Why is the $\ce{I-Cl}$ bond length in $\ce{ICl2-}$ (charge of -1) is longer than in $\ce{ICl2+}$ (charge of +1) ?

Here are a few explanations that I have in mind. Please critique.

  1. I in $\ce{ICl2-}$ is $sp^3d$ hybridized. $\ce{I}$ in $\ce{ICl+}$ is $sp^3$ hybridized. Since additional promotion energy is required to form the $sp^3d$ hybrid compared to the $sp^3$ hybrid, the bond lengths in $\ce{ICl2-}$ is longer.

  2. For $\ce{ICl2-}$, if we let $\ce{I}$ be $sp^2$ hybridized such that the 3 lone pairs lie 120 apart in a plane, and then let the 2 $\ce{Cl}$ atoms and the $\ce{I}$ atom lie in a straight line, then we can form a 3-centred, 4 electron system. This is weaker than the 2-centred, 2 electron system in the $\ce{I-Cl}$ bond in $\ce{ICl+}$.

  3. As $\ce{I}$ in $\ce{ICl2+}$ has a higher effective nuclear charge, its valence orbitals are smaller and more compact, hence the $\ce{I-Cl}$ covalent bond is stronger as the electrostatic attraction between the nuclei and the overlap density is stronger.


I’ll agree with ron concerning the non-hybridisation of d-orbitals and the description of $\ce{ICl2-}$. However, I disagree with ron’s description of $\ce{ICl2+}$, which was calculated to have an $\ce{Cl-I-Cl}$ bond angle of 98°.[1]

I also believe that the contribution of the s-orbital to any bonding, non-bonding or anti-bonding orbitals in the iodine atom is neglegible, the reasoning being that the energy difference between iodine’s s- and p-orbitals is larger than for the atoms of the second period.

The $\ce{ICl2+}$ ion can thus be explained by assuming one double-filled s- and one double filled p-orbital plus two half-filled p-orbitals on iodine. Each of the two half-filled orbitals then create a normal $\sigma$-bonds to a corresponding p-orbital of one of the chlorine atoms, predicting a bond angle of 90° (closer to 98° than if we assumed sp³-hybridisation which would suggest 109.5°).

This means the following:

  • In $\ce{ICl2+}$, we have to bonds with a bond order of 1 to each of the chlorines. Compared to a bond order of ½ for $\ce{ICl2-}$ this means a shorter bond length.

  • The ion’s charge, which is located mostly on the iodine (electronegativity!) means that the iodine atom is contracted and the two chlorines can move closer. This effect is not present in $\ce{ICl2-}$, where the charge is distributed evenly across the two chlorine atoms.

For the reference and on the topic of predictions, I would predict the bond length of $\ce{ICl}$ to be comparable but slightly longer than that of $\ce{ICl2+}$, while still being notably shorter than that of $\ce{ICl2-}$.

[1]: F. Bailly, P. Barthen, H.-J. Frohn, M. Köckerling, Z. anorg. allgem. Chem. 2000, 626, 2419.

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The iodine atom in $\ce{ICl2}$ is what is termed hypercoordinated (or hypervalent). Basically, this means that the 3 atoms combine to produce 3 molecular orbitals (the 3-center 4-electron model). There is no need to invoke d-orbitals.

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In neutral $\ce{ICl2}$ there are three electrons spread across the two $\ce{I-Cl}$ bondss. Placing those electrons in our molecular orbitals, 2 go into the bonding orbital and one into the nonbonding orbital.

In $\ce{ICl2^{+}}$ we have removed the one nonbonding electron. As a consequence there will be less electron-electron repulsion and only the bonding molecular orbital is occupied, because of this we would expect shorter bond lengths in $\ce{ICl2^{+}}$ compared to $\ce{ICl2}$.

In $\ce{ICl2^{-}}$ we have added another electron and it will go into the nonbonding orbital. This will increase the electron-electron repulsion in the molecule without increasing bonding, because of this we would expect longer bond lengths in $\ce{ICl2^{-}}$ compared to $\ce{ICl2}$.

These arguments tell us that we would have shorter bond lengths in $\ce{ICl2^{+}}$ compared to $\ce{ICl2^{-}}$.

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  • $\begingroup$ This paper suggests a $C_{2\mathrm{v}}$ symmetry for $\ce{ICl2+}$ with a bond angle of 98° contradicting the assumption that any kind of 4e-3c bond be present. (F. Bailly, P. Barthen, H.-J. Frohn, M. Köckerling, Z. anorg. allgem. Chem. 2000, 626, 2419.) $\endgroup$ – Jan May 26 '15 at 15:01
  • $\begingroup$ @Jan Thanks for pointing out the theoretical geometry for the cation. What level of computation did they use? +1 for your nice answer based on that geometry. $\endgroup$ – ron May 26 '15 at 16:48
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    $\begingroup$ It’s in the linked paper, table 1, saying LANL2DZ/RHF. It was ‘only’ the first paper I could find entering $\ce{ICl2+}$ into SciFinder and gathering references with crystal structure although there is no crystal structure easily found. I would love to find one though to verify either of our claims $\endgroup$ – Jan May 26 '15 at 17:13
  • $\begingroup$ The article is behind a pay wall, I can only see the abstract. I'm not convinced that a 15 year old calculation has predicted the correct geometry. Why have the chlorines so close together (steric and electronic repulsion) when it is not necessary? The 3-center 3-electron bond model (for the cation) would have the chlorines further apart and still have the lone pairs in low energy sp or (s and p) orbitals. Perhaps you or @Martin-マーチン could perform a better calculation? $\endgroup$ – ron May 26 '15 at 21:33
  • $\begingroup$ Unfortunately I’m not in a group that does calculations any more. I could ask an ex-colleague, though. However, I remember our first inorganic chemistry lecture talk about how to predict interhalogen compounds and their ions with the explicit step of giving the central atom (iodine) the charge, drawing 2-electron 2-centre bonds until octet and filling in the remaining atoms via 4-electron 3-centre bonds — using that technique, I arrive at the same structure. Additionally, the same professor presented us with the structure of the related cation … $\endgroup$ – Jan May 27 '15 at 14:03

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