Hypervalency: I(CN)7 is not hypervalent!

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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8 Responses to “Hypervalency: I(CN)7 is not hypervalent!”

  1. […] Henry Rzepa Chemistry with a twist « Hypervalency: I(CN)7 is not hypervalent! […]

  2. Henry Rzepa says:

    I noted above that the calculation for I(CN)7 revealed I-C bond lengths of ~2.1 (axial) and ~2.2 (equatorial), this at a reasonably good basis set level. What does the Cambridge crystal database have to say about C-I lengths? Well, there are around 11 structures in which iodomethane is a component, and the C-I bond in these is around 2.1Å. There are four structures containing I-CN motifs (TOQCEI, TOQCIM, TOQCOS, YUHLOD). There the average I-C length is ~2.25Å. So I(CN)7 appears to have normal I-C single bonds, judged purely by their length. What does this suggest?

    Well, perhaps one cannot use purely bond lengths to decide if a compound is hypervalent or not.

    There is nevertheless a contradiction. The Wiberg bond order (which treats the filled iodine 5d shell as non-bonding) suggests that the C-I bond orders in I(CN)7 are very much less than one. On the other hand, the ELF analysis clearly reveals a disynaptic basin in the I-C region which integrates to ~2.5 electrons (formally a 1.25 bond order), ie slightly strong than a single bond. It seems that this picture matches the bond lengths rather better. The implication is that the 5d electrons are not entirely passive.

    So there are certainly questions which need answers for this type of molecule.

  3. Henry Rzepa says:

    Analysing bonding is a bit like describing the cheshire cat, it can often be just a smile, without the substance of a body. In the post above, I described the ELF results for I(CN)7 in which a valence shell of 17.5 electrons were distributed in 7 apparently disynaptic basins. However I mis-labelled my shells! In the calculation used (basis set Def2-TZVPP) an effective iodine core of 28 electrons includes an argon core (18) and ten from the 4d shell. A further 25 electrons bring the count up to 53, and these are distributed 8 in 4s/4p, 10 in 5d and 7 in 5s/5p. The ELF for iodine (showed) 17.78 of these in a monosynaptic basin, and a further 17.5 electrons in what are labelled disynaptic basins connecting I and C (of the CN). In fact, the distinction between a mono and a disynaptic basin in ELF can be a knife-edged one. If its disynaptic, then these 17.5 electrons must populate as 8 in the iodine 5s/5p shell, and 10 in the iodine 6d shell (and not the 5d shell as I wrote above). If the seven basins are monosynaptic, then they would correspond to lone pairs on the carbon. In other words, the cyano group would in fact be a cyanide anion, CN. This returns us to the problem I discussed in the blog; that of distinguishing between an ionic bond (with monosynaptic basins) and a covalent bond (with disynaptic basins). So, like the smile on the Cheshire cat, we are back to arguing whether I(CN)7 is ionic or covalent. If the latter, the ELF analysis (if believable) still suggests that d-electrons might be participating. If that were to be the case, then the iodine would have to be treated as a transition series element, and not a main group element. Quite a change in the way we think about it.

    So this leads us inexorably to finding a molecule where these issues are less ambiguous. I will shortly document I8, i.e. II7 and As8, on the assumption that if any bond is going to be clearly covalent, it should be one comprising homonuclear (i.e. the same) atoms.

  4. Matthew says:

    Hello Henry,
    I’m curious about the Rydberg orbitals that some of the electron occupation is partitioned into in this case and some of your other IX7 calculations. Is the partial electron occupation in one Rydberg orbital or smeared out? Why do you think this happens? If you look in the Bond orbital section, and go to the first RY orbitals with occupation what is the orbital in terms of atomic contributions and were is it located in the molecule? Also why does the Rydberg(s) get more occupation in the ICN7 case opposed to as in your IH7 calculation? Thanks, Matthew

  5. Henry Rzepa says:

    Interesting questions Matthew, to which I do not have an immediate answer. I did include a link to the checkpoint file, so you can have a look at the wavefunction yourself if you wish. If you manage to provide answers, I would be very interested!

  6. Matthew says:

    Hi Henry,
    thanks, I took a quick look. It seems most of the 0.7 electron occupancy is mostly (~70%) smeared out and therefore is not really prone to a simple mechanistic type of explanation. The largest single contribution to the Rydberg occupation that I noticed was the following: each carbon has ~0.02 electron occupancy in a Ry 3s/3p orbital, which for all the ligands is a small chunk (0.14 e) of the total Ry occupation (0.7 e). I am thinking that perhaps this might be helpful in bonding to the much more diffuse iodine center. The iodine has around 0.02 e in two different d orbitals, which might help in bonding so many ligands. Perhaps this is due to a donor-acceptor interaction? The nitrogens utilize a pure 3s orbital with 0.01 electrons which I’m guessing because of their anionic and thus diffuse nature. Cheers, Matthew

  7. Matthew says:

    I am wondering how much of those Rydberg (0.7) electrons are just an artifact of using a large diffuse basis set? The total Rydberg population seems to be very basis set dependent. But is this an artifact or reality? Normally one would expect better results with larger basis sets, but I’ve read that diffuse basis sets can actually “confuse” wave function based population schemes like the NBO analysis ( DOI: 10.1021/cr200148b ) but other sources seem to indicate that in general, NBO results are relatively basis set independent. Of course who knows what exactly “NBO results” and “relative” mean without specific references…

  8. Henry Rzepa says:

    And as you showed, a diffuse basis leads to diffuse Rydberg electrons. On a different matter, Peter Gill has a fascinating new way of thinking about density functional theory. A boundary of conventional DFT is the “uniform electron gas”, which of course extends to infinity. In other words, it is “diffuse”. Peter instead constructs the density on a 3-sphere (a 4 dimensional object) which is not infinitely “diffuse”. One can only wonder whether such an approach leads to entirely new ways of thinking about diffuse Rydberg populations, and whether non zero occupancy of these Rydberg orbitals means hypervalency or not. See 10.1063/1.3665393 or 10.1007/s00214-011-1069-7 for more.

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