`AlCl_(3)` forms a dimer but `BC l_(3)` cannot-explain.

Explanation

The size of Al is much larger than that of B. hence Al can easily accommodate 4Cl-atoms around it. In `AlCl_(3)`, as there are 6 electrons around the Alatom, it completes its octet by accepting a lone pair of electrons from Cl-atom of octet by accepting a lone pair fo electrons from Cl-atom of an adjacent molecule, As a result, `AlCl_(3)` exists as a chlorine-bridged dimer forming `Al_(2)Cl_(6)`. on the other hand, B is comparatively smaller in size. though it has an incomplete octet in `BC l_(3)`, it cannot accommodate the fourth chlorine atom around it, owing to the large size of the Cl-atom. thus, `BC l_(3)` does not exist as a dimer.