Question

How many $d_{\pi }-p_{\pi }$ bonds are present in $Cl{O}_3^{-}$ ?

Answer 1

Hybridization of molecule is given by the formula,

$H=\frac{1}{2}[V+M-C+A]$

H = number of orbitals involved in hybridization

V = valence electrons of central atom

M = number of monovalent atoms linked to central atom

C = charge on the cation

A = charge on the anion

Value of H and hybridization, $[H=2(sp),3(s{p}^2),4(s{p}^3),5(s{p}^{3}d\left)\right]$

Hybridization of $Cl{O}_3^{-}$ is,

$H=\frac{1}{2}[7+1]=4$ $\left(s{p}^3\right)$ hybridized

Electronic configuration of chlorine - $\left[Ne\right]3s^{2}3p^5$

Chlorine in ground state: $3s^{2}3p^5$

Chlorine in excited state: $3s^{2}3p^{3}3d^2$

In excited state 2 electrons from $3p$ are promoted to $3d$. And chlorine forms 3 $\sigma$ bond with oxygen. The 2 electrons in $3d$ forms two $\pi$ bond with two oxygen as shown in diagram.

Hence, $Cl{O}_3^{-}$ has two $d_{\pi }-p_{\pi }$ bonds.