NEET Practice Questions (MCQs) with Answers & Solutions

(d) According to Fajan's rule,

Covalent character $$\begin{gathered} \propto \frac{1}{\mathrm{size} \mathrm{of} \mathrm{cation}} \\ \propto \mathrm{size} \mathrm{of} \mathrm{anion} \end{gathered}$$

In the given options, cation is same but anions are different. Among halogens the order of size is

                      F < Cl < Br < I

$\therefore$Order of covalent character is

                   MI > MBr > MCl > MF

(a) Isostructural Compounds having same structure and same hybridisation are known as isostructural species. e.g. XeF₂ and IF₂^-   are sp³d hybridised and both have linear shape.

F-I-F    F-Xe-F

(b) Molecules with zero bond order, do not exist.

According to molecular orbital theory,

$$\begin{gathered} \left(\mathrm{a}\right) {\mathrm{Be}}_2^{+} (4+4-1=7) \\ = \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^1 \\ \mathrm{Bond} \mathrm{order} \left(\mathrm{BO}\right) = \frac{4-3}{2} = 0.5 \\ \left(\mathrm{b}\right) {\mathrm{Be}}_2(4+4=8)= \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2 \\ \mathrm{BO} = \frac{4-4}{2}=0 \\ \left(\mathrm{c}\right) {\mathrm{B}}_2(5+5=10) = \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^1 \approx \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1 \\ \left(\mathrm{d}\right) {\mathrm{Li}}_2 (3+3=6) = \mathrm{\sigma }1{\mathrm{s}}^2, \stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2 \\ \mathrm{BO} = \frac{4-2}{2}=1 \\ \mathrm{Thus}, {\mathrm{Be}}_2 \mathrm{does} \mathrm{not} \mathrm{exist} \mathrm{under} \mathrm{normal} \mathrm{conditions}. \end{gathered}$$

(c) Paramagnetic character is shown by those atoms or molecules which have unpaired electrons.

In the given compounds CO is not paramagnetic since, it does not have unpaired electrons. The configuration of CO molecule is 

CO(14) = $\sigma 1s^2, \stackrel{*}{\sigma }1s^2, \sigma 2s^2, \stackrel{*}{\sigma }2s^2, \sigma 2p_x^2, \pi 2p_y^2 \approx \pi 2p_z^2$

(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)

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