Question

The magnetic moment of a transition metal ion is found to be \( 5.92 \, \text{BM} \). The number of unpaired electrons present in the ion is:

• A. 3
• B. 4
• C. 5
• D. 6

Hint:

Remember common magnetic moments: \[ n=1 \rightarrow 1.73\,BM \] \[ n=2 \rightarrow 2.83\,BM \] \[ n=3 \rightarrow 3.87\,BM \] \[ n=4 \rightarrow 4.90\,BM \] \[ n=5 \rightarrow 5.92\,BM \]

Answer 1

The Correct Option is C

Concept:
The magnetic moment of transition metal ions is calculated using the spin-only formula: \[ \mu = \sqrt{n(n+2)} \] where:
• \( \mu \) = magnetic moment in Bohr Magneton (BM)
• \( n \) = number of unpaired electrons This formula helps determine the electronic configuration and magnetic behavior of coordination compounds and transition metal ions.

Step 1: Write the given magnetic moment.
\[ \mu = 5.92 \, \text{BM} \] Using: \[ \mu = \sqrt{n(n+2)} \] Squaring both sides: \[ (5.92)^2 = n(n+2) \] \[ 35.04 \approx n(n+2) \]

Step 2: Find suitable integer value of \(n\).
Check values: For \( n=5 \): \[ 5(5+2)=5\times7=35 \] \[ \sqrt{35}=5.92 \] This matches the given magnetic moment.

Step 3: Interpretation.
The ion contains: \[ \boxed{5 \text{ unpaired electrons}} \] Hence the ion is strongly paramagnetic.