Boron atom being small in size is unable to accommodate four large-sized halogen atoms (except `F`) around it. However, because of its large size, `Al` can easily accommodate four large-sized `Cl` atoms around it. Since in both `BX_(3)` and `AlCl_(3)`, there are only six electrons present in the valence shell of `B` and `Al`, both are electron deficient compound. `BX_(3)` compounds reduce their electron deficiency by accepting a pair of electron from the filled `np` orbital of the halogen atom in the vacant `2p` orbital of `B`, thus forming `p pi- p pi` back bonding.
However, `Al` accepts a pair of electron from `3p` orbital of `Cl` (from another `AlCl_(3)` molecule) in its vacant `3p` orbital. As a result, `AlCl_(3)` exist as a chlorine-bridged dimeric structure.
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