Question:
In which one of the following, $sp^2$ hybridisation is involved in the central atom?
In which one of the following, $sp^2$ hybridisation is involved in the central atom?
Updated On: Apr 4, 2025
- $NH_3$
$PCl_3$
- $CIF_3$
$ BCl_3$
- $PH_3$
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The Correct Option is D
Solution and Explanation
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.
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