Spin magnetic moment of `X^(n+) (Z=26)` is `sqrt(24) B.M.` Hence number of unpaired electrons and value of `n` respectively are:

Options

A. 4,2
B. 2,4
C. 3,1
D. 0,2

Answer

To solve the problem, we need to determine the number of unpaired electrons and the charge (n) of the ion \(X^{n+}\) with atomic number \(Z = 26\) (which corresponds to iron, Fe). The spin magnetic moment is given as \(\sqrt{24} \, \text{B.M.}\).

### Step-by-Step Solution:

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

2. **Setting Up the Equation**:
We know from the problem that:
\[ \mu = \sqrt{24} \, \text{B.M.} \]
Therefore, we can equate:
\[ \sqrt{n(n + 2)} = \sqrt{24} \]

3. **Squaring Both Sides**:
To eliminate the square root, we square both sides:
\[ n(n + 2) = 24 \]

4. **Rearranging the Equation**:
Rearranging gives us a quadratic equation:
\[ n^2 + 2n - 24 = 0 \]

5. **Factoring the Quadratic Equation**:
We can factor the quadratic:
\[ (n + 6)(n - 4) = 0 \]
This gives us two possible solutions for \(n\):
\[ n = -6 \quad \text{or} \quad n = 4 \]
Since \(n\) must be a non-negative integer, we take:
\[ n = 4 \]

6. **Determining the Charge (n)**:
The atomic number \(Z = 26\) corresponds to iron (Fe), whose electronic configuration is:
\[ \text{Fe: } [\text{Ar}] \, 4s^2 \, 3d^6 \]
In the case of the \(Fe^{2+}\) ion, two electrons are removed (typically from the 4s orbital first), leading to:
\[ \text{Fe}^{2+}: [\text{Ar}] \, 4s^0 \, 3d^6 \]
In this configuration, the 3d subshell has 6 electrons, which can be arranged to show 4 unpaired electrons.

7. **Conclusion**:
Thus, the number of unpaired electrons is \(4\) and the charge \(n\) of the ion is \(2\) (since it is \(Fe^{2+}\)).

### Final Answer:
The number of unpaired electrons and the value of \(n\) respectively are:
- Number of unpaired electrons: \(4\)
- Value of \(n\): \(2\)