2018 H2 Chem P1 Q2 — Chemical bonding Solution

The answer is D.
Both \(\text{Al}_2\text{Cl}_6\) and gaseous \(\text{AlCl}_3\) are simple covalent molecules.
• In \(\text{AlCl}_3\), Al forms three covalent bonds to Cl and has only 6 electrons in its valence shell, so it is electron-deficient.
• In \(\text{Al}_2\text{Cl}_6\), each bridging Cl donates a lone pair into the empty orbital of the other Al atom, forming two dative bonds. A dative arrow points from the donor (Cl) to the acceptor (Al).
In D both dative arrows point from the bridging Cl to Al, and \(\text{AlCl}_3\) is drawn as a covalent molecule.
• A is wrong: the dative arrows point from Al to Cl, and \(\text{AlCl}_3\) is drawn as ionic \(\text{Al}^{3+}3\text{Cl}^-\).
• B is wrong: the dimer is correct, but \(\text{AlCl}_3\) is drawn as ionic. The small, highly charged \(\text{Al}^{3+}\) polarises the electron cloud of \(\text{Cl}^-\) so strongly that the bonding is covalent.
• C is wrong: \(\text{AlCl}_3\) is correct, but the dative arrows point from Al to Cl. Al has no lone pair to donate; it is the acceptor.