JEE Main 2026 — Coordination Compounds Question with Solution

The oxidation state of $Ni$ in all three complexes is $+2$. The electronic configuration of $N{i}^{2+}$ is $[Ar]3d^8$.

In $[Ni(en{)}_3{]}^{2+}$ (A), the coordination number is $6$ (octahedral geometry). The $d^8$ configuration in an octahedral field is $t_{2g}^6{e}_g^2$, which has $2$ unpaired electrons.

In $[NiC{l}_4{]}^{2-}$ (B), the coordination number is $4$ with weak field $C{l}^{-}$ ligands (tetrahedral geometry). The $d^8$ configuration in a tetrahedral field is $e^4{t}_2^4$, which has $2$ unpaired electrons.

In $[Ni(N{H}_3{)}_6{]}^{2+}$ (C), the coordination number is $6$ (octahedral geometry). The $d^8$ configuration in an octahedral field is $t_{2g}^6{e}_g^2$, which has $2$ unpaired electrons.

Thus, the number of unpaired electrons in (A), (B), and (C) are $2,2,$ and $2$ respectively.

The frequency of absorption of light is directly proportional to the crystal field splitting energy ($\Delta$).

According to the spectrochemical series, the field strength of the ligands follows the order: $en>N{H}_3>C{l}^{-}$.

Furthermore, the crystal field splitting energy for octahedral complexes is significantly greater than that for tetrahedral complexes (${\Delta }_o=\frac{9}{4}{\Delta }_t$).

Therefore, the order of crystal field splitting energy, and hence the frequency of absorption, is $[Ni(en{)}_3{]}^{2+}>[Ni(N{H}_3{)}_6{]}^{2+}>[NiC{l}_4{]}^{2-}$, which corresponds to $(A)>(C)>(B)$.

Answer: $2,2,2$ and $(A)>(C)>(B)$