Question

Match the following:
$\begin{array}{cccc}\text{(i)}& C{H}_{2}O& \text{(a)}& \text{Central atom is}sp\text{hybridised}\\ \text{(ii)}& BeC{l}_2& \text{(b)}& \text{Central atom is}s{p}^2\text{hybridised}\\ \text{(iii)}& N{H}_4^{+}& \text{(c)}& \text{Central atom is}s{p}^3\text{hybridised}\\ \text{(iv)}& B{F}_4^{-}& \text{(d)}& \text{Central atom is}s{p}^2\text{hybridised and has a linear shape.}\end{array}$

Answer 1

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

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

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

Using approach -1:
$\left(i\right)$ In $C{H}_{2}O$, m = 2 ; v = 4 (for $C$) ; q = 0.
Hence, S = 3 i.e., $s{p}^2$ hybridised. The molecule has trigonal planar geometry.

$\left(ii\right)$ In $BeC{l}_2$, m = 2 ; v = 2 (for $Be$) ; q = 0. So, the steric number of $Be$ is 2. Therefore, the hybridisation of $Be$ in $BeC{l}_2$ is $sp$ and it has a linear shape.

$\left(iii\right)$ In $N{H}_4^{+}$, m = 4 ; v = 5 ; q = +1. Hence, S = 4 i.e. $s{p}^3$ hybridisation.

$\left(iv\right)$ In $B{F}_4^{-}$, m = 4 ; v = 3 ; q = -1.
Hence, S = 4 i.e. $s{p}^3$ hybridisation.

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