Which of the following statements is/are correct?

Options

A. The second ionization enthalpy of oxygen element is greater than that of fluorine element.
B. The third ionization enthalpy of phosphours is greater than that of aluminium.
C. The first ionization enthalpy of aluminium is slightly greater than that of gallium.
D. The second ionization enthalpy of B is greater than that of Be.

Explanation

To determine which statements are correct regarding ionization enthalpies, we will analyze each statement based on the electronic configurations and the stability of the resulting ions after electron removal.

### Step-by-Step Solution:

1. **Statement 1: The second ionization enthalpy of oxygen is greater than that of fluorine.**
- **Electronic Configuration:**
- Oxygen: \(1s^2 2s^2 2p^4\)
- Fluorine: \(1s^2 2s^2 2p^5\)
- **Analysis:**
- For oxygen, removing the first electron gives \(O^+\) with configuration \(1s^2 2s^2 2p^3\) (half-filled p-orbital).
- For fluorine, removing the first electron gives \(F^+\) with configuration \(1s^2 2s^2 2p^4\).
- The second ionization of oxygen involves removing an electron from a half-filled p-orbital, which is stable. Hence, the second ionization enthalpy of oxygen is indeed greater than that of fluorine.
- **Conclusion:** This statement is **true**.

2. **Statement 2: The third ionization enthalpy of phosphorus is greater than that of chlorine and aluminum.**
- **Electronic Configuration:**
- Phosphorus: \(Ne 3s^2 3p^3\)
- Aluminum: \(Ne 3s^2 3p^1\)
- Chlorine: \(Ne 3s^2 3p^5\)
- **Analysis:**
- Removing three electrons from phosphorus results in \(P^{3+}\) with configuration \(3p^0\) (stable).
- For aluminum, removing three electrons leads to a noble gas configuration (\(Ne\)), which is also stable.
- However, removing electrons from phosphorus disrupts a half-filled p-orbital, which is energetically unfavorable, making its third ionization enthalpy higher than that of aluminum and chlorine.
- **Conclusion:** This statement is **true**.

3. **Statement 3: The first ionization enthalpy of aluminum is slightly greater than that of gallium.**
- **Electronic Configuration:**
- Aluminum: \(Ne 3s^2 3p^1\)
- Gallium: \(Ar 3d^{10} 4s^2 4p^1\)
- **Analysis:**
- The outermost electron in aluminum is in the 3p subshell, while in gallium it is in the 4p subshell.
- Electrons in the 3rd shell are closer to the nucleus than those in the 4th shell, resulting in a stronger attraction and higher ionization energy for aluminum.
- **Conclusion:** This statement is **true**.

4. **Statement 4: The second ionization enthalpy of boron is greater than that of beryllium.**
- **Electronic Configuration:**
- Boron: \(1s^2 2s^2 2p^1\)
- Beryllium: \(1s^2 2s^2\)
- **Analysis:**
- The second ionization enthalpy of boron involves removing an electron from the stable \(2s^2\) configuration after the first ionization, while for beryllium, it involves removing an electron from a full \(2s^2\) configuration.
- Removing an electron from beryllium is more difficult because it disrupts a full subshell, making its second ionization enthalpy higher than that of boron.
- **Conclusion:** This statement is **false**.

### Final Conclusion:
The correct statements are **1, 2, and 3**.