3
$\begingroup$

Why is it that the axial bond length is less than equatorial bond length in $\ce{PF2Cl3}$ and $\ce{SF2Cl2}$ even though both have trigonal bi-pyramidal geometry?

I know that in general axial bond length is greater than equatorial bond length in trigonal bi-pyramidal structures (as in the case of $\ce{PCl5}$). (See: PBP vs TBP geometry?)

Improve this question
$\endgroup$
3

1 Answer 1

Reset to default
5
$\begingroup$

For trigonal bipyramidal structure, we know that lone pairs are preferred first to be positioned in the equatorial position. So in $\ce{SF2Cl2}$ lone pairs will be positioned in the equatorial position.

After placing lone pairs in the equatorial position, double bonds are then preferred in the equatorial position but as there are no double bonds in the structures we are considering, so we jump on to the next rule i.e. atoms having higher electronegativity is positioned at axial position and atoms having low electronegativity is positioned at equatorial position. You can also interpret this in another way. You can also say that atoms having higher atomic size is positioned in the equatorial plane and atoms having lower atomic size is positioned in the axial plane.

Now coming back to your structures, $\ce{PF2Cl3}$ and $\ce{SF2CL2}$ we will see structures in this manner:

structure

$\ce{F}$ is more electronegative than $\ce{Cl}$, hence the $\ce{P-F}$ bonds are going to be shorter than $\ce{P-Cl}$ bonds and $\ce{S-F}$ bonds are going to be shorter than $\ce{S-Cl}$ bonds. As $\ce{F}$s (fluorines) are located in the axial position, hence these $\ce{P-F}$ or $\ce{S-F}$ bonds which are actually axial bonds are shorter than that of equatorial bonds (i.e. $\ce{P-Cl}$ or $\ce{S-Cl}$ bonds).

In $\ce{PCl5}$ all the atoms around $\ce{P}$ are basically the same i.e. $\ce{Cl}$. So the electronegativity factor of $\ce{Cl}$ doesn't play any role in reasoning out why axial bonds are longer than equatorial bonds in case $\ce{PCl5}$. The reason is - in $\ce{PCl5}$ the axial bonds suffer repulsion from the equatorial bonds so they become long. The equatorial bonds providing repulsion to axial bonds also happens in case of $\ce{PF3Cl3}$ and $\ce{SF2Cl2}$, however this factor is overlooked/ not considered in cases of $\ce{PF2Cl3}$ and $\ce{SF2Cl2}$ because there this factor/ effect is not dominating. In those cases as the type of atoms attached (i.e. electronegative atoms) are different, hence the effect of bond length shortening due to the electronegative atom attached is the dominating factor.

Improve this answer
$\endgroup$
7
  • $\begingroup$ What about the molecule XeF3Cl4. I have a reference that it has 2 Cl in axial positions and rest atoms in equatorial positions. Our teacher gave us a rule that more electronegative atom is present at a greater distance i.e. the bond which is longer in these molecules and thus explained the structure of XeF3Cl4 by saying the axial bond length is shorter due to less repulsion (angle is 90 compared to 72 in equatorial) and so Cl being less electronegative will occupy axial position and F will take equatorial position (including the remaining Cl atoms). How far is this reasoning true? $\endgroup$
    – Sourav Suman
    Commented Mar 21, 2017 at 9:12
  • $\begingroup$ If you are concerned with octahedral structure, then there is no such axial and equatorial planes because all the planes in octahedral are identical. In case of octahedral structures, you can have the most electronegative atom or lone pair or double bond any where you want. Axial and equatorial terms are only used in the case of trigonal bi pyramidal structures. So if your given molecule (Xef3Cl4) exist, then you can have your fluorine any where you want because most probably i feel your molecule will be octahedral and hence you can have fluorine at any position you want. $\endgroup$
    – Yb609
    Commented Mar 21, 2017 at 10:43
  • $\begingroup$ Can you say me were did you found XeF3Cl4 molecule? If that structure exist then i think it would be in octahedral form and hence you can have fluorine at any position you want because all the positions in octahedral geometry is identical. $\endgroup$
    – Yb609
    Commented Mar 21, 2017 at 10:44
  • $\begingroup$ Well our teacher explained using that molecule as a reference (I still have it in my notebook). But he gave a pentagonal bi-pyramidal structure for it and hence the axial and equatorial bonds. $\endgroup$
    – Sourav Suman
    Commented Mar 21, 2017 at 11:20
  • $\begingroup$ Though now I think it must have been a hypothetical example just for explanation as I cannot find any such molecule over the internet. $\endgroup$
    – Sourav Suman
    Commented Mar 21, 2017 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.