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`

Explanation

To solve the problem, we need to determine the number of unpaired electrons and the oxidation state \( n \) for the ion \( X^{n+} \) where \( Z = 26 \) (atomic number of iron, Fe). The spin magnetic moment is given as \( \sqrt{24} \) Bohr magneton (B.M.).

### Step 1: Use the formula for spin magnetic moment
The formula for the spin magnetic moment \( \mu_s \) is given by:

\[ \mu_s = \sqrt{n(n + 2)} \text{ B.M.} \]

where \( n \) is the number of unpaired electrons.

### Step 2: Set up the equation
Given that \( \mu_s = \sqrt{24} \) B.M., we can set up the equation:

\[ \sqrt{n(n + 2)} = \sqrt{24} \]

### Step 3: Square both sides
Squaring both sides to eliminate the square root gives:

\[ n(n + 2) = 24 \]

### Step 4: Rearrange the equation
Rearranging the equation leads to:

\[ n^2 + 2n - 24 = 0 \]

### Step 5: Factor the quadratic equation
Now we can factor the quadratic equation:

\[ (n + 6)(n - 4) = 0 \]

### Step 6: Solve for \( n \)
Setting each factor to zero gives us:

1. \( n + 6 = 0 \) → \( n = -6 \) (not possible, as \( n \) cannot be negative)
2. \( n - 4 = 0 \) → \( n = 4 \)

Thus, the number of unpaired electrons \( n = 4 \).

### Step 7: Determine the oxidation state
Now, we need to find the oxidation state \( n \). The atomic number \( Z = 26 \) corresponds to iron (Fe), which has the electronic configuration:

\[ \text{Fe: } [Ar] 4s^2 3d^6 \]

To achieve 4 unpaired electrons, we can lose 2 electrons from the 4s orbital, leading to the configuration:

\[ [Ar] 3d^6 \quad \text{(4s is empty)} \]

This corresponds to an oxidation state of +2, hence \( n = 2 \).

### Final Answer
The number of unpaired electrons and the value of \( n \) respectively are:

**4 and 2.**

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