Question

Which one of the following compounds has sp${}^2$-hybridisation?

• A. $C{O}_2$
• B. $S{O}_2$
• C. $N_{2}O$
• D. $CO$

Answer 1

Answer: (B)

When the number ofhybrid orbitals is 3, hybridisation is sp${}^3$
Number of hybrid orbitals,
$H=\frac{1}{2}[V+M-C+A]$
(a) CO${}_2\Rightarrow H=\frac{1}{2}[4+0-0+0]=2$
$\therefore$ sp hybridisation.
$(b)S{O}_2\Rightarrow H=\frac{1}{2}[6+0-0+0]=3$
$\therefore s{p}^2$ hybridisation.
$(c)N_{2}O\Rightarrow H+\frac{1}{2}[5+0-0+0]=2.5$
$\therefore$ sp hybridisation.
$(d)CO\Rightarrow H=\frac{1}{2}[4+0-0+0]=2$
$\therefore$ sp hybridisation.
Thus, SO${}_{2}hass{p}^2$ hybridisation