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The axial bonds of $\ce{PF5}$ are longer than those of the equatorial positions.

One explanation is that because the axial bonds are experiencing more repulsion than those of the equatorial and this leads to longer bonds.

Another explanation is that the axial bonds of p/d hybrids, whereas the equatorial bonds are s/p hybrids.

I am able to comprehend and appreciate the first explanation. However, I am unable to understand how there are differences in the hybrids. I had always considered d hybrids a dubious argument at best, and I believed the hybridization would lead to degenerate orbitals.

However, the explanation provided would suggest otherwise, that there are different hybridized orbitals of different energies at different portions of a single hybridized atom (excluding lone pairs. I understand those being non-hybridized in a hybridized atom).

Thus, my question is how the hybrid argument that was presented works and how would one be able to realize the different hybridized states of an atom in a molecule (as an approximation).

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  • $\begingroup$ I've been thinking and I thought that perhaps rather than hybridization, the p/d explanation can account for the Non-hybridized state as the bonding is not degenerate and could help explain why a single bond is longer. For Fluorine, instead of a hybridized state, it would bond with a p, not sp3. For P, it would bond with a d? (I would say p) instead of p (I would say s). However, this is only a guess and it would be nice to have someone give their ideas. $\endgroup$ – Andy Apr 17 '15 at 3:38
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    $\begingroup$ This earlier SE Chem post answers your question using $\ce{PCl_5}$ as the example, but the same arguments apply to $\ce{PF_5}$ as well $\endgroup$ – ron Apr 17 '15 at 13:52
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Older textbooks decided to use a d-orbital formalism for elements of the third and lower (in the periodic table) periods to explain hyperconjugation. Nowadays, most of these compounds are explained with multi-centre bonds.

I could go stealing pictures and explanations from ron’s great answer, but I’m not going to. Rather, I’m just going to quickly explain the gist and redirect you to that answer.

If you take away two of the fluorines, then you can imagine the $\ce{PF3}$ fragment to be sp²-hybridised with a remaining p-orbital containing two electrons. (This description is wrong, as can be seen from the pyramidial structure of $\ce{PCl3}$ and its analogues, but we’ll keep it as a working basis.) The p-orbital can now interact with two surrounding further fluorine atoms that approach it from above and below with an electron each in the way that is shown in the image in ron’s answer. The resulting three orbitals will thereby create a four-electron three-centre bond. This bond can also be described mesomerically, by assuming a $\ce{P-F}$ single bond, a positive charge on phosphorous and a negative charge on the other fluorine, with a second mesomeric structure as a mirror image of the first.

From this mesomeric consideration you can see that the bond order of the axial $\ce{P-F}$ bonds is not 1 but rather closer to 0.5. The greater the order the shorter the bond so the smaller the order the longer the bond — and there you have your result.

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