Chemical Bonding & Molecular Structure Questions — JEE Chemistry

Q1

The structure of PCl5 in the solid state is :

A square pyramidal

B tetrahedral [PCl4]+ and octahedral [PCl6]–

C square planar [PCl4]+ and octahedral [PCl6]–

D trigonal bipyramidal

Correct Answer

Option B

Solution

In solid state PCl5 exist in Ionpair i.e. [PCl4] + and [PCl6] – [PCl4]+ is tetrahedral. [PCl6]– is octahedral.

Q2

Which of the following are isostructural pairs ? A.

SO_4^{2 - }

and

CrO_4^{2 - }

B. SiCl4, and TiCl4 C. NH3 and NO3- D. BCl3 and BrCl3

A B and C only

B C and D only

C A and B only

D A and C only

Correct Answer

Option C

Solution

Isostructural compounds are those compounds which have same structure as well as same hybridisation.

Formula for find hybridisation : H = (Lone pair + Sigma bond + coordinate bond ) If number of sigma bond ( $\sigma$ ), co-ordinate bond and lone pair are same for given pairs, they are isostructural.

Q3

Identify the species having one

\pi

-bond and maximum number of canonical forms from the following :

A SO3

B O2

C SO2

D CO

_3^{2 - }

Correct Answer

Option D

Solution

Among SO3, O2, SO2 and CO

_3^{2 - }

, only O2 and CO

_3^{2 - }

has only one $\pi$ -bond.

Q4

The bond order and magnetic behaviour of

O_2^ -

ion are respectively :

A 1.5 and paramagnetic

B 1.5 and diamagnetic

C 2 and diamagnetic

D 1 and paramagnetic

Correct Answer

Option A

Solution

O_2^ - = {({\sigma _{1s}})^2}{(\sigma _{1s}^*)^2}{({\sigma _{2s}})^2}{(\sigma _{2s}^*)^2}{({\sigma _{2{p_z}}})^2}

\left( {\pi _{2{p_x}}^2 = \pi _{2{p_y}}^2} \right)\left( {\pi _{2{p_x}}^{*2} = \pi _{2{p_y}}^{*1}} \right)

Bond order =

{{10 - 7} \over 2} = 1.5

and paramagnetic.

Q5

What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species :

\mathrm{{N_2};N_2^ + ;{O_2};O_2^ + }

?

A 0, 1, 0, 1

B 2, 1, 0, 1

C 0, 1, 2, 1

D 2, 1, 2, 1

Correct Answer

Option C

Solution

.tg .tg Molecule $\begin{gathered}\text{ No. of unpaired electron in } \\\text{ highest occupied } \\\text{ molecular orbital }\end{gathered}$ $\mathrm{N}_{2}$ 0 $\mathrm{N}_{2}^{\oplus}$ 1 $\mathrm{O}_{2}$ 2 $\mathrm{O}_{2}^{\oplus}$ 1

Q6

Which of the following conversions involves change in both shape and hybridisation ?

A NH3

\to

NH4+

B CH4

\to

C2H6

C H2O

\to

H3O+

D BF3

\to

BF4

-

Correct Answer

Option D

Solution

BF3 $\to$ BF4 $-$

Q7

Which one of the following molecules has maximum dipole moment?

A

\mathrm{CH}_4

B

\mathrm{NF}_3

C

\mathrm{NH}_3

D

\mathrm{PF}_5

Correct Answer

Option C

Solution

\mathrm{NH}_3 \text{ have more dipole moment than } \mathrm{NF}_3 \text{. }

Q8

Among the following the maximum covalent character is shown by the compound :

A SnCl2

B AlCl3

C MgCl2

D FeCl2

Correct Answer

Option B

Solution

Charge of cation/Size of cation is called polarising power. $\therefore$ (i) Polarising power $\propto$ charge of cation (ii) Polarising power $\propto$ 1 size of cation Here the AlCl3 is satisfying the above two conditions i.e., Al is in +3 oxidation state and also has small size.

So it has more covalent character.

Q9

The covalent alkaline earth metal halide (X = Cl, Br, I) is :

A CaX2

B SrX2

C MgX2

D BeX2

Correct Answer

Option D

Solution

According to Fajan’s rule, when the size of cation is less then the covalent character would be more for the cation.

Since Be2+ has minimum size among given cation, therefore, BeX2 would be most covalent among given alkaline with metal halides

Q10

The type of hybridisation and number of lone pair (s) of electrons of Xe in XeOF4, respectively, are:

A sp3d and 2

B sp3d2 and 2

C sp3d and 1

D sp3d2 and 1

Correct Answer

Option D

Solution

H =

{1 \over 2}

(V + M - c + a) $\therefore$ H =

{1 \over 2}

(8 + 4) = 6 From structure, it is clear that it has five bond pairs and one lone pair.