Question

The magnetic moment of a transition metal ion is \( \sqrt{15} \,\text{BM} \). Therefore, the number of unpaired electrons present in it is

• A. 3
• B. 4
• C. 1
• D. 2

Hint:

If $\mu$ is between 3 and 4 ($\sqrt{15} \approx 3.87$), the number of unpaired electrons is 3.

Answer 1

The Correct Option is A

Step 1: Formula
The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.
Step 2: Calculation
Given $\mu = \sqrt{15}$, so $\sqrt{15} = \sqrt{n(n+2)}$. Squaring both sides: $15 = n^2 + 2n$.
Step 3: Solve for n
$n^2 + 2n - 15 = 0 \Rightarrow (n+5)(n-3) = 0$. Since $n$ must be positive, $n = 3$.
Final Answer: (a)