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)

$SO_2$

Answer 2

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