Question

Which one of the following compounds has $s{p}^2$ hybridisation?

• A. $C{O}_2$
• B. $S{O}_2$
• C. $Xe{F}_4$
• D. $CO$

Answer 1

The correct option is B $S{O}_2$

Steric Number $\left(x\right)$ = $\frac{1}{2}\left(V+M-C+A\right)$.
where, M = number of monovalent surrounding atoms
V = number of valence electrons on the central atom
C = charge (with sign)
A=charge on anion
$H=\frac{1}{2}(V+M-C+A)$

(i) $C{O}_2:H=\frac{1}{2}(4+0-0+0)=2$
$\therefore sp$ hybridisation.

(ii) $S{O}_2:H=\frac{1}{2}(6+0-0+0)=3$
$\therefore s{p}^2$ hybridisation.

(iii) $Xe{F}_4:H=\frac{1}{2}(8+4-0+0)=6$
$\therefore s{p}^3{d}^2$ hybridisation.


(iv) $CO:H=\frac{1}{2}(4+0-0+0)=2$
$\therefore sp$ hybridisation.