How can there be back-bonding in PF3?

  • #1

Main Question or Discussion Point

Phosphorus: sp3 hybridized
Fluorine (each one of three): sp3 hybridized
Back bonding can occur between only PURE orbitals ie those which are unhyridized.
Let's look at e- config. of both of them to be a little clear of what I'm asking:
P:
upload_2018-7-10_21-18-43.png

F:
upload_2018-7-10_21-19-4.png


Since fluorine is sp3 hybridized, the orbitals ie either of px, py or pz (2 of which are filled and are lone pairs on F of the 3 total lone pairs) can't participate in back bonding with the unhiybridized d orbital of Phosphorus.

My question is, then how does it show back bonding? I know it does, but how?
 

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  • #2
HAYAO
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  • #3
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Phosphorus: sp3 hybridized
Fluorine (each one of three): sp3 hybridized
Back bonding can occur between only PURE orbitals ie those which are unhyridized.
Let's look at e- config. of both of them to be a little clear of what I'm asking:
P: View attachment 227890
F:View attachment 227891

Since fluorine is sp3 hybridized, the orbitals ie either of px, py or pz (2 of which are filled and are lone pairs on F of the 3 total lone pairs) can't participate in back bonding with the unhiybridized d orbital of Phosphorus.

My question is, then how does it show back bonding? I know it does, but how?
In general , every compound in which central atom has vacant orbitals and adjacent atoms(or even molecules) have lone pair of electrons back bonding occurs.
Here P has vacant 3d orbitals even when it is sp3 hybridized to form normal covalent bonds with p orbitals of F atoms( note that the hybridization occurs only to the orbitals of the central atom ). Also there are 3 lone-pairs available for each F atom. This satisfies the general condition and thus back bonding is likely to exist in PF3 molecule.
Hope you are clear.
 
  • #4
DrDu
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Phosphorus: sp3 hybridized
Fluorine (each one of three): sp3 hybridized
Back bonding can occur between only PURE orbitals ie those which are unhyridized.
Let's look at e- config. of both of them to be a little clear of what I'm asking:
P: View attachment 227890
F:View attachment 227891

Since fluorine is sp3 hybridized, the orbitals ie either of px, py or pz (2 of which are filled and are lone pairs on F of the 3 total lone pairs) can't participate in back bonding with the unhiybridized d orbital of Phosphorus.

My question is, then how does it show back bonding? I know it does, but how?
Hybridization is not a property of an atom but a way to explain certain types of bonding, and geometries more easily. In fluorine, the energetic difference between s and p orbitals is so high, that hybridization makes no sense and is also not necessary as a linear bond geometry can be explained without hybridization.
What is even more problematic is that bonding and backbonding involving d-orbitals in main group compounds has since long been disproven. Nevertheless chemistry teachers seem still to be widely ignorant about this fact. See for example, https://onlinelibrary.wiley.com/doi/abs/10.1002/anie.198402721
 
  • #5
HAYAO
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As a reference, here's what happens when you calculate PF3 in Gaussian 09W with NBO (natural bonding orbitals). Essentially, NBO analysis gets you as close as possible to the Lewis structure of a molecule, which would help you understand the bonds.

Computational details: DFT (B3LYP), aug-ccPVDZ, Structure optimization, full NBO analysis.

The result yielded that PF3 have 8 core electron pairs, 3 two-center bonds, 0 three-center bonds, and 10 lone pairs.
Lewis occupancy was 99.420% and non-Lewis occupancy was 0.580% (Valence 0.408%, Rydberg 0.173%). Basically, this means that the molecule is 99.420% Lewis structure, roughly speaking.

The 3 two-center bonds (which is what we are looking at) was this:
1st P-F bond:
P (14.56%) s(11.36%) p(84.45%) d (4.20%)
F (85.44%) s(28.28%) p(71.55%) d (0.17%)
1.jpg
2nd P-F bond:
P (14.56%) s(11.36%) p(84.45%) d (4.20%)
F (85.44%) s(28.28%) p(71.55%) d (0.17%)
2.jpg
3rd P-F bond:
P (14.56%) s(11.36%) p(84.45%) d (4.20%)
F (85.44%) s(28.28%) p(71.55%) d (0.17%)
3.jpg

Basically, 14.56% of the electrons (of a single bond) lies on the P atom and 85.44% of the electrons lies on the F atoms.
The percentage of s, p, and d shows the degree of hybridization of each atomic-like orbitals. You can see here that the hybridization of P is not sp3 but something a little more complicated than that. On the other hand hybridization of F seems to be close to being sp3.



As for the backbonding, I have never tried NBO analysis for analyzing backbonding so I do not know precisely of the procedures. I am assuming that the choice of the reference bonds (turning off donor-NBO-to-acceptor-NBO interaction) is crucial in this case to correctly account for how strong the backbonding energy is. I'll try it if I have the time since I am also interested.
 

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  • #6
DrDu
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First I want to mention that this is a density functional study and thus not directly comparable to valence bond. The small d-orbital contribution is not a sign of backbonding. Rather, the d-orbitals serve as polarisation functions. GAMESS contains the possibility to do true VB calculations, however, I don't have it installed. Maybe somebody wants to try out.
 
  • #7
HAYAO
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First I want to mention that this is a density functional study and thus not directly comparable to valence bond. The small d-orbital contribution is not a sign of backbonding. Rather, the d-orbitals serve as polarisation functions. GAMESS contains the possibility to do true VB calculations, however, I don't have it installed. Maybe somebody wants to try out.
Hence the reason why I performed NBO calculation. But yes, of course, DFT is not a valence bond calculation.

I didn't claim that d-orbital contribution is a sign of backbonding. Just a hybridization, or in a sense polarization like you said. Analyzing backbonding requires to turn off certain interactions and is not simple in this case. I think I have written that quite explicitly above...
 

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