Question

The magnetic moment of a divalent ion in aqueous solution is 3.87 BM. The number of unpaired electrons present in it is

• A. 4
• B. 5
• C. 3
• D. 2
• E. 1

Hint:

Common magnetic moments: n=1 → 1.73 BM, n=2 → 2.83 BM, n=3 → 3.87 BM, n=4 → 4.90 BM, n=5 → 5.92 BM.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Magnetic moment \(\mu = \sqrt{n(n+2)}\) BM, where \(n\) = number of unpaired electrons.

Step 2: Detailed Explanation:
Given \(\mu = 3.87\) BM.
\(\sqrt{n(n+2)} = 3.87\)
Squaring both sides: \(n(n+2) = (3.87)^2 \approx 15\)
\(n^2 + 2n - 15 = 0\)
\((n+5)(n-3)=0\)
\(n=3\) (since \(n\) cannot be negative).

Step 3: Final Answer:
The number of unpaired electrons is 3.