Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.6 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

Question 1.
Solve the following equations by the method of inversion:
(i) x + 2y = 2, 2x + 3y = 3
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (i)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (i).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (i).2
By equality of matrices,
x = 0, y = 1 is the required solution.

(ii) 2x + y = 5, 3x + 5y = -3
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (ii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (ii).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (ii).2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (ii).3

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

(iii) 2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii).2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii).3
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii).4
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii).5
By equality of matrices,
x = 1, y = 2, z = 1 is the required solution.

(iv) x + y + z = 1, x – y + z = 2 and x + y – z = 3
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iv)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iv).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iv).2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iv).3

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

Question 2.
Express the following equations in matrix form and solve them by method of reduction:
(i) x + 3y = 2, 3x + 5y = 4.
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (i)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (i).1
Hence, x = \(\frac{1}{2}\), y = \(\frac{1}{2}\) is the required solution.

(ii) 3x – y = 1, 4x + y = 6.
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (ii)
By equality of matrices,
12x – 4y = 4 …..(1)
7y = 14 …..(2)
From (2), y = 2
Substituting y = 2 in (1), we get
12x – 8 = 4
∴ 12x = 12
∴ x = 1
Hence, x = 1, y = 2 is the required solution.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

(iii) x + 2y + z = 8, 2x + 3y – z = 11 and 3x – y – 2z = 5.
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (iii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (iii).1
By equality of matrices,
x + 2y + z = 8 ……(1)
-y – 3z = -5 …….(2)
16z = 16 ……….(3)
From (3), z = 1
Substituting z = 1 in (2), we get
-y – 3= -5
∴ y = 2
Substituting y = 2, z = 1 in (1), we get
x + 4 + 1 = 8
∴ x = 3
Hence, x = 3, y = 2, z = 1 is the required solution.

(iv) x + y + z = 1, 2x + 3y + 2z = 2 and x + y + 2z = 4.
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (iv)
By equality of matrices,
x + y + z = 1 ……(1)
y = 0
z = 3
Substituting y = 0, z = 3 in (1), we get
x + 0 + 3 = 1
∴ x = -2
Hence, x = -2, y = 0, z = 3 is the required solution.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

Question 3.
The total cost of 3 T.V. and 2 V.C.R. is ₹ 35000. The shopkeeper wants a profit of ₹ 1000 per T.V. and ₹ 500 per V.C.R. He sells 2 T.V. and 1 V.C.R. and he gets total revenue of ₹ 21500. Find the cost and selling price of T.V. and V.C.R.
Solution:
Let the cost of each T.V. be ₹ x and each V.C.R. be ₹ y.
Then the total cost of 3 T.V. and 2 V.C.R. is ₹ (3x + 2y) which is given to be ₹ 35000.
∴ 3x + 2y = 35000
The shopkeeper wants a profit of ₹ 1000 per T.V. and ₹ 500 per V.C.R.
The selling price of each T.V. is ₹ (x + 1000) and of each V.C.R. is ₹ (y + 500).
∴ selling price of 2 T.V. and 1 V.C.R is
₹ [2(x + 1000) + (y + 500)] which is given to be ₹ 21500.
∴ 2(x + 1000) + (y + 500) = 21500
∴ 2x + 2000 + y + 500 = 21500
∴ 2x + y = 19000
Hence, the system of linear equations is
3x + 2y = 35000
2x + y = 19000
The equations can be written in matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q3
By equality of matrices,
-x = -3000 …….(1)
2x + y = 19000 ……….(2)
From (1), x = 3000
Substituting x = 3000 in (2), we get
2(3000) + y = 19000
∴ y = 19000 – 6000 = 13000
Hence, the cost price of one T.V. is ₹ 3000 and of one V.C.R. is ₹ 13000 and the selling price of one T.V. is ₹ 4000 and of one V.C.R. is ₹ 13500.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

Question 4.
The sum of the cost of one Economics book, one Cooperation book, and one Account book is ₹ 420. The total cost of an Economic book, 2 Cooperation books, and an Account book is ₹ 480. Also, the total cost of an Economic book, 3 Cooperation books, and 2 Account books is ₹ 600. Find the cost of each book.
Solution:
Let the cost of 1 Economic book, 1 Cooperation book and 1 Account book be ₹ x, ₹ y and ₹ z respectively.
Then, from the given information
x + y + z = 420
x + 2y + z = 480
x + 3y + 2z = 600
These equations can be written in matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q4
By equality of matrices,
x + y + z = 420 …….(1)
y = 60
2y + z = 180 ………(2)
Substituting y = 60 in (2), we get
2(60) + z = 180
∴ z = 180 – 120 = 60
Substituting y = 60, z = 60 in (1), we get
x + 60 + 60 = 420
∴ x = 420 – 120 = 300
Hence, the cost of each Economic book is ₹ 300, each Cooperation book is ₹ 60 and each Account book is ₹ 60.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.5 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5

Question 1.
Apply the given elementary transformation on each of the following matrices:
(i) \(\left[\begin{array}{cc}
3 & -4 \\
2 & 2
\end{array}\right]\), R1 ↔ R2
(ii) \(\left[\begin{array}{cc}
2 & 4 \\
1 & -5
\end{array}\right]\), C1 ↔ C2
(iii) \(\left[\begin{array}{ccc}
3 & 1 & -1 \\
1 & 3 & 1 \\
-1 & 1 & 3
\end{array}\right]\) 3R2 and C2 → C2 – 4C1
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q1

Question 2.
Transform \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\) into an upper triangularmatrix by suitable row transformations.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q2.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5

Question 3.
Find the cofactor matrix of the following matrices:
(i) \(\left[\begin{array}{cc}
1 & 2 \\
5 & -8
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
5 & 8 & 7 \\
-1 & -2 & 1 \\
-2 & 1 & 1
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q3
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q3.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q3.2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q3.3

Question 4.
Find the adjoint of the following matrices:
(i) \(\left[\begin{array}{cc}
2 & -3 \\
3 & 5
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q4
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q4.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q4.2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q4.3

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5

Question 5.
Find the inverses of the following matrices by the adjoint mathod:
(i) \(\left[\begin{array}{rr}
3 & -1 \\
2 & -1
\end{array}\right]\)
(ii) \(\left[\begin{array}{cc}
2 & -2 \\
4 & 5
\end{array}\right]\)
(iii) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q5
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q5.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q5.2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q5.3
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q5.4
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q5.5

Question 6.
Find the inverses of the following matrices by the transformation method:
(i) \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q6
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q6.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q6.2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q6.3
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q6.4
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q6.5

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5

Question 7.
Find the inverse of A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) by elementary column transformations.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q7
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q7.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q7.2

Question 8.
Find the inverse of \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\) by the elementary row transformations.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q8
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q8.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q8.2

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5

Question 9.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\), then find matrix X such that XA = B.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q9
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q9.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q9.2

Question 10.
Find matrix X, if AX = B, where A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q10
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.5 Q10.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4

Question 1.
Find AT, if
(i) A = \(\left[\begin{array}{cc}
1 & 3 \\
-4 & 5
\end{array}\right]\)
(ii) A = \(\left[\begin{array}{ccc}
2 & -6 & 1 \\
-4 & 0 & 5
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q1

Question 2.
If A = [aij]3×3 where aij = 2(i – j). Find A and AT. State whether A and AT both are symmetric or skew-symmetric matrices.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q2.1
Hence, A and AT are both skew-symmetric matrices.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4

Question 3.
If A = \(\left[\begin{array}{cc}
5 & -3 \\
4 & -3 \\
-2 & 1
\end{array}\right]\), prove that (AT)T = A.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q3

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\), prove that AT = A.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q4

Question 5.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 1 \\
4 & -1 \\
-3 & 3
\end{array}\right]\), C = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 4 \\
-2 & 3
\end{array}\right]\), then show that
(i) (A + B)T = AT + BT
(ii) (A – C)T = AT – CT
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q5
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q5.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4

Question 6.
If A = \(\left[\begin{array}{cc}
5 & 4 \\
-2 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-1 & 3 \\
4 & -1
\end{array}\right]\), then find CT, such that 3A – 2B + C = I, where I is the unit matrix of order 2.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q6

Question 7.
If A = \(\left[\begin{array}{ccc}
7 & 3 & 0 \\
0 & 4 & -2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
0 & -2 & 3 \\
2 & 1 & -4
\end{array}\right]\), then find
(i) AT + 4BT
(ii) 5AT – 5BT
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q7
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q7.1

Question 8.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
3 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{lll}
2 & 1 & -4 \\
3 & 5 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
0 & 2 & 3 \\
-1 & -1 & 0
\end{array}\right]\), verify that (A + 2B + 3C)T = AT + 2BT + 3CT
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q8
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q8.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4

Question 9.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
-3 & 2 & -3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 & 1 \\
-3 & 2 \\
-1 & 3
\end{array}\right]\), prove that (A + BT)T = AT + B.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q9
From (1) and (2),
(A + BT)T = AT + B.

Question 10.
Prove that A + AT is symmetric and A – AT is a skew-symmetric matrix, where
(i) A = \(\left[\begin{array}{ccc}
1 & 2 & 4 \\
3 & 2 & 1 \\
-2 & -3 & 2
\end{array}\right]\)
(ii) A = \(\left[\begin{array}{ccc}
5 & 2 & -4 \\
3 & -7 & 2 \\
4 & -5 & -3
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q10
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q10.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q10.2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q10.3

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4

Question 11.
Express each of the following matrix as the sum of a symmetric and a skew-symmetric matrix:
(i) \(\left[\begin{array}{ll}
4 & -2 \\
3 & -5
\end{array}\right]\)
(ii) \(\left[\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q11
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q11.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q11.2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q11.3

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4

Question 12.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
3 & -2 \\
4 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
0 & 3 & -4 \\
2 & -1 & 1
\end{array}\right]\), verify that
(i) (AB)T = BTAT
(ii) (BA)T = ATBT
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q12
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q12.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.4 Q12.2

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3

Question 1.
Evaluate:
(i) \(\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
2 & -4 & 3
\end{array}\right]\)
Solution:
\(\left[\begin{array}{l}
3 \\
2 \\
1
\end{array}\right]\left[\begin{array}{lll}
2 & -4 & 3
\end{array}\right]\) = \(\left[\begin{array}{rrr}
6 & -12 & 9 \\
4 & -8 & 6 \\
2 & -4 & 3
\end{array}\right]\)

(ii) \(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\)
Solution:
\(\left[\begin{array}{lll}
2 & -1 & 3
\end{array}\right]\left[\begin{array}{l}
4 \\
3 \\
1
\end{array}\right]\) = [8 – 3 + 3] = [8]

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3

Question 2.
If A = \(\left[\begin{array}{ccc}
-1 & 1 & 1 \\
2 & 3 & 0 \\
1 & -3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{lll}
2 & 1 & 4 \\
3 & 0 & 2 \\
1 & 2 & 1
\end{array}\right]\). State whether AB = BA? Justify your answer.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q2
From (1) and (2), AB ≠ BA.

Question 3.
Show that AB = BA, where A = \(\left[\begin{array}{lll}
-2 & 3 & -1 \\
-1 & 2 & -1 \\
-6 & 9 & -4
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
1 & 3 & -1 \\
2 & 2 & -1 \\
3 & 0 & -1
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q3
From (1) and (2), AB = BA.

Question 4.
Verify A(BC) = (AB)C, if A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
2 & 3 & 0 \\
0 & 4 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & -2 \\
-1 & 1 \\
0 & 3
\end{array}\right]\), and C = \(\left[\begin{array}{rrr}
3 & 2 & -1 \\
2 & 0 & -2
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q4
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q4.1
From (1) and (2), A(BC) = (AB)C.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3

Question 5.
Verify that A(B + C) = AB + AC, if A = \(\left[\begin{array}{cc}
4 & -2 \\
2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 1 \\
3 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 1 \\
2 & -1
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q5
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q5.1
From (1) and (2), A(B + C) = AB + AC.

Question 6.
If A = \(\left[\begin{array}{ccc}
4 & 3 & 2 \\
-1 & 2 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 0 \\
1 & -2
\end{array}\right]\), show that matrix AB is non-singular.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q6
Hence, AB is a non-singular matrix.

Question 7.
If A + I = \(\left[\begin{array}{ccc}
1 & 2 & 0 \\
5 & 4 & 2 \\
0 & 7 & -3
\end{array}\right]\), find the product (A + I)(A – I).
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q7

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3

Question 8.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\), show that A2 – 4A is a scalar matrix.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q8
which is a scalar matrix.

Question 9.
If A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 7
\end{array}\right]\), find k so that A2 – 8A – kI = O, where I is a 2 × 2 unit matrix and O is null matrix of order 2.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q9
By equality of matrices,
-k – 7 = 0
∴ k = -7.

Question 10.
If A = \(\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\), prove that A2 – 5A + 7I = 0, where I is a 2 × 2 unit matrix.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q10

Question 11.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & -2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & a \\
-1 & b
\end{array}\right]\) and if(A + B)2 = A2 + B2, find values of a and b.
Solution:
(A + B)2 = A2 + B2
∴ (A + B)(A + B) = A2 + B2
∴ A2 + AB + BA + B2 = A2 + B2
∴ AB + BA = 0
∴ AB = -BA
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q11
By the equality of matrices, we get
0 = a – 2 ……..(1)
0 = 1 + b ……..(2)
a + 2b = 2a – 4 ……..(3)
-a – 2b = 2 + 2b ……..(4)
From equations (1) and (2), we get
a = 2 and b = -1
The values of a and b satisfy equations (3) and (4) also.
Hence, a = 2 and b = -1.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3

Question 12.
Find k, if A = \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\) and A2 = kA – 2I.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q12
By equality of matrices,
1 = 3k – 2 ……..(1)
-2 = -2k ……..(2)
4 = 4k ……..(3)
-4 = -2k – 2 ……..(4)
From (2), k = 1.
k = 1 also satisfies equation (1), (3) and (4).
Hence, k = 1.

Question 13.
Find x and y, if \(\left\{4\left[\begin{array}{ccc}
2 & -1 & 3 \\
1 & 0 & 2
\end{array}\right]-\left[\begin{array}{ccc}
3 & -3 & 4 \\
2 & 1 & 1
\end{array}\right]\right\}\left[\begin{array}{c}
2 \\
-1 \\
1
\end{array}\right]=\left[\begin{array}{c}
x \\
y
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q13
By equality of matrices,
x = 19 and y = 12.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3

Question 14.
Find x, y, z, if \(\left\{3\left[\begin{array}{ll}
2 & 0 \\
0 & 2 \\
2 & 2
\end{array}\right]-4\left[\begin{array}{cc}
1 & 1 \\
-1 & 2 \\
3 & 1
\end{array}\right]\right\}\left[\begin{array}{l}
1 \\
2
\end{array}\right]=\left[\begin{array}{c}
x-3 \\
y-1 \\
2 z
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q14
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q14.1
By equality of matrices,
-6 = x – 3, 0 = y – 1 and -2 = 2z
∴ x = -3, y = 1 and z = -1.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3

Question 15.
Jay and Ram are two friends. Jay wants to buy 4 pens and 8 notebooks. Ram wants to buy 5 pens and 12 notebooks. The price of one pen and one notebook was ₹ 6 and ₹ 10 respectively. Using matrix multiplication, find the amount each one of them requires for buying the pens and notebooks.
Solution:
The given data can be written in matrix form as:
Number of Pens and Notebooks
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q15
For finding the amount each one of them requires to buy the pens and notebook, we require the multiplication of the two matrices A and B.
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.3 Q15.1
Hence, Jay requires ₹ 104 and Ram requires ₹ 150 to buy the pens and notebooks.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2

Question 1.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 2 \\
2 & 2 \\
0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 3 \\
-1 & 4 \\
-2 & 1
\end{array}\right]\) show that
(i) A + B = B + A
(ii) (A + B) + C = A + (B + C)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q1 (i)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q1 (ii)
From (1) and (2), we get
(A + B) + C = A + (B + C).

Question 2.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & -3 \\
4 & -7
\end{array}\right]\), then find the matrix A – 2B + 6I, where I is the unit matrix of order 2.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2

Question 3.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
-3 & 7 & -8 \\
0 & -6 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
9 & -1 & 2 \\
-4 & 2 & 5 \\
4 & 0 & -3
\end{array}\right]\), then find the matrix C such that A + B + C is a zero matrix.
Solution:
A + B + C = 0
∴ C = -A – B
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q3

Question 4.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
3 & -5 \\
-6 & 0
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & -2 \\
4 & 2 \\
1 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
2 & 4 \\
-1 & -4 \\
-3 & 6
\end{array}\right]\), find the matrix X such that 3A – 4B + 5X = C.
Solution:
3A – 4B + 5X = C
∴ 5X = C – 3A + 4B
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q4

Question 5.
If A = \(\left[\begin{array}{ccc}
5 & 1 & -4 \\
3 & 2 & 0
\end{array}\right]\), find (AT)T.
Solution:
A = \(\left[\begin{array}{ccc}
5 & 1 & -4 \\
3 & 2 & 0
\end{array}\right]\)
∴ AT = \(\left[\begin{array}{rr}
5 & 3 \\
1 & 2 \\
-4 & 0
\end{array}\right]\)
∴ (AT)T = \(\left[\begin{array}{ccc}
5 & 1 & -4 \\
3 & 2 & 0
\end{array}\right]\) = A

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2

Question 6.
If A = \(\left[\begin{array}{ccc}
7 & 3 & 1 \\
-2 & -4 & 1 \\
5 & 9 & 1
\end{array}\right]\), find (AT)T.
Solution:
A = \(\left[\begin{array}{ccc}
7 & 3 & 1 \\
-2 & -4 & 1 \\
5 & 9 & 1
\end{array}\right]\)
∴ AT = \(\left[\begin{array}{rrr}
7 & -2 & 5 \\
3 & -4 & 9 \\
1 & 1 & 1
\end{array}\right]\)
∴ (AT)T = \(\left[\begin{array}{ccc}
7 & 3 & 1 \\
-2 & -4 & 1 \\
5 & 9 & 1
\end{array}\right]\) = A

Question 7.
Find a, b, c if \(\left[\begin{array}{ccc}
1 & \frac{3}{5} & a \\
b & -5 & -7 \\
-4 & c & 0
\end{array}\right]\) is a symetric matrix.
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & \frac{3}{5} & a \\
b & -5 & -7 \\
-4 & c & 0
\end{array}\right]\)
Since, A is a symmetric matrix, aij = aji for all i and j
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q7

Question 8.
Find x, y, z if \(\left[\begin{array}{ccc}
0 & -5 i & x \\
y & 0 & z \\
\frac{3}{2} & -\sqrt{2} & 0
\end{array}\right]\) is a skew symmetric matrix.
Solution:
Let A = \(\left[\begin{array}{ccc}
0 & -5 i & x \\
y & 0 & z \\
\frac{3}{2} & -\sqrt{2} & 0
\end{array}\right]\)
Since, A is skew-symmetric matrix,
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q8
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q8.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2

Question 9.
For each of the following matrices, find its transpose and state whether it is symmetric, skew-symmetric or neither:
(i) \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
Then AT = \(\left[\begin{array}{rrr}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
Since, A = AT, A is a symmetric matrix.

(ii) \(\left[\begin{array}{ccc}
2 & 5 & 1 \\
-5 & 4 & 6 \\
-1 & -6 & 3
\end{array}\right]\)
Solution:
Let B = \(\left[\begin{array}{ccc}
2 & 5 & 1 \\
-5 & 4 & 6 \\
-1 & -6 & 3
\end{array}\right]\)
Then BT = \(\left(\begin{array}{rrr}
2 & -5 & -1 \\
5 & 4 & -6 \\
1 & 6 & 3
\end{array}\right)\)
∴ B ≠ BT
Also,
-BT = \(\left(\begin{array}{rrr}
2 & -5 & -1 \\
5 & 4 & -6 \\
1 & 6 & 3
\end{array}\right)=\left(\begin{array}{rrr}
-2 & 5 & 1 \\
-5 & -4 & 6 \\
-1 & -6 & -3
\end{array}\right)\)
∴ B ≠ -BT
Hence, B is neither symmetric nor skew-symmetric matrix.

(iii) \(\left[\begin{array}{ccc}
0 & 1+2 i & i-2 \\
-1-2 i & 0 & -7 \\
2-i & 7 & 0
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q9
Hence, C is a skew-symmetric matrix.

Question 10.
Construct the matrix A = [aij]3×3, where aij = i – j. State whether A is symmetric or skew-symmetric.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q10

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2

Question 11.
Solve the following equations for X and Y, if 3X – Y = \(\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\) and X – 3Y = \(\left[\begin{array}{ll}
0 & -1 \\
0 & -1
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q11
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q11.1

Question 12.
Find matrices A and B, if 2A – B = \(\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) and A – 2B = \(\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q12
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q12.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q12.2

Question 13.
Find x and y, if \(\left[\begin{array}{ccc}
2 x+y & -1 & 1 \\
3 & 4 y & 4
\end{array}\right]+\left[\begin{array}{ccc}
-1 & 6 & 4 \\
3 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q13
By equality of matrices, we get
2x + y – 1 = 3 ……..(1)
and 4y = 18 ……….(2)
From (2), y = \(\frac{9}{2}\)
Substituting y = \(\frac{9}{2}\) in (1), we get
2x + \(\frac{9}{2}\) – 1 = 3
∴ 2x = 3 – \(\frac{7}{2}\) = \(\frac{-1}{2}\)
∴ x = \(\frac{-1}{4}\)
Hence, x = \(\frac{-1}{4}\) and y = \(\frac{9}{2}\).

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2

Question 14.
If \(\left[\begin{array}{cc}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\), find a, b, c and d.
Solution:
\(\left[\begin{array}{cc}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\)
By equality of matrices,
2a + b = 2 ….. (1)
3a – b = 3 …… (2)
c + 2d = 4 …… (3)
2c – d = -1 …… (4)
Adding (1) and (2), we get
5a = 5
∴ a = 1
Substituting a = 1 in (1), we get
2(1) + b = 2
∴ b = 0
Multiplying equation (4) by 2, we get
4c – 2d = -2 …… (5)
Adding (3) and (5), we get
5c = 2
∴ c = \(\frac{2}{5}\)
Substituting c = \(\frac{2}{5}\) in (4), we get
2(\(\frac{2}{5}\)) – d = -1
∴ d = \(\frac{4}{5}\) + 1 = \(\frac{9}{5}\)
Hence, a = 1, b = 0, c = \(\frac{2}{5}\) and d = \(\frac{9}{5}\).

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2

Question 15.
There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects – Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B:
July sales (in Rupees), Physics, Chemistry, Mathematics
A = \(\left[\begin{array}{lll}
5600 & 6750 & 8500 \\
6650 & 7055 & 8905
\end{array}\right]\) First Row Suresh / Second Row Ganesh
August Sales (in Rupees), Physics, Chemistry, Mathematics
B = \(\left[\begin{array}{ccc}
6650 & 7055 & 8905 \\
7000 & 7500 & 10200
\end{array}\right]\) First Row Suresh / Second Row Ganesh
(i) Find the increase in sales in Z from July to August 2017.
(ii) If both book shops get 10% profit in the month of August 2017,
find the profit for each bookseller in each subject in that month.
Solution:
The sales for July and August 2017 for Suresh and Ganesh are given by the matrices A and B as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q15
(i) The increase in sales (in ₹) from July to August 2017 is obtained by subtracting the matrix A from B.
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q15.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q15.2
Hence, the increase in sales (in ₹) from July to August 2017 for:
Suresh book shop: ₹ 1050 in Physics, ₹ 305 in Chemistry, and ₹ 405 in Mathematics.
Ganesh book shop: ₹ 350 in Physics, ₹ 445 in Chemistry, and ₹ 1295 in Mathematics.
(ii) Both the book shops get 10% profit in August 2017,
the profit for each bookseller in each subject in August 2017 is obtained by the scalar multiplication of matrix B by 10%,
i.e. \(\frac{10}{100}=\frac{1}{10}\)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.2 Q15.3
Hence, the profit for Suresh book shop are ₹ 665 in Physics, ₹ 705.50 in Chemistry and ₹ 890.50 in Mathematics and for Ganesh book shop are ₹ 700 in Physics, ₹ 750 in Chemistry and ₹ 1020 in Mathematics.

Maharashtra Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.3

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.3

Question 1.
Write the negation of each of the following statements:
(i) All men are animals.
Solution:
Some men are not animals.

(ii) 3 is a natural number.
Solution:
-3 is not a natural number.

Maharashtra Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.3

(iii) It is false that Nagpur is the capital of Maharashtra.
Solution:
Nagpur is the capital of Maharashtra.

(iv) 2 + 3 ≠ 5.
Solution:
2 + 3 = 5.

Question 2.
Write the truth value of the negation of each of the following statements:
(i) √5 is an irrational number.
Solution:
Let p : √5 is an irrational number.
The truth value of p is T.
Therefore, the truth value of ~p is F.

(ii) London is in England.
Solution:
Let p : London is in England.
The truth value of p is T.
Therefore, the truth value of ~p is F.

Maharashtra Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.3

(iii) For every x ∈ N, x + 3 < 8.
Solution:
Let p : For every x ∈ N, x + 3 < 8.
The truth value of p is F.
Therefore, the truth value of ~p is T.

Maharashtra Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.2

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.2

Question 1.
Express the following statements in symbolic form:
(i) e is a vowel or 2 + 3 = 5.
Solution:
Let p : e is a vowel.
q: 2 + 3 = 5.
Then the symbolic form of the given statement is p ∨ q.

(ii) Mango is a fruit but potato is a vegetable.
Solution:
Let p : Mango is a fruit.
q : Potato is a vegetable.
Then the symbolic form of the given statement is p ∧ q.

Maharashtra Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.2

(iii) Milk is white or grass is green.
Solution:
Let p : Milk is white.
q : Grass is green.
Then the symbolic form of the given statement is p ∨ q.

(iv) I like playing but not singing.
Solution:
Let p : I like playing.
q : I am not singing.
Then the symbolic form of the given statement is p ∧ q.

(v) Even though it is cloudy, it is still raining.
Solution:
The given statement is equivalent to:
It is cloudy and it is still raining.
Let p : It is cloudy.
q : It is still raining.
Then the symbolic form of the given statement is p ∧ q.

Question 2.
Write the truth values of the following statements:
(I) Earth is a planet and Moon is a star.
Solution:
Let p : Earth is a planet.
q : Moon is a star.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. …[T ∧ F ≡ F]

(ii) 16 is an even number and 8 is a perfect square.
Solution:
Let p : 16 is an even number.
q : 8 is a perfect square.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. ….[T ∧ F ≡ F]

Maharashtra Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.2

(iii) A quadratic equation has two distinct roots or 6 has three prime factors.
Solution:
Let p : A quadratic equation has two distinct roots.
q : 6 has three prime factors.
Then the symbolic form of the given statement is p ∨ q.
The truth values of both p and q are F.
∴ the truth value of p ∨ q is F. …..[F ∨ F ≡ F]

(iv) The Himalayas are the highest mountains but they are part of India in the northeast.
Solution:
Let p : the Himalayas are the highest mountains.
q : They are part of India in the northeast.
Then the symbolic form of the given statement is p ∧ q.
The truth values of both p and q are T.
∴ the truth value of p ∧ q is T. …..[T ∧ T ≡ T]

Maharashtra Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

State which of the following sentences are statements. Justify your answer. In case of statements, write down the truth value:

Question (i).
A triangle has ‘ n’ sides.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (ii).
The sum of interior angles of a triangle is 180°.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (iii).
You are amazing!
Solution:
It is an exclamatory sentence, hence it is not a statement.

Question (iv).
Please grant me a loan.
Solution:
It is an imperative sentence, hence it is not a statement.

Maharashtra Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question (v).
√-4 is an irrational number.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (vi).
x2 – 6x + 8 = 0 implies x = -4 or x = -2.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (vii).
He is an actor.
Solution:
It is an open sentence, hence it is not a statement.

Question (viii).
Did you eat lunch yet?
Solution:
It is an interrogative sentence, hence it is not a statement.

Question (ix).
Have a cup of cappuccino.
Solution:
It is an imperative sentence, hence it is not a statement.

Maharashtra Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question (x).
(x + y)2 = x2 + 2xy + y2 for all x, y ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is ‘T’.

Question (xi).
Every real number is a complex number.
Solution:
It is a statement that is true, hence its truth value is ‘T.

Question (xii).
1 is a prime number.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (xiii).
With the sunset, the day ends.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xiv).
1! = 0.
Solution:
It is a statement that is false, hence its truth value is

Question (xv).
3 + 5 > 11.
Solution:
It is a statement that is false, hence its truth value is ‘F’.

Question (xvi).
The number π is an irrational number.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Maharashtra Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question (xvii).
x2 – y2 = (x + y)(x – y) for all x, y ∈ R.
Solution:
It is a mathematical identity that is true, hence its truth value is ‘T’.

Question (xviii).
The number 2 is only even a prime number.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xix).
Two coplanar lines are either parallel or intersecting.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xx).
The number of arrangements of 7 girls in a row for a photograph is 7!
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xxi).
Give me a compass box.
Solution:
It is an imperative sentence, hence it is not a statement.

Question (xxii).
Bring the motor car here.
Solution:
It is an imperative sentence, hence it is not a statement.

Maharashtra Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question (xxiii).
It may rain today.
Solution:
It is an open sentence, hence it is not a statement.

Question (xxiv).
If a + b < 7, where a ≥ 0 and b ≥ 0, then a < 7 and b < 7.
Solution:
It is a statement that is true, hence its truth value is ‘T’.

Question (xxv).
Can you speak English?
Solution:
It is an interrogative sentence, hence it is not a statement.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 15 Hydrocarbons Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 15 Hydrocarbons

Question 1.
What are unsaturated and saturated hydrocarbons?
Answer:
Hydrocarbons which contain carbon-carbon multiple bond (C=C or C≡C) are called unsaturated hydrocarbons, whereas those which contain carbon-carbon single bond (C-C) are called saturated hydrocarbons.

Question 2.
How are hydrocarbons classified?
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 1

Question 3.
Define alkanes. Write general formula of alkanes.
Answer:

  1. Alkanes are aliphatic saturated hydrocarbons containing carbon-carbon and carbon-hydrogen single
    covalent bonds.
  2. They have a general formula CnH2n+2 where, ‘n’ stands for number of carbon atoms in the alkane molecule.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 4.
Give information about isomerism in alkanes. Write the all possible structural isomers of a saturated hydrocarbon containing four carbon atoms along with their IUPAC names.
Answer:
i. Alkanes with more than three carbon atoms generally exhibit, structural isomerism and in particular, the chain isomerism.
ii. The number of possible structural isomers increase rapidly with the number of carbon atoms.
iii. Structural isomers of a saturated hydrocarbon containing four carbon atoms along with their IUPAC names.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 2

Question 5.
Write all the possible structural isomers of a saturated hydrocarbon having molecular formula C5H12.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 3

Question 6.
Define sigma bond.
Answer:
A single covalent bond formed by the coaxial overlap of orbitals is called sigma (σ) bond.

Question 7.
i. Why do C – C bonds in alkanes undergo rotation?
ii. What are conformations?
Answer:
i. a. Alkanes have single covalent bonds (sigma bonds) formed by the coaxial overlap of orbitals.
b. As a direct consequence of coaxial overlap of orbitals, a sigma bond is cylindrically symmetrical and the extent of orbital overlap is unaffected by rotation about the single bond and therefore, C – C bonds undergo rotation.

ii. a. In alkanes, the atoms bonded to one carbon of a C – C single bond change their relative position with reference to the atoms on the other carbon of that bond on rotation of that C – C single bond.
b. The resulting arrangements of the atoms in space about the C – C single bond are called conformations or conformational isomers. Innumerable conformations result on complete rotation of a C – C single bond through 360°.

Question 8.
i. What is conformational isomerism?
ii. Name the two extreme conformations shown by ethane molecule.
Answer:
i. The phenomenon of existence of conformation is a type of stereoisomerism and is known as conformational isomerism.
ii. Ethane molecule shows the following two extreme conformations:

  • Staggered conformation
  • Eclipsed conformation

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 9.
Draw structures representing staggered and eclipsed conformations of ethane using:
i. Sawhorse projection
ii. Newman projection
Answer:
i. Sawhorse projection of ethane:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 4

ii. Newman projection of ethane:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 5

Question 10.
How are alkanes obtained from crude oil?
Answer:
Alkanes are obtained by fractional distillation of crude oil in oil refineries.

Question 11.
How are alkanes obtained from alkenes and alkynes?
OR
How are alkanes obtained from catalytic hydrogenation of alkenes and alkynes?
Answer:
i. Catalytic hydrogenation of alkenes or alkynes with dihydrogen gas gives corresponding alkanes.
ii. Finely divided powder of platinum (Pt) or palladium (Pd) catalyse the hydrogenation of alkenes and alkynes at room temperature.
iii. Relatively high temperature and pressure are required with finely divided nickel as the catalyst.
e.g. a. Propene to propane:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 6
b. Ethyne to ethane:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 7

Question 12.
Write the general reactions for the catalytic hydrogenation of alkenes and alkynes.
Answer:
General reaction for catalytic hydrogenation of alkenes:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 8
General reaction for catalytic hydrogenation of alkynes:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 9

Question 13.
Write the structures of alkenes that on catalytic hydrogenation give n-butane.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 10

Question 14.
Explain the preparation of alkanes by reduction of alkyl halides with the help of an example.
Answer:
i. Alkanes can be prepared by reduction of alkyl halides using zinc and dilute hydrochloric acid.
ii. The reduction of alkyl halides is due to nascent hydrogen obtained from the reaction between reducing agent Zn and dilute HCl.
e.g. Reduction of methyl iodide to methane.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 11

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 15.
How are alkanes obtained by Wurtz reaction?
Answer:
Alkyl halides on treatment with reactive sodium metal in dry ether, gives higher alkanes having double the number of carbon atoms. This is called as Wurtz coupling reaction.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 12

Question 16.
How will you convert ethyl chloride into n-butane?
Answer:
Ethyl chloride on heating with sodium metal in presence of dry ether gives n-butane.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 13

Question 17.
Write chemical equations for reactions that take place on treating ethereal solutions of:
i. Methyl iodide with sodium metal
ii. Ethyl iodide with sodium metal
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 14

Question 18.
Explain the preparation of Grignard reagents.
OR
What is Grignard reagent? Explain its preparation.
Answer:
Grignard reagent are alkyl magnesium halides obtained by treating alkyl halides with dry magnesium metal in the presence of dry ether.

General Reaction:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 15

Question 19.
State the action of water on methyl magnesium bromide in dry ether with the help of a chemical reaction.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 16

Question 20.
Write the reagents involved in the following conversions.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 17
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 18

Question 21.
Straight chain alkanes have higher melting and boiling points as compared to branched isomeric alkanes. Give reason.
Answer:
i. The electronegativity of carbon and hydrogen is nearly the same. Therefore, C-H and C-C bonds are nonpolar covalent bonds and hence, alkanes are nonpolar.
ii. Alkane molecules are held together by weak intermolecular van der Waals forces.
iii. Larger the surface area of molecules, stronger are such intermolecular van der Waals forces.
iv. In straight chain alkane molecules, surface area is relatively larger as compared to branched chain alkanes and as a result, the intermolecular forces are relatively stronger in straight chain alkanes than in branched chain alkanes.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 19
Hence, straight chain alkanes have higher melting and boiling points as compared to branched alkanes.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 22.
State physical properties of alkanes.
Answer:

  • Alkanes are colourless and odourless.
  • At room temperature, the first four alkanes are gases, alkanes having 5 to 17 carbon atoms are liquids while the rest all are solids.
  • Alkanes are readily soluble in organic solvents such as chloroform, ether or ethanol while they are insoluble in water.
  • Alkanes have low melting and boiling points which increases with an increase in the number of carbon atoms for straight chain molecules. But for branched chain molecules, more the number of branches, lower is the boiling/melting point.

Note: [Melting and boiling points of alkanes]
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 20

Question 23.
Define substitution reactions.
Answer:
The reactions in which an atom or group of atoms in a molecule is replaced by another atom or group of atoms is called as substitution reactions.
e.g. Halogenation of alkanes.

Question 24.
i. What is halogenation of alkanes?
ii. Write the order of reactivity of halogens towards alkanes.
Answer:
i. Substitution of H atoms of alkanes by X (halogen, X = Cl, Br, I and F) atom is called halogenation of alkanes.
ii. The reactivity of halogens toward alkanes follows the order: F2 > Cl2 > Br2 > I2
[Note: The ease of replacement of hydrogen atoms from the carbon in alkanes is in the order: 3 > 2 > 1.]

Question 25.
Explain reactions involved in chlorination of methane.
Answer:
Alkanes react with chlorine gas in presence of UV light or diffused sunlight or at a high temperature (573-773 K) to give a mixture of alkyl halides.

Chlorination of methane:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 21
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 22
Tetrachloromethane is a major product when excess of chlorine is used. Chloromethane is obtained as major product when excess of methane is employed.

Question 26.
Predict the products in the following set of reactions.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 23
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 24

Question 27.
What is the action of Cl2 and Br2 on 2-methylpropane?
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 25
[Note: In bromination, there is high degree of selectivity as to which hydrogen atoms are replaced.]

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 28.
Explain mechanism of halogenation of alkanes.
Answer:
i. Halogenation of alkanes follows the free radical mechanism.
ii. Homolysis of halogen molecule (X2) generates halogen atoms, i.e., halogen free radicals.
iii. The mechanism of the first step of chlorination of methane is shown below:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 26

Question 29.
Why are alkanes used as fuels?
Answer:
On heating in the presence of air or dioxygen, alkanes are completely oxidized to carbon dioxide and water with the evolution of a large amount of heat. Hence, alkanes are used as fuels.

Question 30.
What is combustion of alkanes? Write a general equation for alkane combustion.
Answer:
On heating in the presence of air or dioxygen, alkanes are completely oxidized to carbon dioxide and water with the evolution of a large amount of heat. This is known as combustion reaction of alkanes.

General representative equation for combustion is:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 27

Question 31.
Write chemical equations for combustion of butane and methane.
Answer:
i. Combustion of butane:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 28
ii. Combustion of methane:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 29

Question 32.
Write a short note on pyrolysis of alkanes.
Answer:
Alkanes on heating at higher temperature in absence of air decompose to lower alkanes, alkenes and hydrogen, etc. This is known as pyrolysis or cracking.
e.g. Pyrolysis of hexane
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 30

Question 33.
Explain aromatization reaction of alkanes. Give its one application.
Answer:
i. Straight chain alkanes containing 6 to 10 carbon atoms are converted to benzene and its homologues, on heating under 10 to 20 atm pressure at about 773 K in the presence of V2O5, Cr2O3, MO2O3, etc. supported over alumina.

ii. The reaction involves simultaneous dehydrogenation and cyclization. This reaction is known as aromatization or reforming.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 31
This process is used in refineries to produce high quality gasoline which is used in automobiles as fuel.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 34.
Collect the information on CNG and LPG with reference to the constituents and the advantages of CNG over LPG.
Answer:
Constituents of CNG (Compressed Natural Gas):
It mainly consists of methane compressed at a pressure of 200-248 bar.
Constituents of LPG (Liquefied Petroleum Gas):
It contains a mixture of propane and butane liquefied at 15°C and a pressure of 1.7 – 7.5 bar.

Advantages of CNG over LPG:

  • CNG is cheaper and cleaner than LPG.
  • CNG produces less pollutants than LPG.
  • It does not evolve gases containing sulphur and nitrogen.
  • Octane rating of CNG is high, hence thermal efficiency is more.
  • Vehicles powered by CNG produces less carbon monoxide and hydrocarbon emission.

[Note: Students are expected to collect additional information on their own.]

Question 35.
Write the uses of alkane.
Answer:
Uses of alkanes:

  • First four alkanes are used as a fuel mainly for heating and cooking purpose. For example, LPG and CNG.
  • CNG, petrol and diesel are used as fuel for automobiles.
  • Lower liquid alkanes are used as solvent.
  • Alkanes with more than 35 C atoms (tar) are used for road surfacing.
  • Waxes are high molecular weight alkanes. They are used as lubricants. They are also used for the preparation of candles and carbon black that is used in manufacture of printing ink, shoe polish, etc.

Question 36.
i. Write the general molecular formula of alkenes.
ii. Why are alkenes also known as olefins?
Answer:
i. Alkenes have general formula CnH2n, where, n = 2,3,4… etc.
ii. Alkenes are also known as olefins because the first member ethene/ethylene reacts with chlorine to form oily substance.
[Note: Alkenes with one carbon-carbon double bond, contain two hydrogen atoms less than the corresponding alkanes.]

Question 37.
Define alkadienes and alkatrienes. Give one example for each.
Answer:
i. The aliphatic unsaturated hydrocarbons containing two carbon-carbon double bonds are called as alkadienes.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 32

ii. The aliphatic unsaturated hydrocarbons containing three carbon-carbon double bonds are called as alkatrienes.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 33

Question 38.
Explain structural isomerism in alkenes by giving an example.
Answer:
Alkenes with more than three carbon atoms show structural isomerism.
e.g. Alkene with molecular formula C4H8 is butene. The structural formulae for C4H8 can be drawn in three different ways:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 34

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 39.
Draw structures of chain isomers of butene.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 35

Question 40.
Draw structures of position isomers of butene.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 36

Question 41.
Define geometrical isomerism.
Answer:
The isomerism which arises due to the difference in spatial arrangement of atoms or groups about doubly bonded carbon (C=C) atoms is called geometrical isomerism.

Question 42.
Explain geometrical isomerism using a general example.
Answer:
i. If the two atoms or groups bonded to each end of the C=C double bond are different, then the molecule can be represented by two different special arrangements of the groups as follows:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 37
ii. In structure (A), two identical atoms or groups lie on the same side of the double bond.
The geometrical isomer in which two identical or similar atoms or groups lie on the same side of the double bond is called cis-isomer.
iii. In structure (B), two identical atoms or groups lie on the opposite side of the double bond.
The geometrical isomer in which two identical or similar atoms or groups lie on the opposite side of the double bond is called trans-isomer.
iv. Due to different arrangement of atoms or groups in space, these isomers differ in their physical properties like melting point, boiling point, solubility, etc.

Question 43.
i. Define cis- and trans-isomer.
ii. Draw geometrical isomers of but-2-ene.
Answer:
i. a. Cis isomer: The geometrical isomer in which two identical or similar atoms or groups lie on the same side of the double bond is called a cis-isomer.
b. Trans isomer: The geometrical isomer in which two identical or similar atoms or groups lie on the opposite side of the double bond is called a trans-isomer.

ii. Geometrical or cis-trans isomers of but-2-ene are represented as:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 38

Question 44.
State whether the following alkenes can exhibit geometrical (or cis-trans) isomerism or not. Give reason for the answer.
i. CH3 – CH2 – CH2 – CH = CH2
ii. CH3 – CH2 – CH = C(CH3)2
Answer:
Both the alkenes (i) and (ii) cannot exhibit geometrical isomerism, since 1 alkene is a terminal alkene (containing two H-atoms on the same side of the double bond) while the 2nd alkene is a 1,1-disubstituted alkene (containing two identical alkyl groups on the same side of the double bond).

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 45.
Write the general formulae of alkenes which exhibit cis-trans isomerism.
Answer:
Alkenes having the following general formulae exhibit cis-trans isomerism:
RCH=CHR, R1R2C=CR1R3, R1CH=CR1R2, R1CH=CR2R3, R1CH=CHR2 and R1R2C=CR3R4

Question 46.
Draw structures of cis-trans isomers for the following:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 39
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 40
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 41

Question 47.
Which of the following compounds will show geometrical isomerism?
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 42
Answer:
Compounds (III), (IV) and (V) will show geometrical isomerism as they have each of the doubly bonded carbon atoms in their structures, attached to different atoms/groups of atoms.

Question 48.
Alkenes can be obtained from which industrial sources?
Answer:

  1. The most important alkenes for chemical industry are ethene, propene and buta-1,3-diene.
  2. Alkenes containing up to four carbon atoms can be obtained in pure form from the petroleum products.
  3. Ethene is produced from natural gas and crude oil by cracking.

Question 49.
What is β-elimination reaction? Explain in brief.
Answer:
The reactions in which two atoms or groups are eliminated from adjacent carbon atoms are called 1,2-elimination reactions. Since the atom/group is removed from β-carbon atom (β to the leaving group) it is called as β-elimination reaction.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 43
The hybridization of each C in the reactant is sp3 while that in the product is sp2. This means elimination reactions cause change in hybridization state while forming multiple bonds from single bond.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 50.
i. What is dehydrohalogenation reaction?
ii. How is it carried out. Explain with an example.
Answer:
i. a. The reactions in which there is removal of hydrogen (H) atom and halogen (X) atom from adjacent carbon atoms are known as dehydrohalogenation reactions.
b. The carbon carrying X is called α-carbon atom. The hydrogen atom from adjacent carbon called β-carbon atom, is removed and hence, the reaction is known as β-elimination.

ii. When an alkyl halide is boiled with a hot concentrated alcoholic solution of a strong base like KOH or NaOH, alkene is formed with removal of water molecule.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 44
[Note: The ease of dehydrohalogenation of alkyl halides is in the order 3 > 2 > 1.]

Question 51.
State Saytzeff rule.
Answer:
In dehydrohalogenation the preferred product is the alkene that has the greater number of alkyl groups attached to doubly bonded carbon atoms.

Question 52.
Write and explain dehydrohalogenation reaction of 2-chlorobutane.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 45
In dehydrohalogenation of 2-chlorobutane, but-2-ene (disubstituted alkene) is the preferred product because it is formed faster than but-1-ene (monosubstituted alkene) which is in accordance with Saytzeff rule.

Question 53.
Write the CORRECT order of stability of alkenes with respect to Saytzeff rule.
R CH = CH2, CH2 = CH2, R2C = CH2, R2C = CR2, RCH = CHR, R2C = CHR
Answer:
R2C = CR2 > R2C = CHR > R2C = CH2, RCH = CHR > RCH = CH2 > CH2 = CH2

Question 54.
Explain dehydration reaction of alcohols.
Answer:
i. Alcohols on heating with sulphuric acid form alkenes with elimination of water molecule. The reaction is known as catalysed dehydration of alcohols.
ii. The exact conditions of dehydration depend upon the alcohol.
iii. Dehydration of alcohol is an example of β-elimination since -OH group from α-carbon along with H-atom from β-carbon is removed.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 46
The ease of dehydration of alcohol is in the order 3° > 2° > 1°.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 55.
Write dehydration reaction of 1°, 2° and 3° alcohols giving one example for each.
Answer:
The ease of dehydration of alcohols is in the order 3° > 2° > 1°.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 47
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 48

Question 56.
Explain isomerism with structure in the product obtained by acid catalysed dehydration of pentan-2-ol.
Answer:
i. Pentan-2-ol on acid catalysed dehydration, forms the following isomers.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 49
ii. A and B are position isomers.
iii. Pent-2-ene has the following geometrical isomers:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 50

Question 57.
What is dehalogenation? Write the general reaction for dehalogenation of vicinal dihalides.
Answer:
i. Removal of two halogen atoms from adjacent carbon atoms is called dehalogenation.
ii. The dihalides of alkane in which two halogen atoms are attached to adjacent carbon atoms are called vicinal dihalides.
iii. Vicinal dihalides on heating with zinc metal form an alkene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 51

Question 58.
How is propene obtained by dehalogenation reaction?
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 52

Question 59.
How are geometrical isomers of alkenes obtained from alkynes?
Answer:
Alkenes are obtained by partial reduction of alkynes wherein C = C triple bond of alkynes is reduced to a C = C double bond by:
i. using calculated quantity of dihydrogen in presence of Lindlar’s catalyst (palladised charcoal deactivated partially with quinoline or sulphur compound) to give the cis-isomer of alkene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 53
ii. using sodium in liquid ammonia to give trans-isomer of alkene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 54

Question 60.
Write physical properties of alkenes.
Answer:

  • Alkenes are nonpolar or weakly polar compounds that are insoluble in water, and soluble in nonpolar solvents like benzene, ether, chloroform.
  • They are less dense than water.
  • The boiling point of alkene rises with increasing number of carbons.
  • Branched alkenes have lower boiling points than straight chain alkenes.
  • The boiling point of alkene is very nearly the same as that of alkane with the same carbon skeleton.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 61.
Arrange the following alkenes in increasing order of their boiling points.
But-1-ene, 2,3-dimethylbut-2-ene, 2-methylpropene, propene, 2-methylbut-2-ene.
Answer:
Propene < 2-methylpropene < but-1-ene < 2-methylbut-2-ene < 2,3-dimethylbut-2-ene.
Note: Melting points and boiling points of alkenes.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 55

Question 62.
What kind of reactions do alkenes undergo? Give reason.
Answer:
Alkenes undergo electrophilic addition reactions since they are unsaturated and contain pi (π) electrons.

Question 63.
Write a note on halogenation of alkenes.
OR
Explain the formation of vicinal dihalides from alkenes with the help of examples.
Answer:
Alkenes are converted into the corresponding vicinal dihalides by addition of halogens (X2 = Cl2 or Br2).
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 56
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 57

Question 64.
How is carbon-carbon double bond in a compound detected by bromination?
Answer:
When an alkene like ethene is treated with bromine in presence of CCl4, the red-brown colour of bromine disappears due to following reaction.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 58
Hence, decolourisation of bromine is used to detect the presence of C = C bond in unknown compounds.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 65.
Explain the formation of alkyl halides from alkenes.
Answer:
i. Alkenes react with hydrogen halides (HX) like hydrogen chloride, hydrogen bromide and hydrogen iodide to give corresponding alkyl halides (haloalkanes). This reaction is known as hydrohalogenation of alkenes.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 59
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 60
ii. The order of reactivity of halogen acids is HI > HBr > HCl.

Question 66.
State Markovnikov’s rule and explain it with the help of an example.
Answer:
i. Markovnikov’s rule: When an unsymmetrical reagent is added to an unsymmetrical alkene, the negative part (X-) of the reagent gets attached to the carbon atom which carries less number of hydrogen atoms.
ii. For example, addition of HBr to unsymmetrical alkenes yield two isomeric products.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 61
iii. Experimentally it has been found that 2-Bromopropane is the major product.
[Note: Addition of HBr to symmetrical alkenes yields only one product.]

Question 67.
Explain Anti-Markovnikov’s addition or peroxide effect or Kharasch-Mayo effect.
Answer:
In 1933, M. S. Kharasch and F. R. Mayo discovered that the addition of HBr to unsymmetrical alkene in the presence of organic peroxide (R-O-O-R) takes place in the opposite orientation to that suggested by Markovnikov’s rule.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 62
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 63

Question 68.
Write the structure of major alkyl halide obtained by the action of HCl on pent-1-ene
i. in presence of peroxide
ii. in absence of peroxide.
Answer:
The structures of alkyl halides obtained by the action of hydrogen bromide on pent-1-ene are as follows:
i. In presence of peroxide:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 64
ii. In absence of peroxide:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 65
[Note: Presence/absence of peroxide has no effect on addition of HCl or HI.]

Question 69.
Explain the formation of alcohols from alkenes using conc. sulphuric acid with the help of an example.
Answer:
i. Alkenes react with cold concentrated sulphuric acid to form alkyl hydrogen sulphate (ROSO3H). The addition takes place according to Markovnikov’s rule as shown in the following steps.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 66
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 67
ii. If alkyl hydrogen sulphate is diluted with water and heated, then an alcohol having the same alkyl group as the original alkyl hydrogen sulphate is obtained.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 68
iii. This is an excellent method for the large-scale manufacture of alcohols.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 70.
What is hydration of alkenes?
Answer:
i. Reactive alkenes on adding water molecules in the presence of concentrated sulphuric acid, form alcohol.
ii. The addition of water takes place according to Markovnikov’s rule. This reaction is known as hydration of alkenes.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 69

Question 71.
Complete the following conversion.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 70
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 71

Question 72.
But-1-ene and 2-methylpropene are separately treated with following reagents. Predict the product/products. Indicate major/minor product,
i. HBr
ii. H2SO4 / H2O
Answer:
i. HBr:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 72
ii. H2SO4 / H2O:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 73

Question 73.
Explain: Ozonolysis
Answer:
i. The C = C double bond in alkenes, gets cleaved on reaction with ozone followed by reduction.
ii. The overall process of formation of ozonide by reaction of ozone with alkene in the first step and then decomposing it to the carbonyl compounds by reduction in the second step is called ozonolysis.
iii. When ozone gas is passed into solution of the alkene in an inert solvent like carbon tetrachloride, unstable alkene ozonide is obtained.
iv. This is subsequently treated with water in the presence of a reducing agent zinc dust to form carbonyl compounds, namely, aldehydes and/or ketones.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 74

Question 74.
Write reactions for the ozonolysis of the following alkenes:
i. Ethene
ii. Propene
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 75

Question 75.
What is the role of zinc dust, in ozonolysis reaction?
Answer:
In ozonolysis, the role of zinc dust is to prevent the formation of hydrogen peroxide which oxidizes aldehydes to corresponding acids.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 76.
State TRUE or FALSE. If false, correct the statement.
i. In the cleavage products of ozonide, a carbonyl group (C=O) is formed at each of the original doubly bonded carbon atoms.
ii. In ozonolysis, the structure of original alkene reactant cannot be identified by knowing the number and arrangement of carbon atoms in aldehydes and ketones produced.
iii. Ozonolysis reaction is used to locate the position and determine the number of double bonds in alkenes.
Answer:
i. True
ii. False
In ozonolysis, knowing the number and arrangement of carbon atoms in aldehydes and ketones produced, we can identify the structure of original alkene.
iii. True

Question 77.
Identify the alkene which produces a mixture of methanal and propanone on ozonolysis. Write the reactions involved.
Answer:
i. The structure of alkene which produces a mixture of methanol and propanone on ozonolysis is
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 76

ii. Reactions:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 77

Question 78.
Explain the process of hydroboration-oxidation of alkenes.
Answer:
i. Alkenes with diborane in tetrahydrofuran (THF) solvent undergo hydroboration to form trialkylborane, which on oxidation with alkaline peroxide forms primary alcohol.
ii. The overall reaction gives anti-Markovnikov’s product from unsymmetrical alkenes.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 78

Question 79.
Write reactions for the following conversion by hydroboration-oxidation reaction.
Ethene to ethanol
Answer:
Ethene to ethanol:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 79

Question 80.
Define: Polymerization
Answer:
The process in which large number of small molecules join together and form very large molecules with repeating units is called polymerization.

Question 81.
What is the difference between monomer and polymer?
Answer:
The compound having very large molecules made of large number of repeating small units is called polymer while the simple compound forming the repeating units in the polymer is called monomer.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 82.
How is ethene converted to polyethene?
Answer:
Ethene at high temperature and under high pressure interacts with oxygen, and undergoes polymerization giving high molecular weight polymer called polyethene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 80
Here, n represents the number of repeating units and is a large number.

Question 83.
Explain the process of hydroxylation of alkenes.
OR
What is the action of alkaline KMnO4 on alkenes?
Answer:
Alkenes react with cold and dilute alkaline potassium permanganate to form glycols.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 81

Question 84.
Explain Baeyer’s test giving one example.
Answer:
i. During hydroxylation of alkenes the purple colour of KMnO4 disappears.
ii. Hence, such reaction serves as a qualitative test for detecting the presence of double bond in the sample compound. This is known as Baeyer’s test.
e.g. As propene contains a double bond, it reacts with alkaline KMnO4 to give colourless propane-1,2-diol as product. Therefore, the purple colour of alkaline KMnO4 disappears.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 82

Question 85.
What is the action of following reagents on but-1-ene and but-2-ene?
i. Bromine
ii. Cold and dilute alkaline KMnO4.
Answer:
i. Action of Br2:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 83
ii. Action of cold and dilute alkaline KMnO4:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 84

Question 86.
Describe the action of acidic potassium permanganate on alkenes.
Answer:
Acidic potassium permanganate or acidic potassium dichromate oxidizes alkenes to ketones or acids depending upon the nature of the alkene and the experimental conditions. This is called oxidative cleavage of alkenes.
e.g.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 85

Question 87.
Complete the following conversions.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 86
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 87

Question 88.
State some important uses of alkenes.
Answer:

  • Alkenes are used as starting materials for preparation of alkyl halides, alcohols, aldehydes, ketones, acids, etc.
  • Ethene and propene are used to manufacture polythene, polypropylene which are used in polyethene bags, toys, bottles, etc.
  • Ethene is used for artificial ripening of fruits, such as mangoes.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 89.
What are alkynes? Write their general formula.
Answer:

  • Alkynes are aliphatic unsaturated hydrocarbons containing at least one C = C.
  • Their general formula is CnH2n-2.

Question 90.
Explain position isomerism in alkyne.
Answer:
Alkynes show position isomerism which is a type of structural isomerism.
e.g. But-1-yne and but-2-yne, both are represented by C4H6, however, both of them differ in position of triple bond in them.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 88
[Note: 1-Alkynes are also called terminal alkynes.]

Question 91.
Draw the structural isomers of isomers of C5H8. Identify position isomers amongst them.
Answer:
i. Structural isomers of C5H10 (fourth member of homologous series of alkynes):
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 89
ii. The compounds pent-1-yne and pent-2-yne are position isomers of each other.

Question 92.
What are alkadiynes and alkatriynes? Give one example of each.
Answer:
The aliphatic unsaturated hydrocarbons containing two and three carbon-carbon triple bonds in their structure are called alkadiynes and alkatriynes, respectively.
e.g. CH ≡ C – CH2 – C ≡ CH
Alkadiyne (Penta-1,4-diyne)

HC ≡ C- C ≡ C- C ≡ CH
Alkatriyne (Hexa-1,3,5-triyne)

Question 93.
Complete the following table:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 90
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 91

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 94.
How is acetylene prepared from the following compounds?
i. Methane
ii. Calcium carbide
Answer:
i. From methane: Ethyne is industrially prepared by controlled, high temperature, partial oxidation of methane.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 92
ii. From calcium carbide: Industrially, ethyne is prepared by reaction of calcium carbide with water.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 93

Question 95.
How are alkynes prepared by dehydrohalogenation of vicinal dihalides? Write general reaction and explain it using an example.
Answer:
Vicinal dihalides react with alcoholic solution of potassium hydroxide to form alkenyl halide which on further treatment with sodamide forms alkyne.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 94

Question 96.
Convert 1,2-dichloropropane to propyne.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 95

Question 97.
i. What are terminal alkynes?
ii. How are they converted to higher nonterminal alkynes? Give one example.
Answer:
i. Terminal alkynes are the compounds in which hydrogen atom is directly attached to triply bonded carbon atom.
ii. a. A smaller terminal alkyne first reacts with a very strong base like lithium amide to form metal acetylide (lithium amide is easier to handle than sodamide).
b. Higher alkynes are obtained by reacting metal acetylides (alkyn-1-yl lithium) with primary alkyl halides.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 96
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 97

Question 98.
How is pent-2-yne prepared from propyne?
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 98

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 99.
Enlist physical properties of alkenes.
Answer:
The physical properties of alkynes are similar to those of alkanes and alkenes.

  • They are less dense than water.
  • They are insoluble in water and quite soluble in less polar organic solvents like ether, benzene, carbon tetrachloride.
  • The melting points and boiling points of alkynes increase with an increase in molecular mass.

Question 100.
Lithium amide (LiNH2) is very strong base and it reacts with terminal alkynes to form lithium acetylides with the liberation of hydrogen indicating acidic nature of terminal alkynes. Why is it so?
Answer:

  • The hydrogen bonded to C ≡ C triple bond has acidic character.
  • In terminal alkynes, hydrogen atom is directly attached to sp hybridized carbon atom.
  • In sp hybrid orbital, the percentage of s-character is 50%. An electron in s-orbital is very close to the nucleus and is held tightly.
  • The sp hybrid carbon atom in terminal alkynes is more electronegative than the sp2 carbon in ethene or the sp3 carbon in ethane.
  • Due to high electronegative character of carbon in terminal alkynes, hydrogen atom can be given away as proton (H+) to very strong base.

Question 101.
Give reason: Acidic nature of alkynes is used to distinguish between terminal and non-terminal alkynes.
Answer:

  • Acidic alkynes react with certain heavy metal ions like Ag+ and Cu+ and form insoluble acetylides.
  • On addition of acidic alkyne to the solution of AgNO3 in alcohol, it forms a precipitate, which indicates that the hydrogen atom is attached to triply bonded carbon.

Hence, this reaction is used to differentiate terminal alkynes and non-terminal alkynes.

Question 102.
Predict the product in the following reactions.
\(\mathbf{H C} \equiv \mathbf{C H}+\mathbf{2 B r}_{2} \stackrel{\mathrm{CCl}_{4}}{\longrightarrow} ?\)
Answer:
Ethyne reacts with bromine in inert solvent such as carbon tetrachloride to give tetrabromoethane.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 99

Question 103.
Write the general reaction for addition of halogens to alkynes.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 100

Question 104.
Explain the addition of hydrogen halides to alkynes using a general reaction.
Answer:
i. Hydrogen halides (HCl, HBr and HI) add to alkynes across carbon-carbon triple bond in two steps to form geminal dihalides (in which two halogen atoms are attached to the same carbon atom).
ii. The addition of HX in both the steps takes place according to Markovnikov’s rule as shown in below.
General reaction:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 101
iii. The order of reactivity of hydrogen halides is HI > HBr > HCl.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 105.
State the action of HBr on acetylene and methyl acetylene.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 102

Question 106.
Explain reactions of alkynes with water using general reaction.
Answer:
Alkynes react with water in presence of 40% sulphuric acid and 1% mercuric sulphate to form aldehydes or ketones, i.e., carbonyl compounds.
General reaction:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 103

Question 107.
Predict the products when ethyne and propyne are treated with 1% mercuric sulphate in H2SO4.
Answer:
i. Ethyne:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 104
ii. Propyne:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 105a

Question 108.
Convert:
i. But-1-yne to butan-2-one
ii. Hex-3-yne to hexan-3-one
Answer:
i. But-l-yne to butan-2-one:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 106
ii. Hex-3-yne to hexan-3-one:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 107

Question 109.
What are products obtained on hydration of but-1-yne and but-2-yne? Are they same or different? Explain.
Answer:
i. Hydration of but-1-yne:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 108
Hydration of but-2-yne:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 109
The products obtained on hydration of but-1-yne and but-2-yne are same i.e., butan-2-onc since the addition of water to alkyncs takes place according to Markovnikovs rule.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 110.
How is ethylene converted into ethylidene dichloride?
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 110

Question 111.
Write some important uses of acetylene.
Answer:

  • Ethyne (acetylene) is used in preparation of ethanal (acetaldehyde), propanone (acetone), ethanoic acid (acetic acid).
  • It is used in the manufacture of polymers, synthetic rubber, synthetic fibre, plastic, etc.
  • For artificial ripening of fruits.
  • In oxy-acetylene (mixture of oxygen and acetylene) flame for welding and cutting of metals.

Question 112.
Many organic compounds obtained from natural sources such as resins, balsams, oil of wintergreen, etc. have pleasant fragrance or aroma. Such compounds are named as aromatic compounds.
i. Name the simplest aromatic compound.
ii. Write the names of any two aromatic compounds.
Answer:
i. Benzene is the simplest aromatic hydrocarbon.
ii. Toluene and naphthalene

Question 113.
Draw structures of any four aromatic hydrocarbons.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 111

Question 114.
Write the molecular formula of benzene. Give its boiling point.
Answer:
The molecular formula for benzene is C6H6. Its boiling point is 353 K.

Question 115.
State TRUE or FALSE. Correct the false statement.
i. Aromatic hydrocarbons are also called as arenes.
ii. Toluene is a non-aromatic hydrocarbon.
iii. Benzene is colourless liquid having characteristic odour.
Answer:
i. True
ii. False
Toluene is an aromatic hydrocarbon.
iii. True

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 116.
Name any two large-scale sources of benzene.
Answer:
Coal-tar and petroleum are the two large-scale sources of benzene.
[Note: Other aromatic compounds like toluene, phenol, naphthalene, etc. are also obtained from coal-tar and petroleum.]

Question 117.
Draw the structure of an aromatic compound that resembles benzene but does not have pleasant odour.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 112

Question 118.
Name and draw the structures of any three compounds that have pleasant odour but do not resemble benzene.
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 113

Question 119.
Differentiate between aromatic and aliphatic compounds.
Answer:
Aromatic compounds:

  • Aromatic compounds contain higher percentage of carbon.
  • They bum with sooty flame.
  • They are cyclic compounds with alternate single and double bonds.
  • They are not attacked by normal oxidizing and reducing agents.
  • They do not undergo addition reactions easily. They do not decolourise dilute alkaline aqueous KMnO4 and Br2 in CCl4, though double bonds appear in their structure.
  • They prefer substitution reactions.

Aliphatic compounds:

  • Aliphatic compounds contain lower percentage of carbon.
  • They bum with non-sooty flame.
  • They are open chain compounds.
  • They are easily attacked by oxidizing and reducing agents.
  • Unsaturated aliphatic compounds undergo addition reactions easily. They decolourise dilute aqueous alkaline KMnO4 and Br2 in CCl4.
  • The saturated aliphatic compounds give substitution reactions.

Question 120.
Benzene cannot have open chain structure. Explain this statement.
Answer:

  • The molecular formula of benzene is C6H6. This indicates high degree of unsaturation.
  • Open chain or cyclic structure having double and triple bonds can be written for C6H6.
  • However, benzene does not behave like alkenes or alkynes. This indicates that benzene cannot have the open chain structure.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 121.
Compare the reactivity of benzene and alkenes with the following reagents:
i. Dilute alkaline KMnO4
ii. Br6 in CCl4
iii. H6O in acidic medium
Answer:

Reagent Alkenes Benzene
Dilute alkaline aqueous KMnO4 Decolourisation of KMnO4 No decolourisation
Br2 in CCl4 Decolourisation of red brown colour of bromine No decolourisation
H2O in acidic medium Addition of H2O molecule No reaction

Question 122.
Give the evidence for the cyclic structure of benzene.
Answer:
Evidence for the cyclic structure of benzene:
i. Benzene yields only one and no isomeric monosubstituted bromobenzene (C6H5Br) when treated with equimolar bromine in FeBr3. This indicates that all six hydrogen atoms in benzene are identical.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 114
ii. This is possible only if benzene has cyclic structure of six carbons bound to one hydrogen atom each.
iii. Benzene on catalytic hydrogenation gives cyclohexane.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 115
This confirms the cyclic structure of benzene and three C = C in it.

Question 123.
Write a short note on the Kekule structure of benzene.
Answer:
Kekule structure of benzene:
i. August Kekule in 1865 suggested the structure for benzene having a cyclic planar ring of six carbon atoms with alternate single and double bonds and hydrogen atom attached to each carbon atom.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 116
ii. The Kekule structure indicates the possibility of two isomeric 1,2-dibromobenzenes. In one of the isomers, the bromine atoms would be attached to the doubly bonded carbon atoms whereas in the other, they would be attached to single bonded carbons.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 117
iii. However, benzene was found to form only one ortho-disubstituted benzene. This problem was overcome by Kekule by suggesting the concept of oscillating nature of double bonds in benzene as given below.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 118
iv. Even with this modification, Kekule structure of benzene failed to explain unusual stability and preference to substitution reactions rather than addition reactions, which was later explained by resonance.

Question 124.
Explain the resonance phenomenon with respect to benzene.
OR
Explain the resonance hybrid structure of benzene.
Answer:

  • Benzene is a hybrid of various resonance structures. The two structures, (A) and (B) given by Kekule are the main contributing structures.
  • The resonance hybrid is represented by inserting a circle or a dotted circle inscribed in the hexagon as shown in (C).
  • The circle represents six electrons delocalized over the six carbon atoms of benzene ring.
  • A double headed arrow between the resonance structures is used to represent the resonance phenomenon.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 119

Question 125.
Why does benzene not prefer to undergo addition reactions?
Answer:

  • Benzene is highly unsaturated molecule but despite of this feature, it does not give addition reaction.
  • The actual structure of benzene is represented by the resonance hybrid which is the most stable form of benzene than any of its resonance structures.
  • This stability due to resonance (delocalization of π electrons) is so high that π-bonds of the molecule becomes strong and thus, resist breaking.

Thus, benzene does not prefer to undergo addition reactions.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 126.
Explain the resonance structures of benzene using the orbital overlap concept.
Answer:
The structure of benzene can be better explained by the orbital overlap concept,
i. All six carbon atoms in benzene are sp2 hybridized. Two sp2 hybrid orbitals of each carbon atom overlap and form carbon-carbon sigma (σ) bond and the remaining third sp2 hybrid orbital of each carbon overlaps with s orbital of a hydrogen atom to form six C – H sigma bonds.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 120

ii. The unhybridized p orbitals of carbon atoms overlap laterally forming π bonds. There are two possibilities of forming three π bonds by overlap of p orbitals of C1 – C2, C3 – C4, C5 – C6 or C2 – C3, C4 – C5, C6 – C1, respectively, as shown in the following figure. Both the structures are equally probable.

According to resonance theory, these are two resonance structures of benzene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 121

Question 127.
Explain the structure of benzene with respect to molecular orbital theory.
Answer:
i. According to molecular orbital (MO) theory, the six p orbitals of six carbons give rise to six molecular orbitals of benzene.
ii. Shape of the most stable MO is as show in the figure below. Three of these π molecular orbitals lie above and the other below those of free carbon atom energies.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 122
iii. The six electrons of the p orbitals cover all the six carbon atoms and are said to be delocalized. Delocalization of π electrons results in stability of benzene molecule.

Question 128.
Give the carbon-carbon bond length in benzene. Explain why benzene shows unusual behaviour.
Answer:
i. X-ray diffraction data indicate that all C – C bond lengths in benzene are equal (139 pm) which is an intermediate between C – C (154 pm) and C = C bond (133 pm).

ii. Thus, absence of pure double bond in benzene accounts for its reluctance to addition reactions under normal conditions, which explains unusual behaviour of benzene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 123

Question 129.
Write a short note on aromaticity.
Answer:
i. All aromatic compounds undergoes substitution reactions rather than addition reactions and this property is referred to as aromaticity or aromatic character.
ii. The aromatic character of benzene is correlated to its structure.
iii. Aromaticity is due to extensive cyclic delocalization of p electrons in the planar ring structure.
iv. Three rules of aromaticity that is used for predicting whether a particular compound is aromatic or non-aromatic are as follows:

  • Aromatic compounds are cyclic and planar (all atoms in ring are sp2 hybridized).
  • Each atom in aromatic ring has a p orbital. The p orbitals must be parallel so that continuous overlap is possible around the ring.
  • Huckel rule: The cyclic π molecular orbital formed by overlap of p orbitals must contain (4n + 2) p electrons, where n = integer 0, 1, 2, 3, … etc.

Question 130.
State and explain the Huckel rule of aromaticity.
Answer:
Huckel rule: The cyclic π molecular orbital formed by overlap of p orbitals must contain (4n + 2) p electrons, where n = integer 0, 1,2,3, … etc.

Explanation:
According to Huckel rule, a cyclic and planar compound is aromatic if it the number of π electrons is equal to (4n + 2), where n = integer 0, 1, 2, 3, … etc.

n Number of π electrons
n = 0 (4 × 0) – 2 = 2
n = 1 (4 × 1) + 2 = 6
n = 2 (4 × 2) + 2 = 10

e.g. Consider benzene molecule:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 124
Benzene has 6π electrons. According to Huckel rule, if n = 1, then (4n + 2)π = 6π electrons. Hence, benzene is aromatic.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 131.
By using the rules of aromaticity, explain whether the following compounds are aromatic or non-aromatic.
i. Benzene
ii. Naphthalene
iii. Cycloheptatriene
Answer:
i. Benzene:
a. It is cyclic and planar.
b. It has three double bonds and six π electrons.
c. It has a p orbital on each carbon of the hexagonal ring. Hence, a continuous overlap above and below the ring is possible.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 125
d. According to Huckcl rule, this compound is aromatic if, 4n + 2 = Number of π electrons.
4n + 2 = 6,
∴ 4n = 6 – 2 = 4
n = 4/4 = 1, here, ‘n’ comes out to be an integer.
Hence, benzene is aromatic.

ii. Naphthalene:
a. It is cyclic and planar.
b. It has 5 double bonds and 10 n electrons.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 126
c. It has a p orbital on each carbon atom of the ring. Hence, a continuous overlap around the ring is possible.
d. According to Huckel rule, this compound is aromatic if, 4n + 2 = Number of π electrons.
4n + 2 = 10,
∴ 4n = 10 – 2 = 8
n = 8/4 = 2, Here ‘n’ comes out to be an integer.
Hence, naphthalene is aromatic.

iii. Cycloheptatriene:
a. It is cyclic and planar.
b. It has three double bonds and 6 π electrons.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 127
c. But one of the carbon atoms is saturated (sp3 hybridized) and it does not have a p orbital.
d. Hence, a continuous overlap around the ring is not possible in cycloheptatriene. Hence, it is non-aromatic.

Question 132.
How does Huckel rule help in determining the aromaticity of pyridine?
Answer:
i. Pyridine has three double bonds and 6 π electrons.
ii. The six p orbitals containing six electrons form delocalized π molecular orbital.
iii. The unused sp2 hybrid orbital of nitrogen containing two non-bonding electrons is as it is.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 128
iv. According to Huckel rule, this compound is aromatic if, 4n + 2 = Number of π electrons
4n + 2 = 6,
∴ 4n = 6 – 2 = 4
n = 4/4 = 1, here ‘n’ comes out to be an integer. Hence, pyridine is aromatic.

Question 133.
How is benzene prepared from ethyne/acetylene?
Answer:
From ethyne (By trimerization): Alkynes when passed through a red hot iron tube at 873 K, polymerize to form aromatic hydrocarbons. Ethyne when passed through a red hot iron tube at 873 K undergoes trimerization to form benzene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 129

Question 134.
How is benzene prepared from sodium benzoate?
OR
Explain preparation of benzene by decarboxylation.
Answer:
From sodium benzoate (by decarboxylation): When anhydrous sodium benzoate is heated with soda lime, it undergoes decarboxylation and gives benzene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 130
[Note: This reaction is useful for decreasing the length of a carbon chain by one C-atom]

Question 135.
How will you convert phenol to benzene?
Answer:
From phenol (By reduction): When vapours of phenol are passed over heated zinc dust, it undergoes reduction and gives benzene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 131

Question 136.
Enlist physical properties of benzene.
Answer:
Physical properties of benzene:

  • Benzene is a colourless liquid.
  • Its boiling point is 353 K and melting point is 278.5 K.
  • It is insoluble in water. It forms upper layer when mixed with water.
  • It is soluble in alcohol, ether and chloroform.
  • Its vapours are highly toxic which on inhalation lead to unconsciousness.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 137.
What is the action of chlorine on benzene in the presence of UV light?
Answer:
Addition of chlorine: When benzene is treated with chlorine in the presence of bright sunlight or UV light, three molecules of chlorine gets added to benzene to give benzene hexachloride.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 132

Question 138.
Name the γ-isomer of benzene hexachloride which is used as insecticide.
Answer:
The γ-isomer of benzene hexachloride which is used as insecticide is called as gammaxene or lindane.

Question 139.
How will you convert benzene to cyclohexane?
Answer:
Addition of hydrogen: When a mixture of benzene and hydrogen gas is passed over heated catalyst nickel at 453 K to 473 K, cyclohexane is formed.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 133

Question 140.
What is the action of ozone on benzene?
Answer:
Addition of ozone: When benzene is treated with ozone in the presence of an inert solvent carbon tetrachloride, benzene triozonide is formed, which is then decomposed by zinc dust and water to give glyoxal.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 134

Question 141.
What are the different types of electrophilic substitution reactions of benzene?
Answer:
i. Benzene shows electrophilic substitution reactions, in which one or more hydrogen atoms of benzene ring are replaced by groups like – Cl, – Br, – NO2, – SO3H, -R, -COR, etc.
ii. Different types of electrophilic substitution reactions of benzene are as follows:

  • Halogenation (chlorination and bromination)
  • Nitration
  • Sulphonation
  • Friedel-Craft’s alkylation and
  • Friedel-Craft’s acylation

Question 142.
Write a short note on chlorination reaction of benzene.
Answer:
Chlorination of benzene:
i. In chlorination reaction, hydrogen atom of benzene is replaced by chlorine atom.
ii. Chlorine reacts with benzene in dark in the presence of iron or ferric chloride or anhydrous aluminium chloride or red phosphorus as catalyst to give chlorobenzene (C6H5Cl).
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 135
iii. Electrophile involved in the reaction: Cl+, chloronium ion,
Formation of the electrophile: Cl – Cl + FeCl3 → Cl+ + [FeCl4]

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 143.
Write a short note on bromination reaction of benzene.
Answer:
Bromination of benzene:
i. In bromination reaction, hydrogen atom of benzene is replaced by bromine atom.
ii. Bromine reacts with benzene in dark in presence of iron or ferric bromide or anhydrous aluminium bromide or red phosphorus as catalyst to give bromobenzene (C6H5Br).
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 136
iii. Electrophile involved in the reaction: Br+
Formation of the electrophile: Br – Br + FeBr3 → Br+ + [FeBr4]

Question 144.
Why direct iodination of benzene is not possible?
Answer:
Direct iodination of benzene is not possible as the reaction is reversible.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 137
[Note: Iodination of benzene can be carried out in the presence of oxidising agents like HIO3 or HNO3.]

Question 145.
How will you convert benzene to hexachlorobenzene?
Answer:
When benzene is treated with excess of chlorine in presence of anhydrous aluminium chloride, it gives hexachlorobenzene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 138

Question 146.
State true or false. Correct the false statement.
i. In halogenation reaction, hydrogen atom of benzene ring is replaced by halogen atom.
ii. The molecular formula of hexachlorobenzene is C6H6Cl6.
iii. Benzene forms the lower layer when mixed with water.
Answer:
i. True
ii. False
The molecular formula of hexachlorobenzene is C6Cl6
iii. False
Benzene forms the upper layer when mixed with water.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 147.
Explain the nitration reaction of benzene.
Answer:
Nitration of benzene:
i. When benzene is heated with a mixture of concentrated nitric acid and concentrated sulphuric acid (nitrating mixture) at about 313 K to 333 K, it gives nitrobenzene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 139
ii. Electrophile involved in the reaction: \(\mathrm{NO}_{2}^{+}\), nitronium ion
Formation of the electrophile: HO – NO2 + 2H2SO4 ⇌ \(2 \mathrm{HSO}_{4}^{-}\) + H3O+ + \(\mathrm{NO}_{2}^{-}\)

Question 148.
Write a short note on sulphonation of benzene.
Answer:
Sulphonation of benzene:
i. When benzene is heated with fuming sulfuric acid (oleum) at 373 K, it gives benzene sulfonic acid.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 140
ii. Electrophile involved in the reaction: SO3, free sulphur trioxide
Formation of the electrophile: 2H2SO4 → H3O+ + \(\mathrm{HSO}_{4}^{-}\) + SO3

Question 149.
Write a short note on Friedel-Craft’s alkylation reaction of benzene.
Answer:
Friedel-Craft’s alkylation reaction of benzene:
i. When benzene is treated with an alkyl halide like methyl chloride in the presence of anhydrous aluminium chloride, it gives toluene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 141
ii. Electrophile involved in the reaction: R+
Formation of the electrophile: R – Cl + AlCl3 → R+ + \(\mathrm{AlCl}_{4}^{-}\)
iii. Friedel-Craft’s alkylation reaction is used to extend the chain outside the benzene ring.

Question 150.
Explain Friedel-Craft’s acylation reaction of benzene. Give example reactions.
Answer:
Friedel-craft’s acylation reaction of benzene:
i. When benzene is heated with an acyl halide or acid anhydride in the presence of anhydrous aluminium chloride, it gives corresponding acyl benzene.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 142
ii. Electrophile involved in the reaction: R – C- = O, acylium ion
Formation of the electrophile: R – COCl + AlCl3 → R – C+ = O + \(\mathrm{AlCl}_{4}^{-}\)

Question 151.
Write the general combustion reaction for hydrocarbons.
Answer:
General combustion reaction for any hydrocarbon (CxHy) can be represented as follows:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 143

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 152.
Write the combustion reaction of benzene.
Answer:
When benzene is heated in air, it bums with sooty flame forming carbon dioxide and water.
C6H6 + \(\frac {15}{2}\)O2 → 6CO2 + 3H2O

Question 153.
Write a note on the directive influence of substituents (functional groups) in monosubstituted benzene.
Answer:
i. In benzene, all hydrogen atoms are equivalent and so, when it undergoes electrophilic substitution reactions, only one monosubstituted product is possible.
Monosubstituted benzene:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 144
ii. When monosubstituted benzene undergoes further electrophilic substitution, the second substituent (electrophile, E) can occupy any of the five positions available and give three disubstituted products.
But these disubstituted products are not formed in equal amounts.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 145
iii. The position of second substituent (E) is determined by the nature of substituent (S) already present in the benzene ring and not on the nature of second substituent (E).
iv. The groups which direct the incoming group to ortho and para positions are called ortho and para directing groups. The groups which direct the incoming group to meta positions are called meta directing groups. Thus, depending on the nature of the substituent (S) either ortho and para products or meta products are formed as major products.

Question 154.
What are ortho and para directing groups? Enlist few ortho and para directing groups.
Answer:
The groups which direct the incoming group to ortho and para positions are called ortho and para directing groups.
Ortho and para directing groups:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 146

Question 155.
Explain the directive influence of ortho, para directing groups in monosubstituted benzene using suitable example.
OR
Explain the directive influence of -OH group in benzene.
Answer:
i. The directive influence of ortho, para directing groups can be explained with the help of inductive and resonance effects.
ii. phenol has the following resonating structures:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 147
iii. It can be seen from the above resonating structures, that the ortho (o-) and para (p-) positions have a greater electron density than the meta positions.
iv. Therefore, -OH group activates the benzene ring for the attack of second substituent (E) at these electron rich centres. Thus, phenolic -OH group is activating and ortho, para-directing group.
v. In phenol, -OH group has electron withdrawing inductive (-I) effect which slightly decreases the electron density at ortho positions in benzene ring. Thus, resonance effect and inductive effect of -OH group act opposite to each other. However, the strong resonance effect dominates over inductive effect.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Question 156.
Explain the o, p-directive effect of methyl group.
Answer:

  • All ortho and para directing groups possess nonbonding electron pair on the atom which is directly attached to the aromatic ring; however, methyl group is an exception.
    Methyl (or alkyl groups) is ortho and para directing, although it has no nonbonding electron pair on the key atom. This is explained on the basis of special type of resonance called hyperconjugation or no bond resonance.

Question 157.
Explain why halide group is an ortho and para directing group.
Answer:
i. In aryl halides, halogens are moderately deactivating. Because of their strong -I effect, overall electron density on the benzene ring decreases, which makes the electrophilic substitution difficult.
ii. However, halogens are ortho and para directing. This can be explained by considering resonance structures.
iii. e.g. Chlorobenzene has the following resonating structures:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 148
iv. Due to resonance, the electron density on ortho and para positions is greater than meta positions and hence, -Cl is ortho and para directing.

Question 158.
What are meta directing groups? Enlist few of them.
Answer:
The groups which direct the incoming group to meta positions are called meta directing groups.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 149
[Note: All meta directing groups have positive (or partial positive) charge on the atom which is directly attached to an aromatic ring.]

Question 159.
Explain the directive influence of nitro group in nitrobenzene.
OR
Explain why nitro group is a meta-directing group.
Answer:
i. Meta directing group withdraws electrons from the aromatic ring by resonance, making the ring electron-deficient. Therefore, meta groups are ring deactivating groups.
ii. Due to -I effect, -NO2 group reduces electron density in benzene ring on ortho and para positions. So, the attack of incoming group becomes difficult at ortho and para positions. The incoming group can attack on meta positions more easily.
iii. The various resonance structures of nitrobenzene are as shown below:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 150
iv. It is clear from the above resonance structures that the ortho and para positions have comparatively less electron density than at meta positions. Hence, the incoming group/electrophile attacks on meta positions.

Question 160.
What are polycyclic aromatic compounds? How are they produced?
Answer:

  • Polycyclic aromatic compounds are the hydrocarbons containing more than two benzene rings fused together.
  • They are produced by incomplete combustion of tobacco, coal and petroleum.

Question 161.
Write the harmful effects of benzene.
Answer:

  • Benzene is both, toxic and carcinogenic (cancer causing).
  • In fact, it might be considered “the mother of all carcinogens” as a large number of carcinogens have structures that include benzene rings.
  • In liver, benzene is oxidized to an epoxide and benzopyrene is converted into an epoxy diol. These substances are carcinogenic and can react with DNA and thus, can induce mutation leading to uncontrolled growth of cancer cells.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

Multiple Choice Questions

1. Alkanes are represented by the general formula ………….
(A) CnH2n-2
(B) CnH2n+2
(C) CnH2n
(D) CnHn
Answer:
(B) CnH2n+2

2. Which of the following compound is alkanes?
(A) C5H10
(B) C10H22
(C) C15H28
(D) C9H16
Answer:
(B) C10H22

3. Alkanes are commonly called …………
(A) arenes
(B) paraffins
(C) olefins
(D) acetylenes
Answer:
(B) paraffins

4. Every carbon atom in alkanes is …………..
(A) sp hybridized
(B) sp2 hybridized
(C) sp3 hybridized
(D) sp3d hybridized
Answer:
(C) sp3 hybridized

5. Isomerism is the phenomenon in which two or more organic compounds have ………….
(A) same molecular formula but different structural formula
(B) same structural formula but different molecular formula
(C) same general formula, but different structural formula
(D) same empirical formula, same structural formula
Answer:
(A) same molecular formula but different structural formula

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

6. Pentane exhibits …………. chain isomers.
(A) two
(B) three
(C) four
(D) five
Answer:
(B) three

7. Which of the following is NOT an isomer of hexane?
(A) 2-Methylpentane
(B) 2,2-Dimethylbutane
(C) 2,2-Dimethylpentane
(D) 3-Methylpentane
Answer:
(C) 2,2-Dimethylpentane

8. Alkanes can be prepared by ………… of unsaturated hydrocarbons.
(A) hydrogenation
(B) oxidation
(C) hydrolysis
(D) cracking
Answer:
(A) hydrogenation

9. Catalytic hydrogenation of ethene or acetylene gives …………..
(A) ethane
(B) propylene
(C) methane
(D) propane
Answer:
(A) ethane

10. Ethyl iodide when reduced by zinc and dilute HCl, leads to the formation of …………..
(A) Methane
(B) Ethane
(C) Ethylene
(D) Butane
Answer:
(B) Ethane

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

11. The reaction of alkyl halides with sodium in dry ether to give higher alkanes is called ………..
(A) Wurtz reaction
(B) Kolbe’s reaction
(C) Frankland’s reaction
(D) Williamson’s reaction
Answer:
(A) Wurtz reaction

12. Methane is ………… molecule.
(A) polar
(B) nonpolar
(C) highly polar
(D) none of these
Answer:
(B) nonpolar

13. Alkanes are ………… in water.
(A) soluble
(B) sparingly soluble
(C) insoluble
(D) none of these
Answer:
(C) insoluble

14. As branching increases, boiling point of alkanes ………….
(A) increases
(B) decreases
(C) remains same
(D) None of these
Answer:
(B) decreases

15. Halogenation of alkane is an example of …………. reaction.
(A) dehydration
(B) substitution
(C) addition
(D) elimination
Answer:
(B) substitution

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

16. Order of reactivity of halogens in halogenation of alkanes is ………….
(A) F2 > Cl2 > Br2 > I2
(B) I2 > Br2 > Cl2 > F2
(C) Br2 < I2 < F2 < Cl2
(D) Cl2 < I2 < Br2 < F2
Answer:
(A) F2 > Cl2 > Br2 > I2

17. The thermal decomposition of alkanes in absence of air to give lower alkanes, alkenes and hydrogen is called ………….
(A) vapour phase nitration
(B) pyrolysis
(C) polymerisation
(D) combustion
Answer:
(B) pyrolysis

18. But-1-ene and But-2-ene are …………
(A) chain isomers
(B) position isomers
(C) geometrical isomers
(D) metamers
Answer:
(B) position isomers

19. Hex-2-ene and 2-Methylpent-2-ene exhibit …………
(A) chain isomerism
(B) position isomerism
(C) geometrical isomerism
(D) optical isomerism
Answer:
(A) chain isomerism

20. Which of the following shows position isomerism?
(A) Propene
(B) Ethene
(C) 2-Methylpropene
(D) Pent-2-ene
Answer:
(D) Pent-2-ene

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

21. When identical atoms or group of atoms are attached to the two carbon atoms on the same side of the double bond, the isomer is called ………… isomer.
(A) cis
(B) trans
(C) position
(D) chain
Answer:
(A) cis

22. Which of the following does NOT exhibit geometrical isomers?
(A) But-2-ene
(B) Pent-2-ene
(C) But-1-ene
(D) Hex-2-ene
Answer:
(C) But-1-ene

23. When ethyl bromide is heated with alcoholic KOH, ………… is formed.
(A) ethane
(B) ethanol
(C) ethene
(D) acetylene
Answer:
(C) ethene

24. Alkenes are insoluble in …………
(A) benzene
(B) water
(C) ether
(D) chloroform
Answer:
(B) water

25. Markownikov’s rule is applicable to …………
(A) symmetrical alkenes
(B) alkanes
(C) unsymmetrical alkenes
(D) alkynes
Answer:
(C) unsymmetrical alkenes

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

26. When propene is treated with HBr in the dark and in absence of peroxide, then the main product formed is …………
(A) 1-bromopropane
(B) 2-bromopropane
(C) 1,2-dibromopropane
(D) 1,3-dibromopropane
Answer:
(B) 2-bromopropane

27. The product formed by the addition of HCl to propene in presence of peroxide is …………..
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 151
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 153

28. Propene reacts with HBr in presence of peroxide, to form …………..
(A) 2-bromopropane
(B) 1-bromopropane
(C) 3-bromopropane
(D) 1,2-dibromopropane
Answer:
(B) 1-bromopropane

29. Markovnikov’s rule is applicable for …………..
(A) CH2 = CH2
(B) CH3CH = CHCH3
(C) CH3CH2CH = CHCH2CH3
(D) (CH3)2C = CH2
Answer:
(D) (CH3)2C = CH2

30. The addition of HCl in presence of peroxide does not follow anti-Markownikov’s rule because …………..
(A) HCl bond is too strong to be broken homolytically
(B) Cl atom is not reactive enough to add on to a double bond
(C) Cl atom combines with H atom to form HCl
(D) HCl is a reducing agent
Answer:
(A) HCl bond is too strong to be broken homolytically

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

31. An alkene on ozonolysis produces a mixture of acetaldehyde and acetone. Identify the alkene.
(A) But-1-ene
(B) But-2-ene
(C) 2-Methylbut-1-ene
(D) 2-Methylbut-2-ene
Answer:
(D) 2-Methylbut-2-ene

32. The ozonolysis of (CH3)2C = C(CH3)2 followed by treatment with zinc and water will give ……………
(A) acetone
(B) acetone and acetaldehyde
(C) formaldehyde and acetone
(D) acetaldehyde
Answer:
(A) acetone

33. The compound which forms only acetaldehyde on ozonolysis is …………..
(A) ethene
(B) propyne
(C) but-1-ene
(D) but-2-ene
Answer:
(D) but-2-ene

34. Treatment of ethylene with ozone followed by decomposition of the product with Zn/H2O gives two moles of ………….
(A) formaldehyde
(B) acetaldehyde
(C) formic acid
(D) acetic acid
Answer:
(A) formaldehyde

35. Ozonolysis of 2,3-Dimethylbut-1-ene followed by reduction with zinc and water gives ………….
(A) methanoic acid and 3-methylbutan-2-one
(B) methanal and 2-methylbutan-2-one
(C) methanal and 3-methylbutan-2-one
(D) methanoic acid and 2-methylbutan-2-one
Answer:
(C) methanal and 3-methylbutan-2-one

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

36. The reaction, CH2 = CH2 + H2O + [O]
Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons 152
is called ……………
(A) hydroxylation
(B) decarboxylation
(C) hydration
(D) dehydration
Answer:
(A) hydroxylation

37. An alkene on vigorous oxidation with KMnO4 gives only acetic acid. The alkene is …………..
(A) CH3CH2CH = CH2
(B) CH3CH = CHCH3
(C) (CH3)2C = CH2
(D) CH3CH = CH2
Answer:
(B) CH3CH = CHCH3

38. Ethylene reacts with Baeyer’s reagent to give a/an ………….
(A) glycol
(B) aldehyde
(C) acid
(D) alcohol
Answer:
(A) glycol

39. Baeyer’s reagent is ………….
(A) aqueous KMnO4
(B) neutral KMnO4
(C) alkaline KMnO4
(D) aqueous bromine water
Answer:
(C) alkaline KMnO4

40. Alkynes have general formula ………….
(A) CnH2n-2
(B) CnH2n
(C) CnH2n+2
(D) CnH2n+1
Answer:
(A) CnH2n-2

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

41. Aliphatic unsaturated hydrocarbons containing two carbon-carbon triple bonds in their structure are called as ………….
(A) alkadiynes
(B) alkatriynes
(C) alkynes
(D) alkanes
Answer:
(A) alkadiynes

42. Acetylene is prepared in the industry by the action of water on ………….
(A) calcium carbonate
(B) calcium carbide
(C) mercuric chloride
(D) calcium oxide
Answer:
(B) calcium carbide

43. The dihalogen derivatives of alkanes when heated with …………. form corresponding alkynes.
(A) alcoholic water
(B) sodamide
(C) zinc
(D) acids
Answer:
(B) sodamide

44. Alkynes readily undergo …………. reaction.
(A) addition
(B) substitution
(C) elimination
(D) rearrangement
Answer:
(A) addition

45. Liquid bromine reacts with acetylene to form ………….
(A) 1,2-dibromoethene
(B) 1,1,2,2-tetrabromoethane
(C) 1,1-dibromoethene
(D) methyl chloride
Answer:
(B) 1,1,2,2-tetrabromoethane

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

46. When acetylene is passed through dil H2SO4 in the presence of 1% mercuric sulphate, the compound formed is ………….
(A) ethanol
(B) acetone
(C) acetaldehyde
(D) acetic acid
Answer:
(C) acetaldehyde

47. The compounds which contain at least one benzene ring are ………….
(A) aliphatic compounds
(B) aromatic compounds
(C) cycloalkanes
(D) both (A) and (B)
Answer:
(B) aromatic compounds

48. Which of the following compounds does NOT contain any benzene rings in their structure?
(A) Benzaldehyde
(B) Benzoic acid
(C) Naphthalene
(D) Furan
Answer:
(D) Furan

49. Benzene undergoes ………….
(A) only addition reaction
(B) only substitution reaction
(C) both addition and substitution reactions
(D) nucleophilic substitution reactions
Answer:
(C) both addition and substitution reactions

50. If the substituents are on the adjacent carbon atoms in the benzene ring, it is called ………….
(A) meta
(B) para
(C) ortho
(D) beta
Answer:
(C) ortho

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

51. How many molecules of acetylene are required to form benzene?
(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(B) 3

52. Which of the following compound on reduction gives benzene?
(A) Sodium benzoate
(B) Acetylene
(C) Cyclohexane
(D) Phenol
Answer:
(D) Phenol

53. X-Ray diffraction reveals that benzene is a …………. structure.
(A) triangular
(B) planar
(C) co-planar
(D) 3D
Answer:
(B) planar

54. γ-isomer of BHC is known as ………….
(A) gammene
(B) gammaxane
(C) chlorobenzene
(D) hexachlorobenzene
Answer:
(B) gammaxane

55. Benzene when treated with ozone forms ………….
(A) glyoxal
(B) acetic acid
(C) formaldehyde
(D) benzaldehyde
Answer:
(A) glyoxal

Maharashtra Board Class 11 Chemistry Important Questions Chapter 15 Hydrocarbons

56. …………. is formed as intermediate product in ozonolysis of benzene.
(A) Benzaldehyde
(B) Phenol
(C) Benzene triozonide
(D) Cyclohexane
Answer:
(C) Benzene triozonide

57. Electrophile in chlorination of benzene is ………….
(A) Cl
(B) Cl+
(C) Cl
(D) Cl2
Answer:
(B) Cl+

58. Benzene when treated with fuming. H2SO4 at 373 K forms ………….
(A) ethylbenzene
(B) toluene
(C) benzene sulphonic acid
(D) acetophenone sulphonic acid
Answer:
(C) benzene sulphonic acid

59. Ethyl chloride reacts with benzene in presence of anhydrous aluminium chloride to form ………….
(A) ethyl benzene
(B) chlorobenzene
(C) toluene
(D) acetophenone
Answer:
(A) ethyl benzene

60. The electrophile in Friedel-Craft’s alkylation reaction is ………….
(A) R+
(B) R
(C) Cl+
(D) RCO+
Answer:
(A) R+

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 11 Adsorption and Colloids Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 1.
Explain the phenomenon of adsorption with the help of examples.
Answer:
Consider the following two examples:

  • Example 1: When a metal spoon is dipped in milk and taken out, it is observed that a film of milk particles covers the spoon surface.
  • Example 2: If a cold water bottle is taken out from the refrigerator and kept on a table for a while, water vapour is seen to condense on the outer surface of the bottle, forming droplets or a film.
  • In the above examples, the milk particles or the water molecules from the air get adsorbed on the surface of the spoon and the bottle, respectively.
  • Similarly, surfaces of many objects around us are exposed to the atmosphere. Water molecules as well as other gas molecules such as N2, O2, from the air form an invisible multimolecular film on these objects.
    This is known as the phenomenon of adsorption.

Question 2.
Why does adsorption occur?
Answer:

  • The adsorption phenomenon is caused by dispersion forces (also known as London dispersion forces or van der Waals forces) which are short range and additive. Adsorption force is the sum of all interactions between all the atoms.
  • The pulling interactions cause the surface of a liquid to tighten like an elastic film.
  • A measure of the elastic force at the surface of a liquid is called surface tension.
  • There is a tendency to have minimum surface tension, i.e., decrease of free energy, which leads to adsorption.

Question 3.
Define surface tension.
Answer:
A measure of the elastic force at the surface of a liquid is called surface tension.
OR
Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area.

Question 4.
Define the following terms.
i. Adsorbent
ii. Adsorbate
Answer:
i. Adsorbent: The material or substance present in the bulk, on the surface of which adsorption takes place is called adsorbent.
ii. Adsorbate: The substance getting adsorbed on the adsorbent is called as adsorbate.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 5.
Give some examples of adsorption.
Answer:
Following are some examples of adsorption:

  • Adsorption of gases like hydrogen and oxygen by finely divided metals, namely, platinum, palladium, copper, nickel, etc.
  • Adsorption of gases like nitrogen and carbon dioxide by activated charcoal.
  • Removal of colouring matter like an organic dye, for example, methylene blue. When charcoal is added to methylene blue solution and shaken, it becomes colourless after some time as dye molecules accumulate on the surface of charcoal.

Question 6.
What is desorption?
Answer:
The process of removal of an adsorbed substance from a surface on which it was adsorbed is called desorption.

Question 7.
Define sorption.
Answer:
When both adsorption and absorption occur simultaneously, it is known as sorption.
e.g. When a chalk is dipped in ink, the ink molecules are adsorbed at the surface of the chalk while the solvent of the ink goes deeper into the chalk due to absorption.

Question 8.
What is physisorption? State its characteristics.
Answer:
When the adsorbent such as gas molecules are accumulated at the surface of a solid on account of weak van der Waals forces, the adsorption is termed as physical adsorption or physisorption.

Characteristics:

  • The van der Waals forces involved in physical adsorption are similar to forces causing condensation of gas into liquid. Thus, heat is released in physisorption.
  • The heat released during physisorption is of the same order of magnitude as heat of condensation.
  • Due to weak nature of van der Waals forces, physisorption is weak in nature.
  • The adsorbed gas forms several layers of molecules at high pressures.
  • The extent of adsorption is large at low temperatures.
  • The equilibrium is attained rapidly.
  • Physisorption is readily reversed by lowering of pressure of gas or by raising temperature.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 9.
Define chemisorption, Write its main features.
Answer:
When the gas molecules accumulate at the surface of a solid or adsorbate by means of chemical bonds (covalent or ionic), the adsorption is termed as chemical adsorption or chemisorption.
Features of chemical adsorption:

  • Chemisorption is specific in nature.
  • Chemisorption involving the gas-solid as the adsorbate and adsorbent is usually exothermic i.e., heat is released during this process (Exception: The adsorption of hydrogen on glass is endothermic).
  • The heat evolved in chemisorption per mole of adsorbate is nearly the same order of magnitude as that accompanying chemical bonding.
  • Chemisorption involves a large energy of activation and hence, it is also referred as activated adsorption.
  • Chemisorption increases with increase in temperature in the beginning, as a greater number of molecules can have activation energy. But after certain temperature chemisorption decreases with increase in temperature as the chemical bonds break.
  • Sometimes at low’ temperature, physisorption occurs which passes into chemisorption as the temperature is raised.
  • Chemisorption is dependent on surface area of the adsorbent.

[Note: Chemisorption was first investigated in 1916 by American Chemist, Irving Langmuir (1881-1957).]

Question 10.
Why is chemisorption also known as activated adsorption?
Answer:
Chemisorption involves a large energy of activation and hence, it is also referred as activated adsorption.

Question 11.
Give reason: Adsorption of hydrogen on glass is an endothermic process.
Answer:
Adsorption of hydrogen on glass is an endothermic process because heat is absorbed during the process due to dissociation of hydrogen.

Question 12.
Explain graphically the effect of the following factors on the adsorption of gases by solids.
i. Temperature of the adsorbent surface
ii. Pressure of the gas (adsorbate)
Answer:
i. Temperature of the adsorbent surface:

  • Adsorption is an exothermic process.
  • According to Te Chatelier’s principle, it is favoured at low temperature.
  • Therefore, the amount of gas adsorbed is inversely proportional to the temperature.
  • The graph given below shows plots of volume of N? adsorbed per unit mass of adsorbent against the pressure of a gas at different temperatures.
  • As temperature increases from 193 K to 273 K at a constant pressure ‘P’, the amount of gas adsorbed decreases.

ii. Pressure of the gas:

  • At any temperature, the extent of gas adsorbed increases with an increase in pressure.
  • The extent of adsorption is directly proportional to pressure of the gas.
  • At high pressures extent of adsorption becomes independent of the pressure. The surface of adsorbent is then almost fully covered by adsorbed gaseous molecules.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 1

Question 13.
What are the applications of adsorption?
Answer:
Following are the various applications of adsorption:
i. Catalysis (Heterogeneous catalysis):

  • The solid catalysts are used in many industrial manufacturing processes.
  • For example, iron is used as a catalyst in manufacturing of ammonia, platinum in manufacturing of sulphuric acid, H2SO4 (by contact process) while finely divided nickel is employed as a catalyst in hydrogenation of oils.

ii. Gas masks:

  • It is a device which consists of activated charcoal or mixture of adsorbents.
  • It is used for breathing in coal mines to avoid inhaling of the poisonous gases.

iii. Control of humidity: Silica and alumina gels are good adsorbents of moisture.

iv. Production of high vacuum:

  • Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
  • A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.

v. Adsorption indicators: The adsorption is used to detect the end point of precipitation titrations. Dyes such as eosin, fluorescein are used as indicators.
e.g.
a. A solution of sodium chloride containing a small amount of fluorescein is titrated against silver nitrate solution.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 2
b. When chloride ions are over, fluorescein is adsorbed on white silver chloride precipitate and hence, red colour is developed.
c. Thus, colour changes from pale yellow to reddish pink at the end point.

vi. Separation of inert gases:

  • In a mixture of noble gases, different gases adsorb to different extent.
  • Due to selective adsorption principle, gases can be separated on coconut charcoal.

vii. Froth floatation process:

  • A low-grade sulphide ore is concentrated by separating it from silica and other earthy matter using pine oil as frothing agent.
  • Hydrophobic pine oil preferentially adsorbs sulphide ore which is taken up in the froth.

viii. Chromatographic analysis:

  • It is based on selective adsorption of ions from solution using powdered adsorbents such as silica or alumina gel.
  • It has several industrial and analytical applications. Other applications include surface area determination, purification of water, etc.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 14.
Explain how high vacuum can be obtained by adsorption.
Answer:

  • Lowering of temperature at a given pressure, increases the rate of adsorption of gases on charcoal powder. By using this principle, high vacuum can be attained by adsorption.
  • A vessel evacuated by vacuum pump is connected to another vessel containing coconut charcoal cooled by liquid air. The charcoal adsorbs the remaining traces of air or moisture to create a high vacuum.

Question 15.
State whether TRUE or FALSE. Correct if false.
i. The rate of adsorption of gases on charcoal powder decreases on lowering of temperature at a given pressure.
ii. Noble gases can be separated from their mixture using the principle of selective adsorption as they adsorb to different extent.
iii. Pine oil is used as frothing agent in froth floatation process.
Answer:
i. False
The rate of adsorption of gases on charcoal powder increases on lowering of temperature at a given pressure.
ii. True
iii. True

Question 16.
Match the following.

Column A Column B
i. Iron a. Hydrogenation of oils
ii. Nickel b. Production of sulphuric acid
iii. Platinum c. Synthesis of ammonia

Answer:
i – c,
ii – a,
iii – b

Question 17.
What is a catalyst?
Answer:
A catalyst is a substance which when added to a reacting system, increases the rate of a reaction without itself undergoing any permanent chemical change.

Question 18.
Explain the importance of catalysts in chemical industries.
Answer:

  • A large number of the chemicals manufactured in industries make use of catalysts to obtain specific products.
  • The use of catalyst lowers the reaction temperature as well as energy costs significantly.
    Due to these advantages, catalysts are of great importance in chemical industry.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 19.
Name two types of catalysis.
Answer:

  1. Homogeneous catalysis
  2. Heterogeneous catalysis

Question 20.
Define homogeneous catalysis and give any two examples.
Answer:
When the reactants and the catalyst are in the same phase, it is said to be homogeneous catalysis.
e.g.
i. Iodide ion (I) is used as homogeneous catalyst in decomposition of aqueous hydrogen peroxide because both I and H2O2 are present in the same aqueous phase.
ii. Hydrolysis of sugar is catalysed by H+ ions furnished by sulphuric acid.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 3
All reactants and catalyst are in same solution phase.
[Note: Enzyme catalysis is also an important type of homogeneous catalysis.]

Question 21.
Justify: Lead chamber process is an example of homogeneous catalysis.
Answer:
i. In the lead chamber process, sulphur dioxide is oxidized to sulphur trioxide with dioxygen (O2) in the presence of nitric oxide as catalyst.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 4
ii. Since all the reactants as well as the catalyst is present in gaseous state. i.e., in same phase, it is a homogeneous catalysis reaction.
Hence, lead chamber process is an example of homogeneous catalysis.

Question 22.
Describe heterogeneous catalysis with the help of one example.
Answer:
i. When the reactants and catalyst are in different phase, it is said to be heterogeneous catalysis.
ii. The heterogeneous catalyst is generally a solid and the reactants may either be gases or liquids.
iii. When the solid catalyst is added to the reaction mixture, it does not dissolve in the reacting system and the reaction occurs on the surface of the solid catalyst.
e.g. Dinitrogen (N2) and dihydrogen (H2) combine to form ammonia in Haber process in the presence of finely divided iron along with K2O and Al2O3.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 5
b. In the above reaction, Al2O3 and K2O are promoters of the Fe catalyst. Al2O3 is added to prevent the fusion of Fe particles. K2O causes chemisorption of nitrogen atoms. Molybdenum is also used as promoter.
c. Since the reactants are present in gaseous phase while the catalyst used is in solid phase, it represents heterogeneous catalysis.

Question 23.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 6
i. State whether the given reaction is an example of heterogeneous or homogeneous catalysis.
ii. What is the role of Fe, K2O and Al2O3 in this reaction?
Answer:
i. This reaction is an example of heterogeneous catalysis.
ii. Fe is used as a catalyst while K2O and Al2O3 are promoters of the Fe catalyst. Al2O3 is used to prevent the fusion of Fe particles while K2O causes chemisorption of nitrogen atoms.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 24.
Describe hydrogenation reaction of vegetable oils.
Answer:
i. Hydrogenation reaction of vegetable oils used in food industry to produce solid fats. The reaction is as follows:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 7
ii. The reaction is catalysed by finely divided metals like Ni, Pd or Pt.
iii. Vegetable oil contains one or more carbon-carbon double bonds (C = C) in its structure.
iv. On hydrogenation, a solid product (which contains only carbon-carbon single bonds) is formed. It is called Vanaspati ghee.
v. The hydrogenation reaction of vegetable oils is an example of heterogeneous catalysis as the reactant and the catalyst are not present in the same phase.

Question 25.
i. Explain the role of catalytic converters in automobile exhaust.
ii. Why do automobiles with catalytic converter require unleaded petrol?
Answer:
i. a. An important application of heterogeneous catalysts is in automobile catalytic converters.
b. In automobile exhaust, large number of air pollutants such as carbon monoxide, nitric oxide, etc. are present.
c. The catalytic converter transforms these air pollutants into carbon dioxide, water, nitrogen and oxygen.
ii. The catalyst used in the catalytic converter gets poisoned by the adsorption of lead (Pb) present in the petrol. Hence, the automobiles with catalytic converter requires unleaded petrol.

Question 26.
What are inhibitors? Explain with an example.
Answer:
Inhibitors are substances that decreases the rate of chemical reactions.
e.g. Chloroform forms poisonous substance, carbonyl chloride, by air oxidation.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 8
When 2% ethanol is added to chloroform, the formation of COCl2 is suppressed because ethanol acts as an inhibitor and retards the above reaction.
[Note: Chloroform was earlier used as an anaesthetic.]

Question 27.
Write decomposition reaction of hydrogen peroxide. Suggest how this decomposition can be prevented.
Answer:
i. Hydrogen peroxide decomposes as,
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 9
ii. The reaction can be inhibited by addition of dilute acid or glycerol as they act as inhibitors.

Question 28.
Explain why 2% ethanol is added to chloroform?
Answer:
Inhibitors are substances that decreases the rate of chemical reactions.
e.g. Chloroform forms poisonous substance, carbonyl chloride, by air oxidation.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 10
When 2% ethanol is added to chloroform, the formation of COCl2 is suppressed because ethanol acts as an inhibitor and retards the above reaction.
[Note: Chloroform was earlier used as an anaesthetic.]

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 29.
Describe the steps involved in heterogeneous catalysis by solid catalyst.
OR
Explain the mechanism involved in catalytic action of a heterogeneous catalyst.
Answer:
The catalytic action of a heterogeneous catalyst occurs on the surface of a catalyst.
The mechanism involves the following five steps.
i. Diffusion of reactants towards the surface of the catalyst.
ii. Adsorption of reactant molecules on the surface of the catalyst.
iii. Occurrence of chemical reaction on the catalyst surface and formation of an intermediate.
iv. Formation of the products.
v. Desorption of reaction products from the catalyst surface. Products leave the catalyst surface in the following steps.
Steps involved in desorption of reaction products:
Diffusion → Adsorption → Intermediate formation → Product formation → Desorption
vi. Fresh reactant molecules can replace the products to start the cycle again as in first step.
vii. This is why catalyst remains unchanged in mass and chemical composition at the end of the reaction.

Question 30.
Write a short note on catalytic activity.
Answer:

  • The catalytic activity of a catalyst depends on the strength of chemisorption.
  • If large number of reactant molecules (gas or liquid) are strongly adsorbed on the surface of solid catalyst, the catalyst is said to be active.
  • However, the adsorption of reactant molecules on the surface, that is, the bond formed between adsorbate and adsorbent surface should not be very strong so that they are not immobilized.
  • d-block metals such as Fe, V and Cr tend to be strongly active towards O2, C2H2, C2H4, CO, H2, CO2, N2, etc.
  • Mn and Cu are unable to adsorb N2 and CO2.
  • The metals Mg and Li adsorb O2 selectively.

Question 31.
Explain catalytic selectivity with suitable examples.
Answer:
i. Some solid catalysts are selective in their action.
ii. The same gaseous reactants produce different products when different catalysts are used.
e.g.
a. The gaseous ethylene and O2 react to produce different products with different catalysts.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 11

b. The gaseous carbon monoxide and H2 produce different products by using different catalysts.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 12

Question 32.
i. What are zeolites?
ii. Zeolites are shape selective catalysts. Explain.
iii. What is the use of a zeolite catalyst ZSM-5 in petroleum industry?
Answer:
i. a. Zeolites are aluminosilicates with three-dimensional network of silicates.
b. Some silicon atoms in this network are replaced by aluminium atoms giving Al – O – Si framework which results in microporous structure.

ii. a. The reactions in zeolites are dependent on the size and shape of reactant or products, b. It also depends on the pores and cavities of zeolites.
b. Therefore, zeolites are shape selective catalysts.

iii. In petroleum industry, zeolite catalyst ZSM-5 converts alcohols directly to gasoline (petrol) by dehydration which gives a mixture of hydrocarbons.

Question 33.
State the importance of colloids in day-to-day life.
Answer:

  • Colloid chemistry is the chemistry of everyday life.
  • A number of substances we use in our day-to-day life are colloids. For example, milk, butter, jelly, whipped cream, mayonnaise.
  • Knowledge of colloid chemistry is essential for understanding about many useful materials like cement, bricks, pottery, porcelain, glass, enamels, oils, lacquers, rubber, celluloid and other plastics, leather, paper, textiles, filaments, crayons, inks, road construction material, etc.
  • In many daily processes like cooking, washing, dyeing, painting, ore floatation, water purification, sewage disposal, smoke prevention, photography, pharmacy, use of colloids is important.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 34.
What are colloids? Explain.
Answer:
i. Colloids are heterogeneous mixtures.
ii. The component of colloid present in the largest proportion is called dispersion medium and the other components are called dispersed phase.
iii. The particles of the dispersed phase are larger than the size of a molecule and smaller than the particles which we can see with naked eye.
e.g.

  • Observe the formation of solution of salt and water. Salt dissolves completely in water and forms homogeneous system.
  • On the other hand, ground coffee or tea leaves with milk form suspension.
  • Between the two extremes of solution and suspension exists a large group of systems called colloidal dispersions or simply colloids.

Question 35.
State the differences between colloids and solutions.
Answer:
Colloids:

  1. Colloids contain particles of dispersed phase with diameters in the range of 2 to 500 nm.
  2. They are translucent to light.
  3. e.g. Milk, fog, etc.

Solutions:

  1. Solutions contain solute particles with diameters in the range of 0.1 to 2 nm.
  2. They are transparent or may be coloured.
  3. e.g. NaCl solution

Question 36.
Explain: Natural phenomena of colloids observed in daily life.
Answer:
Following are some examples of colloids observed in daily life.
i. Blue colour of the sky: The sky appears blue to us because minute dust particles along with minute water droplets dispersed in air scatter blue light which reaches our eyes.
ii. Blood: It is a colloidal dispersion of plasma proteins and antibodies in water arid at the same time blood is also a suspension of blood cells and platelets in water.
iii. Soils: Fertile soils are colloidal in nature where humus acts as a protective colloid. Soil adsorbs moisture and nourishing materials due to its colloidal nature.
iv. Fog, mist and rain:

  • Mist is caused by small droplets of water dispersed in air.
  • Fog is formed whenever there is temperature difference between ground and air.
  • A large portion of air containing dust particles gets cooled below its dew point, the moisture from the air condenses on the surface of these particles which form fine droplets, which are colloidal particles and float in the air as fog or mist.

Question 37.
State different ways to classify colloids.
Answer:
Colloids can be classified in three different ways:

  • Physical states of dispersed phase and dispersion medium
  • Interaction or affinity of phases
  • Molecular size

Question 38.
Name the types of colloids based on the physical states of dispersed phase and dispersion medium. Give two examples of each.
Answer:
There are eight types of colloids based on the physical states of dispersed phase and dispersion medium as given below.

Sr. No. Type of Colloids Examples
i. Solid sol (solid dispersed in solid) Coloured glasses, gemstones
ii. Sols and gels (solid in liquid) Gelatin, muddy water
iii. Aerosol (solid in gas) Smoke, dust
iv. Gel (liquid in solid) Cheese, jellies
v. Emulsion (liquid in liquid) Milk, hair cream
vi. Aerosol (liquid in gas) Fog, mist
vii. Solid sol (gas in solid) Foam rubber, plaster
viii. Foam (gas in liquid) Froth, soap lather

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 39.
Complete the following chart.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 13
Answer:
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 14
[Note: Students can write any one example of the given type of colloids.]

Note: Types of colloids based on the physical states of dispersed phase and dispersion medium.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 15

Question 40.
Describe classification of colloids based on the interaction or affinity of phases.
Answer:
On the basis of interaction or affinity of phases, a colloidal solution is classified as lyophilic and lyophobic.
i. Lyophilic colloids:

  • A colloidal solution in which the particles of dispersed phase have a great affinity for the dispersion medium are lyophilic colloids.
  • If the lyophilic sol is evaporated, the dispersed phase separates. However, if it is remixed with the medium, the sol. can be formed again and hence, such sols are called reversible sols.
  • They are stable and difficult to coagulate.

ii. Lyophobic colloids:

  • Colloidal solution in which the particles of the dispersed phase have no affinity for the dispersion
    medium are called lyophobic colloids.
  • The common examples are Ag, Au, hydroxides like Al(OH)3, Fe(OH)3, metal sulphides.
  • Once precipitated or coagulated they have little tendency or no tendency to revert back to colloidal state.

[Note: Lyo means liquid and philic means loving whereas phobic means fearing and hence liquid hating. If water is the dispersion medium, the terms hydrophilic and hydrophobic are used.]

Question 41.
Give reason: Lyophilic sols are called reversible sols.
Answer:

  • When lyophilic sol is evaporated, the dispersed phase separates.
  • However, if the dispersed phase is remixed with the medium, the sol can be formed again.

Hence, lyophilic sols are called reversible sols.

Question 42.
How are colloids classified based on their molecular size?
Answer:
Colloids are classified into three types based on their molecular size as described below.
i. Multimolecular colloids:

  • In multimolecular colloids, the individual particles consist of an aggregate of atoms or small molecules with size less than 103 pm.
    e.g. Gold sol consists of particles of various sizes having several gold atoms.
  • Colloidal solution in which particles are held together with van der Waals force of attraction is called multimolecular colloid.
    e.g. S8 sulphur molecules

ii. Macromolecular colloids: In this type of colloids, the molecules of the dispersed phase are sufficiently large in size (macro) to be of colloidal dimensions.
e.g. Starch, cellulose, proteins, polythene, nylon, plastics.

iii. Associated colloids or micelles:

  • The substances behave as normal electrolytes at low concentration and associated in higher concentration forming a colloidal solution.
  • The associated particles are called micelles, e.g. Soaps and detergents

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 43.
How can be colloids prepared by chemical methods?
Answer:
i. Colloidal dispersions can be prepared by chemical reactions leading to formation of molecules by double decomposition, oxidation, reduction or hydrolysis.
ii. Molecules formed in these reactions are water-insoluble and thus, they aggregate leading to the formation of colloids.
e.g.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 16

Question 44.
Describe the process involved in peptization?
Answer:

  • During peptization a precipitate is converted into colloidal sol by shaking with dispersion medium in the presence of a small amount of an electrolyte. The electrolyte used is known as peptizing agent.
  • During the process, the precipitate adsorbs one of the ions of the electrolyte on its surface and as a result, positive or negative charge is developed on the precipitate which finally breaks up into small particles of colloidal size.

[Note: This method is generally applied to convert a freshly prepared precipitate into a colloidal sol.]

Question 45.
Why is it necessary to purify colloidal solutions?
Answer:

  • Colloidal solution generally contains excessive amount of electrolytes and some other soluble impurities.
  • A small quantity of an electrolyte is necessary for the stability of colloidal solution, however, a large quantity of electrolyte may result in coagulation.
  • It is also necessary to reduce soluble impurities.

Hence, it is necessary to purify colloidal solutions.

Question 46.
i. What is purification of colloidal solution?
ii. How can a colloidal solution be purified using the method of dialysis?
Answer:
i. The process used for reducing the amount of impurities to a requisite minimum is known as purification of colloidal solution.
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 17

ii. a. Dialysis is a process of removing a dissolved substance from a colloidal solution by diffusion through a suitable membrane.
b. Purification of colloidal solution can be carried out using dialysis by the following method.

  • The apparatus used is dialyser.
  • A bag of suitable membrane containing the colloidal solution is suspended in a vessel through which fresh water is continuously flowing.
  • The molecules and ions diffuse through membrane into the outer water and pure colloidal solution is left behind.

Question 47.
What are the general properties exhibited by colloidal dispersions?
Answer:
General properties exhibited by colloidal dispersions:

  • Colloidal system is heterogeneous and consists of two phases, dispersed phase and dispersion medium.
  • The dispersed phase particles pass slowly through parchment paper or animal membrane, but readily pass through ordinary filter paper.
  • Colloidal particles are usually not detectable by powerful microscope.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 48.
Discuss the factors that influence the colour of colloidal solutions.
Answer:

  • Colour of colloidal solution depends on the wavelength of light scattered by dispersed particles.
  • The colour of colloidal dispersion also changes with the manner in which the observer receives the light.
    e.g. Mixture of a few drops of milk and large amount of water appears blue when viewed by the scattered light and red when viewed by transmitted light.
  • It also depends on size of colloidal particles.
    e.g. Finest gold sol is red in colour whereas with increase in size it appears purple.

Question 49.
Give three examples each:
i. Positively charged sols
ii. Negatively charged sols
Answer:
i. Positively charged sols: Al2O3. xH2O, haemoglobin, TiO2 sol
ii. Negatively charged sols: Au sols, Congo red sol, clay

Note: Some common sols with the nature of charge on the particles are listed in the table below.

Positively charged sols Negatively charged sols
Hydrated metallic oxides: Al2O3.xH2O, CrO3.xH2O, Fe2O3.xH2O. Metals: Cu, Ag. Au sols

Metallic sulphides: As2S3, Sb2S3, CdS

Basic dye stuff, methylene blue sols Acid dye stuff, eosin, Congo red sol
Haemoglobin (blood) Sols of starch, gum
Oxides: TiO2 sol Gelatin, clay, gum sols

Question 50.
Explain the term electroosmosis.
Answer:

  • Movement of dispersed particles can be prevented by suitable means such as use of membrane.
  • On doing so, it is observed that the dispersion medium begins to move in an electric field. This is known as electroosmosis.

Question 51.
What is coagulation?
Answer:
The precipitation of colloids by removal of charge associated with colloidal particles is called coagulation.

Question 52.
How can we bring about precipitation of lyophobic colloids?
Answer:

  • The charge on the colloidal particles is due to the preferential adsorption of ions on their surface.
  • Hence, lyophobic colloids can be precipitated out by removing the charge on the colloidal particles (dispersed phase).

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 53.
Discuss various methods that are used to bring about coagulation of lyophobic sols.
Answer:
Coagulation of the lyophobic sols can be carried out in the following ways.

  • By electrophoresis: The colloidal particles move towards oppositely charged electrodes, get discharged and precipitate.
  • By mixing two oppositely charged sols: Oppositely charged sols when mixed in almost equal proportions neutralize their charges and get precipitated.
    e. g. Mixing of hydrated ferric oxide (positive sol) and arsenious sulphide (negative sol) brings them in the precipitated forms. This type of coagulation is called mutual coagulation.
  • By boiling: When a sol is boiled, the adsorbed layer is disturbed as a result of increased collisions with molecules in the dispersion medium. This reduces charge on the particles and subsequently particles settle down as a precipitate.
  • By persistent dialysis: On prolonged dialysis, traces of the electrolyte present in the sol are removed almost completely. The colloids then become unstable and finally precipitate.
  • By addition of electrolytes: When excess of an electrolyte is added, the colloidal particles are precipitated.

Question 54.
Write Hardy-Schulze rule.
Answer:
Generally, greater the valency of the flocculating ion added, greater is its power to cause precipitation. This is known as Hardy-Schulze rule.

Question 55.
Differentiate between oil in water and water in oil emulsions.
Answer:
Oil in water:

  1. Oil is the dispersed phase and water is the dispersion medium.
  2. If water is added, it will be miscible with the emulsion.
  3. Addition of small amount of an electrolyte makes the emulsion conducting.
  4. Continuous phase is water.
  5. Basic metal sulphates, water soluble alkali metal soaps are used as emulsifiers.

Water in oil:

  1. Water is the dispersed phase and oil is the dispersion medium.
  2. If oil is added, it will be miscible with the emulsion.
  3. Addition of small amount of an electrolyte has no effect on conducting power.
  4. Continuous phase is oil.
  5. Water insoluble soaps such as those of Zn, Al, Fe, alkaline earth metals are used as emulsifiers.

Question 56.
What are the properties of emulsion?
Answer:
Properties of emulsion:

  • Emulsion can be diluted with any amount of the dispersion medium. On the other hand, the dispersed liquid when mixed forms a separate layer.
  • The droplets in emulsions are often negatively charged and can be precipitated by electrolytes.
  • Emulsions show Brownian movement and Tyndall effect.
  • The two liquids in emulsions can be separated by heating, freezing, centrifuging, etc.

Question 57.
Give applications of colloids.
Answer:
Applications of colloids:
i. Electrical precipitation of smoke:

  • Smoke is a colloidal solution of solid particles of carbon, arsenic compound, dust, etc. in the air.
  • When smoke is allowed to pass through chamber containing charged plates, smoke particles lose their charge and get precipitated. The particles then settle down on the floor of the chamber.
  • The precipitator used is called Cottrell precipitator.

ii. Purification of drinking water:

  • Water obtained from natural sources contains colloidal impurities.
  • By addition of alum to such water, colloidal impurities get coagulated and settle down. This makes water potable.

iii. Medicines:

  • Usually medicines are colloidal in nature.
  • Colloidal medicines are more effective owing to large surface area to volume ratio of a colloidal particle and easy assimilation.
    e.g. Argyrol is a silver sol used as an eye lotion. Milk of magnesia, an emulsion is used in stomach disorders.

iv. Rubber industry: Rubber is obtained by coagulation of latex.
v. Cleansing action of soaps and detergents.
vi. Photographic plates, films, and industrial products like paints, inks, synthetic plastics, rubber, graphite lubricants, cement, etc. are colloids.

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

Question 58.
Match column A with column B.

Column A Column B
i. Tyndall effect i. Kinetic property
ii. Electrophoresis ii. Argyrol
iii. Silver sol iii. Optical property
iv. Brownian motion iv. Coagulation

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

Question 59.
In drinking water treatment, often alum is added for the complete removal of suspended impurities. On complete dissolution, alum produces positive charge which neutralizes the charge on the suspended particles and thus, impurities are easily removed.
i. Name and define the process involved due to which charge on particles get neutralized.
ii. What is the role of alum in the above mentioned process?
Answer:
i. a. Charge on particles get neutralized due to coagulation.
b. The precipitation of colloids by removal of charge associated with colloidal particles is called coagulation.
ii. Alum acts as a reagent that helps in coagulation of the suspended particles by the removal of the charge associated with these particles.

Multiple Choice Questions

1. Which of the following is responsible for adsorption phenomenon?
(A) Hydrogen bonding
(R) Dipole-dipole forces
(C) Ion-dipole forces
(D) Dispersion forces
Answer:
(D) Dispersion forces

2. A substance which adsorbs another substance on its surface is called ……………..
(A) adsorbate
(B) absorbate
(C) adsorbent
(D) absorbent
Answer:
(C) adsorbent

3. During adsorption, the molecules of the substance which gets adsorbed are termed as
(A) adsorbent
(B) adsorbate
(C) absorbent
(D) absorbate
Answer:
(B) adsorbate

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

4. in adsorption of acetic acid on charcoal, acetic acid is ……………
(A) adsorhate
(B) adsorbent
(C) absorbent
(D) absorbate
Answer:
(A) adsorhate

5. The process of removal of an adsorbed substance from the surface is known as
(A) sorption
(B) oxidation
(C) reduction
(D) desorption
Answer:
(D) desorption

6. ………….. is the process in which adsorbate molecules are held on the surface of the adsorbent by weak van der Waals forces.
(A) Chemisorption
(B) Absorption
(C) Physisorption
(D) Biosorption
Answer:
(C) Physisorption

7. Which of the following is an example of physical adsorption?
(A) Adsorption of acetic acid in solution by charcoal
(B) Adsorption of O2 on tungsten
(C) Adsorption of N2 on Fe
(D) Adsorption of H2 on Ni
Answer:
(A) Adsorption of acetic acid in solution by charcoal

8. Chemisorption is a slow process because …………….
(A) it forms multimolecular layer
(B) it is reversible
(C) it takes place at normal temperature
(D) it requires high activation energy
Answer:
(D) it requires high activation energy

9. The number of layer(s) formed on adsorbent in chemical adsorption is …………….
(A) one
(B) two
(C) three
(D) many
Answer:
(A) one

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

10. Which of the following statements is CORRECT regarding chemical adsorption?
(A) It is highly specific in nature.
(B) It is relatively strong.
(C) It involves the formation of monolayer of adsorbed particles.
(D) All of these.
Answer:
(D) All of these.

11. Which of the following is adsorbed to maximum extent on charcoal?
(A) H2
(B) N2
(C) Cl2
(D) O2
Answer:
(C) Cl2

12. The relation between the amount of substance adsorbed by an adsorbent and the equilibrium pressure or …………. at any constant temperature is called adsorption isotherm.
(A) surface area
(B) volume
(C) circumference
(D) concentration
Answer:
(D) concentration

13. For equilibrium pressure (P), the mass of gas adsorbed (x) and mass of adsorbent (m) may be expressed as Freundlich adsorption isotherm as ……………
Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids 18
Answer:
(B) \(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kP}^{\frac{1}{\mathrm{n}}}\)

14. When log x/m is plotted against log P, the intercept obtained …………..
(A) on Y axis is equal to log K
(B) on Y axis is equal to K
(C) on X axis is equal to log K
(D) on X axis is equal to K
Answer:
(A) on Y axis is equal to log K

15. The adsorption isotherm tends to saturate at ………….. pressure.
(A) low
(B) moderate
(C) all of these
(D) high
Answer:
(D) high

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

16. In Haber process for manufacture of NH3, the catalyst used is ……………
(A) iron
(B) copper
(C) vanadium pentoxide
(D) nickel
Answer:
(A) iron

17. A substance that decreases the rate of a chemical reaction is called ……………
(A) inhibitor
(B) prohibitor
(C) promoter
(D) reactor
Answer:
(A) inhibitor

18. Whether a given mixture forms a true solution or a colloidal dispersion depends on the …………….
(A) charge of solute particles
(B) size of solvent particles
(C) size of solute particles
(D) charge of solvent particles
Answer:
(C) size of solute particles

19. An aerosol is a dispersion of a ……………
(A) gas in a solid
(B) liquid in a gas
(C) solid in a gas
(D) both (B) and (C)
Answer:
(D) both (B) and (C)

20. The dispersed phase in Pumice stone is ……………
(A) solid
(B) liquid
(C) gas
(D) none of these
Answer:
(C) gas

21. Colloidal solution in which the dispersed phase has little affinity for the dispersion medium is called ………………
(A) lyophobic colloids
(B) lyophilic colloids
(C) hydrophilic colloids
(D) emulsions
Answer:
(A) lyophobic colloids

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

22. Which of the following is NOT an example of macromolecular colloid?
(A) Starch
(B) Proteins
(C) S8 molecules
(D) Nylon
Answer:
(C) S8 molecules

23. Tyndall effect is useful ……………….
(A) to identify colloidal dispersions
(B) to count number of particles in colloidal dispersion.
(C) to determine the size of the colloidal particles
(D) all of these
Answer:
(D) all of these

24. Brownian movement is a ……………… type of property of the colloidal sol.
(A) electrical
(B) optical
(C) kinetic
(D) colligative
Answer:
(C) kinetic

25. The migration of colloidal particles under the influence of an electric field is called …………….
(A) catalysis
(B) Brownian movement
(C) electrophoresis
(D) Tyndall effect
Answer:
(C) electrophoresis

26. The capacity of an ion to coagulate a colloidal solution depends on ……………….
(A) its shape
(B) its valency
(C) the sign of charge
(D) both (B) and (C)
Answer:
(D) both (B) and (C)

Maharashtra Board Class 11 Chemistry Important Questions Chapter 11 Adsorption and Colloids

27. ……………… is an example of water in oil type of emulsion.
(A) Milk
(B) Cod liver oil
(C) Vanishing cream
(D) Paint
Answer:
(B) Cod liver oil

28. Which of the following has highest precipitation power to precipitate negative sol?
(A) Al3+
(B) Mg2+
(C) Na+
(D) K+
Answer:
(A) Al3+