Explanation (Complete Step-by-Step Calculation)

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

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