Question

In which one of the following, $sp^2$ hybridisation is involved in the central atom?

• A. $NH_3$
• B. $PCl_3$
• C. $CIF_3$
• D. $ BCl_3$
• E. $PH_3$

Answer 1

The Correct Option is D

Hybridisation of the central atom depends on the number of electron domains (bond pairs and lone pairs) around it.

Let’s analyze each compound:

• \( NH_3 \): Nitrogen forms 3 bonds and has 1 lone pair → 4 electron domains → \( sp^3 \) hybridisation.
• \( BCl_3 \): Boron forms 3 bonds and has no lone pairs → 3 electron domains → \( sp^2 \) hybridisation.
• \( ClF_3 \): Chlorine forms 3 bonds and has 2 lone pairs → 5 electron domains → \( sp^3d \) hybridisation.
• \( PCl_3 \): Phosphorus forms 3 bonds and has 1 lone pair → 4 electron domains → \( sp^3 \) hybridisation.
• \( PH_3 \): Phosphorus forms 3 bonds and has 1 lone pair → 4 electron domains → \( sp^3 \) hybridisation.

Correct Answer:

\( BCl_3 \) involves \( sp^2 \) hybridisation in the central atom.