How many of the following metal ions have similar value of spin only magnetic moment in gaseous state?
(Given: Atomic number : V, 23; Cr, 24; Fe, 26; Ni, 28)
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(Given: Atomic number : V, 23; Cr, 24; Fe, 26; Ni, 28)
, , ,
Show answer and solution
Answer: 2
Solution
Find unpaired electrons for each ion:
- V³⁺ (23−3=20 electrons): [Ar] 3d² → 2 unpaired electrons
- Cr³⁺ (24−3=21 electrons): [Ar] 3d³ → 3 unpaired electrons
- Fe²⁺ (26−2=24 electrons): [Ar] 3d⁶ → 4 unpaired electrons
- Ni³⁺ (28−3=25 electrons): [Ar] 3d⁷ → 3 unpaired electrons
Spin-only magnetic moment = √(n(n+2)) BM. Cr³⁺ and Ni³⁺ both have 3 unpaired electrons, giving μ = √(3×5) = √15 ≈ 3.87 BM. These two ions share the same spin-only magnetic moment value, so the answer is 2.
- V³⁺ (23−3=20 electrons): [Ar] 3d² → 2 unpaired electrons
- Cr³⁺ (24−3=21 electrons): [Ar] 3d³ → 3 unpaired electrons
- Fe²⁺ (26−2=24 electrons): [Ar] 3d⁶ → 4 unpaired electrons
- Ni³⁺ (28−3=25 electrons): [Ar] 3d⁷ → 3 unpaired electrons
Spin-only magnetic moment = √(n(n+2)) BM. Cr³⁺ and Ni³⁺ both have 3 unpaired electrons, giving μ = √(3×5) = √15 ≈ 3.87 BM. These two ions share the same spin-only magnetic moment value, so the answer is 2.
[JEE Main 2023 · 25 Jan · Shift 1]