JEE Main 2023 · 13 April · evening · Single correct · +4 / −1
Which of the following complexes will exhibit maximum attraction to an applied magnetic field?
- (a) [Ni(H₂ O)₆]²⁺
- (b) [Co(H₂ O)₆]²⁺
- (c) [Co(en)₃]³⁺
- (d) [Zn(H₂ O)₆]²⁺
Answer and solution
Answer (b) [Co(H₂ O)₆]²⁺
Solution
Chemistry ma’am · 3 steps
The pull in a magnetic field grows with unpaired electrons, so count them on each metal ion.
- What decides attraction. Paramagnetic attraction comes from unpaired d electrons. The spin-only moment is μ = √n(n+2) BM, where n is the number of unpaired electrons. More unpaired electrons give a stronger pull.
- Check the ligand. H₂O is a weak-field ligand, so electrons spread out first (high spin). Ethylenediamine (en) is a strong-field ligand, so electrons pair up in the lower t_(2g) set (low spin).
Complex Metal ion Arrangement Unpaired n μ (BM) [Ni(H₂O)₆]²⁺ Ni²⁺, 3d⁸ t_(2g)⁶ e_g² 2 2.83 [Co(H₂O)₆]²⁺ Co²⁺, 3d⁷ high spin, t_(2g)⁵ e_g² 3 3.87 [Co(en)₃]³⁺ Co³⁺, 3d⁶ low spin, t_(2g)⁶ 0 0 [Zn(H₂O)₆]²⁺ Zn²⁺, 3d¹⁰ all filled 0 0 - So. [Co(H₂O)₆]²⁺ has the most unpaired electrons (3) and the largest moment, so it is attracted most strongly. The answer is (b).
The idea to keep. For magnetic questions, find the d-electron count of the ion, decide high or low spin from the ligand, then count unpaired electrons. The most unpaired electrons give the strongest attraction.
Answer from the NTA final answer key. Solution by PrepWiser.