MCQ
The decreasing order of bond angle in the following molecules is
- A$CH _4> H _2 O > NH _3$
- B$NH _3> CH _4> H _2 O$
- ✓$CH _4> NH _3> H _2 O$
- D$H _2 O > NH _3> CH _4$
✓
Answer
Correct option: C.
$CH _4> NH _3> H _2 O$
(c):
$CH _4, NH _3$ and $H _2 O$ involve $s p^3$ hybridisation. The number of lone pair of electrons present on $C , N$ and $O$
in $CH _4, NH _3$ and $H _2 O$ are 0,1 and 2 respectively. Greater the number of lone pairs, greater is the repulsion with the bond pairs and hence, smaller is the bond angle.
$CH _4, NH _3$ and $H _2 O$ involve $s p^3$ hybridisation. The number of lone pair of electrons present on $C , N$ and $O$
in $CH _4, NH _3$ and $H _2 O$ are 0,1 and 2 respectively. Greater the number of lone pairs, greater is the repulsion with the bond pairs and hence, smaller is the bond angle.
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