What is/are the correct option(s) about $C l O_{2}$ ?

It has one odd electron and dimerizes to form

C l_{2} O_{4}

which exists as

C l^{+} C l O_{4}^{-}

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It has bond angle more than

O C l_{2}

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It can be prepared by reaction of

K C l O_{3}

with oxalic acid

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It is

10

times more soluble in water than chlorine because it hydrolyzes more readily than

C l_{2}

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Solution

The correct options are
B It has bond angle more than $O C l_{2}$
C It can be prepared by reaction of $K C l O_{3}$ with oxalic acid
a) It has one odd electron, but it does not dimerize due to delocalization of odd electron in d orbitals. Also, $C l_{2} O_{4}$ is not a dimer which exists as $C l^{+} C l O_{4}^{-} .$ But $C l_{2} O_{4}$ can be formed from $C l O_{2}$ by the photolysis in UV light.
b) In $O C l_{2}$ , $O$ is the central atom, and it has higher electron density.Thus lonepair-lonepair repulsion in $O C l_{2}$ is more than $C l O_{2}$ . Hence $C l O_{2}$ has a higher bond angle than $O C l_{2}$ .
c) $K C l O_{3} + H_{2} C_{2} O_{4} ⟶ 2 C l O_{2} + K_{2} C O_{3} + C O_{2} + H_{2} O$
d) It is 10 times more soluble in water than chlorine but it does not hydrolyze and remains as gas.

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