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}^{-}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

It has bond angle more than

O C l_{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

It can be prepared by reaction of

K C l O_{3}

with oxalic acid

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

It is

10

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

C l_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is 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.

Suggest Corrections