Question

Identify the correct match.
$\begin{array}{cccc}\text{(i)}& Xe{F}_2& \text{(a)}& \text{Central atom is}s{p}^3\text{hybridised and has a bent shape.}\\ \text{(ii)}& N_3^{-}& \text{(b)}& \text{Central atom is}s{p}^2\text{hybridised and has an octahedral shape.}\\ \text{(iii)}& Cl{O}_2^{-}& \text{(c)}& \text{Central atom is}sp\text{hybridised and has a linear shape.}\\ \text{(iv)}& IC{l}_2^{+}& \text{(d)}& \text{Central atom is}s{p}^{3}d\text{hybridised and has a linear shape.}\end{array}$

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

Answer 1

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

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

$Approach-1$ : Using steric number approach which is ,
Steric Number $\left(x\right)$ = $\frac{1}{2}\left(s+v-q\right)$.
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.


$\left(i\right)$ 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 $Xe$.



$\left(ii\right)$ 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.

$\left(iii\right)$ In $Cl{O}_2^{-}$, s = 0 ; v = 7 ; q = -1. Hence, x = 4 i.e. $s{p}^3$ hybridisation. If we arrange this compound in a geometry corresponding to its hybridisation, we get tetrahedral geometry and bent shape with 2 lone pairs on $Cl$.


$\left(iv\right)$ In $IC{l}_2^{+}$, s = 2 ; v = 7 ; q = +1.
Hence, x = 4 i.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.