JEE Main 2025 · 2 April · evening · Numerical answer · +4 / −1
The spin-only magnetic moment value of Mⁿ⁺ ion formed among Ni, Zn, Mn and Cu that has the least enthalpy of atomisation is_________ . (in nearest integer) Here n is equal to the number of diamagnetic complexes among K₂[NiCl₄],[Zn(H₂ O)₆] Cl₂, K₃[Mn(CN)₆] and [Cu(PPh₃)₃ I]
Answer and solution
Answer 0
Solution
Chemistry ma’am · 3 steps
Zinc has the lowest atomisation enthalpy, and Zn²⁺ is d¹⁰, so nothing is unpaired.
- Pick the metal. Enthalpies of atomisation are Ni 430, Cu 339, Mn 281 and Zn 126, all in kJ mol⁻¹. Zinc is lowest because its full 3d¹⁰ shell gives no unpaired d electrons for metallic bonding. So M is Zn.
Complex Metal ion d electrons and field Unpaired Magnetic nature K₂[NiCl₄] Ni²⁺ d⁸, tetrahedral, weak Cl⁻ 2 paramagnetic [Zn(H₂O)₆]Cl₂ Zn²⁺ d¹⁰ 0 diamagnetic K₃[Mn(CN)₆] Mn³⁺ d⁴, strong CN⁻, t_(2g)⁴ 2 paramagnetic [Cu(PPh₃)₃I] Cu⁺ d¹⁰ 0 diamagnetic - Find n. Two complexes are diamagnetic, so n = 2 and the ion is Zn²⁺. Its configuration is [Ar] 3d¹⁰, with every d electron paired.
μ = √nᵤ(nᵤ+2) BM = √(0 × 2) = 0
- So. With no unpaired electrons, the spin-only magnetic moment of Zn²⁺ is 0 BM.
The common slip: 6. Taking Mn as the metal with the lowest enthalpy of atomisation gives Mn²⁺ (d⁵, 5 unpaired) and √35 ≈ 5.92, which rounds to 6. Zinc's value is far lower.
The idea to keep. Any d¹⁰ ion (Zn²⁺, Cu⁺, Cd²⁺) is diamagnetic whatever the ligand. For d⁴ to d⁷ ions, check whether the ligand is strong-field before you count unpaired electrons.
Answer from the NTA final answer key. Solution by PrepWiser.