JEE Main 2023 — Coordination Compounds Question with Solution

In the complex [Co(NH${}_3$)${}_6$]${}^{3+}$, Co is in the +3 oxidation state. This means it has a configuration of $d^6$. NH${}_3$ is a strong field ligand, which causes the d-electrons to pair up. Therefore, this complex is a low-spin (also referred to as spin-paired) complex.

The term "low-spin" means that the electrons prefer to pair up in the lower energy d-orbitals rather than occupy the higher energy orbitals. When all the electrons are paired, the complex is diamagnetic, which means it's not attracted to an external magnetic field. The hybridization of this complex is $d^{2}s{p}^3$, also known as inner orbital complex, because the inner d-orbitals are used in the hybridization.

The complexes [CoF${}_6$]${}^{3-}$ and [CoCl${}_6$]${}^{3-}$ are both octahedral, but they do have unpaired electrons because $F^{-}$ and $C{l}^{-}$ are not strong enough field ligands to cause pairing of all the electrons in the d-orbitals. Therefore, they are not low-spin or diamagnetic.

Lastly, the complex [NiCl${}_4$]${}^{2-}$ is not octahedral. Because $C{l}^{-}$ is a weak field ligand, the electrons in the d-orbitals do not pair up, and the $N{i}^{2+}$ ion, with a $d^8$ configuration, undergoes sp^3 hybridization, resulting in a tetrahedral complex.

Therefore, the only octahedral, diamagnetic, and low-spin complex in the options provided is [Co(NH${}_3$)${}_6$]${}^{3+}$ (Option A).