Question

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

• A. $CO_2$
• B. $SO_2$
• C. $N_2O$
• 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.
$(b) SO_2 \Rightarrow \, H=\frac{1}{2}[6+0-0+0]=3$
$\, \, \therefore \, \, sp^2$ hybridisation.
$(c)N_2O \Rightarrow \, H+\frac{1}{2}[5+0-0+0]=2.5$
$\, \, \therefore \, \, $ sp hybridisation.
$(d) CO \Rightarrow \, \, H=\frac{1}{2}[4+0-0+0]=2$
$\, \, \therefore \, \, $ sp hybridisation.
Thus, SO$_2 \, has \, sp^2$ hybridisation