What is the correct order of spin only magnetic moment (in BM) of `Mn^(+2), Cr^(+2) and V^(+2)`? 

Options

A. `Mn^(+2) gt V^(+2)gt Cr^(+2)`
B. `V^(+2) gtCr^(+2)gt Mn^(+2)`
C. `Mn^(+2) gt Cr^(+2) gt V^(+2)`
D. `Cr^(+2) gt V^(+2)gt Mn^(+2)`

Answer

To determine the correct order of spin-only magnetic moment (in Bohr Magneton, BM) for \( \text{Mn}^{2+} \), \( \text{Cr}^{2+} \), and \( \text{V}^{2+} \), we will follow these steps:

### Step 1: Determine the electronic configuration and number of unpaired electrons for each ion.

1. **For \( \text{Mn}^{2+} \)**:
- Atomic number of Mn = 25
- Electronic configuration of Mn = \( [Ar] 3d^5 4s^2 \)
- For \( \text{Mn}^{2+} \), we remove 2 electrons from the 4s orbital:
- \( \text{Mn}^{2+} \): \( [Ar] 3d^5 \)
- Number of unpaired electrons = 5 (all 5 electrons in the 3d subshell are unpaired).

2. **For \( \text{Cr}^{2+} \)**:
- Atomic number of Cr = 24
- Electronic configuration of Cr = \( [Ar] 3d^5 4s^1 \)
- For \( \text{Cr}^{2+} \), we remove 2 electrons (1 from 4s and 1 from 3d):
- \( \text{Cr}^{2+} \): \( [Ar] 3d^4 \)
- Number of unpaired electrons = 4 (4 electrons in the 3d subshell).

3. **For \( \text{V}^{2+} \)**:
- Atomic number of V = 23
- Electronic configuration of V = \( [Ar] 3d^3 4s^2 \)
- For \( \text{V}^{2+} \), we remove 2 electrons from the 4s orbital:
- \( \text{V}^{2+} \): \( [Ar] 3d^3 \)
- Number of unpaired electrons = 3 (3 electrons in the 3d subshell).

### Step 2: Calculate the spin-only magnetic moment using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

1. **For \( \text{Mn}^{2+} \)**:
- \( n = 5 \)
- \( \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92 \, \text{BM} \)

2. **For \( \text{Cr}^{2+} \)**:
- \( n = 4 \)
- \( \mu = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90 \, \text{BM} \)

3. **For \( \text{V}^{2+} \)**:
- \( n = 3 \)
- \( \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87 \, \text{BM} \)

### Step 3: Arrange the magnetic moments in order.
- \( \text{Mn}^{2+} \approx 5.92 \, \text{BM} \)
- \( \text{Cr}^{2+} \approx 4.90 \, \text{BM} \)
- \( \text{V}^{2+} \approx 3.87 \, \text{BM} \)

### Final Order:
The correct order of spin-only magnetic moment is:
\[ \text{Mn}^{2+} > \text{Cr}^{2+} > \text{V}^{2+} \]