Question

Which one of the following sequences represents the correct increasing order of bond angles in the given molecules ?

• A. $H_{2}O < OF_{2} < OCl_{2} < ClO_{2}$
• B. $OCl_{2} < ClO_{2} < H_{2}O < OF_{2}$
• C. $OF_{2} < H_{2}O < OCl_{2} < ClO_{2} $
• D. $ ClO_{2} < OF_{2} < OCl_{2} < H_{2}O$

Answer 1

Answer: (C)

$H_2O$ is $sp^3$ hybridised with bond angle $104.5^°$ due to presence of two lone pairs. $OF_2$ has structure similar to $H_2O$ with bond angle $103^°$ due to higher electronegativity of fluorine. $OCl_2$ also has similar structure with bond angle $111^°$ because of steric crowding of two chlorine atoms. However $ClO_2$ has $π$- bond character with an odd electron so that bond angle is $118^°$. Thus four compounds can be arranged in order of their bond angle as