In which of the following back - bonding is possible

Options

A. `BF_3`
B. `(SiH_3)_3N`
C. `NH_3`
D. both A and B

Answer

To determine in which of the given compounds back-bonding is possible, we need to analyze each compound based on their electronic configurations and the presence of empty orbitals.

### Step-by-Step Solution:

1. **Identify the Compounds:**
Let's consider the compounds mentioned in the video transcript:
- A: Boron trifluoride (BF₃)
- B: Silicon hydride (SiH₄)
- C: Ammonia (NH₃)

2. **Analyze Boron Trifluoride (BF₃):**
- Boron (B) has an atomic number of 5, with an electronic configuration of 1s² 2s² 2p¹, resulting in 3 valence electrons.
- Fluorine (F) has an atomic number of 9, with an electronic configuration of 1s² 2s² 2p⁵, resulting in 7 valence electrons.
- BF₃ is electron-deficient (6 electrons total), acting as a Lewis acid.
- Fluorine can donate a pair of electrons to boron, allowing back-bonding to occur.

**Conclusion:** Back-bonding is possible in BF₃.

3. **Analyze Silicon Hydride (SiH₄):**
- Silicon (Si) has an atomic number of 14, with an electronic configuration of 1s² 2s² 2p⁶ 3s² 3p², resulting in 4 valence electrons.
- Nitrogen (N) has an atomic number of 7, with an electronic configuration of 1s² 2s² 2p³, resulting in 5 valence electrons.
- In SiH₄, silicon has empty 3d orbitals, and nitrogen can donate a lone pair to silicon, allowing back-bonding to occur.

**Conclusion:** Back-bonding is possible in SiH₄.

4. **Analyze Ammonia (NH₃):**
- In NH₃, nitrogen has a lone pair but hydrogen does not have any empty orbitals to accept electrons.
- Therefore, there is no possibility for back-bonding in ammonia.

**Conclusion:** Back-bonding is not possible in NH₃.

5. **Final Conclusion:**
- Back-bonding is possible in compounds A (BF₃) and B (SiH₄), but not in C (NH₃).
- Therefore, the correct answer is that back-bonding is possible in both A and B.