I have come across a following line in my textbook:

$\ce{PI5}$ does not exist due to steric hindrance.

I searched for the explanation of steric hindrance in Wikipedia but couldn't understand it. Can anyone explain steric hindrance in simpler language?

Wikipedia says:

The existence of $\ce{PI5}$ has been claimed intermittently since the early 1900s. The claim is disputed: "The pentaiodide does not exist (except perhaps as $\ce{PI3\cdot I2}$, but certainly not as $\ce{PI4+I-}$)".

What does this mean?

  • 2
    $\begingroup$ Interestingly, $\ce{PI4^+}$ does exist. So, the questions are 1)why phosphorous avoids forming more than four bonds (just like nitrogen) and 2) why $\ce{PI4^+}$ (and for that matter $\ce{PBr4^+}$) could be a rather strong oxidizer, strong enough to oxidize corresponding anion. $\endgroup$
    – permeakra
    Commented Jul 15, 2015 at 21:00

2 Answers 2


$\ce{PI5}$ versus $\ce{CI4}$

More is involved than steric hindrance in preventing the formation of a $\ce{PI5}$ isomer with pentacoordinate phosphorus. There is a subtle antibonding interaction in this $\ce{PI5}$ isomer which does not occur in $\ce{CI4}$.

This answer describes the hidden antibonding ligand-ligand interactions in the "nonbonding" ligand-based orbitals of "expanded octet" molecules such as $\ce{SF6}$. The ligand-ligand antibonding interaction is small enough to minimize its impact on $\ce{SF6}$. But $\ce{OF6}$, where central atom is smaller and the ligands are closer together, is less fortunate.

Similarly in $\ce{PI5}$ there is one ligand-based orbital that becomes antibonding with the ligand-ligand interactions. Occupation of that iodine-based orbital combined with its crowding around the smaller phosphorus atom thus poses a combined electronic-steric hindrance.

With $\ce{CI4}$, however, there is no such ligand-based "hidden antibonding" orbital because there is no formal octet expansion, and without this electronic factor $\ce{CI4}$ can hold together (at least if it's chilled) despite the unfavorable steric condition.*

*This answer mentions the unusual deep red/purple color of this saturated compound. The relative weakness of the carbon-iodine bond, which moves normally ultraviolet electronic transitions into the visible range, explains the color.


The simplest explanation is that you can not fit five iodine atoms around a phosphorous atom.

Consider the pentachloride equivalent - this is already quite crowded. Since chlorine atoms have a smaller covalent radius than iodine, you can get five of them around the phosphorous. Of course, all this depends on the bond length of P-I and P-Cl and the size of the phosphorous atom.

Interestingly, there seems to be some suggestion of an uranium pentaiodide compound, for example this PubChem compound. However, there are no references on that page, so presumably this has never been synthesised either?

In general, steric hindrance means geometric conformations of a molecule that are impossible (or difficult to achieve) due to clashes between groups and atoms within the molecule.

  • 1
    $\begingroup$ Uranium pentaiodide is also hypothetical. $\endgroup$
    – bon
    Commented Jul 15, 2015 at 17:04
  • $\begingroup$ Quite possibly (the reference on that page is from 1968!) however, there is a tantalum compound that has been prepared (although I don't quite see how the shared edge X2Y10 structures are pentagonal bipyramidal, but whatever). $\endgroup$
    – gilleain
    Commented Jul 15, 2015 at 17:07
  • 2
    $\begingroup$ Sterical hindrance does NOT prevent formation of $\ce{CI4}$ (and similarly crouded $\ce{C6I6}$ for that matter), so it is not sufficient to say that $\ce{PI5}$ is too hindered to form $\endgroup$
    – permeakra
    Commented Jul 15, 2015 at 17:25
  • $\begingroup$ @permeakra Admittedly, I should not have assumed that steric hindrance is truly the main reason for preventing PI5. Are you saying that the carbon-iodine bonds in CI4 are shorter than phosporous-iodine bonds? Presumably CI4 is also tetrahedral, which PI5 could not be. $\endgroup$
    – gilleain
    Commented Jul 15, 2015 at 17:32
  • 3
    $\begingroup$ @gilleain both $\ce{CI4}$ and $\ce{C6I6}$ have short I-I contacts well below $\ce{I}$ van-der-waals radius. In fact it seems that in $\ce{C6I6}$ there is enough overlapping to fully delocalize two positive charges over iodines, forming a rather stable $\ce{[C6I6]^{2+}}$ $\endgroup$
    – permeakra
    Commented Jul 15, 2015 at 20:48

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