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In order to attain stability $\ce{AlCl3}$ dimerises to $\ce{Al2Cl6}$ whose structure is depicted as: enter image description here

I want to understand why one of the bond angles is $79^\circ$ and the other $118^\circ$. Is it possible to justify this using Bent's rule?

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  • $\begingroup$ Related: chemistry.stackexchange.com/questions/34842/… $\endgroup$ Oct 17, 2017 at 7:23
  • $\begingroup$ chemistry.stackexchange.com/questions/19715/… $\endgroup$ Oct 17, 2017 at 7:23
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    $\begingroup$ @abcd I see there were two links, the first one is the one I meant. Here is the important bit for your question: "So after all we have two kinds of bonds in the dimers four terminal X−Cl and four bridging X−μCl bonds. Therefore the most accurate description is with formal charges (also the simplest)." So, if it is correct to describe the dimer with formal charges, then the deviation in bond angles is in perfect compliance with Bent's rule. $\endgroup$
    – Stian
    Oct 17, 2017 at 15:49
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    $\begingroup$ Related: pubs.acs.org/doi/abs/10.1021/jp9842042 $\endgroup$ Feb 27, 2018 at 14:46
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    $\begingroup$ @StianYttervik I think Martin's calculation does not really address why the other Cl-Al-Cl angle is 118. Sure, bent's rule will explain why the inner Cl-Al-Cl is 79 < tetrahedral, it does not explain why the other angle is larger than tetrahedral (and using bent's rule there will mean those Cl will have more negative formal charges than the bridging Cl) $\endgroup$
    – Secret
    Mar 3, 2018 at 17:39

1 Answer 1

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The bonding in aluminum chloride ought to be considered largely ionic. The electronegativity of $\ce{Al}$ is $1.47$ vs $2.83$ for $\ce{Cl}$ (for comparison: $\ce{C}$ $2.50$; $\ce{Mg}$ $1.23$; $\ce{Na}$ $1.01$). Molecular orbitals and covalent bonding can be added for a fuller picture of electron densities.

Aluminum chloride solid (Melting point $\pu{193^\circ C}$, under pressure) has a cubic close-packed array of chlorine atoms with only $2/3$ of the octahedral holes between every other pair of chlorine planes occupied by the aluminum atoms. At low temperatures in the vapor state, aluminum chloride is a tetrahedrally coordinated dimer, and at very high temperatures it is a trigonal monomer. So aluminum can easily have $3$, $4$ or $6$ coordination. Whatever directionality the bonds have from hybridization of the $\ce{3s}$ and $\ce{3p}$ orbitals is mixed with plain electrostatic (ionic) attraction.

Imagine putting just the 6 chlorine atoms together. There will be two tetrahedral holes for aluminum (see image below). The radius ratio $$\frac{\ce{Al^3+}}{\ce{Cl-}} = \frac{\pu{50 pm}}{\pu{181 pm}} = 0.276$$ is well within the stability range ($0.225 - 0.414$) for tetrahedral coordination. So you might expect all the $\ce{Cl-Al-Cl}$ bond angles to be $109^\circ$.

But while the void for the aluminum is big enough, the distance between the internal chlorine atoms would be only $\pu{300 pm}$ in this configuration; the chlorine atoms are squeezed (ionic radius $\pu{181 pm}$). In addition, the $\ce{Al-Al}$ distance is only $\pu{250 pm}$ and the $\ce{Al^3+\bond{-}Al^3+}$ repulsion would be extreme. (In the solid, the octahedral holes are farther apart) So both aluminum atoms and the internal chlorine atoms are pushed apart, consistent with imagining more $\ce{p}$ character in the central bonds and therefore less $\ce{p}$ character in the external chlorine atoms, so $\ce{Al}$ goes to $\ce{sp^2}$ (or nearly so). The internal $\ce{Cl-Al-Cl}$ bond angle is even less than $90^\circ$ (less than you would expect for two $\ce{p}$ orbitals), so π bonding needs to be considered.

Aluminum chloride is a strong Lewis acid. Its solubility in organic solvents is due to formation of complexes. It is not soluble in the way that carbon tetrachloride or hexachloroethane are soluble. The salt-like (ionic) character of solid aluminium chloride is still present in $\ce{Al2Cl6}$; the aluminum atoms are surrounded by a cocoon of chlorides. Solid aluminum chloride can be ground into particles and nano particles; $\ce{Al2Cl6}$ is like a pico particle of salt.

edge-sharing tetraeders in Al2Cl6

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  • $\begingroup$ (1) Did you mean " At low temperatures in the solid state,"? (in the second paragraph) (2) Also, "So aluminum can easily have 3, 4 or 6 coordination. " do you have an example of a 6 coordination aluminium compound? $\endgroup$ Mar 2, 2018 at 11:14
  • $\begingroup$ In solid aluminum chloride, the aluminum is octahedrally coordinated - in the holes between two close packed chlorine atoms. $\endgroup$ Mar 2, 2018 at 16:55
  • $\begingroup$ Well, JD Lee says that the compound is ionic at low temperatures or when dissolved in waer. $\endgroup$
    – Archer
    Oct 7, 2018 at 2:12

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