Unpaired Electrons and Absorption Frequency of Nickel Complexes

To determine the correct option, we need to evaluate the number of unpaired electrons and the crystal field splitting energy ($\Delta$) for each metal complex.

1. Determination of Unpaired Electrons

The central metal ion in all three complexes is Nickel in the $+2$ oxidation state ($\text{Ni}^{2+}$). The atomic number of $\text{Ni}$ is $28$, so the electronic configuration of $\text{Ni}^{2+}$ is: $\text{Ni}^{2+}: [\text{Ar}]\, 3d^8$

• Complex (A): $[\text{Ni(en)}_3]^{2+}$

  • Geometry: Octahedral
  • Ligand: $\text{en}$ (ethylenediamine) is a strong field ligand.
  • For a $d^8$ octahedral complex, the filling of electrons in the $t_{2g}$ and $e_g$ orbitals follows: $t_{2g}^6 e_g^2$
  • Number of unpaired electrons = $2$

• Complex (B): $[\text{NiCl}_4]^{2-}$

  • Geometry: Tetrahedral ($\text{Cl}^-$ is a weak field ligand and coordination number is 4)
  • For a $d^8$ tetrahedral complex, the filling of electrons in the $e$ and $t_2$ orbitals follows: $e^4 t_2^4$
  • Number of unpaired electrons = $2$

• Complex (C): $[\text{Ni(NH}_3)_6]^{2+}$

  • Geometry: Octahedral
  • Ligand: $\text{NH}_3$ (amine ligand)
  • For a $d^8$ octahedral complex, the electronic configuration is: $t_{2g}^6 e_g^2$
  • Number of unpaired electrons = $2$

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

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2. Determination of Absorption Frequency Order

The energy of the light absorbed ($E$) during $d$-$d$ transitions is directly proportional to the crystal field splitting energy ($\Delta$), which in turn is directly proportional to the absorption frequency ($\nu$): $E = h\nu = \Delta$

• Field Strength of Ligands: According to the spectrochemical series, the ligand field strength increases in the order: $\text{Cl}^- < \text{NH}_3 < \text{en}$

• Effect of Geometry and Ligand Field:

  1. $[\text{Ni(en)}_3]^{2+}$ (A) and $[\text{Ni(NH}_3)_6]^{2+}$ (C) are both octahedral complexes. Since $\text{en}$ is a stronger field ligand than $\text{NH}_3$, we have: $\Delta_o(\text{en}) > \Delta_o(\text{NH}_3) \implies \nu(\text{A}) > \nu(\text{C})$
  2. $[\text{NiCl}_4]^{2-}$ (B) is a tetrahedral complex with a weak field ligand ($\text{Cl}^-$). Since crystal field splitting in tetrahedral fields ($\Delta_t$) is inherently much smaller than in octahedral fields ($\Delta_t \approx \frac{4}{9}\Delta_o$), complex (B) has the lowest splitting energy: $\Delta(\text{A}) > \Delta(\text{C}) > \Delta(\text{B})$

Therefore, the order of absorption frequency is: $(\text{A}) > (\text{C}) > (\text{B})$

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Conclusion

• Number of unpaired electrons: $2, 2, 2$
• Order of frequency of absorption: $(\text{A}) > (\text{C}) > (\text{B})$

This corresponds to Option A.