JEE Main 2021 · 20 July · morning · Single correct · +4 / −1
The correct order of intensity of colors of the compounds is :
- (a) [Ni(CN)₄]²⁻ > [NiCl₄]²⁻ > [Ni(H₂O)₆]²⁺
- (b) [Ni(H₂O)₆]²⁺ > [NiCl₄]²⁻ > [Ni(CN)₄]²⁻
- (c) [NiCl₄]²⁻ > [Ni(H₂O)₆]²⁺ > [Ni(CN)₄]²⁻
- (d) [NiCl₄]²⁻ > [Ni(CN)₄]²⁻ > [Ni(H₂O)₆]²⁺
Answer and solution
Answer (c) [NiCl₄]²⁻ > [Ni(H₂O)₆]²⁺ > [Ni(CN)₄]²⁻
Solution
Chemistry ma’am · 3 steps
Intensity depends on how allowed the d–d jump is, so first ask which complex has no centre of symmetry.
- Selection rule. A d–d transition is formally forbidden (Laporte rule) when the complex has a centre of symmetry, so its absorption is weak. Without a centre of symmetry the rule is relaxed, and the same transition absorbs light much more strongly.
Complex Ligand field and geometry Centre of symmetry Colour intensity [NiCl₄]²⁻ weak field Cl⁻, sp³ tetrahedral, 2 unpaired electrons absent strongest [Ni(H₂O)₆]²⁺ weak field H₂O, octahedral, 2 unpaired electrons present moderate (pale green) [Ni(CN)₄]²⁻ strong field CN⁻, dsp² square planar, 0 unpaired electrons present weakest (pale yellow) - Why cyanide is last. In [Ni(CN)₄]²⁻ all 8 d electrons are paired and only the high-energy d_(x²-y²) orbital is empty. The splitting is so large that the d–d absorption sits near the ultraviolet, and the complex is also centrosymmetric. Very little visible light is absorbed.
- So. The order is [NiCl₄]²⁻ > [Ni(H₂O)₆]²⁺ > [Ni(CN)₄]²⁻, which is option (c).
The idea to keep. For colour intensity, tetrahedral beats octahedral because it has no centre of symmetry. A low-spin complex with a very large splitting absorbs near the UV, so it is the palest.
Answer from the NTA final answer key. Solution by PrepWiser.