To solve the question, we need to analyze both statements regarding the `F2` molecule and determine their validity.
### Step-by-Step Solution:
1. **Understanding the Molecular Orbital Configuration of `F2`:**
- The total number of electrons in a `F2` molecule is 18 (since each fluorine atom contributes 9 electrons).
- The molecular orbital (MO) configuration for `F2` can be written as:
- σ1s², σ*1s², σ2s², σ*2s², σ2p_z², π2p_x², π2p_y², π*2p_x², π*2p_y².
- Here, the asterisk (*) indicates antibonding orbitals.
2. **Counting Electrons in Bonding and Antibonding Orbitals:**
- **Bonding Electrons:**
- From the configuration, the bonding MOs are σ1s, σ2s, σ2p_z, π2p_x, and π2p_y.
- Total bonding electrons = 2 (σ1s) + 2 (σ2s) + 2 (σ2p_z) + 2 (π2p_x) + 2 (π2p_y) = 10 electrons.
- **Antibonding Electrons:**
- The antibonding MOs are σ*1s, σ*2s, π*2p_x, and π*2p_y.
- Total antibonding electrons = 2 (σ*1s) + 2 (σ*2s) + 2 (π*2p_x) + 2 (π*2p_y) = 8 electrons.
3. **Calculating the Bond Order:**
- The bond order (B.O.) is calculated using the formula:
\[
\text{Bond Order} = \frac{(\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons})}{2}
\]
- Substituting the values:
\[
\text{Bond Order} = \frac{(10 - 8)}{2} = \frac{2}{2} = 1
\]
- Therefore, the bond order of `F2` is indeed 1.
4. **Analyzing the Second Statement:**
- The second statement claims that in `F2`, the number of electrons in the antibonding M.O. is two less than in the bonding M.O.
- We found that the number of bonding electrons is 10 and the number of antibonding electrons is 8.
- The difference is indeed:
\[
10 - 8 = 2
\]
- Thus, the second statement is also true.
5. **Conclusion:**
- Both statements are true, and the second statement correctly explains the first statement.
### Final Answer:
- **Statement 1:** True (The bond order of `F2` is 1.)
- **Statement 2:** True (In `F2`, the number of electrons in the antibonding M.O. is two less than in the bonding M.O.)
