A compound of a metal ion `M^(X+)(z=24)` has a spin only magnetic moment of `sqrt(15)B.M.` . The number of unpaired electrons in the compound are

Options

A. 2
B. 4
C. 5
D. 3

Answer

To solve the problem, we need to determine the number of unpaired electrons in a metal ion \( M^{X+} \) with a given spin-only magnetic moment of \( \sqrt{15} \) Bohr magneton (B.M.).

### Step-by-Step Solution:

1. **Understand the Formula for Spin-Only Magnetic Moment:**
The spin-only magnetic moment (\( \mu \)) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

2. **Set Up the Equation:**
We know from the problem that:
\[ \mu = \sqrt{15} \text{ B.M.} \]
Therefore, we can set up the equation:
\[ \sqrt{n(n + 2)} = \sqrt{15} \]

3. **Square Both Sides:**
To eliminate the square root, we square both sides:
\[ n(n + 2) = 15 \]

4. **Rearrange the Equation:**
Rearranging gives us a quadratic equation:
\[ n^2 + 2n - 15 = 0 \]

5. **Factor the Quadratic Equation:**
We can factor this equation:
\[ (n + 5)(n - 3) = 0 \]
This gives us two possible solutions:
\[ n + 5 = 0 \quad \Rightarrow \quad n = -5 \quad (\text{not a valid solution}) \]
\[ n - 3 = 0 \quad \Rightarrow \quad n = 3 \quad (\text{valid solution}) \]

6. **Conclusion:**
The number of unpaired electrons in the compound is:
\[ n = 3 \]

### Final Answer:
The number of unpaired electrons in the compound is **3**.