Question

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

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

Answer 1

The correct option is B

$S{O}_2$

Let's take each compound one by one.
→ By VSEPR, we can get the structure of each of the compounds

Now, hybrid oribtals

Number of hybrid orbitals = Number of bonded atoms + number of lone pairs.

$\to$ $C{O}_2$ $\Rightarrow$ 2+ 0 $\Rightarrow$ $sp$

$\to$ $S{O}_2$ $\Rightarrow$ 2+ 1 $\Rightarrow$ ${sp}^2$

$\to$ $N_{2}O$ $\Rightarrow$ 2+ 0 $\Rightarrow$ $sp$

$\Rightarrow$ $CO$ $\Rightarrow$ 1 + 1 $\Rightarrow$ $sp$

$\to$ By VSEPR, we can get the structure of each of the compounds

Therefore, $S{O}_2$ has ${sp}^2$ hybridization.

[Because sulphur has extended octet (d-orbitals) for hybridization]