The correct of decreasing second ionisation enthalpy of `Ti(22),V(23),Cr(24)` and `Mn(25)` is

Options

A. `Cr gt Mn gt V gt Ti`
B. `V gt Mn gt Cr gt Ti`
C. `Mn gt Cr gt Ti gt V`
D. `Ti gt V gt Cr gt Mn`

Explanation

To determine the correct order of decreasing second ionization enthalpy for Titanium (Ti), Vanadium (V), Chromium (Cr), and Manganese (Mn), we will analyze the electronic configurations and stability of their respective ions after the removal of the first electron.

### Step 1: Understand the concept of second ionization enthalpy
The second ionization enthalpy is the energy required to remove a second electron from a positively charged ion. This means we need to consider the electronic configurations of the ions formed after the first ionization.

### Step 2: Write the electronic configurations
1. **Titanium (Ti, atomic number 22)**:
- Neutral Ti: [Ar] 3d² 4s²
- After first ionization (Ti⁺): [Ar] 3d² 4s¹
- After second ionization (Ti²⁺): [Ar] 3d²

2. **Vanadium (V, atomic number 23)**:
- Neutral V: [Ar] 3d³ 4s²
- After first ionization (V⁺): [Ar] 3d³ 4s¹
- After second ionization (V²⁺): [Ar] 3d³

3. **Chromium (Cr, atomic number 24)**:
- Neutral Cr: [Ar] 3d⁵ 4s¹
- After first ionization (Cr⁺): [Ar] 3d⁵
- After second ionization (Cr²⁺): [Ar] 3d⁴

4. **Manganese (Mn, atomic number 25)**:
- Neutral Mn: [Ar] 3d⁵ 4s²
- After first ionization (Mn⁺): [Ar] 3d⁵ 4s¹
- After second ionization (Mn²⁺): [Ar] 3d⁵

### Step 3: Analyze the stability of the resulting ions
- **Mn²⁺** has a half-filled d-subshell (3d⁵), which is particularly stable.
- **Cr²⁺** has a 3d⁴ configuration, which is less stable than Mn²⁺ but still stable due to the half-filled nature of the original configuration.
- **V²⁺** has a 3d³ configuration, which is less stable than Cr²⁺.
- **Ti²⁺** has a 3d² configuration, which is the least stable among the four.

### Step 4: Determine the order of second ionization enthalpy
Since stability is directly related to ionization enthalpy, we can conclude that:
- Mn²⁺ (most stable) > Cr²⁺ > V²⁺ > Ti²⁺ (least stable)

Thus, the order of decreasing second ionization enthalpy is:
**Mn > Cr > V > Ti**

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