MCQ
Which one of the following compounds has bond angle as nearly ${90^o}$
- A$N{H_3}$
- ✓${H_2}S$
- C${H_2}O$
- D$C{H_4}$
✓
Answer
Correct option: B.
${H_2}S$
b
$H _2 S$ is $sp ^3$ hybridized with $2$ bond pair and $2$ lone pair. But due to lone pair lone pair repulsions, the bond angle decreases from $109.5^{\circ}$ to $92^{\circ}$ approx. Ammonia and water are $sp ^3$ hybridized. $NH _3$ has angle around $107$ and $H _2 O$ has angle around $104.5$.
$H _2 S$ is $sp ^3$ hybridized with $2$ bond pair and $2$ lone pair. But due to lone pair lone pair repulsions, the bond angle decreases from $109.5^{\circ}$ to $92^{\circ}$ approx. Ammonia and water are $sp ^3$ hybridized. $NH _3$ has angle around $107$ and $H _2 O$ has angle around $104.5$.
Oxygen being highly electronegative than sulphur, keeps the bond pair electrons of its bond with hydrogen near itself, due to which the bond angle is higher than $H _2 S$. In $SF _6$, we have 6 bond pairs in $sp ^3 d ^2$ hybridization and the bond angle is $90^{\circ}$.
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