"hybridization of central atom does not always change due to back bonding". This statement is valid for which of the following compounds?
(i)`CCl_(3)^(-)` (ii)`CCl_(2)` (iii)`O(SiH_(3))_(2)` (iv)`N(SiH_(3))_(3)`

Options

A. (i) and (ii)
B. (i) and (iii)
C. (ii) and (iii)
D. All

Answer

To determine which of the given compounds validates the statement "hybridization of the central atom does not always change due to back bonding," we need to analyze each compound in terms of back bonding and hybridization changes.

### Step-by-Step Solution:

1. **Understanding Back Bonding:**
- Back bonding occurs when a central atom donates a lone pair of electrons to a side atom that has a vacant orbital.
- For back bonding to occur, the central atom must belong to the second period, while the side atom can belong to the second or third period.

2. **Analyzing Each Compound:**
- **(i) CCl₃⁻ (Trichloromethane anion):**
- Central atom: Carbon (C).
- Carbon does not have a lone pair to donate, and chlorine (Cl) is in the third period. Therefore, back bonding cannot occur here.
- **Hybridization:** Carbon has 3 sigma bonds and 1 lone pair (due to the negative charge), leading to a steric number of 4 (sp³). No change in hybridization due to back bonding.

- **(ii) CCl₂ (Dichloromethane):**
- Central atom: Carbon (C).
- Similar to CCl₃⁻, carbon does not have a lone pair to donate, and chlorine is in the third period. Thus, back bonding is not applicable.
- **Hybridization:** Carbon has 2 sigma bonds and 2 lone pairs (inferred from the absence of hydrogen), leading to a steric number of 4 (sp³). No change in hybridization due to back bonding.

- **(iii) O(SiH₃)₂ (Disiloxane):**
- Central atom: Oxygen (O).
- Oxygen has 2 lone pairs and can donate one to silicon (Si), which has a vacant orbital. Here, back bonding occurs.
- **Initial Hybridization:** Oxygen has 2 sigma bonds and 2 lone pairs, leading to a steric number of 4 (sp³).
- **After Back Bonding:** One lone pair is donated, resulting in 2 sigma bonds and 1 lone pair, leading to a steric number of 3 (sp²). Hybridization changes from sp³ to sp².

- **(iv) N(SiH₃)₃ (Silazane):**
- Central atom: Nitrogen (N).
- Nitrogen has 1 lone pair and can donate it to silicon, which has vacant orbitals. Back bonding occurs here as well.
- **Initial Hybridization:** Nitrogen has 3 sigma bonds and 1 lone pair, leading to a steric number of 4 (sp³).
- **After Back Bonding:** One lone pair is donated, resulting in 3 sigma bonds and no lone pairs, leading to a steric number of 3 (sp²). Hybridization changes from sp³ to sp².

3. **Conclusion:**
- The statement "hybridization of the central atom does not always change due to back bonding" is valid for compounds (i) CCl₃⁻ and (ii) CCl₂, as back bonding does not occur in these cases.
- For compounds (iii) O(SiH₃)₂ and (iv) N(SiH₃)₃, back bonding occurs, and hybridization changes.

### Final Answer:
The statement is valid for compounds **(i) CCl₃⁻ and (ii) CCl₂**.