To solve the question, we need to determine the value of 'x' in the ion \( M^{(x+)} \) given that the magnetic moment is 2.84 BM and the atomic number \( Z \) of metal \( M \) is 23.
### Step-by-Step Solution:
1. **Understanding Magnetic Moment**:
The magnetic moment (μ) is given as 2.84 BM. We know that the magnetic moment can be expressed in terms of the number of unpaired electrons (n) using the formula:
\[
\mu = \sqrt{n(n + 2)}
\]
Setting this equal to 2.84, we can square both sides:
\[
2.84^2 = n(n + 2)
\]
\[
8.0656 = n(n + 2)
\]
2. **Solving for n**:
Rearranging the equation gives us:
\[
n^2 + 2n - 8.0656 = 0
\]
We can use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2, c = -8.0656 \):
\[
n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-8.0656)}}{2 \cdot 1}
\]
\[
n = \frac{-2 \pm \sqrt{4 + 32.2624}}{2}
\]
\[
n = \frac{-2 \pm \sqrt{36.2624}}{2}
\]
\[
n = \frac{-2 \pm 6.02}{2}
\]
This gives us two possible values for n:
\[
n = \frac{4.02}{2} = 2.01 \quad \text{(approximately 2)}
\]
or
\[
n = \frac{-8.02}{2} \quad \text{(not valid since n cannot be negative)}
\]
Thus, we conclude that \( n = 2 \).
3. **Determining the Electron Configuration**:
The atomic number \( Z \) of metal \( M \) is 23. The electron configuration for \( M \) is:
\[
1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 4s^2 \, 3d^3
\]
This shows that there are 3 electrons in the 3d subshell and 2 in the 4s subshell.
4. **Ionization Process**:
To achieve 2 unpaired electrons, we need to remove electrons from the outermost shell. The 4s electrons are removed first:
- Removing 2 electrons from 4s gives us \( 3d^3 \, 4s^0 \).
- To have 2 unpaired electrons, we must remove one more electron from the 3d subshell:
- This results in \( 3d^2 \, 4s^0 \), which has 2 unpaired electrons.
5. **Conclusion**:
Since we removed a total of 3 electrons (2 from 4s and 1 from 3d), the charge of the ion is \( x = 3 \). Therefore, the ion is \( M^{3+} \).
### Final Answer:
Thus, the value of \( x \) is **3**.
