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In an earlier SE Chem post, the structure and bonding in the trigonal bipyramid molecule $\ce{PCl_5}$ was discussed. The molecular hybridization is pictured below. The axial bonds are hypervalent (this concept is explained in the earlier post) and they are constructed from p orbitals. The equatorial bonds are constructed from $\ce{sp^2}$ orbitals. General Rule #2 suggests that the equatorial bonds will be shorter than the axial bonds since the equatorial bonds contain more s-character.

Someone, after reading that last paragraph, is going to notice that there are 5 equatorial fluorines, but only 4 bonding sites with the two p orbitals - what gives? First off, note that the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals on iodine form hypervalent bonds with the p orbital on fluorine and so yes, there are only 4 equatorial points of attachment. But note that the abstract also mentioned that the equatorial bonds are semi-ionic. This suggests that we have a resonance structure where we have an [$\ce{IF_{6}^{+}}$][$\ce{F^{-}}$] ionic contributor. We can draw 4 more resonance structures just like this for the other 4 equatorial $\ce{I-F}$ bonds. This means that mixed in with our hypervalent p-p bonds is some ionic character - this allows us to make 5 equivalent equatorial bonds. Said differently, our 5 equatorial bonds are all identical resonance hybrids obtained by mixing 2 hypervalent p bonds (each one providing bonding for 2 fluorines) with 1 ionic bond (providing bonding for 1 fluorine).

In an earlier SE Chem post, the structure and bonding in the trigonal bipyramid molecule $\ce{PCl_5}$ was discussed. The molecular hybridization is pictured below. The axial bonds are hypercoordinate (this concept is explained in the earlier post) and they are constructed from p orbitals. The equatorial bonds are constructed from $\ce{sp^2}$ orbitals. General Rule #2 suggests that the equatorial bonds will be shorter than the axial bonds since the equatorial bonds contain more s-character.

Someone, after reading that last paragraph, is going to notice that there are 5 equatorial fluorines, but only 4 bonding sites with the two p orbitals - what gives? First off, note that the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals on iodine form hypercoordinate bonds with the p orbital on fluorine and so yes, there are only 4 equatorial points of attachment. But note that the abstract also mentioned that the equatorial bonds are semi-ionic. This suggests that we have a resonance structure where we have an [$\ce{IF_{6}^{+}}$][$\ce{F^{-}}$] ionic contributor. We can draw 4 more resonance structures just like this for the other 4 equatorial $\ce{I-F}$ bonds. This means that mixed in with our hypercoordinate p-p bonds is some ionic character - this allows us to make 5 equivalent equatorial bonds. Said differently, our 5 equatorial bonds are all identical resonance hybrids obtained by mixing 2 hypercoordinate p bonds (each one providing bonding for 2 fluorines) with 1 ionic bond (providing bonding for 1 fluorine).

In an earlier SE Chem post, the structure and bonding in the trigonal bipyramid molecule $\ce{PCl_5}$ was discussed. The molecular hybridization is pictured below. The axial bonds are hypervalent (this concept is explained in the earlier post) and they are constructed from p orbitals. The equatorial bonds are constructed from $\ce{sp^2}$ orbitals. General Rule #2 suggests that the equatorial bonds will be shorter than the axial bonds since the equatorial bonds contain more s-character.

Here is a link to an article discussing the structure and bonding in $\ce{IF_7}$. Only the abstract is freely available. For completeness, I'll mention that there was an earlier SE Chem question related to $\ce{IF_7}$, but it doesn't seem to add much to the conversation. Returning to the published abstract, the key part of the abstract states,

In an earlier SE Chem post, the structure and bonding in the trigonal bipyramid molecule $\ce{PCl_5}$ was discussed. The molecular hybridization is pictured below. The axial bonds are hypervalent (this concept is explained in the earlier post) and they are constructed from p orbitals. The equatorial bonds are constructed from $\ce{sp^2}$ orbitals. General Rule #2 suggests that the equatorial bonds will be shorter than the axial bonds since the equatorial bonds contain more s-character.

Here is a link to an article discussing the structure and bonding in $\ce{IF_7}$. Only the abstract is freely available. For completeness, I'll mention that there was an earlier SE Chem question related to $\ce{IF_7}$, but it doesn't seem to add much to the conversation. Returning to the published abstract, the key part of the abstract states,

As the following structuraL diagram shows, this is indeed the case.

As you note in your question, the equatorial bonds are now longer than the axial bonds. Based on what we've discussed thus far, we might make an initial "guess" that in $\ce{IF_7}$ the axial bonds have more s-character than the equatorial bonds. As it turns out, this is correct.

The abstract tells us that the equatorial bonds are formed from the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals; the axial bonds are formed from the $\ce{sp_z}$ hybrid orbitals (after using two p orbitals for equatorial bonding we are left with one p orbital and 1 s orbital on the central iodine, hence, when they combine two sp hybridized orbitals result). That is the answer to the question. As we guessed, there is more s-character in the axial bonds than the equatorial bonds. Hence, the axial bonds are now shorter than the equatorial bonds.

As the following structural diagram shows, this is indeed the case.

As you noted in your question, the equatorial bonds are now longer than the axial bonds. Based on what we've discussed thus far, we might make an initial "guess" that in $\ce{IF_7}$ the axial bonds have more s-character than the equatorial bonds. As it turns out, this is correct.

The abstract tells us that the equatorial bonds are formed from the $\ce{p_{x}}$ and $\ce{p_{y}}$ orbitals; the axial bonds are formed from the $\ce{sp_{z}}$ hybrid orbitals (after using two p orbitals for equatorial bonding we are left with one $\ce{p_{z}}$ orbital and one s orbital on the central iodine, hence, when they combine two sp hybridized orbitals result). That is the answer to the question. As we guessed, there is more s-character in the axial bonds than the equatorial bonds. Hence, the axial bonds are now shorter than the equatorial bonds.