Correct order of second ionization energy among the following is:

Options

A. `Li^+gt C^+gt B^+gt Be^+`
B. `C^+ gt B^+ gt Be^+ gt Li^+`
C. `B^+gt Be^+gt C^+gt Li^+`
D. `Li^+ gt B^+ gt C^+ gt Be^+ `

Answer

To determine the correct order of second ionization energy among the ions \( \text{Li}^+, \text{B}^+, \text{C}^+, \text{Be}^+ \), we need to analyze the electronic configurations of these ions and understand the stability associated with removing a second electron.

### Step-by-Step Solution:

1. **Identify the Electronic Configurations**:
- \( \text{Li}^+ \): After losing one electron, the configuration is \( 1s^2 \) (helium configuration).
- \( \text{B}^+ \): After losing one electron, the configuration is \( 1s^2 2s^2 2p^1 \) (removal from \( 2p \)).
- \( \text{C}^+ \): After losing one electron, the configuration is \( 1s^2 2s^2 2p^2 \) (removal from \( 2p \)).
- \( \text{Be}^+ \): After losing one electron, the configuration is \( 1s^2 2s^1 \) (removal from \( 2s \)).

2. **Stability Considerations**:
- **Lithium Ion (\( \text{Li}^+ \))**: The configuration \( 1s^2 \) is stable (noble gas configuration). Removing another electron would require a lot of energy because it would disrupt this stable configuration.
- **Boron Ion (\( \text{B}^+ \))**: It has a configuration of \( 1s^2 2s^2 2p^1 \). Removing an electron from \( 2p \) will lead to a stable \( 1s^2 2s^2 \) configuration, which is relatively stable.
- **Carbon Ion (\( \text{C}^+ \))**: The configuration is \( 1s^2 2s^2 2p^2 \). Removing an electron from \( 2p \) will lead to \( 1s^2 2s^2 2p^1 \), which is less stable than \( \text{B}^+ \) after the second ionization.
- **Beryllium Ion (\( \text{Be}^+ \))**: The configuration is \( 1s^2 2s^1 \). Removing the second electron will lead to a stable \( 1s^2 \) configuration, making it easier to remove the second electron.

3. **Order of Second Ionization Energy**:
- The second ionization energy is highest for the most stable configuration after the removal of the second electron.
- Therefore, the order of the second ionization energy from highest to lowest is:
- \( \text{Li}^+ \) (most stable, noble gas configuration)
- \( \text{B}^+ \) (stable after removing an electron)
- \( \text{C}^+ \) (less stable than \( \text{B}^+ \))
- \( \text{Be}^+ \) (least stable after removal, as it achieves noble gas configuration)

Thus, the correct order of second ionization energy is:
\[ \text{Li}^+ > \text{B}^+ > \text{C}^+ > \text{Be}^+ \]