JEE Main 2023 — Coordination Compounds Question with Solution

Option A : $${\mathrm{Na}}_3\left[{\mathrm{CoCl}}_6\right]$$

This is an octahedral complex, but it is not diamagnetic. Co in this complex is in +3 oxidation state, and the d-electron configuration is $t_{2g}^6{e}_g^0$, leading to two unpaired electrons.

Option B : $$\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_2$$

This is also an octahedral complex, but it is not diamagnetic. The Co ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^5{e}_g^2$, leading to 3 unpaired electrons.

Option C : $${\mathrm{K}}_3\left[\mathrm{Co}(\mathrm{CN}{)}_6\right]$$

This complex is both octahedral and diamagnetic. Co in this complex is in +3 oxidation state and the d-electron configuration is $t_{2g}^6{e}_g^0$, which means no unpaired electrons. This complex is more stable due to a strong field ligand (CN^-) which leads to a greater splitting of d-orbitals and more pairing of electrons.

Option D : $$\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]{\mathrm{Cl}}_2$$

This is an octahedral complex, and it is diamagnetic. The Ni ion is in +2 oxidation state with a d-electron configuration of $t_{2g}^8{e}_g^0$, leading to no unpaired electrons. However, NH₃ is a weaker field ligand compared to CN^-, so the complex is less stable than Option C.

Hence, the complex that is octahedral, diamagnetic, and the most stable among the given options is $${\mathrm{K}}_3\left[\mathrm{Co}(\mathrm{CN}{)}_6\right]$$.