Question

Identify the correct match.

(i)	$Xe{F}_2$	(a)	Central atom is $s{p}^3$ hybridised and has a bent shape.
(ii)	$N_3^{-}$	(b)	Central atom is $s{p}^3{d}^2$ hybridised and has an octahedral shape.
(iii)	$PC{l}_6^{-}$	(c)	Central atom is $sp$ hybridised and has a linear shape.
(iv)	$IC{l}_2^{+}$	(d)	Central atom is $s{p}^{3}d$ hybridised and has a linear shape.

• A. (i - a), (ii - b), (iii - c), (iv - d)
• B. (i - d), (ii - b), (iii - d), (iv - c)
• C. (i - b), (ii - c), (iii - a), (iv - d)
• D. (i - d), (ii - c), (iii - b), (iv - a)

Answer 1

The correct option is D (i - d), (ii - c), (iii - b), (iv - a)

Generally, we have two approaches to calculate the hybridisation for different structures:

Approach - 1: Using steric number approach which is,
$\text{Steric Number (x)}=\frac{1}{2}(s+v-q)$.
where, s = number of monovalent surrounding atoms
v = number of valence electrons on the central atom
q = charge (with sign)

Approach - 2: Counting the of total number of sigma bonds and lone pairs around a particular atom.

(i) In $Xe{F}_2,s=2;v=8;q=0$.
Hence, s =5 i.e., $s{p}^{3}d$ hybridised. The molecule is linear shaped with three lone pairs occupying the equatorial positions on $Xe$.


(ii) In $N_3^{-}$, we determine the hybridisation by counting the number of sigma bonds and lone pairs around the central atom. Hence, the central nitrogen has 2 $\sigma$ bonds and no lone pairs. So, the steric number of $N$ in $N_3^{-}$ is 2. Therefore, the hybridisation of $N$ in $N_3^{-}$ is $sp$ and it has a linear shape.
$N^{-}=N^{+}=N^{-}$

(iii) In $PC{l}_6^{-},s=6;v=5;q=-1$. Hence, $x=6i.e.S{p}^3{d}^2$ hybridisation. If we arrange this compound in a geometry corresponding to its hybridisaiton, we get octahedral shape with no lone pairs on $P$.


(iv) In $IC{l}_2^{+},s=2;v=7;1=+1$.
Hence, $x=4i.e.s{p}^3$ hybridisation. If we arrange this compound in a geometry corresponding to its hybridisation, we get bent shape with two lone pairs on $I$.


Hence, (i - d), (ii - c), (iii - b), (iv - a) is the right option.