Hybridisation in $S O_{2}$ molecule is:

s p

No worries! We‘ve got your back. Try BYJU‘S free classes today!

s p^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

s p^{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

s p^{3} d

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $s p^{2}$
In Sulphur Dioxide, the $S$ atom has $2$ bond pairs and $1$ lone pair, so it requires $3$ hybrid orbitals. So the hybridisation is $s p^{2}$ .

Suggest Corrections

Similar questions

C (1) H_{2} = C (2) H - C (3) \equiv C H

s p^{3}, s p^{2}, s p^{2}, s p^{3}, s p^{2}, s p^{2}, s p, a n d s p

A l H_{3}

A l H_{4}^{-}

s p^{3}, s p^{2}, s p^{2}, s p^{3}, s p^{2}, s p^{2}, s p, a n d s p

In the molecule CH ≡ C -CH = $C H_{2}$ , the hybridisation of C -Cbond is

Join BYJU'S Learning Program