-1

Do $\ce{PH3F2}$ and $\ce{XeS2}$ exist, in accordance with Valence Bond Theory?

According to Drago's Rule, if the central atom of a compound is an element of the third period, and the attached atom has an electronegativity lesser than 2.5, then hybridisation does not occur, and bonding occurs solely in atomic orbitals.

In the case of a compound like $\ce{PH3F2}$, the fluorine atoms occupy the axial positions, while the hydrogen atoms occupy the equatorial positions. However, the electronegativity of fluorine is 4, hence the EN. difference with phosphorous is larger, making the axial bonds very strong, meanwhile the electronegativity of hydrogen is 2.1 (below 2.5), hence the equatorial bonds are very weak, and hybridisation shouldn't occur ie. $\ce{PH3F2}$ shouldn't exist, according to Valence Bond Theory.

Searching for answers online just led me to this Quora post, contradicting the answer given, which I found inconclusive.

And I couldn't find any explanation for $\ce{XeS2}$ not existing, at all.

This doubt stemmed from the following JEE (competitive exam) question:

Identify the number of molecules which do not exist.

$\ce{SBr6}$, $\ce{PH5}$, $\ce{XeF^-_3}$, $\ce{PH4F}$, $\ce{PH3F2}$, $\ce{OF4}$, $\ce{XeF^-_5}$, $\ce{XeO4}$, $\ce{XeS2}$

The answer given was: (7)

$\ce{SBr6}$, $\ce{PH5}$, $\ce{XeF^-_3}$, $\ce{PH4F}$, $\ce{PH3F2}$, $\ce{OF4}$, $\ce{XeS2}$

I understood the rationale behind all, except for $\ce{PH3F2}$ and $\ce{XeS2}$.

Edit: If anyone could answer this question with respect to VBT, VSEPR and hybridisation I'd gladly give you my upvote :)

Bongo Man
  • 361
  • 8
  • A one minute Google search turns up https://pubs.acs.org/doi/pdf/10.1021/ic00322a013 – Ian Bush Oct 31 '23 at 13:59
  • 1
    I suggest you read one of the many answers here that explain why hybridisation should not be used here, or arguably anywhere. One example https://chemistry.stackexchange.com/questions/18427/why-does-f-replace-the-axial-bond-in-pcl5 – Ian Bush Oct 31 '23 at 14:06
  • For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations. – Buttonwood Oct 31 '23 at 15:37
  • @Poutnik Considering that Chemistry is infamous for exceptions, I've gotten used to expecting the unexpected. Why don't you leave an answer explaining the rationale of your "chemist guts" (even if it doesn't incorporate theory)? – Bongo Man Oct 31 '23 at 15:45
  • https://chemistry.stackexchange.com/questions/64613/what-is-dragos-rule-does-it-really-exist – Mithoron Oct 31 '23 at 16:05
  • I don't see why $\ce{PH4^+ F-}$ wouldn't exist. Iodide does https://en.wikipedia.org/wiki/Phosphonium_iodide – Mithoron Oct 31 '23 at 16:10
  • There's no way you get Xe compound without electron withdrawing groups. Compare electronegativity of Xe and S for starters. – Mithoron Oct 31 '23 at 16:14
  • @Mithoron This is perplexing. I did some research online, and found this: [https://www.quora.com/How-does-PH3F2-exist-whereas-PH4F-does-not] I know that Quora is known for being unreliable, but the majority of the answers have the same point (PH3F2 exists, but PH4F doesn't), and are given by graduates from some of the top Indian universities, so I trust them. – Bongo Man Oct 31 '23 at 16:17
  • @Mithoron The OP probably means pentavalent PH4F like PF5, not the phosphonium fluoride. – Poutnik Oct 31 '23 at 16:36
  • @Poutnik So I highlighted its structure. With some "PH3F2", even being ionic wouldn't help because of this remaining P-F bond. – Mithoron Oct 31 '23 at 16:40
  • @Mithoron I do not object. – Poutnik Oct 31 '23 at 16:45
  • @Poutnik Not sure $\ce{PH4^+F^-}$ can make it. Fluoride might accept a proton displacing the extremely weak base phosphine. Substituted phosphonium fluorides are known. – Oscar Lanzi Oct 31 '23 at 17:57
  • 1
    The notion stated in the "Edit-statement" is simply not true and quite rude actually. Hybridisation is a necessary construct of Valence Bond Theory, which is an actual theory and not that handwavy stuff that is written in some textbooks. As such, hybridisation is not a concept that is frowned upon, at least when you try to really understand it. Sure, there are other ways of looking at things and you might not need the concept. Anyway, there is a time to use the concept of hybridisation and then there is a time where it simply is wrong. This has nothing to do with preference. – Martin - マーチン Nov 07 '23 at 17:25
  • @Martin-マーチン Thanks for your support. People on this site don't often side with newbies like me +1. – Bongo Man Nov 08 '23 at 11:07
  • I don't understand your comment. I'm saying that the last paragraph of your question is in my opinion wrong and rude. I'm also not supporting your question. All of the referenced molecules should not be explained with hybridisation. And what you call Valence bond theory is unfortunately just a simplified and broken down view of the real theory. I'm sorry to burst your bubble, but the jee stuff is notoriously bad at teaching the right tools to understand the world. – Martin - マーチン Nov 08 '23 at 12:42
  • @Martin-マーチン For now, I just want to get into a good college and make my parents proud ;), I can delay a little on understanding the world. But very well, I'm a man of honour, deleted the edit. – Bongo Man Nov 08 '23 at 18:18

2 Answers2

3

In theory, all of the mentioned compounds in the answer don't exist expect for $\ce{PH3F2}$. Difluorophosphorane has been characterized through spectroscopical method long time ago. Here is an excerpt from a paper1:

$\ce{PH3F2}$, first reported in 1971 by Seel and Velleman, is now accessible in pure form and has been unambiguously characterized by high resolution infrared spectroscopy. At present it marks the borderline between the well-established fluorophosphoranes $\ce{H_{n}PF_{5−n}, n = 2−5}$, and the unknown species $\ce{PH4F}$ and $\ce{PH5}$.

I am going to also shamelessly plug the paper2 mentioned by @IanBush in the comments.

Regarding $\ce{XeS2}$, it doesn't exist. However, oxosulfides of form $\ce{XeO_{n}S_{m}}$ are known3. Also, excimer $\ce{XeS}$* is known4.

References

  1. Gas-phase structure of difluorophosphorane determined by high resolution infrared spectroscopy, H. Beckers, H. Bürger, A. Rahner, Vol.54, 1991, DOI: 10.1016/S0022-1139(00)83877-4
  2. Preparation and spectroscopic characterization of difluorophosphorane, PH3F2. Phosphorus-31 NMR spectrum of protonated diphosphine, P2H5+, Rolf Minkwitz and Andreas Liedtke, Inorganic Chemistry 1989 28 (23), 4238-4242 DOI: 10.1021/ic00322a013
  3. Xenon oxides, sulfides, and oxysulfides. A theoretical ab initio investigation, Aristotle Papakondylis, Vol. 1015, 2013, DOI: 10.1016/j.comptc.2013.03.030
  4. Xenon oxide and xenon sulfide emission systems at 234 and 227 nm, J. Xu, D.W. Setser, J.K. Ku, Vol. 132, 1986, DOI: 10.1016/0009-2614(86)80640-6
Oscar Lanzi
  • 56,895
  • 4
  • 89
  • 175
Nilay Ghosh
  • 25,509
  • 26
  • 91
  • 197
  • Nice to have an Indian, who understands what a fellow JEE aspirant is going through, answer my question without relentlessly bashing me. +1 and marked as answer. – Bongo Man Nov 17 '23 at 06:19
2

As if to refute the hybridization theory once and for all, $\ce{PH3F2}$ (difluorophosphorane) exists! Andrews and Withnall[1] produced it and trapped it in an argon matrix by cocondensing phosphine and elemental fluorine into this matrix. The authors also mention that their result is the first experimental evidence for $\ce{PH2F}$, which "should be" more stable and easier to access according to the hybridization-based argument.

Reference

  1. Lester Andrews and Robert Withnall (1989). "Cocondensation reaction between phosphine and fluorine: matrix infrared spectra of difluorophosphorane, difluorophosphine and fluorophosphine". Inorg. Chem. 28, 3, 494–499. https://doi.org/10.1021/ic00302a023
Oscar Lanzi
  • 56,895
  • 4
  • 89
  • 175