Question

In $Cl{O_2}$, unpaired electron reside in:
A.$s{p^2}$ hybridised orbital
B. $s{p^3}$ hybridised orbital
C. p orbital
D. d orbital

Answer 1

To solve this kind of problem, we need to find out the hybridisation of the molecule. Then the hybridised orbital corresponding to the hybridisation will be the orbital in which the unpaired electron resides.

Complete step by step answer:
We know that the electronic configuration of chlorine is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^5}$. The outermost shell contains seven electrons which participates in hybridisation. Out of the seven electrons in the valence shell, two pairs of electrons are bonded with two oxygen atoms to form double bonds and one pair of electrons exist as lone pairs. The remaining one unpaired electron will exist as separately. So, the steric number of this molecule is four since two bonded pairs, one lone pair and one odd electron. Therefore, the hybridisation of the molecule will be $s{p^3}$. So, the unpaired electron will reside in one of the $s{p^3}$ hybridised orbits. Also, the geometry of the molecule will be tetrahedral.

Additional Information:
The chlorine dioxide has many uses like bleaching, disinfection of drinking water. This molecule will be used to treat the drinking water in order to destroy the taste and odour due to phenolic compounds.

Note: $Cl{O_2}$ is very rarely exist and highly unstable. Therefore, it is a very reactive molecule. Normally, the molecule exists as $Cl{O_2}^ -$ with two unpaired electrons. The bond angle of $Cl{O_2}$ will be greater than the bond angle of $Cl{O_2}^ -$ since the repulsion will be greater in the case of $Cl{O_2}^ -$.