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 can dimerize by forming three-center four-electron (3c-4e) bonds, we will analyze each molecule one by one.

### Step 1: Analyze BeCl₂
- **Structure**: BeCl₂ has a linear structure with beryllium in the center and two chlorine atoms bonded to it.
- **Hybridization**: The hybridization of Be in BeCl₂ is sp.
- **Dimerization**: When two BeCl₂ molecules come together, the lone pairs on the chlorine atoms can donate electrons to the beryllium atom, forming a dimer. This results in a three-center four-electron bond involving two beryllium atoms and three chlorine atoms.
- **Conclusion**: BeCl₂ can dimerize by forming a 3c-4e bond.

### Step 2: Analyze AlCl₃
- **Structure**: AlCl₃ has a trigonal planar structure with aluminum in the center and three chlorine atoms bonded to it.
- **Hybridization**: The hybridization of Al in AlCl₃ is sp².
- **Dimerization**: AlCl₃ can also dimerize. The vacant p orbital on aluminum can accept lone pairs from the chlorine atoms of another AlCl₃ molecule. This results in a structure where aluminum and chlorine form a three-center four-electron bond.
- **Conclusion**: AlCl₃ can dimerize by forming a 3c-4e bond.

### Step 3: Analyze BH₃
- **Structure**: BH₃ has a trigonal planar structure with boron in the center and three hydrogen atoms.
- **Hybridization**: The hybridization of B in BH₃ is sp².
- **Dimerization**: BH₃ does not dimerize to form a 3c-4e bond. Instead, it forms a three-center two-electron bond (banana bond) when it interacts with other BH₃ molecules.
- **Conclusion**: BH₃ does not dimerize by forming a 3c-4e bond.

### Step 4: Analyze BeH₂
- **Structure**: BeH₂ has a linear structure similar to BeCl₂.
- **Hybridization**: The hybridization of Be in BeH₂ is sp.
- **Dimerization**: BeH₂ can also form a polymeric structure but does not form a stable 3c-4e bond in the same way as BeCl₂.
- **Conclusion**: BeH₂ does not dimerize by forming a 3c-4e bond.

### Step 5: Analyze ICl₃
- **Structure**: ICl₃ has a trigonal bipyramidal structure with iodine in the center and three chlorine atoms.
- **Hybridization**: The hybridization of I in ICl₃ is sp³d.
- **Dimerization**: ICl₃ can dimerize and form a three-center four-electron bond due to the presence of lone pairs on the iodine atom.
- **Conclusion**: ICl₃ can dimerize by forming a 3c-4e bond.

### Step 6: Analyze CH₃COOH
- **Structure**: CH₃COOH (acetic acid) can form dimers through hydrogen bonding.
- **Dimerization**: The dimerization occurs due to hydrogen bonding, not through the formation of a 3c-4e bond.
- **Conclusion**: CH₃COOH does not dimerize by forming a 3c-4e bond.

### Final Conclusion
The molecules that can dimerize by forming three-center four-electron bonds are:
- BeCl₂
- AlCl₃
- ICl₃

### Summary of Results:
- **Dimerizing Molecules**: BeCl₂, AlCl₃, ICl₃
- **Total Count**: 3 molecules