Spin-only magnetic moment formula:
\[ \mu = \sqrt{n(n+2)} \; BM \]
Where n = number of unpaired electrons.
1) Co²⁺
Co: 27 → [Ar] 3d⁷4s²
Co²⁺ → 3d⁷
High spin octahedral → d⁷ → 3 unpaired electrons
\[ \mu = \sqrt{3(3+2)} = \sqrt{15} = 3.87 \, BM \]
Greater than 3.0 BM ✔
2) Ni²⁺
Ni: 28 → 3d⁸4s²
Ni²⁺ → 3d⁸
High spin octahedral → 2 unpaired electrons
\[ \mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83 \, BM \]
Less than 3.0 BM ✖
3) Fe²⁺
Fe: 26 → 3d⁶4s²
Fe²⁺ → 3d⁶
High spin octahedral → 4 unpaired electrons
\[ \mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90 \, BM \]
Greater than 3.0 BM ✔
4) V³⁺
V: 23 → 3d³4s²
V³⁺ → 3d²
High spin octahedral → 2 unpaired electrons
\[ \mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83 \, BM \]
Less than 3.0 BM ✖
5) Ti²⁺
Ti: 22 → 3d²4s²
Ti²⁺ → 3d²
High spin octahedral → 2 unpaired electrons
\[ \mu = \sqrt{2(2+2)} = \sqrt{8} = 2.83 \, BM \]
Less than 3.0 BM ✖
Metal ions with μ > 3.0 BM:
Co²⁺ → 3 unpaired electrons
Fe²⁺ → 4 unpaired electrons
Sum of unpaired electrons:
\[ 3 + 4 = 7 \]
Final Answer = 7