Question

Among \( [\text{Ni(CN)}_4]^{2-} \), \( [\text{NiCl}_4]^{2-} \), and \( [\text{Ni(CO)}_4] \), which one has the following:

• A. \( [\text{NiCl}_4]^{2-} \) is square planar, and \( [\text{Ni(CN)}_4]^{2-} \), \( [\text{Ni(CO)}_4] \) are tetrahedral.
• B. \( \text{Ni(CO)}_4 \) is square planar, and \( [\text{Ni(CN)}_4]^{2-} \), \( [\text{NiCl}_4]^{2-} \) are tetrahedral.
• C. \( [\text{Ni(CN)}_4]^{2-} \) is square planar, while \( [\text{NiCl}_4]^{2-} \) and \( \text{Ni(CO)}_4 \) are tetrahedral.
• D. None of these

Hint:

The geometry of the complex depends on the ligands and their electronic effects, such as the ligand field theory for metal-ligand interactions.

Answer 1

The Correct Option is C

\( [\text{Ni(CN)}_4]^{2-} \) and \( \text{Ni(CO)}_4 \) are tetrahedral complexes because the ligands (CN⁻ and CO) are small and can form stable tetrahedral coordination structures with the central metal ion, nickel (Ni). On the other hand, \( [\text{NiCl}_4]^{2-} \) adopts a square planar geometry. This difference in coordination geometry is due to the nature of the ligands and the electronic configuration of the central metal ion, which influences the overall geometry of the complex.

Therefore, the correct option is (3).