JEE Main 2023 · 8 April · morning · Single correct · +4 / −1
The correct order of spin only magnetic moments for the following complex ions is
- (a) [Fe(CN)₆]³⁻<[CoF₆]³⁻<[MnBr₄]²⁻<[Mn(CN)₆]³⁻
- (b) [MnBr₄]²⁻<[CoF₆]³⁻<[Fe(CN)₆]³⁻<[Mn(CN)₆]³⁻
- (c) [Fe(CN)₆]³⁻<[Mn(CN)₆]³⁻<[CoF₆]³⁻<[MnBr₄]²⁻
- (d) [CoF₆]³⁻<[MnBr₄]²⁻<[Fe(CN)₆]³⁻<[Mn(CN)₆]³⁻
Answer and solution
Answer (c) [Fe(CN)₆]³⁻<[Mn(CN)₆]³⁻<[CoF₆]³⁻<[MnBr₄]²⁻
Solution
Chemistry ma’am · 3 steps
Find each metal's d count, decide high or low spin from the ligand, then count unpaired electrons.
- Find the d count. Work out the oxidation state from the charge: Fe³⁺ is d⁵, Mn³⁺ is d⁴, Co³⁺ is d⁶ and Mn²⁺ is d⁵.
- Decide the spin. CN⁻ is a strong-field ligand, so it pairs electrons into t_(2g) (low spin). F⁻ and Br⁻ are weak-field ligands, so electrons stay unpaired (high spin). A tetrahedral splitting is also small, so [MnBr₄]²⁻ is high spin.
μ = √n(n+2) BM
Complex Ion and field Unpaired n μ (BM) [Fe(CN)₆]³⁻ d⁵, low spin, t_(2g)⁵ 1 1.73 [Mn(CN)₆]³⁻ d⁴, low spin, t_(2g)⁴ 2 2.83 [CoF₆]³⁻ d⁶, high spin, t_(2g)⁴ e_g² 4 4.90 [MnBr₄]²⁻ d⁵, tetrahedral, high spin 5 5.92 - So. The order is [Fe(CN)₆]³⁻ < [Mn(CN)₆]³⁻ < [CoF₆]³⁻ < [MnBr₄]²⁻, which is option (c).
The idea to keep. For spin-only moments, only the unpaired electron count matters. Get the d count from the oxidation state, then let the ligand decide pairing: strong field pairs, weak field and tetrahedral do not.
Answer from the NTA final answer key. Solution by PrepWiser.