Assertion: The first ionisation energy of `Be` is greater than that of `B`.
Reason: 2p-orbital is lower in energy than 2s-orbital.

Options

A. Both Statement I and Statement II are correct, Statement II is the correct explanation of Statement I
B. Both Statement I and Statement II are correct, Statement II is not the correct explanation of Statement I
C. Statement I is correct, Statement II is the incorrect
D. Statement I is incorrect, Statement II is the correct

Explanation

To solve the assertion and reason question regarding the first ionization energy of beryllium (Be) and boron (B), we will analyze both statements step by step.

### Step 1: Understand the Assertion
The assertion states that the first ionization energy of beryllium (Be) is greater than that of boron (B).

- **Ionization Energy**: This is the energy required to remove an electron from an atom in its gaseous state. A higher ionization energy means it is more difficult to remove an electron.

### Step 2: Compare the Electronic Configurations
- Beryllium (Be) has an atomic number of 4, and its electronic configuration is:
\[ \text{Be: } 1s^2 \, 2s^2 \]
- Boron (B) has an atomic number of 5, and its electronic configuration is:
\[ \text{B: } 1s^2 \, 2s^2 \, 2p^1 \]

### Step 3: Analyze the Stability of the Configurations
- Beryllium has a completely filled 2s subshell, which is a stable configuration.
- Boron has one electron in the 2p subshell. When boron loses this electron, it achieves a stable configuration of:
\[ \text{B}^+: 1s^2 \, 2s^2 \]
This configuration is stable because the 2s subshell is fully filled.

### Step 4: Evaluate the Ionization Energies
- Since boron can easily lose its single 2p electron to achieve a stable configuration, it has a lower ionization energy compared to beryllium, which does not have a favorable configuration to lose an electron easily.
- Therefore, the assertion that the first ionization energy of Be is greater than that of B is **correct**.

### Step 5: Analyze the Reason
The reason states that the 2p orbital is lower in energy than the 2s orbital. This statement is **incorrect**.

- In atomic structure, the 2s orbital is lower in energy than the 2p orbital. This is a fundamental principle of quantum mechanics and electron configuration.
- Therefore, the reason provided does not support the assertion.

### Conclusion
- The assertion is **true**: The first ionization energy of Be is greater than that of B.
- The reason is **false**: The 2p orbital is not lower in energy than the 2s orbital.

### Final Answer
- Assertion: True
- Reason: False