Three-Centre Two-Electron Bonding and Lewis Acidity in Group 13 Halides and Hydrides

To determine the correct statement(s), let us analyze each option individually:

1. Analysis of Option [A]: Trimethylaluminium, $\text{Al(CH}_3\text{)}_3$, exists in the vapor state and non-polar solvents as a dimer, $\text{Al}_2\text{(CH}_3\text{)}_6$. Since the methyl group ($-\text{CH}_3$) does not possess any lone pair of electrons, the bridge bonding cannot occur via coordinate covalent bonding. Instead, each bridging methyl group shares its bonding electron pair with two aluminium atoms, forming two three-centre two-electron ($3c\text{-}2e$) $\text{Al}-\text{C}-\text{Al}$ bonds. Therefore, Option [A] is correct.

2. Analysis of Option [B]: Borane ($\text{BH}_3$) is an electron-deficient species and dimerizes to form diborane ($\text{B}_2\text{H}_6$). In the structure of $\text{B}_2\text{H}_6$, four terminal $\text{B}-\text{H}$ bonds are standard two-centre two-electron ($2c\text{-}2e$) covalent bonds, while the two bridging hydrogen atoms form two three-centre two-electron ($3c\text{-}2e$) $\text{B}-\text{H}-\text{B}$ bridge bonds (often called banana bonds). Therefore, Option [B] is correct.

3. Analysis of Option [C]: Aluminium chloride ($\text{AlCl}_3$) dimerizes to form $\text{Al}_2\text{Cl}_6$. Unlike the methyl group or hydrogen, the chlorine atom possesses lone pairs of electrons. In $\text{Al}_2\text{Cl}_6$, each of the two bridging chlorine atoms donates a lone pair of electrons to the empty $p$-orbital of an adjacent aluminium atom via a coordinate bond. Each $\text{Al}-\text{Cl}-\text{Al}$ bridge thus involves 4 electrons (a total of two electron pairs), classifying it as a three-centre four-electron ($3c\text{-}4e$) bond rather than a $3c\text{-}2e$ bond. Therefore, Option [C] is incorrect.

4. Analysis of Option [D]: Boron has a smaller covalent radius and higher electronegativity compared to aluminium, giving the boron atom in $\text{BCl}_3$ a higher charge density and a lower-lying vacant $2p$ orbital compared to the $3p$ vacant orbital in $\text{AlCl}_3$. Furthermore, $\text{AlCl}_3$ dimerizes into $\text{Al}_2\text{Cl}_6$ to satisfy its octet via coordinate bonds, whereas $\text{BCl}_3$ remains monomeric. Consequently, $\text{BCl}_3$ acts as a stronger Lewis acid than $\text{AlCl}_3$ towards electron donors. Therefore, Option [D] is correct.

Conclusion: The correct statements are A, B, and D.