Question

Consider the metal complexes \( [Ni(en)_3]^{2+} \) (A), \( [NiCl_4]^{2-} \) (B) and \( [Ni(NH_3)_6]^{2+} \) (C). Choose the CORRECT option by considering the number of unpaired electrons present in (A), (B) and (C) respectively and the order of frequency of absorption.

• A. 2, 2, 2 and (A) > (C) > (B)
• B. 0, 2, 0 and (A) > (C) > (B)
• C. 2, 2, 0 and (B) > (C) > (A)
• D. 2, 2, 2 and (C) > (A) > (B)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Evaluate the oxidation state and d-electron count for Nickel in all three complexes. Use Crystal Field Theory (CFT) to distribute these electrons in octahedral and tetrahedral geometries to find unpaired electrons. Finally, use the spectrochemical series to order the splitting energy ($\Delta$), which is directly proportional to the frequency of absorbed light.
Step 2: Key Formula or Approach:
Frequency of absorption $\nu \propto \Delta E$.
Splitting energy depends on geometry ($\Delta_o>\Delta_t$) and ligand strength (Spectrochemical series: $en>NH_3>Cl^-$).
$\text{Ni}^{2+}$ is a $3d^8$ system.
Step 3: Detailed Explanation:
Unpaired Electrons Analysis:
In all three complexes, the metal ion is $\text{Ni}^{2+}$, which has a $3d^8$ configuration.
(A) $[\text{Ni(en)}_3]^{2+}$: Octahedral complex. The $d^8$ configuration in an octahedral field ($t_{2g}^6 e_g^2$) always has exactly 2 unpaired electrons in the $e_g$ orbitals, regardless of ligand strength.
(C) $[\text{Ni(NH}_3)_6]^{2+}$: Octahedral complex. Similar to the above, $d^8$ in an octahedral field gives 2 unpaired electrons.
(B) $[\text{NiCl}_4]^{2-}$: Tetrahedral complex. Chloride is a weak field ligand. The $d^8$ configuration in a tetrahedral field ($e^4 t_2^4$) yields exactly 2 unpaired electrons in the $t_2$ orbitals.
So, the number of unpaired electrons is 2, 2, 2 respectively.
Absorption Frequency Analysis:
The frequency of light absorbed ($\nu$) corresponds to the crystal field splitting energy ($\Delta$). $\Delta E = h\nu = \Delta_o$ (or $\Delta_t$).
1. Octahedral fields split orbitals much more than tetrahedral fields: $\Delta_o \approx \frac{9}{4} \Delta_t$. So complexes (A) and (C) absorb at a much higher frequency than complex (B).
2. Between the two octahedral complexes (A) and (C), we look at ligand strength. Ethylenediamine (en) is a bidentate ligand and is stronger than Ammonia (NH$_3$).
According to the spectrochemical series: $en>NH_3$.
Thus, $\Delta_o (\text{en})>\Delta_o (\text{NH}_3)$.
Bringing it all together: $\Delta_E(A)>\Delta_E(C)>\Delta_E(B)$.
Therefore, the frequency order is $\nu_{(A)}>\nu_{(C)}>\nu_{(B)}$.
Step 4: Final Answer:
The correct option is 2, 2, 2 and (A)>(C)>(B).