JEE Main 2021 · 17 March · morning · Single correct · +4 / −1
What is the spin-only magnetic moment value (BM) of a divalent metal ion with atomic number 25, in it's aqueous solution?
- (a) zero
- (b) 5.26
- (c) 5.0
- (d) 5.92
Answer and solution
Answer (d) 5.92
Solution
Chemistry ma’am · 3 steps
Mn²⁺ is d⁵, and weak-field water leaves all five electrons unpaired.
- Identify the ion. Atomic number 25 is manganese, [Ar] 3d⁵ 4s². To form Mn²⁺, the two 4s electrons leave first, which gives [Ar] 3d⁵.
- Count unpaired electrons. In water the ion is [Mn(H₂O)₆]²⁺. Water is a weak-field ligand, so the electrons do not pair up. The five 3d electrons sit one in each orbital, so n = 5.
μ = √n(n+2) = √(5 × 7) = √35 ≈ 5.92 BM
- So. The spin-only magnetic moment is 5.92 BM, which is option (d).
The idea to keep. For spin-only moment, count unpaired electrons and use μ = √n(n+2) BM. Remember transition metals lose 4s electrons before 3d ones.
Answer from the NTA final answer key. Solution by PrepWiser.