Chemistry MCQs for NEET — Practice Questions with Answers

(a) SCl₄ (Seesaw structure)  is not isostructural with  Sicl₄ because shows tetrahedral structure due to involvement of repulsion between lone pair and bond pair of electrons.

SO₄^2- shows tetrahedral structure due to sp³ hybridisation.

PO₄^3- shows tetrahedral structure due to sp³ hybridisation.

NH₄⁺ shows tetrahedral structure due to sp³ hybridisation

(c) Molecules having the same number of hybrid orbitals, have same hybridisation and number of hybrid oebitals.

                     $\mathrm{H} = \frac{1}{2}[\mathrm{V} + \mathrm{X} - \mathrm{C} + \mathrm{A}]$

where,

V= number of valance electrons of central atom

X = number of monovalent atoms

C= charge on cation

A = charge on anion

$$\begin{gathered} {\mathrm{SbCl}}_5^{2-} = {\mathrm{sp}}^3{\mathrm{d}}^2, {\mathrm{PCl}}_5 = {\mathrm{sp}}^3\mathrm{d} \\ {\mathrm{SF}}_4= {\mathrm{sp}}^3\mathrm{d}, {\mathrm{I}}_3^{-}={\mathrm{sp}}^3\mathrm{d} \end{gathered}$$

The SO₃^2- ion has a tetrahedral geometry with one S=O double bond and three S-O single bonds. The double bond is formed by pπ-dπ bonding between the 3p orbitals of sulfur and 2p orbitals of oxygen, exhibiting π bond character.

(d)

$$\begin{gathered} \left(\mathrm{a}\right) {\mathrm{SF}}_4 = \mathrm{irregular} \mathrm{tetrahedral} ({\mathrm{sp}}^3\mathrm{d}, \mathrm{one} \mathrm{lone} \mathrm{pair}) \\ {\mathrm{XeF}}_4 = \mathrm{square} \mathrm{planar} ({\mathrm{sp}}^3{\mathrm{d}}^2, \mathrm{two} \mathrm{lone} \mathrm{pairs}) \\ \left(\mathrm{b}\right) {\mathrm{SO}}_3^{2-} = \mathrm{pyramidal} ({\mathrm{sp}}^3, \mathrm{one} \mathrm{lone} \mathrm{pair}) \\ {\mathrm{NO}}_3^{-} = \mathrm{trigonal} \mathrm{planar} \left({\mathrm{sp}}^2\right) \\ \left(\mathrm{c}\right) {\mathrm{BF}}_3 = \mathrm{trigonal} \mathrm{planar} \left({\mathrm{sp}}^2\right) \\ {\mathrm{NF}}_3 = \mathrm{pyramidal} \left({\mathrm{sp}}^3\right) \\ \left(\mathrm{d}\right) {\mathrm{BrO}}_3^{-} = \mathrm{pyramidal} ({\mathrm{sp}}^3, \mathrm{one} \mathrm{lone} \mathrm{pair}) \\ {\mathrm{XeO}}_3 = \mathrm{pyramidal} ({\mathrm{sp}}^3, \mathrm{one} \mathrm{lone} \mathrm{pair}) \end{gathered}$$

(b) BeF₂ - sp hybridisation

       BF₃ - sp₂ hybridisation

     C₂H₂ - sp hybridisation

      NH₃ - sp₃ hybridisation

The correct order of C-O bond length is CO < CO₂ < CO₃^2-. The bond order decreases from CO to CO₂ to CO₃^2-, resulting in an increase in bond length due to weaker bond strength.

(b) According to the molecular orbital theory (MOT),

$$\begin{gathered} {\mathrm{N}}_2(7+7=14) = \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 \\ \approx 2{\mathrm{p}}_{\mathrm{y}}^2, \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \\ \mathrm{Bond} \mathrm{Order} = \frac{10-4}{2} = 3 \\ {\mathrm{N}}_2^{-}(7+7+1=15) =\mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2\approx 2{\mathrm{p}}_{\mathrm{y}}^2, \\ \mathrm{BO}= \frac{10-5}{2}= 2.5 \\ {\mathrm{N}}_2^{2-}(7+7+2=16)=\mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2\approx 2{\mathrm{p}}_{\mathrm{y}}^2,\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1\approx \stackrel{*}{\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1} \\ \mathrm{BO} = \frac{10-6}{2}=2 \end{gathered}$$

Hence, the increasing order of bond order is,

                     ${\mathrm{N}}_2^{2-} < {\mathrm{N}}_2^{-} < {\mathrm{N}}_2$