JEE Main 2024 · 30 January · evening · Single correct · +4 / −1
The orange colour of K₂ Cr₂ O₇ and purple colour of KMnO₄ is due to
- (a) Charge transfer transition in both.
- (b) d → d transitions in KMnO₄ and charge transfer transitions in K₂ Cr₂ O₇.
- (c) d → d transitions in both
- (d) d → d transitions in K₂ Cr₂ O₇ and charge transfer transitions in KMnO₄.
Answer and solution
Answer (a) Charge transfer transition in both.
Solution
Chemistry ma’am · 3 steps
Find the oxidation state first: a d⁰ metal has no d electron to jump, so d–d is ruled out.
- Need for d–d. A d–d transition moves an electron from one d orbital to another. For that, the metal ion must have at least one d electron.
Ion Oxidation state of metal Configuration of metal ion d electrons Cr₂O₇²⁻ Cr: +6 [Ar] 3d⁰ 0 MnO₄⁻ Mn: +7 [Ar] 3d⁰ 0 - Charge transfer. Both ions are still strongly coloured. When light is absorbed, an electron moves from an oxygen ligand to the empty d orbitals of the metal. This is ligand-to-metal charge transfer, and it absorbs strongly in the visible region.
- So. The orange of K₂Cr₂O₇ and the purple of KMnO₄ both come from charge transfer transitions, which is option (a).
The idea to keep. Before you say d–d, count the d electrons. A d⁰ ion such as Cr(VI), Mn(VII) or Ti(IV) that is still coloured gets its colour from ligand-to-metal charge transfer.
Answer from the NTA final answer key. Solution by PrepWiser.