Chemical Bonding(11th And 12th > Chemistry ) Questions and answers for exam Preparation

11th And 12th > Chemistry

CHEMICAL BONDING MCQs

Total Questions : 30 | Page 1 of 3 pages

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 $1.00 g m l^{- 1}$ , methanol $0.7914 g m l^{- 1}$ and that of diethyl ether is $0.7137 g m l^{- 1}$

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 2.

Which of thefollowing is the correct order of boiling points

$H F > H_{2} O > N H_{3} > C H_{4}$
$H_{2} O > H F > N H_{3} > C H_{4}$
$H F > H_{2} O > C H_{4} > N H_{3}$
$H_{2} O > H F > C H_{4} > N H_{3}$

Discuss Question

Answer: Option B. ->

H_{2} O > H F > N H_{3} > C H_{4}

:
B

H_{2} O

has higher b.pt among hydrogen bonded molecules

Question 3.

Which among the following compound have non-zero dipole moment?

Discuss Question

Answer: Option B. ->

:
B

$(μ \neq 0)$ for the cis isomer:

The trans isomer has 0 dipole moment as expected:

Question 4.

$A s s e r t i o n - ------- -$ (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 5.

The pair of substances that have no hydrogen bonding is

$C H_{3} C O O H, N H_{3}$
$C H_{3} F, C H_{4}$
$C_{6} H_{5} N H_{2}, H F$
$C H_{3} O H, H_{2} O$

Discuss Question

Answer: Option B. ->

C H_{3} F, C H_{4}

:
B

F is not linked to directly to hydrogen

Question 6.

The correct order of decreasing polarisability of ion is :

$C l^{-} > B r^{-} > I^{-} > F^{-}$
$F^{-} > I^{-} > B r^{-} > C l^{-}$
$I^{-} > B r^{-} > C l^{-} > F^{-}$
$F^{-} > C l^{-} > B r^{-} > I^{-}$

Discuss Question

Answer: Option C. ->

I^{-} > B r^{-} > C l^{-} > 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 7.

The correct sequence of decrease in the bond angles of the following hydrides is

$N H_{3} > P H_{3} > A s H_{3} > S b H_{3}$
$N H_{3} > A s H_{3} > P H_{3} > S b H_{3}$
$S b H_{3} > A s H_{3} > P H_{3} > N H_{3}$
$P H_{3} > N H_{3} > A s H_{3} > S b H_{3}$

Discuss Question

Answer: Option A. ->

N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

:
A

N H_{3}

has

s p^{3}

hybridisation with bond angle

107^{0}

where as

P H_{3}, A s H_{3}, S b H_{3}

has pure orbitals overlapping with bond angle around

90^{0}

Question 8.

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.