Chemical Bonding(12th Grade > Chemistry ) Questions and answers for exam Preparation

12th Grade > Chemistry

CHEMICAL BONDING MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11. The pair of substances that have no hydrogen bonding is

CH3COOH, NH3
CH3F, CH4
C6H5NH2, HF
CH3OH, H2O

Discuss Question

Answer: Option B. -> CH3F, CH4
:
B
F is not linked to directly to hydrogen

Question 12. Which of thefollowing is the correct order of boiling points

HF> H2O> NH3> CH4
H2O> HF> NH3> CH4
HF> H2O> CH4> NH3
H2O> HF> CH4> NH3

Discuss Question

Answer: Option B. -> H2O> HF> NH3> CH4
:
B

H_{2} O

has higher b.pt among hydrogen bonded molecules

Question 13.

A s s e r t i o n - ------- -

(A): The boiling point of HF is higher than that of

H_{2} O

R e a s o n - ----- -

(R): The hydrogen bond strength in HF is lower than that in

H_{2} O

A and R are true and R is the correct explanation of A
A and R are true and R is not the correct explanation of A
A is true, R is false
A is false but R is true

Discuss Question

Answer: Option D. -> A is false but R is true
:
D
Due to stronger hydrogen bonding in water, it has a higher boilind point.

Question 14. The correct order of hybridisation of the central atom in the following species

N H_{3}

{[P t C l_{4}]}^{2 -}

P C l_{5}

and

B C l_{3}

dsp2 , dsp3 , sp2,sp3
sp3, dsp2, sp3d , sp2
dsp2,sp2,sp3,dsp3
dsp2,sp3,sp2,dsp3

Discuss Question

Answer: Option B. -> sp3, dsp2, sp3d , sp2
:
B

P t

would form an inner orbital d complex and

P

would form an outer orbital d complex.
This eliminates all the other options and you have b as the correct option.

Question 15. The correct order of decreasing polarisability of ion is :

Cl−>Br−>I−>F−
F−>I−>Br−>Cl−
I−>Br−>Cl−>F−
F−>Cl−>Br−>I−

Discuss Question

Answer: Option C. -> I−>Br−>Cl−>F−
:
C
The polarisability of an anion also depends on its size and charge – the larger the negative ion and the larger its charge the more polarisable it becomes, that is, more is the size of anion more will be the polarisability.

Question 16. The boiling point of methanol, water and dimethyl ether are respectively

65^{0} C, 100^{0} C a n d {34.5}^{0} C

. Which of the following best explains these wide variations in b.p.?

The molecular mass increases from water (18) to methanol (32) to diethyl ether (74)
The extent of H-bonding decreases from water to methanol while it is absent in ether
The extent of intramolecular H-bonding decreases from ether to methanol to water
The density of water is 1.00 g ml−1, methanol 0.7914 g ml−1 and that of diethyl ether is 0.7137 g ml−1

Discuss Question

Answer: Option B. -> The extent of H-bonding decreases from water to methanol while it is absent in ether
:
B
Water has strongest hydrogen bond of the three

Question 17. The state of hybridization of S in

S O_{2}

is similar to that of

C in C2H2
C in C2H4
C in CH4
C in CO2

Discuss Question

Answer: Option B. -> C in C2H4
:
B

S O_{2}

C_{2} H_{4}

have

{s p}^{2}

hybridisation.
The Carbon atoms in

C_{2} H_{2}

and

C O_{2}

has sphybridisation.
The Carbon atom in

C H_{4}

has

{s p}^{3}

hybridisation.

Question 18.

The order of dipole moment of the above molecules is

I > II = III > IV
II > I > III > IV
II > III > I > IV
II > I = III > IV

Discuss Question

Answer: Option D. -> II > I = III > IV
:
D
Para compound has zero dipole moment and ortho compound has maximum value.
If we calculate the vector addition, magnitudes of dipole moment for I and III turn
out to be the same.

Question 19. The types of bonds present in