# Why does AlCl3 dimerise to Al2Cl6 at low temperatures?

I know that it has something to do with HOMOs and LUMOs (I can't what the HOMO/LUMO for AlCl3 is, though).

• At low temprature, $\ce{AlCl3}$ exists as a close packed lattice of $\ce{Cl-}$ with $\ce{Al^3+}$ occupying octahedral holes. On heating $\ce{Al2Cl6}$ species are formed. – Kishore S Shenoy Dec 15 '16 at 6:47
• I agree with @KishoreSShenoy, OP has it the other way around, but why doesn't it dimerize to form a macromolecular structure? – Andrea Rowlatt Nov 1 at 12:57

There are certain stable compounds where the octet rule is not obeyed, that is, their valence shells contain more or less than 8 electrons. The compounds containing less than 8 electrons in the valence shells are called electron deficient compounds such as $\ce{AlCl3}$, where the aluminum atom has 6 valence electrons. It has been suggested that two aluminum atoms complete their octet by accepting a pair of electrons from chlorine atoms:

So, that is why $\ce{AlCl3}$ dimerises to $\ce{Al2Cl6}$ at low temperatures.

• This does not answer why this happens only at low temperatures. – Jori Nov 19 '14 at 15:27
• But why at low temperatures? Also; does it have anything to do with HOMOs or LUMOs at all? Thank you, though. – Mick Jenkins Nov 19 '14 at 15:44
• I think that the highest occupied molecular orbital (HOMO) of one AlCl3 (consisting of non-bonding electron pair on chloride) interacts with the lowest unoccupied molecular orbital (LUMO) of the other molecule AlCl3 (consisting of the unoccupied case on Al). This interaction leads to the formation of a sigma bond between Al and Cl. – Yomen Atassi Nov 19 '14 at 19:43
• The increase of temperature may decrease the gap between the HOMO and LUMO, and this may lowers the interaction and disfavor the bond formation. – Yomen Atassi Nov 19 '14 at 21:02
• Why isn't this dimerization macromolecular? It seems like it should be – Andrea Rowlatt Nov 1 at 12:55

Perhaps, we can employ the Gibbs free energy equation for spontaneous reactions...

$$\Delta G=\Delta H-T\cdot\Delta S$$

Since the dimerisation process involves the formation of bonds, $\Delta H$ should be negative, while $\Delta S$ will be negative too, since two $\ce{AlCl3}$ molecules become one $\ce{Al2Cl6}$. As such, when temperature is low, the magnitude of $T\cdot\Delta S$ will be smaller, and $\Delta G$ will be negative or more negative, thus favouring the dimerisation process.