Question

Among the compounds $B{F}_3,NC{l}_3,H_{2}S,S{F}_4\text{and}BeC{l}_2$ identify the ones in which the central atom has the same type of hybridisation.

• A. $NC{l}_3\text{and}{H}_{2}S$
• B. $H_{2}S\text{and}BeC{l}_2$
• C. $S{F}_4\text{and}BeC{l}_2$
• D. $B{F}_3\text{and}NC{l}_3$

Answer 1

The correct option is A $NC{l}_3\text{and}{H}_{2}S$

$\text{Hybridization bond = Bond pair + lone pair}$
$B{F}_3:3\text{Bond pairs}+0\text{lone pair}=s{p}^2$
$NC{l}_3:3\text{Bond pairs}+1\text{lone pair}=s{p}^3$
$H_{2}S:2\text{Bond pairs}+2\text{lone pairs}=s{p}^3$
$S{F}_4:4\text{Bond pairs}+1\text{lone pair}=s{p}^{3}d$
Hence, $NC{l}_3\text{and}{H}_{2}S$ have same hybridisation.