Why does $\ce{F}$ replace an axial bond in $\ce{PCl5}$? I realize that it would be more stable there than at equatorial bond, but what is the reason of its stability? Similarly in $\ce{AB4}$ type of molecules with $\ce{sp^3d}$ hybridization (4 bond pairs and 1 lone pair) why is the geometry of the molecule that of a see-saw, where the lone pair is at equatorial position rather than at axial one?

My book states the reason to be "due to Bent's rule" but I have difficulty linking these two.
Also Bent's rule states that the hybridized orbitals of equivalent energy of central atom tend to give more %s character to the electropositive atom attached to it rather than the electronegative one. (e.g. - in $\ce{CH3F}$) Am I right? What else is more to this rule?

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    $\begingroup$ Just as a very general remark on the topic: The involvement of d-orbitals is marginal (usually below 1%), hence hybridisation involving these orbitals is not very likely. $\endgroup$ Commented Oct 14, 2014 at 9:29
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    $\begingroup$ It is not what I meant. The hypothesis of a $\ce{sp^3d}$ orbital has been disproven. Hypercoordination can and should be explained different. $\endgroup$ Commented Oct 14, 2014 at 9:44
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    $\begingroup$ Ohh! I am aware that VBT (& MOT) is a basic theory and isn't fully applicable to all cases and many of its concepts have become obsolete but I have only these two in my curriculum so an answer within the context of these will be more helpful than using other complex theories. Thanks for mentioning though! :) $\endgroup$
    – Shubham
    Commented Oct 14, 2014 at 10:07
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    $\begingroup$ Oh on the contrary, Valence Bond Theory and of course Molecular Orbital theory are not basic theories. They are the foundation of modern quantum chemistry and most molecules can be sufficiently explained by these theories. Just the assumption of the involvement of d orbitals is wrong. (I do unfortunately not have the time to write up an answer myself at the moment, but I am confident that someone else will do it.) $\endgroup$ Commented Oct 14, 2014 at 10:10
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    $\begingroup$ 1) Lone pair behaves is a very big ligand, and chlorine is bigger than fluorine. Angles to neigbors in axial position are ~90, while in equatorial ~120, so larger ligands tend to occupy equatorial positions. 2) Also, axial bonds are actually ~ 1/2 order with more significant negative charge on ligands, so more electronegative atoms should go there. 3) +1 to martin, d-orbitals are not involved, see here chemistry.stackexchange.com/questions/15172/… $\endgroup$
    – permeakra
    Commented Oct 17, 2014 at 7:48

1 Answer 1


Recently, there has been a lot of discussion of Bent's rule (see for example "What is Bent's rule?") here in SE Chem. Simply stated, the rule suggests that $\mathrm{p}$-character tends to concentrate in orbitals directed at electronegative elements.

Why does $\ce{F}$ replace an axial bond in $\ce{PCl5}$?

In order to answer this question, we need to start by understanding the bonding in $\ce{PCl5}$ and its fluorinated isomers. In introductory courses and texts, it is usually stated that $\ce{AX5}$ type molecules adopt a trigonal bipyramid geometry and are $\mathrm{sp^{3}d}$ hybridized. As the comments by Martin and permeakra point out, and as is learned in more advanced classes, this hybridization scheme is likely incorrect. There are several reasons that argue against $\mathrm{sp^{3}d}$ hybridization including: 1) the fact that $\mathrm{d}$ orbitals are relatively high in energy compared to $\mathrm{s}$ and $\mathrm{p}$ orbitals and therefore it is energetically quite costly to involve them in bonding; and 2) $\mathrm{d}$ orbitals in non-metals are very diffuse leading to poor overlap with other orbitals and any resulting bonds would be very weak.

A reasonable hybridization alternative involves what is termed hypercoordinated bonding; where 3 center, 4 electron bonds are involved. Applying this concept to $\ce{PCl5}$ we would say that the central phosphorus atom is $\mathrm{sp^2}$ hybridized. Thus, there would be 3 $\mathrm{sp^2}$ orbitals (these will be used to create the equatorial bonds) and a $\mathrm{p}$ orbital (this will be used to create our axial bonds) emanating from the central phosphorus atom.

Hybridisation of phosphorus in PCl5

The $\mathrm{p}$ orbital contains 2 electrons and will form bonds to 2 ligands (chlorine or fluorine in the case at hand). For simplicity, let's say that these ligands also use $\mathrm{p}$ orbitals for bonding (but it could be any type of orbital, $\mathrm{sp^3}$ or whatever is appropriate) and each of these orbitals contains one electron for bonding. This is our 3-center-4-electron bond and its MO diagram is pictured below. Notice how the four electrons are distributed - there are two electrons in the HOMO which is a non-bonding M.O., so the bond order in this bond is reduced.

MO diagram of Cl–P–Cl 3c4e bonding

This reduced bond order in the bond using the phosphorus $\mathrm{p}$ orbital explains why the axial bonds in $\ce{AX5}$ type molecules are longer than the equatorial bonds.

P-Cl bond lengths

Now that we understand the bonding in $\ce{PCl5}$ we can consider the case of $\ce{PCl4F}$ and how Bent's rule applies to the situation. First, note that fluorine is more electronegative than chlorine. As stated above, Bent's rule suggests that more electronegative ligands prefer to form bonds with orbitals that are high in $\mathrm{p}$-character. Why is this? $\mathrm{s}$-Orbitals are lower in energy than $\mathrm{p}$-orbitals. Therefore electrons are more stable (lower energy) when they are in orbitals with more $\mathrm{s}$-character. The two electrons in the $\ce{P-F}$ bond will spend more time around the electronegative fluorine and less time around phosphorus. If that's the case (and it is), why "waste" precious, low-energy, $\mathrm{s}$-orbital character in an orbital that doesn't have much electron density to stabilize. Instead, save that $\mathrm{p}$-character for use in phosphorus hybrid orbitals that do have more electron density around phosphorus (like the $\ce{P-Cl}$ bonds). So, as a consequence of Bent's rule, we would expect phosphorus to use the orbital with lowest $\mathrm{s}$-character, the axial $\mathrm{p}$-orbital, to form the $\ce{P-F}$ bond; and the orbitals with more $\mathrm{s}$-character, the equatorial $\mathrm{sp^2}$ orbitals, to form $\ce{P-Cl}$ bonds.

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    $\begingroup$ May I just clarify? Is it always true that the filling of nonbonding MOs lead to a reduction in bond order? Is it also possible to quantify this reduction in bond order? I was thinking another explanation as to why there is a longer bond in the axial positions is due to the decreased effectiveness of the p-p-p overlap. $\endgroup$ Commented Nov 23, 2017 at 8:36
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    $\begingroup$ Because p orbitals are more diffuse as compared to the sp2-hybrids used in the bonds in the equatorial positions. $\endgroup$ Commented Nov 23, 2017 at 8:41
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    $\begingroup$ @TanYongBoon Is it always true that...? I think so, but feel free to ask that and your following question as a new question. As to your argument about the decreased effectiveness of p-p-p overlap, I'm not sure. I think that a p-orbital is less diffuse then an sp2 orbital, but again that might be another good question to ask. $\endgroup$
    – ron
    Commented Nov 23, 2017 at 14:39
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    $\begingroup$ Another question I have is: It is not necessary for the two peripheral atoms involved in a 3-c 4-e bond to be the same right? In this case, the PCl4F would have a Cl-P-F 3-c 4-e bond. $\endgroup$ Commented Nov 25, 2017 at 1:32
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    $\begingroup$ @Abcd Why not? Two large, equivalent lobes are exposed and available. This is the basis for hypercoordinate bonding. You've seen it before in the common example of the SN2 transition state. The hydrogen 1s orbital can also form 2 bonds. In that case it is a 3-center 2-electron bond (search that term on Chem SE for more detail). Common examples include the bridging hydrogen in $\ce{B2H6}$ and in nonclassical carbocations. $\endgroup$
    – ron
    Commented Mar 24, 2018 at 19:24

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