The molecule , which are electron deficient having tendency to form back bonding, in which transfer of lone pair of electron from filled orbital of an atom to the vacant orbital of the adjacent atom. While some electron deficient species which are not capable to form back bonding, then driving forces occurs for dimerisation/polymerisation to achieve octet temporily or permanently by the monomeric species.
Which of the following matching is INCORRECT between molecule and the orbitals of atoms involved in bridge bond.

Options

A. `B_(2)H_(6) (vap.) tosp^(3) -s -sp^(3)`
B. `Al_(2)(CH_(3))_(6) (vap)to sp^(3) -sp^(3) -sp^(3)`
C. `BeH_(2)(s) to sp^(3)-s-sp^(3)`
D. `Be_(2)H_(4) (vap) to sp^(3) -s-sp^(3)`

Answer

To solve the question regarding the incorrect matching between molecules and the orbitals of atoms involved in bridge bonding, we will analyze each option step by step.

### Step-by-Step Solution:

1. **Understanding Back Bonding:**
- Back bonding occurs when an electron-deficient atom (A) has a vacant orbital and an adjacent atom (B) donates a lone pair from its filled orbital to A's vacant orbital. This helps A to complete its octet.

2. **Analyzing Each Option:**
- We need to evaluate the hybridizations and the types of bonding in each molecule mentioned in the options.

3. **Option 1: B2S6**
- In the vapor phase, B2S6 exists as a dimer, and in the solid phase, it forms a polymer.
- The hybridization of boron in BH3 is sp2, and when forming a bridge bond, it can be considered sp3 due to the presence of four effective bonds (three from B-H and one from the bridge bond).
- **Conclusion:** This option is correct.

4. **Option 2: Al2(CH3)6**
- Aluminum in this compound can also form bridge bonds with the methyl groups.
- Each aluminum atom forms four bonds (three from CH3 and one from the bridge bond), leading to sp3 hybridization.
- **Conclusion:** This option is correct.

5. **Option 3: BeH2 (solid)**
- In the solid state, BeH2 forms a polymer where each beryllium atom forms four bonds (two from hydrogen and two from bridging hydrogens).
- This gives it an sp3 hybridization.
- **Conclusion:** This option is correct.

6. **Option 4: BeH2 (vapor)**
- In the vapor phase, BeH2 exists as a dimer. Each beryllium atom forms three sigma bonds (two from hydrogen and one from the bridge bond).
- This results in sp2 hybridization, not sp3 as stated in the option.
- **Conclusion:** This option is incorrect.

### Final Answer:
The incorrect matching is found in **Option 4: BeH2 (vapor)**, where the hybridization is incorrectly stated as sp3 instead of sp2.