Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Chemical Bonding and Molecular Structure

Multiple Choice Questions

101.

Which of the following pair have identical bond order?

CN^- and NO⁺
CN^- and O $_{2}^{-}$
CN^- and CN⁺
NO⁺ and O $_{2}^{-}$

102.

Hybridisation of carbon is CH $_{3}^{-}$

sp²
sp³
sp³d
sp³d²

103.

The common features among CO, CN^- and NO $_{2}^{+}$ are

Bond order three and isoelectronic.
Bond order three and weak field ligands.
Bond order two and $π$ -acceptors.
Bond order three and $π$ -donors.

104.

Identify the T-shaped molecule from the following.

BF₃
NH₃
NF₃
CIF₃

105.

Which one of the following molecules has the least dipole moment?

H₂O
BeF₂
NH₃
NF₃

106.

Dipole-dipole interaction energy between polar molecules in solids depends on the radius of the molecule (r) and it is directly proportional to

(1/r³)
(1/r⁶)
(1/r)
(1/r⁵)

107.

Which of the following changes in the respective bond order values are caused by removal of an electron from N₂ and F₂ molecules?

Decrease by 0.5 in both
Increase by 0.5 in both
Increase by 0.5 in the former and decrease by 0.5 in the latter
Decrease by 0.5 in the former and increase by 0.5 in the latter

108.

The hybridised state of bromine in bromine pentafluoride is

sp³d
dsp³
d²sp³
sp³d²

109.

The correct ascending order of ligand field strength of the given ligands is

F^- < I^- < CN^- < H₂O < CO
I^- < F^- < H₂O < CO < CN^-
I^- < F^- < H₂O < CN^- < CO
F^- < H₂O < I^- < CN^- < CO

110.
Arrange the following species in the correct order of their stability
C₂, Li₂, O $_{2}^{+}$ , He $_{2}^{+}$

Li₂ < He $_{2}^{+}$ < O $_{2}^{+}$ < C₂

C₂ < O $_{2}^{+}$ < Li₂ < He $_{2}^{+}$

He $_{2}^{+}$ < Li₂ < C₂ < O $_{2}^{+}$

O $_{2}^{+}$ < C₂ < Li₂ < He $_{2}^{+}$

He $_{2}^{+}$ < Li₂ < C₂ < O $_{2}^{+}$

Higher the bond order, more is the stability. The bond order of the given species can be calculated as-

MO configuration of C₂ = (6 + 6 = 12)

= σ1s², σ*1s², σ2s², σ*2s², $π 2 p_{x}^{2} \approx π 2 p_{y}^{2}$

BO = $\frac{N_{b} - N_{a}}{2} = \frac{8 - 4}{2} = \frac{4}{2} = 2$

MO configuration of Li₂ = (3 + 3 = 6)

= σ1s², σ*1s², σ2s²

BO = $\frac{4 - 2}{2}$ = 1

MO configuration of O₂ = (8 + 8 -1 = 15)

= σ1s², σ*1s², σ2s², σ*2s², σ2p $_{z}^{2}$ , $π 2 p_{x}^{2} \approx π 2 p_{y}^{2}, π * 2 p_{x}^{1}$

BO = $\frac{2 - 1}{2} = \frac{1}{2} = 0.5$

Thus, the order of bond orders of the given species is

He $_{2}^{+}$ < Li₂ < C₂ < O $_{2}^{+}$

Since, bond order α stability.

Thus, order of stability will also be the same.

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

110. Arrange the following species in the correct order of their stabilityC2, Li2, O2+, He2+ Li2 < He2+ < O2+ < C2 C2 < O2+ < Li2 < He2+ He2+ < Li2 < C2 < O2+ O2+ < C2 < Li2 < He2+

110.
Arrange the following species in the correct order of their stability
C₂, Li₂, O $_{2}^{+}$ , He $_{2}^{+}$

Li₂ < He $_{2}^{+}$ < O $_{2}^{+}$ < C₂

C₂ < O $_{2}^{+}$ < Li₂ < He $_{2}^{+}$

He $_{2}^{+}$ < Li₂ < C₂ < O $_{2}^{+}$

O $_{2}^{+}$ < C₂ < Li₂ < He $_{2}^{+}$