Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
The intensity of color in transition metal complexes arises from d-d electronic transitions. The magnitude of crystal field splitting ($$\Delta$$) determines the energy of the absorbed light, but the actual intensity (molar absorptivity) of the color depends on several factors including the symmetry of the complex and the nature of the ligands.
For tetrahedral complexes, d-d transitions are LaPorte allowed (approximately) compared to octahedral complexes where they are strictly LaPorte forbidden (g → g forbidden), which makes tetrahedral complexes generally more intensely colored than octahedral ones.
Now consider each complex: $$[\text{NiCl}_4]^{2-}$$ is tetrahedral with a weak-field ligand Cl⁻. $$[\text{Ni(H}_2\text{O)}_6]^{2+}$$ is octahedral with a weak-field ligand H₂O. $$[\text{Ni(CN)}_4]^{2-}$$ is square planar with a strong-field ligand CN⁻.
For $$[\text{Ni(CN)}_4]^{2-}$$: Ni²⁺ has the configuration d⁸. With CN⁻ as a strong-field ligand in a square planar geometry, all 8 d electrons pair up. The d-d transitions in square planar complexes are formally LaPorte forbidden and the strong field causes very large $$\Delta$$, pushing absorption to UV region. Thus, $$[\text{Ni(CN)}_4]^{2-}$$ appears colorless (or very pale yellow), with very low visible color intensity.
For $$[\text{Ni(H}_2\text{O)}_6]^{2+}$$: This is octahedral with a weak-field ligand. The d-d transitions are LaPorte forbidden (g → g), resulting in weak intensity. It appears green/blue with weak color intensity.
For $$[\text{NiCl}_4]^{2-}$$: This is tetrahedral. In tetrahedral complexes, d-d transitions are LaPorte "allowed" (the center of inversion is absent, so the g → g rule doesn't apply), giving much higher molar absorptivity. Therefore, $$[\text{NiCl}_4]^{2-}$$ is more intensely colored than the octahedral $$[\text{Ni(H}_2\text{O)}_6]^{2+}$$.
Thus, the order of color intensity is: $$[\text{NiCl}_4]^{2-} > [\text{Ni(H}_2\text{O)}_6]^{2+} > [\text{Ni(CN)}_4]^{2-}$$
This corresponds to Option 3.
Create a FREE account and get:
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.