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

Options

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

Answer

To determine the correct order of decreasing second ionization enthalpy for Titanium (Ti), Vanadium (V), Chromium (Cr), and Manganese (Mn), we will follow these steps:

### Step 1: Write the electronic configurations of the elements.
- **Titanium (Ti, Z=22)**: The electronic configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2\).
- **Vanadium (V, Z=23)**: The electronic configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^3 4s^2\).
- **Chromium (Cr, Z=24)**: The electronic configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1\).
- **Manganese (Mn, Z=25)**: The electronic configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^2\).

### Step 2: Determine the electronic configuration after the first ionization.
- **Ti**: After losing one electron, \(Ti^+\) has the configuration \(Ar 3d^2 4s^1\).
- **V**: After losing one electron, \(V^+\) has the configuration \(Ar 3d^3 4s^1\).
- **Cr**: After losing one electron, \(Cr^+\) has the configuration \(Ar 3d^5 4s^0\).
- **Mn**: After losing one electron, \(Mn^+\) has the configuration \(Ar 3d^5 4s^1\).

### Step 3: Analyze the stability of the resulting configurations for the second ionization.
- **Chromium (Cr)**: The \(3d^5\) configuration is half-filled, which is particularly stable. Removing another electron from this stable configuration requires a significant amount of energy.
- **Manganese (Mn)**: The \(3d^5 4s^1\) configuration is also stable due to the half-filled \(3d\) subshell. However, removing the \(4s\) electron is easier than removing from the \(3d\) subshell.
- **Vanadium (V)**: The configuration \(3d^3 4s^1\) is less stable than that of Mn and Cr, so the second ionization enthalpy will be lower than those two.
- **Titanium (Ti)**: The configuration \(3d^2 4s^1\) is the least stable among these, so it will have the lowest second ionization enthalpy.

### Step 4: Write the final order of decreasing second ionization enthalpy.
Based on the analysis:
1. **Cr** (highest second ionization enthalpy)
2. **Mn**
3. **V**
4. **Ti** (lowest second ionization enthalpy)

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