Consider the metal complexes [Ni(en)_3]^ 2+ (A), [NiCl_4]^ 2- (B)...

The oxidation state of $Ni$ in all three complexes is $+2$. The electronic configuration of $Ni^{2+}$ 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 $[NiCl_4]^{2-}$ (B), the coordination number is $4$ with weak field $Cl^{-}$ ligands (tetrahedral geometry). The $d^{8}$ configuration in a tetrahedral field is $e^{4} t_{2}^{4}$, which has $2$ unpaired electrons.

In $[Ni(NH_3)_6]^{2+}$ (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: $en > NH_3 > Cl^{-}$.

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

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

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