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Oxides of Sulphur

The sequence ...

Question

The sequence that correctly represents the increasing order of bond angles in $O F_{2}, O C l_{2}, C l O_{2}$ and $H_{2} O$ is :

O F_{2} < H_{2} O < O C l_{2} < C l O_{2}

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H_{2} O < F_{2} O < C l_{2} O < C l O_{2}

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O F_{2} < C l_{2} O < H_{2} O < C l O_{2}

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C l O_{2} < C l_{2} O < O F_{2} < H_{2} O

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Solution

The correct option is A $O F_{2} < H_{2} O < O C l_{2} < C l O_{2}$
Among $O F_{2}$ , $H_{2} O$ and $O C l_{2}$ bond angle will depend upon electronegativity of surrounding atom as central atom is same and their hybridisation is also same . $O C l_{2}$ has unexpectedly greater angle due to bulky surrounding group Cl. Angle of $O F_{2}$ is smaller than that of $H_{2} O$ as bond angle is inversely to electronegativity.
All these are $s p^{3}$ hybridized. And have 2 lone pairs but $C l O_{2}$ has only 3 free electrons. Therefore, has highest bond angle.
Therefore, the bond angle order is: $C l O_{2}$ > $O C l_{2}$ > $H_{2} O$ > $O F_{2}$ .

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