JEE Main 2023 · 10 April · evening · Single correct · +4 / −1
The correct order of the number of unpaired electrons in the given complexes is
A. [Fe(CN)₆]³⁻
B. [Fe F₆]³⁻
C. [CoF₆]³⁻
D. [Cr (oxalate)₃]³⁻
E. [Ni(CO)₄]
Choose the correct answer from the options given below:
- (a) E < A < B < D < C
- (b) A < E < D < C < B
- (c) A < E < C < B < D
- (d) E < A < D < C < B
Answer and solution
Answer (d) E < A < D < C < B
Solution
Chemistry ma’am · 3 steps
Find each metal's d-electron count, then let the ligand decide high spin or low spin.
- Oxidation state first. Work out the metal's oxidation state and its d-electron count. Fe is +3 (d⁵), Co is +3 (d⁶) and Cr is +3 (d³). In Ni(CO)₄ nickel is 0, and the CO ligands push its 4s electrons into 3d, giving d¹⁰.
- Ligand strength. CN⁻ and CO are strong-field ligands, so electrons pair up in the lower set (low spin). F⁻ is a weak-field ligand, so electrons spread out first (high spin). A d³ ion has 3 unpaired electrons with any ligand.
Complex Metal ion Field and spin Unpaired electrons A: [Fe(CN)₆]³⁻ Fe³⁺, d⁵ strong, low spin: t_(2g)⁵ 1 B: [FeF₆]³⁻ Fe³⁺, d⁵ weak, high spin: t_(2g)³ e_g² 5 C: [CoF₆]³⁻ Co³⁺, d⁶ weak, high spin: t_(2g)⁴ e_g² 4 D: [Cr(ox)₃]³⁻ Cr³⁺, d³ t_(2g)³ either way 3 E: [Ni(CO)₄] Ni⁰, d¹⁰ all paired 0 - So. The counts are 0 < 1 < 3 < 4 < 5, which is E < A < D < C < B, option (d).
The idea to keep. Same metal ion, different ligand, different answer: a strong-field ligand pairs electrons up, a weak-field ligand spreads them out. A d³ ion always has 3 unpaired electrons in an octahedral field.
Answer from the NTA final answer key. Solution by PrepWiser.