How many of these molecules get dimerise by 3c - 4e bonds
`BeCl_(2), AlCl_(3), BH_(3), BeH_(2), Icl_(3), CH_(3)COOH`

Answer

To determine how many of the given molecules dimerize by forming 3-center 4-electron bonds, we will analyze each molecule step by step.

### Step 1: Analyze BeCl₂ (Beryllium Dichloride)
1. **Valence Electrons**: Beryllium has 2 valence electrons, and each chlorine has 7, giving a total of \(2 + 2 \times 7 = 16\) valence electrons.
2. **Hybridization**: Beryllium in BeCl₂ is sp hybridized, meaning it has 2 hybrid orbitals and 2 vacant p orbitals.
3. **Dimerization**: BeCl₂ dimerizes to form a structure where two beryllium atoms are connected by chlorine atoms. Each beryllium atom shares its vacant p orbitals with chlorine, forming a 3-center 4-electron bond.
4. **Conclusion**: BeCl₂ forms a 3-center 4-electron bond upon dimerization.

### Step 2: Analyze AlCl₃ (Aluminum Chloride)
1. **Valence Electrons**: Aluminum has 3 valence electrons, and each chlorine has 7, giving a total of \(3 + 3 \times 7 = 24\) valence electrons.
2. **Hybridization**: AlCl₃ is sp² hybridized, indicating one vacant p orbital.
3. **Dimerization**: AlCl₃ dimerizes to form Al₂Cl₆, where each aluminum atom shares its vacant p orbital with chlorine atoms, forming a 3-center 4-electron bond.
4. **Conclusion**: AlCl₃ forms a 3-center 4-electron bond upon dimerization.

### Step 3: Analyze BH₃ (Boron Hydride)
1. **Valence Electrons**: Boron has 3 valence electrons, and each hydrogen has 1, giving a total of \(3 + 3 \times 1 = 6\) valence electrons.
2. **Hybridization**: BH₃ is sp² hybridized.
3. **Dimerization**: BH₃ does not dimerize to form a 3-center 4-electron bond; instead, it forms a 3-center 2-electron bond (banana bond).
4. **Conclusion**: BH₃ does not form a 3-center 4-electron bond.

### Step 4: Analyze BeH₂ (Beryllium Hydride)
1. **Valence Electrons**: Beryllium has 2 valence electrons, and each hydrogen has 1, giving a total of \(2 + 2 \times 1 = 4\) valence electrons.
2. **Hybridization**: BeH₂ is also sp hybridized.
3. **Dimerization**: BeH₂ forms a polymeric structure with 3-center 2-electron bonds.
4. **Conclusion**: BeH₂ does not form a 3-center 4-electron bond.

### Step 5: Analyze ICl₃ (Iodine Trichloride)
1. **Valence Electrons**: Iodine has 7 valence electrons, and each chlorine has 7, giving a total of \(7 + 3 \times 7 = 28\) valence electrons.
2. **Hybridization**: ICl₃ is sp³d hybridized.
3. **Dimerization**: ICl₃ dimerizes to form a structure where the chlorine atoms donate lone pairs to the vacant orbitals of iodine, forming a 3-center 4-electron bond.
4. **Conclusion**: ICl₃ forms a 3-center 4-electron bond upon dimerization.

### Step 6: Analyze CH₃COOH (Acetic Acid)
1. **Valence Electrons**: Acetic acid has 4 (from C) + 1 (from H) + 4 (from O) = 9 valence electrons.
2. **Dimerization**: Acetic acid primarily forms dimers through hydrogen bonding, not through 3-center 4-electron bonds.
3. **Conclusion**: CH₃COOH does not form a 3-center 4-electron bond.

### Final Conclusion
The molecules that dimerize to form 3-center 4-electron bonds are:
1. BeCl₂
2. AlCl₃
3. ICl₃

Thus, the total number of molecules that form 3-center 4-electron bonds is **3**.