pℼ – dℼ bonding is present in which molecule

The correct option is SO₃²⁻.

Let us understand why pπ – dπ bonding is present in the sulfite ion (SO₃²⁻) by examining the valence shell electronic configurations and hybridization of the central atoms in each of the given options.

1. Understanding pπ – dπ Bonding:
A pπ – dπ bond is a type of covalent pi (π) bond formed by the lateral overlap of a filled or half-filled p-orbital of one atom with a half-filled d-orbital of another atom. For this type of bonding to occur, at least one of the atoms involved in the pi bond must have access to d-orbitals in its valence shell (which requires it to belong to Period 3 or higher in the periodic table).

2. Analyzing the central atoms in the given options:
In all the given options, oxygen (O) is the surrounding atom. Oxygen belongs to Period 2 and has the electronic configuration:
$1s^{2}2s^{2}2p^4$
Thus, oxygen only has 2s and 2p valence orbitals available for bonding and lacks d-orbitals. Therefore, any pi bond involving oxygen must utilize one of its 2p orbitals.

Now let's look at the central atoms of each molecule/ion:

- In CO₃²⁻ (Carbonate ion):
The central atom is carbon (C), which belongs to Period 2. Its valence configuration is:
$2s^{2}2p^2$
Since carbon does not have vacant d-orbitals in its valence shell, any pi bond formed between C and O must be a pπ – pπ bond.

- In NO₃⁻ (Nitrate ion):
The central atom is nitrogen (N), which belongs to Period 2. Its valence configuration is:
$2s^{2}2p^3$
Nitrogen does not have d-orbitals. Therefore, the pi bond between N and O can only be a pπ – pπ bond.

- In BO₃³⁻ (Borate ion):
The central atom is boron (B), which belongs to Period 2. Its valence configuration is:
$2s^{2}2p^1$
Like carbon and nitrogen, boron has no d-orbitals. Any pi bonding present is strictly of the pπ – pπ type.

- In SO₃²⁻ (Sulfite ion):
The central atom is sulfur (S), which belongs to Period 3. Its ground-state valence configuration is:
$3s^{2}3p^4$
Since sulfur is in Period 3, it has empty 3d orbitals available in its valence shell. In the sulfite ion (SO₃²⁻), sulfur undergoes sp³ hybridization to form three sigma (σ) bonds with three oxygen atoms, leaving one lone pair on sulfur. To form the double bond (pi bond) with one of the oxygen atoms, sulfur extends its valency by exciting an electron to a 3d orbital. The lateral overlap then occurs between a filled 2p orbital of oxygen and a half-filled 3d orbital of sulfur, resulting in a pπ – dπ bond.