JEE Main 2025 · 2 April · evening · Single correct · +4 / −1
The type of hybridization and the magnetic property of [MnCl₆]³⁻ are,
- (a) s p³ d², paramagnetic with four unpaired electrons.
- (b) d² s p³, paramagnetic with four unpaired electrons.
- (c) sp³ d², paramagnetic with two unpaired electrons.
- (d) d² sp³, paramagnetic with two unpaired electrons.
Answer and solution
Answer (a) s p³ d², paramagnetic with four unpaired electrons.
Solution
Chemistry ma’am · 4 steps
Find the d-count of Mn, then let the weak-field Cl⁻ decide whether those electrons pair up.
- Oxidation state. Six Cl⁻ give -6 and the complex carries -3, so Mn is +3. Mn is [Ar] 3d⁵ 4s², so Mn³⁺ is [Ar] 3d⁴.
- Weak-field ligand. Cl⁻ is a weak-field ligand, so it cannot force the 3d electrons to pair. The four electrons stay in four separate 3d orbitals, which gives 4 unpaired electrons.
- Choose orbitals. That leaves only one empty 3d orbital, but d²sp³ needs two. So Mn uses the outer orbitals instead: one 4s, three 4p and two 4d, which is sp³d² (an outer-orbital, high-spin complex).
μ = √n(n+2) = √(4 × 6) ≈ 4.9 BM
- So. [MnCl₆]³⁻ is sp³d² hybridised and paramagnetic with four unpaired electrons, which is option (a).
The idea to keep. For octahedral complexes, count the free inner d orbitals after deciding on pairing. Two empty 3d orbitals give d²sp³. Fewer than two, as with weak-field ligands on d⁴ to d⁹, push you to sp³d².
Answer from the NTA final answer key. Solution by PrepWiser.