Magnetic moment of `V(Z = 23),Cr(Z = 24),and Mn(Z= 25)` are ` x,y,z ` respectively hence

Options

A. `x = y = z`
B. `x lt y lt z`
C. `x lt z lt y`
D. `z lt y lt x`

Explanation

To determine the magnetic moments of Vanadium (V), Chromium (Cr), and Manganese (Mn) and their order, we will follow these steps:

### Step 1: Understand the Formula for Magnetic Moment
The magnetic moment (μ) can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

### Step 2: Determine the Electronic Configuration
1. **Vanadium (V, Z = 23)**:
- Atomic number: 23
- Electronic configuration: \([Ar] 4s^2 3d^3\)
- Unpaired electrons: 3 (in the 3d subshell)

2. **Chromium (Cr, Z = 24)**:
- Atomic number: 24
- Electronic configuration: \([Ar] 4s^1 3d^5\) (exceptional case)
- Unpaired electrons: 6 (5 in the 3d subshell and 1 in the 4s subshell)

3. **Manganese (Mn, Z = 25)**:
- Atomic number: 25
- Electronic configuration: \([Ar] 4s^2 3d^5\)
- Unpaired electrons: 5 (in the 3d subshell)

### Step 3: Calculate the Magnetic Moment for Each Element
1. **For Vanadium**:
\[ \mu_V = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87 \text{ Bohr magneton} \]

2. **For Chromium**:
\[ \mu_{Cr} = \sqrt{6(6 + 2)} = \sqrt{6 \times 8} = \sqrt{48} \approx 6.93 \text{ Bohr magneton} \]

3. **For Manganese**:
\[ \mu_{Mn} = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92 \text{ Bohr magneton} \]

### Step 4: Compare the Magnetic Moments
Now we have:
- \( \mu_V \approx 3.87 \)
- \( \mu_{Cr} \approx 6.93 \)
- \( \mu_{Mn} \approx 5.92 \)

### Step 5: Order the Magnetic Moments
From the calculated values:
- \( \mu_V < \mu_{Mn} < \mu_{Cr} \)
- Therefore, the order is \( x < z < y \) where \( x \) is for Vanadium, \( z \) is for Manganese, and \( y \) is for Chromium.

### Final Answer
The correct order of magnetic moments is:
\[ x < z < y \]