JEE Main 2023 · 24 January · morning · Single correct · +4 / −1
The magnetic moment of a transition metal compound has been calculated to be 3.87 B.M. The metal ion is
- (a) Ti²⁺
- (b) V²⁺
- (c) Cr²⁺
- (d) Mn²⁺
Answer and solution
Answer (b) V²⁺
Solution
Chemistry ma’am · 3 steps
Turn 3.87 B.M. into a count of unpaired electrons, then find the ion with exactly that many.
- Spin-only formula. For 3d ions the magnetic moment comes from unpaired electrons alone. Here n is the number of unpaired electrons.
μ = √n(n+2) B.M.
- Find n. Squaring gives n(n+2) = 3.87² ≈ 15. Since 3 × 5 = 15, the ion has n = 3 unpaired electrons.
Ion Configuration Unpaired electrons μ (B.M.) Ti²⁺ 3d² 2 √8 = 2.83 V²⁺ 3d³ 3 √15 = 3.87 Cr²⁺ 3d⁴ 4 √24 = 4.90 Mn²⁺ 3d⁵ 5 √35 = 5.92 - So. Vanadium (Z = 23) loses its two 4s electrons to give V²⁺, which is 3d³ with three unpaired electrons. The answer is (b).
The idea to keep. Learn the spin-only values for 1 to 5 unpaired electrons: 1.73, 2.83, 3.87, 4.90, 5.92 B.M. Then a magnetic moment question becomes a d-electron count.
Answer from the NTA final answer key. Solution by PrepWiser.