Magnetic moment of `M^(x+)` is `sqrt(24)BM`. The number of unpaired electrons in `M^(x+)` is.

Explanation

To find the number of unpaired electrons in the ion \( M^{x+} \) given that its magnetic moment is \( \sqrt{24} \, BM \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Magnetic Moment**:
The magnetic moment (\( \mu \)) is given by the formula:
\[ \mu = \sqrt{s(s + 2)} \, BM \]
where \( s \) is the total spin quantum number.

2. **Set Up the Equation**:
We know from the problem that:
\[ \mu = \sqrt{24} \, BM \]
Therefore, we can equate the two expressions:
\[ \sqrt{s(s + 2)} = \sqrt{24} \]

3. **Square Both Sides**:
Squaring both sides to eliminate the square root gives:
\[ s(s + 2) = 24 \]

4. **Rearrange the Equation**:
This can be rearranged into a standard quadratic equation:
\[ s^2 + 2s - 24 = 0 \]

5. **Solve the Quadratic Equation**:
We can use the quadratic formula \( s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -24 \):
\[ s = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} \]
\[ s = \frac{-2 \pm \sqrt{4 + 96}}{2} \]
\[ s = \frac{-2 \pm \sqrt{100}}{2} \]
\[ s = \frac{-2 \pm 10}{2} \]

This gives us two possible values for \( s \):
\[ s = \frac{8}{2} = 4 \quad \text{(valid since s must be non-negative)} \]
\[ s = \frac{-12}{2} = -6 \quad \text{(not valid)} \]

6. **Determine the Number of Unpaired Electrons**:
The total spin quantum number \( s \) corresponds to the number of unpaired electrons. The number of unpaired electrons is given by \( s \):
\[ \text{Number of unpaired electrons} = s = 4 \]

### Final Answer:
The number of unpaired electrons in \( M^{x+} \) is **4**.