How would you explain the fact that the first ionization enthalpy of Lithium is lesser than that of Beryllium but its second ionization enthalpy is greater than that of Beryllium?

Answer

To explain why the first ionization enthalpy of Lithium (Li) is less than that of Beryllium (Be), but its second ionization enthalpy is greater than that of Beryllium, we can break down the explanation into several steps.

### Step-by-Step Solution:

1. **Understanding the Elements**:
- Lithium (Li) and Beryllium (Be) are both in the second period of the periodic table.
- Lithium has an atomic number of 3, with an electronic configuration of \(1s^2 2s^1\).
- Beryllium has an atomic number of 4, with an electronic configuration of \(1s^2 2s^2\).

2. **First Ionization Enthalpy**:
- The first ionization enthalpy is the energy required to remove the outermost electron from an atom.
- For Lithium, removing the \(2s^1\) electron results in a stable noble gas configuration (\(He\)).
- Beryllium has a \(2s^2\) configuration, and removing one of these electrons does not lead to a noble gas configuration, but rather a \(2s^1\) configuration.

3. **Factors Affecting First Ionization Energy**:
- **Atomic Size**: Lithium is larger than Beryllium, which means the outer electron is further from the nucleus and experiences less nuclear attraction.
- **Effective Nuclear Charge**: Beryllium has a higher effective nuclear charge due to having more protons (4) compared to Lithium (3), which increases the attraction on the outer electrons in Beryllium.
- **Stable Configuration**: Beryllium has a completely filled \(2s\) subshell, making it more stable and requiring more energy to remove an electron compared to Lithium.

4. **Conclusion for First Ionization Enthalpy**:
- Due to these factors, Lithium has a lower first ionization enthalpy than Beryllium.

5. **Second Ionization Enthalpy**:
- The second ionization enthalpy is the energy required to remove a second electron after the first has already been removed.
- After losing one electron, Lithium becomes \(Li^+\) with a configuration of \(1s^2\) (noble gas configuration).
- Beryllium after losing one electron becomes \(Be^+\) with a configuration of \(1s^2 2s^1\).

6. **Factors Affecting Second Ionization Energy**:
- **Noble Gas Configuration**: \(Li^+\) has a stable noble gas configuration, which means it is more stable and requires more energy to remove the second electron.
- **Cation Size**: The size of cations is smaller than their respective neutral atoms. \(Li^+\) is smaller than \(Li\), and \(Be^+\) is smaller than \(Be\), but \(Li^+\) has a greater effective nuclear charge (3 protons attracting 2 electrons) compared to \(Be^+\) (4 protons attracting 3 electrons).

7. **Conclusion for Second Ionization Enthalpy**:
- Thus, the second ionization enthalpy of Lithium is greater than that of Beryllium due to the stable noble gas configuration and the effective nuclear charge acting on fewer electrons in \(Li^+\).

### Final Summary:
- The first ionization enthalpy of Lithium is less than that of Beryllium due to its larger size, lower effective nuclear charge, and the fact that it achieves a stable configuration upon losing one electron.
- The second ionization enthalpy of Lithium is greater than that of Beryllium because \(Li^+\) has a stable noble gas configuration, while \(Be^+\) does not, and the effective nuclear charge is more favorable for Lithium.