Magentic moment of `Fe^(a+)(Z= 26 )` is `sqrt 24 BM`. Hence muber of unpaired electron and value lf a respectively are :

Options

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

Explanation

To solve the problem of finding the number of unpaired electrons and the value of 'a' for \( Fe^{a+} \) (where \( Z = 26 \)) given that the magnetic moment is \( \sqrt{24} \) BM, we can follow these steps:

### Step 1: Understand the Magnetic Moment Formula
The magnetic moment (\( \mu \)) is given by the formula:
\[ \mu = \sqrt{n(n + 2)} \text{ BM} \]
where \( n \) is the number of unpaired electrons.

### Step 2: Set Up the Equation
We know from the problem that:
\[ \mu = \sqrt{24} \text{ BM} \]
Thus, we can set up the equation:
\[ \sqrt{n(n + 2)} = \sqrt{24} \]

### Step 3: Square Both Sides
To eliminate the square root, we square both sides:
\[ n(n + 2) = 24 \]

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

### Step 5: Solve the Quadratic Equation
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:
\[ b^2 - 4ac = 2^2 - 4 \times 1 \times (-24) = 4 + 96 = 100 \]
Now substituting into the quadratic formula:
\[ n = \frac{-2 \pm \sqrt{100}}{2 \times 1} = \frac{-2 \pm 10}{2} \]
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 Oxidation State 'a'
Iron has an atomic number of 26, and its electronic configuration in the neutral state is:
\[ [Ar] 3d^6 4s^2 \]
For \( Fe^{2+} \), the configuration becomes:
\[ 3d^6 \]
In this configuration, there are 4 unpaired electrons. Thus, the oxidation state \( a \) is 2.

### Final Answers
- Number of unpaired electrons: \( n = 4 \)
- Value of \( a = 2 \)