MCQ
The bond angle in ammonia molecule is
- A${91^o}8'$
- B${93^o}3'$
- ✓${106^o}45'$
- D${109^o}28'$
✓
Answer
Correct option: C.
${106^o}45'$
(c) In $N{H_3},s{p^3}$-hybridization is present but bond angle is ${106^o}45'$ because Nitrogen has lone pair of electron according to $VSEPR$ theory due to $bp-lp$ repulsion bond angle decreases from ${109^o}45'$ to ${106^o}45'$.
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