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Following one’s nose: a quadruple bond to carbon. Surely I must be joking!

Thursday, December 16th, 2010

Do you fancy a story going from simplicity to complexity, if not absurdity, in three easy steps? Read on! The following problem appears in one of our (past) examination questions in introductory organic chemistry. From relatively mundane beginnings, one can rapidly find oneself in very unexpected territory.

How would one make 3-nitrobenzonitrile?

One teaches how to disconnect each group, identifying that both are meta-directing towards electrophiles, and hence asking what an appropriate electrophile might be? The “correct” answer is a nitronium cation (nitric + sulfuric acids) acting upon m-directing benzonitrile. But (sacrilege), why not a “cyonium” cation (CN+) on m-directing nitrobenzene? Well, as a tutor one would normally swat it away on the grounds it has never been previously observed (or that cyanide is always seen as an anion, not a cation). But then one (or a student) asks, why not? How about generating it from e.g. TfOCN. TfO– is a jolly good leaving group, one of the best. Well, this precursor truly appears never to have been made (or even calculated!). By now (if encountered in a tutorial), most chemistry students would be rather bemused. So the process of following one’s nose (more accurately, my nose) continues in the peace and quiet of a blog, where a rather different readership might be bemused (or inflamed).

A Quadruple CN bond?

One might start the same place a student would. How would one represent this diatomic with bonds? How about the above? It has the merit that both atoms are associated with a (shared) octet of electrons, in the form of a quadruple bond. I did show this (briefly) to a colleague, but they recoiled in horror, although it has to be said they were slightly at a loss to actually explain their horror.

Well, time for calculations. How about CCSD/aug-cc-pVTZ (DOI: 10042/to-6261) or B3LYP/aug-cc-pVTZ (DOI: 10042/to-6255). The latter allows a so-called Wiberg bond index to be computed (a reasonably accepted index). This comes out at 3.55, well on the way to being quadruple. An NBO analysis (NBO 5.9) identifies FOUR NBO orbitals with an occupancy of ~2.0, all designated BD (rather than e.g. Lp). What are these NBOs like? (as it turns out, they are almost identical to the MOs for this molecule).

Orbital 7. Click for 3D

Orbital 6 (π)

Orbital 5 (π)

Orbital 4. Click for 3D

Orbital 3. Click for 3D

Orbitals 5 & 6 are standard π orbitals with no mystery (and 8 & 9, not shown, are the matching π* pair). Orbital 3 results from the overlap of two 2s AOs (but note the curious little toroid at the carbon end). Orbitals 4 and 7 (the LUMO) are the interesting ones. Nominally, the result of overlapping two 2px AOs to give what should be a bonding and antibonding pair, they both appear to be bonding in the C-N region! Perhaps the quadruple bond is not looking quite so unlikely after all (comprising ~double occupancy of orbitals  3-6)!

What about those stalwarts I often use in these blogs, QTAIM and ELF? The former  (using the CCSD natural orbitals) has a ρ(r) of 0.346 and a ∇2ρ(r) of +2.01 at the bond-critical point (BCP). The former is certainly a high value, although no calibration exists to compare it to a quadruple bond. The Laplacian has a positive value at this point, possibly an indication of a charge-shift bond (see this and this blog, although more likely due to the adjacency of the bond critical point to the core shell of the carbon atom). ELF (also using natural orbitals) declares the presence of TWO disynaptic basins, with integrations of 5.39e and 2.44 (totalling 7.83e). The basins will each take the form of a torus (see DOI: 10.1021/ct100470g). Hm, perhaps, on reflection, this paragraph might not be entirely suitable for an introductory tutorial to organic chemistry. The density of mumbo-jumbo is rather high!

So starting from a simple retrosynthetic analysis of a simple aromatic molecule, in which the less obvious route is at least considered, one derives a “new” reagent, the cyonium cation CN+. In a effort to analyse its bonding, one concludes that a quadruple bond needs to be taken at least seriously. I would note as a warning that these diatomic species can be really tricky to pin down, and the iso-electronc C2 is a good example of that. But C2 has all sorts of issues, some of which are avoided with CN+. So the last word is hardly written, but not a bad outcome, I venture to suggest, of following one’s nose in a tutorial.

I have appended to this post a 3D exploration of the ELF function, showing the two torus basins referred to above.

 

ELF function for CN+. Click for 3D

Henry Rzepa, URL:http://www.ch.imperial.ac.uk/rzepa/blog/?p=3065. Accessed: 2011-06-04. (Archived by WebCite® at http://www.webcitation.org/5zBSjBjhM)

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Following one's nose: a quadruple bond to carbon. Surely I must be joking!

Thursday, December 16th, 2010

Do you fancy a story going from simplicity to complexity, if not absurdity, in three easy steps? Read on! The following problem appears in one of our (past) examination questions in introductory organic chemistry. From relatively mundane beginnings, one can rapidly find oneself in very unexpected territory.

How would one make 3-nitrobenzonitrile?

One teaches how to disconnect each group, identifying that both are meta-directing towards electrophiles, and hence asking what an appropriate electrophile might be? The “correct” answer is a nitronium cation (nitric + sulfuric acids) acting upon m-directing benzonitrile. But (sacrilege), why not a “cyonium” cation (CN+) on m-directing nitrobenzene? Well, as a tutor one would normally swat it away on the grounds it has never been previously observed (or that cyanide is always seen as an anion, not a cation). But then one (or a student) asks, why not? How about generating it from e.g. TfOCN. TfO– is a jolly good leaving group, one of the best. Well, this precursor truly appears never to have been made (or even calculated!). By now (if encountered in a tutorial), most chemistry students would be rather bemused. So the process of following one’s nose (more accurately, my nose) continues in the peace and quiet of a blog, where a rather different readership might be bemused (or inflamed).

A Quadruple CN bond?

One might start the same place a student would. How would one represent this diatomic with bonds? How about the above? It has the merit that both atoms are associated with a (shared) octet of electrons, in the form of a quadruple bond. I did show this (briefly) to a colleague, but they recoiled in horror, although it has to be said they were slightly at a loss to actually explain their horror.

Well, time for calculations. How about CCSD/aug-cc-pVTZ (DOI: 10042/to-6261) or B3LYP/aug-cc-pVTZ (DOI: 10042/to-6255). The latter allows a so-called Wiberg bond index to be computed (a reasonably accepted index). This comes out at 3.55, well on the way to being quadruple. An NBO analysis (NBO 5.9) identifies FOUR NBO orbitals with an occupancy of ~2.0, all designated BD (rather than e.g. Lp). What are these NBOs like? (as it turns out, they are almost identical to the MOs for this molecule).

Orbital 7. Click for 3D

Orbital 6 (π)

Orbital 5 (π)

Orbital 4. Click for 3D

Orbital 3. Click for 3D

Orbitals 5 & 6 are standard π orbitals with no mystery (and 8 & 9, not shown, are the matching π* pair). Orbital 3 results from the overlap of two 2s AOs (but note the curious little toroid at the carbon end). Orbitals 4 and 7 (the LUMO) are the interesting ones. Nominally, the result of overlapping two 2px AOs to give what should be a bonding and antibonding pair, they both appear to be bonding in the C-N region! Perhaps the quadruple bond is not looking quite so unlikely after all (comprising ~double occupancy of orbitals  3-6)!

What about those stalwarts I often use in these blogs, QTAIM and ELF? The former  (using the CCSD natural orbitals) has a ρ(r) of 0.346 and a ∇2ρ(r) of +2.01 at the bond-critical point (BCP). The former is certainly a high value, although no calibration exists to compare it to a quadruple bond. The Laplacian has a positive value at this point, possibly an indication of a charge-shift bond (see this and this blog, although more likely due to the adjacency of the bond critical point to the core shell of the carbon atom). ELF (also using natural orbitals) declares the presence of TWO disynaptic basins, with integrations of 5.39e and 2.44 (totalling 7.83e). The basins will each take the form of a torus (see DOI: 10.1021/ct100470g). Hm, perhaps, on reflection, this paragraph might not be entirely suitable for an introductory tutorial to organic chemistry. The density of mumbo-jumbo is rather high!

So starting from a simple retrosynthetic analysis of a simple aromatic molecule, in which the less obvious route is at least considered, one derives a “new” reagent, the cyonium cation CN+. In a effort to analyse its bonding, one concludes that a quadruple bond needs to be taken at least seriously. I would note as a warning that these diatomic species can be really tricky to pin down, and the iso-electronc C2 is a good example of that. But C2 has all sorts of issues, some of which are avoided with CN+. So the last word is hardly written, but not a bad outcome, I venture to suggest, of following one’s nose in a tutorial.

I have appended to this post a 3D exploration of the ELF function, showing the two torus basins referred to above.

 

ELF function for CN+. Click for 3D

Henry Rzepa, URL:http://www.ch.imperial.ac.uk/rzepa/blog/?p=3065. Accessed: 2011-06-04. (Archived by WebCite® at http://www.webcitation.org/5zBSjBjhM)

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Posted in Hypervalency, Interesting chemistry | 15 Comments »

Hypervalency: I(CN)7 is not hypervalent!

Sunday, October 17th, 2010

In the , IH7 was examined to see if it might exhibit true hypervalency. The iodine, despite its high coordination, turned out not to be hypervalent, with its (s/p) valence shell not exceeding eight electrons (and its d-shell still with 10, and the 6s/6p shells largely unoccupied). Instead, the 14 valence electrons (7 from H, 7 from iodine) fled to the H…H regions. Well, perhaps H is special in its ability to absorb electrons into the H…H regions. So how about I(CN)7? (the species has not hitherto been reported in the literature according to CAS). The cyano group is often described as a pseudohalide, but the advantage of its use here is that it is about the same electronegativity as I itself, and hence the I-C bond is more likely to be covalent (than for example an I-F bond). As noted in the earlier blog, if the potentially hypervalent atom is very ionic, it can be difficult to know whether the electrons are truly associated with that atom, or whether they are in fact in lone pairs associated with the other electronegative atom (e.g. F). It is also important to avoid large substituents, otherwise steric interactions will cause problems around the equator.

I(CN)7. Click for 3D

The calculated (B3LYP/Def2-TZVPP) geometry for I(CN)7 is similar to IH7, having essentially D5h symmetry. The C-I bond lengths range from 2.20Å (equatorial) to 2.10Å (axial); the Wiberg bond orders for these are respectively 0.482 and 0.609. The total bond orders are 3.94 (iodine), 3.91 (carbon) and 3.14 (nitrogen). The total carbon bond order for e.g. atom 2 is made up of 0.482 to I, 2.939 to N, 0.110 to C6, C7, 0.049 to C5, C8 and 0.040 to C3, C4. As with IH7, the erstwhile hypervalent iodine electrons have in fact departed from that atom, and taken up residence in the C…C regions. The NBO analysis confirms the electrons as originating from an effective iodine core (28), explicit I,C,N cores (46), 69.3 valence and 0.7 Rydberg (outer shell) electrons. The molecular orbitals are shown in this post.

Finally, for good measure, ELF analysis (on top of an effective core of 28) integrates to an outer core of 17.78 on iodine and a valence shell which includes 17.5 electrons distributed in seven explicit C-I disynaptic basins of ~2.5 electrons each. These 17.5 electrons can be considered as originating from ~10 (non-bonding?) electrons corresponding to the filled iodine 5d-shell, and ~7.5 shared bonding electrons in the iodine 5s/5p shell (the ELF procedure cannot distinguish between the 5d and 5s/5p electrons). There is no indication from these integrations that the iodine valence shells are expanded (i.e. from 10 for the 5d or from 8 for the 5s/5p).

As with IH7, this molecule shows absolutely no evidence of being hypervalent! So, if hypervalency is to survive as a concept, the hunt must surely be on for one unambiguous, as yet to be found, example of the phenomenon in the main group.

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Hypervalency: Is it real?

Saturday, October 16th, 2010

The Wikipedia page on hypervalent compounds reveals that the concept is almost as old as that of normally valent compounds. The definition there,  is “a molecule that contains one or more main group elements formally bearing more than eight electrons in their valence shells” (although it could equally apply to e.g. transition elements that would contain e.g. more than 18 electrons in their valence shell). The most extreme example would perhaps be of iodine (or perhaps xenon). The normal valency of iodine is one (to formally complete the octet in the valence shell) but of course compounds such as IF7 imply the valency might reach 7 (and by implication that the octet of electrons expands to 14). So what of IF7? Well, there is a problem due to the high electronegativity of the fluorine. One could argue that the bonds in this molecule are ionic, and hence that the valence electrons really reside in lone pairs on the F. Thus the apparently hypervalent PF5 could be written PF4+…F–, in which case the P is not really hypervalent after all. We need a compound with un-arguably covalent bonds. Well, what about IH7? One might probably still argue about ionicity (for example H+…IH6–) but that puts electrons on I and not H, and hence does not change any hypervalency on the iodine. Surely, if hypervalency is a real phenomenon, it should manifest in IH7?

IH7. A true hypervalent molecule? Click for 3D

A reasonably high level calculation (B3LYP/Def2-TZVPP with pseudopotential on I to absorb relativistic effects) shows the molecule to be a minimum, with D5h symmetry. An NBO analysis shows 46 core (i.e. non-valence) electrons. These arise from a Kr core (=36) plus a filled iodine 5d shell of 10 electrons. So, only the 5s/5p orbitals can be used for valence bonding. The total minimal basis which one can construct valence molecular orbitals from is thus I 5s, 5p and H 1s, a total of eleven AOs, into which 14 electrons must pair up into 7 doubly-occupied molecular orbitals. Now, the real issue is whether this occupancy corresponds to seven I-H covalent single bonds, each with a two-electron Lewis pair. If so, the molecule is hypervalent! The occupied MOs are shown below.

MO 16 (E2'). Click for 3D

MO 15 (E2')

MO 14 (A1')

MO 13 (E1')

MO 12 (E1')

MO 11 (A2'')

MO 10 (A1')

One notices that whilst orbitals 10-14 are clearly bonding in the I-H region, orbitals 15-16 seem antibonding in that region (there is a node along the I-H bond). We are seeing much the same phenomenon that occurs when the bond order of 3 in N2 is reduced to 1 in F2 due to occupancy of anti-bonding orbitals. Can this be quantified? NBO (5.9) analysis reveals the following.

  1. The effective (pseudopotential) core has 28 electrons, and the outer core 18. The valence orbitals contain 13.79, and only 0.21 electrons are Rydberg (higher shell). So little occupancy of e.g. 6s/6p then!
  2. The Wiberg bond index indicates each H has a total bond order of very close to 1 (its natural valence state), whilst I is 3.45. Remember that the maximum total bond index of a covalently bound atom using a pure octet of valence electrons is 4 (8/2, think carbon). The iodine is NOT hypervalent!
  3. So why, if its not hypervalent, is it so strongly hypercoordinate? Well, there are 14 valence electrons, but they do NOT all occupy the I-H regions, which have bond orders between 0.46 (equatorial) or 0.57 (axial). The only other place they can be is in the H-H regions! Consider the bond order values between say hydrogen 2 to the four other equatorial atoms 4,5,7,8. They are respectively 0.05, 0.15, 0.05, 0.15. So each hydrogen has a total bond order of 1, but only slightly more than half of this comes from the I-H, the rest comes from H-H. To put it another (approximate) way, of the 14 valence electrons, ~8 might be considered to be associated with I-H bonds, and ~6 with H-H bonds. The hypervalency has been in effect absorbed into the H-H regions. This means no atom in this molecule is at all hypervalent.
  4. Well, if the iodine is not hypervalent, and some of its valence electrons occupy I-H anti-bonding orbitals, why is it stable at all (in the sense that all the vibrations are real, and it’s clearly a minimum in the potential energy surface). Here, I merely speculate. Iodine has a large core charge, and hence the inner core electrons are starting to exhibit relativistic contractions. This effect stabilizes the outer 5s/5p electrons, and so occupancy of  anti-bonding MOs generated from such AOs is not so unfavourable as one might expect. Perhaps, the  hypercoordination shown by  IH7 is after all a relativistic effect rather than a hypervalent effect?  In which case, what will  AsH7 show?

So, what might have been an archetypal covalent hypervalent molecule is no such! IH7 shows entirely normal valencies, one for H and the iodine does not even reach four (if anything, its sub-valent rather than hypervalent). One may be entitled to ask if ANY main group element exhibits hypervalency!

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Quintuple bonds

Tuesday, February 16th, 2010

Climbers scale Mt. Everest, because its there, and chemists have their own version of this. Ever since G. N. Lewis introduced the concept of the electron-pair bond in 1916, the idea of a bond as having a formal bond-order has been seen as a useful way of thinking about molecules. The initial menagerie of single, double and triple formal bond orders (with a few half sizes) was extended in the 1960s to four, and in 2005 to five. Since then, something of a race has developed to produce the compound with the shortest quintuple bond. One of the candidates for this honour is shown below (2008, DOI: 10.1002/anie.200803859) which is a crystalline species (a few diatomics which exist in the gas phase are also candidates; for other reviews of the topic see 10.1038/nchem.359, 10.1021/ja905035f and 10.1246/cl.2009.1122).

A molecule with a Quintuple-bond

(OK, its shown as a quadruple bond, but Chemdraw cannot handle five!). The Cr…Cr length is 1.74Å (R=aryl). It was also reported that DFT calculations (BP86/triple-ζ) reproduce this length well. The five highest occupied molecular orbitals are all centred around the Cr-Cr region, and the bonding is formally described as five pairs of electrons filling 1σ, 2π, and 2δ type molecular orbitals.

So the electron pair bond, approaching its 100th birthday, is alive and well? But it does seem worth asking if those ten electrons really do cram together to occupy the region between the two Cr atoms. The stalwarts in these blog posts, AIM and ELF will be deployed to see if they too verify this simple concept. Firstly, AIM (calculated at the BP86/6-311G(d) level, DOI: 10042/to-4181 for a model system with R=H).

Quintuple bond complex, AIM analysis. Click for 3D

The Cr…Cr region has the requisite bond critical point, and the value of ρ(r) at this point has the large value (for Cr) of 0.313 au, indeed hinting at a large bond order. The Laplacian ∇2ρ(r) has the more extraordinary value of +1.45 at this point, which makes it the strongest charge-shift bond ever noted (typically, ∇2ρ(r) is ~+0.5 for other examples of homonuclear charge-shift bonds, see DOI: 10.1038/nchem.327).

This charge-shift character perhaps hints that this quintuple bond is no ordinary bond. Charge-shift bonds are characterized by valence bond structures where the covalent form may actually be repulsive, and the bond is stabilized instead by resonance with charge-shifted ionic valence bond forms. So given this, the ELF perhaps comes as no surprise.

Quintuple bond. ELF analysis

This diagram needs some explanation. The colour code is as follows: purple spheres represent the centroids of conventional disynaptic ELF basins. The only interesting ones are the four connecting the nitrogens to the Cr (21-24) which integrate to 3.35 electrons each. The cyan spheres (shown as 3,4 above) are the inner core-electrons of the Cr atoms (10.2 electrons of a neon core) and surrounding them are five further basins for each Cr integrating to 12 electrons per Cr. These include 8 of the outer-core (3s,3p) and four of the valence (3d, 4s) electrons, leaving ~2 valence Cr electrons not accounted for. Some of these final electrons are to be found in the basins represented by red spheres. The very diffuse (39, 40) basins far from the centre have a tiny electron integration (~0.003) and more missing Cr valence electrons are found in the bridging basins (32,36; 0.56 and 0.25 electrons each). Added to the 2*3.35 electrons found in the Cr-N bond, this suggests the 3d/4s shell of the Cr is occupied by ~11.5 electrons. The 3d-shell is thus full, and the system is indeed an 18-electron (8+10) system with some occupancy of the 4s shell as well. An alternative view of the ELF surface can be seen below, showing the unusual environment surrounding the Cr pair.

Quintuple bond, showing ELF isosurface. Click for 3D

It seems that AIM (the topology of the electron density) and ELF (the topology of the electron localization function) are giving us quite different pictures of the quintuple bond. The latter does seem to indicate that the conventional covalent shared electron pair picture of this bond is not really what is going on, and that the idea of a quintuple bond as sharing five electron pairs in the bonding region between the two Cr atoms is not really realistic. It may be of course that the ELF concept also is not really applicable for such bonds (it is after all essentially an empirical function, the deeper significance of which is debatable).  Nonetheless, the quintuple bond clearly has some surprises for us, and it would itself be no surprise to find out that controversy about the meaning of such a bond continues apace.

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Clar islands in a π Cloud

Wednesday, December 9th, 2009

Clar islands are found not so much in an ocean, but in a type of molecule known as polycyclic aromatic hydrocarbons (PAH). One member of this class, graphene, is attracting a lot of attention recently as a potential material for use in computer chips. Clar coined the term in 1972 to explain the properties of PAHs, and the background is covered in a recent article by Fowler and co-workers (DOI: 10.1039/b604769f). The concept is illustrated by the following hydrocarbon:

Clar islands in a polybenzenoid hydrocarbon

Clar islands in a polybenzenoid hydrocarbon

The Clar islands are shown in red, and represent in effect the resonance form of this species which maximises the number of aromatic electronic sextets possible to achieve via a cyclohexatriene resonance form. It encapsulates the concept that maximum stabilization is achieved when the π-electrons in the molecule cluster together (or localize) in cyclic groups of six (rather than eg other allowed values as predicted by the 4n+2 rule of aromaticity). As a historical note, although Clar popularized the concept in the 1970s, the (C) representation had in fact been introduced almost one hundred years earlier, by Henry Armstrong (DOI: 10.1039/PL8900600095). Many demonstrations that Clar islands are reasonably based in quantum mechanical reality have been made; a very graphical and convincing one is that by Fowler and coworkers in the reference noted above, using the calculated magnetic response property known as π current densities (although this shows that the six outer islands tend merge into a single continuous outer periphery).

Current density maps showing Clar islands (taken from DOI: 10.1039/b604769f
Current density maps showing Clar islands for the molecule above (taken from DOI: 10.1039/b604769f)

Previous posts on this blog have mentioned the application of another computed quantum mechanical property known as ELF, the electron localization function introduced by Becke and Edgecombe in 1990 (DOI: 10.1063/1.458517 ) and subsequently adapted for use with DFT-based wavefunctions. ELF is normally applied to help analyze the bonding in a molecule; the value of the function is normalized to lie between 1.0 (a simple interpretation is that this is the value associated with a perfectly localized electron pair) and 0.0. ELF has no association with magnetic response (the latter being an excitation phenomenon), but since the Clar islands can also be considered a localizing property of the π electrons, it is tempting to ask whether the ELF function can also reveal their characteristics (this question was first posed in DOI: 10.1039/b810147g).

The ELF function, as isosurfaces contoured at various thresholds

The ELF function, as isosurfaces contoured at various thresholds. Click for 3D

The diagram above shows the ELF function computed for the π-electrons of the molecule above (B3LYP/6-31G(d), as isosurfaces contoured at various values. At the value of 1.0, no features are discernible, but at 0.95 features which resemble basins associated with each atom centre have appeared, in the region of the 2p-valence atomic orbital on each carbon atom we regard as contributing the π-electron to the system. As the ELF threshold is reduced, these objects start to merge into what are called valence basins associated with bonds in the molecule. The outer periphery is the first to start coalescing. By a value of 0.75 (click on the diagram above to see a 3D model) the basins have merged to form seven clear-cut rings which happen to coincide exactly with the Clar islands. This feature persists down to a threshold of 0.55. Below this value, the seven individual basins merge into a single basin contiguous across the top (and bottom) surfaces of the molecule. One can also conceptualize the journey in the other direction. At low ELF values, the function is continuous, but as the threshold increases, it starts to bifurcate into separated basins. The first clear-cut bifurcation is indeed into the Clar islands, and this persists across a relatively wide range of ELF values, which suggests it is a significant feature. What is somewhat surprising is the close apparent correspondence of this way of analysing the electronic properties of the π electrons with their magnetic response computed via current densities. But association with aromaticity has previously been made (DOI: 10.1063/1.1635799). Thus Santos and co-workers have shown that the value of the ELF function at the point where it bifurcates from a ring into discrete valence or atomic basins can be related to other metrics of aromaticity. Here, that value is around 0.75 for the Clar basins, which is also within the range of values that Santos et al associate with prominent aromaticity (benzene itself has a value around  0.95).

A C114 PAH

A C114 PAH

The ELF function for the 114-carbon unit shown above again reveals prominent Clar islands, the inner heptet being very similar to the picture painted using current densities.

Clar islands in the ELF function for a C114 carbon PAH

Clar islands in the ELF function for a C114 carbon PAH

The final example involves diboranyl isophlorin (DOI: 10.1002/chem.200700046), a 20 π-electron antiaromatic system. Such systems are particularly prone to forming locally aromatic Clar islands as an alternative to global antiaromaticity (DOI: 10.1039/b810147g).

A Diborinyl system.

A Diboranyl isophlorin.

The ELF function is shown for both the neutral diboranyl system and its (supposedly more aromatic) dication. Here a mystery forms. No Clar islands are seen, and instead it is the periphery that bifurcates, at ELF thresholds of 0.5 for the neutral and 0.7 for the dication. The latter value clearly is that of an aromatic species, but the former is somewhat in no-man’s land, but certainly less aromatic that the dication. One for further study I fancy!

ELF Function for diboranyl molecules (red=neutral, green=dication). Click for 3D

ELF Function for diboranyl molecules (red=neutral, green=dication). Click for 3D

Does the ELF function have any possible advantage over the use of current density methods for analysing aromaticity? Well, the latter is normally applied to flat systems with planes of symmetry defining the π-system, and with respect to which an applied magnetic field is oriented. How to orient this magnetic field is not so obvious for prominently non-planar or helical molecules. Since the ELF function does not depend on the orientation of an applied magnetic field, it may be a useful adjunct for studying the properties of π-electrons in non-planar systems.

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Towards the ultimate bond!

Monday, August 24th, 2009


The 100th anniversary of G. N. Lewis’ famous electron pair theory of bonding is rapidly approaching in 2016 (DOI: 10.1021/ja02261a002). He set out a theory of bond types ranging from 1-6 electrons. The strongest bond recognized by this theory was the 6-electron triple bond, a good example of which occurs in dinitrogen, N2. In terms of valence electrons, nitrogen has an atomic configuration of 2s2, 2p3. Each atom has five electrons in total, some or all of which in principle could be used for forming bonds. An exploration of this motif across the entire periodic table is presented in part one of this blog.
Elements in Groups 5/15 of the Periodic Table.
Nitrogen is in the main group 15, and the element at the bottom of this group is Bismuth (also with the same atomic configuration). We can then move to the corresponding column of the transition series, this time occupying group 5. The first examplar in this set, Vanadium has an atomic configuration of 3d3, 4s2, again five valence electrons, but now utilizing the d- rather than the p-shell of valence atomic orbitals (AOs). The final forage across the period table would land us with Pr and Pa, which occupy the lanthanide and actinide series respectively, and which have atomic configurations of 4f3, 6s2 and 5f2, 6d1 and 7s2 respectively. You can now see the theme developing; how does the bonding develop between two atoms that between them have ten valence electrons occupying molecular orbitals constructed from s, and then either p, d or f atomic orbitals. The next in that series, g atomic orbitals, are thought unlikely to have any chemical significance in the presently known periodic table.

We are in fact going to study the diatomic molecule comprising two atoms of each element, and further we are going to protonate this species on one of these atoms resulting in the molecule HX2+. So let us start with the two systems HN2+ (Figure 1, 1a-1e) and HBi2+ (Figure 1, 2a-2e).

Figure 1. The molecular orbitals of HN2(+) and HBi2(+)

Figure 1. The molecular orbitals of HN2(+) and HBi2(+)

The two most stable valence molecular orbitals (MOs) for each system (1a and 2a) are the symmetric and antisymmetric combinations of (assumed pure) s-AOs, each populated with two electrons. For Bi (2a/2b) it is fairly clear that this bonding and anti-bonding combination cancel almost exactly. The bond order resulting from these four valence electrons is therefore close to zero. But N is in fact rather odder (and in part, the reason for protonating the systems was to tease out this oddity)! The apparently antibonding 2s-2s combination (1b) actually has electron density along the N-N bond, and the node occurs not along the bond, but at the actual nitrogen atom. So for N, the bonding and anti-bonding σ-bond combinations do not cancel, and the sum of these two may actually lead to a non-zero bond order.

With Bi, the next most stable MO results from the overlap of two 6p-AOs end on (2c); the anti-bonding combination of these AOs is in fact the 2e (with no occupying electrons). Result: bond order of +1, and this bond is called a σ-bond. The final MO shown (2d) is in fact one of a pair of MOs (only one of which is shown here), resulting from parallel overlap of two 6p-AOs. Result: bond order of +2, and we call these two p-π-bonds. The total bond order is +3, comprising 1 σ-bond and two p-π-bonds. No surprises here yet! For N, the relative order of the π- (1c) and the σ-bonds (1d) is swapped compared to Bi (which might be due to relativistic effects, see below), but one can again argue that these three orbitals together contribute 1 σ-bond and two p-π-bonds to the total bonding. So, to summarize, HBi2+ exhibits a classical triple Bi…Bi bond; HN2+ in contrast may actually exceed the bond order of +3, and a case could be made for arguing it is an abnormally strong bond (there is evidence that the N…N stretching frequency in the protonated species is significantly higher than the simple diatomic nitrogen gas).

Let us now move to the combination HV2+ (3) and HTa2+ (4, Figure 2). There are two major differences for V compared to N. Firstly, a 3d rather than 2p -AO is used for the bonding. Secondly, the 3d-AO is actually lower in energy than the 4s-AO, the reverse of the 2s/2p order for eg nitrogen. So what effect does this have on the resulting molecular orbitals?


Figure 2. Valence molecular orbitals for HV2(+) and HTa2(+)

Figure 2. Valence molecular orbitals for HV2(+) and HTa2(+). Click image for 3D of Delta orbital


The difference between N and V turns out to be spectacular, in several regards! The most stable MO (3a) now turns out to be composed of end-on overlaps of two 3d-AOs. There are in fact two of these orbitals (only one is shown in Figure 2), and together they form two 3d-π bonds. The next MO (3b) involves the end-on overlap of a 3dz2-AO, this time forming one 3d-σ-bond. Finally, the 4s-AOs get in on the act, forming now one bonding s-σ-bond (3c). So far, eight of the ten valence electrons have been consumed; two more to go. But now we have a problem. The next MO is formed by parallel overlap of two 3d-AOs (Figure 2, 3d), and in fact there is a pair of these combinations (again, only one is shown in the figure), which are in fact equal (degenerate) in energy. Because they are equal in energy, they must both be populated by an equal number of electrons, but since there are only two valence electrons left, we end up with one electron in each of these two orbitals, resulting in a triplet spin state. Exactly the same phenomenon is responsible for diatomic oxygen also adopting a triplet rather than singlet spin state. This parallel overlap of two 3d-AOs is said to form a 3d-δ-bond. The total bond order in molecule 3 therefore comprises two 3d-π-bonds, one 3d-σ-bond, one 4s-σ and two half 3d-δ-bonds, i.e. five in total and enforces a triplet spin state. So in the sense of formal bond order, the V-V bond in 3 is greater (and perhaps even stronger) than in 1. What a difference from nitrogen!

How about Ta? This has an atomic electronic configuration of 5d3, 6s2. The molecular orbitals are shown in Figure 2 (right). The significant difference in this region of the periodic table is that so-called relativistic effects start to influence the relative ordering of the atomic orbitals. This so-called relativistic contraction impacts upon s-AOs far more than p, or d or f. Thus orbital 4a (Figure 2, right) comprising a symmetric combination of Ta 5s-AOs is more compact than the corresponding V orbital (3c), and relatively more stable. Next come the end-on overlaps of two 5d-AOs (4b), of which there are two (degenerate) combinations (only one is shown in Figure 2). MOs 4c and 4d are again best described as originating from the 6s Ta AO, but with significant contributions from the 5d-AOs (sd-hybrids if you will). Significantly, the relativistic effect means that the δ-bond formed by parallel overlap of two 5d-AOs (4e) is left vacant. Thus the bonding in 4 comprises an 6s-σ-bond, two 5d-π-bonds, and two sd-σ-bonds (one of which does look rather anaemic) but NO 5d-δ-bond! The consequences of the relativistic effect is the relegation of the δ-bond to unoccupied status and the formation of a singlet rather than a triplet spin ground state.

Given the big differences in bonding which occur upon changing a 2p-valence atomic orbital to 3-5d-AOs, one wonders what will happen with 4-5f-AOs? A π-bond can be formed by the parallel overlap of two p-AOs, or end-on overlap of two d-AOs. A δ-bond can be formed by the parallel overlap of two d-AOs. Could then a δ-bond also be formed by the end-on overlap of two f-AOs? Would a φ-bond be formed by the parallel overlap of two f-AOs? The latter might look something like that shown in Figure 3, shown in two different orientations (these diagrams were obtained by inspecting the unfilled MOs in the V/Ta examples shown here; notice that although the φ bond appears to be bonding, the system has chosen not to occupy it with any electrons!).

Figure 3. The "phi" bond.

Figure 3. Two views of a φ bonding MO. Click image for 3D
Figure 3. The "phi" antibond.

Figure 4. A φ* antibonding MO. Click image for 3D

This blog will end by posing the question “can any molecule be devised which supports one or more φ-bonds”, or will the relativistic contraction always scupper such efforts by depriving such bonds of electrons (e.g. Ta above)? Are the systems reported in DOI: 10.1021/ja067281g examples of a 5f-φ bond for uranium (the claim is made in a very low key manner)? This will be investigated in the follow up to this post!

(For serious geeks/computational chemists only, N was computed at the B3LYP/6-31G(d) level, V at the ROHF/6-31G(d) level, and Bi/Ta at a triple-ζ-pseudopotential level (which incorporates some of the relativistic effects).

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Molecular toys: Tetrahedral cavities

Saturday, July 4th, 2009


An earlier post described how a (spherical) halide anion fitted snugly into a cavity generated by the simple molecule propanone, itself assembled by sodium cations coordinating to the oxygen. A recent elaboration of this theme, reminiscent of the children’s toys where objects have to be fitted into the only cavity that matches their shape, Nitschke and co-workers report the creation of a molecule with a tetrahedral rather than a spherical cavity (DOI: 10.1126/science.1175313 ), into which another but much smaller tetrahedral molecule is fitted.  The small molecule is P4, in which each of the three valencies of the P atom is directed to a corner of the tetrahedron. The large molecule  comprises four Fe atoms. These are each octahedrally coordinated with six ligand sites, three of which mimic the P atoms in also being directed towards the remaining three vertices of a tetrahedron.

P4 inside a Tetrahedral cavity.

Needless to say, the properties of the P4 molecule when entrained into this larger container are nothing like that of the free molecule. Now it is quite inert, but this is due purely to the snug fit. For example, the normal reaction of this molecule is to oxidize in air. But such oxidation would now produce a molecule too big to fit into the cavity. Hence no reaction!

So, now the search is on for a cubic container to include a cubic molecule!

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