33
$\begingroup$

What makes dimerization possible in $\ce{AlCl3}$? Are there 3c-2e bonds in $\ce{Al2Cl6}$ as there are in $\ce{B2H6}$?

$\endgroup$
  • 2
    $\begingroup$ Possible duplicate of Why does AlCl3 dimerise to Al2Cl6 at low temperatures? $\endgroup$ – Mithoron May 7 '16 at 15:51
  • 2
    $\begingroup$ I think the question is related, but I think it is distinctively different also. On the other hand, the answer at the other question does not really explain why the boron compound does not dimerise. $\endgroup$ – Martin - マーチン May 7 '16 at 16:39
  • $\begingroup$ Extra links in decreasing order of relevance: 1, 2, 3, 4, 5, 6. $\endgroup$ – Linear Christmas Dec 19 '16 at 19:36
28
+400
$\begingroup$

Introduction

The bonding situation in $\ce{(AlCl3)2}$ and $\ce{(BCl3)2}$ is nothing trivial and the reason why aluminium chloride forms dimers, while boron trichloride does not, cannot only be attributed to size.

In order to understand this phenomenon we need to look at both, the monomers and the dimers, and compare them to each other.
Understanding the respective bonding situation of the monomers, is key to understand which deficiencies lead to dimerisations.

Computational details

Since I was unable to find any compelling literature on the subject, I ran some calculations of my own. I used the DF-M06L/def2-TZVPP for geometry optimisations. Each structure has been optimised to a local minimum in their respective symmetry restrictions, i.e. $D_\mathrm{3h}$ for the monomers and $C_\mathrm{2v}$ for the dimers.
Analyses with the Natural Bond Orbital model (NBO6 program) and the Quantum Theory of Atoms in Molecules (QTAIM, MultiWFN) have been run on single point energy calculations at the M06/def2-QZVPP//DF-M06-L/def2-TZVPP level of theory.
A rudimentary energy decomposition analysis has been done on that level, too.

Energy decomposition analysis

The dissociation energy of the dimers $\ce{(XY3)2}$ to the monomers $\ce{XY3}$ is defined as the difference of the energy of the dimer $E_\mathrm{opt}[\ce{(XY3)2}]$ and double the energy of the monomer $E_\mathrm{opt}[\ce{XY3}]$ at their optimised (relaxed) geometries $\eqref{e-diss-def}$.
The interaction energy is defined as the difference of energy of the relaxed dimer and double the energy of the monomers in the geometry of the dimer $E_\mathrm{frag}[\ce{(XY3)^{\neq}}]$ $\eqref{e-int-def}$. That basically means breaking the molecule in two parts, but keeping these fragments in the same geometry.
The deformation energy (or preparation energy) is defined as the difference of the energy of the optimised and the non-optimised monomer $\eqref{e-def-def}$. This is the energy required to distort the monomer (in its ground state) to the configuration it will have in the dimer. $$\begin{align} E_\mathrm{diss} &= E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{opt}[\ce{XY3}] \tag1\label{e-diss-def}\\ E_\mathrm{int} &= E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{frag}[\ce{(XY3)^{\neq}}] %\ddag not implemented \tag2\label{e-int-def}\\ E_\mathrm{def} &= E_\mathrm{frag}[\ce{(XY3)^{\neq}}] - E_\mathrm{opt}[\ce{XY3}] \tag3\label{e-def-def}\\ E_\mathrm{diss} &= E_\mathrm{int} + 2E_\mathrm{def}\tag{1'} \end{align}$$

Results & Discussion

The Monomers $\ce{XCl3; X{=}\{B,Al\}}$.

Let's just get the obvious out of the way: Boron is (vdW-radius 205 pm) smaller than aluminium (vdW-radius 240 pm). For comparison chlorine has a vdW-radius of 205 pm, too. That is pretty much reflected in the bond lengths and the chlorine-chlorine distance. \begin{array}{llrrr}\hline &\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline \mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.0 &173.6&--\\ \mathbf{d}(\ce{Cl\bond{~}Cl'})&/\pu{pm} & 356.8 & 300.6 & --\\\hline \mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240 & 205 & 205\\ \mathbf{r}_\mathrm{sing}&/\pu{pm} & 126 & 85 & 99\\ \mathbf{r}_\mathrm{doub}&/\pu{pm} & 113 & 78 & 95\\\hline \end{array}

From this data we can draw certain conclusions without further looking. The boron monomer is much more compact than the aluminium monomer. When we compare the bond lengths to the covalent radii (Pyykkö and Atsumi) we find that the boron chloride bond is about the length that we would expect from a double bond ($\mathbf{r}_\mathrm{doub}(\ce{B}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 173~\pu{pm}$). While the aluminium chloride bond is still significantly shorter than a single bond ($\mathbf{r}_\mathrm{sing}(\ce{Al}) + \mathbf{r}_\mathrm{sing}(\ce{Cl}) = 225~\pu{pm}$), it is still also much longer than a double bond ($\mathbf{r}_\mathrm{doub}(\ce{Al}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 191~\pu{pm}$).

This itself offers compelling evidence, that there is more π-backbonding in $\ce{BCl3}$ than in $\ce{AlCl3}$. Molecular orbital theory offers more evidence for this. In both compounds is a doubly occupied π orbital. The following pictures are for a contour value of 0.05; aluminium (left/top) and boron (right/bottom)

aluminium boron

In numbers, the main contributions are as follows (this is just a representation, not the actual formula): $$\begin{align} \pi(\ce{BCl3}) &= 21\%~\ce{p_{$z$}-B} + \sum_{i=1}^3 26\%~\ce{p_{$z$}-Cl^{$(i)$}}\\ \pi(\ce{AlCl3}) &= 13\%~\ce{p_{$z$}-Al} + \sum_{i=1}^3 29\%~\ce{p_{$z$}-Cl^{$(i)$}} \end{align}$$

There is still some more evidence. The natural atomic charges (NPA of NBO6) fairly well agree with that assesment; aluminium is far more positive than boron. $$\begin{array}{lrr} & \ce{AlCl3} & \ce{BCl3}\\\hline \mathbf{q}(\ce{X})~\text{[NPA]} & +1.4 & +0.3 \\ \mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5 & -0.1 \\\hline %\mathbf{q}(\ce{X})~\text{[QTAIM]} & +2.4 & +2.0 \\ %\mathbf{q}(\ce{Cl})~\text{[QTAIM]} & -0.8 & -0.7 \\\hline \end{array}$$

The analysis in terms of QTAIM also shows that the bonds in $\ce{AlCl3}$ they are predominantly ionic (left/top) while in $\ce{BCl3}$ are predominantly covalent (right/bottom).

aluminium boron

One final thought on the bonding can be supplied with a natural resonance theory analysis (NBO6). I have chosen the following starting configurations and let the program calculate their contribution.

resonance structures

The overall structures in terms of resonance are the same for both cases, that is if you force resonance treatment of the aluminium monomer. Structure A does not contribute, while the others contribute to about 31%. However, when not forced into resonance, structure A is the best approximation of the bonding situation for $\ce{AlCl3}$. In the case of $\ce{BCl3}$ the algorithm finds a hyperbond between the chlorine atoms, a strongly delocalised bond between multiple centres. In this case these are 3-centre-4-electron bonds between the chlorine atoms, resulting from the higher lying degenerated π orbitals.

pi-orbitals of bcl3

This all is quite good evidence that the monomer of boron chloride should be more stable towards dimerisation than the monomer of aluminium.

The Dimers $\ce{(XCl3)2; X{=}\{B,Al\}}$.

The obvious change is that the co-ordination of the central elements goes from trigonal planar to distorted tertrahedral. A look at the geometries will give us something to talk about.

\begin{array}{llrrr}\hline &\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline \mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.7 &175.9&--\\ \mathbf{d}(\ce{X-{\mu}Cl})&/\pu{pm} & 226.1 &198.7&--\\ \mathbf{d}(\ce{Cl\bond{~}{\mu}Cl})&/\pu{pm} & 354.1 & 308.0 & --\\ \mathbf{d}(\ce{{\mu}Cl\bond{~}{\mu}Cl'})&/\pu{pm} & 323.6 & 287.3 & --\\ \mathbf{d}(\ce{B\bond{~}B'})&/\pu{pm} & 315.7 & 274.7 & --\\\hline \mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240 & 205 & 205\\ \mathbf{r}_\mathrm{sing}&/\pu{pm} & 126 & 85 & 99\\ \mathbf{r}_\mathrm{doub}&/\pu{pm} & 113 & 78 & 95\\\hline \end{array}

In principle nothing much changes other than the expected elongation of the bonds that are now bridging. In case of aluminium the stretch is just below 10% and for boron it is slightly above 14%, having a bit more impact. In the boron dimer also the terminal bonds are slightly (> +1%) affected, while for aluminium there is almost no change.

The charges are not really a reliable tool, especially when they are that close to zero as they are for boron. In both cases one can see that charge density is transferred from the bridging chlorine to the central $\ce{X}$.

$$\begin{array}{lrr} & \ce{(AlCl3)2} & \ce{(BCl3)2}\\\hline \mathbf{q}(\ce{X})~\text{[NPA]} & +1.3 & +0.2 \\ \mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5 & -0.1 \\\hline \mathbf{q}(\ce{{\mu}Cl})~\text{[NPA]} & -0.4 & +0.1 \\\hline \end{array}$$

A look at the central four-membered-ring of in terms of QTAIM offers that the overall bonding does not change. In aluminium they get a little more ionic, while in boron they stay largely covalent.

aluminium dimer boron dimer

The NBO analysis offers a maybe quite surprising result. There are no hyperbonds in any of the dimers. While a description in these terms is certainly possible, after all it is just an interpretation tool, it is completely unnecessary. So after all we have two kinds of bonds in the dimers four terminal $\ce{X-Cl}$ and four bridging $\ce{X-{\mu}Cl}$ bonds. Therefore the most accurate description is with formal charges (also the simplest). The notation with the arrows is not wrong, but it does not represent the fact that the bonds are equal for symmetry reasons alone.

lewis structur for dimer

To make this straight: There are no hyperbonds in $\ce{(XCl3)2; X{=}\{B,Al\}}$; this includes three-centre-two-electron bonds, and three-centre-four-electron bonds. And deeper insight to those will be offered on another day.

The differentiation between a dative bond and some other for of bond does not make sense, as the bonds are equal and only introduced by a deficiency of the used description model.

A natural resonance theory for $\ce{(BCl3)2}$ gives us a overall contribution of the main (all single bonds) structure of 46%; while all other structure do contribute, there are too many and their contribution is too little (> 5%). I did not run this analysis for the aluminium case as I did not expect any more insight and I did not want to waste calculation time.

Dimerisation - yes or no

The energies offer us a clear trend. Aluminium likes to dimerise, boron not. However, there are still some things to discuss. I am going to argue for the reaction $$\begin{align} \ce{2XCl3 &-> (XCl3)2}& \Delta E-\mathrm{diss}/E_\mathrm{o}/H/G&. \end{align};$$ therefore if reaction energies are negative the dimerisation is favoured.

The following table includes all calculated energies, including the energy decomposition analysis mentioned at the beginning. All energies are given in $\pu{kJ mol^-1}$. \begin{array}{lrcrcrcrr} \Delta & E_\mathrm{diss} &(& E_\mathrm{int} &+2\times& E_\mathrm{def}&)& E_\mathrm{o} &H &G\\\hline \ce{Al} & -113.5 &(& -224.2 &+2\times& 55.4&)& -114.7 & -60.4 & -230.4\\ \ce{B} & 76.4 &(& -111.2 &+2\times& 93.8&)& 82.6 & -47.1 & 152.5\\\hline \end{array}

The result is fairly obvious at first. The association for aluminium is strongly exergonic, while for boron it is strongly endergonic. While both reactions should be exothermic, stronger for aluminium, the trend for the observed electronic energies ($E_\mathrm{o}$ including the zero-point energy correction) and the (electronic) dissociation energies reflect the overall trend for the Gibbs enthalpies.

While it is fairly surprising how strongly entropy favours association of $\ce{AlCl3}$, it is also surprising how it strongly disfavours it for $\ce{BCl3}$.
A look at the decomposed electronic energy offers great insight into the reasons why one dimer is stable and the other not (at room temperature).
The interaction energy of the fragments is double for aluminium than it is for boron. This can be traced back to the very large difference in the atomic partial charges. One could expect that the electrostatic energy is a lot more attractive for aluminium than it is for boron.
The deformation energy on the other hand clearly reflects the changes in the geometry discussed above. For aluminium there is a smaller penalty resulting from the elongation of the $\ce{Al-Cl}$ bond and pyramidalisation. For boron on the other hand this has a 1.5 times larger effect. The distortion also weakens the π-backbonding, which the additional bonding would need to compensate.
The four-membered-ring is certainly not an ideal geometry and the bridging chlorine atoms come dangerously close.

Conclusion, Summary and TL;DR:

The distortion of the geometry of the monomer $\ce{BCl3}$ cannot be compensated by the additional bonding between the two fragments. Therefore the monomers are more stable than the dimer. Additionally entropy considerations at room temperature favour the monomer, too.

On the other hand, the distortion of the molecular geometry in $\ce{AlCl3}$ is less severe. The gain in interaction energy of the two fragments well overcompensates for the change. Entropy also favours the dimerisation.

While size of the central atom is certainly the distinguishing factor, its impact is only severe on the electronic structure. Steric crowding would not be a problem when the interaction energy would compensate for that. This is quite evident because $\ce{BCl3}$ is still a very good Lewis acid and forms stable compounds with much larger moieties than itself.

References

The used van der Waals radii were taken from S. S. Batsanov Inorg. Mat. 2001, 37 (9), 871-885. And the covalent radii have been taken from P. Pyykkö and M. Atsumi Chem. Eur. J. 2009, 15, 12770-12779.

Computations have been carried out using Gaussian 09 rev D.01 with NBO 6.0. Additional analyses have been performed with MultiWFN 3.3.8. Orbital pictures were generated with the incredible ChemCraft.

$\endgroup$
  • 1
    $\begingroup$ Nice. Although the entropy does intrigue me - that $-T\Delta S$ contribution is huge. I find it counter-intuitive that the entropy change can be so drastically different for two reactions that are fundamentally the same: $\ce{2A <=> A2}$. (Also would have expected the dimers to have $D_\mathrm{2h}$ symmetry.) $\endgroup$ – orthocresol Dec 16 '16 at 13:14
  • 2
    $\begingroup$ Martin, It would be good to perform similar analisys for boron trifluoride. It also is monomeric, by fluorine is oh so smaller than chlorine, so other factors should dominate. $\endgroup$ – permeakra Dec 16 '16 at 13:30
  • 2
    $\begingroup$ @ortho (D2h is correct, don't know where my brains went wrong - it's a long post.) There seems to be a slight numerical error in the calculation which causes (AlCl3)2 to have a tiny imaginary mode. While this usually has no effect, it can lead to wrong corrections and this is more predominant in such small systems. Gaussian has a paper on how they do thermochemistry. I'll look into this further when I get the chance next week, maybe lifting the symmetry restriction will run into the minimum. $\endgroup$ – Martin - マーチン Dec 16 '16 at 13:50
  • 1
    $\begingroup$ @permeakra This is indeed an interesting question, and when I get the chance I'll have a look. I think the reasoning will be very similar as the pi bonding should also increase with a smaller bond length. But we'll see. $\endgroup$ – Martin - マーチン Dec 16 '16 at 13:56
  • 1
    $\begingroup$ @orthocresol entropy is hugely influenced by vibrational zero-energy and it falls rapidly with lowered mass of the atoms involved. $\endgroup$ – permeakra Dec 16 '16 at 18:12
14
+200
$\begingroup$

TL;DR

  • $\ce{BCl3}$ does not dimerize to $\ce{B2Cl6}$ due to a conflict between the short $\ce{B-Cl}$ bond length in $\ce{BCl3}$ $(1.74~\mathring{\mathrm{A}}$ in my calculations$)$ and the long $\ce{B-Cl}$ bond lengths $(\sim 2~\mathring{\mathrm{A}})$ that would be required for the $\ce{B2Cl2}$ core of a $\ce{B2Cl6}$ dimer.

  • $\ce{B2Cl6}$ does appear to have a positive (albeit small) dissociation energy, however, forming a van der Waals complex at a $\ce{B-B}$ distance of $3.63~\mathring{\mathrm{A}}$. Thus, this weakly-bound dimeric form may have the potential to form at very low temperatures.

  • $\ce{Al2Cl6}$ does not possess 2e-3c $\ce{Al-Cl-Al}$ bonds, in contrast to $\ce{B2H6}$. Instead, each of the four $\ce{Al-Cl}$ bonds in the $\ce{Al2Cl2}$ core is moderately 'sub-valent,' carrying $\sim1.5~e^–$.

  • The valence orbital localization and orbital shapes of an optimized $\ce{B2Cl6}$ system with one $\ce{B-Cl}$ bond on each boron constrained to $2.08~\mathring{\mathrm{A}}$ are very similar to those of the fully optimized $\ce{Al2Cl6}$, in agreement with Martin’s results.

These conclusions are based upon ORCA v3.0.3 DFT calculations and subsequent MultiWFN 3.3.7 electron density and ELF topology analysis on the three dimeric systems: $\ce{B2H6}$, $\ce{B2Cl6}$, and $\ce{Al2Cl6}$. Where used, "$\mathrm{E_h}$" indicates the Hartree energy unit.


As a first entry into the question, I performed relaxed surface scans on each dimer system, adjusting the $\ce{B-B}$/$\ce{Al-Al}$ distance from $9.0~\mathring{\mathrm{A}}$ down to $2.0~\mathring{\mathrm{A}}$ ($1.5~\mathring{\mathrm{A}}$ for $\ce{B2H6}$). A representative ORCA input file for these scans is:

! RKS PBE0 def2-SVP def2-SVP/J RIJCOSX D3BJ
! GRID4 GRIDX5 TIGHTOPT
! PRINTBASIS

%output  PrintLevel Small  end

%geom
   Scan
      B 0 4 = 9.0, 1.5, 16
      end
   end

* xyz 0 1
  B -4.91130   1.95714  -6.10613
  H -3.80330   1.95714  -6.10613
  H -5.46530   2.39076  -5.25014
  H -5.46530   1.52351  -6.96212
  B -2.06337  -4.02491   0.00000
  H -0.95537  -4.02491   0.00000
  H -2.61737  -4.79680  -0.57003
  H -2.61737  -3.25303   0.57003
*

Note that unlike in Martin’s calculations, I did not impose any symmetry constraints.

The figure below shows the potential energy curves obtained for these scans (scale increased in the right panel to show $\ce{B2Cl6}$ more clearly; click to enlarge):

dissociation curves

The solid curves are the best-fit Morse potentials, determined by minimizing the sum-squared deviation of the fit from the data:

$$ \begin{array}{ccccc} \hline & D_e~(\mathrm{E_h}) & D_e~\left(\mathrm{kcal\over mol}\right) & a~\left(\mathring{\mathrm{A}}^{-1}\right) & r_e~\left(\mathring{\mathrm{A}}\right) \\ \hline \ce{B2H6} & 0.0867 & 54.41 & 2.3304 & 1.7025 \\ \ce{B2Cl6} & 0.00599 & 3.76 & 0.79832 & 3.9632 \\ \ce{Al2Cl6} & 0.0532 & 33.36 & 0.97947 & 3.0259 \\ \hline \end{array} $$

To note, these data likely do not run along any particular internal reaction coordinate. The images below are of the optimized geometries (left: $\ce{B2H6}$; center: $\ce{B2Cl6}$; right: $\ce{Al2Cl6}$); click each to display an animated GIF of the optimized geometries at each step of the respective scan, which show the irregularities in the progression of the scans (all GIFs pause at the step closest to the optimized geometry):

B2H6   B2Cl6   Al2Cl6

Despite these irregularities, the energy profiles seem qualitatively reasonable. Of particular interest is the fact that $\ce{B2Cl6}$ may actually form a van der Waals complex, enabling it to be bound at a sufficiently low temperature (plotted at the $0.5$ isosurface of the reduced density gradient):

NCI plot for B2Cl6

So, these computations agree with the experimental observation that $\ce{BCl3}$ does not dimerize in the way that $\ce{BH3}$ and $\ce{AlCl3}$ do.


In the course of exploring the $\ce{B2Cl6}$ system, I performed a constrained optimization where I locked one $\ce{B-Cl}$ bond on each boron atom to be considerably longer than its equilibrium value, starting from the above optimized geometry. I arbitrarily chose $2.08~\mathring{\mathrm{A}}$, which is the length of the terminal $\ce{Al-Cl}$ bonds in $\ce{Al2Cl6}$. Once the energetic cost is paid to stretch the two $\ce{B-Cl}$ bonds, an appreciable minimum in the energy profile is exhibited:

B-B scan for relaxed and 'extended' B2Cl6 forms

The data in red is the same $\ce{B2Cl6}$ data from above; that in yellow is with the two $\ce{B-Cl}$ bonds constrained at $2.08~\mathring{\mathrm{A}}$. All energies are relative to the relaxed system at a $9~\mathring{\mathrm{A}}$ $\ce{B-B}$ separation.

The resulting optimized structure under this constraint is below; as above, click the image to view an animated GIF of the progression of the constrained optimization:

'extended' B2Cl6

I suspect that the symmetry-constrained $\ce{B2Cl6}$ species studied by Martin is very similar to this constrained-optimized structure, and if he had relaxed the symmetry constraint he would have obtained the van der Waals complex above. Based on his data, though, it appears my choice of $2.08~\mathring{\mathrm{A}}$ for the stretched $\ce{B-Cl}$ bonds was too large by $0.1~\mathring{\mathrm{A}}$ or so.

In any event, the fictional "pseudo-dissociation curve" for this "extended $\ce{B-Cl}$" system follows much the same form as that for $\ce{Al2Cl6}$:

'extended' B2Cl6 vs Al2Cl6

The '$\ce{B2Cl6}$-extended' data in the above chart was shifted vertically so that $E[9~\mathring{\mathrm{A}}] = 0$, in order to illustrate its qualitative similarity to the $\ce{Al2Cl6}$ data.

So, simply stretching one $\ce{B-Cl}$ bond at each boron atom is sufficient to induce $\ce{B2Cl6}$ to behave in substantially the same fashion as $\ce{Al2Cl6}$.


$\ce{B2H6}$ does show the expected 2e-3c bonding behavior.

ORCA localization of the $\ce{B2H6}$ valence orbitals (omitting the $\ce{B}$ core orbitals) indicates their presence (the bridging hydrogens are 1H and 5H):

----------------------------------------
LOCALIZED MOLECULAR ORBITAL COMPOSITIONS
----------------------------------------
The Mulliken populations for each LMO on each atom are computed
The LMO`s will be ordered according to atom index and type
  (A) Strongly localized MO`s have populations of >=0.950 on one atom
  (B) Two center bond orbitals have populations of >=0.850 on two atoms
  (C) Other MO`s are considered to be `delocalized`
FOUND  -   0 strongly local MO`s
       -   4 two center bond MO`s
       -   2 significantly delocalized MO`s
Bond-like localized orbitals:
MO   5:   7H  -   0.494574  and   4B  -   0.525955
MO   4:   6H  -   0.494586  and   4B  -   0.525936
MO   3:   3H  -   0.494588  and   0B  -   0.525939
MO   2:   2H  -   0.494573  and   0B  -   0.525953
More delocalized orbitals:
MO   7:  0B - 0.269 1H - 0.478 4B - 0.269
MO   6:  0B - 0.269 4B - 0.269 5H - 0.478

Both the localized orbitals and the QTAIM LCP paths are consistent with the 2e-3c bond paradigm, and ELF basin analysis finds attractors at the bridging hydrogen atoms, whose basins are calculated to contain $1.94~e^-$ each, very close to the expected value of $2.0$.

B2H6 banana bond orbital   B2H6 QTAIM LCP paths

Orbital plots here and below are at the $\phi = \pm 0.02$ isosurfaces.

On the other hand, $\ce{Al2Cl6}$ does not show multicenter bonding upon localization of the valence orbitals (the bridging atoms here are 7Cl and 1Cl):

FOUND  -   8 strongly local MO`s
       -  16 two center bond MO`s
       -   0 significantly delocalized MO`s
Rather strongly localized orbitals:
MO  47:   7Cl -   0.955366
MO  46:   7Cl -   1.001974
MO  45:   6Cl -   1.007096
MO  44:   5Cl -   1.007111
MO  43:   3Cl -   1.007113
MO  42:   2Cl -   1.007094
MO  41:   1Cl -   0.955367
MO  40:   1Cl -   1.001973
Bond-like localized orbitals:
MO  63:   7Cl -   0.844289  and   4Al -   0.149451
MO  62:   7Cl -   0.843809  and   0Al -   0.150019
MO  61:   6Cl -   0.932028  and   4Al -   0.074001
MO  60:   6Cl -   0.938027  and   4Al -   0.063923
MO  59:   6Cl -   0.746625  and   4Al -   0.250592
MO  58:   5Cl -   0.931694  and   4Al -   0.074421
MO  57:   5Cl -   0.938034  and   4Al -   0.063923
MO  56:   5Cl -   0.746451  and   4Al -   0.250727
MO  55:   3Cl -   0.931694  and   0Al -   0.074422
MO  54:   3Cl -   0.938034  and   0Al -   0.063923
MO  53:   3Cl -   0.746450  and   0Al -   0.250728
MO  52:   2Cl -   0.938028  and   0Al -   0.063922
MO  51:   2Cl -   0.932027  and   0Al -   0.074001
MO  50:   2Cl -   0.746627  and   0Al -   0.250589
MO  49:   1Cl -   0.843808  and   0Al -   0.006200
MO  48:   1Cl -   0.844290  and   0Al -   0.149450

The shape of the bond orbitals in the $\ce{Al2Cl2}$ core is consistent with strictly two-center bonding:

An Al2Cl2 core bond orbital   Another Al2Cl2 core bond orbital

Interestingly, the valence orbital localization pattern and core bond orbital structures in the 'extended $\ce{B-Cl}$' system above are very similar to $\ce{Al2Cl6}$:

FOUND  -   8 strongly local MO`s
       -   16 two center bond MO`s
       -   0 significantly delocalized MO`s
Rather strongly localized orbitals:
MO  39:   7Cl -   0.967482
MO  38:   7Cl -   1.006838
MO  37:   6Cl -   1.008398
MO  36:   5Cl -   1.008336
MO  35:   3Cl -   0.967481
MO  34:   3Cl -   1.006837
MO  33:   2Cl -   1.008334
MO  32:   1Cl -   1.008398
Bond-like localized orbitals:
MO  55:   7Cl -   0.830908  and   4B  -   0.171645
MO  54:   7Cl -   0.769240  and   0B  -   0.232367
MO  53:   6Cl -   0.941563  and   4B  -   0.066433
MO  52:   6Cl -   0.931032  and   4B  -   0.073365
MO  51:   6Cl -   0.658532  and   4B  -   0.343433
MO  50:   5Cl -   0.941702  and   4B  -   0.066259
MO  49:   5Cl -   0.930994  and   4B  -   0.073395
MO  48:   5Cl -   0.658437  and   4B  -   0.343525
MO  47:   4B  -   0.232364  and   3Cl -   0.769241
MO  46:   3Cl -   0.830907  and   0B  -   0.171646
MO  45:   2Cl -   0.941703  and   0B  -   0.066256
MO  44:   2Cl -   0.930995  and   0B  -   0.073394
MO  43:   2Cl -   0.658441  and   0B  -   0.343521
MO  42:   1Cl -   0.931031  and   0B  -   0.073367
MO  41:   1Cl -   0.941565  and   0B  -   0.066432
MO  40:   1Cl -   0.658533  and   0B  -   0.343432

an 'extended B-Cl' bond orbital   another 'extended B-Cl' bond orbital

Of particular interest, in both chlorine systems, all of the ELF basins with attractors lying along the $\ce{B-Cl}$ / $\ce{Al-Cl}$ bonds each have an integrated electron density of $\sim 1.5~e^-$. Thus, taken together, the four bonds in each rectangular core comprise six electrons in total. While I agree with Martin that there are no multicenter bonds here, there does appear to be an interesting sort of resonance phenomenon at play.

In sum: $\ce{B2H6}$ does display 2e-3c bonding; $\ce{Al2Cl6}$ does not.

$\endgroup$
  • $\begingroup$ "Once the energetic cost is paid to stretch the two B−Cl bonds, an appreciable binding energy is exhibited." At no point in that plot is the extended $\ce{(BCl3)2}$ bound (it has a positive energy at all points), yet in the next plot you show it as being bound. $\endgroup$ – pentavalentcarbon Dec 19 '16 at 17:03
  • $\begingroup$ @pentavalentcarbon Mm, did I not emphasize enough that the dissociation energy curve for the "extended" system is fictional? I can add that.... $\endgroup$ – hBy2Py Dec 19 '16 at 17:48
  • 1
    $\begingroup$ I get what you're saying, that there is a minimum along that coordinate, but to call it bound would be wrong. Fixing the wording still doesn't correct that either the first plot or second plot is wrong in some way (look at the energy scale, then look at the "extended" curves). $\endgroup$ – pentavalentcarbon Dec 19 '16 at 19:08
2
$\begingroup$

What makes dimerization possible in AlCl3?

Aluminium trichloride is an electron deficient molecule and it tends to get stability by accepting a lone pair of electron from Cl atom of another AlCl3 molecule so that it now has an octet in outer shell.

Are there 3c-2e bonds in Al2Cl6 as there are in B2H6?

No, the Cl forms dative bond.

Why is BCl3 a monomer whereas AlCl3 exists as a dimer?

STABILITY, As it is seen in the following figure

enter image description here

The bond length and atomic size of boron in BCl3 are smaller than in AlCl3 so this leads to interatomic repulsion so BCl3 could not accept another BCl3 via a dative bond because it leads to more repulsion and finally breaking of dimer. Also some degree of pi backbonding is present which may make Cl unable to form dative bond.

One may argue that BCl4- exists but in this cases size of 4th entity which binds to Boron will be smaller than a BCl3 monomer.

$\endgroup$
  • 5
    $\begingroup$ I find your last sentence hard to believe. The immediate surrounding of the boron atom is identical in $\ce{BCl4-}$ and $\ce{B2Cl6}$. The extra boron atom on the far side doesn’t interact with the first. $\endgroup$ – Jan Dec 13 '16 at 23:12
  • $\begingroup$ @Jan but that would prevent dimerisation by straining the bonds. $\endgroup$ – JM97 Dec 15 '16 at 1:15
  • $\begingroup$ I think your answers to the first two questions are spot on, from Cotton and Wilkinson "formation of dimers is attributable to the tendency of the metal atoms to complete their octets". AlCl3 also exists as a monomer in the gas phase near the b.p., the dH for disassociation is between 46-63 kJ/mol. For the third question, I think the MO diagram of B2H4 vs AlCl3 would help explain why the MO levels of B and H allow 3C-2E bonds while Al and Cl do not. $\endgroup$ – Efram Goldberg Dec 15 '16 at 6:33
0
$\begingroup$

In $\ce{AlCl3}$, $\ce{Al}$ has 6 valence $\ce{e-}$ now to complete its octet it needs 2 more $\ce{e-}$. So it accepts a pair of electron from the $\ce{Cl}$ of the neighboring $\ce{AlCl3}$. This donation of electron results in the formation of dimer.

enter image description here

But in $\ce{BCl3}$, due to the small size and the extent of π-bonding in $\ce{B}$ results in the short bond length, which also prevents $\ce{Cl}$ to form coordinate bond. And due to this short bond length when they dimerises (like in the above structure) chlorine atoms are closer than that in $\ce{AlCl3}$ and it breaks so it exist as monomer

$\endgroup$
  • 5
    $\begingroup$ $\ce{BCl4^-}$ does exist, so the size argument is void. $\endgroup$ – permeakra Dec 13 '16 at 11:02
  • 2
    $\begingroup$ The second boron atom in the ring would strain the bonds, so a direct comparison to BCl4- cannot really be made. $\endgroup$ – Efram Goldberg Dec 15 '16 at 7:54
0
$\begingroup$

It is said that $\ce{AlCl_3}$ exist as tetra-coordinate aluminium when as dimer, meaning the aluminium has an $sp^3$ like structure with a coordinate bond provided by a $\ce{Cl}$ atom.

The $\ce{BX3}$ halides attain an octet by dative $\pi_{p-p}$ interaction, forming a four-centre $\pi$ molecular orbital covering all four atoms which contains two bonding electrons. This is not feasible in case of other elements in the group. They have larger atoms and cannot get effective $\pi$ overlap, so they polymerize to remedy the electron deficiency.

In case of $\ce{B2H6}$, 3 centred-2 electron bonds are formed because boron as well as hydrogen do not have enough electrons to form a classic single bond. But here, chlorine does. So there is no 3-c 2-e bonds and instead there are coordinate covalent bonds.

As dimer, they exist as 3 centred-4 electron non-hypervalent bond, forming a delocalized dative bond as shown: $\ce{Al2Cl6}$ structure

Also note that similar structure is shown by $\ce{(BeCl2)_n}$ polymer, with beryllium atom having $sp^3$ hybridisation.

This type of 3c - 4e bonds are not possible in case of boron because boron is too small and the chlorine atoms will not fit in between them to form the bridge. Thus boron handles it through back bonding.

$\endgroup$
  • 2
    $\begingroup$ I dont think the AlCl3 dimer is considered a 3C-4E bond? I think itbis more dative or backbonding of non bonding Cl electrons to fill an Al oxtet, the multi center bond requires a MO orbital with equal distribution? Do you have a source for that? $\endgroup$ – Efram Goldberg Dec 15 '16 at 6:36
  • $\begingroup$ Bond length of $\ce{Al-Cl}$ bridges are equal to $2.3a$ with $2e^-$ in each bond. They all have the same bond orders (Mulliken), $\approx 0.66$. Visit Schupf Computational Chemistry Lab $\endgroup$ – Kishore S Shenoy Dec 15 '16 at 7:12
  • $\begingroup$ The definition of a multicenter multielectron bond is 3-center orbitals one bonding, one non bonding, and antibonding. An open 3 center bond is called 3 center 4 electron, where in addition to uniform electron distribution in the bonding orbital, there are a pair of electrons in the non bonding orbitals, ENTIRELY concentrated on the end atoms, giving a relatively polar bond. I do not think Al2Cl3 meets those requirements. (From Cotton&Wilkerson verbatim) $\endgroup$ – Efram Goldberg Dec 15 '16 at 7:43
  • $\begingroup$ In Al2Cl6 you have 2 electrons in 2 Al-Cl bridges for a total of 4 electrons, but those 4 electrons are not distributed relative to eachother in a 3 centered orbital with 2 electrons in a bonding, and the other two removed from the center orbital in non bonding orbitals. $\endgroup$ – Efram Goldberg Dec 15 '16 at 7:46
  • 2
    $\begingroup$ @Efram The Al-µCl-Al are 3c-4e-bonds, because symmetry dictates so. Cotton and Wilkerson are oversimplifying and the concept of the dative bond is just a book keeping tool. $\endgroup$ – Martin - マーチン Dec 15 '16 at 8:24
-1
$\begingroup$

A planar four membered ring with equal sides should form in both cases. The optimum bond angles would be 90 degrees within the ring. (This is independent of the bond lengths.)

Both the B and the Al atoms are four coordinate and thus would prefer a tetrahedral arrangement (109.5 degrees). 90 degrees is not too bad for the Cl-Al-Cl angle, (see H2S). However the ring is too strained in the case of boron, the Cl-B-Cl angle is too strained.

It is another question why B and C atom show much higher strain compared to heavier counterparts when distorted from the tetrahedral arrangement.

$\endgroup$
  • $\begingroup$ Actual bond angles in Al2Cl6 are 93, 87, 110, and 122 far from tetrahedral. $\endgroup$ – Efram Goldberg Dec 15 '16 at 7:30
  • $\begingroup$ 110 is not far actually from tetrahedral, but the argument was about the angles in the ring. These are close to 90 in Al2Cl6, which is tolerable for elements not in the 2nd period, but the same would cause too much strain in B2Cl6. $\endgroup$ – Stapke Feb 19 '17 at 19:35
-5
$\begingroup$

BH3 and AlCl3 each form "dimers" but by different processes.

  1. AlCl3 is trigonal planar but in vapor phase at certain temperatures, it can dimerize to Al2Cl6 with a dH of ~ 46-63 kJ/mol. A few degrees above the boiling point, and the monomer is predominant. This happens by change of geometry from trigonal planar to tetrahedral and backbonding of Cl pi electrons to fill the octet of Al giving a dimer. This is possible due to the low relative energy difference between Cl and Al valence orbitals in the MO diagram. See http://pages.swcp.com/~jmw-mcw/Parsing%20the%20Bonding%20in%20the%20Aluminum%20Trichloride%20Dimer.htm for a good overview of bond angles and lengths.

  2. BH3 is tetrahedral and forms a dimer B2H6 but through a different process. The bonding is described by Cotton&Wilkerson as: "each boron atom uses two electrons and two roughly sp3 orbitals to form 2 center 2 electron (covalent) bonds to two hydrogen atoms. The boron atom in each BH2 group still has one electron and two hybrid orbitals for use in further bonding. The plane of the two remaining orbitals is perpendicular to the BH2 plane. Thus when two BH2 radicals approach eachother with the with the hydrogen atoms also in the plane, a B-H-B 3 center 2 electron bond is formed." The MO orbital diagram for the 3C2E bond requires close energy of the three atomic orbitals which is met between boron and hydrogen.

  3. BCl3 does not form dimers as others have mentioned above, the difference in atomic radii between B and Cl explain the instability of the potential dimer. Additionally, larger difference in orbital energy would make such bonding unfavorable.

$\endgroup$
  • 1
    $\begingroup$ This is as wrong as most of the textbooks on this topic. Your cited source in 1 calls three-centre bonds an aberration, clearly an indicator that they have not understood the first thing about bonding. They try to explain their reasoning by using the hybridisation model completely and utterly wrong. The dimerisation process (the bonding?) is pretty much the same for BH3 and AlCl3. The size argument is only a small part of a more complicated matter. All in all, this answer just repeats what most of the others already stated. $\endgroup$ – Martin - マーチン Dec 15 '16 at 8:38
  • 4
    $\begingroup$ The website you cited is borderline pseudoscience. The author published that web page (which cites - amongst other articles by himself - Wikipedia) in a non-peer-reviewed online "journal" and is most "famously" known for arguing against the spdf model of atomic orbitals in favour of his own construction. One downvote doesn't do justice to any answer that uses him as a reference. cc @Martin $\endgroup$ – orthocresol Dec 15 '16 at 11:26
  • $\begingroup$ @orthocresol I was only citing the website for the bond angles, I did not read the rest of it. I actually use Cotton & Wilkerson's definition for multicenter multielectron bonding. I think my answer adds some thermodynamic values to help show the difference and I think is the only answer to mention MO theory. Either way, if the other responses are adequate, then should the question be considered answered? $\endgroup$ – Efram Goldberg Dec 15 '16 at 23:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.