Question

Which of the following has $p\pi - d\pi$ bonding?
A. $NO_3^ -$
B. $SO_3^{ - 2}$
C. $BO_3^{ - 3}$
D. $CO_3^{ - 2}$

Answer 1

The atom contains electron configuration. The two atoms are bonding where one atom has one vacant orbital and another is having one lone pair of electrons. If the electron is donated to the respective vacant orbital, this bonding is called $p\pi - d\pi$ .

Complete step by step answer:
When metal complexes contain halide ligands, it has significance of $p\pi - d\pi$ bonding. In this complexes filled $p\pi$ orbital on the ligand donates electron density to unfilled $d\pi$ orbital.
As we know that, $SO_3^{ - 2}$ ion contains $s{p^3}$ hybridization. so, S has three p-orbitals that form bonds with three oxygen atoms and are unhybridized. Orbital involves bond formation.
Here an electron configuration of O and S is given as below:
$_{16}O = 1{s^2}2{s^2}2p_x^22p_y^12p_z^1$
$_{14}S = 1{s^2}2{s^2}2p_x^22p_y^22p_z^23{s^2}3p_x^23p_y^13p_z^1$
As we see oxygen has two unpaired electrons in p-orbital, in which one has $\partial$ -bond and other is used in $\pi$ bond formation.
Hence , in $SO_3^{ - 2}$ , $p\pi - p\pi$ orbitals are involved for the formation of $p\pi - d\pi$ .
In $SO_3^{ - 2}$ , the central atom is Sulphur and oxygen acts as a ligand. An element of Sulphur can form $p\pi - d\pi$ bonding. In this bonding d-orbital overlap with p-orbital of oxygen.
All the bonds depend on the transfer of electrons and bonding with each other by unpaired electrons.
$SO_3^{ - 2}$ has $p\pi - d\pi$ bond.

Therefore, the correct answer is option (B).

Note:
Nitrogen, Bromine and carbon, which are the second period of elements. These elements have no vacant d-orbital. So they are not able to form $p\pi - d\pi$ bonding. The atoms containing 3d orbitals are able to form this bonding. In this case, the d-orbitals overlap with p-orbitals of atoms.