What will be the charge on `Fe^(x +)` if the megnetic moment is `sqrt(24)` ?

Options

A. `+ 2`
B. `+ 3`
C. Zero
D. None of these

Explanation

To determine the charge on \( \text{Fe}^{(x +)} \) given that the magnetic moment is \( \sqrt{24} \), we can follow these steps:

### Step 1: Use the formula for magnetic moment
The magnetic moment (\( \mu \)) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

### Step 2: Set up the equation
Given that the magnetic moment is \( \sqrt{24} \), we can set up the equation:
\[ \sqrt{24} = \sqrt{n(n + 2)} \]

### Step 3: Square both sides
Squaring both sides to eliminate the square root gives:
\[ 24 = n(n + 2) \]

### Step 4: Rearrange the equation
Rearranging the equation, we have:
\[ n^2 + 2n - 24 = 0 \]

### Step 5: Solve the quadratic equation
Now, we can solve this quadratic equation using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 2, c = -24 \).

Calculating the discriminant:
\[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-24) = 4 + 96 = 100 \]
Now substituting back into the formula:
\[ n = \frac{-2 \pm \sqrt{100}}{2 \cdot 1} = \frac{-2 \pm 10}{2} \]
This gives us two potential solutions:
\[ n = \frac{8}{2} = 4 \quad \text{and} \quad n = \frac{-12}{2} = -6 \]
Since \( n \) must be a non-negative integer, we take \( n = 4 \).

### Step 6: Determine the electronic configuration of iron
The atomic number of iron (Fe) is 26, and its electron configuration in the ground state is:
\[ \text{Fe}: [\text{Ar}] 3d^6 4s^2 \]
Given that there are 4 unpaired electrons, we need to find the oxidation state of iron that results in this configuration.

### Step 7: Find the oxidation state
When iron loses electrons to form cations, the 4s electrons are lost first, followed by the 3d electrons. If we assume that the oxidation state is \( +2 \), the configuration would be:
\[ \text{Fe}^{2+}: [\text{Ar}] 3d^6 \]
This configuration has 4 unpaired electrons, which matches our earlier calculation.

### Conclusion
Thus, the charge on \( \text{Fe}^{(x +)} \) is:
\[ x = +2 \]