The `IE_(1)` of Be is greater than that of B. [T/F]

Answer

To determine whether the statement "The first ionization energy (IE₁) of Be is greater than that of B" is true or false, we can analyze the ionization energies of beryllium (Be) and boron (B) step by step.

### Step-by-Step Solution:

1. **Identify the Elements and Their Groups**:
- Beryllium (Be) is in Group 2 (alkaline earth metals).
- Boron (B) is in Group 13 (p-block element).

2. **Understand Ionization Energy Trend**:
- Ionization energy generally increases across a period (from left to right) due to increased nuclear charge, which holds the electrons more tightly.
- However, there are exceptions due to electron configurations.

3. **Determine the Electron Configurations**:
- Beryllium (atomic number 4):
- Electron configuration: 1s² 2s²
- It has 2 electrons in the outermost shell (2s).
- Boron (atomic number 5):
- Electron configuration: 1s² 2s² 2p¹
- It has 2 electrons in the 2s and 1 electron in the 2p.

4. **Analyze Stability of Electron Configurations**:
- Beryllium has a completely filled 2s subshell (2 electrons), which is a stable configuration.
- Boron has a partially filled 2p subshell (1 electron), which is less stable compared to the filled 2s subshell.

5. **Consider the Energy Required for Ionization**:
- Removing an electron from a stable, filled subshell (like 2s in Be) requires more energy than removing an electron from a less stable, partially filled subshell (like 2p in B).
- Therefore, the first ionization energy of beryllium (Be) is expected to be higher than that of boron (B).

6. **Conclusion**:
- Based on the analysis, the statement "The first ionization energy of Be is greater than that of B" is **True**.