Part 3: On the nature of bonding in aggregates of borane

Borane, BH₃, as we looked at it in Part 1, is an electron deficient molecule - a strong Lewis acid, by virtue of the vacant p_z orbital. In a sample of borane, we do not actually have free BH₃, but a dimer forms between two molecule of BH₃ to form diborane, B₂H₆. Hence, free BH₃ is difficult to study, and this therefore demonstrates the power of computational methods. The bonding in B₂H₆ is unusual, involving bridging H atoms between two B centres - 3c2e bonds. We shall now use the computational methods we have learned, to apply them to the problem of B₂H₆.

We shall use computation to help construct the MO diagram for B₂H₆. As we shall see, when we are dealing with a multi-centre molecule, it can be difficult to judge the overall bonding / antibonding nature of a given orbital, relative to others, so we shall use the form of the calculated orbitals to guide us. Then we shall look at the vibrations of diborane.

Then, we shall consider other aggregates of borane, formed again through 3c-2e bonds where H atoms bridge the boron atoms, to understand why they do or don't form, comparing the stability of isomers. We shall construct the molecular orbital diagram of a planar D₃h BH₃ trimer, and again look at its vibrational spectrum.

B₂H₆

Optimisation

A molecule of diborane, B₂H₆, was created using the Gaussview GUI. This was done by first placing a carbon tetrahedral fragment, then changing the C centre to B, and placing in the plane of one BH₂ group another B atom, and in the plane of the other BH₂ unit two hydrogen atoms. The bond angles for the freely placed atoms were set to tetrahedral, and the dihedral angles set to 180^o across the HBBH units, i.e to make two tetrahedral BH⁴ units, bridged by two H atoms. The geometry was optimised using a DFT/B3LYP/6-311G(d) calculation. Initially, the B atoms push away from each other (Vnn repulsion) and the terminal H-B-H angles widen from 109^o to 122.3^o. The internal H-B-H bond angles lessened to 95.5^o. We shall return to discuss this geometry when we look at the molecular orbitals of the molecule.

The optimised geometry has the geometrical parameters shown. The terminal BH distances were 1.19Å, as those for BH₃ above were. The internal BH distances were calculated to be 1.32Å. The BB distance was 1.77Å. We can compare these to literature^[1] and find them to be 1.19, 1.33 and 1.77Å respectively. The calculation is in very good agreement with experiment. The terminal H-B-terinal H angle is calculated to be 122.4^o. Comparing to literature, we find it to be the same. Likewise, the internal H-B-H angle is calcualted to be 95.6^o, and reported to be 97.0^o. The calculation is in very good agreement with experiment.

BH3 Optimised Geometry

A frequency calculation was carried out on the optimised structure. Reading from the log file, we find two negative frequencies of -0.0005cm^-1 and -0.0002cm^-1. This is a valid stable conformation.

We said that BH₃ dimerises to reduce its electron deficiency. Let us see if that is the case by analysing the partial charges. Before, B was highly positively charged, and the three H were electron rich. In the dimer, the case is reversed. The two B atoms are slightly electron rich now, with a partial negative charge of -0.12. The internal H are slightly positively charged, with a partial charge of +0.11. Finally, the terminal H and almost neutral. Hence, we can indeed say that BH₃ dimerises to reduce the electron deficiency at the B centre. Because B is more electronegative than H, the B atom has the slight delta negative.

When we compare the energy of the diborane molecule to two molecules of borane, we find, on dimerisation, the energy is reduced by 159Kjmol^-1. This stability gain explains the observed behavior of a sample to dimerise totally.

Molecular Orbitals

We shall construct the molecular orbital diagram for the B₂H₆ molecule, using two BH₂ fragments, adapted from the MO diagram for water. These will be combined to make a B₂H₄ fragment, which will then be mixed with a H₂ fragment, in the plane perpendicular to the B₂H₄ unit. However, it is difficult to determine exactly how the MOs will be ordered in the molecule, based upon our simple analysis, because there are many supporting and conflicting interactions in a given orbital, in some cases. Using our chemical intuition, we may expect to draw the MO diagram as shown here. Draw for bonding and non-bonding BH₂ fragment orbitals.

The overlap of two BH₂ fragments means the SAOs of the fragment will form symmetry allowed combination with themselves on the other fragment, since that are at the same energy, and of the same symmetry. The BH₂ 1a₁ fragment overlaps in an s-s manner, so we expect a strong splitting. The 2a₁ orbitals overlap in p-σ type interaction, which is less effective than the s-s, so hence the splitting is less. This though is stronger than the p-π overlap seen in the 1b₁ and 1b₂ BH2 orbitals. Hence these are split less. As always, an antibonding orbital of a bond/antibond pair is always more destablised than the bonding is stabilised.

The H₂ fragment is split less than a H₂ molecule because the H...H distance is further, so a less effective overlap. On combination with the B₂H₄ fragment, we would predict the lower energy a_g B₂H₄ orbital to be too low in energy to interact with the H₂ a_g orbital, so it is left non-bonding. There are many B₂H₄ orbitals which have no symmetry partner in H₂, so are left non-bonding. The overlap of the BH₄ 2a_g orbital and the H₂ 1a_g orbital is allowed, and the energies are close, but not very close. The resulting 3a_g and 4a_g orbitals I would predict are split little because the interaction is not strongly directional, and the introduction of the H₂ fragment would stabilise the B₂H₄ fragment, but not by a great deal. The overlap between the B₂H₄ and H₂ b_3u fragment though are close in energy, and more directional, and hence more split. The final position of the MOs is based upon energy considerations from the number of nodes. The two overlaps we just discussed are those responsible for the 3c-2e bonds. There are two such bonding orbitals which are occupied. We expect no mixing because there are no like-symmetry Mos close in energy.

This explains why the terminal HBH angles were wider than the internal HBH angles. The widening external angle makes for a more effective overlap in the BH₂ b₁ orbital, and the narrower the internal angles, the closer the H₂ fragment is to each other, making for a better overlap, and also the closer to the B₂H₄ a_g and b_3u orbital like-lobes.

B₂H₆ MO diagram, prediction based upon interaction considerations

Energy not to scale

Now, we shall use Gaussian to build the orbital coefficients. This will tell us the position of the energy levels. Let us compare the prediction above and the calculated result. Isovalue 0.02, rendered in in Avagadro 1.1. Unfortunately, ChemDraw will not save orbital images in a png format, so I have taken screenshots of the MO, but had to split it into three parts. Depending upon your screen size, the three images may not line up over each other. A link to a tiff of this while MO: File:Pm08B2H6MOcalculation.tif

B₂H₆ MO diagram, energy levels position from calculation

Energy not to scale

We see we wrongly predicted the MO energy level order, and the splitting of the 3c-2e orbitals. Also, more worryingly, we predicted the HOMO and LUMO wrongly. We can say that for simple molecule like BH3, we can use group theory predictions to make judgments above chemical structure and reactivity. However, when orbitals become more complex, we have to back up our predictions with calculation, to get the right order of molecular energy levels. However, the form of the MOs predicted matches well to those seen.

Vibrations

We want to predict the number, and symmetry of vibrational modes of the B₂H₆ molecule. We form a total basis, by placing three axis vectors on each atom, and see how they transform under the operations of D₂h. Then we form our reducible representation for that basis:

Red arrows count 1(unchanged), blue arrows count 0(reversed). Now, we treat our representation as we did before, using the reduction formula to find the component irreducible representations. But first, we have to subtract the 6 rotational and vibrational modes:

n A_g = (1x21x1)+(1x-1x1)+(1x1x1)+(1x-1x1)+(1x-3x1)+(1x7x1)+(1x5x1)+(1x3x1) = 32/8 = 4

n B_1g = (1x21x2)+(1x-1x1)+(1x1x-1)+(1x-1x-1)+(1x-3x1)+(1x7x1)+(1x5x-1)+(1x3x-1) = 16/8 = 2

n B_2g = (1x21x1)+(1x-1x-1)+(1x1x1)+(1x-1x-1)+(1x-3x1)+(1x7x-1)+(1x5x1)+(1x3x-3) = 16/8 = 2

n B_3g = (1x21x1)+(1x-1x-1)+(1x-1x1)+(1x-1x1)+(1x-3x1)+(1x7x-1)+(1x5x-1)+(1x3x1) = 8/8 = 1

n A_u = (1x21x1)+(1x-1x1)+(1x1x1)+(1x-1x1)+(1x-3x-1)+(1x7x-1)+(1x5x-1)+(1x3x-1) = 8/8 = 1

n B_1u = (1x21x1)+(1x-1x1)+(1x1x-1)+(1x-1x-1)+(1x-3x-1)+(1x7x-1)+(1x5x1)+(1x3x1) = 24/8 = 3

n B_2u = (1x21x1)+(1x-1x-1)+(1x1x1)+(1x-1x-1)+(1x-3x-1)+(1x7x1)+(1x5x-1)+(1x3x1) = 32/8 = 4

n B_3u = (1x21x1)+(1x-1x-1)+(1x1x-1)+(1x-1x1)+(1x-3x-1)+(1x7x1)+(1x5x1)+(1x3-1) = 32/8 = 4

Thus, in total, our 3N-6 reducible representation is a sum of: Γ(3N-6) = 4A_g + 2B_1g + 2B_2g + B_3g + A_u + 3B_1u + 4B_2u + 4B_3u

We calculate the vibrational modes of the molecule, and animate the vibrations, to test our prediction:

Calculated vibrational modes of B₂H₆

Displacement

Frequency

Intensity

Symmetry

BH Stretching modes

Vibration

2700

185

B_2u

Vibration

2686

0

B_1g

Vibration

2609

0

A_g

Vibration

2594

153

B_3u

Bridging H Displacement

Vibration

2155

0

A_g

Vibration

1983

9

B_1u

Vibration

1834

0

B_2g

Vibration

1696

489

B_3u

BH Bending modes

Vibration

1193

0

A_g

Vibration

1185

77

B_3u

Overall rocking modes

Vibration

1039

0

B_3g

Vibration

984

17

B_1u

Vibration

955

1

B_2u

Vibration

937

0

B_1g

Vibration

889

0

B_2g

Vibration

849

0

A_u

Vibration

792

0

A_g

Vibration

349

16

B_2u

We seem to have a slight problem when we add up these. We have three too few. I accidently forgot to subtract the translational vectors from the reducible representation above. That meant we predicted those again when we carried out the reduction. Hence, we found three less actual vibrations, listed above. Taking this into account, we find we have managed to exactly calculate the number and type of the vibrational modes of diborane, from group theory considerations. This shows the power of these tools. I'm sorry, there wasn't time to change the representation above.

Higher Aggregates?

Borane Trimer and Tetramer Optimisation

Whilst looking at diborane, and the formation of 3c-2e orbitals to stabilise a B₂H₄ unit, I began to wonder if it was possible for higher aggregates of borane to exist, by similar 3c-2e bonding. Using Gaussview, a planar, D₃h triborane species was created, and similarly, a cubic, D₂d tetraborane species. The triborane was created by taking a benzene fragment, to create a planar 6-membered ring, then substituting in alternating B and H atoms. Then two H atoms per B atom were added. The tetraborane was created by taking a cubane molecule, optimised for a previous project, and again replacing C by B and H, and adjusting the valency.

The two molecules were optimised, using a DFT/B3LYP/6-311G(d) Opt calculation. The resulting geometries are shown below.

BH3 Optimised Geometry

Before we get excited at the possibility of new forms of borane (though this did go through my head!) we need to check the geometry. As we have said before, we may have found a transition state. Running a frequency calculation with the same route/method/basis, we get the following results. For the trimer, 2 negative frequencies were reported, of -429cm^-1 and -298cm^-1. These correspond to three vibrational modes of A₂" and E" symmetry. For the tetramer, 3 imaginary frequencies were found, of -2369cm^-1, -945cm^-1 and -890cm^-1, corresponding to four modes of A₂, E and B₁ symmetry.

What does this mean? We see that the curvature of the tetramer transition state is steeper than that of the trimer transition state. This means that the slope to reach the tetramer is higher, and therefore, the tetramer TS is higher in energy than the trimer TS. But these are transition states of what? Let us have a look at these vibrations, to try to determine this. Firstly, for the tetramer:

Borane Tetramer TS Imaginary Frequencies

Vibration

The vibrations involve the displacement of the bridging H atoms from close one BH₂ unit to another. May I therefore make a suggestion that this is the transition state through which two molecules of diborane pass, during which the individual borane units swap over.

Such a reaction would show a symmetric reaction profile about the transition state. This is an interesting posibility. We saw in the MO diagram for B₂H₆ that the inner H₂ fragment acts to stabilise the B₂H₄ fragment, but that the bonding is non-directional, since the H₂ fragment uses a linear combination of 1s AOs to form its SAOs, so could swap between B₂H₄ units without having to rotate. The BH₂ fragments would be required to rotate, breaking the directional sp²-sp² overlap, which explains why this transition state is high energy.

If we look at the calculated energies for diborane and this tetraborane transition state, we find that the transition state is 713.4Kjmol^-1 higher in energy than two molecules of isolated diborane. This is immensely high, so would likley never happen, unless at a very high temperature, under high pressure.

Looking now at the triborane transiton state. Using the same logic, we could imagine this to the the transition state for the sawpping of a BH₃ unit of diborane with another.

Looking again at the energies of this transition state, compared to a molecule of diborane and a molecule of borane, we find the transition state is only 12.6Kjmol^-1 higher in energy. This is a small barrier and might occur in appreciable amounts at room temperature or pressure. We looked before on the energy gain by dimerisation of borane, and found that diborane is 159.1kjmol^-1 lower in energy than two molecule of borane. Therefore, the likleyhood of having free borane in a sample is very low, so this swapping via a D₃h 6-membered trimer transition state may not be observed. I however propose an experiment where this swapping could be tested.

Boron is present in two natural, stable isotopes, ¹⁰B, 19.9% and ¹¹B, 80.1% abundances. If a sample of isotopically pure diborane was prepared, by the reduction of BF₃ by, say LiAlH₄, to give BH₃, then the subsequent dimerisation, then quench to remove excess LiAlH4. Then, we would add an isotopically pure sample of BF₃, of the other boron isotope to that which was used to prepare the diborane, to our sample of diborane, followed by another reduction. Then, transiently, as the BF₃ was reduced to free BH₃, we may expect some swappping to occur, and we would be able to tell by analysis of the isotopic ratio of the resulting diborane, by Mass spectrometry.

The difference in energy of the two aggregates is reflected in the magnitude of the imaginary frequencies, as we said.

Borane Trimer TS Imaginary Frequencies

Vibration

Borane Trimer TS MO diagram

Why is the energy of the trimer TS only 12.6Kjmol^-1 higher than the diborane and borane molecules? To andswer this question, I am going to construct the molecular orbital diagram of this B₃H₉ D₃h trimer to try to answer this question.

This will also serve as the ultimate test for our group theory predictions, backed up with computationally derived orbitals to determine the order of the energy levels. The molecular orbital diagram shows many more energy levels than is required to answer this question, but it serves as another test of the group theoy prediction as to the form of the orbitals.

Three BH₂ fragments were simultaneously mixed, using the projection operator to find the form of the B₃H₆ fragment orbitals. A general rule followed, that the three BH₃ SAOs mixed to form a non-degenerate and a doubly degenerate pair, of higher energy. These overlap with a H3 fragment. The energy gao between the HOMO and LUMO is large in this molecule, because all the low energy orbitals are filled. The LUMO is made up of opposite phase overlap of the BH₂ b₂ FO. This situation is mirrored in organic aromatic molecules. The situation there is of 4n+2 pi electrons, which is the correct number to fill totally the low energy orbitals, leaving the high energy orbitals empty. This is a similar story here, though the energy gap will not be so large. This filling of the lower energy orbitals explains why the transition state energy is not much higher than the isolated molecules.

B3H9 Mo diagram, with orbital position from calculation

Once again, even in this complex, mixed system, the group theory predictions, with a little help from computation are able to show the individual orbital contribution to the molecular orbitals.