JEE Main 2026 — Chemistry Inorganic Chemistry

The oxidation state of $Ni$Ni in all three complexes is $+2$+2. The electronic configuration of $Ni^{2+}$Ni2+ is $[Ar] 3d^{8}$[Ar]3d8.

In $[Ni(en)_3]^{2+}$[Ni(en)3​]2+ (A), the coordination number is $6$6 (octahedral geometry). The $d^{8}$d8 configuration in an octahedral field is $t_{2g}^{6} e_{g}^{2}$t2g6​eg2​, which has $2$2 unpaired electrons.

In $[NiCl_4]^{2-}$[NiCl4​]2− (B), the coordination number is $4$4 with weak field $Cl^{-}$Cl− ligands (tetrahedral geometry). The $d^{8}$d8 configuration in a tetrahedral field is $e^{4} t_{2}^{4}$e4t24​, which has $2$2 unpaired electrons.

In $[Ni(NH_3)_6]^{2+}$[Ni(NH3​)6​]2+ (C), the coordination number is $6$6 (octahedral geometry). The $d^{8}$d8 configuration in an octahedral field is $t_{2g}^{6} e_{g}^{2}$t2g6​eg2​, which has $2$2 unpaired electrons.

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

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

According to the spectrochemical series, the field strength of the ligands follows the order: $en > NH_3 > Cl^{-}$en>NH3​>Cl−.

Furthermore, the crystal field splitting energy for octahedral complexes is significantly greater than that for tetrahedral complexes ($\Delta_o = \dfrac{9}{4}\Delta_t$Δo​=49​Δt​).

Therefore, the order of crystal field splitting energy, and hence the frequency of absorption, is $[Ni(en)_3]^{2+} > [Ni(NH_3)_6]^{2+} > [NiCl_4]^{2-}$[Ni(en)3​]2+>[Ni(NH3​)6​]2+>[NiCl4​]2−, which corresponds to $(A) > (C) > (B)$(A)>(C)>(B).

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