In any transition series, from left to right, the d-orbitals are progressively filled and their properties vary accordingly.
Q. Which of the following is the correct order of second ionisation energy ?

Options

A. `VgtCrgtMn`
B. `VltCrltMn`
C. `VltCrgtMn`
D. `VgtCrltMn`

Explanation

To determine the correct order of the second ionization energy for vanadium (V), chromium (Cr), and manganese (Mn), we need to analyze their electronic configurations and the stability of the resulting ions after the first ionization.

### Step-by-Step Solution:

1. **Identify the elements and their atomic numbers:**
- Vanadium (V) - Atomic number 23
- Chromium (Cr) - Atomic number 24
- Manganese (Mn) - Atomic number 25

2. **Write the electronic configurations:**
- Vanadium: \( \text{[Ar]} 3d^3 4s^2 \)
- Chromium: \( \text{[Ar]} 3d^5 4s^1 \)
- Manganese: \( \text{[Ar]} 3d^5 4s^2 \)

3. **Consider the first ionization:**
- When the first electron is removed:
- Vanadium becomes \( \text{[Ar]} 3d^3 4s^1 \)
- Chromium becomes \( \text{[Ar]} 3d^5 \)
- Manganese becomes \( \text{[Ar]} 3d^5 4s^1 \)

4. **Analyze the second ionization:**
- For Vanadium: Removing another electron from \( 3d^3 4s^1 \) (removing from 4s is easier).
- For Chromium: Removing an electron from \( 3d^5 \) (this is a half-filled subshell, which is stable, making it harder to remove an electron).
- For Manganese: Removing from \( 3d^5 4s^1 \) (removing from 4s is easier).

5. **Compare the stability and ionization energy:**
- Chromium has a half-filled \( 3d^5 \) configuration, which is particularly stable. Thus, it requires more energy to remove an electron from chromium compared to vanadium and manganese.
- Vanadium and manganese have their second electron removed from the 4s subshell, which is easier than removing from the stable half-filled \( 3d^5 \) of chromium.

6. **Conclusion:**
- The order of second ionization energy from highest to lowest is:
- Chromium (Cr) > Manganese (Mn) > Vanadium (V)

### Final Answer:
The correct order of second ionization energy is: **Cr > Mn > V**