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MP Board Class 12th Chemistry Important Questions Chapter 15 Polymers

Polymers Important Questions

Polymers Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
The polymerization in which two or more chemically different monomers take part is called :
(a) Addition polymerization
(b) Copolymerization
(c) Chain polymerization
(d) Homogeneous polymerization.
Answer:
(b) Copolymerization

Question 2.
Natural rubber is mainly a polymer of:
(a) Chloroprene
(b) Neoprene
(c) Isoprene
(d) Butadiene.
Answer:
(c) Isoprene

Question 3.
Which is a heat resistant polymer :
(a) P.V.C.
(b) P.V.A.
(c) Bakelite
(d) Rubber.
Answer:
(c) Bakelite

Question 4.
Is not a polymer :
(a) Orlon
(b) Teflon
(c) Neoprene soprene.
(d) Isoprene
Answer:
(d) Isoprene

Question 5.
Which among the following is a thermosetting polymer :
(a) P.V.C.
(b) P.V.A.
(c) Bakelite
(d) Perspex.
Answer:
(c) Bakelite

Question 6.
Which substance is used as ‘non stick’ in cooking utencils :
(a) P.V.C.
(b) Polystyrene
(c) Poly ethylene Pterephthalate
(d) Poly tetrafluoro ethylene.
Answer:
(d) Poly tetrafluoro ethylene.

Question 7.
Out of the following which polymer contain nitrogen :
(a) Nylon
(b) Polythene
(c) P.V.C.
(d) Terylene.
Answer:
(a) Nylon

Question 8.
Is a natural polymer :
(a) Starch
(b) Nylon
(c) Teflon
(d) Buna – S – Rubber
Answer:
(a) Starch

Question 9.
Is a polymer :
(a) Macro molecule
(b) Micromolecule
(c) Submicro molecule
(d) None of these
Answer:
(a) Macro molecule

Question 10.
P.V.C is a polymer of the following :
(a) CH₂ = CH₂
(b) CH₂ = CHCl
(c) ClCH₂ = CH₂Cl
(d) Cl – C ≡ C – Cl
Answer:
(b) CH₂ = CHCl

Question 11.
Teflon is a polymer of:
(a) Vinyl chloride
(b) Ethylene
(c) Acetylene
(d) Tetrafluroethene
Answer:
(d) Tetrafluroethene

Question 12.
Example of condensation polymeris :
(a) Polythene
(b) P.V.C.
(c) Orion
(d) Terylene
Answer:
(d) Terylene

Question 13.
Intermolecular force in elastomer is :
(a) Not present
(b) Weak
(c) Strong
(d) Extremely strong.
Answer:
(b) Weak

Question 14.
Complete hydrolysis of cellulose gives :
(a) D – Fructose
(b) D – Ribose
(c) D – Glucose
(d) L – Glucose
Answer:
(c) D – Glucose

Question 15.
Cellulose is a :
(a) Protein
(b) Fat
(c) Hormone
(d) Polysaccharide
Answer:
(d) Polysaccharide

Question 16.
Which of the following is a natural polymer :
(a) Starch
(b) Nylon
(c) Teflon
(d) Buna – s – Rubber
Answer:
(a) Starch

Question 17.
Nylon is an example of:
(a) Polyamide
(b) Polythene
(c) Polyester
(d) Polysaccharide
Answer:
(a) Polyamide

Question 18.
Nylon 6,6 is not a :
(a) Condensation polymer
(b) Co – Polymer
(c) Polyamide Bakelite is a polymer of:
(d) Homopolymer
Answer:
(d) Homopolymer

Question 19.
Bakelite is a polymer of :
(a) HCHO and acetic acid
(b) HCHO and phenol
(c) C₂H₅ – OH and phenol
(d) CH₃ – COOH and benzene.
Answer:
(b) HCHO and phenol

Question 20.
Which of the following is a biodegradable polymer :
(a) Cellulose
(b) Polythene
(c) Polyvinyl chloride
(d) Nylon 6.
Answer:
(a) Cellulose

Question 2.
Fill in the blanks :

1. ……………… is used for the preparation of chloroprene.
2. Charge on polymers is ………………
3. Polymer ……………… the light.
4. Molecular mass of polymers is ………………
5. Glucose is a monomer of ………………
6. Cellulose is a ……………… polymer. (MP 2015)
7. Polymer of ethylene glycol and phthalic acid is ………………
8. Rubber is a ……………… polymer.
9. Vulcanisation of rubber is an example of ………………
10. Bakelite is a ……………… polymer.
11. Nylon 6 is also called ……………… (MP 2011)
12. Teflon is a polymer of ……………… (MP 2103)

Answer:

1. Synthetic rubber
2. Nil (zero)
3. Scatter
4. High
5. Cellulose and starch
6. Natural
7. Glyptal
8. Natural
9. Elastomer
10. Heat resistant
11. Perlon – L
12. Tetra fluoro ethylene.

Question 3.
Make correct pairs :

Answer:

1. (d)
2. (a)
3. (c)
4. (f)
5. (b)
6. (g)
7. (e)
8. (h)

Question 4.
Answer in one word / sentence :

1. Give two examples of natural polymer.
2. Give two examples of addition polymer.
3. Give two examples of condensation polymer.
4. Write chemical name of Buna rubber.
5. Give an example of synthetic rubber.
6. Monomer of polythene is. (MP 2011)
7. Name the polymer which is formed by condensation of ethylene glycol and dimethyl teraphthalic acid. (MP 2011)
8. Name the polymerisation which takes place by addition of two or more than two different monomers. (MP 2010)
9. Give the name of polymer used for formation of tyre thread. (MP 2010)
10. Which polymer obtained by polymerisation of caprolactum?

Answer:

1. Natural polymer – Rubber, starch
2. Polythene, Polypropylene
3. Nylon – 6, Bakelite
4. Styrene Butadiene rubber
5. Styrene Butadiene rubber (S.B.R.)
6. Ethylene
7. Terylene
8. Copolymerisation
9. Nylon – 6
10. Nylon – 6

MP Board Class 12th Chemistry Important Questions

MP Board Class 12th Maths Important Questions Chapter 3 Matrices

Matrices Important Questions

Matrices Objective Type Questions:

Question 1.
If A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\) and A + A’ = I, then the value of α is:
(a) \(\frac { \pi }{ 6 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) π
(d) \(\frac { 3\pi }{ 2 } \)
Answer:
(b) \(\frac { \pi }{ 3 } \)

Question 2.
If A = \(\left[\begin{array}{lll}
{2} & {0} & {0} \\
{0} & {2} & {0} \\
{0} & {0} & {2}
\end{array}\right]\), then A⁵ is equal to:
(a) 5 A
(b) 10 A
(c) 16 A
(d) 32 A
Answer:
(c) 16 A

Question 3.
If a matrix is both symmetric and skew – symmetric, then:
(a) A is a diagonal matrix
(b) A is zero matrix
(c) A is a square matrix
(d) None of these
Answer:
(b) A is zero matrix

Question 4.
If A = \(\begin{bmatrix} \alpha & \beta \\ \lambda & -\alpha \end{bmatrix}\) is such that A² = I, then:
(a) 1 + α² + βλ = 0
(b) 1 – α² + βλ = 0
(c) 1 – α² – βλ = 0
(d) 1 + α² – βλ = 0
Answer:
(c) 1 – α² – βλ = 0

Question 5.
If A = \(\begin{bmatrix} 2 & -1 \\ 3 & -2 \end{bmatrix}\), then Aⁿ = ………………………:
(a) A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), if n is even natural number
(b) A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), if n is odd natural number
(c) A = \(\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\), if n ∈ N
(d) None of these
Answer:
(a) A = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), if n is even natural number

Question 2.
Fill in the blanks:

1. If A = \(\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}\) and B = \(\begin{bmatrix} 1 & 3 \\ 2 & 5 \end{bmatrix}\), then AB = …………………………….
2. If A = diag [1, -1, 2] and B = diag [2, 3, -1], then value of 3A + 4B will be ……………………
3. A matrix A is said to be idempotent if ………………………
4. A matrix A is said to be orthogonal if ……………………..
5. If [x, 1] \(\begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix}\) = 0, then find the value of x will be …………………………….

Answer:

1. \(\begin{bmatrix} 10 & 26 \\ 7 & 19 \end{bmatrix}\)
2. diag [11, 9, 2]
3. A² = A
4. AA’ = A’A = I
5. x = 2

Question 3.
Write True/False:

1. Multiplication of matrix is always commutative?
2. Two matrix are said to be comparable if they have same number of rows and columns?
3. If A is a square matrix, then A. adj A = |A| I?
4. A square matrix A is said to be symmetric if A = – A^T?
5. Matrix A and B are inverse of each other if AB = BA?

Answer:

1. False
2. True
3. True
4. False
5. False

Question 4.
Match the Column:

Answer:

1. (d)
2. (e)
3. (a)
4. (b)
5. (c)

Question 5.
Write the answer in one word/sentence:

1. If A and B are two square matrix of same order, then what is the value of Adj (AB)?
2. A square matrix A is said to be Involountary matrix if?
3. If A = \(\begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}\), then find the value of A²?
4. If A = [1, 2, 3], then find the value of AA^T?
5. If X + Y = \(\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}\) and X – Y = \(\begin{bmatrix} 3 & 2 \\ -1 & 0 \end{bmatrix}\), then find the value of X?

Answer:

1. Adj.(AB) = (Adj B). (Adj A)
2. A² = I
3. -I
4. [1, 4]
5. \(\begin{bmatrix} 2 & 0 \\ 1 & 2 \end{bmatrix}\)

Matrices Short Answer Type Questions

Question 1.
If A = \(\begin{bmatrix} a^{ 2 }+b^{ 2 } & b^{ 2 }+c^{ 2 } \\ a^{ 2 }+c^{ 2 } & a^{ 2 }+b^{ 2 } \end{bmatrix}\) and B = \(\begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix}\), then find the value of A + B? (NCERT)
Solution:
A + B =

Question 2.
If A = \(\begin{bmatrix} cos^{ 2 }x & sin^{ 2 }x \\ sin^{ 2 }x & cos^{ 2 }x \end{bmatrix}\) and B = \(\begin{bmatrix} sin^{ 2 }x & cos^{ 2 }x \\ cos^{ 2 }x & sin^{ 2 }x \end{bmatrix}\), then find A + B? (NCERT)
Solution:
A + B = \(\begin{bmatrix} cos^{ 2 }x & sin^{ 2 }x \\ sin^{ 2 }x & cos^{ 2 }x \end{bmatrix}\) + \(\begin{bmatrix} sin^{ 2 }x & cos^{ 2 }x \\ cos^{ 2 }x & sin^{ 2 }x \end{bmatrix}\)

⇒ A + B = \(\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\)

Question 3.
If A = and B = then find 3A – 5B? (NCERT)
Solution:

Question 4.
Simplify: cos θ \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\) + sinθ \(\begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}\)
Solution:
cos θ \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\) + sinθ \(\begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}\)

Question 5.
From the following equation find the value of x and y?
2 \(\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix}\) + \(\begin{bmatrix} 3 & -4 \\ 1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 7 & 6 \\ 15 & 14 \end{bmatrix}\)? (NCERT)
Solution:

From defnition of matrix,
2x + 3 = 7
⇒ 2x = 4 ⇒ x = 2
⇒ 2y – 4 = 14
⇒ 2y = 18 ⇒ y = 9
∴x = 2, y = 9.

Question 6.
Find the value of X and Y if X + Y = \(\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}\) and X – Y = \(\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}\)? (NCERT)
Solution:
Given X + Y = \(\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}\) ……………….. (1)
and X – Y = \(\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}\) ………………….. (2)
adding eqns. (1) and (2),
2X = \(\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}\) + \(\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}\)
⇒ 2X = \(\begin{bmatrix} 5+3 & 2+6 \\ 0+0 & 9-1 \end{bmatrix}\)
⇒ 2X = \(\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}\)
⇒ X = \(\frac{1}{2}\) \(\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}\) = \(\begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}\)
Substracting eqn. (2) from eqn. (1),
2Y = \(\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}\) – \(\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}\)
⇒ 2Y = \(\begin{bmatrix} 5-3 & 2-6 \\ 0-0 & 9+1 \end{bmatrix}\)
⇒ Y = \(\frac{1}{2}\) \(\begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix}\) = \(\begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}\)

Question 7.
Find the value of x and y
2 \(\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}\) + \(\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\) (NCERT)
Solution:
Given:
2 \(\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}\) + \(\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}\) + \(\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 2+y & 6+0 \\ 0+1 & 2x+2 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}\)
By defnition of matrix,
2 + y = 5 ⇒ y = 3
2x + 2 = 8
⇒ x + 1 = 4
⇒ x = 3
∴ x = 3, y = 3.

Question 8.
If \(\left[\begin{array}{c}
{x+y+z} \\
{x+z} \\
{y+z}
\end{array}\right]\) = [ \(\begin{matrix} 9 \\ 5 \\ 7 \end{matrix}\) ] find the value of x, y, and z?
Solution: Given \(\left[\begin{array}{c}
{x+y+z} \\
{x+z} \\
{y+z}
\end{array}\right]\) = [ \(\begin{matrix} 9 \\ 5 \\ 7 \end{matrix}\) ]
By defnition of matrix,
x + y + z = 9
x + z = 5
y + z = 7
From eqns. (1) and (2),
x + y + z = 9
⇒ 5 + y = 9 ⇒ y = 4
From eqns. (1) and (3),
x + (y + z) = 9
⇒ x + 7 = 9
⇒ x = 2
Putting the value of x in eqn. (2),
2 + z = 5
⇒ z = 3
∴ x = 2, y = 4, z = 3.

Question 9.
If \(\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}\) = \(\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}\), then fund the value of x, y and z? (NCERT)
Solution:
Given:
\(\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}\) = \(\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}\)
By defnition of matrix,
x + y = 6 ……………….. (1)
xy = 8 ………………. (2)
5 + z = 5
⇒ z = 0
From eqn. (1), y = 6 – x
xy = 8
⇒ 6x – x² = 8
⇒ x² – 6x + 8 = 0
⇒ x² – 4x – 2x + 8 = 0
⇒ x (x – 4) – 2 ( x – 4) = 0
⇒ ( x – 2) (x – 4) = 0
⇒ x = 2, 4
When x = 2 then y = 6 – 2 = 4
When x = 4 then y = 6 – 4 = 2
So x = 2, y = 4, z = 0
x = 4, y = 2, z = 0.

Question 10.
If A = \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\) then prove that A² – 4A + 5I = 0?
Solution:
Given:
A = \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)
A² = A.A = \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\) \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)
= \(\begin{bmatrix} 1.1+(-1)2 & 1(-1)+(-1).3 \\ 2.1+3.2 & 2(-1)+3.3 \end{bmatrix}\)
= \(\begin{bmatrix} 1-2 & -1-3 \\ 2+6 & -2+9 \end{bmatrix}\)
= \(\begin{bmatrix} -1 & -4 \\ 8 & 7 \end{bmatrix}\)
4A = 4\(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\) = \(\begin{bmatrix} 4 & -4 \\ 8 & 12 \end{bmatrix}\)
5I = 5\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) = \(\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}\)
= \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
= 0. Proved.

Question 11.
If A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) then prove that A² – 6A + 17I = 0?
Solution:
A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\)
A² = A.A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\)
= \(\begin{bmatrix} 4-9 & -6-12 \\ 6+12 & -9+16 \end{bmatrix}\) = \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\)
∴ L.H.S = A² – 6A + 17I
= \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\) – 6 \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) + 17 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\) – \(\begin{bmatrix} 12 & -18 \\ 18 & 24 \end{bmatrix}\) + \(\begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix}\)
= \(\begin{bmatrix} -5-12+17 & -18+18+0 \\ 18-18+0 & 7-24+17 \end{bmatrix}\)
= \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\) = 0 = R.H.S. proved.

Question 12.
If A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\) then prove that A² – 5A + 7I = 0. (NCERT)
Solution:
A² = A.A = \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\) × \(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)
⇒ A² = \(\begin{bmatrix} 3\times 3-1\times 1 & 3\times 1+1\times 2 \\ -1\times 3+2\times -1 & -1\times 1+2\times 2 \end{bmatrix}\)
⇒ A² = \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\)
5A = 5\(\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\)
⇒ 5A = \(\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}\)
7I = 7\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
⇒ 7I = \(\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\)
∴ A² – 5A + 7I = \(\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}\) – \(\begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}\) + \(\begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}\)
⇒ A² – 5A + 7I = \(\begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix}\)
⇒ A² – 5A + 7I = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
⇒ A² – 5A + 7I = 0.

Question 13.
If A = \(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\) and I = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) then find the value of k if A² = kA – 2I? (NCERT)
Solution:
Given:
A² = kA – 2I
⇒ kA = A² + 2I
⇒ k \(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\) = \(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\) × \(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\) + 2 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}\) = \(\begin{bmatrix} 3\times 3-2\times 4 & 3\times -2+2\times 2 \\ 4\times 3-2\times 4 & 4\times -2+2\times 2 \end{bmatrix}\) + \(\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}\) = \(\begin{bmatrix} 1 & -2 \\ 4k & -4 \end{bmatrix}\) + 2\(\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}\) = \(\begin{bmatrix} 1+2 & 2+0 \\ 4+0 & -4+2 \end{bmatrix}\)
⇒ \(\begin{bmatrix} 3k & -2k \\ 4k & -2k \end{bmatrix}\) = \(\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\)
⇒ 2k = 2
⇒ k = 1.

Question 14.
If f(x) = x² – 2x – 3, then find the value of f(A) if A = \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\)
Solution:
f(x) = x² – 2x – 3
∴ f(A) = A² – 2A – 3I
⇒ f(A) = \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\) × \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\) – 2 \(\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}\) – 3 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)

⇒ f(A) = \(\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}\) – \(\begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix}\) – \(\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\)
⇒ = \(\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}\) – \(\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
⇒ = 0.

Question 15.
If matrix A = \(\left[\begin{array}{ccc}
{0} & {a} & {-3} \\
{2} & {0} & {-1} \\
{b} & {1} & {0}
\end{array}\right]\) skew – symmetric matrix, then find the value of ‘a’ and ‘b’? (CBSE 2018)
Solution:
If A is skew symmmetric matrix then A’ = -A
Given:
A = \(\left[\begin{array}{ccc}
{0} & {a} & {-3} \\
{2} & {0} & {-1} \\
{b} & {1} & {0}
\end{array}\right]\)
A’ = \(\left[\begin{array}{ccc}
{0} & {2} & {b} \\
{a} & {0} & {1} \\
{-3} & {-1} & {0}
\end{array}\right]\)
– A = \(\left[\begin{array}{ccc}
{0} & {-a} & {3} \\
{-2} & {0} & {1} \\
{-b} & {-1} & {0}
\end{array}\right]\)
∵ A’ = – A
\(\left[\begin{array}{ccc}
{0} & {2} & {b} \\
{a} & {0} & {1} \\
{-3} & {-1} & {0}
\end{array}\right]\) = \(\left[\begin{array}{ccc}
{0} & {-a} & {3} \\
{-2} & {0} & {1} \\
{-b} & {-1} & {0}
\end{array}\right]\)
∴ 2 = -a or a = -2
-3 = -b or b = 3.

Question 16.
If A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\), then prove that AA^-1 = I?
Solution:
Given:
A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)
A^-1 = \(\frac { adjA }{ |A| } \)
|A| = |\(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)|
= cos²α – (-sin² α)
= cos²α + sin²α = 1 ………………………… (1)
∴|A| = 1
We know adj A = \(\begin{bmatrix} C_{ 11 } & C_{ 21 } \\ C_{ 12 } & C_{ 22 } \end{bmatrix}\)
Where C₁₁ = (-1)² cos α = cos α
C₁₂ = (-1)³ sin α = -sin α
C₂₁ = (-1)³ (-sin α ) = sin α
C₂₂ = (-1)⁴cos α = cos α
∴ adj A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\) …………….. (2)
∴ A^-1 = \(\frac { adjA }{ |A| } \) = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)
∴ A.A^-1 = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\) \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)

= \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) = I. Proved.

Question 17.
If A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\) then prove that:
A. (Adj A) = |A| I?
Solution:
Given A = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)
A₁₁ = cos α, A₁₂ = sin α, A₂₁ = – sin α, A₂₂ = cos α
Adj A = \(\begin{bmatrix} A_{ 11 } & A_{ 21 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}\) = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)
A.(Adj A) = \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\) \(\begin{bmatrix} cos\alpha & -sin\alpha \\ sin\alpha & cos\alpha \end{bmatrix}\)

∴ A.(Adj A) = I = |A| I. Proved.

Question 18.
Prove that the square matrix A = \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\) is orthogonal matrix?
Solution:
Given:
A = \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\)
∴ A’ = \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\)
Now A.A’ = \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\) \(\begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}\)

= \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) = I
Similarly A’.A = I
Then A.A’ = A’A = I
So A is orthogonal matrix.
Proved.

Question 19.
If A = \(\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\) and B = \(\begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix}\), then find the value of (AB)^-1?
Solution:
Given:
A = \(\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\), B = \(\begin{bmatrix} 6 & 7 \\ 8 & 9 \end{bmatrix}\)
= \(\begin{bmatrix} 3.6+2.8 & 3.7+2.9 \\ 7.6+5.8 & 7.7+5.9 \end{bmatrix}\) = \(\begin{bmatrix} 34 & 39 \\ 82 & 94 \end{bmatrix}\)
|AB| = \(\begin{bmatrix} 34 & 39 \\ 82 & 94 \end{bmatrix}\)
= \(\begin{bmatrix} 34 & 39 \\ 82 & 94 \end{bmatrix}\) = 3196 – 3198 = -2 ≠ 0
AB₁₁ = 94, AB₁₂ = -82, AB₂₁ = -39, AB₂₂ = 34
adj AB = \(\begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix}\)
(AB)^-1 = \(\frac { adjAB }{ |AB| } \) = \(\frac{1}{ -2}\) \(\begin{bmatrix} 94 & -39 \\ -82 & 34 \end{bmatrix}\)
= \(\begin{bmatrix} -47 & \frac { 39 }{ 2 } \\ 41 & -17 \end{bmatrix}\)

Question 20.
If A = \(\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}\) then prove that:
2A^-1 = 9I – A? (CBSE 2018)
Solution:
Given:
A = \(\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}\)
∴ A₁₁ = 7, A₁₂ = – (- 4), A₂₁ = – (- 3), A₂₂ = 2
∴ adj A = \(\begin{bmatrix} A_{ 11 } & A_{ 21 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}\) = \(\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}\)
Now |A| = \(\begin{vmatrix} 2 & -3 \\ -4 & 7 \end{vmatrix}\) = 14 – 12 = 2
∴ A^-1 = \(\frac { adjA }{ |A| } \) = \(\frac{1}{2}\) \(\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}\)
⇒ 2A^-1 = \(\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}\)
Again, 9I – A = 9 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\) – \(\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}\)
= \(\begin{bmatrix} 9 & 0 \\ 0 & 9 \end{bmatrix}\) – \(\begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix}\) = \(\begin{bmatrix} 9-2 & 0+3 \\ 0+4 & 9-7 \end{bmatrix}\)
∴ 9I – A = \(\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix}\)
From eqns. (1) and (2),
2A^-1 = 9I – A. proved.

Question 21.
For matrix A and B prove that (AB)’ = B’A’ where A = [ \(\begin{matrix} 1 \\ -4 \\ 3 \end{matrix}\) ], B = [-1 2 1]? (NCERT)
Solution:
AB = [ \(\begin{matrix} 1 \\ -4 \\ 3 \end{matrix}\) ]_3×1 [-1 2 1]_1×3
AB = \(\left[\begin{array}{rrr}
{-1 \times 1} & {1 \times 2} & {1 \times 1} \\
{-4 \times-1} & {-4 \times 2} & {-4 \times 1} \\
{-3 \times 1} & {3 \times 2} & {3 \times 1}
\end{array}\right]\)
⇒ AB = [ \(\begin{matrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{matrix}\) ]
(AB)’ = [ \(\begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{matrix}\) ] …………………… (1)
B = [ -1 2 1]’
B’ = [ \(\begin{matrix} -1 \\ 2 \\ 1 \end{matrix}\) ]
A’ = [ \(\begin{matrix} -1 \\ -4 \\ 3 \end{matrix}\) ]
A’ = [1 -4 3]_{1 ×3}
B’A’ = [ \(\begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{matrix}\) ]
From eqns. (1) and (2),
(AB)’ = B’A’. proved.

Matrices Long Answer Type Questions – II

Question 1.
If A = \(\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}\) then prove that:
A.adj A = (adj A). A = |A| I?
Solution:
Given:
A = \(\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}\)
Then, |A| = \(\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}\) = 5 – 12 = -7
A₁₁ = 5, A₁₂ = -3, A₂₁ = -4, A₂₂ = 1
∴ adj A = \(\begin{bmatrix} A_{ 11 } & A_{ 21 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}\) = \(\begin{bmatrix} 5 & -4 \\ -3 & 1 \end{bmatrix}\)
⇒ A. adj A = \(\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}\) × \(\begin{bmatrix} 5 & -4 \\ -3 & 1 \end{bmatrix}\)

⇒ A.adj A = \(\begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}\) = -7\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
∴ A.adjA = |A| I
and (adj A) . A = \(\begin{bmatrix} 5 & -4 \\ -3 & 1 \end{bmatrix}\) × \(\begin{bmatrix} 1 & 4 \\ 3 & 5 \end{bmatrix}\)

⇒ (adj A).A = \(\begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix}\)
⇒ (adj A).A = – 7 \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
⇒ (adj A).A = |A| I
From eqns. (1) and (2),
A.adj A = (adj A) .A = |A| I. Proved.

Question 2.
If A = \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) then prove that:
A.(adj A) = (adj A). A = |A| I?
Solution:
Solve like Q.No. 1.

Question 3.
If matrix A = [ \(\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}\) ] then prove that: A^-1 = A?
Solution:
Given:
A = [ \(\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}\) ]
By formula A^-1 = \(\frac { adjA }{ |A| } \)
Then, |A| = [ \(\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}\) ] = 1.(0-1) = -1
A₁₁ = (-1)² (0 – 0) = 0
A₁₂ = (-1)³ (0 – 0) = 0
A₁₃ = (-1)⁴ ( 0 – 1) = -1, A₂₁ = (-1)³ (0 – 0) = 0
A₂₂ = (-1)⁴ ( 0 – 1) = -1, A₂₃ = (-1)⁵ (0 – 0) = 0
A₃₁ = (-1)⁴ ( 0 – 1) = -1, A₃₂ = (-1)⁵ (0 – 0) = 0
A₃₃ = (-1)⁶ ( 0 – 0) = 0
adj A = \(\left[\begin{array}{ccc}
{A_{11}} & {A_{21}} & {A_{31}} \\
{A_{12}} & {A_{22}} & {A_{32}} \\
{A_{13}} & {A_{23}} & {A_{33}}
\end{array}\right]\) = [ \(\begin{matrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{matrix}\) ]
= [- \(\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}\) ]
∴ image 14 = [ \(\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}\) ]
∴ A^-1 = A.

Question 4.
Matrix A = [ \(\begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}\) ], find inverse of matrix A?
Solution:
Given A = [ \(\begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix}\) ]
∴ |A| = 2 (4 × 2 – 7 × 1) – 3 (3 × 2 – 3 × 1) + 1 (3 × 7 – 3 × 4)
= 2 ( 8 – 7) – 3 ( 6 – 3) + 1 (21 – 12)
= 2(1) – 3(3) + 1(9)
= 2 – 9 + 9 = 2
If |A| ≠ 0 then A^-1 will exist.
Now, A₁₁ = + (8 – 7) = 1, A₁₂ = – (6 – 3) = -3
A₁₃ = + (21 – 12), A₂₁ = – (6 – 7) = 1
A₂₂ = +(4 – 3) = 1, A₂₃ = – (14 – 9) = -5
A₃₁ = + (3 – 4) = -1, A₃₂ = – (2 – 3) = 1
A₃₃ = + (8 – 9) = -1
∴ adj A = \(\left[\begin{array}{ccc}
{A_{11}} & {A_{21}} & {A_{31}} \\
{A_{12}} & {A_{22}} & {A_{32}} \\
{A_{13}} & {A_{23}} & {A_{33}}
\end{array}\right]\) = [ \(\begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix}\) ]
∵ A^-1 = \(\frac { adjA }{ |A| } \)
∴A^-1 = \(\frac{1}{2}\) [ \(\begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix}\) ]
⇒ A^-1 = \(\left[\begin{array}{ccc}
{\frac{1}{2}} & {\frac{1}{2}} & {\frac{-1}{2}} \\
{\frac{-3}{2}} & {\frac{1}{2}} & {\frac{1}{2}} \\
{\frac{9}{2}} & {\frac{-5}{2}} & {\frac{-1}{2}}
\end{array}\right]\)

Question 5.
If A = [ \(\begin{matrix} 1 & 2 & 3 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}\) ], then find the value of A^-1?
Solution:
Solve like Q.No.4.
Answer:
[ \(\begin{matrix} 1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0 \end{matrix}\) ]

Question 6.
If A = [ \(\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}\) ], then find the value of A^-1?
Solution:
Given:
A = [ \(\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}\) ]
|A| = [ \(\begin{matrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{matrix}\) ]
⇒ |A| = 1 (1 – 4) + 2 ( 4 – 2) + 2 ( 4 – 2)
= – 3 + 4 + 4 = 5
If |A| ≠ 0, then A^-1 exists
A₁₁ = \(\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}\) = 1 – 4 = -3
A₁₂ = – \(\begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix}\) = – ( 2 – 4) = 2
A₁₃ = \(\begin{vmatrix} 2 & 1 \\ 2 & 2 \end{vmatrix}\) = 4 – 2 = 2
A₂₁ = – \(\begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix}\) = – ( 2 – 4) = 2
A₂₂ = \(\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}\) = 1 – 4 = – 3
A₂₃ = – \(\begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix}\) = – ( 2 – 4) = 2
A₃₁ = \(\begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix}\) = 4 – 2 = 2
A₃₂ = – \(\begin{vmatrix} 1 & 2 \\ 2 & 2 \end{vmatrix}\) = – (2 – 4) = 2
A₃₃ = \(\begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix}\) = 1 – 4 = -3
∴adj A = \(\left[\begin{array}{ccc}
{A_{11}} & {A_{21}} & {A_{31}} \\
{A_{12}} & {A_{22}} & {A_{32}} \\
{A_{13}} & {A_{23}} & {A_{33}}
\end{array}\right]\)
= [ \(\begin{matrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{matrix}\) ]
∴A^-1 = \(\frac { adjA }{ |A| } \) = \(\frac{1}{5}\) [ \(\begin{matrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{matrix}\) ]

Question 7.
If A = \(\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}\) then prove that: A² – 2A + 3I = 0?
Solution:
Given:
A = \(\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}\)
A² = A.A = \(\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}\) \(\begin{bmatrix} 2 & 3 \\ -1 & 0 \end{bmatrix}\)
= \(\begin{bmatrix} 2.2+3(-1) & 2.3+3.0 \\ -1.2+0(-1) & -1.3+0.0 \end{bmatrix}\)
= \(\begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix}\)
2A = \(\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}\)
3I = \(\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\)
∴ A² – 2A + 3I = \(\begin{bmatrix} 1 & 6 \\ -2 & -3 \end{bmatrix}\) – \(\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}\) + \(\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\)
= \(\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}\) – \(\begin{bmatrix} 4 & 6 \\ -2 & 0 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
= 0. Proved.

Question 8.
If A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) then prove A² – 6A + 17I = 0 and find A^-1?
Solution:
Given:
A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\)
A² = A.A = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\)
= \(\begin{bmatrix} 4-9 & -6-12 \\ 6+12 & -9+16 \end{bmatrix}\) = \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\)
∴L.H.S = A² – 6A + 17I
= \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\) – 6 \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) + 17\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\)
= \(\begin{bmatrix} -5 & -18 \\ 18 & 7 \end{bmatrix}\) – \(\begin{bmatrix} 12 & -18 \\ 18 & 24 \end{bmatrix}\) + \(\begin{bmatrix} 17 & 0 \\ 0 & 17 \end{bmatrix}\)
= \(\begin{bmatrix} -5-12+17 & -18+18+0 \\ 18-18+0 & 7-24+17 \end{bmatrix}\) = \(\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
= 0 = R.H.S. Proved.
Then |A| = \(\begin{bmatrix} 2 & -3 \\ 3 & 4 \end{bmatrix}\) = 8 + 9 = 17
∴ A₁₁ = (-1)¹⁺¹ 4 = 4
A₁₂ = (-1)¹⁺² (3) = -3
A₂₁ = (-1)²⁺¹ (-3) = 3
A₂₂ = (-1)²⁺² (2) = 2.
and adj A = \(\begin{bmatrix} A_{ 11 } & A_{ 12 } \\ A_{ 12 } & A_{ 22 } \end{bmatrix}\) = \(\begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix}\)
∴A^-1 = \(\frac { adjA }{ |A| } \)
= \(\frac{1}{17}\) \(\begin{bmatrix} 4 & 3 \\ -3 & 2 \end{bmatrix}\)

Question 9.
If A = \(\begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}\), then prove A² + 4A – 42 I = 0 and find A^-1?
Solution:
Solve like Q.No. 8.
Answer:
A^-1 = \(\frac{1}{42}\) \(\begin{bmatrix} -4 & 5 \\ 2 & 0 \end{bmatrix}\).

Question 10.
(A) Solve the following equations by matrix method:
x + y + z = 3
2x – y + z = 2
x – 2y + 3z = 2.
Solution:
If A = [ \(\begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & -2 & 3 \end{matrix}\) ], X = [ \(\begin{matrix} x \\ y \\ z \end{matrix}\) ] and B = [ \(\begin{matrix} 3 \\ 2 \\ 2 \end{matrix}\) ]
|A| = \(\left|\begin{array}{ccc}
{1} & {1} & {1} \\
{2} & {-1} & {1} \\
{1} & {-2} & {3}
\end{array}\right|\)
= 1( – 3 + 2) – 2 (3 + 2) + 1(1 + 1)
= -1 – 10 + 2 = -9
A₁₁ = \(\begin{vmatrix} -1 & 1 \\ -2 & 3 \end{vmatrix}\) = – 3 + 2 = – 1
A₁₂ = – \(\begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix}\) = – ( 6 – 1) = – 5
A₁₃ = \(\begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix}\) = – 4 + 1 = – 3
A₂₁ = – \(\begin{vmatrix} 1 & 1 \\ -2 & 3 \end{vmatrix}\) = – (3 + 2) = – 5
A₂₂ = \(\begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix}\) = 3 -1 = 2
A₂₃ = – \(\begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix}\) = – ( -2 -1) = 3
A₃₁ = \(\begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix}\) = 1 + 1 = 2
A₃₂ = – \(\begin{vmatrix} 1 & 1 \\ 2 & 1 \end{vmatrix}\) = – (1 – 2) = 1
A₃₃ = \(\begin{vmatrix} 1 & 1 \\ 2 & -1 \end{vmatrix}\) = -1 -2 = -3
and adj A = [ \(\begin{matrix} -1 & -5 & 2 \\ 5 & 2 & 1 \\ -3 & 3 & -3 \end{matrix}\) ]
∴ A^-1 = \(\frac { adjA }{ |A| } \)
⇒ A^-1 = \(\frac{-1}{9}\) [ \(\begin{matrix} -1 & -5 & 2 \\ 5 & 2 & 1 \\ -3 & 3 & -3 \end{matrix}\) ]
⇒ A^-1 = image 14
X = A^-1B

∴ x = 1, y = 1, z = 1.

Question 10.
(B) Solve the following equations by matrix method:
x + y + z = 6
x + 2y = 3z = 14
x + 4y + 9z = 36
Solution:
x + y + z = 6
x + 2y + 3z = 14
x + 4y + 9z = 36.
Where
A = [ \(\begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{matrix}\) ], B = [ \(\begin{matrix} 6 \\ 14 \\ 36 \end{matrix}\) ] , X = [ \(\begin{matrix} x \\ y \\ z \end{matrix}\) ]
= 6 – 6 + 2 = 2


X = A^-1 B
⇒ [ \(\begin{matrix} x \\ y \\ z \end{matrix}\) ] = [ \(\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}\) ]
∴ x = 1, y = 2, z = 3.

Question 11.
If A’ = \(\left[\begin{array}{rr}
{3} & {4} \\
{-1} & {2} \\
{0} & {1}
\end{array}\right]\) and B = \(\left[\begin{array}{rrr}
{-1} & {2} & {1} \\
{1} & {2} & {3}
\end{array}\right]\) then prove the following:
(i) (A + B)’ = A’ + B’
(ii) (A – B)’ = A’ – B’. (NCERT)
Solution:
(i) Given

Question 12.
If A = \([latex]\left[\begin{array}{rrr}
{-1} & {2} & {3} \\
{5} & {7} & {9} \\
{-2} & {1} & {1}
\end{array}\right]\)[/latex] and B = \(\left[\begin{array}{rrr}
{-4} & {1} & {-5} \\
{1} & {2} & {0} \\
{1} & {3} & {1}
\end{array}\right]\) then prove that:
(i) (A + B)’ = A’ + B’
(ii) (A – B)’ = A’ – B’. (NCERT)
Solution:
solve like Q.No.11.

Question 13.
Express matrix A = \(\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}\) as sum of a symmetric and a skew symmetric matrix? (NCERT)
Solution:

Given:

Eqn. (1) and symmetric matrix and eqn. (2) is skew symmetric matrix.

Question 14.
(A) By using elementary operations, find the inverse of matrix A = \(\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}\)
Solution:
Using A = AI

Question 14.
(B) By using elementary operation, find the inverse of matrix A = \(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\)?
Solution:
Solve like Q.No. 14 (A).
Answer:
\(\begin{bmatrix} 7 & -10 \\ -2 & 3 \end{bmatrix}\)

Question 15.
Solve the following system of equations by using matrix method: (NCERT, CBSE 2011)
\(\frac{2}{x}\) + \(\frac{3}{y}\) + \(\frac{10}{z}\) = 4
\(\frac{4}{x}\) – \(\frac{6}{y}\) + \(\frac{5}{z}\) = 1
\(\frac{6}{x}\) + \(\frac{9}{y}\) – \(\frac{20}{z}\) = 2, x, y, z, ≠ 0.
Solution:
Let \(\frac{1}{x}\) = u,
\(\frac{1}{y}\) = v
\(\frac{1}{z}\) = w, then
2u + 3v + 10w = 4
4u – 6v + 5w = 1
6u + 9v – 20 w = 2
Applying formula AX = B
where

⇒ |A| = 2 × (120 – 45) -3 (- 80 – 30) + 10 (36 + 36)
⇒ |A| = 150 + 330 + 720 = 1200
⇒ |A| ≠ 0, hence A^-1 exists.
Applying formula
X = A^-1B


As \(\frac{1}{x}\) = u, \(\frac{1}{y}\) = v, and \(\frac{1}{z}\) = \(\frac{1}{z}\) = w
∴\(\frac{1}{x}\) = \(\frac{1}{2}\), \(\frac{1}{y}\) = \(\frac{1}{3}\) and \(\frac{1}{z}\) = \(\frac{1}{5}\)
⇒ x = 2, y = 3 and z = 5. is required solution.

Question 16.
Find A^-1 where A = \(\left[\begin{array}{ccc}
{1} & {2} & {-3} \\
{2} & {3} & {2} \\
{3} & {-3} & {-4}
\end{array}\right]\), corresponding equation is:
x + 2y – 3z = -4
2x + 3y + 2z = 2
3x – 3y – 4z = 11. (CBSE 2008, 10, 12)
Solution:
Given:
A = \(\left[\begin{array}{ccc}
{1} & {2} & {-3} \\
{2} & {3} & {2} \\
{3} & {-3} & {-4}
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{ccc}
{1} & {2} & {-3} \\
{2} & {3} & {2} \\
{3} & {-3} & {-4}
\end{array}\right]\)
⇒ |A| = 1(- 12 + 6) – 2 (- 8 – 6) – 3 (- 6 – 9)
⇒ |A| = 1 (-12 + 6) -2 (-8 -6) -3 (-6 -9)
⇒ |A| ≠ 0
⇒ Hence A^-1 exists.
Hence



Equation of above matrix
x + 2y – 3z = – 4
2x + 3y + 2z = 2
3x – 3y – 4z = 11
Solution of above equations
AX = B
Where A = \(\left[\begin{array}{ccc}
{1} & {2} & {-3} \\
{2} & {3} & {2} \\
{3} & {-3} & {-4}
\end{array}\right]\), X = [ \(\begin{matrix} x \\ y \\ z \end{matrix}\) ] , B = [ \(\begin{matrix} -4 \\ 2 \\ 11 \end{matrix}\) ]
Applying formula
X = A^-1B

Question 17.
Find the product of matrix \(\left[\begin{array}{ccc}
{-4} & {4} & {4} \\
{-7} & {1} & {3} \\
{5} & {-3} & {-1}
\end{array}\right]\) \(\left[\begin{array}{ccc}
{1} & {-1} & {1} \\
{1} & {-2} & {-2} \\
{2} & {1} & {3}
\end{array}\right]\) and with the help of product of matrix solve the equations? (CBSE 2012)
x – y + z = 4
x – 2y – 2z = 9
2x + y + 3z = 1.
Solution:
Let

Writing above equation in matrix form
AX = C
Where


⇒ x = 3, y = -2, z = -1, (by equating of two matrix)

Question 18.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method?
Solution:
Let the cost of 1 kg onion = Rs. x
1 kg wheat = Rs. y
and 1 kg rice = Rs. z
According to equation
4x + 3y + 2z = 60
2x + 4y + 6z = 90
6x + 2y + 3z = 70
Matrix form will be
AX = B
Where

⇒ Hence A^-1 exist.
Applying formula
X = A^-1 B



∴ x = Rs. 5, y = Rs. 8, z = Rs. 8.

MP Board Class 12 Maths Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 14 Biomolecules

Biomolecules Important Questions

Biomolecules Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
Which protein transports oxygen in blood flow :
(a) Haemoglobin
(b) Insulin
(c) Albumin
(d) Myoglobin.
Answer:
(a) Haemoglobin

Question 2.
Enzyme which enhances the conversion of glucose to ethanol is :
(a) Zymase
(b) Invertase
(c) Maltase
(d) Diastase.
Answer:
(a) Zymase

Question 3.
In human body carbohydrate is stored :
(a) In the form of glucose
(b) In the form of glycogen
(c) In the form of starch
(d) In the form of fructose.
Answer:
(b) In the form of glycogen

Question 4.
Change in optical rotation of a freshly prepared solution of sugar after some time is called :
(a) Optical activity
(b) Inversion
(c) Specific rotation
(d) Mutation.
Answer:
(b) Inversion

Question 5.
Formula of most familiar disachharide is :
(a) C₁₀H₁₈O₉
(b) C₁₀H₁₂O₁₀
(c) C₁₈H₂₂O₁₁
(d) C₁₂H₂₂O₁₁.
Answer:
(d) C₁₂H₂₂O₁₁.

Question 6.
The following statement is false in relation to Ribose :
(a) It is a polyhydroxy compound
(b) It is a aldehydic sugar
(c) It contain 6 carbon atoms
(d) It has optical rotation.
Answer:
(c) It contain 6 carbon atoms

Question 7.
How many subunits are present in haemoglobin :
(a) 2
(b) 3
(c) 4
(d) 5.
Answer:
(b) 3

Question 8.
Starch is polymer of :
(a) Glucose
(b) Sucrose
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Glucose

Question 9.
Which sugar is present in maximum amount in human blood :
(a) Fructose
(b) d – glucose
(c) Sucrose
(d) Lactose.
Answer:
(b) d – glucose

Question 10.
Element present in vitamin B₁₂ is :
(a) Pb
(b) Zn
(c) Fe
(d) Co.
Answer:
(d) Co.

Question 11.
Amount of glucose in blood is determined by :
(a) Tollen’s reagent
(b) Benedict’s solution
(c) Alkaline iodine solution
(d) Bromine water.
Answer:
(b) Benedict’s solution

Question 12.
Vitamin B, is : (MP2014)
(a) Riboflavin
(b) Cobaltamine
(c) Thiamine
(d) Pyrimidine.
Answer:
(c) Thiamine

Question 13.
Deficiency of Vitamin C leads to :
(a) Scurvy
(b) Rickets
(c) Pyorrhoea
(d) Anaemia.
Answer:
(a) Scurvy

Question 14.
Most effective energy reservoir in all living cells is :
(a) A.M.P.
(b) A.T.P.
(c) A.D.P.
(d) U.D.P.
Answer:
(b) A.T.P.

Question 15.
Disaccharide present in milk is :
(a) Sucrose
(b) Lactose
(c) Maltose
(d) Cellulose.
Answer:
(b) Lactose

Question 16.
Which is not glyceroid : (MP2018)
(a) Fat
(b) Oil
(c) Phospholipid
(d) Soap.
Answer:
(d) Soap.

Question 17.
Which is not found in R.N.A.: (MP 2016)
(a) Thymine
(b) Uracil
(c) Adenine
(d) Guanine.
Answer:
(a) Thymine

Question 18.
Deficiency of which vitamin causes Rickets :
(a) Vitamin C
(b) Vitamin B
(c) Vitamin A
(d) Vitamin D.
Answer:
(d) Vitamin D.

Question 2.
Fill in the blanks :

1. By the oxidation of glucose ……………… molecules of ATP are produced.
2. The breaking of complex molecules in organisms is known as ………………
3. In hyperglycemia the amount of ……………… in blood increases.
4. Deficiency of ……………… leads to eye disease.
5. Deficiency of iodine leads to ……………… disease.
6. Blood ……………… the temperature of the entire body.
7. ……………… hormone balances the amount of sugar in blood.
8. ……………… is responsible for the clotting of blood.
9. Denaturation does not affect the ……………… structure of protein.
10. Protein is a polymer of ……………… (MP 2018)
11. ……………… is the basic unit of protein.
12. ……………… is not present in D.N.A.
13. Haemoglobin is a ……………… compound of iron.
14. Oils and fats obtained from plants and animals (organisms) are called ………………

Answer:

1. 38
2. Catabolism
3. Sugar
4. Vitamin A
5. Goitre
6. Balance
7. Insulin
8. Vitamin K
9. Primary
10. Amino acids
11. Amino acid
12. Uracil
13. Complex
14. Lipids.

Question 3.
(A) Match the following :

Answer:

1. (f)
2. (c)
3. (b)
4. (e)
5. (d)
6. (a)
7. (h)
8. (g)

Question 4.
Answer in one word/sentence :

1. Write the chemical name of Vitamin C.
2. Tell the source of Vitamin K.
3. Is responsible for clotting of blood?
4. Which bond links amino acids together?
5. How many amino acids are synthesized by human body?
6. Cellulose is a linear polymer of which glucose?
7. In RNA molecule which pyrimidine is present in place of Thymine?
8. Lactose on hydrolysis gives.
9. Glucose contains Pyranose ring whereas Fructose contain.
10. In polysaccharides, monosaccharide units are linked to each other by which bond?
11. Which protein helpful for clotting of blood known as? (MP 2018)
12. Write one example of Monosaccharide Carbohydrate.
13. What is the name of Disaccharides sugar present in milk?

Answer:

1. Ascorbic acid
2. Green leafy vegetables
3. Vitamin K (Phylloquinone)
4. Peptide bond
5. Ten
6. β – glucose
7. Uracil
8. Glucose and Lactose
9. Furanose ring
10. Glycosidic
11. Fibrinogen
12. Glucose or fructose
13. Lactose.

Biomolecules Very Short Answer Type Questions

Question 1.
Write the reaction of hydrolysis of sucrose.
Answer:

Question 2.
What type of compounds are bases?
Answer:
Bases are heterocyclic compounds.

Question 3.
What is the formula of peptide bond?
Answer: Formula of peptide bond is – CO – NH -.

Question 4.
Which are the main biomolecules found in the biological system?
Answer:
Carbohydrates, proteins, nucleic acids and lipids etc. are found in the biological system.

Question 5.
What are oligosaccharides?
Answer:
Carbohydrates that yield 2 to 10 oligosaccharides monosaccharide units are called oligosaccharides.

Question 6.
Write two examples of fibrous protein.
Answer:
Keratin and myosin are the examples of fibrous protein.

Question 7.
What are biomolecules?
Answer:
Molecules which take part in the formation of living system are known as bio – molecules. Like : Proteins, carbohydrates.

Biomolecules Short Answer Type Questions

Question 1.
Where does the water present in the egg go after boiling the egg? (NCERT)
Answer:
When egg is boiled, the denaturation of protein and then coagulation takes place probably through H – bonding. Water present in the egg gets absorbed or adsorbed during denaturation and disappears. In this process, the globular protein in egg changes to insoluble fibrous protein.

Question 2.
Why cannot vitamin C be stored in our body? (NCERT)
Answer:
Vitamin C is soluble in water. It cannot be stored in our body because it is easily excreted in urine.

Question 3.
What are proteins?
Answer:
The word protein is derived from the Greek word protious (Protious = to take the first) i.e. first or very important. Proteins are high molecular mass nitrogen containing complex organic compounds found in the protoplasm of all animal and plants.
Chemically protein is the condensation polymer of α – amino acid.

Question 4.
What are essential and non – essential amino acids? Give two examples of each type. (NCERT)
Answer:
Essential amino acids:
The amino acids which our body cannot make and must be obtained through diet.
Examples : Valine, Isoleucine, Arginine, Lysine, Threonine etc.

Non – essential amino acids:
These are the amino acids which our body can make.
Examples : Glycine, Alanine, Glutamic acid, Aspartic acid, Glutamine, Serine etc.

Question 5.
What are the common types of secondary structure of proteins? (NCERT)
Answer:
The conformation of polypeptide chain assumed as a result of hydrogen bonding is called secondary structure of proteins. The two types of secondary structures are α – helix and β – pleated sheet structure. (For detail refer your NCERT text – book.)

Question 6.
What is the effect of denaturation on the structure of proteins? (NCERT)
Answer:
During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. As a result of denaturation, the globular proteins (soluble in H₂O) are converted into fibrous proteins (insoluble in H₂O) and their biological activity is lost. The coagulation of egg white on boiling is a common example of denaturation.

Question 7.
Differentiate between globular and fibrous proteins. (NCERT)
Answer:
Differences between globular and fibrous protein :
Globular protein:

• They have coiled ball like structure.
• They have three – dimensional structure.
• They are soluble in water and aq. solution of salt and base.
• These proteins are inactive towards temperature and pH value.

Fibrous protein:

• These molecule have long threads like structure.
• They have sheet like structure.
• These are insoluble in water.
• Fibrous protein are active towards temperature and pH value.

Question 8.
What are monosaccharides? (NCERT)
Answer:
Monosaccharides are the carbohydrates which cannot be hydrolysed further to give simpler units of polyhydroxy aldehydes or ketones.

Question 9.
What are disaccharides? Write general formula of disaccharides.
Answer:
Disaccharides are sugar which are formed by combination of two monosaccharides by removal of one molecule of water. Both monosaccharides are of hexose type in which one unit is glucose. Thus, disaccharides are of aldose – aldose or aldose – ketose type. General formula of disaccharides is C₁₂H₂₂O₁₁.
Example : Sucrose, maltose, lactose etc.

Question 10.
What is the basic structural difference between starch and cellulose? (NCERT)
Answer: Starch consists of two components : amylose and amylopectin. Amylose is a long linear polymer of 200 – 1000 α – D – (+) glucose units held by C₁ – C₄ glycosidic linkage. It is soluble in water. Amylopectin is a branched chain polymer of α – D – (+) glucose linkage whereas branching occurs by C₁ – C₆ glycosidic linkage. It is insoluble in water.

On the other hand cellulose is a straight chain polysaccharide composed only of β – D – (+) glucose units which are formed by glycosidic linkage between C₁ of one glucose unit and C₄ of next glucose unit.

Question 11.
What do you understand by the term glycosidic linkage? (NCERT)
Answer:
The oxygen (ethereal) linkage through which two monosaccharide units are joined together by the loss of a water molecule to form a molecule of disaccharide is called glycosidic linkage. For example, sucrose (a disaccharide) is composed of C₁ of α – glucose and C₂ β – fructose through the glycosidic linkage.

Question 12.
What is glycogen? How is it different from starch? (NCERT)
Answer:
1. The carbohydrates are stored in animal body as glycogen. It is present in liver, muscles and brain. Enzymes break the glycogen down to glucose when the body needs glucose.

2. Glycogen is more highly branched than amylopectin (starch) glycogen chain consist of 10 – 14 glucose units, whereas amylopectin (starch) glycogen chain consist of 20 – 25 glucose units.

Question 13.
Write two Differences between α – Amino acid and Protein.
Answer:
Differences between α – Amino acid and Protein :
α – Amino acid:

• They are simple compounds having amino and Carboxylic acid group.
• On combining amino acid gives dipeptide, polypeptide and then protein. example glucose, lysine etc.

Protein:

• Proteins are complex nitrogenous compounds.
• Protein on hydrolysis gives amino acid. example Haemoglobin, casein etc.

Question 14.
What are carbohydrates? Which unit of carbohydrate provide energy to human body?
Answer:
Carbohydrates are compound of carbon, hydrogen and oxygen. General formula of carbohydrate is where x and y are integers. These compounds have ratio of hydrogen and oxygen 2 : 1 like water (H₂O). Therefore the name of these compounds is given carbohydrates. Examples of carbohydrates are glucose (C₆H₁₂O₆), fructose (C₆H₁₂O₆), sucrose (C₆H₁₂O₆) etc.

Glucose is the unit of carbohydrates which is responsible to provide energy. Glucose decomposes slowly with the help of oxydase enzyme present in the body into CO₂ and water. Energy is released in this process.

Question 15.
Define polysaccharides. Give examples also.
Answer:
Polysaccharides are natural isomers which have molecular weight upto many lacks, general formula of polysaccharides is (C₆H₁₀O₅)_nwhere value of n is from 12 to thousands. These are complex material which are formed by condensations of mono-saccharides. These compounds contain glycocydic bonds.
Example : Starch, cellulose etc.

Question 16.
What is invert sugar?
Answer:
Cane – sugar is dextro – rotatory [D or +] which gives equimolar mixture of monosaccharides. This mixture is laevorotatory [L or -]. Hence the mixture of glucose and fructose obtained as a result of hydrolysis of cane sugar is called as invert sugar.

Question 17.
Define the following as related to proteins :

1. Peptide linkage
2. Primary structure and
3. Denaturation.

Answer:
1. Peptide linkage:
A peptide bond is an amide linkage formed between – COOH group of one α – amino acid and NH₂ group of the other α – amino acid by loss of a molecule of water. It units two amino acids unit in a peptide bond (molecule).

2. Proteins are the polymers of aramino acids. These polymers (also known as polypeptide) consist of amino acids linked with each other in a specific sequence. This sequence of amino acids is known as the primary structure of proteins. Any change in this sequence of amino acids (i.e., primary structure) creates a different proteins.

3. A process that changes the physical and biological properties of proteins without affecting the chemical composition of proteins is called denaturation. The denaturation is caused by certain physical and chemical treatment such as changes in pH, temperature, presence of somegalts or certain chemical agents.

Question 18.
Give the names and functions of any four proteins.
Answer:
Proteins and their functions :

1. Haemoglobin : Transport of oxygen from lungs to different tissues of body.
2. Myosin : For motion of muscles
3. Pepsin : As a catalyst in bio-chemical reactions
4. Keratin : Present in hairs, nails and teeth.

Question 19.
What are nucleic acids? Mention their two important functions. (NCERT)
Answer:
Nucleic acids are long chain polymers of nucleotides. They are also called poly-nucleotides. Nucleic acids are mainly of two types, the deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).

Functions :
1. DNA is responsible for transmission of hereditary effects from one generation to another. This is due to unique property of replication, during cell division and two identical DNA strands are transferred to the daughter cells.

2. DNA and RNA are responsible for synthesis of all proteins needed for the growth and maintenance of our body. Actually, the proteins are synthesized by various RNA molecules (m – RNA and f – RNA etc.) in the cell but the message for the synthesis of a particular protein is present in DNA.

Question 20.
What is the difference between a nucleoside and a nucleotide? (NCERT)
Answer:
A nucleoside contains only basic component of nucleic acids namely a pentose sugar and a nitrogenous base. A nucleotide contains all the three basic components of nucleic acids namely a phosphoric acid group, a pentose sugar and a nitrogenous base.

Question 21.
The two strands in DNA are not identical but are complementary. Explain. (NCERT)
Answer:
The two strands in DNA molecule are held together by hydrogen bonds between purine base of one strand and pyrimidine base of the other and vice – versa. Because of different sizes and geometries of the bases, the only possible pairing in DNA are G (guanine) and C (cytosine) through three H – bonds i.e., (C = G) and between A (adenine) and T (thiamine) through two H – bonds i.e., (A = T) (for figure refer your NCERT text – book). Due to this base pairing principle, the sequence of bases in one strand automatically fixes the sequence of bases in the other strand. Thus, the two strands are complimentary and are not – identical.

Question 22.
Write the important structural and functional differences between DNA and RNA. (NCERT)
Answer:
Differences between DNA and RNA:
DNA :

• Occurs mainly in the nucleus of the cell.
• It contains the sugar deoxyribose.
• Does not contain nitrogenous base, uracil.
• It has a double strand helix.
• It is responsible for the transmission of heredity character.
• Alkaline hydrolysis is quite slow.
• Ratio A/T = 1 and G/C = 1.

RNA:

• Occurs in the cytoplasm of the cell.
• It contains the sugar ribose.
• Does not contain nitrogenous base thymine.
• It has double as well as single strand helix.
• It helps in protein biosynthesis.
• Alkaline hydrolysis takes place readily.
• Such ratio is not present.

Question 23.
How are vitamins classified? Name the vitamin responsible for the co – agulation of blood. (NCERT)
Answer:
On the basis of solubility in water or fat, the vitamins are generally classified into two types:
1. Water soluble vitamins:
These include vitamin B – complex. (B₁, B₂, B₃, B₄, B₆, B₁₂ and nicotinic acid etc.) and vitamin C etc.

2. Fat soluble vitamins:
These include vitamins A, D, E and K. Liver cells are rich in fat soluble vitamins. Vitamin K is responsible for coagulation of blood.

Question 24.
What happens when protein is denatured ?
Or, Explain the denaturation of protein. (MP 2016)
Answer:
Denaturation of Protein:
Disruption of tertiary structure of protein is called denaturation. These reactions take place by heating in in presence of acids or highly concentrated salts or heavy metals. Denauration does not affect the primary structure of protein. Denaturation takes place in the rearrangement of secondary and tertiary structures. As a result of this, protein losses its biological actvity.

During denaturation the protein molecule uncoils fro an ordered and specific conformation into a more disordered conformation. Denaturation takes place when proteins are heated or treated with acids, bases, alcohols, KI, urea, acetone etc. or when exposed to UV or X – ray radiations. Denaturation is of two types :

1. Reversible and
2. Irreversible.

Reversible denaturation of proteins takes place in presence of denaturating agents like salts. But on removal of denaturating agent protein acquires its original structure. In irreversible denaturation protein cannot change to its original state. For example, white of example, i.e., yolk is a globulin and soluble protein. On putting it into boiling water it changes to white rubber like solid which is insoluble in water. Similarly, addition of an acid generally lemon juice to milk results in denaturation and milk coagulates to form cheese.

Question 25.
Write functions and sources of the following bio – molecules/elements : (MP 2011)

1. Protein
2. Carbohydrates
3. Fat
4. Calcium.

Answer:
1. Protein : Formation of new tissues and their repairing with the body.
Source : Milk, egg, meat, cheese, fish.

2. Carbohydrates : Carbohydrates provides energy to the body.
Source : Rice, potato, fruits, cane sugar etc.

3. Fats : They provides energy to the body.
Source : Ghee, oils, milk, egg. etc.

4. Calcium : Increase the bones and teeth.
Source : Green leafy vegetables, milk.

Biomolecules Long Answer Type Questions

Question 1.
Write a note on Nucleic acid.
Answer:
Nucleic acid:
It is found in nucleus of the cell. It has large amount of phosphorus. Nucleic acid are polynucleotides which is formed by the combination of various nucleotide units.

Each nucleotide is formed by three chemical components:

1. Phosphate group
2. Pentose ribose sugar or De – oxyribose
3. Heterocyclic base : Like derivative of pyrimidine (Thiamine, uracil, cytosine) and derivatives of purine (Adenine and Guanine).

Nucleic acids are of two types :

1. DNA : De – oxyribonucleic acid.
2. RNA : Ribonucleic acid.

Components of DNA:

(a) De – oxyribose sugar molecule
(b) Phosphoric acid molecule
(c) Nitrogenous base : It is of two types :

• Pyrimidine base : It includes Cytosine (C) and Thymine (T)
• Purine base : It includes Adenine (A) and Guanine (G)

Components of RNA:
RNA contains Ribose and nitrogen base like Adenine (A), Guanine (G), Uracil (U) and Cytosine (C).

Question 2.
What are carbohydrates? Write its classification and four main functions.
Answer:
Definition:
Optically active polyhydroxy aldehydes or ketones or substances which yield these on hydrolysis are known as carbohydrates.
Example : Glucose, starch, cellulose, sucrose etc.

Classification of Carbohydrate:

Functions of Carbohydrates:
1. It is the main structural component of cell.

2. It acts as a bio – fuel and provides energy to organisms for doing work.

3. Carbohydrate is stored in the liver as glycogen, which hydrolysis to provide the energy required.

4. Cellulose is found in grass and plants which provide energy to animals grazing grass because these animals possess specific enzymes which hydrolyses cellulose to glucose.

Question 3.
Write the functions of the following vitamins : (MP 2014)

1. Vitamin – A
2. Vitamin – D
3. Vitamin – E
4. Vitamin – K.

Answer:
Functions caused by the above Vitamins :

1. Vitamin – A : For vision and growth develops resistance against diseases.
2. Vitamin – D : For bones, control of metabolism of calcium and phosphorus.
3. Vitamin – E : Virility in man and reproduction.
4. Vitamin – K : Coagulation of blood.

Question 4.
Give the diseases caused by ascorbic acid, thiamine retinol and nicotinic
Or,
Give the source and diseases caused by Vitamin A, B, C and D.
Answer:
Name of Vitamins deficiency diseases are given in the ahead chart:
Chemical names of Vitamins , their sources and diseases due to deficiency:

Question 5.
Give differences between monosaccharide, disaccharide and polysaccharide.
Answer:
Difference between monosaccharide, disaccharide and polysaccharide :

MP Board Class 12th Chemistry Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 13 Amines

Amines Important Questions

Amines Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
A compound which gives oily nitrosoamine with nitrous acid at low temperature:
(a) Methyl amine
(b) Dimethyl amine
(c) Trimethyl amine
(d) Triethyl amine.
Answer:
(b) Dimethyl amine

Question 2.
Which of the following has strongest basic character :
(a) C₆H₅NH₂
(b) (CH₃)₂NH
(c) (CH₃)₃NH
(d) NH₃.
Answer:
(b) (CH₃)₂NH

Question 3.
Benzene diazonium chloride gives on hydrolysis :
(a) Chlorobenzene
(b) Phenol
(c) Alcohol
(d) Benzene.
Answer:
(b) Phenol

Question 4.
In the reaction C₆H₅CHO + C₆H₅NH₂ → C₆H₅N = CHC₆H₅ + H₂O + C₆H₅N = CHC₆H₅ is known as :
(a) Aldol
(b) Schiff’s reagent
(c) Schiff’s base
(d) Benedict reagent.
Answer:
(c) Schiff’s base

Question 5.
Nitrobenzene gives N – phenyl hydroxyl amine when it reacts with :
(a) Sn/HCl
(b) C₆H₅CH₂NH – CH₃
(c) Zn/NaOH
(d) Zn/NH₄Cl.
Answer:
(c) Zn/NaOH

Question 6.
Which of the mixture when reacts with ale. KOH known as Carbylamine reaction :
(a) Chloroform and Ag powder
(b) Trihalogenated methane and primary amine
(c) Alkyl trihalide and primary amine
(d) Alkyl cyanide and primary amine.
Answer:
(b) Trihalogenated methane and primary amine

Question 7.
Which of the following gas is responsible for Bhopal gas tragedy in 1984 :
(a) CH₃ – N = C = O
(b) CH₃ – N = C = S
(c) CHCl₃
(d) C₆H₅COCl.
Answer:
(a) CH₃ – N = C = O

Question 8.
Aniline reacts with cold nitrous acid (NaNO₂ + HCl) and gives :
(a) C₆H₅ – OH
(b) C₆H₅ – N, – Cl
(C) C₆H₅ – NO₂
(d) C₆H₅ – Cl.
Answer:
(b) C₆H₅ – N, – Cl

Question 9.
The product of mustard oil reaction is : (MP 2013,16)
(a) Alkyl isothiocyanate
(b) Dithiocarbonamide
(c) Dithioethyl acetate
(d) Thioether.
Answer:
(c) Dithioethyl acetate

Question 10.
A nitrogen containing compound, on heating with chloroform and alcoholic KOH gives vapours of disagreeable odour, the compound can be :
(a) Nitrobenzene
(b) Benzamide
(c) N, – N – dimethyl aniline
(d) Aniline.
Answer:
(d) Aniline.

Question 11.
Ethyl amine reacts with nitrous acid to form :
(a) Ammonia
(b) Nitrous oxide
(c) Ethane
(d) Nitrogen.
Answer:
(d) Nitrogen.

Question 12.
Oil of mirbane is :
(a) Aniline
(b) Nitrobenzene
(c) p – Nitroaniline
(d) p – Aminoazobenzene.
Answer:
(b) Nitrobenzene

Question 13.
Aniline is purified by :
(a) Steam distillation
(b) Vacuum distillation
(c) Simple distillation
(d) Solvent extraction.
Answer:
(a) Steam distillation

Question 14.
Amine which will not react with acetyl chloride is :
(a) CH₃ – NH₂
(b) (CH₃)₂NH
(c) (CH₃)₃N
(d) None of these.
Answer:
(c) (CH₃)₃N

Question 15.

(a) Gattermann’s reaction
(b) Sandmeyer’s reaction
(c) Wurtz’s reaction
(d) Frankland reaction.
Answer:
(b) Sandmeyer’s reaction

Question 2.
Fill in the blanks :

1. With transition metal ion amine establish co – ordination and form …………….
2. By reduction cyanides form ……………. and isocyanides form
3. Benzoic acid reacts with hydrazoic acid to form …………….
4. All aliphatic amines are more ……………. than ammonia.
5. 1° and 2° amine react with Grignard reagent to form …………….
6. C₆H₆ – COOH + ……………. → C₆H₅NH₂ + N₂ + CO₂
7. Trinitrotoluene is an ……………. substance. (MP 2011)
8. Methyl amine is ……………. basic than ammonia. (MP 2011,15)
9. Aromatic amines are ……………. in water.
10. Amines are benzolate in presence of NaOH. This reaction is called …………….
11. By reaction with nitrous acid 1° amine forms alcohol and 2° amine form …………….
12. Reaction of 2° amine with nitrous acid represent …………….
13. Basic nature of amine is due to presence of ……………. on nitrogen atom. (MP 2009)
14. Primary amine on heating with ……………. and ……………. form alkyl isocyanides. (MP 2009)
15. Mixture of T.N.T. and ammonium nitrate is known as …………….
16. On reacting aniline with HCl and NaNO₂ at 0°C temperature benzene diazonium chloride is formed. This is called ……………. reaction.
17. On heating alkyl isocyanide at 250°C ……………. is formed. (MP 2014)

Answer:

1. Complex ion
2. Primary amine, secondary amine
3. Aniline
4. Basic
5. Alkane
6. N₃H
7. Explosive
8. More
9. Insoluble
10. Schotten Baumann
11. Nitrosamine
12. Libermann nitroso test
13. Lone electron pair
14. Chloroform and caustic soda
15. Amatol
16. Diazotisation
17. Alkyl isocyanate

Question 3.
Match the followings:
I.

Answer:

1. (e)
2. (d)
3. (b)
4. (c)
5. (a)
6. (g)
7. (f)
8. (h)

II.

Answer:

1. (e)
2. (d)
3. (a)
4. (c)
5. (b)

III.

Answer:

1. (d)
2. (e)
3. (a)
4. (c)
5. (f)
6. (b)

Question 4.
Answer in one word/sentence:

1. Why aniline turns blackish brown in open air? (MP 2013)
2. Tertiary amine does not acetanilised, why?
3. Which isomer of C₃H₉N is least basic and having lowest b.p.?
4. Which amine gives diazotization reaction?
5. The compound obtained when primary aromatic amine when heated with CHCl₃ and ale. KOH.
6. Secondary amine can be identified by.
7. Primary nitroalkane reacts with nitrous acid to form which compound?
8. What is the nature of amines? (MP 2010)
9. Write the name and formula of Hinsberg’s reagent. (MP 2011)
10. What is nitrating mixture?
11. Nitrobenzene is known as.
12. 
    Write the name of reaction.
13. What is the name of reaction for preparation of methyl isocyanide?
14. What do 1° and 2° amine form on reaction with phosgene?
15. Write the name of reaction :
    
16. 
17. What is obtained by reacting amine with chloroform?
18. In which form is amine used in organic synthesis ?
19. What does ethylamine form on oxidation in the presence of KMnO₄?
20. On adding Br₂ water in aqueous solution of C₆H₅NH₂ which precipitate is obtained?
21. Which amine is obtained by the reduction of cyanide in the presence of Pt or Ni?
22. 
    Write the formula of x.
23. 
    Write the formula of x.
24. Write the formula of Benzene diazonium chloride.

Answer:

1. Aniline gets oxidised by air
2. Active hydrogen is absent
3. Tertiary amine
4. All primary aromatic amines
5. Phenyl isocyanide
6. By Liebermann’s nitroso test
7. Nitrolic acid
8. Basic
9. Benzene sulphonyl chloride (C₆H₅SO₂Cl)
10. Conc. HNO₃ + conc. H₂SO₄
11. Oil of mirbane
12. Diazotisation
13. Carbyl – amine reaction
14. Substituted urea
15. Carbylamine reaction
16. Sand – meyer’s reaction
17. Alkyl isocyanide
18. Reagent alkane
19. Aldehyde
20. Symmetrical tribromoaniline
21. 1°amine
22. C₆H₅NH₂
23. Schmidt reaction
24. C₆H₅ – N₂ – Cl.

Amines Very Short Answer Type Questions

Question 1.
What is Hinsberg reagent?
Answer:
Aryl sulphonyl chloride like Benzene sulphonyl chloride (C₆H₅SO₂Cl) is called Hinsberg reagent.

Question 2.
Which group linked in Aniline increases basic strength?
Answer:
Groups like – OCH₃, – CH₃ etc. linked in aniline increases the basic strength.

Question 3.
Write the structure of Phthalimide.
Answer:

Question 4.
How is the structure of amines and why?
Answer:
Structure of amines is pyramidal because of the presence of lone electron pair on nitrogen.

Question 5.
What is the use of Benadryl? Tell the functional group present in it.
Answer:
Benadryl is used as an antihistamine and tertiary amine group is present in it.

Question 6.
Why are amines soluble in dilute mineral acids?
Answer:
Amines form ionic crystalline salts in dilute mineral acids.
R – NH₂ + HCl > [R – \(\overset { + }{ N }\) H₃s]Cl^–

Question 7.
Write the formula and IUPAC name of Sulphonic acid.
Answer:

Question 8.
In what are amines found in nature?
Answer:
In nature, amines are found in proteins, vitamins, alkaloids and hormones.

Question 9.
What is the bond angle in trimethyl amine?
Answer:
Bond angle in trimethyl amine (CH₃)₃N is 108°.

Question 10.
Write the equation of formation of iodobenzene from aniline.
Answer:

Amines Short Answer Type Questions

Question 1.
Aniline is insoluble in water but soluble in HCl. Explain.
Answer:
Being basic nature, aniline forms soluble salts with strong acids like HCl while with water no such salt is formed. Therefore, aniline is insoluble in water but soluble in HCl.

Question 2.
Write a short note on Schotten Baumann reaction.
Answer:
Aromatic acid chloride reacts with phenol and aniline in presence of aqueous NaOH or pyridine. The reaction is known as Schotten Baumann reaction.

Question 3.
How will you convert: (NCERT)
(i) Benzene into aniline
(ii) Benzene into N, N – dimethylaniline
(iii) Cl – (CH₂)₄ – Cl into hexane – 1, 6 – diamine.
Answer:

Question 4.
Arrange the following in increasing order of their basic strength: (NCERT)

1. C₂H₅NH₂,C₆H₅NH₂,NH₃,C₆H₅CH₂NH₂ and (C₂H₅)₂NH
2. C₂H₅NH₂,(C₂H₅)₂NH,(C₂H₅)₃N, C₆H₅NH₂
3. CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂NH₂.

Answer:

1. C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH
2. C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH
3. C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH.

Question 5.
Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved. (NCERT)
Answer:
Hinsberg’s test:
This is a excellent test for distinguishing between primary, secondary and tertiary amines. The amine is treated with benzene sulphonyl chloride (Hinsberg’s reagent) in presence of excess of aqueous potassium hydroxide solution. (Refer text for details)

Question 6.
What is Mendius reaction?
Answer:
Reduction of alkyl cyanides by sodium and alcohols yield primary amine. This reaction is called Mendius reaction.

Question 7.
Ethyl amine is more basic than ammonia, why?
Answer:
The value of K_a = 4.5 x 10^-4 for ethyl amine and for ammonia it is 1.8 x 10^-5. Larger is the K_b value, more basic is the amine and vice – versa. In ethyl amine the availability of lone pair of electrons on nitrogen atom increases due to the +I inductive effect of the ethyl group. Hence, this lone pair of electrons can easily accept a proton. This explains why ethyl amine is more basic than ammonia.

Question 8.
Write short notes on :

1. Chmidt reaction
2. Mustard oil reaction.

Answer:
1. Schmidt reaction:
When hydrazoic acid dissolved in chloroform or benzene, react with mono carboxylic acid in presence of H₂SO₄ at 55°C, primary amine is obtained.

Mustard oil reaction:
When aliphatic primary amine is heated with carbon disulphide and HgCl₂, alkyl isothiocyanate is formed, which has smell like mustard oil. Therefore, this reaction is called mustard oil reaction.

Question 9.
Give a Hinsberg method to identify primary, secondary and tertiary amines.
Answer:
Hinsberg method:
This method is capable to differentiate primary, secondary and tertiary amines. Amines are heated with benzene sulphonyl chloride (Hinsberg reagent) and various products are obtained.

1. Primary amine : These form sulphonamide which are soluble in KOH.

2. Secondary amine : Secondary amine also form sulphonamide which are insoluble in KOH.

3. Tertiary amine : Tertiary amine does not react.
C₆H₅SO₂Cl + R₃N → No reaction.

Question 10.
Write the points of difference between Ethyl amine and Aniline.
Answer:
Differences between Ethyl amine and Aniline:

Question 11.
Why aniline is less basic than ethyl amine?
Answer:
Aniline is less basic than ethyl amine as due to resonance of benzene nucleus, the lone pair of electron of nitrogen atom is attracted towards nucleus and gets delocalised in the ring Thus, electron pair is liberated with difficulty in aniline than ethyl amine. Hence, its basic property is less than ethyl amine.

The delocalisation of electron as a result of resonance is shown as follows :

Question 12.
Convert: (NCERT)

1. 3 – Methylaniline into 3 – nitrotoluene
2. Aniline into 1,3,5 – tribromobenzene.

Answer:

Amines Long Answer Type Questions

Question 1.
Write reduction reactions of nitro – benzene :

1. In acid medium
2. Neutral medium
3. In basic medium.

Or,
Describe reduction reactions of nitro – benzene in different conditions.
Answer:
Reduction of Nitro – benzene:
Nitro – benzene is readily reduced. This gives different compounds under different conditions depending upon the pH of the medium and the nature of the reducing agent. The reduction takes place in three steps :

(a) Acidic medium : When reduced with Sn + HCl or Fe + HCl, gives aniline.

(b) Neutral medium : When reduced with aluminium mercury couple or zinc dust and ammonium chloride, phenyl hydroxyl amine is formed.

(c) In basic medium (Alkaline medium):
(i) Reduction with alkaline sodium arsenite : Azoxy benzene is formed.

(ii) Reduction with zinc dust and caustic soda : Hydrozo benzene is formed.

Question 2.
Distinguish among p, s and t amines under the following points:

1. Reaction with HNO₂
2. Carbyl amine reaction
3. Mustard oil reaction
4. Reaction with acid halide
5. Reaction with alkyl halide
6. Reaction with C₆H₅SO₂Cl

Answer:
Differences among p, s and t amines :

Question 3.
Explain the laboratory method of preparation of ethyl amine under the following heads:

1. Procedure
2. Equation of reaction
3. Diagram
4. Physical properties.

Answer:
Laboratory method of preparation of ethyl amine:
In laboratory ethyl amine is prepared by Hofmann Bromoamide reaction. In this process reaction of propanamide with bromine and caustic potash solution takes place. All steps in the process are following:

C₂H₅CONH₂ + Br₂ → C₂H₅CONHBr + HBr
KOH + HBr → KBr + H₂O
C₂H₅CONHBr + KOH → C₂H₅NCO + KBr + H₂O
C₂H₅NCO + 2KOH → C₂H₅NH₂ + K₂CO₃
C₂H₅CONH₂ + Br₂ + 4KOH → C₂H₅NH₂ + 2KBr + K₂CO₃ + 2H₂O

Procedure:
In a round bottom distillation flask equivalent quantities of bromine and propanamide is taken and 10% KOH solution is added in it. Now 50% KOH solution is added in excess and the flask is heated on water bath upto 57.67°C. When the solution becomes colourless, ethyl amine starts to be distilled which is absorbed in dil. HCl.

Physical properties:
It is a colourless liquid with an odour like ammonia which is soluble in water and organic solvent. It is a combustible substance. Its boiling point is 19°C.

Question 4.
Write laboratory method of preparation of aniline. Give all chemical equations of this process.
Or,
Describe commercial method of preparation of aniline.
Answer:
Laboratory Method for the Preparation of Aniline:
Aniline is prepared in the laboratory by the reduction of nitro – benzene with tin and hydrochloric acid.
C₆H₅NOZ +6[H] → C₆H₅NH₂ + 2H₂O

Experiment:
Nitrobenzene (20 g) and granulated tin (40 g) are taken in a round – bottom flask and a reflux condenser is fitted. Now hydrochloric acid (100 ml) is added in small amounts (10 ml at a time). The flask is shaken after each addition of acid and the temperature is not allowed to rise above 90°C. The flask is heated on a water bath till the smell of nitro – benzene disappears. On cooling the flask, a solid mass of molecular formula
(C₆H₅NH₂.HCl)₂.SnCl₄ separates out.

Sn + 4HCl → SnCl₄ +4H
C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O

The solid mass is treated with concentrated NaOHsolution. So that a clear alkaline solution is obtained. Aniline is liberated and floats as a dark brown coloured oil.

From this aniline is obtained by steam distillation.

Question 5.
Write down the chemical equation representing preparation of nitrobenzene in laboratory and give the following chemical reaction of nitrobenzene:

1. Nitration
2. Sulphonation.

Answer:
Preparation of nitrobenzene in laboratory : Nitrobenzene is prepared in laboratory by treating benzene with a mixture of cone. HNO₃and cone. H₂SO₄at 60°C.

Reaction of nitrobenzene:
1. Nitration : Nitrobenzene forms m – dinitrobenzene when heated with fuming HNO₃ and cone. H₂SO₄.

2. Sulphonation : Nitrobenzene forms m – nitrobenzene sulphonic acid on heating with fuming H₂SO₄.

Question 6.
Write short notes on the following : (NCERT)

1. Carbylamine reaction
2. Diazotisation
3. Hofmann’s bromamide reaction,
4. Coupling reaction
5. Ammonolysis
6. Acetylation
7. Gabriel phthalimide synthesis.

Answer:
1. Carbylamine reaction:
Primary aliphatic amine or aniline when warmed with chloroform in presence of alcoholic KOH gives isocyanide or carbylamine, a compound with disagreeable odour. This reaction is known as Carbylamine reaction.

2. Diazotisation:
By action of NaNO₂ solution in ice cooled solution of aromatic amine formed inorganic acid, diazonium salts are obtained.

Diazonium salt contains diazo group (- N = N -) therefore, this process is called diazotization.

3. Hofmann’s bromamide reaction:
Primary aliphatic and aromatic amines can be prepared from amides by treatment with Br₂ and KOH. The amine formed contains one carbon atom less than the parent amide. Therefore, this method is used for stepping down the series in organic conversion. Due to this reason it is also known as Hofmann degradation.

4. Coupling reaction:
Aniline reacts with diazonium chloride at ice cold temperature to giveoright orangered dye.

5. Ammonolysis:
It is a process of replacement of either halogen atom in alkyl halides (or aryl halides) or hydroxyl group in alcohols (or phenols) by amino group. The reagent used for ammonolysis is alcoholic ammonia. Generally, a mixture of primary, secondary and tertiary amine is formed.

6. Acetylation:
Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. This reaction is considered as replacement of hydrogen atom of – NH₂ or >NH group by acyl group.

This reaction is known as acylation. The reaction is carried out in the presence of a base stronger than amine like pyridine, which removes HCl so formed and shift the equilibrium towards right hand side. The product obtained by acylation reaction is known as amides.

7. Gabriel phthalimide synthesis:
In this reaction, potassium phthalimide react with alkyl halide to form N – methyl phthalimide which on hydrolysis give primary amine.

Question 7.
Complete the following reactions
(i) C₆H₅NH₂+ CHCl₃+ (alc.)KOH →
(ii) C₆H₅N₂Cl + H₃PO₂+H₂O →
(iii) C₆H₅NH₂+ H₂SO₄(Conc.) →
(iv) C₆H₅N₂Cl + C₂H₅OH →
(v) C₆H₅NH₂+Br_2(aq) →
(vi) C₆H₅NH₂+ (CH3CO)₂O →
(viii)
Answer:

MP Board Class 12th Chemistry Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Aldehydes, Ketones and Carboxylic Acids Important Questions

Aldehydes, Ketones and Carboxylic Acids Short Answer Type Questions

Question 1.
Arrange the following compounds in increasing order of their boiling points :
CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃. (NCERT)
Answer:
CH₃CH₂CH₃ < CH₃O CH₃ < CH₃CHO < CH₃CH₂OH.
This order can be predicted on the basis of inter-molecular force operating between them, these are having comparable molecular mass. CH₃CH₂OH undergoes the strongest H – bonding. In CH₃OCH₃ and CH₃CHO dipole – dipole attraction is more in CH₃CHO, since CH₃CHO is more polar than CH₃OCH₃ therefore its boiling point is more than CH₃ – O – CH₃. Propane being non – polar therefore, weak van der Waals’ forces exist between them.

Question 2.

1. Why ketones are less reactive than aldehydes?
2. Benzaldehyde is less reactive than Acetaldehyde. Why?

Answer:
1. Ketones are less reactive than aldehydes because in ketones there are two alky group attached with carbonyl group, due to the positive inductive effect (+I) of both the alkyl group the positive charge on carbon atom decreases. Hence, the sensitivity of ketones to the nucleophilic reagents decreases. In aldehydes, they have only one alkyl group so they are more reactive than ketones.

2. – CHO group of benzaldehyde becomes stable due to resonance with benzene ring whereas resonance is not found in acetaldehyde. Benzaldehyde is aromatic and alde – hyde is aliphatic.

Question 3.
How is urotropine obtained from formaldehyde? Write its chemical name and structural formula.
Answer:
When formaldehyde is treated with ammonia, urotropine is formed. Its chemical name is hexamethylene tetra ammine or hexa ammine.

Question 4.
Write a short note on Tollen’s reagent.
or,
What is Tollen’s reagent? Write its reaction with acetaldehyde.
Answer:
Tollen’s Reagent:
Ammoniacal silver nitrate solution is known as Tollen’s reagent. When Tollen’s reagent is heated with aldehyde, aldehyde reduces Ag+ to Ag and forms a bright silver mirror on the wall.

Ketones do not give this test.

Question 5.
Why boiling point of carboxylic acid is higher than alcohols having same molecular mass?
Answer:
Carboxylic acid exist as dimer due to hydrogen bond. These bonds are more stronger in acids compared to alcohols, therefore boiling point of carboxylic acid is higher than alcohol.

Question 6.
Explain Fehling reaction with equation.
Answer:
Fehling Reaction:
Sodium, Potassium tartarate associated with alkaline CuSO₄ is known as Fehling solution. When aldehyde is heated with Fehling solution, then aldehyde is oxidized and red precipitate of cuprous oxide is obtained. This is known as Fehling test.

Ketones do not give this test.

Question 7.
Why is the boiling point of ketone little higher than its corresponding isomeric aldehyde?
Answer:
Ketones are comparatively more polar than their corresponding isomeric alde – hyde because the >C=O group in ketone is linked with two electron releasing alkyl group. Thus, the dipole attractive force of ketone is comparatively higher. This is the reason that the boiling point of ketone is comparatively higher than its corresponding isomeric aldehyde.

Question 8.
Among formaldehyde, acetaldehyde and acetone which is more reactive and why? Explain.
Answer:
Among HCHO, CH₃CHO and CH₃COCH₃, HCHO is more reactive. This can be explained on the basis of:
1. Electron releasing effect:
Alkyl groups are electron releasing in nature due to which magnitude of positive charge on carbonyl carbon decreases and hence it becomes less susceptible to nucleophilic attack.

2. Steric effect:
The bulkier groups in ketones hinders approach of the nucleophile to the carbonyl carbon. This is known as steric effect. Thus, HCHO with negligible electron releasing effect as well as steric effect is more reactive.

Question 9.
Compare acidic strength of acetic acid, formic acid and chloroacetic acid.
Answer:
Chlorine atom present in chloroacetic acid has strong negative inductive effect (-I). Due to this, electrons of O – H bond easily displaced towards oxygen and it releases H⁺ easily. CH₃ group present in CH₃COOH which produces (+I) effect causes decrease in acidic nature.

In formic acid there is no such group which produces (+1) or (-1) effect. Hence, formic acid is stronger than acetic acid and chloroacetic acid is stronger than acetic acid. In short chloroacetic acid is stronger than formic acid and formic acid is stronger than acetic acid.

Question 10.

1. What is Hell – Volhard – Zelinsky (HVZ) reaction?
2. What happens when formic acid is heated?

Answer:
1. Hell – Volhard – Zelinsky Reaction:
When carboxylic acid is treated with Cl₂ or Br₂ in presence of phosphorus, α – halogenated carboxylic acid is formed. This reaction is known as Hell – Volhard – Zelinsky reaction (HVZ).

2. When formic acid is heated to 160°C it dissociates into CO and H₂O.

Question 11.
Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why? (NCERT)
Answer:

In carboxylate ion negative charge is delocalised over two oxygen atoms which are highly electronegative whereas in phenoxide ion negative charge is delocalised over only one oxygen atom. Carboxylate ion is more stable than phenoxide ion that is why carboxylic acid is more acidic than phenols.

Question 12.
Give chemical equation of the following :

1. Acetaldehyde from formaldehyde.
2. Formaldehyde from acetaldehyde.
3. Acetic acid from formic acid.

Answer:
1. Acetaldehyde from formaldehyde :

2. Formaldehyde from acetaldehyde :

3.

Question 13.
Write down the difference between formic acid and acetic acid on the basis of following points :

1. Effect of heat.
2. Reaction with acidified KMnO₄.
3. Distillation of Ca salt.
4. Reaction with ammoniacal silver nitrate solution.
5. Reaction with PCl₅.

Answer:
Differences between Formic Acid and Acetic Acid :

Question 14..
Explain Stephen’s reaction and Benzoin condensation with example.
Answer:
stephen’s reaction:
Alkyl cyanide on reduction with acidified stannous chloride (i.e., SnCl₂ + HCl) at room temperature forms aldimine hydrochloride, which on hydrolysis with boiling water gives aldehyde. This specific type of reduction of cyanide is known as Stephen’s reaction.
SnCl₂ + 2HCl → SnCl₄ + 2H

Benzoin condensation:
Two molecules of benzaldehyde in presence of alcoholic KCN or NaCN condenses to form benzoin.

Question 15.

1. Write a short note on Perkin’s reaction.
2. What happens when acetone is heated with H₂SO₄?

Answer:
1. Perkin’s reaction:
When aromatic aldehyde is heated in presence of sodium salt of aliphatic acid with anhydride of aliphatic acid, then α, β unsaturated acid is obtained.

2. In presence of H₂SO₄ three molecules of acetone get condensed and form mesitylene.

Aldehydes, Ketones and Carboxylic Acids Long Answer Type Questions

Question 1.
Write down the difference between compounds containing aldehydic group and ketonic group.
Answer:
Differences between Aldehydic group and Ketonic group :

Question 2.
Describe the following: (NCERT)

1. Acetylation
2. Cannizzaro reaction
3. Cross – aldol condensation
4. Decarboxylation.

Answer:
1. Acetylation:
The introduction of an acetyl functional group into an organic compound is known as acetylation. It is usually carried out in the presence of a base such as pyridine, dimethylaniline, etc. This process involves the substitution of an acetyl group for an active hydrogen atom. Ecetyl chloride and acetic anhydride are commonly used as acety – lating agents.
For example, acetylation of ethanol produces ethyl acetate.
CH₂CH₂OH + CH₃COCl → CH₃COOC₂H₅ + HCl

2. Cannizzaro reaction:
Aldehydes which do not contain α – hydrogen like HCHO, C₆H₅CHO react with cone. NaOH solution to form methyl alcohol and formic acid. This reaction is called Cannizzaro reaction.

3. Cross – aldol condensation:
When aldol condensation is carried out between different aldehydes or two different ketones or an aldehyde and a ketone, then the reaction is called a Cross – aldol condensation. If both the reactants contain α – hydrogens, four compounds are obtained as products.

4. Decarboxylation:
Decarboxylation refers to the reaction in which carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with soda – lime.

Decarboxylation also takes place when aqueous solutions of alkali metal salts of carboxylic acids are electrolysed. This electrolytic process is known as Kolbe’s electrolysis.

Question 3.
Describe the laboratory method of preparation of acetone. Draw labelled diagram and write down chemical equations.
Answer:
In laboratory, acetone is prepared by dry distillation of anhydrous calcium acetate.

Method 30 – 40 gm of calcium acetate mixed with equal amount of sodium acetate is heated in a glass retort fitted with a condenser and receiver. Acetone is collected in the receiver. The acetone so obtained is not pure. To purify this, it is shaken with saturated solution of sodium bisulphite then crystals of acetone sodium bisulphite salt separate out. The crystal is washed and heated with sodium carbonate and then dried over anhydrous and CaCl₂ then distilled at 56°C to get pure acetone.

(CH₃)₂C = O + NaHSO₃ → (CH₃)₂C(OH)SO₃Na
2(CH₃)₂C(OH)SO₃Na + Na₂CO₃ → 2(CH₃)₂C = O + 2Na₂SO₃ + H₂O + CO₂.

Question 4.
Write down the following reaction giving example and equation :

1. Iodoform reaction
2. Tischenko reaction
3. Gattermann – Koch synthesis
4. Rosenmund’s reaction.

Answer:
1. Iodoform (Haloform) reaction:
Acetaldehyde or methyl ketone reacts with iodine in presence of alkali to form yellow coloured iodoform. This reaction is known as Iodoform test.
2NaOH + I₂ → NaI + NaOI + H₂O

2. Tischenko reaction:
Two molecules of benzaldehyde is coupled together in presence of aluminium ethoxide or isopropoxide then benzylbenzoate (ester) is formed.

3. Gattermann – Koch synthesis:
Mixture of CO and HCl bubbled through a solution of aromatic hydrocarbon in ether solution in the presence of anhydrous AlCl₃, then benzaldehyde is formed.

4. Rosenmund’s reaction:
Aldehydes are obtained by the reduction of acid chloride with hydrogen in boiling xylene in presence of a catalyst Pd suspended in BaSO₄.

This reaction is called Rosenmund reaction.

Question 5.
Give quick vinegar method of preparation of acetic acid. Give its reaction with phosphorus pentaoxide and phosphorus pentachloride and write its two uses.
Answer:
In this process, a dilute aqueous solution of ethyl alcohol is oxidized in presence of enzyme Mycoderma aceti.

In this process, a wooden vat is fitted with two wooden plates having holes. Between these plates is filled by beech wood savings, moistened with old vinegar solution which is the chief source of Mycoderma aceti. A 10% aqueous solution of ethyl alcohol is dropped slowly from the top of the vat and air is passed at a controlled rate through the holes near the bottom of the vat. Ethyl alcohol is oxidized to acetic acid. This process is called quick vinegar process because vinegar is formed very quickly.

(a) Reaction with P₂O₅:

(b) Reaction with PCl₅:
CH₃CHOOH + PCl₅ → CH₃COCl + POCl₃ + HCl

Uses:

1. As a reagent and solvent in the lab.
2. As vinegar in the preparation of pickel, chutney etc.
3. In the preparation of methyl acetate, ethyl acetate and other esters.

Question 6.
Write a brief note on :

1. Claisen condensation
2. Benzoin condensation.

Answer:
1. Claisen condensation:
When aromatic aldehyde reacts with aliphatic alde – hyde or ketone with α – hydrogen, in presence of weak base, α, β unsaturated aldehyde or ketone is formed. This type of condensation is called Claisen condensation.

2. Benzoin condensation:
Two molecules of benzaldehyde in presence of alcoholic KCN or NaCN condenses to form benzoin.

Question 7.
How will you convert ethanal into the following compounds : (NCERT)
(i) Butan – 1,3 – diol
(ii) But – 2 – enal
(iii) But – 2 – enoic acid.

Answer:

Question 8.
What happens when, (only equation):

1. On reacting acetone with Grignard reagent?
2. Reaction of acetone with chloroform in presence of KOH?
3. Benzaldehyde reacts with aniline?
4. On heating sodium salt of carboxylic acid with soda lime?
5. Benzene reacts with acetyl chloride in presence of anhydrous AlCl₃?

Answer:
1. Reaction of acetone with Grignard reagent.

2. Reaction of acetone with chloroform.

3. Reaction of benzaldehyde with aniline

4. Reaction of sodium salt of carboxylic acid with soda lime.

5. Reaction of benzene with CH₃COCl.

Question 9.
How will you obtain the following from acetic acid (Give only equations):

1. Acetamide
2. Ethyl acetate
3. Acetic Anhydride
4. Trichloro acetic acid.

Answer:
1. Acetamide:

2. Ethyl acetate:

3. Acetic anhydride:

4. Trichloro acetic acid:
CH₃ – COOH + 3Cl₂ → CCl₃ – COCH + 3HCl

MP Board Class 12th Chemistry Important Questions

ΨMP Board Class 12th Maths Important Questions Chapter 2 Inverse Trigonometric Functions

Inverse Trigonometric Functions Important Questions

Inverse Trigonometric Functions Objective Type Questions:

Question 1.
Choose the correct answer:

Question 1.
If sin^-1x – cos ^-1 x = \(\frac { \pi }{ 6 } \), then the value of x is equal to:
(a) \(\frac{1}{2}\)
(b) \(\frac { \sqrt { 3 } }{ 2 } \)
(c) \(\frac{-1}{2}\)
(d) None of these
Answer:
(a) \(\frac{1}{2}\)

Question 2.
If tan^-13 + tan^-1 8, then the value of x is equal to:
(a) \(\frac { \pi }{ 2 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) \(\frac { \pi }{ 4 } \)
(d) \(\frac { -3\pi }{ 4 } \)
Answer:
(b) \(\frac { \pi }{ 3 } \)

Question 3.
tan^-1 \(\frac{x}{y}\) + tan^-1 \(\frac{x-y}{x+y}\) is equal to:
(a) \(\frac { \pi }{ 2 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) \(\frac { \pi }{ 4 } \)
(d) \(\frac { -3\pi }{ 4 } \)
Answer:
(c) \(\frac { \pi }{ 4 } \)

Question 4.
The value of 2 tan^-1 {cosec (tan^-1 x ) – tan (cot^-1 x)} is equal to:
(a) cot^-1x
(b) cot^-1\(\frac{1}{x}\)
(c) tan^-1x
(d) tan^-1\(\frac{1}{x}\)
Answer:
(c) tan^-1x

Question 5.
The value of tan{cos^-1\(\frac { 1 }{ 5\sqrt { 2 } } \) – sin^-1 \(\frac { 4 }{ \sqrt { 17 } } \)} is equal to:
(a) \(\frac { \sqrt { 29 } }{ 3 } \)
(b) \(\frac{29}{3}\)
(c) \(\frac { \sqrt { 3 } }{ 29 } \)
(d) \(\frac{3}{29}\)
Answer:
(d) \(\frac{3}{29}\)

Question 2.
Fill in the blanks:

1. tan^-1(1) + tan^-1(2) + tan^-1 (3) = …………………………..
2. tan^-1(2) – tan^-1 (1) = …………………………..
3. cot^-1 3 + cosec^-1 \(\sqrt { 5 } \) = ……………………………
4. sin(sin^-1 x + 2 cos^-1 x) = ……………………………….
5. If sin^-1(\(\frac { 2a }{ 1+a^{ 2 } } \)) + sin^-1 (\(\frac { 2b }{ 1+b^{ 2 } } \)) = 2 tan^-1 x, then x = ……………………….
6. If tan^-1 \(\frac{1-x}{1+x}\) = \(\frac{1}{2}\) tan^-1 x ( When x > 0), then x = ………………………..
7. tan^-1\(\frac{a-b}{1+ab}\) + tan^-1\(\frac{b-c}{1+bc}\) + tan^-1c = ………………………….

Answer:

1. π
2. tan^-1\(\frac{1}{3}\)
3. \(\frac{π}{4}\)
4. x
5. \(\frac{a+b}{1-ab}\)
6. \(\frac { 1 }{ \sqrt { 3 } } \)
7. tan^-1 (a)

Question 3.
Write True/False:

1. tan^-1x + tan^-1 y = tan^-1\(\frac{x+y}{1-xy}\)
2. cos^-1(-x) = – cos^-1 x
3. sin^-1(3x – 4x³) = sin^-1 \(\frac{x}{3}\)
4. cos^-1 (\(\frac { 1-x^{ 2 } }{ 1+x^{ 2 } } \)) = 2 tan^-1x
5. sin^-1x – sin^-1[xy – \(\sqrt { 1-x^{ 2 } } \) \(\sqrt { 1-y^{ 2 } } \)]

Answer:

1. True
2. False
3. False
4. True
5. False

Question 4.
Match the column:

Answer:

1. (b)
2. (e)
3. (f)
4. (a)
5. (c)
6. (d)

Question 5.
Write the answer in one word/sentence:

1. Find the value of tan^-1\(\frac{1}{2}\) + tan^-1\(\frac{3}{2}\)
2. Find the value of tan^-1\(\frac { x }{ \sqrt { 1-x^{ 2 } } } \)
3. Find the value of sin ( 2 sin^-1\(\frac{3}{5}\))
4. Solve the equation: sin^-1\(\frac{x}{5}\) + cosec ^-1\(\frac{5}{4}\) = \(\frac { \pi }{ 2 } \)
5. Write the principal value of cos^-1 (cos \(\frac { 7\pi }{ 6 } \))
6. If cos^-1 (\(\frac{1}{x}\)) = θ, then find the value of tan θ

Answer:

1. tan^-1 8
2. sin^-1x
3. \(\frac{24}{25}\)
4. x = 3
5. \(\frac { 5\pi }{ 6 } \)
6. \(\sqrt { x^{ 2 }-1 } \)

Inverse Trigonometric Functions Very Short Answer Type Questions

Question 1.
Find the principle value of the following?

1. tan^-1[sin(- \(\frac { \pi }{ 2 } \)) ] (CBSE 2014)
2. cot [ \(\frac { \pi }{ 2 } \) – 2 cot^-1 \(\sqrt{3}\) ] (CBSE 2014)
3. tan^-1(- \(\sqrt{3}\) )
4. sec^-1 ( \(\frac{-2}{3}\) \(\sqrt{3}\) ) (NCERT)
5. cosec^-1(2) (NCERT)

Solution:
1. Let, tan^-1[ sin(- \(\frac { \pi }{ 2 } \) ) = θ
⇒ tan^-1 [-sin \(\frac { \pi }{ 2 } \) ] = θ
⇒ tan^-1 (-1) = θ
⇒ tan θ = 1
⇒ tan θ = – tan \(\frac { \pi }{ 4 } \)
⇒ tan θ = tan ( \(\frac { -\pi }{ 4 } \) )
θ = \(\frac { -\pi }{ 4 } \)
∴ The principle value is \(\frac { -\pi }{ 4 } \)

2. cot(\(\frac { -\pi }{ 2 } \) – cot^-1 \(\sqrt{3}\))
Let cot^-1 \(\sqrt{3}\) = θ
⇒ cot θ = \(\sqrt{3}\)
⇒ cot θ = cot \(\frac { \pi }{ 6 } \)
∴ θ = \(\frac { \pi }{ 6 } \)
∴ cot ( \(\frac { \pi }{ 2 } \) – cot^-1\(\sqrt{3}\) ) = cot (\(\frac { \pi }{ 2 } \) – 2 × \(\frac { \pi }{ 6 } \))
= cot ( \(\frac { \pi }{ 2 } \) – \(\frac { \pi }{ 3 } \) )
= cot \(\frac { \pi }{ 6 } \)
= \(\sqrt{3}\)
∴ The principal value is \(\sqrt{3}\)

3. Let tan(- \(\sqrt{3}\)) = θ
⇒ tan θ = – \(\sqrt{3}\)
⇒ tan θ = – tan (\(\frac { \pi }{ 3 } \))
⇒ tan θ = tan (- \(\frac { \pi }{ 3 } \))
⇒ θ = – \(\frac { \pi }{ 3 } \)
Hence the principle value is – \(\frac { \pi }{ 3 } \)

4. Let sec^-1( \(\frac{-2}{3}\) \(\sqrt{3}\) ) = θ
⇒ sec^-1( \(\frac { -2 }{ \sqrt { 3 } } \) ) = θ
⇒ sec θ = \(\frac { -2 }{ \sqrt { 3 } } \)
⇒ sec θ = – sec ( \(\frac { \pi }{ 6 } \) )
⇒ sec θ = sec (π – \(\frac { \pi }{ 6 } \) )
⇒ sec θ = sec \(\frac { 5\pi }{ 6 } \)
∴θ = \(\frac { 5\pi }{ 6 } \)
The principle value is \(\frac { 5\pi }{ 6 } \)

5. Let cosec^-1(2) = θ
⇒ cosec θ = 2
⇒ cosec θ = cosec \(\frac { \pi }{ 6 } \) θ ∈ [- \(\frac { \pi }{ 2 } \), \(\frac { \pi }{ 2 } \) ]
The principle value is \(\frac { \pi }{ 6 } \).

Question 2.
Prove the following:

1. 2 cos^-1(\(\frac{4}{5}\)) = cos^-1( \(\frac{7}{25}\) )
2. 2 sin^-1( \(\frac{5}{13}\) ) = sin^-1( \(\frac{120}{169}\) )
3. 2 sin^-1( \(\frac{3}{5}\) ) = sin^-1( \(\frac{24}{25}\) )

Solution:
1. 2 cos^-1( \(\frac{4}{5}\) ) = cos^-1( \(\frac{7}{25}\) )
Formula 2 cos^-1 x = cos^-1(2x² – 1)
∴ L.H.S = 2 cos^-1( \(\frac{4}{5}\) )
= cos^-1 (2 \(\frac{16}{25}\) – 1)
= cos^-1 ( \(\frac{32}{25}\) – 1)
= cos^-1( \(\frac{32-25}{25}\) )
= cos^-1\(\frac{7}{25}\)
= R.H.S.

2. 2 sin^-1\(\frac{3}{5}\) = sin^-1\(\frac{24}{25}\)
Formula 2 sin^-1(x) = sin^-1(2x\(\sqrt { 1-x^{ 2 } } \))
∴ 2 sin^-1 \(\frac{3}{5}\) = sin^-1[2. \(\frac{3}{5}\) \(\sqrt { 1-\frac { 9 }{ 25 } } \)]
= sin^-1[ \(\frac{6}{5}\) \(\sqrt { \frac { 16 }{ 25 } } \) ]
= sin^-1[ \(\frac{6}{5}\) . \(\frac{4}{5}\) ]
= sin^-1[ \(\frac{24}{25}\) ]
= R.H.S. Proved.

3. 2 sin^-1(\(\frac{5}{13}\)) = sin^-1\(\frac{120}{169}\)
Solve like Q.2(b)

Question 3.
Find the value of tan^-1{2 cos(2 sin^-1\(\frac{1}{2}\)} (CBSE 2013, NCERT)
Solution:
tan^-1[2 cos(2 sin^-1\(\frac{1}{2}\)) ]
= tan^-1[ 2 cos (2 sin^-1 sin \(\frac { \pi }{ 6 } \)) ]
= tan^-1[ 2 cos (2. \(\frac { \pi }{ 6 } \)) ]
= tan^-1[ 2 cos \(\frac { \pi }{ 3 } \) ]
= tan^-1 [ 2. \(\frac{1}{2}\) ]
= tan^-1 1
= \(\frac { \pi }{ 4 } \).

Question 4.
Find the value of sin [ \(\frac { \pi }{ 3 } \) – sin^-1( \(\frac{-1}{2}\) ) ]? [CBSE 2008, 2013]
Solution:
sin[ \(\frac { \pi }{ 3 } \) – sin^-1(\(\frac{-1}{2}\) ) ] = sin^-1 [ \(\frac { \pi }{ 3 } \) – ( – sin^-1\(\frac{1}{2}\) ) ]
= sin^-1 [ \(\frac { \pi }{ 3 } \) + sin^-1 sin\(\frac { \pi }{ 6 } \) ]
= sin^-1 [ \(\frac { \pi }{ 3 } \) + \(\frac { \pi }{ 6 } \) ]
= sin^-1 ( \(\frac { \pi }{ 2 } \) )
= 1.

Question 5.
Prove that:
2 tan^-1\(\frac{1}{5}\) = tan^-1 ( \(\frac{5}{12}\) )
Solution:
2 tan^-1( \(\frac{1}{5}\) ) = tan^-1 ( \(\frac{5}{12}\) )
L.H.S. = 2 tan^-1 ( \(\frac{1}{5}\) )

= tan^-1 [ \(\frac{2×25}{5×24}\) ]
= tan^-1 [ \(\frac{5}{12}\) ]
= R.H.S. Proved.

Question 6.
Find the value of tan [ 2 tan^-1 \(\frac{1}{5}\) – \(\frac { \pi }{ 4 } \) ]?
solution:

= tan tan^-1( \(\frac{-7}{17}\) )
= \(\frac{-7}{17}\).

Question 7.
Prove that: 3 sin^-1 x = sin^-1 (3x – 4x³)? (NCERT, CBSE 2018)
Solution:
Let sin^-1 x = θ
⇒ x = sin θ
We know that sin 3θ = 3 sinθ – 4 sin³ θ
= 3x – 4x³
⇒ 3θ = sin^-1 ( 3x – 4x³)
⇒ 3.sin^-1 x = sin^-1(3x – 4x³). proved.

Question 8.
Prove that: 3 cos^-1 x = cos^-1 (4x³ – 3x)? (NCERT)
Solution:
Let cos^-1 x = cos θ
⇒ x = cos θ
We know that cos 3θ = 4 cos³θ – 3 cos θ
= 4x³ – 3x
⇒ 3θ = cos^-1 (4x³ – 3x)
⇒ 3 cos^-1x = cos^-1 (4x³ – 3x). Proved.

Question 9.
Prove the following:

1. tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{3}\) = \(\frac { \pi }{ 4 } \)
2. cos^-1 \(\frac{12}{13}\) = tan^-1 \(\frac{5}{12}\)
3. cos^-1 \(\frac{3}{5}\) = sin^-1 \(\frac{4}{5}\)

Solution:
1. tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{3}\) = \(\frac { \pi }{ 4 } \)
L.H.S = tan^-1 \(\frac{1}{2}\) + tan^-1 \(\frac{1}{3}\)

∴ A = tan^-1 \(\frac{5}{12}\)
\(\frac { \pi }{ 4 } \) = R.H.S. Proved

2. cos^-1 \(\frac{12}{13}\) = tan^-1 \(\frac{5}{12}\)
Let cos^-1 \(\frac{12}{13}\) = A
\(\frac{12}{13}\) = cos A
sin A = \(\sqrt { 1-cos^{ 2 }A } \) = \(\sqrt { 1-\frac { 144 }{ 169 } } \)
= \(\sqrt { \frac { 25 }{ 169 } } \) = \(\frac{5}{13}\)
tan A = \(\frac { sinA }{ cosA } \) = \(\frac { 5/13 }{ 12/13 } \) = \(\frac{5}{12}\)
A = tan^-1 \(\frac{5}{12}\)
From eqns. (1) and (2), L.H.S = R.H.S. Proved.

3. cos^-1 \(\frac{3}{5}\) = sin^-1 \(\frac{4}{5}\)
Let cos^-1 \(\frac{3}{5}\) = A ……………… (1)
⇒ cos A = \(\frac{3}{5}\)
⇒ sin A = \(\sqrt { 1-cos^{ 2 }A } \)
= \(\sqrt { \frac { 16 }{ 25 } } \) = \(\frac{4}{5}\) ……………… (2)
A = sin^-1 \(\frac{4}{5}\).
From eqns. (1) and (2), L.H.S = R.H.S. Proved.

Question 10.
Prove that:

1. sec^-1 x + cosec^-1 x = \(\frac { \pi }{ 2 } \)
2. sin^-1x + cos^-1x = \(\frac { \pi }{ 2 } \)
3. tan^-1x + cot^-1x = \(\frac { \pi }{ 2 } \)

Solution:
1. sec^-1 x + cosec^-1x = \(\frac { \pi }{ 2 } \)
Let sec ^-1 x = θ
∴x = sec θ
⇒ x = cosec ( \(\frac { \pi }{ 2 } \) – θ)
⇒ cosec ^-1 x = \(\frac { \pi }{ 2 } \) – θ. Proved.

2. sin^-1 x + cos^-1 x = \(\frac { \pi }{ 2 } \)
Let sin^-1 x = θ ……………………. (1)
⇒ x = sin θ
⇒ x = cos ( \(\frac { \pi }{ 2 } \) – θ)
⇒ cos ^-1 x = \(\frac { \pi }{ 2 } \) – θ ………………. (2)
Adding eqns (1) and (2),
sin ^-1 x + cos ^-1 x = θ + \(\frac { \pi }{ 2 } \) – θ
⇒ sin ^-1 x + cos^-1 x = \(\frac { \pi }{ 2 } \) Proved.

3. tan ^-1 x + cot^-1 x = \(\frac { \pi }{ 2 } \)
Let tan ^-1 x = θ
⇒ x = tan θ
⇒ x = cot ( \(\frac { \pi }{ 2 } \) – θ)
⇒ cot ^-1 x = \(\frac { \pi }{ 2 } \) – θ
Adding eqns. (1) and (2),
tan ^-1 x + cot ^-1 x = \(\frac { \pi }{ 2 } \). Proved.

Question 11.
Prove the following:

1. tan^-1 5 – tan^-1 3 = tan^-1 \(\frac{1}{8}\)
2. tan^-1 3 – tan^-1 2 = tan^-1 \(\frac{1}{7}\)
3. tan^-1 7 – tan^-1 5 = tan^-1 3 = tan^-1 \(\frac{1}{18}\)

Solution:
1. tan^-1 5 – tan^-1 3 = tan^-1 \(\frac{1}{8}\)
L.H.S. = tan^-1 5 – tan^-1 3
= tan^-1 \(\frac{5-3}{1+5.3}\) = tan^-1 \(\frac{2}{16}\) = tan^-1 \(\frac{1}{8}\)
= R.H.S. Proved.

2. Solve like Q.No. 11 (A).

3. Solve like Q.No. 11(A).

Question 12.
Prove that:

1. tan^-1 \(\frac{4}{7}\) – tan^-1 \(\frac{1}{5}\) = tan^-1 \(\frac{1}{3}\)
2. tan^-1 \(\frac{1}{2}\) – tan^-1 \(\frac{2}{9}\) = tan^-1 \(\frac{1}{4}\)
3. tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{8}\) = tan^-1 \(\frac{3}{11}\)

Solution:
1. tan^-1 \(\frac{4}{7}\) – tan^-1 \(\frac{1}{5}\) = tan^-1 \(\frac{1}{3}\)

2. Solve like Q.No. 12 (A).

3. Solve like Q.No. 12 (A).

Question 13.
tan^-1 1 + tan^-1 2 + tan^-1 3 = π?
Solution:
L.H.S. = tan^-1 1 + (tan^-1 2 + tan^-1 3)
= tan^-1 (1) + π + tan^-1 ( \(\frac{2+3}{1-2×3}\) ),
[∵ tan^-1 x + tan^-1 y = π + tan^-1 \(\frac{x+y}{1-xy}\), if x > 0, y > 0, xy > 1 Here xy = 6 > 1]
= tan^-1 )1) + π + tan^-1( \(\frac{5}{1-6}\) )
= tan^-1 (1) + π + tan^-1 (-1)
= tan^-1 (1) + π – tan^-1 (1), [∵tan^-1 (-x) = – tan^-1 x]
= π = R.H.S. Proved.

Question 14.
(A) If tan^-1 ( \(\frac{1}{2}\) ) + tan^-1 ( \(\frac{1}{k}\) ) = \(\frac { \pi }{ 4 } \) then find the value of k?
Solution:
tan^-1 ( \(\frac{1}{2}\) ) + tan^-1 ( \(\frac{1}{k}\) ) = \(\frac { \pi }{ 4 } \)

⇒ \(\frac{k+2}{2k – 1}\) = 1
⇒ k + 2 = 2k – 1
⇒2 + 1 = 2k – k
⇒ k = 3.

(B) If tan ^-1 ( \(\frac{1}{2}\) ) + tan^-1 ( \(\frac{1}{k}\) ) = \(\frac { \pi }{ 4 } \) then find the value of k?
Solution:
Solve like Q.No. 14 (A).
[Answer: k = -1]

(C) If tan ^-1 ( \(\frac{4}{5}\) ) + tan^-1 ( \(\frac{1}{k}\) ) = \(\frac { \pi }{ 4 } \) then find the value of k?
Solution:
Solve like Q.No. 14 (A).
[Answer: k = 9]

Question 15.
Prove that:
tan ^-1 \(\sqrt { x } \) = \(\frac{1}{2}\) cos^-1 ( \(\frac{1-x}{1+x}\) )?
Solution:
R.H.S = \(\frac{1}{2}\) cos ^-1 ( \(\frac{1-x}{1+x}\) )
Let \(\sqrt { x } \) = tan θ
⇒ x = tan² θ
⇒ \(\frac{1-x}{1+x}\) = \(\frac { 1-tan^{ 2 }\theta }{ 1+tan^{ 2 }\theta } \) = cos 2θ
∴ R.H.S. = \(\frac{1}{2}\) cos ^-1(cos 2θ)
= \(\frac{1}{2}\) × 2θ = θ
= tan^-1 ( \(\sqrt { x } \) ) [∵\(\sqrt { x } \) = tan θ ⇒ tan^-1 ( \(\sqrt { x } \) ) = θ]
= L.H.S. Proved.

Question 16.
Prove that:
sin (cos^-1 x ) = cos (sin^-1 x)?
Solution:
L.H.S. = sin(cos^-1 x)
= sin [ \(\frac { \pi }{ 2 } \) – sin ^-1 x],
[∵ sin^-1 x + cos^-1x = \(\frac { \pi }{ 2 } \) , cos^-1x = \(\frac { \pi }{ 2 } \) – sin^-1 x]
= cos (sin^-1 x), [ ∵sin (90° – θ) = cos θ ]
= R.H.S. Proved.

Question 17.
(A) Prove that:
tan^-1 ( \(\frac{b-c}{1+bc}\) ) + tan^-1 \(\frac{b-c}{1+bc}\) + tan^-1 c = tan^-1 b?
Solution:
L.H.S = tan^-1 ( \(\frac{b-c}{1+bc}\) ) + tan^-1 \(\frac{c-a}{1+ca}\) + tan^-1 a
= (tan^-1 a – tan^-1 b ) + (tan^-1 b – tan^-1 c) + tan^-1 c
= tan^-1 b – tan^-1 c + tan^-1 c – tan^-1 a + tan^-1 a
= tan^-1 b = R.H.S. Proved.

(B) Prove that:
tan^-1 ( \(\frac{a-b}{1+ab}\) ) + tan^-1 \(\frac{b-c}{1+bc}\) + tan^-1 c = tan^-1 a?
Solution:
L.H.S = tan^-1 ( \(\frac{a-b}{1+ab}\) ) + tan^-1 \(\frac{b-c}{1+bc}\) + tan^-1 c
= (tan^-1 a – tan^-1 b) + (tan^-1 b – tan^-1c) + tan^-1 c
= tan^-1 a = R.H.S. Proved.

(C) Prove that:
tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\) = tan^-1 \(\frac{2}{9}\)?
Solution:
tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\) = tan^-1 \(\frac{2}{9}\)
L.H.S = tan^-1 \(\frac{1}{7}\) + tan^-1 \(\frac{1}{13}\)

= tan^-1 \(\frac{20}{91}\) × \(\frac{91}{90}\) = tan^-1 \(\frac{20}{90}\) = tan^-1 \(\frac{2}{9}\)
= R.H.S. Proved.

Question 18.
Solve the equation:
sin^-1 \(\frac { 2a }{ 1+a^{ 2 } } \) + sin^-1 \(\frac { 2b }{ 1+a^{ 2 } } \) = 2 tan^-1x?
Solution:
sin^-1 \(\frac { 2a }{ 1+a^{ 2 } } \) + sin^-1 \(\frac { 2b }{ 1+a^{ 2 } } \) = 2 tan^-1x, (given)
⇒ 2 tan^-1 a + 2 tan^-1 b = 2 tan^-1 x [∵sin^-1 \(\frac { 2x }{ 1+a^{ 2 } } \) = 2 tan^-1 x]
⇒ tan^-1 a + tan^-1 b = tan^-1 x
⇒ tan^-1 \(\frac{a+b}{1-ab}\) = tan^-1 x
∴ x = \(\frac{a+b}{1-ab}\).

Question 19.
solve the equation:
cos^-1 ( \(\frac { 1-a^{ 2 } }{ 1+a^{ 2 } } \) ) – cos^-1 ( \(\frac { 1-b^{ 2 } }{ 1+b^{ 2 } } \) ) = 2 tan^-1x?
Solution:
cos^-1 ( \(\frac { 1-a^{ 2 } }{ 1+a^{ 2 } } \) ) – cos^-1 ( \(\frac { 1-b^{ 2 } }{ 1+b^{ 2 } } \) ) = 2 tan^-1x, (given)
⇒ 2 tan^-1 a – 2 tan^-1 b = 2 tan^-1 x
⇒ tan^-1a – tan^-1 b = tan^-1 x
⇒ tan^-1 \(\frac{a-b}{1+ab}\) = tan^-1 x
∴ x = \(\frac{a-b}{1+ab}\).

Question 20.
(A) Prove the following:
2 tan^-1 \(\frac{1}{4}\) = tan^-1 \(\frac{8}{15}\)?
Solution:
∵ 2 tan^-1 x = tan^-1 ( \(\frac { 2x }{ 1-x^{ 2 } } \) )

L.H.S = tan^-1 \(\frac{16}{2.5}\)
= tan^-1 \(\frac{8}{15}\) = R.H.S. Proved.

(B) Prove the following:
2 tan^-1 \(\frac{1}{2}\) = tan^-1 \(\frac{4}{3}\)?
Solution:
We know that 2 tan^-1 x = tan^-1 ( \(\frac { 2x }{ 1-x^{ 2 } } \) )

= tan^-1 ( \(\frac{4}{3}\) )
= R.H.S. Proved.

Question 21.
Write in simplest form
tan^-1 \(\sqrt { \frac { 1-cosx }{ 1+cosx } } \)?
Solution:

Question 22.
Write in simplest form:
cos^-1 \(\sqrt { \frac { 1 }{ 2 } (1+cosx) } \)?
Solution:
cos^-1 \(\sqrt { \frac { 1 }{ 2 } (1+cosx) } \) = cos^-1 \(\sqrt { \frac { 1 }{ 2 } .2cos^{ 2 }\frac { x }{ 2 } } \)
= cos^-1 (cos \(\frac{x}{2}\) ) = \(\frac{x}{2}\).

Question 23.
If tan^-1 a + tan^-1 b + tan^-1 c = \(\frac { \pi }{ 2 } \) then prove that ab + bc + ca = 1?
Solution:
tan^-1 a + tan^-1 b + tan^-1 c = \(\frac { \pi }{ 2 } \) , given
⇒ tan^-1 a + tan^-1 b + tan ^-1 c = tan^-1 a + cot^-1 a, [∵tan^-1 a + cot ^-1 a = \(\frac { \pi }{ 2 } \) ]
⇒ tan^-1 b + tan^-1c = cot^-1 a
⇒ tan^-1 ( \(\frac{b+c}{1+bc}\) ) = \(\frac{1}{a}\)
⇒ ab + ca = 1 – bc
⇒ ab + bc + ca = 1. Proved.

Question 24.
Prove that:
tan^-1 \(\frac{2}{11}\) + cot^-1 \(\frac{24}{7}\) = tan^-1 \(\frac{1}{2}\)?
Solution:
L.H.S. = tan^-1 \(\frac{2}{11}\) + cot^-1 \(\frac{7}{24}\)
= tan^-1 \(\frac{2}{11}\) + tan^-1 \(\frac{7}{24}\)

tan^-1 = \(\frac{125}{250}\) = tan^-1 \(\frac{1}{2}\) = R.H.S.

Question 25.
Prove that:
cos^-1 x = 2 cos^-1 \(\sqrt { \frac { 1+x }{ 2 } } \)?
Solution:
R.H.S. = 2 cos^-1 \(\sqrt { \frac { 1+x }{ 2 } } \)
= 2 cos^-1 \(\sqrt { \frac { 1+cos\theta }{ 2 } } \)

= cos^-1 x
= L.H.S. Proved.

Question 26.
Prove that:
cos^-1 x = 2 tan^-1\(\sqrt { \frac { 1-x }{ 1+x } } \)?
Solution:
R.H.S. = 2 tan^-1 \(\sqrt { \frac { 1-x }{ 1+x } } \)
= 2 tan^-1 \(\sqrt { \frac { 1-cos\theta }{ 1+cos\theta } } \) (putting x = cos θ)

= 2. \(\frac { \theta }{ 2 } \) = θ = cos^-1 x = L.H.S. Proved.

Question 27.
tan^-1 ( \(\frac{a}{b}\) ) – tan^-1 ( \(\frac{a-b}{a+b}\) ) = \(\frac { \pi }{ 4 } \)?
Solution:
tan^-1 ( \(\frac{a}{b}\) ) – tan^-1 ( \(\frac{a-b}{a+b}\) ) = \(\frac { \pi }{ 4 } \)

⇒ 1 = tan\(\frac { \pi }{ 4 } \)
or tan\(\frac { \pi }{ 4 } \) = 1. Proved.

Inverse Trigonometric Functions Long Answer Type Questions – I

Question 1.
(A) Prove that:
sin^-1 \(\frac { 1 }{ \sqrt { 5 } } \) + sin^-1 \(\frac { 1 }{ \sqrt { 10 } } \) = \(\frac { \pi }{ 4 } \)?
Solution:
Let sin^-1 \(\frac { 1 }{ \sqrt { 5 } } \) = A, sin^-1 \(\frac { 1 }{ \sqrt { 10 } } \) = B
∴ A + B = \(\frac { \pi }{ 4 } \)
⇒ sin (A + B) = sin\(\frac { \pi }{ 4 } \)
⇒ sin (A + B) = sin\(\frac { \pi }{ 4 } \)
⇒ sin A cos B + cos A sin B = \(\frac { 1 }{ \sqrt { 2 } } \)
L.H.S = sin A \(\sqrt { 1-sin^{ 2 }B } \) + \(\sqrt { 1-sin^{ 2 }A } \). sin B
= \(\frac { 1 }{ \sqrt { 5 } } \). \(\sqrt { 1-\frac { 1 }{ 10 } } \) + \(\sqrt { 1-\frac { 1 }{ 5 } } \). \(\frac { 1 }{ \sqrt { 10 } } \)
= \(\frac { 3 }{ \sqrt { 5.\sqrt { 10 } } } \) + \(\frac { 2 }{ \sqrt { 5.\sqrt { 10 } } } \)
= \(\frac { 5 }{ \sqrt { 5.\sqrt { 10 } } } \) = \(\sqrt { \frac { 5 }{ 10 } } \) = \(\frac { 1 }{ \sqrt { 2 } } \)
= R.H.S. Proved.

(B) Solve the following equation:
sin^-1 x + sin^-1 (1 – x) = sin^-1 \(\sqrt { 1-x^{ 2 } } \)?
Solution:
Let sin^-1 x = α ∴ x = sin α
Here α + sin^-1 ( 1 – sin α) = sin^-1 \(\sqrt { 1-sin^{ 2 }\alpha } \)
⇒ α + sin^-1 ( 1 – sin α) = sin^-1 cos α
⇒ α + sin^-1 ( 1 – sin α) = sin^-1. sin ( \(\frac { \pi }{ 2 } \) – α)
⇒ α + sin^-1 ( 1 – sin α) = \(\frac { \pi }{ 2 } \) – α
⇒ sin^-1 ( 1 – sin α) = \(\frac { \pi }{ 2 } \) – 2α
⇒ 1 – sin α = sin ( \(\frac { \pi }{ 2 } \) – 2α)
⇒ 1 – sin α = cos 2α
⇒ 1 – cos 2α = sin α
⇒ 2 sin²α = sin α
⇒ sin α = \(\frac{1}{2}\) ∴α = \(\frac { \pi }{ 6 } \)
or x = \(\frac { \pi }{ 6 } \)

Question 2.

1. tan^-1 \(\frac{x+1}{x}\) – tan^-1 \(\frac{1}{2x+1}\) = \(\frac { \pi }{ 4 } \)?
2. If tan^-1 x + tan^-1 y + tan^-1 z = π then prove that x + y + z = xyz?
3. If tan^-1 x + tan^-1 y + tan^-1 z = \(\frac { \pi }{ 2 } \) then prove that xy + yz + zx = 1?

Solution:
1. tan^-1 \(\frac{x+1}{x}\) – tan^-1 \(\frac{1}{2x+1}\) = \(\frac { \pi }{ 4 } \)

= R.H.S.

2. tan^-1 x + tan^-1 y + tan^-1 z = π
⇒ tan^-1 \(\frac{x+y}{1-xy}\) + tan^-1 z = π

⇒ x + y + z – xyz = 0, [∵ tan π = 0]
∴ x + y + z = xyz. Proved.

3. Solve like Q.2(B), take tan \(\frac { \pi }{ 4 } \) = ∞ = \(\frac{1}{0}\)

Question 3.
Write in simplest form:
tan^-1 [ \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \) ]?
Solution:
tan^-1 [ \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \) ]
Let x = tan θ

Question 4.
(A) Prove the following:
\(\frac{1}{2}\) sin^-1 x = cot^-1 [ \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \) ]?
Solution:
R.H.S = cot^-1 [ \(\frac { \sqrt { 1+x^{ 2 }-1 } }{ x } \) ]

= L.H.S. Proved.

(B) Prove that:
\(\frac{1}{2}\) cot^-1 x = cot^-1 ( \(\sqrt { 1+x^{ 2 }+x } \) )?
Solution:
\(\frac{1}{2}\) cot^-1 x = cot^-1 ( \(\sqrt { 1+x^{ 2 }+x } \) )
R.H.S = cot^-1 ( \(\sqrt { 1+x^{ 2 }+x } \) )
Let x = cos θ
R.H.S = cot^-1 ( \(\sqrt { 1+cot^{ 2 }\theta } \) + cot θ )
= cot^-1 ( \(\sqrt { cosec^{ 2 }\theta } \) + cot θ )
= cot^-1 (cosec θ + cot θ)

= \(\frac{1}{2}\) cot^-1 x
= L.H.S. Proved.

Question 5.
Solve the following equation:
tan^-1 (x + 1) + tan^-1 (x – 1) = tan^-1 ( \(\frac{6}{17}\) )?
Solution:
Given: tan^-1 (x + 1) + tan^-1 (x – 1) = tan^-1 ( \(\frac{6}{17}\) )
⇒ tan^-1 (x + 1) + tan^-1 (x – 1) = tan^-1 ( \(\frac{6}{17}\) )

⇒ 17 x = 6 – 3x²
⇒ 3x² + 17x – 6 = 0
⇒ 3x² + 18x – x – 6 = 0
⇒ 3x (x + 6) – 1 (x + 6) = 0
⇒ (x+6) (3x – 1) = 0
∴ x = – 6, x = \(\frac{1}{3}\)

Question 6.
Prove that cos^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{4}{5}\) = \(\frac { \pi }{ 2 } \)?
Solution:
L.H.S = cos^-1 \(\frac{3}{5}\) + cos^-1 \(\frac{4}{5}\),

= \(\frac { \pi }{ 2 } \) = R.H.S Proved.

Question 7.
If cos^-1 x + cos^-1 y + cos^-1 z = π then prove that:
x² + y² + z² + 2xyz = 1?
Solution:
Given: cos^-1 x + cos^-1 z = π
⇒ cos^-1 x + cos^-1 y = π- cos^-1 z

Squaring on both sides
x²y² + z² + 2xyz = (1 – x²) (1 – y²)
⇒ x²y² + z² + 2xyz = 1 – y² – x² + x²y²
⇒ z² + 2xyz = 1 – y² – x²
⇒ x² + y² + z² + 2xyz = 1. Proved.

Question 8.
If sin^-1 \(\frac { 2a }{ 1+a^{ 2 } } \) – cos^-1 \(\frac { 1-b^{ 2 } }{ 1+b^{ 2 } } \) = tan^-1 \(\frac { 2x }{ 1-x^{ 2 } } \) then prove that:
x = \(\frac{a-b}{1+ab}\)?
Solution:


⇒ sin^-1 (sin 2θ) – cos^-1 (cos 2ϕ) = tan^-1 (tan 2Ψ)
⇒ 2θ – 2ϕ = 2Ψ
⇒ θ – ϕ = Ψ
⇒ tan^-1 a – tan^-1 b = tan^-1 x
⇒ tan^-1 ( \(\frac{a-b}{1+ab}\) ) = tan^-1 x
⇒ x = \(\frac{a-b}{1+ab}\). Proved.

Question 9.
Prove the following

Solution:
Let x = cos θ, then θ = cos^-1 x.

\(\frac { \pi }{ 4 } \) – \(\frac{1}{2}\) cos^-1 x.

Question 10.
Write in simplest form:
tan^-1 ( \(\frac { x }{ \sqrt { 1+x^{ 2 }-1 } } \) )?
Solution:
tan^-1 ( \(\frac { x }{ \sqrt { 1+x^{ 2 }-1 } } \) )
Putting x = tan θ, we get

Question 11.
Prove that:
tan^-1 \(\sqrt{x}\) = \(\frac{1}{2}\) cos^-1 ( \(\frac{1-x}{1+x}\) )? (NCERT)
Solution:
tan^-1 \(\sqrt{x}\) = \(\frac{1}{2}\) cos^-1 ( \(\frac{1-x}{1+x}\) )
R.H.S = \(\frac{1}{2}\) cos^-1 ( \(\frac{1-x}{1+x}\) )

Let tan^-1 \(\sqrt{x}\) = θ
\(\sqrt{x}\) = tan θ
R.H.S. = \(\frac{1}{2}\) cos^-1(cos 2θ)
= \(\frac{1}{2}\). 2θ
= θ
= tan^-1 \(\sqrt{x}\)
= L.H.S. Proved.

Question 12.
Prove that:

Solution:


= tan^-1 (tan 3θ) – tan^-1 (tan 2θ)
= 3θ – 2θ
= θ
= tan^-1 2x
= R.H.S. Proved.

Question 13.
Prove that:
cos^-1 \(\frac{4}{5}\) + cos^-1 \(\frac{12}{13}\) = cos^-1 \(\frac{33}{65}\)?
Solution:
Let cos^-1 \(\frac{4}{5}\) = A
∴ \(\frac{4}{5}\) = cos A
∴ sin A = \(\sqrt { 1-cos^{ 2 }A } \) = \(\sqrt { 1-\frac { 16 }{ 25 } } \) = \(\sqrt { \frac { 9 }{ 25 } } \)
⇒ sin A = \(\frac{3}{5}\)
Let cos^-1 \(\frac{12}{13}\) = B
⇒ \(\frac{12}{13}\) = B
∴ sin B = \(\sqrt { 1-cos^{ 2 }B } \) = \(\sqrt { 1-\frac { 144 }{ 169 } } \) = \(\sqrt { \frac { 25 }{ 169 } } \)
⇒ sin B = \(\frac{5}{13}\)
A + B = cos^-1 \(\frac{33}{65}\)
⇒ cos (A + B) = \(\frac{33}{65}\)
⇒ cos A.cos B – sin A.sin B = \(\frac{33}{65}\)
L.H.S. = \(\frac{4}{5}\). \(\frac{12}{13}\) – \(\frac{3}{5}\). \(\frac{5}{13}\)
= \(\frac{48}{65}\) – \(\frac{15}{65}\)
= \(\frac{33}{65}\)
= R.H.S. Proved.

MP Board Class 12 Maths Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 11 Alcohols, Phenols and Ethers

Alcohols, Phenols and Ethers Important Questions

Alcohols, Phenols and Ethers Very Short Answer Type Questions

Question 1.
What is Ether?
Answer:
Compounds formed by the substitution of hydrogen atom of hydrocarbon by alkoxy group are called Ethers.

Question 2.
What will be the type of alcohol formed by the hydration of propene in the presence of acid?
Answer:

Question 3.
What is the special name of Phenol and from what was it first isolated?
Answer:
Phenol is also known as carbolic and it was first isolated from coal tar.

Question 4.
Picric acid is a strong acid. Why?
Answer:
In picric acid, acidic character increases due to the presence of three electron attracting – NO₂ groups because these groups are helpful in the release of H⁺. Thus picric acid is a strong acid.

Question 5.
Write the reagent required for the preparation of tertiary butyl alcohol starting from propanone.
Answer:
Methyl magnesium bromide.

Question 6.
What type of isomerism is exhibited between alcohol and ether?
Answer:
Alcohol and ether exhibit Functional isomerism.

Question 7.
Write the equation of catalytic reduction of Butanols.
Answer:

Question 8.
Write IUPAC name of the following compound.

Answer:

IUPAC name of this compound is 2,2, 3 – trimethyl pentan – 1 – ol.

Question 9.
Why do ethers have low boiling points?
Answer:
Molecules of ether do not possess H – bonding, therefore boiling points of ethers are low.

Alcohols, Phenols and Ethers Short Answer Type Questions

Question 1.
What is Lucas reagent? How are primary, secondary and tertiary alcohol identified by it? Explain.
Answer:
Mixture of anhydrous ZnCl₂ and cone. HCl is known as Lucas reagent.
1. Tertiary Alcohol : On adding Lucas reagent in alcohol at normal temperature, immediately white oily precipitate of Alkyl chlorides is formed, then it is tertiary alcohol.

2. Secondary Alcohol : If on adding Lucas reagent in alcohol, at normal temperature, a white oily precipitate of alkyl chloride is obtained after 5 minutes, then it is secondary alcohol.

3. Primary Alcohol : Primary alcohol does not show any reaction with Lucas reagent at normal temperature.

Question 2.
Why the b.p. of alcohol are higher than ethers and alkene?
Or C₃H₅OH and CH₃OCH₃ both have same molecular formula (C₂H₆O) but the b.p. of alcohol is 78.4°C and b.p. of ether is – 240°C. Explain the reason.
Answer:
In case of C₂H₅OH there is strong intermolecular hydrogen bonding between the molecules of alcohol. So alcohols (C₂H₅OH) required much energy to evaporate than ether molecules. In other words, we can say that the C₂H₅OH molecules are in associated form due to H – bonding so the b.p. of C₂H₅OH is very higher than ether and alkene.

Question 3.
Boiling point of alcohol higher than corresponding alkane. Why?
Answer:
Boiling point of alcohols is much higher than hydrocarbons of nearly similar molecular mass due to inter molecular hydrogen bond. Alcohols molecules associate kilo calories mole^-1. Thus, extra energy is required for the separation of these molecules, which lead to increase in boiling point. Hydrocarbons do not form hydrogen bond, thus their boiling point is comparatively less.

Question 4.

1. How can we obtain phenol from benzene diazonium chloride?
2. What is the reaction of diethyl ether with HI acid?

Answer:
1. Phenols are prepared by hydrolysis of diazonium salts by water, dil. acids etc.

2. The reaction of diethyl ether with cone. HI acid, on heating gives one molecule of ethyl iodide and one molecule of ethyl alcohol.

Diethyl ether Ethyl iodide Ethyl alcohol

Question 5.
Write the equations involved in the following reactions :

1. Reimer – Tiemann reaction (NCERT, MP2018)
2. Kolbe’s reaction.

Answer:
1. Reimer – Tiemann reaction:
When phenol is treated with chloroform in presence of aqueous sodium hydroxide at 60°C, oHydroxy benzaldehyde (Salicylaldehyde) and p – Hydroxy benzaldehyde are formed. The ortho – isomer is the major product. This reaction is called Reimer-Tiemann reaction.

If carbon tetrachloride is used in place of chloroform, salicylic acid is obtained as the main product.

2. Kolbe – Schmidt reaction:
When sodium salt of a phenol is heated with CO₂ at 130°C. (403K) and 4 – 7 atm pressure, sodium salicylate is formed. This on acidification gives salicylic acid.

At high temperature p – derivative is formed.

Question 6.
Name the reagents used in the following reactions: (NCERT)

1. Oxidation of a primary alcohol to carboxylic acid.
2. Oxidation of a primary alcohol to aldehyde.
3. Bromination of phenol to 2,4,6 – tribromo – phenol.
4. Benzyl alcohol to benzoic acid.
5. Dehydration of propan – 2 – ol to propene.
6. Butan – 2 – one to butan – 2 – ol.

Answer:

1. Acidified K₂Cr₂O₇ or neutral acidic or alkaline KMnO₄.
2. Pyridinium chlorochromate (pcc) in CH₂Cl₂ or Cu at 573K.
3. Bromine water (Br₂/H₂O)
4. Acidified or alkaline KMnO₄
5. Conc. H₂SO₄ at 443K or 85% phosphoric acid at 443K.
6. Ni/H₂ or NaBH₄ or LiAlH₄.

Question 7.
Explain, how does the – OH group attached to a carbon of benzene ring activate it towards electrophilic substitution? (NCERT)
Answer:
The – OH group exerts +R effect on the benzene ring under the effect of attacking electrophile. As a result, there is an increase in the electron density in the ring particularly at ortho and para positions, therefore electorphilic substitution occurs mainly at o – and p – positions.

Question 8.
Write Victor Meyer method to distinguish primary, secondary and tertiary alcohol.
Answer:
Victor Meyer’s method:

1. The given alcohol is converted into an iodide by concentrated HI or red phosphorus and iodine.

2. The iodide is treated with silver nitrite to form nitroalkane.

3. Nitroalkane is finally treated with nitrous acid (NaNO₂ + H₂SO₄) and made alkaline with KOH.

• If a blood red colour is obtained, the original alcohol is primary.
• If a blue colour is obtained, the alcohol is secondary.
• If no colour is produced, the alcohol is tertiary.

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Answer:

Question 10.
Differentiate between Phenol and Alcohol and write Libermann’s reaction related to phenol.
Answer:
Differences between Phenol and Alcohol:
Phenol:

• Physical properties – Characteristic phenolic odour, sparingly soluble in water.
• It is acidic and dissolve in bases to form salt.
• On oxidation, hybrid coloured product is formed.
• Produce characteristic colour with Ferric chloride.
• It does not react with halogen acid.
• With PC₁₅, mainly form triaryl phosphate.

Alcohol:

• Pleasant odour, fairly soluble in water.
• It is neutral and do not reacts with bases.
• It can easily oxidize to Aldehydes and ketones.
• It does not reacts with ferric chloride.
• Forms Alkyl halide.
• Alkyl chloride are formed.

Libermann’s Reaction:
On adding few drops of concentrated sulphuric acid and little sodium nitrite in phenol first dark blue colour is produced on adding water colour becomes red and on adding an alkali red colour again changes to blue colour.

Question 11.
Give equations for the preparation of ethyl alcohol by starch and write name of enzymes.
Answer:

Enzymes:

• Diastase
• Maltase
• Zymase.

Question 12.
Explain, why propanol has higher boiling point than that of the hydrocarbon, butane? (NCERT)
Answer:
The molecules of butane are held together by weak van der Waals’ force of attraction while those of propanol are held together by stronger intermolecular hydrogen bonding.

Therefore, the b.p. of propanol is much higher than that of butane.

Question 13.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact. (NCERT)
Answer:
Alcohols can form hydrogen bonds with water and break the H – bond exist between water molecules. Hence, they are soluble in water.

On the other hand, hydrocarbons cannot form hydrogen bonds with water molecules and hence are insoluble in water.

Question 14.
What is meant by hydroboration – oxidation reaction? Illustrate it with an example. (NCERT)
Answer:
The addition of diborane to alkene to form trialkyl boranes followed by their oxidation with alkaline hydrogen peroxide to form alcohol is called hydroboration – oxidation. For example:

The alcohols obtained by this process appear to have been formed by direct addition of water to the alkene against Markownikoffs rule.

Question 15.
Ethyl alcohol and phenol both contain – OH group. What is the reason that phenoffs acidic and alcohol has alkaline effect? (MP 2015)
Or
Ethyl alcohol and phenol both contain – OH group. What is the reason that phenol is acidic and alcohol is neutral in nature? (MP 2013)
Answer:
Explanation of acidic nature of phenol:
One possible explanation why phenols are stronger acids as compared to alcohols is that phenols exist as a resonance hybrid.

Due to resonance, the oxygen atom gets a positive charge and attracts the electron pair of the O – H bond and thus facilitates the release of a proton. The phenoxide ion formed after the release of a proton is also stabilized by resonance.

In alcohols, no resonance is possible hence the hydrogen atom is more firmly linked to the oxygen.

Question 16.
Pure phenol is a colourless solid but why it is converted into pink after some time?
Or
What change in colour is observed in phenol in presence of oxygen? Explain with reaction. (MP2011)
Answer:
In the presence of air pure phenol oxidises into quinone.

This quinone again combines with two molecules of phenol by H – bond and gives pink phenoquinone.

Question 17.
Explain the manufacture of CH₃OH by water gas.
Answer:
From water gas : Steam is passed over red hot coke when water gas is formed.

Water gas is mixed with half its volume of hydrogen, compressed to about 200 atm and passed over a catalyst which is a mixture of oxides of copper, zinc and chromium at 300°C.

Alcohols, Phenols and Ethers Long Answer Type Questions

Question 1.
What is Williamson continuous etherification process? Is it a continuous process? Explain. Give labelled diagram.
Or,
Describe the laboratory method of preparation of diethyl ether. How ether thus obtained is purified?
Answer:
Laboratory Method for the Preparation of Diethyl Ether (Sulphuric Ether):
Diethyl ether is prepared in the laboratory and industry by the Williamson continuous etherification process, i.e., by heating ethanol (in excess) with concentrated sulphuric acid.

Sulphuric acid is regenerated in the reaction hence, it appears as if only a small amount of acid may convert an excess of alcohol into ether. So, this method is called Williamson continuous etherification process but actually we cannot get ether continuously.

This is due to the following two reasons :

• Water formed in the reaction dilutes the acid and its reactivity decreases.
• A part of sulphuric acid is reduced by alcohol into sulphur dioxide.

Method:
Ethanol and H₂SO₄ (2:1) are taken in a flask and heated on sand bath at 140°C. Ethanol is added at the same rate at which ether distilled over and is collected in a receiver cooled in ice – cold water.

Purification:
Ether contain ethanol, water and sulphuric acid as impurities. It is washed with NaOH to remove sulphuric acid and then agitated with 50% solution of calcium chloride to remove alcohol. It is then washed with water, dried over anhydrous calcium chloride and redistilled.

Question 2.
How can you change the following:

1. Methanol to ethanol
2. Ethanol to methanol.

Answer:
1. Methanol to ethanol :

2. Ethanol to methanol :

Question 3.
Differentiate primary, secondary and tertiary alcohol by oxidation and dehydrogenations method. (MP 2011)
Answer:
1. Oxidation:
The oxidizing agents generally used for oxidation of alcohols are acid dichromate, acid or alkaline KMnO₄ and dilute HNO₃.
(i) A primary alcohol is easily oxidized to an aldehyde and then to an acid both containing the same number of carbon atoms as the original alcohol.

(ii) A secondary alcohol on oxidation gives a ketone with the same number of carbon atoms as the original alcohol, ketones are oxidized with difficulty but prolonged action of oxidizing agents produce carboxylic acids containing fewer number of carbon atoms than the original alcohol.

(iii) A tertiary alcohol is resistant to oxidation in neutral or alkaline solutions but is readily oxidized by an acid oxidizing agent giving a mixture of ketone and acid each having lesser number of carbon atoms than the original alcohol.

2. Dehydrogenation (Action of hot reduced copper at 300 °C):
Different types of alcohols give different products when their vapours are’passed over Cu gauze at 300°C.

Primary alcohols lose hydrogen and yield an aldehyde.

Secondary alcohols lose hydrogen and yield a ketone.

Tertiary alcohols are not dehydrogenated but lose a water molecule to give alkenes.

Question 4.
Explain the mechanism of dehydration of alcohol.
Answer:
Dehydration of alcohol :
(i) When ethyl alcohol is heated in excess of cone. H₂SO₄ molecule of water is eliminated and alkene is formed.

Mechanism :
(i) Protonation of alcohol by H₂SO₄

(ii) Removal of water

(iii) Elimination of β – hydrogen in the form of proton by base (bisulphate ion)

Stability of the carbocation (I) determines the case of dehydration and order of stability of carbocation is:
CH₃ < C₂H₅ < Isopropyl < Tertiary butyl

Question 5.
How is ethyl alcohol obtained by molasses? Explain in brief.
Or,
What are molasses? How is alcohol obtained by fermentation? Explain. Tell favourable conditions of fermentation. Draw labelled diagram of coffee still.
Answer:
From molasses:
Molasses is the syrupy solution of sugar left after the separation of cane sugar or beet sugar crystals from the concentrated juice.

The different steps of the manufacture processes are :
1. Dilution:
The molasses is diluted with water so that a concentration of 8 – 10 percent sugar is obtained in solution. This is acidified with dilute sulphuric acid to retard other bacterial growth. A solution of ammonium salts is also added which acts as food for the ferment.

2. Alcoholic fermentation:
The dilute solution obtained above [From step (a)] is taken in big fermentation tanks and some yeast is added. The mixture is kept for a few days and the temperature is maintained at about 30°C. The fermentation reaction starts and the enzyme Invertase (From Yeast) converts sucrose into glucose and fructose which are then converted into

The fermentation is completed in about 3 days. The carbon dioxide is collected as a by product.

3. Distillation:
The fermented liquor is technically called wash or wort which contains about 9 – 10 percent ethanol. It is then distilled in a continuous still called Coffey’s still. It consists of two tall fractionating columns which are called analyser and the rectifier. It works on the counter current principle and the steam and wash travel in opposite directions through the still.

The steam goes upwards in the analyser and takes away the alcohol vapours from the downcoming dilute alcohol. The mixture leaves the analyser from the top and enters the rectifier at the base. Here it heats the wash flowing through the pipes on its way to the analyser. Most of the steam condenses and the alcohol vapours condenses in the condenser. The distillate contain 90% alcohol.

4. Rectification : Wash is rectified by fractional distillation.

Question 6.
Give equations for three methods of preparation of phenol.
Answer:
Methods of preparation of phenol:
1. By the hydrolysis of Benzene diazonium salts:
Benzene diazonium salt is formed by aromatic primary amine (aniline) with nitrous acid at 0 – 5°C. On boiling aqueous solution of this salt phenol is formed.

2. By alkaline fusion of sodium benzene sulphonate:
On fusing sodium benzene sulphonate with NaOH, sodium phenoxide is formed which on acidification forms phenol.

3. Rasching method:
On heating benzene with mixture of HCl and air to 230°C in the presence of Cu catalyst chlorobenzene is formed which on hydrolysis form phenol.

Question 7.
Write IUPAC names of the following compounds: (NCERT)

Answer:

1. 2,2,4 – Trimethyl pentan – 3 – ol
2. 5 – Ethylheptane – 2,4 – diol
3. Butan – 2,3 – diol
4. Propane – 1,2,3 – triol
5. 2 – Methylphenol
6. 4 – Methylphenol
7. 2,5 – Dimethylphenol
8. 2,6 – Dimethylphenol
9. 1 – Methoxy – 2 – methylpropane
10. Ethoxybenzene
11. 1 – Phenoxyheptane
12. 2 – Ethoxybutane.

Question 8.
How can you obtained following compounds from phenol: (MP 2012; Supp. 14,16)

1. 4, 6 – Tribromophenol
2. Picric acid
3. Aniline
4. Benzene
5. Phenolp – hthalene
6. p – cresol, o – cresol.

Answer:
1. Phenol to Tribromophenol : Phenols readily react with halogens to give polyhalogen substituted compounds. Phenol gives white precipitate of 2, 4, 6 – tribromo – phenol with bromine water.

2. Phenol to Picric acid : Nitration : On nitration, phenols give a variety of products depending upon the conditions.

Nitration of phenol with cone. HNO₃ in presence of cone. H₂SO₄ gives, 2,4,6 – Trini – trophenol (Picric acid). (MP 2014)

3. Phenol to Aniline:

4. Phenol to Benzene :

5. Phenol to Phenolphthalene : Phenol condenses with phthalic anhydride in presence of cone. H₂SO₄ to give phenolphthalein which is an indicator for acid – base titrations and is used as a laxative in medicine.

(vi) Phenol to para cresol :

MP Board Class 12th Chemistry Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes

Haloalkanes and Haloarenes Important Questions

Haloalkanes and Haloarenes Very Short Answer Type Questions

Question 1.
What is plane polarized light?
Answer:
Light which vibrates in one specific plane is known as plane polarized light. On passing normal light through a Nicol prism, plane polarized light is obtained.

Question 2.
Write an example of 3° alkyl chloride.
Answer:
Tertiary butyl chloride

(2-chloro-2-methyl propane)

Question 3.
What is the order of boiling points of alkyl halides for the same alkyl group?
Answer:
Decreasing order of boiling points : RI > RBr > RCl > RF.

Question 4.
Write the equation of Swart’s reaction.
Answer:

Question 5.
What are Enantiomers?
Answer:
The stereoisomers related to each other as non – superimposable mirror images are called Enantiomers.

Question 6.
What are polyhalogen compounds? Give two examples.
Answer:
Carbon compounds which contain more than one halogen atom are known as polyhalogen compounds like CHCl₃, CCl₄ etc.

Haloalkanes and Haloarenes Short Answer Type Questions

Question 1.

1. Write Iodoform reaction.
2. Iodoform gives yellow ppt. with AgNO₃ solution but chloroform doesn’t Why?
3. What happens when ethyl bromide is heated with alcoholic KOH?

Answer:
1. Iodoform reaction:
When ethyl alcohol or acetone is heated with iodine and NaOH, yellow crystals of iodoform are formed.
C₂H₅OH + 4I₂ + 6NaOH → 5Naf + HCOONa + 5H₂O + CHI₃

2. When iodoform is heated with AgNO₃ solution a yellow ppt. (Agl) is obtained but chloroform doesn’t give this reaction because in chloroform C – CI bond is more stable than C – I bondin iodoform.

3. On boiling Ethyl bromide with alcoholic KOH ethylene is formed.

Question 2.
Explain the Sandmeyer reaction with example.
Answer:
Decomposition of diazonium salts (Sandmeyer reaction):
When a diazonium salt solution is added to a solution of cuprous halide dissolved in the corresponding halogen acid, the diazo group is replaced by a halogen atom.

Question 3.
Give chemical reaction between chlorobenzene and chloral in presence of cone. H₂SO₄.
Or, How is D.D.T. formed? Write its one application.
Answer:
DDT (Dichlorodiphenyl trichloroethane) is formed by the condensation of one molecule of chloral with two molecules of chlorobenzene in presence of cone, H₂SO₄.

Application : It is an important pesticide.

Question 4.
What are Gem – dihalide and Vicinal – dihalide?
Answer:
When both halogen atoms are linked to one carbon atom of hydrocarbon then it is known as Gem-dihalide. Gem means geminal i. e., same position.

When both halogen atoms are connected to two different neighbouring carbon atoms, then it is known as vicinal dihalide. Vic means vicinal which means adjacent position.

Question 5.
What is Lucas reagent? Give its application.
Answer:
Solution of ZnCl₂ in cone. HCl is known as Lucas reagent.
Application:
It is used to differentiate primary, secondary and tertiary alcohols.
On adding the alcohol to Lucas reagent, a tertiary alcohol reacts immediately forming a ppt. of alkyl chloride. If the ppt. appears after few minutes, then the alcohol is secondary. If no ppt. is obtained in cold the alcohol is primary.

Question 6.
Explain, Carbylamine reaction and give one application of this reaction. (MP 2018)
Answer:
Carbylamine reaction: On heating chloroform with primary amine (e.ganiline) and alcoholic KOH solution, phenylisocyanide or carbylamine is formed which has a very bad smell and is poisonous.

Application : Chloroform and primary amine can be tested by this reaction.

Question 7.
What is 666 (lindane)? Explain its preparation and use in agriculture.
Answer:
It is 1, 2, 3, 4, 5, 6 – HexachIorocyciohexane. It is obtained by heating benzene with chlorine in presence of sunlight.

Preparation : It is prepared by the chlnncriUiv benzene in the presence of ultraviolet light.

1,2, 3,4, 5, 6 – Hexachlorocyclohexane (B.H.C.)

Uses:
Benzene hexachloride is an addition compound and its γ – isomer is called gammexane. It is an important pesticide used in agriculture. It is also called lindane or 666.

Question 8.
Explain the following reaction of chlorobenzene:

1. Ration with chlorine in the presence of FeCI3 in dark
2. Fittig reaction.

Answer:
1. When Chlorobenzene reacts with Cl₂ hi the presence of FeCl₃ in dark. o – dichlorobenzene and p – dichlorobenzene is obtained.

2. Fittig reaction : When Chlorobenzene is heated at 200°C with Cu powder in a sealed tube Diphenyl is formed.

When two molecules of aryl halide reacts with sodium metal in presence of dry ether, then diphenyl is formed. This reaction is known as Fittig reaction.

Question 9.
Write short notes on :

1. Hunsdiecker method and
2. Raschig process.

Answer:
1. Hunsdiecker method : When silver salt of a carboxylic acid is heated with bromine, in the presence of an inert solvent like CCl₄, aryl bromide is formed.

This method is called Hunsdiecker method.

2. Raschig process : When benzene vapours mixed with air and HCl gas is passed over CuCl₂ (catalyst) at 230°C, chlorobenzene is formed.

Question 10.
Write method of preparation, properties and uses of Freon.
Answer:
Freon : Dichloro, Difluoro methane.
it is formed by the action of SbF₃ with CCl₃ in presence of SbCl₅.

It has very low boiling point due to which by increasing the pressure at room temperature it can be liquefied. It is a non – poisonous, non – combustible and inactive substance which is used as a cooling agent in the refrigerator. It is used in aerosol and foam.

Question 11.

1. B.P. of ethyl iodide is higher than b.p. of ethyl bromide. Give reason.
2. Explain why the m.p. of para dichlorobenzene is higher than its ortho arid meta derivatives.

Answer:

1. In alkyl halides containing same alkyl group boiling point increases with increase in atomic weights of halogen atoms. Molecular weight of ethyl iodide is more than ethyl bromide and therefore boiling point of ethyl iodide is also high.
2. Para derivatives of dichlorobenzene is more symmetrical than its ortho and meta derivatives therefore its m.p. is higher.

Question 12.
Explain Friedel – Craft’s reaction with chemical equation,
Answer:
Friedel – Craft’s reaction (alkylation) : Alkyl halides react with benzene in presence of anhydrous aluminium chloride to give alkyl benzene.

Acetylation:
Acetyl chloride reacts with benzene in presence of anhydrous aluminium chloride to give acetophenone.

Question 13.
Give the uses of freon – 12, D.D.T., Carbon tetrachloride and Iodoform. (NCERT)
Answer:
It is an important pesticide.
Freon – 12:
Since freons have been found to be one of the factors responsible for the depletion of ozone layer, they are being replaced by other harmless compounds in many countries.

D.D.T:
DDT is highly toxic and has strong insecticidal properties and thus it was widely used as an insecticide and pesticide.

Carbon Tetrachloride:
It is produced in large quantities for use in the manufacture of refrigerants and propellants for aerosol cans. It is used as feed stock in the synthesis of chlorofluorocarbons and other chemicals, pharmaceutical manufacturing and general solvents use. Until the mid 1960’s, it was widely used as a cleaning fluid both in industry as a degreasing agent and in the home, as a spot remover and as fire extinguisher.

Iodoform:
It was earlier used as an antiseptic but the antiseptic properties are due to the liberation of free iodine and not due to iodoform itself. Due to its objectionable smell, it has been replaced by other formulations containing iodine.

Question 14.
Name of the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides : (NCERT)

1. (CH₃)₂CHCH(CI)CH₃
2. CH₃CH₂CH(CH₃)CH(C₂H)CI
3. CH₃CH₂C(CH₃)₂CH₂I
4. (CH₃)₃CCH₂CH(Br)C₆H₅
5. CH₃CH(CH₃)CH(Br)CH₃
6. CH₃C(C₂H₅)₂CH₂Br
7. CH₃C(CI)(C₂H₅)CH₂CH₃
8. CH₃CHC(Cl)CH₂CH(CH₃)₂
9. CH₃CH=CHC(Br)(CH₃)₂
10. p – ClC₆H₄CH₂CH(CH₃)₂
11. m – ClCH₂C₆H₄CH₂C(CH₃)₃
12. o – Br – C₆H₄CH(CH₃)CH₂CH₃.

Answer:

1. 2 – Chloro – 3 – methylbutane (2° alkyl)
2. 3 – Chloro – 4 – methyIhexane (2° alkyl)
3. 1 – Iodo – 2,2 – dimethylbutane (1° alkyl)
4. 1 – Bromo – 3,3 – dimethyl – 1 – phenylbutane (2° benzylic)
5. 2 – Bromo – 3 – methylbutane (2° alkyl)
6. 3 – Bromomethyl – 3 – methylpentane (1° alkyl)
7. 3 – Chloro – 3 – methylpentane (3° alkyl)
8. 3 – Chloro – 5 – methylhex – 2 – ene (vinyl)
9. 4 – Bromo- 4 – methylpent – 2 – ene (allylic)
10. 1 – Chloro – 4 – (2′-methylpropyl) benzene (aryl) or p – Chloro isobutyl benzene
11. 1 – Chloromethyl-3-(2’2′-diethylpropyl) benzene (benzylic) or m – Neopentyl benzyl chloride
12. 1 – Bromo – 2 – (l’ – methylpropyl) benzene (aryl).

Question 15.
Identify ‘A’, ‘B’, ‘C’ and ‘D’.

Answer:

Question 16.
Write the equation of following reactions of chlorobenzene : (MP 2015)

1. Halogenation
2. Nitration
3. Sulphonation
4. Alkylation.

Answer:
1. Halogenation : Haloarene reacts with halogen in presence of halogen carrier like FeCl₃ to form ortho and para substituted dihaloarene.

2. Nitration : Haloarenes react with nitrating mixture to form o – nitro and p – nitro substituted haloarenes.

3. Sulphonation : On heating with cone. H₂SO₄, 2 – Chlorobenzene sulphonic acid and 4-Chlorobenzene sulphonic acid are formed.

4. Alkylation : Alkylation takes place with alkyl halide in presence of anhydrous A1C13.

Haloalkanes and Haloarenes Long Answer Type Questions

Question 1.
Explain the nucleophilic substitution reaction in alkyl halide by S_N1 and S_N2 medianism.
Answer:
Nucleophilic substitution reaction:
In the carbon halogen bond ofhaloalkane. halogen atom is more electronepativ0 as compared to the carbon atom hence, the shared pair of electrons between carbon and halogen is more attracted by the halogen atom. As a result a small negative charge and an equivalent positive charge develops on halogen atom and carbon atom respectively.

Nucleophile attacks the electron deficient carbon due to the presence of partial positive charge on it and replaces the weaker nucleophilic ion i.e. the halide ion. Thus, the reaction is known as nucleophilic substitution reaction.

The order of reactivity of different alkyl halide towards nucleophilic substitution reaction is:
RI > RBr > RC1 > RF

Mechanism of Nucleophilic substitution reactions :
Nucleophilic substitution reaction occurs through two different mechanism :

(1) S_N1 Mechanism (Unimolecular nucleophilic substitution) : In this mechanism following steps are involved :
(a) Formation of carbocation by dissociation of substrate i.e., reactant molecule.

(b) Attack of nucleophile on carbocation forming the product.

(2) S_N2 mechanism (Bimolecular nucleophilic substitution):
Reactions of this type occur in one step i.e. they are concerted reactions. These reaction nucleophilic attack results in a transition state in which both the reactant molecules are partially bonded to each other and then the halide ion escapes out forming the product.
ROH^– + CH₃X → CH₃OH + RX

Rate of reaction = K[(RX)OH^–]
Order of reactivity of alkyl halide is : Primary > Secondaxy > Tertiary.

Question 2.
Draw labelled diagram of laboratory method for preparation of iodoform from akfohol. Write related chemical equation.
Answer:

Chemical reaction:

Question 3.
Write the structure of the major organic product in each of the following reactions: (NCERT)

Answer:

Question 4.
Give the laboratory method for the preparation of chloroform. Describe the formation of chloroform by ethanol with labelled diagram, equation and principle.
Answer:
Laboratory method : Chloroform is prepared in the laboratory by the action of water and bleaching powder on ethyl alcohol or acetone.

Method:
About 100 gm of bleaching powder made into a paste by adding about 200 ml of water and taken in a flask fitted with a condenser. Now, 25 ml of alcohol or acetone is added and the mixture is distilled, chloroform collects as a heavy liquid under water.

It is washed with dilute NaOH solution then with water, dried over fused calcium chloride and redistilled. The available chlorine of bleaching powder acts as oxidising as well as chlorinating agent during the preparation of chloroform from alcohol and acetone.
CaOCl₂ + H₂O → Ca(OH)₂ + Cl₂

The chemistry involved in the conversion of alcohol and acetone into chloroform is as shown below:

(A) From alcohol: The steps involved are :
(i) Ethyl alcohol is oxidized by chlorine to acetaldehyde.

(ii) Acetaldehyde reacts with chlorine to give chloral, i.e. trichloro acetaldehyde.

(iii) Two moles of chloral react with one mole of calcium hydroxide to produce chloroform.

Calcium hydroxide Chloral Chloroform Calcium formate.

Question 5.
MaloaIkanes are more reactive than haloarenes. Give reason.
or,
Why, aryl halides are less reactive than alkyl halides?
Answer:
In aryl halides, halogen atom is attached more strongly to the nucleus therefore the nucleophilic substitution takes slowly than alkyl halides. There is two reasons for the less reactivity of aryl halides.
(i) In aryl halides sp³hybridization takes place whereas in alkyl halides sp² hybridization is present due to sp² hybridization in haloarenes the halogen are attached to nucleus more strongly.

(ii) Due to the presence of resonance in aryl halides there is some double bond character in C – Cl bond. Thus, the bond length of C – Cl bond is lesser than C – Cl bond in haloalkanes. Therefore, it is difficult to replace the halogen of haloarenes.

Question 6.
What product is formed by the reduction of chloroform? Give the chemical equation when it reacts to nitric acid and acetone.
Answer:
Reduction:
1. On heating with Zn and HCl, it reduces to form methylene dichloride.

2. On heating with zinc dust and water, methane is formed.

Reactions of Chloroform :
1. With Conc.HNO₃:
On treating chloroform with concentrated nitric acid, the hydrogen atom of chloroform is replaced by nitro group and nitro chloroform (or chloropicrin) is formed. It is a liquid (b.p. 112°C) which is used in war as a poisonous gas.

2. With Acetone:
Chloroform condenses with acetone in presence of sodium hydroxide to form chloretone which is a hypnotic (sleep inducing drug) of high grade.

Question 7.
Give the Chemical reaction when chloroform reacts with following:

1. Oxidation
2. Carbylamine reaction
3. Ag powder
4. Nitration
5. Reimer – Tiemann reaction.

Or
How will you obtain the following from chloroform:

1. Carbonyl chloride
2. Acetylene
3. Chloropicrin
4. Phenyl isocyanide
5. Chloretone
6. Salicylaldehyde.

How trichloro methane reacts with :

1. Atmospheric air
2. Aniline and ale. KOH
3. Ag powder
4. Cone. HNO₃
5. Phenol
6. Acetone.

Answer:
(a) Action of air and light (Oxidation):
Chloroform oxidizes in presence of sunlight and air and forms a poisonous gas, phosgene (carbonyl chloride).

Ordinary chloroform contains phosgene gas and is used as a solvent. Pure chloroform which is used as an anaesthetic does not contain even traces of phosgene. While preserving chloroform which is to be used as an anaesthetic, the following precautions are taken:
(i) The chloroform is filled in blue or brown coloured bottle up to the neck. After putting a stopper, the bottle is kept in dark. As there is no empty space in the bottle, it is also to from air

(ii) One percent ethyl alcohol is added in the bottle. If phosgene gas is formed, alcohol reacts with it to form diethyl carbonate, a non – toxic substance, i.e. (C₂H₅)₂CO₃

(b) Carbylamine reaction:
On heating chloroform with primary amine (example aniline) and alcoholic KOH solution, phenylisocyanide or carbylamine is formed which has a very bad smell and is poisonous.

(c) Reaction with Ag powder or Dehalogenation: On heating chloroform with silver powder, pure acetylene gas is formed.
CHC₁₃ + 6Ag + CI3CH → HC ≡ CH + 6 AgCl

(d) Nitration:
Haloarenes react with nitrating mixture to form o – nitro and p – nitro substituted haloarenes.

(e) Reimer – Tiemann reaction:
On heating chloroform with concentrated alkali and phenol at 60 – 70°C, o – hydroxy benzaldehyde (salicylaldehyde) is formed. Traces of p- hydroxybenzaldehyde are also formed.

(f) Acetone :
Chloroform condenses with acetone in presence of NaOH to form chloretone which is a hypnotic of high grade.

Question 8.
An alcohol ‘A’ on reaction with cone. H₂SO₄ gives an alkene ‘B’. ’B’ after bromination with sodamide gives dehydrogenated compound ‘C’. ‘C’ on reaction of H₂SO₄ in presence of H₂SO₄ gives ‘D’. Identify ‘A’, ‘B% ‘C’, and ‘D’. (MP 2017)
Answer:

MP Board Class 12th Chemistry Important Questions

MP Board Class 12th Maths Important Questions Chapter 1 Relations and Functions

Relations and Functions Important Questions

Relations and Functions Objective Type Questions:

Question 1.
Choose the correct answer:

Question 1.
Let f : R → R te a function defined as f(x) = 8x:
(a) f is one – one onto (c) / is one-one but not onto
(b) f is many – one onto
(c) f is one – one but not onto
(d) f is neither one – one nor onto
Answer:
(a) f is one – one onto (c) / is one-one but not onto

Question 2.
Let f : R → R be a function is defined as f (x) = \(\frac { e^{ x^{ 2 } }-e^{ -x^{ 2 } } }{ e^{ x^{ 2 } }+e^{ -x^{ 2 } } } \):
(a) f is one – one but not onto
(b) f is one – one onto
(c) f is neither one – one nor onto
(d) f is many one onto
Answer:
(a) f is one – one but not onto

Question 3.
If f : R → R be given by f(x) = \((3-x^{ 3 })^{ 1/3 }\) then (fof) x is:
(a) x \(x^{ 1/3 }\)
(b) x\(x^{ 3 }\)
(c) x
(d) \((3-x^{ 3 })\)
Answer:
(c) x

Question 4.
Consider a binary operation * on N defined as a*b = \(a^{ 3 }\) + \(b^{ 3 }\):
(a) Is * both associative and commulative
(b) Is * commutative but not associative
(c) Is * associative but not commutative
(d) Is * neither commutative nor associative
Answer:
(b) Is * commutative but not associative

Question 5.
Number of binary operations on the set {a,b} are:
(a) 10
(b) 16
(c) 20
(d) 8
Answer:
(b) 16

Question 2.
Fill in the blanks:

1. Number of equivalance relation in the set {1, 2, 3} containing (1,2) is …………………………….
2. If f : R → R, when f(x) = 5x – 7,∀x∈R, then value of \(f^{ -1 }\) (7) is ……………………….
3. If f : R → R, be defined by f(x) = x² – 3x + 2, then f {f(x)} ……………………………
4. If f : R → R, be defined by f(x) = 2x + 5, then \(f^{ -1 }\) (y) ………………………………
5. If f : R → R, be defined by f(x) = x³ the f is ………………………………….

Answer:

1. 2
2. 0
3. x⁴ – 6x³ + 10x² – 3x
4. \(\frac{1}{2}\) (y – 5)
5. Is one – one

Question 3.
Write True/False:

1. Let R = {(1, 3), (3, 1), (3, 3)} be a relation defined on A = {1, 2, 3}. Then R is symmetric, transitive but not reflexive.
2. If f : A → B is bijective function, then the inverse of \(f^{ -1 }\) of f is unique.
3. The composition of funtion is commutative
4. Every function is invertible
5. Let a binary operation * on the set Q⁺ of positive rational numbers be defined by a*b = \(\frac{ab}{3}\), ∀a,b ∈ Q⁺. Then the inverse of 4*6 is \(\frac{9}{8}\)

Answer:

1. False
2. True
3. False
4. False
5. True

Question 4.
Write the answer in one word/sentence:

1. Write the identity relation on the set A = {a, b, c}
2. What is the range of the function f(x) = \(\frac{|x – 1|}{x – 1}\)?
3. Write the domain of real valued function f defined by f(x) = \(\sqrt { 25-x^{ 2 } } \)
4. Let * be a binay operation defined by a*b = 3a + 4b – 2, then find 4*5
5. Let fg : R → R, defined f(x) = 2x + 1 and g(x) = x² – 2, ∀x ∈ R respectively. Then find fog.

Answer:

1. {(a,a), (b,b), (c,c)}
2. {-1,1}
3. [-5,5]
4. 30
5. 4x² + 4x – 1

Relations and Functions Long Answer Type Questions – I

Question 1.
Show that the relation R in set Z of integers given by R = {(a, b) : 2 divides a – b} is an equivalence relation? (NCERT)
Solution:
R is reflexive as 2 divides (a – a) for all Z.
Let (a,b) ∈ R ⇒ 2, divides (a – b)
⇒ 2,divides – (b – a)
(a,b) e R ⇒ (b,a) ∈ R
∴R is symmetric.
Let (a,b) ∈ R ⇒ 2, divides (a -b)
(b,c) ∈ R ⇒ 2, divides (b – c)
(a – b) and (b – c), are divisible by 2
(a – b) + (b – c), are divisible by 2
Hence (a – c) is also divisible by 2
∴ (a,b) ∈ R,(b,c) ∈ R ⇒ (a,c) ∈ R
∴ R is transitive.
∴ R is reflexive, symmetric and transitive. Therefore R is an equivalence relation in Z. Proved.

Question 2.
Let R be the relation defined in the set A = {1,2,3,4, 5,6,7} by R = {(a, b): both a and b are either odd or even}
Show that R is an equivalence relation. Further, show that all the elements of the subset {1,3,5,7} are related to each other and all the elements of the subset {2,4,6} are related to each other but no element of the subset {1,3,5, 7} is related to any element of the subset {2,4, 6}? (NCERT)
Solution:
Given any element a in A, both a and a must be either odd or even. So that (a, a) ∈ R .
Further (a,b) ∈ R ⇒ both a and b must be either odd or even (a,b) ∈ R
Similarly (a,b) ∈ R (b,c) ∈ R ⇒ all the elements of the a, b, c must be either even or odd simultaneously ⇒ (a,c) ∈ R .
Hence R is an equivalence relation.
All elements of {1,3,5,7} are related to each other, as all the elements of this subset are odd. Similarly, all the elements of the subset {2, 4, 6} are related to each other(MPBoardSolutions.com) as all of them are even. Also no elements of the subset{ 1, 3, 5, 7} can be related to any elements of {2,4, 6} as elements of {1, 3, 5, 7} are odd, while elements of {2, 4, 6} are even.

Question 3.
Let N is set of natural numbers. If R is a relation defined in set N × N such that (a, b) R (c, d). If ad (b + c) = bc (a + d). Prove that R is an equivalence? (CBSE 2015)
Solution:
Reflexive: For every (a,b) ∈ N × N
ab (b + a) = ba (a + b)
⇒ (a, b) R {a, b)
∴ R is reflexive.
Symmetric: Let (a,b) (c,d) ∈ N × N
(a,b) R (c,d) ⇒ ab(b + c) = bc (a + d)
⇒ bc (a + d) = ab(b + c)
(a,b) R (c,d) ⇒ (c,d) R (a,b)
∴ R is symmetric.
Transitive:
Let (a, b) R(c,d) and (c, d) R (e,f)
⇒ ad (b + c) = bc (a + d)
and cf (d + e) = de (c+f)
⇒ \(\frac{ad (b+c)}{abcd}\) = \(\frac{bc (a+d)}{abcd}\)
and \(\frac{cf(d+e)}{cdef}\) = \(\frac{de(c+f)}{cdef}\)
⇒ \(\frac{1}{c}\) + \(\frac{1}{b}\) = \(\frac{1}{d}\) + \(\frac{1}{a}\)
and \(\frac{1}{e}\) + \(\frac{1}{d}\) = \(\frac{1}{f}\) + \(\frac{1}{c}\)
⇒ \(\frac{1}{c}\) + \(\frac{1}{b}\) + \(\frac{1}{e}\) + \(\frac{1}{d}\) = \(\frac{1}{d}\) + \(\frac{1}{a}\) + \(\frac{1}{f}\) + \(\frac{1}{c}\), (by adding)
⇒ \(\frac{1}{b}\) + \(\frac{1}{e}\) = \(\frac{1}{a}\) + \(\frac{1}{f}\)
⇒ \(\frac{b+e}{be}\) = \(\frac{a+f}{af}\)
⇒ af (b+e) = be (a+f)
(a,b) R (c,d) R (e,f) ⇒ (a,b) R (e,f)
∴ R is transitive
∴ R is reflecxive, symmetric and transitive, hence, equivalence.

Question 4.
Let A = {1,2,3,4,5} and R = {(a, b) : |a – b| is divided by 2}. Prove that R is an equivalence relation. Also form equivalence class?
Solution:
A = { 1, 2, 3,4, 5 } and R = {(a, b ) : |(a – b) is divisible by 2}
R = {(1, 1), (2,2), (3,3), (4,4), (5,5), (1,3), (1,5), (2,4), (3,5), (3,1) (5,1), (4,2),(5,3)}
∀a eA(a,a) E R R is reflexive.
Since (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) ∈ R
(a,b) ∈ R ⇒ (b,a) ∈ R
(1, 3), (1, 5), (2,4), (3, 5), (3, 1), (5, 1), (4, 2), (5, 3) ∈ R
∴ R is symmetric.
∀(a,b) ∈ R, (b,c) ∈ R ⇒ (a,c) ∈ R
Since (1,3), (3,1) ∈ R ⇒ (1,1) ∈ R
R is transitive.
R is reflexive, symmetric and transitive, hence R is equivalence. Proved.
Equivalence class:
[1] = { a : a and 2 is divided by |a – 1|}
[1] = {a : a ∈ A and a – 1 = 2 k}
[1] = {1, 3, 5}
[2] = { a : a and 2, is divided by |a – 2 |}
[2] = {a : a and a – 2 = 2k}
[2] = {2,4}. Proved.

Question 5.
Let for all n ∈ N

defines a function f : N → N.
Tell, is the function f one – one onto? Justify your answer also? (NCERT)
Solution:
In f : N → N

f (1) = \(\frac{1+1}{2}\) = 1
f (2) = \(\frac{2}{2}\) = 1
Here, f (1) = f (2) ⇒ 1 ≠ 2.
In co – domain there is only one image 1 of the two elements 1 and 2 of domain. Hence, f is not an one – one function.

Case I.
When n is odd
n = 2r + l, r ∈ N
Then, 4r + 1 ∈ N exists such that,
f (4r + 1) = \(\frac{4r+1+1}{2}\) = \(\frac{4r+2}{2}\) = 2r + 1.
Clearly, there is a pre – image in domain of the each element of co – domain. Hence, f is onto function.

Case II.
When n = 2r (even number)
Then, 4r ∈ N exists such that,
f (4r) = \(\frac{4r}{2}\) = 2r
Clearly, there is a pre – image in domain of the each element of co – domain.
Hence, f is onto function.
Thus, f is one – one onto function

Question 6.
Let A = R – {3} and B = R – {1}. Discuss the function f : A → B defined by f (x) = \(\frac{x-2}{x-3}\), is the function is one – one and onto? Justify your answer also? (NCERT)
A = R – {3} and B = R – {1}.
Solution:
f: A → B, f(x) = \(\frac{x-2}{x-3}\) , A = R – {3} and B = R – {1}.
Let x,y ∈ A is such that,
f (x) = f (y)
⇒ \(\frac{x-2}{x-3}\) = \(\frac{y-2}{y-3}\)
⇒ (x – 2) (y – 3) = (y – 2) (x – 3)
⇒ xy – 3x – 2y + 6 = xy – 3y – 2x + 6
⇒ – 3x – 2y = – 3y – 2x
⇒ x = y
Here, f(x) = f(y) ⇒ x = y.
∴ f is one – one function.
Let y ∈ B = R – {1}
f is onto if, x ∈ A is such that,
f (x) = y
\(\frac{x-2}{x-3}\) = y
⇒ x – 2 = y (x-3)
⇒ x – 2 = xy – 3y
⇒ xy – x = xy – 3y – 2
⇒ x (y – 1) = 3y – 2
⇒ x = \(\frac{3y-2}{y-1}\) ∈ A
For each y ∈ B, x ∈ A, then

f (x) = y
∴ f(x) is onto function,
∴ f is one – one function.

Question 7.
Prove that function defined below f : N → N is one – one and onto: (NCERT)

Solution:
Let f(x₁) = f(x₂)
If x₁ is odd and x₂ is even, then
x₁ + 1 = x₂ – 1
x₁ – x₂ = – 2
Which is impossible.
There is no chance that x₁ is even and x₂ is odd.
Hence either x₁ and x₂ both are even or odd.
Let x₁, x₂ both are odd.
f(x₁) = f(x₂)
⇒ x₁ + 1 = x₂ + 1
⇒ x₁ = x₂
Let x₁ and x₂ both are even.
f(x₁) = f(x₂)
⇒ x₁ – 1 = x₂ – 1
⇒ x₁ = x₂
∴ f is one – one
Odd number 2r + 1 of co – domain N is image of number 2r + 2 of N and any even number of co – domain N. i.e; 2r is image of number 2r – 1 of N.
∴ f is onto.

Question 8.
Discuss on function given by f(x) = 4x + 3, f : R →R. Prove that f is invertible. Find also the inverse of f? (NCERT)
Solution:
f : R → R, f(x) = 4x + 3
Domain and co – domain of the function is R.
Let, x,y ∈R are such that,
f(x) = f(y)
4x + 3 = 4y + 3
⇒ 4x = 4y
∴ x = y
Thus, f(x) = f(y) ⇒ x = y
f is one – one.
Let y is any element of co – domain R.
∴ y = f(x)
y = 4x + 3
⇒ 4x = y – 3
⇒ x = \(\frac{y – 3}{4}\)
y is real. Then, x = \(\frac{y – 3}{4}\) is also real. i.e., there is pre – image in domain of each of the element in co – domain.
i.e., range = co – domain
f is onto.
Hence, f is one – one and onto.
Let f – image of x is y. Then,
y = f(x)
y = 4x + 3, y ∈R
4x = y – 3, [f(x) = y ∴x \(f^{ -1 }\) (y)]
⇒ \(f^{ -1 }\) (y) = \(\frac{y – 3}{4}\)
∴ \(f^{ -1 }\) (x) = \(\frac{x – 3}{4}\)

Question 9.
Let y = {n²: n ∈ N] ⊂ N consider f : N → y as f(n) = n² Show that/is invertible. Find the inverse of f? (NCERT)
Solution:
y = f(n) = n²
n = \(\sqrt { y } \)
From g(y) = \(\sqrt { y } \) is defined g : y → N.
gof (n) = g[f(n)]
= g[n²]
= \(\sqrt { n^{ 2 } } \). [ ∵g (n) = \(\sqrt { n } \) ⇒ g(n²) = \(\sqrt { n^{ 2 } } \) = n]
(gof) n = n
and (fog) y = f [g(y)]
= f \(\sqrt { y } \), [ ∵f (n) = n² ⇒ f ( \(\sqrt { y } \)) = ( \(\sqrt { y } \))² = y]
= y
Clearly, gof = I_n and fog = I_y
Hence, f is invertibel and f^-1 = g. Proved.

Question 10.
If f : R → R and g : R → R are defined as f(x) = cosx and g(x) = 3x². Find gof and fog. Prove that gof ≠ fog? (NCERT)
Solution:
Given:
f(x) = cosx
g(x) = 3x²
(gof) x = g[ f(x)]
(gof) x = g [cosx]
Given: f(x) = cos x …………….. (1)
g (x) = 3x²
g(cosx) = 3 cos²x
From eqns. (1) and (2)
(gof) x = 3 cos²x ………………. (2)
(fog) = f [g(x)]
= f [3x²] …………………….. (3)
Given: g(x) = 3x²
f (x) = cosx
f [3x²] = cos3x² ………………………… (4)
From eqns. (3) and (4),
(fog)x = cos 3x²
For x = 0
3 cos²x ≠ cos 3x²
Hence, gof ≠ fog.

Question 11.
Let f : {1,2,3} → {a,b,c} given by f(1) = a, f(2) = b, f(3) = c. Find f^-1 and show that (f^-1)^-1 = f?
Solution:
Given:
f : {1,2,3} → {a,b,c}
f(1) = a, f(2) = b, f(3) = c
Let g: {a,b,c} → {1,2,3}
g(a) = 1, g(b) = 2, g(c) = 3
(fog)a = f[g(a)]
= f[1] = a
(fog)b = f[g(b)]
= f(2) = b
(fog)c = f[g(b)]
= f(3) = c
and (gof) (1) = g[f(1)]
= g(a) = 1
(gof) (2) = g[f(2)]
= g(b) = 2
(gof) (3) = g[f(3)]
= g(c) = 3
Hence gof = I_x and fog = I_y
Where x = {1,2,3} and y = {a,b,c}
Inverse of f exists
and f^-1 = g
∴ f^-1 → {a,b,c} {1,2,3}
f^-1(a) = 1, f^-1(b) = 2, f^-1(c) = 3
Now, find inverse of f^-1 i.e.,g.
Let h : {1,2,3} → {a,b,c}
h(1) = a, h(2) = b, h(3) = c
(goh) 1 = g [h(1)] = g(a) = 1
(goh) 2 = g [h(2)] = g(b) = 2
(goh) 3 = g [h(3)] = g(c) = 3
and (hog) a = h[g(a)] = h(1) = a
(hog) b = h [g(b)] = h(2) = b
(hog) c = h [g(c)] = h(3) = c
∴ goh = I_x and hog = I_y
Where, x = {1,2,3} and y = {a,b,c}
Inverse of g exists and g^-1 = h = (f^-1)^-1 = f
∴ h = f
∴ (f^-1)^-1 = f. Proved.

Question 12.
Show that the relation defined in the set A of all polygons as R = {(P₁,P₂): P₁ and P₂ have same number of sides } is an equivalence relation. (MPBoardSolutions.com) What is the set of all elements in A related of the right angle triangle T with sides 3,4 and 5? (NCERT)
Solution:
Given, A = set of all polygons
R = {(P₁, P₂): P₁ and P₂ have same number of sides}
In each polygon P the number of sides of polygon P are equal.
(P,P) ∈ R, ∀P ∈ A
Let (P₁,P₂) ∈ R
⇒ The number of sides in polygon P₁ and Polygon P₂ are same.
⇒ The sides of polygon P₂ and polygon P₁ are same.
(P₁,P₂) ∈ R ⇒ P₂P₁ ∈ R
∴ R is a symmetric relation.
Let (P₁,P₂) ∈ R and P₂,P₃) ∈ R.
The number of sides of polygon P₁ and P₂) are same.
∴ (P₁,P₂) ∈ R , (P₂,P₃) ⇒ (P₁,P₃) ∈ R
∴ Relation R is transitive.
The number of polygon realated with right angle traingle with sides 3, 4, 5 be three. Hence polygon realted with right angles traingle with sides 3, 4, 5 is a traingle. Proved.

Question 13.
If f(x) = \(\frac{4x+3}{6x-4}\) , x ≠ \(\frac{2}{3}\) then, prove that for all x ≠ \(\frac{2}{3}\), fof(x) = x. What is inverse function of f?
Solution:

Let, inverse function of f^-1 (x) = y .
Then f(y) = x
∴ \(\frac{4y+3}{6y-4}\) = x
⇒ 4y +3 = 6xy – 4x
⇒ 6xy – 4y = 3 + 4x
⇒ y (6x – 4) = 3 + 4x
⇒ y = \(\frac{3+4x}{6x-4}\)
⇒ f^-1 (x) = \(\frac{3+4x}{6x-4}\) = f(x)
∴ f^-1 = f.

Question 14.
Prove that the function given by f : [-1,1] → R, f(x) = \(\frac{x}{x+2}\) is one – one. Find the inverse function of function f : [-1,1] → (Range of f).
Solution:
f(x) = \(\frac{x}{x+2}\)
f : [-1,1] → R
Here f(x) = f(y)
⇒ \(\frac{x}{x+2}\) = \(\frac{y}{y+2}\)
⇒ xy + 2x = xy + 2y
⇒ 2x = 2y
⇒ x = y
∴ f is one – one
Let f^-1(x) = y
∴ f(y) = x
⇒ \(\frac{y}{y+2}\) = x
⇒ y = xy + 2x
⇒ y (1 – x) = 2x
∴y = \(\frac{2x}{1-x}\)
⇒ f^-1(x) = \(\frac{2x}{1-x}\).

Question 15.
If f(x) = \(\frac{x}{1+|x|}\), ∀x ∈ R and g(x) = \(\frac{x}{1-|x|}\) , ∀x ∈ R where -1 < x < 1, then find gof and fog? Show that fog = gof?
Solution:
Given: f(x) = \(\frac{x}{1+|x|}\)


∴ gof (x) = x
∴ From eqns. (1) and (2),
fog = gof.

Question 16.
Consider f : N → N, g : N → N and h : N → R defined as f(x) = 2x, g(y) = 3y + 4 and h(z) = sin z ∀ x, y and z ∈ N. Show that ho(gof) = (hog)of? (NCERT)
Solution:
ho(gof) x = h[gof(x)]
= h[g[f(x)]
= h(g(2x))
= h[3(2x) + 4]
= h[6x + 4]
= sin (6x + 4) ……………………. (1)
Similarly, ((hog)of)x = (hog) f(x)
= (hog) 2x
= h(g(2x))
= h[3(2x) + 4]
= h [6x + 4]
= sin (6x + 4) ……………………… (2)
From eqns. (1) and (2),
ho(gof) = (hog)of. Proved.

MP Board Class 12 Maths Important Questions

MP Board Class 12th Chemistry Important Questions Chapter 9 Coordination Compounds

Coordination Compounds Important Questions

Coordination Compounds Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
The correct structural formula of Zeise’s salt is :
(a) K⁺ [PtCl₃(C₂H₄)]^–
(b) K⁺[PtCl₂(η² – C₂H₄)^–]Cl^–
(c) K⁺[PtCl₃(η² – C₂H₄)]
(d) K⁺PtCl₃(η² – C₂H₄)]^–
Answer:
(a) K⁺ [PtCl₃(C₂H₄)]^–

Question 2.
AgCl is soluble in aqueous ammonia due to formation of:
(a) [Ag(NH₃)₂]²⁺
(b) [Ag(NH₄)₂]⁺
(c) [Ag(NH₃)₄]⁺
(d) [Ag(NH₃)₂]⁺.
Answer:
(d) [Ag(NH₃)₂]⁺.

Question 3.
Which of the following gives a white precipitate in aqueous solution with silver nitrate:
(a) [Cr(NH₃)₅Cl](NO₂)₂
(b) [Pt(NH₃)₃Cl₂]
(c) [Pt(CN)₂Cl₂]
(d) [Pt(NH₃)₂]Cl₂.
Answer:
(d) [Pt(NH₃)₂]Cl₂.

Question 4.
Correct nomenclature of Fe₄[Fe(CN)₆]₃ is :
(a) Ferroso ferric cyanide
(b) Ferric ferrous hexacyanate
(c) Iron (III) Hexacyanoferrate
(d) Hexacyanoferrate (III-II).
Answer:
(c) Iron (III) Hexacyanoferrate

Question 5.
In which of the following compounds oxidation state of metal is zero :
(a) [Pt(NH₃)₂Cl₂]
(b) [Cr(CO)₆]
(c) [Cr(NH₃)₃Cl₃]
(d) [Cr(CN)₂Cl₂]
Answer:
(b) [Cr(CO)₆]

Question 6.
Example of dsp² hybridization is :
(a) [Fe(CN)₆]^-3
(b) [Ni(CN)₄]^-2
(c) [Zn(NH₃)₄]⁺²
(d) [FeF₆]^-3
Answer:
(b) [Ni(CN)₄]^-2

Question 7.
Which of the following complex is used as anti-cancer agent:
(a) Trsns[CO(NH₃)₂Cl₃]
(b) cis[Pt(NH₃)₂Cl₂]
(c) cisK₂[PtCl₂Br₂]
(d) Na₂CO₃.
Answer:
(b) cis[Pt(NH₃)₂Cl₂]

Question 8.
Oxidation state of Fe in [Fe(CO)₃] complex is :
(a) -1
(b) +2
(c) + 4
(d) 0
Answer:
(d) 0

Question 9.
Grignard reagent is :
(a) Organometallic compound
(b) Complex compound
(c) Double salt
(d) Neutral compound.
Answer:
(c) Double salt

Question 10.
Structure of complex salt was proposed by :
(a) Berzelius
(b) Werner
(c) Raoult
(d) Faraday.
Answer:
(b) Werner

Question 11.
Mohr’s salt is :
(a) Double salt
(b) Complex salt
(c) Neutral salt
(d) Reagent.
Answer:
(a) Double salt

Question 12.
The formula of nitroprusside is :
(a) Na₄[Fe(CN)₅NO₅]
(b) Na₂[Fe(CN)₅NO]
(c) NaFe[Fe(CN)₆]
(d) Na₂[Fe(CN)₆NO₂].
Answer:
(b) Na₂[Fe(CN)₅NO]

Question 13.
Which of the following is not an organometallic compound : (MP 2018)
(a) C₂H₅MgBr
(b) (C₂H₅)₄Pb
(c) C₂H₅ONa
(d) (CH₃)₄Al.
Answer:
(b) (C₂H₅)₄Pb

Question 14.
Zeigler Natta catalyst is :
(a) (Ph₃P)₃RhCl
(b) K[PtCl₃(C₂H₄)]
(c) [Al₂(C₂H₆)₆]
(d) [Fe(C₂H₅)₂].
Answer:
(a) (Ph₃P)₃RhCl

Question 15.
The I.U.P.A.C. name of [Ni(CO)4] is :
(a) Tetracarbonyl nickelate (0)
(b) Tetracarbonyl nickelate (II)
(c) Tetracarbonyl nickel (0)
(d) Tetracarbonyl nickel (II)
Answer:
(c) Tetracarbonyl nickel (0)

Question 2.
Fill in the blanks :

1. Cis [Pt(NH₃)₂Cl₂] complex is used as an ……………. agent.
2. Haemoglobin is a ……………. compound of iron.
3. Geometrical isomerism is found in both ……………. and ……………. complexes.
4. Oxidation state of Ni in Ni(CO)₄ is …………….
5. Diethyl zinc is a ……………. compound.
6. The correct I.U.P.A.C. name of K₄[Fe(CN)₆] is …………….
7. Oxidation state of Co in [Co (E.D.T.A)] is …………….
8. The formula of dibromo chlorotriaquo chromium (III) is …………….
9. [COF₆]^-3 is a ……………. spin complex.
10. The formula of antiknock organometallic substance is …………….
11. E. D. T. A is ……………. ligand.
12. Example of hexadentate ligand is …………….

Answer:

1. Anti – cancer
2. Complex
3. Tetrahedral, Octahedral
4. Zero
5. Organometallic compound
6. Potassium hexacyano ferrate (II)
7. +3,8. [Cr(H₂O)₃Cl Br₂]
8. High
9. Tetraethyl lead (C₂H₅)₄Pb
10. Hexadentate
11. E.D.T.A.

Question 3.
Match the following :

Answer:

1. (h)
2. (g)
3. (f)
4. (e)
5. (b)
6. (d)
7. (a)
8. (c)
9. (i)

Question 4.
Answer in one word/sentence :

1. Which type of isomerism is found in [CO(NH₃)₅Br]SO₄ and [CO(NH₃)₅SO₄]Br?
2. Name the organometallic compound which is used as an antiknock compound in petrol. (MP2018)
3. Complexes of E.D.T.A. formed with calcium is used to remove the poisoning caused due to which metal?
4. How is the structure of dibenzene?
5. What type of hybridization is found in [Ni(CO)₄]?
6. Which type of isomerism is represented by [Cr(H₂O)₅SCN]²⁺ and [Cr(H₂O)₅NCS]²⁺?

Answer:

1. Ionization isomerism
2. Tetraethyl lead
3. Lead
4. Sandwich
5. sp³
6. Linkage isomerism.

Coordination Compounds Very Short Answer Type Questions

Question 1.
Among the following ions, whose magnetic moment value will be maximum. (NCERT)

1. [Cr(H₂O)₆]³⁺
2. [Fe(H₂O)₆]²⁺
3. [Zn(H20)6]²⁺.

Answer:
2. [Fe(H₂O)₆]²⁺.

Question 2.
Write two examples of monodentate ligands.
Answer:
NO⁺ (Nitrosonium), NH₂NH₃ (Hydrazinium).

Question 3.
What will be the geometry of [Cr(NH₃)₆]³⁺ complex ion?
Answer:
Tetrahedral.

Question 4.
Write IUPAC name of the complex : [Pt(NH₃)₄][PtCl₄].
Answer:
Tetraammine platinum (II) tetrachloridoplatinate (II).

Question 5.
What is the hybridization of Fe in the complex ion [Fe(CN)₆]^3-?
Answer:
Hybridization of Fe in the complex [Fe(CN)₆]^3- is d²sp³.

Question 6.
State the full name of EDTA.
Answer:
Ethylene diammine tetraacetate ion.

Question 7.
Which are eg orbitals?
Answer:
d_{x² – y²} and d_z² are e_g orbitals.

Question 8.
Give an example of a neutral bidentate ligand.
Answer:
Ethylene diamine

Question 9.
Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligand co – ordinated with the central metal ion?
Answer:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom are the some with respect to each other.

Question 10.
State the magnetic property and hybridization of [Ni(CO)₄J.
Answer:
Diamagnetic (All electrons paired) and sp³ hybridization.

Coordination Compounds Short Answer Type Questions

Question 1.
Explain double salt and complex salt. Give one – one example of each.
Answer:
Double salt: Double salts are additive compounds which are stable in the crystal lattice but when dissolved in water break into different compounds.
Example : Ferrous ammonium sulphate is a double salt which ionise in water as
FeSO₄ (NH₄)₂ SO₄ . H₂O ⇌ Fe²⁺ + SO₄^2- + 2(NH₄)⁺ + SO₄^2- + H₂O

Complex salt:
The compounds in which ligand with lone-pair electron are linked with any metal atom or metal ion by coordinate bonds, are called coordination compounds. In these compounds metal ion and ligand in combined state act as complex ion and thus these compounds are also known as complex compounds.
Example: K₄[Fe(CN)₆].

Question 2.
What do you understand by ligand ? Explain giving example.
Answer:
Ligand:
Any atom, ion or molecule which can donate electron pair to central ion and forms co – ordinate bond are called ligand.
Example : In K₄[Fe(CN)₆], CN is ligand. (MPBoardSolutions.com) In ligand the specific atom which donates the electron pair is known as donor atom.
On the basis of number of donor atoms in a ligand, they are classified as monodentate, bidentate, tridentate, polydentate ligands. Some such ligands are given below :

In the ligand the asterisk atom is donor atom.

Question 3.
What is meant by the chelate effect? Give an example. (NCERT)
Answer:
When a ligand attaches to the metal ion in a manner that forms a ring, then the metal – ligand association is found to be more stable. In other words, we can say that complexes containing chelate rings are more stable than complexes without rings. This is known as the chelate effect for example :

Question 4.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain. (NCERT)
Answer:
H₂O is a weak ligand [Ni(H₂O)₆]²⁺ is a outer orbital complex. The complex has two unpaired electron. The d – d transition is possible. It absorbs red light and complementary green light is emitted.

CN^–is a strong ligand. The unpaired electrons are paired up. The central atoms undergoes dsp² hybridization. Square planar complex is formed. No unpaired electrons are present i.e. d – d transition is not possible, hence the complex is colourless.

Question 5.
[Fe(CN)₆]⁴⁺ and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions, why? (NCERT)
Answer:
In both the complexes, Fe is in +2 state with the configuration 3d⁶ i.e., it has four unpaired electrons. As the ligand H₂O and CN^– possess different crystal field splitting energy (Δ₀), they absorb different components of the visible light (VIBGYOR) for d – d transition. Hence, the transmitted colours are different.

Question 6.
Discuss the nature of bonding in metal carbonyls. (NCERT)
Answer:
The metal carbon bonds in metal carbonyls have both s nad p characters. M – C σ – bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. M – C π – bond is formed by the donation of a pair of electrons from the filled metal d orbital into the vacant anti – bonding π * orbital of carbon monoxide. (MPBoardSolutions.com)
This is also known as back bonding of the carbonyl group. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal. This synergic effect strengthens the bond between CO and the metal.

Question 7.
Give evidence that [CO(NH₃)₅Cl]SO₄ and [CO(NH₃)₅SO₄]Cl are ionization isomers. (NCERT)
Answer:
Ionisation isomers when dissolved in water furnish different ions which can be tested.

White precipitate of AgCl indicates that the isomer has Cl ion outside the co – ordination sphere.

White precipitate of BaSO₄ indicates the isomer has SO₄^2- ion outside the co – ordination sphere.

Question 8.
Explain optical isomerism in co – ordination compounds.
Answer:
Optical isomerism:
This type of isomerism is observed in such similar compounds which are the mirror images of each other and are non- superimposable. They rotate the path of plane polarized light to the left (l) or to the right (d).

Question 9.
What is complex ion?
Answer:
Complex Ion:
A complex ion is an electrically charged redical or a species, carrying either positive or negative charge, in which the central metal ion is surrounded through co – ordinate bond by a suitable number of ligands (neutral molecules or negative ions).
Example:
Complex ferrocyanide ion[Fe(CN)₆]^4- is formed by the union of six CN^–ions with one Fe²⁺ ion. While writing the formula of a complex ion, the co – ordinating groups are written inside the bracket ( ), and the whole of the complex ion in a square bracket [ ]. The net charges is written on right hand top comer of the square bracket. The square bracket is known as co – ordination sphere.

Question 10.
Give example of bidentate and hexadentate ligand.
Answer:
Bidentate ligand :

Hexadentate ligand:

Question 11.
What is co – ordination number? Give any two examples.
Answer:
Co – ordination number:
Number of ligands which are directly linked with central metal or metal ion by co – ordinate bonds, are co – ordination number.
Example:

• [CO(NH₃)₆]Cl₃ co – ordination number of CO³⁺ ion is 6.
• In [Ag(CN)₂] co – ordination number of Ag is 2.

Question 12.
What are organometallic compounds? Write its two applications
Answer:
Those compounds in which the carbon atom of organic groups are directly bonded to metal atoms are called organometallic compounds. The compound of element such as boron, phosphorus, silicon, germanium and antimony with organic groups are also included in the organometallics.

Applications of organometallic compounds :

1. Tetraethyl lead (C₂H₅)₄Pb is used as an antiknock compound.
2. Ziegler – Natta catalysis is used in the polymerization of ethylene or other alkenes.
3. Ethyl mercuric chloride (C₂H₅HgCl) is used in agriculture as an insecticide.
4. Willkinson catalyst is used in the hydrogenation of some alkenes.

Question 13.
Explain geometrical isomerism with an example.
Answer:
Geometrical isomerism:
When the ligands are situated at different position around the metal it gives rise to geometrical isomerism. The isomer in which similar groups are in adjacent position (making an angle 90° with metal ion) is called cis – isomer. The isomer in which the similar group occupy the opposite position (making an angle of 180° with metal ion) is called transisomer. This type of isomerism is observed in square planar (CN = 4) and octahedral (CN = 6) type of complexes.

Question 14.
[NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why? (NCERT)
Answer:
[NiCl₄]^2- has 2 unpaired electrons and is paramagnetic (See Q. No. 4 for details). Like CN^–, CO is also strong field ligand, same like CO, it causes pairing of electrons. No unpaired electrons left, hence it is diamagnetic.

Question 15.
[Fe(H₄O)₆ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic. Explain. (NCERT)
Answer:
In both the complexes Fe is in +3 oxidation state with the configuration 3d⁵. CN^– is a strong ligand. In its presence, 3d – electrons pair up leaving only one unpaired electron. The hybridization is d²sp³ forming inner orbital complex. H₄O is a weak ligand. In its presence, 3d – electrons do not pair up. The hybridization is sp³ d² forming an outer orbital complex containing five unpaired electrons. Hence, it is strongly paramagnetic.

Question 16.
Explain [CO(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex. (NCERT)
Answer:
In [CO(NH₃)₆]³⁺, CO is in +3 oxidation state with the configuration 3d⁶. In the presence of NH₃, 3d – electrons pair up leaving two d – orbitals empty. Hence, the hybridization is d²sp³ forming an inner orbital complex.

In [Ni(NH₃)₆]²⁺, Ni is in +2 oxidation state with the configuration 3d⁶. In presence of NH₃, the 3d – electrons do not pair up. The hybridization involved is sp³d² forming outer orbital complex.

Question 17.
Write the IUPAC names of the following co – ordination compounds : (NCERT)

1. [CO(NH₃)₆]C₁₃
2. [CO(NH3)5CI]C12
3. K₃[Fe(CN)₆]
4. K₃[Fe(C₂O₄)₃]
5. K₂[PdCl₄]
6. [Pt(NH₃)₂Cl(NH₂CH₃)]Cl.

Answer:

1. Hexaammine cobalt(III) chloride
2. Pentaamminechloridocobalt(III) chloride
3. Potassium hexacyanoferrate(III)
4. Potassium trioxalatoferrate(III)
5. Potassium tetrachloridopalladate(II)
6. Diamminechlorido(methanamine)platinum(II) chloride.

Question 18.
Explain the structure of [Ni(CO)₄] on the basis of valence bond theory.
Answer:
Complexes in which the metal ion is sp³ hybridized represent tetrahedral geometry.
For example : Formation of tetracarbonylnickel (0) Can be explained by sp³ hybridization.
In [Ni(CO)₄], Nickel is in zero oxidation state. Thus, outer electronic configuration of nickel is 3d⁸ 4s³. CO is a strong ligand, thus due to the effect of ligand the 4s electrons move to 3d and make all the 3d electrons paired. This way, one 4s and three 4p become vacant and intermix to form equivalent

sp³ hybridization of [Ni(CO)₄]

sp³ hybrid orbitals which are tetrahedrally oriented. These four hybrid orbitals overlap with lone electron pairs of
4CO and form sigma bond.

[Ni(CO)₄] does not contain unpaired electron. Thus, it is diamagnetic and tetrahedral.

Question 19.
Explain the structure of [Zn(NH₃)₄]²⁺ on the basis of valence bond theory.
Answer:
Structure of [Zn(NH₃)₄]²⁺ Tetraammine zinc (II) ion:
Outer electronic configuration of zinc (Z = 30) in ground state is 3d¹⁰4s². In this complex zinc is in +2 oxidation state with the outer electronic configuration of 3d¹⁰.
The 3d orbital being completely filled does not take part in hybridization. The vacant 4s and 4p orbitals hybridize and form four hybridized sp³orbitals directed

towards the four comers of a tetrahedron. The four lone pairs from four NH₃ overlap with these sp³ orbitals and forms four σ – bonds. The compound is diamagnetic as it does not contain unpaired electrons.

Question 20.
Write IUPAC name of the following :

1. [Pt(NH₃)₃ Cl₂]
2. K₃[Fe(CN)₆]
3. [CO(NH₃)₆] Cl₃
4. Pt[(NH₃)₆] Cl₄
5. CuCl₃.

Answer:

1. Dichlorodiammine platinum (II)
2. Potassium hexacyanoferrate (III)
3. Hexaammine cobalt (III) chloride
4. Hexaammine platinum (IV) chloride
5. Tetrachlorocuprate (II)

Question 21.
1. Write I.U.P.A.C. name of

2. Write the ligands and co – ordination number of the complex [Cr (NH₃)₄(ONO) CI]NO₃.
3. Write chemical formula of carbonate pentaammine cobalt (III) chloride.
Answer:
1. µ – amido – µ -hydroxobis – (tetraammine cobalt) (III) ion.
2. Ligands :

• NH₃
• ONO
• Cl

Thus, number of ligands are 3 and co – ordination number is 6.

3. [CO(NH₃)₅CO₃]Cl.

Question 22.
Using IUPAC norms write the formulas for the following :

1. Tetrahydroxozincate(II)
2. Potassium tetrachloridopalladate(II)
3. Diamminedichloridoplatinum(II)
4. Potassium tetracyanonickelate(II)
5. Pentaamminenitrito-o-cobalt(III)
6. Hexaamminecobalt(III) sulphate
7. Potassium tri-(oxalato)chromate(III)
8. HexaamminepIatinum(IV)
9. Tetrabromidocuprate(II)
10. Pentaamminenitrito-N-cobalt(III).

Answer:

1. [Zn(OH]²
2. K₂[PdCl4]
3. [Pt(NH₃)₃Cl₂]
4. K₂[Ni(CN)₄]
5. [CO(NH₃)₅(ONO)]²⁺
6. [CO(NH₃)₆]₂ (SO₄)₃
7. K₃[Cr(C₂O₄)₃]
8. [Pt(NH₃)₆]⁴⁺
9. [Cu(Br)₄]^2-
10. [CO(NH₃)₅(NO₂)]²⁺.

Question 23.
Write the I.U.P.A.C. names of the following compounds:
(i)
(a) [HgI₄]^2-
(b) [Ag (CN)₂]^–
(c) [Fe (C₅H₅)₂]
(d) K [Ag (CN)₂].

(ii) What is Zeise’s Salt and Ferrocene? Explain with structure.
Answer:
(i)
(a) Tetraiodomercurate (II) ion.
(b) Dicyanoargentate (I) ion.
(c) Bis (cyclopentadienyl) iron (II).
(d) Potassium dicyano argentate (I).

(ii)
(a) Zeise’s salt K [PtCl₃ – η²(C₂ H₄ ) :
This salt was prepared by Danish pharmacist Zeise in 1830. It is one of the compound of transition metals which was prepared earlier. The plane of ethylene molecule and the C = C axis are perpendicular to the expected bond direction of the central atom.

(b) Ferrocene Fe (η⁵ – C₅H₅)₂:
It is an orange yellow coloured compound. Kealy and Pauson reported it in 1951. It has sandwich structure in which iron atom is in between two cyclopentadienyl rings. The planes of the rings are parallel so that all the carbon atoms are equidistant from iron atom.

Question 24.
Explain Linkage and Ionisation isomerism with example.
Answer:
Linkage isomerism:
Linkage isomerism occurs when different atoms of the ligand are attached to the central metal ion. The structure obtained are called linkage isomers. Such type of ligands are called ambidentate ligand.

In structure I Ni²⁺ is linked by thiocyanate sulphur and in structure II it is linked by nitrogen atom.

Ionization Isomerism:
This type of isomerism is shown by such compounds which have same composition but liberate different ions in solution.
[CO(NH₃)₅Br] SO₄ – It liberates SO^2- ions.
[CO(NH₃)₅SO₄]Br – It liberates Br^– ions.

Question 25.
What is the difference between primary and secwraary valency? Give example also.
Answer:
Primary valency is ionisable while secondary valency does not ionise. Primary valency is represented in figure by solid lines and secondary valency by broken or dotted lines.

Example:
[CO (NH₃)₆] Cl₃, primary valency is 3 and secondary valency is 6. Co – Central metal, NH₃, Cl – Ligand.

Question 26.
What is effective atomic number (EAN)?
Answer:
In co – ordination compounds, sum of electrons present in central metal ion and electrons received in bond formation is called effective atomic number.
EAN = Atomic number – Electrons lost in ion formation + Electrons obtained in bond formation
In K₄[Fe(CN)₆] EAN for Fe = 26 – 2 + 12 = 36.

Coordination Compounds Long Answer Type Questions

Question 1.
Explain the bonding in co – ordination compounds in terms of Werner’s postulates. (NCERT)
Answer:
Werner’s co – ordination theory:
Alfred Werner gave his co – ordination theory in 1893. The important postulates of this theory are :
1. All metals in atomic or ionic form exhibit two types of valencies in co – ordination compounds :

• Primary or principal or ionic valency ( ….. )
• Secondary or auxiliary or non-ionic valency (-).

The primary valency is ionizable and it is shown by dotted lines. The secondary valency is non – ionizable and it is shown by continuous line.

2. Primary valency represents oxidation states of metal atom or ion and secondary valency represents co – ordination number of metal ion which is fixed for a particular atom.

3. The primary valencies are satisfied by negative ions whereas the secondary valencies may be satisfied either by negative ions (Example Cl^–, Br^–, CN^– etc.) or neutral molecules (Example H₂O).

4. Secondary valencies are directed towards fixed position in space.

5. Every element tends to satisfy both its primary and secondary valencies. For this
purpose a negative ion may often act a dual behaviour i.e., it may satisfy primary as well as secondary valency ( ).

Example:
Luteo cobaltic chloride COCl₃.6NH₃ or [CO(NH₃)₆]Cl₃.
Purpureo cobaltic chlorideq COCl₃.5NH₃ or [CO(NH₃)₅Cl]Cl₂

Question 2.
List various types of isomerism possible for co – ordination compounds, giving an example of each. (NCERT)
Answer:
Isomerism in Co – ordination Compounds:
Two or more compounds having the same molecular formula but different arrangement of atoms are called isomers.
Isomerism in Co – ordination Compounds is given below:

1. Structural Isomerism
This type of isomerism arises due to the difference in structures of co-ordination compounds. It is further sub divided into four different types as explained below :

(a) Ionization Isomerism:
This type of isomerism is shown by such compounds which have same composition but liberate different ions in solution.
[CO(NH₃)₅Br] SO₄ – It liberates SO^2- ions.
[CO(NH₃)₅SO₄]Br – It liberates Br^– ions.

(b) Co – ordination isomerism:
This type of isomerism is shown by such compound which contain both cationic and anionic species and it arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex.
For example, [CO(NH₃)₆][Cr(CN)₆ and [Cr(NH₃)₆][CO(CN)₆]
hexaamminecobalt (III) hexacyano chromate (III) and hexaamminechromium (III) hexacyanocobalt (III).

(c) Linkage isomerism:
Linkage isomerism occurs when different atoms of the ligand are attached to the central metal ion. The structure obtained are called linkage isomers. Such type of ligands are called ambidentate ligand.

(d) Hydrate Isomerism:
This isomerism arises when different number of water molecules are present within and outside the co-ordination sphere. For example, three hydration isomer of CrCl₃.6H₂O are :

(i) [Cr(H₂O)₆]Cl₃ All the six water molecules act as ligand
(Violet)
(ii) [Cr(H₂O)₅Cl]Cl₂.H₂O Five water molecules acts as ligand while one molecule
(Green)
of water as crystallization.
(iii) [Cr(H₂O)₄Cl₂]Cl.2H₂O Four water molecules acts as ligand, while the two molecules of water as crystallization.
(Green)

2. Stereoisomerism:
Two compounds are called stereoisomers when they contain the same ligand in their co – ordination sphere but differ in their spatial arrangement. Stereoisomerism is further classified as geometrical and optical isomerism.

(a) Geometrical Isomerism:
When the ligands are situated at different position around the metal it gives rise to geometrical isomerism. The isomer in which similar groups are in adjacent position (making an angle 90° with metal ion) is called cis – isomer. (MPBoardSolutions.com) The isomer in which the similar group occupy the opposite position (making an angle of 180° with metal ion) is called transisomer. This type of isomerism is observed in square planar (CN = 4) and octahedral (CN = 6) type of complexes.

(b) Optical Isomerism:
This type of isomerism is observed in such similar compounds which are the mirror images of each other and are non- superimposable. They rotate the path of plane polarized light to the left (l) or to the right (d).

Question 3.
Explain crystal field theory.
Answer:
This theory was proposed by Bethe and Von Black. According to this theory bonding between central metal ion and ligand is due to pure electrostatic attraction. If ligand is anion then attraction towards cation, is just like the attraction between two oppositely charged particles. If ligand is neutral molecule then the anionic end of this dipole is attracted to central positive ion. Thus, bonding between them is due to ion – ion attraction or ion dipole attraction.

In crystal field theory due to approaching ligands the d – orbitals split into different energy levels on the basis of extent of splitting (depend on metal ion and nature of ligand) structure and properties of complex can be discussed. (MPBoardSolutions.com) Colour of transition metal complex is due to absorption of visible light which leads to excitation of electron from one d – orbital to the other d – orbital (d – d transition). This way this theory is easy and successfully explains maximum properties of the complexes.

MP Board Class 12th Chemistry Important Questions