No. 112)

Question 1.
\(2 \mathrm{Na}_{(\mathrm{s})}+\mathrm{H}_{2(\mathrm{~g})} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaH}_{(\mathrm{s})}\)
In the above chemical reaction which element does undergo oxidation and which does undergo reduction?
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na_(s) + H_2(g) → 2Na⁺ + 2H^–
ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and each hydrogen atom gains one electron to form H. This can be represented as follows:

iii. Na is oxidised to NaH by loss of electrons while the elemental hydrogen is reduced to NaH by a gain of electrons.

Can you recall? (Textbook Page No. 113)

Question i.
What is the name of the family of reactive metals having valency one?
Answer:
The family of reactive metals having valency one is known as alkali metals (group 1).

Question ii.
What is the name of the family of reactive metals having valency two?
Answer:
The family of reactive metals having valency two is known as alkaline earth metals (group 2).

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 7 Modern Periodic Table Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 7 Modern Periodic Table

1. Explain the following

Question A.
The elements Li, B, Be and N have the electronegativities 1.0, 2.0, 1.5, and 3.0, respectively on the Pauling scale.
Answer:

• Li, B, Be and N belong to the same period.
• As we move across a period from left to right in the periodic table, the effective nuclear charge increases steadily and therefore, electronegativity increases.

Hence, the elements Li, B, Be and N have the electronegativities 1.0, 2.0, 1.5, and 3.0, respectively on the Pauling scale.

Question B.
The atomic radii of Cl, I and Br are 99, 133 and 114 pm, respectively.
Answer:

• Cl, I and Br belong to group 17 (halogen group) in the periodic table.
• As we move down the group from top to bottom in the periodic table, a new shell gets added in the atom of the elements.
• As a result, the effective nuclear charge decreases due to increase in the atomic size as well as increased shielding effect.
• Therefore, the valence electrons experience less attractive force from the nucleus and are held less tightly resulting in the increased atomic radius.
• Thus, their atomic radii increases in the following order down the group.
  Cl (99 pm) < Br (114 pm) < I (133 pm)

Hence, the atomic radii of Cl, I and Br are 99, 133 and 114 pm, respectively.

Question C.
The ionic radii of F^– and Na⁺ are 133 and 98 pm, respectively.
Answer:

• F^– and Na⁺ are isoelectronic ions as they both have 10 electrons.
• However, the nuclear charge on F^– is +9 while that of Na⁺ is +11.
• In isoelectronic species, larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.

Thus, F^– has larger ionic radii (133 pm) than Na⁺ (98 pm).

Question D.
₁₃Al is a metal, ₁₄Si is a metalloid and ₁₅P is a nonmetal.
Answer:

• Electronic configuration of Al is [Ne] 3s² 3p¹, ₁₄Si is [Ne] 3s² 3p² and that of ₁₅P is [Ne] 3s² 3p³.
• Metals are characterized by the ability to form compounds by loss of valence electrons.
• ‘Al’ has 3 valence electrons, thus shows tendency to lose 3 valence electrons to complete its octet. Hence, Al is a metal.
• Nonmetals are characterized by the ability to form compounds by gain of valence electrons in valence shell.
• ‘P’ has 5 valence electrons thus, shows tendency to gain 3 electrons to complete its octet. Hence, ‘P’ is a nonmetal.
• Si has four valence electrons, thus it can either lose/gain electrons to complete its octet. Hence, behaves as a metalloid.

Question E.
Cu forms coloured salts while Zn forms colourless salts.
Answer:

• Electronic configuration of ₂₉CU is [Ar] 3d¹⁰4s¹ while that of Zn is [Ar] 3d¹⁰4s².
• Electronic configuration of Cu in its +1 oxidation state is [Ar] 3d¹⁰ while that in +2 oxidation state is [Ar] 3d⁹.
• Therefore, Cu contains partially filled d orbitals in +2 oxidation state and thus, Cu²⁺ salts are coloured.
• However, Zn has completely filled d orbital which is highly stable and hence, it does not form coloured ions.

Hence, Cu forms coloured salts while Zn forms colourless salts.

2. Write the outer electronic configuration of the following using orbital notation method. Justify.
A. Ge (belongs to period 4 and group 14)
B. Po (belongs to period 6 and group 16)
C. Cu (belongs to period 4 and group 11)
Answer:
A. a. Ge belongs to period 4. Therefore, n = 4.
b. Group 14 indicates that the element belongs to the p-block of the modem periodic table.
c. The general outer electronic configuration of group 14 elements is ns² np².
d. Thus, the outer electronic configuration of Ge is 4s² 4p².

B. a. Po belongs to period 6. Therefore, n = 6.
b. Group 16 indicates that the element belongs to the p-block of the modem periodic table.
c. The general outer electronic configuration of group 16 elements is ns² np⁴.
d. Thus, the outer electronic configuration of Po is 6s² 6p⁴.

C. a. Cu belongs to period 4. Therefore, n = 4.
b. Group 11 indicates that the element belongs to the d-block of the modem periodic table.
c. The general outer electronic configuration of the d-block elements is ns^0-2(n-1)d^1-10.
d. The expected configuration of Cu is 4s²3d⁹. However, the observed configuration of Cu is 4s¹3d¹⁰. This is due to the extra stability associated with completely filled d-subshell. Thus, the outer electronic configuration of Cu is 4s¹3d¹⁰.

3. Answer the following

Question A.
La belongs to group 3 while Hg belongs to group 12 and both belong to period 6 of the periodic table. Write down the general outer electronic configuration of the ten elements from La to Hg together using orbital notation method.
Answer:
i. La and Hg both belongs to period 6. Therefore, n = 6.
ii. Elements of group 3 to group 12 belong to the d-block of the modem periodic table.
iii. The general outer electronic configuration of the d-block elements is ns^0-2 (n -1 )^1-10.
iv. Therefore, the outer electronic configuration of all ten elements from La to Hg is as given in the table below.

[Note: There are 14 elements between La and Hf which are called lanthanides. Therefore, after La, electrons are filled in 4f subshell of lanthanide elements. Once all the 14 elements of lanthanide series are filled, next electron enters 5d subshell of Hf. Hence, the outer electronic configurations of Hf to Hg often include completely filled 4f subshell. For example, the electronic configuration of Hf ‘5d²6s²’ can also be written as ‘4f¹⁴5d²6s²’.]

Question B.
Ionization enthalpy of Li is 520 kJ mol^-1 while that of F is 1681 kJ mol^-1. Explain.
Answer:

• Both Li and F belong to period 2.
• Across a period, the screening effect is the same while the effective nuclear charge increases.
• As a result, the outer electron is held more tightly and therefore, the ionization enthalpy increases across a period.
• Hence, F will have higher ionization enthalpy than Li.

Thus, ionization enthalpy of Li is 520 kJ mol^-1 while that of F is 1681 kJ mol^-1.

Question C.
Explain the screening effect with a suitable example.
Answer:
i. In a multi-electron atom, the electrons in the inner shells tend to prevent the attractive influence of the nucleus from reaching the outermost electron.
ii. Thus, they act as a screen or shield between the nuclear attraction and outermost or valence electrons. This effect of the inner electrons on the outer electrons is known as screening effect or shielding effect.
iii. Across a period, screening effect due to inner electrons remains the same as electrons are added to the same shell.
iv. Down the group, screening effect due to inner electrons increases as a new valence shell is added.
e.g. Potassium (19K) has electronic configuration 1s²2s²2p⁶3s²3p⁶4s¹.
K has 4 shells and thus, the valence shell electrons are effectively shielded by the electrons present in the inner three shells. As a result of this, valence shell electron (4s¹) in K experiences much less effective nuclear charge and can be easily removed.

Question D.
Why the second ionization enthalpy is greater than the first ionization enthalpy ?
Answer:
The second ionization enthalpy (Δ_iH₂) is greater than the first ionization enthalpy (Δ_iH₁) as it involves removal of electron from the positively charged species.

Question E.
Why the elements belonging to the same group do have similar chemical properties ?
Answer:

• Chemical properties of elements depend upon their valency.
• Elements belonging to the same group have the same valency.

Hence, the elements belonging to the same group show similar chemical properties.

Question F.
Explain : electronegativity and electron gain enthalpy. Which of the two can be measured experimentally?
Answer:
i. The ability of a covalently bonded atom to attract the shared electrons toward itself is called electronegativity (EN). Electronegativity cannot be measured experimentally. However, various numerical scales to express electronegativity were developed by many scientists. Pauling scale of electronegativity is the one used most widely.

ii. Electron gain enthalpy is a quantitative measure of the ease with which an atom adds an electron forming the anion and is expressed in kJ mol^-1. Thus, it is an experimentally measurable quantity.

4. Choose the correct option

Question A.
Consider the elements B, Al, Mg and K predict the correct order of metallic character :
a. B > Al > Mg > K
b. Al > Mg > B > K
c. Mg > Al > K > B
d. K > Mg > Al > B
Answer:
d. K > Mg > Al > B

Question B.
In modern periodic table, the period number indicates the :
a. atomic number
b. atomic mass
c. principal quantum number
d. azimuthal quantum number
Answer:
c. principal quantum number

Question C.
The lanthanides are placed in the periodic table at
a. left hand side
b. right hand side
c. middle
d. bottom
Answer:
d. bottom

Question D.
If the valence shell electronic configuration is ns²np⁵, the element will belong to
a. alkali metals
b. halogens
c. alkaline earth metals
d. actinides
Answer:
b. halogens

Question E.
In which group of elements of the modern periodic table are halogen placed ?
a. 17
b. 6
c. 4
d. 2
Answer:
a. 17

Question F.
Which of the atomic number represent the s-block elements ?
a. 7, 15
b. 3, 12
c. 6, 14
d. 9, 17
Answer:
b. 3, 12

Question G.
Which of the following pairs is NOT isoelectronic ?
a. Na⁺ and Na
b. Mg²⁺ and Ne
c. Al³⁺ and B³⁺
d. P3^– and N^3-
Answer:
b. Mg²⁺ and Ne

Question H.
Which of the following pair of elements has similar properties ?
a. 13, 31
b. 11, 20
c. 12, 10
d. 21, 33
Answer:
a. 13, 31

5. Answer the following questions

Question A.
The electronic configuration of some elements are given below:
a. 1s²
b. 1s²2s²2p⁶
In which group and period of the periodic table they are placed ?
Answer:
a. 1s²
Here n = 1. Therefore, the element belongs to the 1st period.
The outer electronic configuration 1s² corresponds to the maximum capacity of 1s, the complete duplet. Therefore, the element is placed at the end of the 1st period in the group 18 of inert gases in the modem periodic table,

b. 1s²2s²2p⁶
Here n = 2. Therefore, the element belongs to the 2nd period.
The outer electronic configuration 2s²2p⁶ corresponds to complete octet. Therefore, the element is placed in the 2nd period of group 18 in the modem periodic table.

Question B.
For each of the following pairs, indicate which of the two species is of large size :
a. Fe²⁺ or Fe³⁺
b. Mg²⁺ or Ca²⁺
Answer:
a. Fe²⁺ has a larger size than Fe³⁺.
b. Ca²⁺ has a larger size than Mg²⁺.

Question C.
Select the smaller ion form each of the following pairs:
a. K⁺, Li⁺
b. N^3-, F^–
Answer:
i. Li⁺ has smaller ionic radius than K⁺
ii. F^– has smaller ionic radius than N^3-.

Question D.
With the help of diagram answer the questions given below:

a. Which atom should have smaller ionization enthalpy, oxygen or sulfur?
b. The lithium forms +1 ions while berylium forms +2 ions ?
Answer:
Sulfur should have smaller ionization energy than oxygen.
a. Lithium has electronic configuration 1s²2s¹ while that of beryllium is 1s²2s².
b. Li can achieve a noble gas configuration by losing one electron while Be can do so by losing two electrons. Hence, lithium forms +1 ions while beryllium forms +2 ions.

Question E.
Define : a. Ionic radius
b. Electronegativity
Answer:
a. Ionic radius: Ionic radius is defined as the distance of valence shell of electrons from the centre of the nucleus in an ion.

b. Electronegativity: The ability of a covalently bonded atom to attract the shared electrons toward itself is called electronegativity (EN).

Question F.
Compare chemical properties of metals and non-metals.
Answer:
i. Metals (like alkali metals) react vigorously with oxygen to form oxides which reacts with water to form strong bases.
e. g. Sodium (Na) reacts with oxygen to form Na₂O which produces NaOH on reaction with water.

ii. Nonmetals (like halogens) react with oxygen to form oxides which on reaction with water form strong acids.
e.g. Chlorine reacts with oxygen to form Cl₂O₇ which produces HClO₄ on reaction with water.

Question G.
What are the valence electrons ? For s-block and p-block elements show that number of valence electrons is equal to its group number.
Answer:

• Electrons present in the outermost shell of the atom of an element are called valence electrons.
• 3Li is an s-block element and its electronic configuration is 1s²2s¹. Since it has one valence electron, it is placed in group 1.
• Therefore, for s-block elements, group number = number of valence electrons.
• However, for p-block elements, group number = 18 – number of electrons required to attain complete octet.
• 7N is a p-block element and its electronic configuration is 1s²2s²2p³. Since it has five electrons in its valence shell, it is short of three electrons to complete its octet.
• Therefore, its group number = 18 – 3 = 15.

Question H.
Define ionization enthalpy. Name the factors on which ionisation enthalpy depends? How does it vary down the group and across a period?
Answer:
i. The energy required to remove an electron from the isolated gaseous atom in its ground state is called ionization enthalpy (Δ_iH).
Ionization enthalpy is the quantitative measure of tendency of an element to lose electron and expressed in kJ mol^-1.

ii. Ionization energy depends on the following factors

• Size (radius) of an atom
• Nuclear charge
• The shielding or screening effect of inner electrons
• Nature of electronic configuration

iii. Variation of ionization energy down the group: On moving down the group, the ionization enthalpy decreases. This is because electron is to be removed from the larger valence shell. Screening due to core electrons goes on increasing and the effective nuclear charge decreases down the group. As a result, the removal of the outer electron becomes easier down the group.

iv. Variation of ionization energy across a period: The screening effect is the same while the effective nuclear charge increases across a period. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period. Therefore, the alkali metal shows the lowest first ionization enthalpy while the inert gas shows the highest first ionization enthalpy across a period.

Note: First ionization enthalpy values of elements of group 1.

Note: First ionization enthalpy values of elements of period 2.

Question I.
How the atomic size vary in a group and across a period? Explain with suitable example.
Answer:
i. Variation in atomic size down the group:
a. As we move down the group from top to bottom in the periodic table, the atomic size increases with the increase in atomic number.
b. This is because, as the atomic number increases, nuclear charge increases but simultaneously the number of shells in the atoms also increases.
c. Asa result, the effective nuclear charge decreases due to increase in the size of the atom and shielding effect increases down the group. Thus, the valence electrons experience less attractive force from nucleus and are held less tightly.
d. Hence, the atomic size increases in a group from top to bottom.

e. g.

• In group 1, as we move from top to bottom i.e., from Li to Cs, a new shell gets added in the atom of the elements and the electrons are added in this new shell.
• As a result of this, the effective nuclear charge goes on decreasing and screening effect goes on increasing down a group.
• Therefore, the atomic size is the largest for Cs and is the smallest for Li in group 1.

[Note: Atomic radii of Li and Cs are 152 pm and 262 pm respectively.]

ii. Variation in atomic size across a period:
a. As we move across a period from left to right in the periodic table, the atomic size of an element decreases with the increase in atomic number.
b. This is because, as the atomic number increases, nuclear charge increases gradually but addition of electrons takes place in the same shell.
c. Therefore, as we move across a period, the effective nuclear charge increases but screening effect caused by the core electrons remains the same.
d. As a result of this, attraction between the nucleus and the valence electrons increases. Therefore, valence electrons are more tightly bound and hence, the atomic radius goes on decreasing along a period resulting in decrease in atomic size.

e. g.

• In the second period, as we move from left towards right i.e., from Li to F, the electrons are added in the second shell of all the elements in second period (except noble gas Ne).
• As a result of this, the effective nuclear charge goes on increasing from Li to F, however, screening effect remains the same.
• Therefore, the atomic size is the largest for Li (alkali metal) and is the smallest for F (halogen).

[Note: Atomic radii of Li and F are 152 pm and 64 pm respectively.]

Question J.
Give reasons.
a. Alkali metals have low ionization energies.
b. Inert gases have exceptionally high ionization energies.
c. Fluorine has less electron affinity than chlorine.
d. Noble gases possess relatively large atomic size.
Answer:
a. i. Across a period, the screening effect is the same while the effective nuclear charge increases.
ii. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period.
iii. Since the alkali metals are present in the group 1 of the modem periodic table, they have low ionization energies.

b. i. Across a period, the screening effect is the same and the effective nuclear charge increases.
ii. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period.
iii. Inert gases are present on the extreme right of the periodic table i.e., in group 18. Also, inert gases have stable electronic configurations i.e., complete octet or duplet. Due to this, they are extremely stable and it is very difficult to remove electrons from their valence shell.
Hence, inert gases have exceptionally high ionization potential.

c. The less electron affinity of fluorine is due to its smaller size. Adding an electron to the 2p orbital in fluorine leads to a greater repulsion than adding an electron to the larger 3p orbital of chlorine.
Hence, fluorine has less electron affinity than chlorine.

d. i. Noble gases have completely filled valence shell i.e., complete octet (except He with complete duplet).
ii. Since their valence shell contains eight electrons, they experience greater electronic repulsion and this results in increased atomic size (atomic radii) of the noble gas elements.
Hence, noble gases possess

Question K.
Consider the oxides Li₂O, CO₂, B₂O₃.
a. Which oxide would you expect to be the most basic?
b. Which oxide would be the most acidic?
c. Give the formula of an amphoteric oxide.
Answer:
a. Li₂O is the most basic oxide.
b. CO₂ is the most acidic oxide.
c. Formula of an amphoteric oxide: Al₂O₃.
[Note: Both B₂O₃ and CO₂ are acidic oxides. But CO₂ is more acidic oxide as compared to B₂O₃. Hence, CO₂ is most acidic oxide amongst the given.]

Activity :

Question 1.
Prepare a wall mounting chart of the modern periodic table.
Answer:
Students can scan the adjacent Q.R. Code to visualise the modern periodic table and are expected to prepare the chart on their own.

11th Chemistry Digest Chapter 7 Modern Periodic Table Intext Questions and Answers

Can you recall? (Textbook Page No. 93)

Question 1.
What was the basis of classification of elements before the knowledge of electronic structure of atom?
Answer:
Elements were classified on the basis of their physical properties before the knowledge of electronic structure of atom.

Question 2.
Name the scientists who made the classification of elements in the nineteenth century.
Answer:
Dmitri Mendeleev, John Newlands and Johann Doberiener were the scientists who made the classification of elements based on their atomic mass in the nineteenth century.

Question 3.
What is Mendeleev’s periodic law?
Answer:
Mendeleev’s periodic law: “The physical and chemical properties of elements are the periodic function of their atomic masses

Question 4.
How many elements are discovered until now?
Answer:
Including manmade elements, total 118 elements are discovered until now.

Question 5.
How many horizontal rows and vertical columns are present in the modern periodic table?
Answer:
The modem periodic table consists of seven horizontal rows called periods numbered from 1 to 7 and eighteen vertical columns called groups numbered from 1 to 18.

Just think. (Textbook Page No. 93)

Question 1.
How many days pass between two successive full moon nights?
Answer:
29.5 days i.e., approximately 30 days pass between two successive full moon nights.

Question 2.
What type of motion does a pendulum exhibit?
Answer:
A pendulum exhibits periodic motion since it traces the same path after regular interval of time.

Question 3.
Give some other examples of periodic events.
Answer:
Following are some other examples of periodic events:

• Motion of earth around the sun.
• Rotation of earth around its own axis.
• Day and night.

Can you recall? (Textbook Page No. 95)

Question i.
What does the principal quantum number ‘n’ and azimuthal quantum number ‘l’ of an electron belonging to an atom represent?
Answer:
The principal quantum number ‘n’ represents the outermost or valence shell of an element (which corresponds to period number) while azimuthal quantum number ‘l’ constitutes a subshell belonging to the shell for the given ‘n’.

Question ii.
Which principle is followed in the distribution of electrons in an atom?
Answer:
The distribution of electrons in an atom is according to the following three principles:

1. Aufbau principle
2. Pauli’s exclusion principle
3. Hund’s rule of maximum multiplicity

[Note: According to aufbau principle, electrons are filled in the subshells in the increasing order of their energies which follows the following order: s < p < d < f.]

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 6 Redox Reactions Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 6 Redox Reactions

1. Choose the most correct option

Question A.
Oxidation numbers of Cl atoms marked as Cl^a and Cl^b in CaOCl₂ (bleaching powder) are

a. zero in each
b. -1 in Cl^a and +1 in Cl^b
c. +1 in Cl^a and -1 in Cl^b
d. 1 in each
Answer:
b. -1 in Cl^a and +1 in Cl^b

Question B.
Which of the following is not an example of redox reacton ?
a. CuO + H₂ → Cu + H₂O
b. Fe₂O₃ + 3CO₂ → 2Fe + 3CO₂
c. 2K + F₂ → 2KF
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl
Answer:
d. BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Question C.
A compound contains atoms of three elements A, B and C. If the oxidation state of A is +2, B is +5 and that of C is -2, the compound is possibly represented by
a. A₂(BC₃)₂
b. A₃(BC₄)₂
c. A₃(B₄C)₂
d. ABC₂
Answer:
b. A₃(BC₄)₂

Question D.
The coefficients p, q, r, s in the reaction
\(\mathrm{pCr}_{2} \mathrm{O}_{7}^{2-}\) + q Fe^2⊕ → r Cr^3⊕ + s Fe^3⊕ + H₂O respectively are :
a. 1, 2, 6, 6
b. 6, 1, 2, 4
c. 1, 6, 2, 6
d. 1, 2, 4, 6
Answer:
c. 1, 6, 2, 6

Question E.
For the following redox reactions, find the correct statement.
Sn^2⊕ + 2Fe^3⊕ → Sn^4⊕ + 2Fe^2⊕
a. Sn^2⊕ is undergoing oxidation
b. Fe^3⊕ is undergoing oxidation
c. It is not a redox reaction
d. Both Sn^2⊕ and Fe^3⊕ are oxidised
Answer:
a. Sn^2⊕ is undergoing oxidation

Question F.
Oxidation number of carbon in H₂CO₃ is
a. +1
b. +2
c. +3
d. +4
Answer:
d. +4

Question G.
Which is the correct stock notation for magenese dioxide ?
a. Mn(I)O₂
b. Mn(II)O₂
c. Mn(III)O₂
d. Mn(IV)O₂
Answer:
d. Mn(IV)O₂

Question I.
Oxidation number of oxygen in superoxide is
a. -2
b. -1
c. –\(\frac {1}{2}\)
d. 0
Answer:
c. –\(\frac {1}{2}\)

Question J.
Which of the following halogens does always show oxidation state -1 ?
a. F
b. Cl
c. Br
d. I
Answer:
a. F

Question K.
The process SO₂ → S₂Cl₂ is
a. Reduction
b. Oxidation
c. Neither oxidation nor reduction
d. Oxidation and reduction.
Answer:
a. Reduction

2. Write the formula for the following compounds :
A. Mercury(II) chloride
B. Thallium(I) sulphate
C. Tin(IV) oxide
D. Chromium(III) oxide
Answer:
i. HgCl₂
ii. Tl₂SO₄
iii. SnO₂
iv. Cr₂O₃

3. Answer the following questions

Question A.
In which chemical reaction does carbon exibit variation of oxidation state from -4 to +4 ? Write balanced chemical reaction.
Answer:
In combustion of methane, carbon exhibits variation from -4 to +4. The reaction is as follows:
CH₄ + 2O₂ → CO₂ + 2H₂O
In CH₄, the oxidation state of carbon is -4 while in CO₂, the oxidation state of carbon is +4.

Question B.
In which reaction does nitrogen exhibit variation of oxidation state from -3 to +5 ?
Answer:

C. Calculate the oxidation number of underlined atoms.
a. H₂SO₄
b. HNO₃
c. H₃PO₃
d. K₂C₂O₄
e. H₂S₄O₆
f. Cr₂O₇^2-
g. NaH₂PO₄
Answer:
i. H₂SO₄
Oxidation number of H = +1
Oxidation number of O = -2
H₂SO₄ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of H₂SO₄ = 0
∴ 2 × (Oxidation number of H) + (Oxidation number of S) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + (Oxidation number of S) + 4 × (-2) = 0
∴ Oxidation number of S + 2 – 8 = 0
∴ Oxidation number of S in H₂SO₄ = +6

ii. HNO₃
Oxidation number of H = +1
Oxidation number of O = -2
HNO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms of HNO₃ = 0
∴ (Oxidation number of H) + (Oxidation number of N) + 3 × (Oxidation number of O) = 0
∴ (+1) + (Oxidation number of N) + 3 × (-2) = 0
∴ Oxidation number of N + 1 – 6 = 0
∴ Oxidation number of N in HNO₃ = +5

iii. H₃PO₃
Oxidation number of O = -2
Oxidation number of H = +1
H₃PO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of P) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of P) + 3 × (-2) = 0
∴ Oxidation number of P + 3 – 6 = 0
Oxidation number of P is H₃PO₃ = +3

iv. K₂C₂O₄
Oxidation number of K = +1
Oxidation number of O = -2
K₂C₂O₄ is a neutral molecule.
∴ Sum of the oxidation number of all atoms = 0
∴ 2 × (Oxidation number of K) + 2 × (Oxidation number of C) + 4 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of C) + 4 × (-2) = 0
∴ 2 × (Oxidation number of C) + 2 – 8 = 0
∴ 2 × (Oxidation number of C) = + 6
∴ Oxidation number of C = +\(\frac {6}{2}\)
∴ Oxidation number of C in K₂C₂O₄ = +3

v. H₂S₄O₆
Oxidation number of H = +1
Oxidation number of O = -2
H₂S₄O₆ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of H) + 4 × (Oxidation number of S) + 6 × (Oxidation number of O) = 0
∴ 2 × (+1) + 4 × (Oxidation number of S) + 6 × (-2) = 0
∴ 4 × (Oxidation number of S) + 2 – 12 = 0
∴ 4 × (Oxidation number of S) = + 10
∴ Oxidation number of S = +\(\frac {10}{4}\)
∴ Oxidation number of S in H₂S₄O₆ = +2.5

vi. Cr₂O₇^2-
Oxidation of O = -2
Cr₂O₇^2- is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 2
∴ 2 × (Oxidation number of Cr) + 7 × (Oxidation number of O) = -2
∴ 2 × (Oxidation number of Cr) + 7 × (-2) = – 2
∴ 2 × (Oxidation number of Cr) – 14 = – 2
∴ 2 × (Oxidation number of Cr) = – 2 + 14
∴ Oxidation number of Cr = +\(\frac {12}{2}\)
∴ Oxidation number of Cr in Cr₂O₇^2- = +6

vii. NaH₂PO₄
Oxidation number of Na = +1
Oxidation number of H = +1
Oxidation number of O = -2
NaH₂PO₄ is a neutral molecule
Sum of the oxidation numbers of all atoms = 0
(Oxidation number of Na) + 2 × (Oxidation number of H) + (Oxidation number of P) + 4 × (Oxidation number of O) = 0
(+1) + 2 × (+1) + (Oxidation number of P) + 4 × (-2) = 0
(Oxidation number of P) + 3 – 8 = 0
Oxidation number of P in NaH₂PO₄ = +5

Question D.
Justify that the following reactions are redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which act as a reductant.
a. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
b. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
c. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
Answer:
i. 2Cu₂O_(s) + Cu₂S_(s) → 6Cu_(s) + SO_2(g)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from -2 to +4 and that of Cu decreases from +1 to 0. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and it itself is oxidised. On the other hand, the oxidation number of Cu decreases by gain of electrons and therefore, Cu is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agents (Reduced species): Cu₂O/ Cu₂S
3. Reductant/reducing agent (Oxidised species): Cu₂S

[Note: Cu in both Cu₂O and Cu₂S undergoes reduction. Hence, both Cu₂O and Cu₂S can be termed as oxidising agents in the given reaction.]

ii. HF_(aq) + OH^–_(aq) → H₂O_(l) + F^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Since, the oxidation numbers of all the species remain same, this is NOT a redox reaction. Result:
The given reaction is NOT a redox reaction.

iii. I_2(aq) + 2 S₂O₃^2-_(aq) → S₄O₆^2-_(aq) + 2I^–_(aq)
a. Write oxidation number of all the atoms of reactants and products.

b. Identify the species that undergoes change in oxidation number.

c. The oxidation number of S increases from +2 to +2.5 and that of I decreases from 0 to -1. Because oxidation number of one species increases and that of the other decreases, the reaction is a redox reaction.
d. The oxidation number of S increases by loss of electrons and therefore, S is a reducing agent and itself is oxidised. On the other hand, the oxidation number of I decreases by gain of electrons and therefore, I is an oxidising agent and itself is reduced.

Result:

1. The given reaction is a redox reaction.
2. Oxidant/oxidising agent (Reduced species): I₂
3. Reductant/reducing agent (Oxidised species): S₂O₃^2-

Question E.
What is oxidation? Which one of the following pairs of species is in its oxidized state ?
a. Mg / Mg²⁺
b. Cu / Cu²⁺
c. O₂ / O^2-
d. Cl₂ / Cl^–
Answer:
a. Mg / Mg²⁺
Here, Mg loses two electrons to form Mg²⁺ ion.
\(\mathrm{Mg}_{(\mathrm{s})} \longrightarrow \mathrm{Mg}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Mg / Mg²⁺ is an oxidized state.

b. Cu/Cu²⁺
Here, Cu loses two electrons to form Cu²⁺ ion.
\(\mathrm{Cu}_{(\mathrm{s})} \longrightarrow \mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-}\)
Hence, Cu/Cu²⁺ is in an oxidized state.

c. O₂ / O^2-
Here, each O gains two electrons to form O^2- ion.
\(\mathrm{O}_{2(\mathrm{~g})}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}_{(\mathrm{aq})}^{2-}\)
Hence, O₂ / O^2- is in a reduced state.

d. Cl₂ / Cl^–
Here, each Cl gains one electron to form Cl^– ion.
\(\mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cl}_{(\mathrm{aq})}^{-}\)
Hence, Cl₂ / Cl^– is in a reduced state.

Question F.
Justify the following reaction as redox reaction.
2 Na_2(s) + S_(s) → Na₂S_(s)
Find out the oxidizing and reducing agents.
Answer:
i. Redox reaction can be described as electron transfer as shown below:
2Na_(s) + S_(s) → 2Na⁺ + S^2-
ii. Charge development suggests that each sodium atom loses one electron to form Na⁺ and sulphur atom gains two electrons to form S^2-. This can be represented as follows:

iii. When Na is oxidised to Na₂S, the neutral Na atom loses electrons to form Na⁺ in Na₂S while the elemental sulphur gains electrons and forms S^2- in Na₂S.
iv. Each of the above steps represents a half reaction which involves electron transfer (loss or gain).
v. Sum of these two half reactions or the overall reaction is a redox reaction.
vi. Oxidising agent is an electron acceptor and hence, S is an oxidising agent. Reducing agent is an electron donor and hence, Na is a reducing agent.

Question G.
Provide the stock notation for the following compounds : HAuCl₄, Tl₂O, FeO, Fe₂O₃, MnO and CuO.
Answer:

Question H.
Assign oxidation number to each atom in the following species.
a. Cr(OH)₄^–
b. Na₂S₂O₃
c. H₃BO₃
Answer:
i. Cr(OH)₄^–
Oxidation number of O = -2
Oxidation number of H = +1
Cr(OH)₄^– is an ionic species.
∴ Sum of the oxidation numbers of all atoms = – 1
∴ Oxidation number of Cr + 4 × (Oxidation number of O) + 4 × (Oxidation number of H) = – 1
∴ Oxidation number of Cr + 4 × (-2) + 4 × (+1) = – 1
∴ Oxidation number of Cr – 8 + 4 = – 1
∴ Oxidation number of Cr – 4 = – 1 –
∴ Oxidation number of Cr = – 1 + 4
∴ Oxidation number of Cr in Cr(OH)₄^– = +3

ii. Na₂S₂O₃
Oxidation number of Na = +1
Oxidation number of O = -2
Na₂S₂O₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 2 × (Oxidation number of Na) + 2 × (Oxidation number of S) + 3 × (Oxidation number of O) = 0
∴ 2 × (+1) + 2 × (Oxidation number of S) + 3 × (-2) = 0
∴ 2 × (Oxidation number of S) + 2 – 6 = 0
∴ 2 × (Oxidation number of S) = + 4
∴ Oxidation number of S = +\(\frac {4}{2}\)
∴ Oxidation number of S in Na₂S₂O₃ = +2

iii. H₃BO₃
Oxidation number of H = +1
Oxidation number of O = -2
H₃BO₃ is a neutral molecule.
∴ Sum of the oxidation numbers of all atoms = 0
∴ 3 × (Oxidation number of H) + (Oxidation number of B) + 3 × (Oxidation number of O) = 0
∴ 3 × (+1) + (Oxidation number of B) + 3 × (-2) = 0
∴ Oxidation number of B + 3 – 6 = 0
∴ Oxidation number of B in H₃BO₃ = +3

Question I.
Which of the following redox couple is stronger oxidizing agent ?
a. Cl₂ (E⁰ = 1.36 V) and Br₂ (E⁰ = 1.09 V)
b. \(\mathrm{MnO}_{4}^{\Theta}\) (E0 = 1.51 V) and \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\) (E⁰ = 1.33 V)
Answer:
a. Cl₂ has a larger positive value of E⁰ than Br₂. Thus, Cl₂ is a stronger oxidizing agent than Br₂.
b. \(\mathrm{MnO}_{4}^{\Theta}\) has larger positive value of E⁰ than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\). Thus, \(\mathrm{MnO}_{4}^{\Theta}\) is stronger oxidizing agent than \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2 \Theta}\)

Question J.
Which of the following redox couple is stronger reducing agent ?
a. Li (E⁰ = – 3.05 V) and Mg(E⁰ = – 2.36 V)
b. Zn(E⁰ = – 0.76 V) and Fe(E⁰ = – 0.44 V)
Answer:
a. Li has a larger negative value of E⁰ than Mg. Thus, Li is a stronger reducing agent than Mg.
b. Zn has a larger negative value of E⁰ than Fe. Thus, Zn is a stronger reducing agent than Fe.

4. Balance the reactions/equations :

Question A.
Balance the following reactions by oxidation number method

Answer:
i. \(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+\mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-} \quad(\text { acidic })\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(a q)}^{2-}+\mathrm{SO}_{3(a)}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+\mathrm{SO}_{4(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation number to Cr and S. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of S and 2 atoms of Cr. (There are already 2 Cr atoms.)
Step 3: Balance ‘O’ atoms by adding 4H₂O to the right-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is acidic. To make the charges and hydrogen atoms on the two sides equal, add 8H on the left-hand side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7(\mathrm{aq})}^{2-}+3 \mathrm{SO}_{3(\mathrm{aq})}^{2-}+8 \mathrm{H}_{(\mathrm{aq})}^{+} \longrightarrow 2 \mathrm{Cr}_{(\mathrm{aq})}^{3+}+3 \mathrm{SO}_{4(\mathrm{aq})}^{2-}+4 \mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation:

ii. \(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{(a q)} \quad \text { (basic) }\)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3}^{-}{ }_{(\mathrm{aq})}\)
Step 2: Assign oxidation number to Mn and Br. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 2 atoms of Mn.
\(2 \mathrm{MnO}_{4(\mathrm{aq})}^{-}+\mathrm{Br}_{(\mathrm{aq})}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3(\mathrm{aq})}^{-}\)
Step 3: Balance ‘O’ atoms by adding H₂O to the right-hand side.
\(2 \mathrm{MnO}_{4(a q)}^{-}+\mathrm{Br}_{(2 q)}^{-} \longrightarrow 2 \mathrm{MnO}_{2(\mathrm{~s})}+\mathrm{BrO}_{3 \text { (aq) }}^{-}+\mathrm{H}_{2} \mathrm{O}_{(l)}\)
Step 4: The medium is basic. To make the charges and hydrogen atoms on the two sides equal, add 2H⁺ on the left-hand side.

iii. H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l) (acidic)
Step 1: Write skeletal equation and balance the elements other than O and H.
H₂SO_4(aq) + C_(s) → CO_2(g) + SO_2(g) + H₂O_(l)
Step 2: Assign oxidation number to S and C. Calculate the increase and decrease in the oxidation number and make them equal.


To make the net increase and decrease equal, we must take 2 atoms of S.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)
Step 3: Balance ‘O’ atoms by adding H2O to the right-hand side.
2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l) + H₂O_(l)
Step 4: The medium is acidic. There is no charge on either side. Hydrogen atoms are equal on both side.
2H₂SO_4(aq) + C_(s) → CO₂ + 2SO_2(g) + H₂O_(l)
Step 5: Check two sides for balance of atoms and charges.
Hence, balanced equation: 2H₂SO_4(aq) + C_(s) → CO_2(g) + 2SO_2(g) + H₂O_(l)

iv. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\) (basic)
Step 1: Write skeletal equation and balance the elements other than O and H.
\(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{Sn}(\mathrm{OH})_{3(\mathrm{aq})}^{-} \longrightarrow \mathrm{Bi}_{(\mathrm{s})}+\mathrm{Sn}(\mathrm{OH})_{6(\mathrm{aq})}^{2-}\)
Step 2: Assign oxidation numbers to Bi and Sn. Calculate the increase and decrease in the oxidation number and make them equal.

To make the net increase and decrease equal, we must take 3 atoms of Sn and 2 atoms of Bi.

Step 4: The medium is basic. To make hydrogen atoms on the two sides equal, add 3W on the right-hand side.

Question B.
Balance the following redox equation by half reaction method

Answer:
i. H₂C₂O_4(aq) + \(\mathrm{MnO}_{4(a q)}^{-}\) → CO_2(g) + \(\mathrm{Mn}_{(\mathrm{aq})}^{2+}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance the atoms except O and H in each half equation. Balance half equation for O atoms by adding 4H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 8H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 5 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 5 and reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

ii. \(\mathrm{Bi}(\mathrm{OH})_{3(\mathrm{~s})}+\mathrm{SnO}_{2(\mathrm{aq})}^{2-} \longrightarrow \mathrm{SnO}_{3(\mathrm{aq})}^{2-}+\mathrm{Bi}_{(\mathrm{s})}\)
Step 1: Write unbalanced equation for the redox reaction. Assign oxidation number to all the atoms in reactants and products. Divide the equation into two half equations.

Step 2: Balance half equations for O atoms by adding H₂O to the side with less O atoms. Add 1H₂O to left side of oxidation half equation and 3H₂O to the right side of reduction half equation.

Step 3: Balance H atoms by adding H⁺ ions to the side with less H. Hence, add 2H⁺ ions to the right side of oxidation half equation and 3H⁺ ions to the left side of reduction half equation.

Step 4: Now add 2 electrons to the right side of oxidation half equation and 3 electrons to the left side of reduction half equation to balance the charges.

Step 5: Multiply oxidation half equation by 3 reduction half equation by 2 to equalize number of electrons in two half equations. Then add two half equation.

Reaction occurs in basic medium. However, H⁺ ions cancel out and the reaction is balanced. Hence, no need to add OH^– ions. The equation is balanced in terms of number of atoms and the charges.
Hence, balanced equation:

5. Complete the following table :

Assign oxidation number to the underlined species and write Stock notation of compound

Compound	Oxidation number	Stock notation
AuCl3	……………..	……………..
SnCl2	……………..	……………..
\(\underline{\mathrm{V}}_{2} \mathrm{O}_{7}^{4-}\)	……………..	……………..
\(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\)	……………..	……………..
H3AsO3	……………..	……………..

Answer:

Compound	Oxidation number	Stock notation
AuCl3	+3	Au(III)Cl3
SnCl2	+2	Sn(II)Cl2
\(\underline{\mathrm{V}}_{2} \mathrm{O}_{7}^{4-}\)	+5	V2(V)\(\mathrm{O}_{7}^{4-}\)
\(\underline{\mathrm{Pt}} \mathrm{Cl}_{6}^{2-}\)	+4	Pt(IV)\(\mathrm{Cl}_{6}^{2-}\)
H3AsO3	+3	H3As(III)O3

11th Chemistry Digest Chapter 6 Redox Reactions Intext Questions and Answers

Can you tell? (Textbook Page No. 81)

Question i.
Why does cut apple turn brown when exposed to air?
Answer:
Cut apple turns brown when exposed to air because polyphenols are released. These polyphenols undergo oxidation in the presence of air and impart brown colour.

Question ii.
Why does old car bumper change colour?
Answer:
Car bumper is made of iron which undergoes rusting over a period of time. Hence, old car bumper changes colour.

Question iii.
Why do new batteries become useless after some days?
Answer:
Batteries generate electricity by redox reactions. Once the chemicals taking part in redox reaction are used up, the battery cannot generate power. Hence, new batteries become useless after some days.

Can you recall? (Textbook Page No. 81)

Question i.
What is combustion reaction?
Answer:
Combustion is a process in which a substance combines with oxygen.

Question ii.
Write an equation for combustion of methane.
Answer:
Combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O + Heat + Light

Question iii.
What is the driving force behind reactions of elements?
Answer:
The ability of element to combine with other element or the ability of element to replace other element in compound is the driving force behind the reactions. This may involve formation of precipitates, formation of water, release of gas, etc.

Try this. (Textbook Page No. 82)

Question 1.
Complete the following table of displacement reactions. Identify oxidising and reducing agents involved.

Reactants	Products
Zn(s) + ————(aq)	————-(aq) + Cu(s)
Cu(s) + 2Ag+(aq)	—————– + ————–
———– + ————-	\( \mathrm{Co}_{(\mathrm{aq})}^{2+}\) + Ni(s)

Answer:

Try this (Textbook Page No. 88)

Question 1.
Classify the following unbalanced half equations as oxidation and reduction.

Answer:

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 5 Chemical Bonding Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 5 Chemical Bonding

1. Select and write the most appropriate alternatives from the given choices.

Question A.
Which molecule is linear?
a. SO₃
b. CO₂
c. H₂S
d. Cl₂O
Answer:
b. CO₂

Question B.
When the following bond types are listed in decreasing order of strength (strongest first). Which is the correct order?
a. covalent > hydrogen > van der waals
b. covalent > vander waal’s > hydrogen
c. hydrogen > covalent > vander waal’s
d. vander waal’s > hydrogen > covalent.
Answer:
a. covalent > hydrogen > van der waals

Question C.
Valence Shell Electron Pair repulsion (VSEPR) theory is used to predict which of the following :
a. energy levels in an atom
b. the shapes of molecules and ions.
c. the electron negetivities of elements.
d. the type of bonding in compounds.
Answer:
b. the shapes of molecules and ions.

Question D.
Which of the following is true for CO₂?

	C=O bond	CO2 molecule
A	polar	non-polar
B	non-polar	polar
C	polar	polar
D	non-polar	non-polar

Answer:

	C=O bond	CO2 molecule
A	polar	non-polar

Question E.
Which O₂ molecule is pargmagnetic. It is explained on the basis of :
a. Hybridisation
b. VBT
c. MOT
d. VSEPR
Answer:
c. MOT

Question F.
The angle between two covalent bonds is minimum in:
a CH₄
b. C₂H₂
c. NH₃
d. H₂O
Answer:
d. H₂O

2. Draw

Question A.
Lewis dot diagrams for the folowing
a. Hydrogen (H₂)
b. Water (H₂O)
c. Carbon dioxide (CO₂)
d. Methane (CH₄)
e. Lithium Fluoride (LiF)
Answer:

[Note: H atom in H₂ and Li atom in LiF attain the configuration of helium (a duplet of electrons).]

Question B.
Diagram for bonding in ethene with sp² Hybridisation.
Answer:

Question C.
Lewis electron dot structures of
a. HF
b. C₂H₆
c. C₂H₄
d. CF₃Cl
e. SO₂
Answer:

Question D.
Draw orbital diagrams of
a. Fluorine molecule
b. Hydrogen fluoride molecule
Answer:
a.

b.

3. Answer the following questions

Question A.
Distinguish between sigma and pi bond.
Answer:

σ (sigma) bond	π (pi) bond
1. It is formed when atomic orbitals overlap along internuclear axis.	1. It is formed when atomic orbitals overlap side-ways (laterally).
2. Electron density is high along the axis of the molecule (i.e., internuclear axis).	2. Electron density is zero along the axis of the molecule (i.e., internuclear axis).
3. In the formation of sigma bond, the extent of overlap is greater, hence, more energy is released.	3. In the formation of pi bond, the extent of overlap is less, hence, less energy is released.
4. It is a strong bond.	4. It is a weak bond.
5. Formation of sigma bonds involves s-s, s-p, p-p overlap and overlap between hybrid orbitals.	5. Formation of pi bonds involves p-p or d-d overlap. The overlap between hybrid orbitals is not involved.

Question B.
Display electron distribution around the oxygen atom in water molecule and state shape of the molecule, also write H-O-H bond angle.
Answer:
Electron distribution around oxygen atom in water molecule:
Shape of water molecule: Angular or V shaped H-O-H bond angle = 104°35′

Question C.
State octet rule. Explain its inadequecies with respect to
a. Incomplete octet
b. Expanded octet
Answer:
Statement: During the formation of chemical bond, atom loses, gains or shares electrons so that its outermost orbit (valence shell) contains eight electrons. Therefore, the atom attains the nearest inert gas electronic configuration.

a. Molecules with incomplete octet: e.g. BF₃, BeCl₂, LiCl
In these covalent molecules, the atoms B, Be and Li have less than eight electrons in their valence shell but these molecules are stable.
Li in LiCl has only two electrons, Be in BeCl₂ has four electrons while B in BF₃ has six electrons in the valence shell.

b. Molecules with expanded octet: Some molecules like SF₆, PCl₅, H₂SO₄ have more than eight electrons around the central atom.

Question D.
Explain in brief with one example:
a. Ionic bond
b. covalent bond
c. co-ordinate bond
Answer:
a. Formation of calcium chloride (CaCl₂):
i. The electronic configurations of calcium and chlorine are:
Na (Z = 11): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² or (2, 8, 8, 2)
Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵ or (2, 8, 7)
ii. Calcium has two electrons in its valence shell. It has tendency to lose two electrons to acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iii. Chlorine has seven electrons in its valence shell. It has tendency to gain one electron and thereby acquire the electronic configuration of the nearest inert gas, argon (2, 8, 8).
iv. During the combination of calcium and chlorine atoms, the calcium atom transfers its valence electrons to two chlorine atoms.
v. Calcium atom changes into Ca²⁺ ion while the two chlorine atoms change into two Cl^– ions. These ions are held together by strong electrostatic force of attraction.
vi. The formation of ionic bond(s) between Ca and Cl can be shown as follows:

b. Formation of Cl₂ molecule:
i. The electronic configuration of Cl atom is [Ne] 3s² 3p⁵.
ii. It needs one more electron to complete its valence shell.
iii. When two chlorine atoms approach each other at a certain internuclear distance, they share their valence electrons. In the process, both the atoms attain the valence shell of octet of nearest noble gas, argon.
iv. The shared pair of electrons belongs equally to both the chlorine atoms. The two atoms are said to be linked by a single covalent bond and a Cl₂ molecule is formed.

c. co-ordinate bond:
i. A coordinate bond is a type of covalent bond where both of the electrons that form the bond originate from the same atom
ii. An atom with a lone pair of electrons (non-bonding pair of electrons) is capable of forming a coordinate bond.
iii. For example, reaction of ammonia with boron trifluoride: Before the reaction, nitrogen (N) in ammonia has eight valence electrons, including a lone pair of electrons. Boron (B) in boron trifluoride has only six valence electrons, so it is two electrons short of an octet. The two unpaired electrons form a bond between nitrogen and boron, resulting in complete octets for both atoms. A coordinate bond is represented by an arrow. The direction of the arrow indicates that the electrons are moving from nitrogen to boron. Thus, ammonia forms a coordinate bond with boron trifluoride.

iv. Once formed, a coordinate covalent bond is the same as any other covalent bond.

Question E.
Give reasons for need of Hybridisation.
Answer:
The concept of hybridization was introduced because the valence bond theory failed to explain the following points:
i. Valencies of certain elements:
The maximum number of covalent bonds which an atom can form equals the number of unpaired electrons present in its valence shell. However, valence bond theory failed to explain how beryllium, boron and carbon forms two, three and four covalent bonds respectively.
a. Beryllium: The electronic configuration of beryllium is 1s² 2s². The expected valency is zero (as there is no unpaired electron) but the observed valency is 2 as in BeCl₂.
b. Boron: The electronic configuration of boron is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\). The valency is expected to be 1 but it is 3 as in BF₃.
c. Carbon: The electronic configuration of carbon is 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) . The valency is expected to be 2, but observed valency is 4 as in CH₄.

ii. The shapes and geometry of certain molecules:
The valence bond theory cannot explain shapes, geometries and bond angles in certain molecules,
e.g. a. Tetrahedral shape of methane molecule.
b. Bond angles in molecules like NH₃ (107°18′) and H₂O (104°35′).
However, the valency of the above elements and the observe structural properties of the above molecules can be explained by the concept of hybridization. These are the reasons for need of the concept of hybridization.

Question F.
Explain geometry of methane molecule on the basis of Hybridisation.
Answer:
Formation of methane (CH₄) molecule on the basis of sp³ hybridization:
i. Methane molecule (CH₄) has one carbon atom and four hydrogen atoms.
ii. The ground state electronic configuration of C (Z = 6) is 1s² \(2 \mathrm{p}_{\mathrm{x}}^{1}\) \(2 \mathrm{p}_{\mathrm{y}}^{1}\) \(2 \mathrm{p}_{\mathrm{z}}^{1}\);
Electronic configuration of carbon:

iii. In order to form four equivalent bonds with hydrogen, the 2s and 2p orbitals of C-atom undergo sp³ hybridization.
iv. One electron from the 2s orbital of carbon atom is excited to the 2pz orbital. Then the four orbitals 2s, px, py and pz mix and recast to form four new sp³ hybrid orbitals having same shape and equal energy. They are maximum apart and have tetrahedral geometry with H-C-H bond angle of 109°28′. Each hybrid orbital contains one unpaired electron.
v. Each of these sp³ hybrid orbitals with one electron overlap axially with the 1s orbital of hydrogen atom to form one C-H sigma bond. Thus, in CH₄ molecule, there are four C-H bonds formed by the sp³-s overlap.
Diagram:

Question G.
In Ammonia molecule the bond angle is 107°18 and in water molecule it is 104°35′, although in both the central atoms are sp³ hybridized Explain.
Answer:
i. The ammonia molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 107°28′. It is due to the following reasons.

• One lone pair and three bond pairs are present in ammonia molecule.
• The strength of lone pair-bond pair repulsion is much higher than that of bond pair-bond pair repulsion.
• Due to these repulsions, there is a small decrease in bond angle (~2°) from 109°28′ to 107°18′.

ii. The water molecule has sp³ hybridization. The expected bond angle is 109°28′. But the actual bond angle is 104°35′. It is due to the following reasons.

• Two lone pairs and two bond pairs are present in water molecule.
• The decreasing order of the repulsion is Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair.
• Due to these repulsions, there is a small decrease in bond angle (~5°) from 109°28′ to 104°35′.

Question H.
Give reasons for:
a. Sigma (σ) bond is stronger than Pi (π) bond.
b. HF is a polar molecule
c. Carbon is a tetravalent in nature.
Answer:
a. i. The strength of the bond depends on the extent of overlap of the orbitals. Greater the overlap, stronger is the bond.
ii. A sigma bond is formed by the coaxial overlap of the atomic orbitals which are oriented along the internuclear axis, hence the extent of overlap is maximum.
iii. A pi bond is formed by the lateral overlap of the atomic orbitals which are oriented perpendicular to the internuclear axis, hence the extent of orbital overlapping in side wise manner is less.
Hence, sigma bond is stronger than pi bond.

b. i. When a covalent bond is formed between two atoms of different elements that have different electronegativities, the shared electron pair does not remain at the centre. The electron pair is pulled towards the more electronegative atom resulting in the separation of charges.
ii. In H-F, fluorine is more electronegative than hydrogen. Therefore, the shared electron pair is pulled towards fluorine and fluorine acquires partial -ve charge and simultaneously hydrogen acquires partial +ve charge. This gives rise to dipole and H-F bond becomes polar. Hence, H-F is a polar molecule.

c. The electronic configuration of carbon is:
1s² 2s² 2p_x¹ 2p_y¹
One electron from ‘2s’ orbital is promoted to the empty ‘2p’ orbital.
Thus, in excited state, carbon has four half-filled orbitals.

Hence, carbon can form 4 bonds and is tetravalent in nature.

Question I.
Which type of hybridization is present in ammonia molecule? Write the geometry and bond angle present in ammonia.
Answer:
The type of hybridization present in ammonia (NH₃) molecule is sp³.
Geometry of ammonia molecule is pyramidal or distorted tetrahedral.
Bond angle in ammonia molecule is 107°18′.

Question J.
Identify the type of orbital overlap present in
a. H₂
b. F₂
c. H-F molecule.
Explain diagramatically.
Answer:
i. s-s σ overlap:
a. The overlap between two half-filled s orbitals of two different atoms containing unpaired electrons with opposite spins is called s-s overlap.
e.g. Formation of H₂ molecule by s-s overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹. The 1s¹ orbitais of two hydrogen atoms overlap along the internuclear axis to form a σ bond between the atoms in H₂ molecule.
b. Diagram:

ii. p-p σ overlap:
a. This type of overlap takes place when two p orbitals from different atoms overlap along the internuclear axis.
e.g. Formation of F₂ molecule by p-p overlap:
Fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\).
During the formation of F₂ molecule, half-filled 2p_z orbital of one F atom overlaps with similar half-filled 2p_z orbital containing electron with opposite spin of another F atom axially and a p-p σ bond is formed.
b. Diagram:

iii. s-p σ overlap:
a. In this type of overlap one half filled s orbital of one atom and one half filled p orbital of another orbital overlap along the internuclear axis.
e.g. Formation of HF molecule by s-p overlap:
Hydrogen atom (Z = 1) has electronic configuration: 1s¹ and fluorine atom (Z = 9) has electronic configuration 1s² 2s² \(2 \mathrm{p}_{\mathrm{x}}^{2}\) \(2 \mathrm{p}_{\mathrm{y}}^{2}\) \(2 \mathrm{p}_{\mathrm{z}}^{2}\). During the formation of HF molecule, half-filled Is orbital of hydrogen atom overlaps coaxially with half-filled 2p_z orbital of fluorine atom with opposite electron spin and an s-p σ bond is formed.
b. Diagram:

Question K.
F-Be-F is a liner molecule but H-O-H is angular. Explain.
Answer:
i. In the BeF₂ molecule, the central beryllium atom undergoes sp hybridization giving rise to two sp hybridized orbitals placed diagonally opposite with an angle of 180°. Thus, F-Be-F is a linear molecule.

ii. In the H₂O molecule, the central oxygen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. There are two lone pairs of electrons in two of the sp³ hybrid orbitals of oxygen. The lone pair-lone pair repulsion distorts the structure. Hence, H-O-H is angular or V-shaped.

Question L.
BF₃ molecule is planar but NH₃ pyramidal. Explain.
Answer:
i. In the BF₃ molecule, the central boron atom undergoes sp² hybridization giving rise to three sp² hybridized orbitals directed towards three comers of an equilateral triangle. Thus, the geometry is trigonal planar.

ii. In the NH₃ molecule, the central nitrogen atom undergoes sp³ hybridization giving rise to four sp³ hybridized orbitals directed towards four comers of a tetrahedron. The expected geometry of NH₃ molecule is regular tetrahedral with bond angle 109°28′. There is one lone pair of electrons in one of the sp³ hybrid orbitals of nitrogen. The lone pair-bond pair repulsion distorts the bond angle. Hence, the structure of NH₃ is distorted and it has pyramidal geometry.

Question M.
In case of bond formation in Acetylene molecule :
a. How many covalend bonds are formed ?
b. State number of sigma and pi bonds formed.
c. Name the type of Hybridisation.
Answer:
a. In acetylene molecule, there are five covalent bonds.
b. In acetylene molecule, there are three sigma bonds and two pi bonds.
c. In acetylene molecule, each carbon atom undergoes sp hybridization.

Question N.
Define :
a. Bond Enthalpy
b. Bond Length
Answer:
a. Bond Enthalpy:
Bond enthalpy is defined as the amount of energy required to break one mole of a bond of one type, present between two atoms in a gaseous state.

b. Bond Length:
Bond length is defined as the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule.

Question O.
Predict the shape and bond angles in the following molecules:
a. CF₄
b. NF₃
c. HCN
d. H₂S
Answer:
a. CF₄: There are four bond pairs on the central atom. Hence, shape of CF₄ is tetrahedral and F-C-F bond angle is 109° 28′.
b. NF₃: There are three bond pairs and one lone pair on the central atom. Hence, shape of NF₃ is trigonal pyramidal and F-N-F bond angle is less than 109° 28′.
c. HCN: There are two bond pairs on the central atom. Hence, shape of HCN is linear and H-C-N bond angle is 180°.
d. H₂S: There are two bond pairs and two lone pairs on the central atom. Hence, shape of H₂S is bent or V-shaped and H-S-H bond angle is slightly less than 109° 28′.

4. Using data from the Table, answer the following :

a. What happens to the bond length when unsaturation increases?
b. Which is the most stable compound?
c. Indicate the relation between bond strength and Bond enthalpy.
d. Comment on overall relation between Bond length, Bond Enthalpy and Bond strength and stability.
Answer:
a. When unsaturation increases, the bond length decreases.
b. The stable compound is ethyne (C₂H₂).
c. Bond strength ∝ Bond enthalpy
Larger the bond enthalpy, stronger is the bond.
d. As bond length decreases, bond enthalpy, bond strength and stability increase.

5. Complete the flow chart

Answer:

6. Complete the following Table

Answer:

7. Answer in one sentence:

Question A.
Indicate the factor on which stalility of ionic compound is measured?
Answer:
The stability of an ionic compound is measured by the amount of energy released during lattice formation.

Question B.
Arrange the following compounds on the basis of lattice energies in decreasing (descending) order: BeF₂, AlCl₃, LiCl, CaCl₂, NaCl.
Answer:
AlCl₃ > BeF₂ > CaCl₂ > LiCl > NaCl

Question C.
Give the total number of electrons around sulphur (S) in SF₆ compound.
Answer:
The total number of electrons around sulphur (S) in SF₆ is 12.

Question D.
Covalant bond is directional in nature. Justify.
Answer:
Covalent bond is formed by the overlap of two half-filled atomic orbitals. The atomic orbitals are oriented in specific directions in space (except s-orbital which is spherical). Hence, covalent bond is directional in nature.

Question E.
What are the interacting forces present during formation of a molecule of a compound ?
Answer:
a. Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
b. Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

Question F.
Give the type of overlap by which pi (π) bond is formed.
Answer:
The type of overlap by which pi (π) bond is formed is p-p lateral overlap.

Question G .
Mention the steps involved in Hybridization.
Answer:
The steps involved in hybridization are:

• formation of the excited state and
• mixing and recasting of orbitals.

Question H.
Write the formula to calculate bond order of molecule.
Answer:
Bond order of a molecule = \(\frac{\mathrm{N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}}{2}\)
where, N_b is the number of electrons present in bonding MOs and N_a is the number of electrons present in antibonding MOs.

Question I.
Why is O₂ molecule paramagnetic?
Answer:
The electronic configuration of O₂ molecule is (σ1s)² (σ*1s)² (σ2s)² (σ*2s)² (σ2p_z)² (π2p_x)² (π2p_y)² (π*2p_x)¹ (π*2p_y)¹
Since the oxygen molecule contains two unpaired electrons, it is paramagnetic.

Question J.
What do you mean by formal charge ? Explain its significance with the help of suitable example.
Answer:
Formal charge is the charge assigned to an atom in a molecule, assuming that all electrons are shared equally between atoms, regardless of their relative electronegativities.

Structure (I):

Structure (II):

Structure (III):

While determining the best Lewis structure per molecule, the structure is chosen such that the formal charge is as close to zero as possible. The structure having the lowest formal charge has the lowest energy.

In structure (I), the formal charge on each atom is 0 while in structures (II) and (III) formal charge on carbon is 0 while oxygens have formal charge -1 or +1. Hence, the possible structure with the lowest energy will be structure (I). Thus, formal charges help in the selection of the lowest energy structure from a number of possible Lewis structures for a given species.

11th Chemistry Digest Chapter 5 Chemical Bonding Intext Questions and Answers

(Textbook Page No. 55)

Question 1.
Why are atoms held together in chemical compounds?
Answer:
Atoms are held together in chemical compounds due to chemical bonds.

Question 2.
How are chemical bonds formed between two atoms?
Answer:
There are two ways of formation of chemical bonds:

1. by loss and gain of electrons
2. by sharing a pair of electrons between the two atoms.

In either process of formation of chemical bond, each atom attains a stable noble gas electronic configuration.

Question 3.
Which electrons are involved in the formation of chemical bonds?
Answer:
The electrons present in the outermost shell of an atom are involved in the formation of a chemical bond.

Internet my friend (Textbook Page No. 55)

Question 1.
Search more atoms, which complete their octet during chemical combinations.
Answer:
In compounds like KCl, MgCl₂, CaO, NaF, etc, the constituent atoms complete their octet by lose or gain of electrons.
e.g. K → K⁺ + e^–
Cl + e^– → Cl^–
K⁺ + Cl^– → KCl
[Note: Students are expected to search more atoms on their own.]

Use your brainpower. (Textbook Page No. 60)

Question 1.
Which atom in \(\mathrm{NH}_{4}^{+}\) will have formal charge +1?
Answer:
In \(\mathrm{NH}_{4}^{+}\), nitrogen atom (N) will have formal charge of+1.

Use your brainpower. (Textbook Page No. 61)

Question 1.
How many electrons will be around I in the compound IF₇?
Answer:
Lewis structure of IF₇ is:

In IF₇, iodine (I) atom will be surrounded by 14 electrons.

Question 2.
Why is H₂ stable even though it never satisfies the octet rule?
Answer:
The valence shell configuration of hydrogen atom is 1s¹. Two hydrogen atoms approach each other and share their valence electrons. By having two electrons in its valence shell, H atom attains the nearest noble gas configuration of He. H₂ molecule attains stability due to duplet formation. Hence, H₂ is stable even though it never satisfies the octet rule.

(Textbook Page No. 64)

Question 1.
Lowering of energy takes during bond formation. How does this happen?
Answer:
i. When two combining atoms approach each other to form a covalent bond, the following interacting forces come into play.

• Forces of attraction: The nucleus of one atom attracts the electrons of the other atom and vice-versa.
• Forces of repulsion: The electron of one atom repels the electron of the other atom and vice-versa (as electrons are negatively charged). There is repulsion between the two nuclei (as the nuclei are positively charged).

ii. The balance between attractive and repulsive forces decide whether the bond will be formed or not.
iii. When the magnitude of attractive forces is more than the magnitude of repulsive forces, the energy of the system decreases and a covalent bond is formed.
iv. When the magnitude of repulsive forces becomes more than that of attraction, the total energy of the system increases, and a covalent bond is not formed.
Hence, lowering of energy takes during bond formation.

Can you tell? (TextBook Page No. 76)

Question 1.
Which molecules are polar?
H-I, H-O-H, H-Br, Br₂, N₂, I₂, NH₃
Answer:
i. H-I: Polar
ii. H-O-H: Polar
iii. H-Br: Polar
iv. Br₂: Nonpolar
v. N₂: Nonpolar
vi. I₂: Nonpolar
vii. NH₃: Polar

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 4 Structure of Atom Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 4 Structure of Atom

1. Choose the correct option.

Question A.
The energy difference between the shells goes on ……….. when moved away from the nucleus.
a. Increasing
b. decreasing
c. equalizing
d. static
Answer:
b. decreasing

Question B.
The value of Plank’s constant is
a. 6.626× 10^-34 Js
b. 6.023× 10^-24 Js
c. 1.667 × 10^-28 Js
d. 6.626× 10^-28 Js
Answer:
a. 6.626× 10^-34 Js

Question C.
p-orbitals are ……. in shape.
a. spherical
b. dumbbell
c. double dumbbell
d. diagonal
Answer:
b. dumbbell

Question D.
“No two electrons in the same atoms can have an identical set of four quantum numbers”. This statement is known as
a. Pauli’s exclusion principle
b. Hund’s rule
c. Aufbau rule
d. Heisenberg uncertainty principle
Answer:
a. Pauli’s exclusion principle

Question E.
Principal Quantum number describes
a. shape of orbital
b. size of the orbital
c. spin of electron
d. orientation of in the orbital electron cloud
Answer:
b. size of the orbital

2. Make the pairs:

	A		B
a.	Neutrons	i.	six electrons
b.	p-orbital	ii.	-1.6 × 10-19 C
c.	charge on electron	iii.	Ultraviolet region
d.	Lyman series	iv.	Chadwick

Answer:
a – iv,
b – i,
c – ii,
d – iii

3. Complete the following information about the isotopes in the chart given below :

(Hint: Refer to Periodic Table if required)
Answer:

4. Match the following :

Answer:
a – iv,
b – iii,
c – ii,
d – i

5. Answer in one sentence :

Question A.
If an element ‘X’ has mass number 11 and it has 6 neutrons, then write its representation.
Answer:
The representation of the given element is \({ }_{5}^{11} \mathrm{X}\).

Question B.
Name the element that shows simplest emission spectrum.
Answer:
The element that shows simplest emission spectrum is hydrogen.

Question C.
State Heisenberg uncertainty principle.
Answer:
Heisenberg uncertainty principle states that “It is impossible to determine simultaneously, the exact position and exact momentum (or velocity) of an electron”.

Question D.
Give the names of quantum numbers.
Answer:
The four quantum numbers are: principal quantum number (n), azimuthal or subsidiary quantum number (l), magnetic quantum number (m_l) and electron spin quantum number (m_s).

Question E.
Identify from the following the isoelectronic species:
Ne, O^2-, Na⁺ OR Ar, Cl^2-, K⁺
Answer:
Atoms and ions having the same number of electrons are isoelectronic.

Species	No. of electrons
Ne	10
O2-	8 + 2 = 10
Na+	11 – 1 = 10
Ar	18
Cl2-	17 + 2 = 19
K+	19 – 1 = 18

Hence, Ne, O^2-, Na⁺ are isoelectronic species.

6. Answer the following questions.

Question A.
Differentiate between Isotopes and Isobars.
Answer:

No.	Isotopes	Isobars
i.	Isotopes are atoms of same element.	Isobars are atoms of different elements.
ii.	They have same atomic number but different atomic mass number.	They have same atomic mass number but different atomic numbers.
iii.	They have same number of protons but different number of neutrons.	They have different number of protons and neutrons.
iv.	They have same number of electrons.	They have different number of electrons.
V.	They occupy same position in the modem periodic table.	They occupy different positions in the modem periodic table.
vi.	They have similar chemical properties.	They have different chemical properties.
e.g.	\({ }_{6}^{12} \mathrm{C}\) and \({ }_{6}^{14} \mathrm{C}\)	\({ }_{6}^{14} \mathrm{C}\) and \({ }_{7}^{14} \mathrm{~N}\)

Question B.
Define the terms:
i. Isotones
ii. Isoelectronic species
iii. Electronic configuration
Answer:
i. Isotones: Isotones are defined as the atoms of different elements having same number of neutrons in their nuclei. e.g. \({ }_{5}^{11} \mathrm{B}\) and \({ }_{6}^{12} \mathrm{C}\) having 6 neutrons each are isotones.

ii. Isoelectronic species:
soelectronic species are defined as atoms and ions having the same number of electrons.
e. g. Ar, Ca²⁺ and K⁺ containing 18 electrons each.

iii. Electronic configuration:
Electronic configuration of an atom is defined as the distribution of its electrons in orbitals.

Question C.
State and explain Pauli’s exclusion principle.
Answer:
Pauli’s exclusion principle:
i. Statement: “No two electrons in an atom can have the same set of four quantum numbers”. OR “Only two electrons can occupy the same orbital and they must have opposite spins. ”
ii. The capacity of an orbital to accommodate electrons is decided by Pauli’s exclusion principle.
iii. According to this principle, for an electron belonging to the same orbital, the spin quantum number must be different since the other three quantum numbers are the same.
iv. The spin quantum number can have two values: +\(\frac {1}{2}\) and –\(\frac {1}{2}\).
v. Example, consider helium (He) atom with electronic configuration 1 s².
For the two electrons in Is orbital, the four quantum numbers are as follows:
Electron number Quantum number Set of values of quantum numbers

Thus, in an atom, any two electrons can have the same three quantum numbers, but the fourth quantum number must be different.
vi. This leads to the conclusion that an orbital can accommodate maximum of two electrons and if it has two electrons, they must have opposite spin.

Question D.
State Hund’s rule of maximum multiplicity with suitable example.
Answer:
Hund’s rule of maximum multiplicity:
i. Statement: “Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each.”
ii. Example, according to Hund’s rule, each of the three-degenerate p-orbitals must get one electron of parallel spin before any one of them receives the second electron of opposite spin. Therefore, the configuration of four electrons occupying p-orbitals is represented as

iii. As a result of Hund’s rule, the atom with fully filled and half-filled set of degenerate orbitals has extra stability.

Question E.
Write the drawbacks of Rutherford’s model of an atom.
Answer:
Drawbacks of Rutherford’s model of an atom:
i. Rutherford’s model of an atom resembles the solar system with the nucleus playing the role of the massive sun and the electrons are lighter planets. Thus, according to this model, electrons having negative charge revolve in various orbits around the nucleus. However, the electrons revolving about the nucleus in fixed orbits pose a problem. Such orbital motion is an accelerated motion accompanied by a continuous change in the velocity of electron as noticed from the continuously changing direction. According to Maxwell’s theory of electromagnetic radiation, accelerated charged particles would emit electromagnetic radiation. Hence, an electron revolving around the nucleus should continuously emit radiation and lose equivalent energy. As a result, the orbit would shrink continuously and the electron would come closer to the nucleus by following a spiral path. It would ultimately fall into the nucleus. Thus, Rutherford’s model has an intrinsic instability of atom. However, real atoms are stable.

ii. Rutherford’s model of an atom does not describe the distribution of electrons around the nucleus and their energies.

Question F.
Write postulates of Bohr’s Theory of hydrogen atom.
Answer:
Postulates of Bohr’s theory of hydrogen atom:
i. The electron in the hydrogen atom can move around the nucleus in one of the many possible circular paths of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states. These orbits are arranged concentrically around the nucleus in an increasing order of energy.

ii. The energy of an electron in the orbit does not change with time. However, the electron will move from a lower stationary state to a higher stationary state if and when the required amount of energy is absorbed by the electron. Energy is emitted when electron moves from a higher stationary state to a lower stationary state. The energy change does not take place in a continuous manner.

iii. The frequency of radiation absorbed or emitted when transition occurs between two stationary states that differ in energy by ΔE is given by the following expression:
ν = \(\frac{\Delta E}{h}=\frac{E_{2}-E_{1}}{h}\) ………….(1)
Where E₁ and E₂ are the energies of the lower and higher allowed energy states respectively. This expression is commonly known as Bohr’s frequency rule.

iv. The angular momeñtum of an electron in a given stationary state can be expressed as mvr = n × h/2π
where, n 1,2, 3
Thus, an electron can move only in those orbits for which its angular momentum is integral multiple of h/2π.
Thus, only certain fixed orbits are allowed.

Question G.
Mention demerits of Bohr’s Atomic model.
Answer:
Demerits of Bohr’s atomic model:

• Bohr’s atomic model (theory) failed to account for finer details of the atomic spectrum of hydrogen as observed in sophisticated spectroscopic experiments.
• Bohr’s atomic model (theory) was unable to explain the spectrum of atoms other than hydrogen.
• Bohr’s atomic model (theory) could not explain the splitting of spectral lines in the presence of a magnetic field (Zeeman effect) or electric field (Stark effect).
• Bohr’s atomic model (theory) failed to explain the ability of atoms to form molecules by chemical bonds.

Question H.
State the order of filling atomic orbitals following Aufbau principle.
Answer:
Aufbau principle:
i. Aufbau principle gives the sequence in which various orbitals are filled with electrons.
ii. In the ground state of an atom, the orbitals are filled with electrons based on increasing order of energies of orbitals, Pauli’s exclusion principle and Hund’s rule of maximum multiplicity.
iii. Increasing order of energies of orbitals:

• Orbitals are filled in order of increasing value of (n + l)
• In cases where the two orbitals have same value of (n + l), the orbital with lower value of n is filled first.

iv. The increasing order of energy of different orbitals in a multi-electron atom is:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s and so on.

Question I.
Explain the anomalous behavior of copper and chromium.
Answer:
i. Copper:

• Copper (Cu) has atomic number 29.
• Its expected electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁹.
• The 3d orbital is neither half-filled nor fully filled. Hence, it has less stability.
• Due to interelectronic repulsion forces, one 4s electron enters into 3d orbital. This makes 3d orbital completely filled and 4s orbital half-filled which gives extra stability and the electronic configuration of Cu becomes, 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰.

ii. Chromium:

• Chromium (Cr) has atomic number 24.
• Its expected electronic configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²3d⁴.
• The 3d orbital is less stable as it is not half-filled.
• Due to inter electronic repulsion forces, one 4s electron enters into 3d orbital. This makes 4s and 3d orbitals half-filled which gives extra stability and the electronic configuration of Cr becomes, 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵.

Question J.
Write orbital notations for electrons in orbitals with the following quantum numbers.
a. n = 2, l =1
b. n = 4, l = 2
c. n = 3, l = 2
Answer:
i. 2p
ii. 4d
iii. 3d

Question K.
Write electronic configurations of Fe, Fe²⁺, Fe³⁺
Answer:

Species	Orbital notation
Fe	1s2 2s2 2p63s2 3p6 4s2 3d6 OR [Ar] 4s2 3d6
Fe2+	Is2 2s2 2p6 3s2 3p6 3d6 OR [Ar] 3d6
Fe3+	1s2 2s2 2p6 3s2 3p6 3d5 OR [Ar] 3d5

Question L.
Write condensed orbital notation of electonic configuration of the following elements:
a. Lithium (Z = 3)
b. Carbon (Z=6)
c. Oxygen (Z = 8)
d. Silicon (Z = 14)
e. Chlorine (Z = 17)
f. Calcium (Z = 20)
Answer:

No.	Element	Condensed orbital notation
i.	Lithium (Z = 3)	[He] 2s1
ii.	Carbon (Z = 6)	[He] 2s2 2p2
iii.	Oxygen (Z = 8)	[He] 2s2 2p4
iv.	Silicon (Z = 14)	[Ne] 3s2 3p2
v.	Chlorine (Z = 17)	[Ne] 3s2 3p5
vi.	Calcium (Z = 20)	[Ar] 4s2

Question M.
Draw shapes of 2s and 2p orbitals.
Answer:
2s orbital:

2p orbital:

Question N.
Explain in brief, the significance of azimuthal quantum number.
Answer:
Azimuthal quantum number (l):

• Azimuthal quantum number is also known as subsidiary quantum number and is represented by letter l.
• It represents the subshell to which the electron belongs. It also defines the shape of the orbital that is occupied by the electron.
• Its value depends upon the value of principal quantum number ‘n’. It can have only positive values between 0 and (n – 1).
• Atomic orbitals with the same value of ‘n’ but different values of ‘l’ constitute a subshell belonging to the shell for the given ‘n’ The azimuthal quantum number gives the number of subshells in a principal shell. The subshells have l to be 0, 1, 2,3 … which are represented by symbols s, p, d, f, … respectively.

Question O.
If n = 3, what are the quantum number l and m_l?
Answer:
: For a given n, l = 0 to (n – 1) and for given l, m_l = -l……, 0…….. + l
Therefore, the possible values of l and m_l for n = 3 are:

Question P.
The electronic configuration of oxygen is written as 1s² 2s² 2p_x² 2p_y¹ 2p_z¹ and not as 1s² 2s² 2p_x² 2p_y² 2p_z⁰. Explain.
Answer:

• According to Hund’s rule of maximum multiplicity “Pairing of electrons in the orbitals belonging to the same subshell does not occur unless each orbital belonging to that subshell has got one electron each.”
• Oxygen has 8 electrons. The first two electrons will pair up in the Is orbital, the next two electrons will pair up in the 2s orbital and this leaves 4 electrons, which must be placed in the 2p orbitals.
• Each of the three degenerate p-orbitals must get one electron of parallel spin before any one of them receives the second electron of opposite spin. Therefore, two p orbitals have one electron each and one p-orbital will have two electrons.

Thus, the electronic configuration of oxygen is written as 1s² 2s² 2p_x² 2p_y¹ 2p_z¹ and not as 1s² 2s² 2p_x² 2p_y² 2p_z⁰.

Question Q.
Write note on ‘Principal Quantum number.
Answer:
Principal quantum number (n):
i. Principal quantum number indicates the principal shell or main energy level to which the electron belongs.
ii. It is denoted by ‘n’ and is a positive integer with values 1, 2, 3, 4, 5, 6, ….
iii. A set of atomic orbitals with given value of ‘n’ constitutes a single shell. These shells are also represented by the letters K, L, M, N, etc.
iv. With increase of ‘n’, the number of allowed orbitals in that shell increases and is given by n².
v. The allowed orbitals in the first four shells are given below:

vi. As the value of ‘n’ increases, the distance of the shell from the nucleus increases and the size of the shell increases. Its energy also goes on increasing.

Question R.
Using concept of quantum numbers, calculate the maximum numbers of electrons present in the ‘M’ shell. Give their distribution in shells, subshells and orbitals.
Answer:
i. Each main shell contains a maximum of 2n² electrons.
For ‘M’ shell, n = 3.
Therefore, the maximum numbers of electrons present in the ‘M’ shell = 2 × (3)² = 18.

ii. The distribution of these electrons in shells, subshells and orbitals can be given as follows:

Note: Orbital distribution in the first four shells:

Question S.
Indicate the number of unpaired electrons in :
a. Si (Z = 14)
b. Cr (Z = 24)
Answer:
i. . Si (Z = 14): 1s² 2s² 2p⁶ 3s² 3p²
Orbital diagram:

Number of unpaired electrons = 2

ii. Cr (Z = 24): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
Orbital diagram:

Number of unpaired electrons = 6

Question T.
An atom of an element contains 29 electrons and 35 neutrons. Deduce-
a. the number of protons
b. the electronic configuration of that element.
Answer:
a. In an atom, number of protons is equal to number of electrons.
The given atom contains 29 electrons.
∴ Number of protons = 29

b. The electronic configuration of an atom of an element containing 29 electrons is:
1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰
[Note: Given element is copper (Cu) with Z = 29]

11th Chemistry Digest Chapter 4 Structure of Atom Intext Questions and Answers

Can you recall? (Textbook Page No. 35)

Question i.
What is the smallest unit of matter?
Answer:
The smallest unit of matter is atom.

Question ii.
What is the difference between molecules of an element and those of a compound?
Answer:
The molecules of an element are made of atoms of same element while the molecules of a compound are made of atoms of different elements.

Question iii.
Does an atom have any internal structure or is it indivisible?
Answer:
Yes, an atom has internal structure. Different subatomic particles such as protons, electrons and neutrons constitute an atom. So, it is divisible.

Question iv.
Which particle was identified by J. J. Thomson in the cathode ray tube experiment?
Answer:
Electron was identified by J.J. Thomson in the cathode ray tube experiment.

Question v.
Which part of an atom was discovered by Ernest Rutherford from the experiment of scattering of α-particles by gold foil?
Answer:
Nucleus of an atom was discovered by Ernest Rutherford from the experiment of scattering of α-particles by gold foil.

Just Think (Textbook Page No. 41)

Question 1.
What does the negative sign of electron energy convey?
Answer:
Negative sign for the energy of an electron in any orbit in a hydrogen atom indicates that the energy of the electron in the atom is lower than the energy of a free electron at rest. A free-electron at rest is an electron that is infinitely far away from the nucleus and is assigned the energy value of zero.

As the electron gets close to the nucleus, value of ‘n’ decreases and E_n becomes large in absolute value and more negative. The negative sign corresponds to attractive forces between electron and nucleus.

Internet my friend (Textbook Page No. 44)

Question 1.
Collect information about the structure of atom.
Answer:
Students can use links given below as references and collect information about structure of atom on their own.
https://www.livescience.com/65427-fundamental-elementary-particles.html http://www.chemistryexplained.com/Ar-Bo/Atomic-Structure.html
https://www.thoughtco.com/basic-model-of-the-atom-603799

Balbharti Maharashtra State Board 11th Chemistry Textbook Solutions Chapter 3 Basic Analytical Techniques Textbook Exercise Questions and Answers.

Maharashtra State Board 11th Chemistry Solutions Chapter 3 Basic Analytical Techniques

1. Choose the correct option

Question A.
Which of the following methods can be used to separate two compounds with different solubilities in the same solvent?
a. Fractional crystallization
b. Crystallization
c. Distillation
d. Solvent extraction
Answer:
a. Fractional crystallization

Question B.
Which of the following techniques is used for the separation of glycerol from soap in the soap industry?
a. Distillation under reduced pressure
b. Fractional distillation
c. Filtration
d. Crystallization
Answer:
a. Distillation under reduced pressure

Question C.
Which technique is widely used in industry to separate components of the mixture and also to purify them?
a. Steam distillation
b. Chromatography
c. Solvent extraction
d. Filtration
Answer:
b. Chromatography

Question D.
A mixture of acetone and benzene can be separated by the following method :
a. Simple distillation
b. Fractional distillation
c. Distillation under reduced pressure
d. Sublimation
Answer:
b. Fractional distillation

Question E.
Colourless components on chromatogram can not be observed by the following :
a. Using UV light
b. Using iodine chamber
c. Using the spraying reagent
d. Using infrared light
Answer:
d. Using infrared light

2. Answer the following

Question A.
Which of the following techniques is used for purification of solid organic compounds?
a. Crystallisation
b. Distillation
Answer:
Solid (crude/impure) organic compounds can be purified by crystallization.

Question B.
What do you understand by the terms
a. residue
b. filtrate.
Answer:
a. Residue: In the process of filtration, the insoluble (undissolved) impurities which remain on the filter paper are called residue.

b. Filtrate: In the process of filtration, the liquid which pass through the filter paper and collected in the beaker is called filtrate.

Question C.
Why is a condenser used in distillation process?
Answer:
In the process of distillation, a liquid is converted into its vapour and the vapour is then condensed back to liquid on cooling. The condenser has a jacket with two outlets through which water is circulated. Hence, to provide efficient cooling, a condenser is used.

Question D.
Why is paper moistened before filtration?
Answer:
Before filtration, filter paper is moistened with appropriate solvent to ensure that it sticks to the funnel and does not let the air to pass through the leaks.

Question E.
What is the stationary phase in Paper Chromatography?
Answer:
Paper chromatography is a type of partition chromatography in which a special quality paper, namely Whatman paper 1 is used. The water trapped in the fibres of the paper acts as stationary phase.

Question F.
What will happen if the upper outlet of the condenser is connected to the tap instead of the lower outlet?
Answer:

• If water enters through upper outlet of condenser, the water will quickly flow down under the influence of gravity. This allows only a small section of the condenser to be cooled enough.
• If water enters through lower outlet of condenser, the entire condenser will be filled with water before it leaves out providing maximum cooling to the condenser. This results in maximum recovery of purified liquid.

Hence, water must be allowed to enter through lower outlet of condenser during distillation process.

Question G.
Give names of two materials used as stationary phase in chromatography.
Answer:

1. Alumina
2. Silica gel

Question H.
Which properties of solvents are useful for solvent extraction?
Answer:

• Organic compound must be more soluble in the organic solvent, than in water.
• Solvent should be immiscible with water and be able to form two distinct layers.

Question I.
Why should spotting of mixture be done above the level of mobile phase ?
Answer:

• If spotting of a mixture is done at the level of mobile phase, then solvent will come in contact with the sample spot.
• Sample spot will dissolve in the mobile phase and its components will move all over the plate resulting in no distinct separation.

Hence, spotting of mixture should be done above the level of mobile phase.

Question J.
Define : a. Stationary phase b. Saturated solution
Answer:
a. Stationary phase:
Stationary phase is a solid or a liquid supported on a solid which remains fixed in a place and on which different solutes are adsorbed to a different extent.

b. Saturated solution:
A saturated solution is a solution which cannot dissolve additional quantity of a solute.

Question K.
What is the difference between simple distillation and fractional distillation?
Answer:

No.	Simple distillation	Fractional distillation
i.	If in a mixture the difference in boiling points of two liquids is appreciable/large, they are separated from each other using the simple distillation.	If in a mixture the difference in boiling points of two liquids is not appreciable/large, they are separated from each other using the fractional distillation.
ii.	Simple distillation assembly is used.	fractionating column is fitted in distillation assembly.
e.g.	Mixture of acetone (b.p. 329 K) and water (b.p. 373 K) can be separated by this method.	Mixture of acetone (b.p. 329 K) and methanol (b.p. 337.7 K) can be separated by this method.

Question L.
Define a. Solvent extraction
b. Distillation.
Answer:
a. Solvent extraction:
Solvent extraction is a method used to separate an organic compound present in an aqueous solution, by shaking it with a suitable organic solvent in which the compound is more soluble than water.

b. Distillation:
The process in which liquid is converted into its vapour phase at its boiling point and the vapour is then condensed back to liquid on cooling is known as distillation.

Question M.
List the properties of solvents which make them suitable for crystallization.
Answer:
The solvent to be used for crystallization should have following properties:

• The compound to be crystallized should be least or sparingly soluble in the solvent at room temperature but highly soluble at high temperature.
• Solvent should not react chemically with the compound to be purified.
• Solvent should be volatile so that it can be removed easily.

Question N.
Name the different types of Chromatography and explain the principles underlying them.
Answer:
Depending on the nature of the stationary phase i.e., whether it is a solid or a liquid, chromatography is classified into adsorption chromatography and partition chromatography.
i. Adsorption chromatography: This technique is based on the principle of differential adsorption. Different solutes are adsorbed on an adsorbent to different extent.

Adsorption chromatography is further classified into two types:

1. Column chromatography
2. Thin-layer chromatography

ii. Partition chromatography: This technique is based on continuous differential partitioning of components of a mixture between stationary and mobile phases. For example, paper chromatography

Question O.
Why do we see bands separating in column chromatography?
Answer:

• In column chromatography, the solutes get adsorbed on the stationary phase and depending on the degree to which they are adsorbed, they get separated from each other.
• The component which is readily adsorbed are retained on the column and others move down the column to various distances forming distinct bands.

Hence, we see bands separating in column chromatography.

Question P.
How do you visualize colourless compounds after separation in TLC and Paper Chromatography?
Answer:
i. Thin-layer chromatography (TLC): If components are colourless but have the property of fluorescence then they can be visualized under UV light, or the plate can be kept in a chamber containing a few iodine crystals. The iodine vapours are adsorbed by the components and the spots appear brown. Also, spraying agent like ninhydrin can also be used (for amino acids).

ii. Paper Chromatography: The spots of the separated colourless components may be observed either under ultra-violet light or by the use of an appropriate spraying agent.

Question Q.
Compare TLC and Paper Chromatography techniques.
Answer:

Chromatography technique	TLC	Paper chromatography
Principle	It is based on the principle of differential adsorption. Different solutes are adsorbed on an adsorbent to different extent.	It is based on continuous differential partitioning of components of a mixture between stationary and mobile phases.
Stationary phase	Solid (adsorbent like silica gel or alumina over a glass plate)	Liquid (water trapped in the fibres of a Paper)
Mobile phase	Liquid (single solvent/mixture of solvents)	Liquid (single solvent/mixture of solvents)
Visualization of components of a mixture	Similar to TLC the coloured components are visible as coloured spots and the colourless components are observed under UV light or using a spraying agent.

3. Label the diagram and explain the process in your words.

Answer:
When filtration is carried out using a vacuum pump it is called filtration under suction. It is a faster and more efficient technique than simple filtration. The diagram is as follows:

ii. Procedure:

• The assembly for filtration under suction consists of a thick wall conical flask with a sidearm (Buchner flask).
• The flask is connected to a safety bottle by rubber tube through the side arm.
• Buchner funnel (a special porcelain funnel with a porous circular bottom) is fitted on the conical flask with the help of a rubber cork.
• A circular filter paper of correct size is placed on the circular porous bottom of the Buchner funnel and the funnel is placed on the flask.
• Filter paper is moistened with a few drops of water or solvent.
• Suction is created by starting the pump and filtration is carried out.

iii. Crystals are collected on the filter paper and filtrate in the flask.