Question

Compound ${\text{BC}}{{\text{l}}_{\text{3}}}$does not exist as dimer but ${\text{B}}{{\text{H}}_{\text{3}}}$exist as dimer (${{\text{B}}_{\text{2}}}{{\text{H}}_{\text{6}}}$) because:
A. chlorine is more electronegative than hydrogen
B. there is ${{\text{P}}_\pi } - {{\text{P}}_\pi }$ back bonding in ${\text{BC}}{{\text{l}}_{\text{3}}}$but ${\text{B}}{{\text{H}}_{\text{3}}}$does not contain such multiple bonding.
C. large-sized chlorine atoms do not fit in between the small boron atoms whereas small-sized hydrogen atoms fit between boron atoms
D. none of these

Answer 1

To answer we should know the reason for dimerization and we should also know the requirements for dimerization. As during dimerization two molecules combine to form dimer so, the steric hindrance is very important. We will check the size of atoms and steric hindrance to find the reason.

Complete step-by-step answer:
The formation of the dimer is known as dimerization.
The reason for dimerization:
Every atom forms a bond to complete its octet. Sometimes even after the formation of a molecule, the octet of the atom remains incomplete so the molecule dimerizes to complete its octet.
Requirements for dimerization:
The atom of the molecule should have a vacant orbital to gain electrons.
The formation of the dimer should not cause steric hindrance.
Electronegativity does not affect the formation of dimer so, the electronegativity of chlorine is not the reason for the non-existence of ${\text{BC}}{{\text{l}}_{\text{3}}}$dimer.
Chlorine atoms require only one electron to complete its p-orbital which is donated by boron. So, chlorine has vacant d-orbitals for the back bonding. So, ${{\text{P}}_\pi } - {{\text{P}}_\pi }$ back bonding is not the reason for non-existence of ${\text{BC}}{{\text{l}}_{\text{3}}}$ dimer.
Hydrogen is of small size, so it easily forms a bridge without causing steric hindrance whereas the size of chlorine is larger than boron so it is not able to form a bridge. So, due to the large size of chlorine ${\text{BC}}{{\text{l}}_{\text{3}}}$does not exist as a dimer.
So, ${\text{BC}}{{\text{l}}_{\text{3}}}$does not exist as dimer but ${\text{B}}{{\text{H}}_{\text{3}}}$exists as a dimer (${{\text{B}}_{\text{2}}}{{\text{H}}_{\text{6}}}$) because of large size of chlorine.

Therefore, option (C) large-sized chlorine atoms do not fit in between the small boron atoms whereas small-sized hydrogen atoms fit between boron atoms, which is correct.

Note: The absence of back bonding is also a requirement for dimerization. Back bonding is also done to complete the octet. For dimerization molecules should not have back bonding. After the formation of the sigma bond, the donation of an electron from the central atom to the surrounding or the surrounding to the central atom is known as back bonding. Boron halides show back bonding. In boron halides, boron trifluoride shows strong back bonding.