JEE Chemistry 2023 Question with Solution

Given: The compound is potassium ferrocyanide, $K_4[Fe(CN)_6]$.

Find: The number of electron pairs in the $t_{2g}$ set of orbitals.

Strong field ligands like $CN^-$ cause maximum pairing of electrons in the lower energy $t_{2g}$ orbitals, resulting in a low-spin octahedral configuration.

From the extracted solution:

• In $K_4[Fe(CN)_6]$, iron is in the $+2$ oxidation state.
• $CN^-$ is a strong field ligand.
• It causes pairing of electrons in the $d$-orbitals.

Thus, the $d$-orbital configuration is

$$t_{2g}^6 e_g^0$$

The $t_{2g}$ set contains $6$ electrons, so the number of electron pairs in $t_{2g}$ is $3$.

Therefore, the required number of electron pairs is $3$.