Incorrect order about bond angle is:

H_{2} O > H_{2} S > H_{2} S e > H_{2} T e

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C_{2} H_{2} > C_{2} H_{4} > C H_{4} > N H_{3}

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S F_{6} < N H_{3} < H_{2} O < O F_{2}

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C l O_{2} > H_{2} O > H_{2} S > S F_{6}

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Solution

The correct option is C $S F_{6} < N H_{3} < H_{2} O < O F_{2}$
(a) According to VSEPR, as electronegativity of central atom decreases, bond angle decreases. So bond angle will be in order:
$H_{2} O > H_{2} S > H_{2} S e > H_{2} T e$
(B) Due to bond pair-lone pair repulsion, bond angle of these molecules will be: $C_{2} H_{2} s p > C_{2} H_{4} s p^{2} > C H_{4} s p^{3} > N H_{3} s p^{3}$
(c) $S F_{6} < N H_{3} < H_{2} O < O F_{2}$
In this case bond angle of $N H_{3}$ is highest because $l p - l p$ repulsion is absent in it.
(d) $C l O_{2} > H_{2} O > H_{2} S > S F_{6}$
$C l O_{2}$ bond angle is highest due to its $s p^{2}$ hybridised, more repulsion in double bond electrons.

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