If Hind's rule is not followed , magnetic moment of `Fe^(2+) , Mn^(+) ` and Cr all having `24` electron will be in order

Options

A. `Fe^(2+) lt Mn^("+) lt Cr`
B. `Fe^(2+) = Cr lt Mn^(+)`
C. `Fe^(2+) = Mn^(+) lt Cr`
D. `Mn^(+) =Cr lt Fe^(+2)`

Explanation

To solve the question regarding the magnetic moments of \( \text{Fe}^{2+} \), \( \text{Mn}^{+} \), and \( \text{Cr} \) (all having 24 electrons) when Hund's rule is not followed, we will follow these steps:

### Step 1: Determine the electronic configurations

1. **For \( \text{Fe}^{2+} \)**:
- Atomic number of Fe = 26
- Electronic configuration of Fe: \( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^6 \, 4s^2 \)
- For \( \text{Fe}^{2+} \), we remove 2 electrons (from 4s and 3d):
- Configuration: \( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^6 \)

2. **For \( \text{Mn}^{+} \)**:
- Atomic number of Mn = 25
- Electronic configuration of Mn: \( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^5 \, 4s^2 \)
- For \( \text{Mn}^{+} \), we remove 1 electron (from 4s):
- Configuration: \( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^5 \)

3. **For \( \text{Cr} \)**:
- Atomic number of Cr = 24
- Electronic configuration of Cr: \( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^5 \, 4s^1 \)
- Since we are considering the case where Hund's rule is not followed, we will assume the configuration is \( 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^4 \, 4s^2 \).

### Step 2: Count the number of unpaired electrons

1. **For \( \text{Fe}^{2+} \)**:
- Configuration: \( 3d^6 \)
- Unpaired electrons = 4 (since the configuration would be \( \uparrow \downarrow \, \uparrow \, \uparrow \, \uparrow \, \uparrow \))

2. **For \( \text{Mn}^{+} \)**:
- Configuration: \( 3d^5 \)
- Unpaired electrons = 5 (all electrons in 3d are unpaired)

3. **For \( \text{Cr} \)**:
- Configuration: \( 3d^4 \)
- Unpaired electrons = 4 (since the configuration would be \( \uparrow \, \uparrow \, \uparrow \, \uparrow \))

### Step 3: Calculate the magnetic moment

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

1. **For \( \text{Fe}^{2+} \)**:
- \( n = 4 \)
- \( \mu = \sqrt{4(4 + 2)} = \sqrt{24} \)

2. **For \( \text{Mn}^{+} \)**:
- \( n = 5 \)
- \( \mu = \sqrt{5(5 + 2)} = \sqrt{35} \)

3. **For \( \text{Cr} \)**:
- \( n = 4 \)
- \( \mu = \sqrt{4(4 + 2)} = \sqrt{24} \)

### Step 4: Order the magnetic moments

Now we can summarize the magnetic moments:
- \( \text{Fe}^{2+} \): \( \sqrt{24} \)
- \( \text{Mn}^{+} \): \( \sqrt{35} \)
- \( \text{Cr} \): \( \sqrt{24} \)

### Final Order of Magnetic Moments
The order of magnetic moments from highest to lowest is:
1. \( \text{Mn}^{+} \) (highest)
2. \( \text{Fe}^{2+} \) = \( \text{Cr} \) (lowest, equal)