$\ce{I2Cl6}$ is a dimer of $\ce{ICl3}$. $\ce{ICl3}$ is T-shaped, with $\ce{I}$ being $\mathrm{sp^3d}$ and having two lone pairs.

I read in Inorganic Reaction Mechanisms by R. K. Sharma that $\ce{I2Cl6}$ is formed by the extra two $\ce{Cl-I}$ bridge bonds. What I don't understand is, why does the bridge bond form? And is it a coordinate bond or something like a banana bond (as in $\ce{B2H6}$)?

Source: Sharma, R. K. Inorganic Reaction Mechanisms; Discovery Publishing House: 2007.


1 Answer 1


The bonding situation in the compound $\ce{(ICl3)2}$ is by far more complex than what is depicted in this book. The molecule itself has very high symmetry, i.e. $D_\mathrm{2h}$, that needs to be satisfied in the molecular orbital/ valence bond description.[1,2] That is why one cannot distinguish between the bridging chlorine bonds.

Analysing this bonding situation in the simple concept of Lewis-structures is not trustworthy and will lead to wrong conclusions. I have performed a short gas-phase optimisation at the DF-BP86/def2-SVP level of theory and analysed the resulting wave function with the Natural Bond Orbital theory. Covalent treatment of any of the bonds seems to be wrong. In this case I assumed all single bonds. Due to symmetry reasons, we only need to be concerned with two bonds. The outer $\ce{I-Cl}$ bond is polarised 93% towards the chlorine, but only occupied by 1.25 electrons. The remaining 0.75 electrons can be found in anti-bonding lone pair orbitals. This is not even close to expected Lewis structure behaviour. The bridging $\ce{I-\mu{}Cl}$ bond is still 81% polarised towards the chlorine, and has an occupation of 1.92 electrons. This only tells us, that this bond has a higher covalent character.

The analysis with the Quantum Theory of Atoms In Molecules (QTAIM) gives us more tools to analyse the electron density. In here the picture is quite obvious. All of the bonds in the molecule are of highly ionic character. The electron density at the bond critical point of the outer $\ce{I-Cl}$ is with $0.08~\mathrm{e\,a.u.^{-3}}$ very low and the Laplacian is positive at this point. The electron density at the bridging bond is even lower with $0.04~\mathrm{e\,a.u.^{-3}}$, again with a positive Laplacian.

QTAIM Laplacian plot of (ICl3)2
(Laplacian distribution, solid blue lines indicate charge depletion $∇^2ρ<0$, dashed blue lines indicate charge accumulation $∇^2ρ>0$, red spheres are bond critical points, pink spheres are ring critical points, black lines are bond paths, dark red lines are zero flux surfaces.)

This finding is fairly consistent with the observation, that the melt of $\ce{(ICl3)2}$ conducts electricity, which according to Wikipedia is a hint for partial ionic dissociation. $$\ce{(ICl3)2 <=> ICl2+ + ICl4- }$$

However, the planarity of this compound in solid state can be attributed to the possibility of having a somewhat quite extensive pi-bonding system as part of the small covalent bonding. For final conclusions about this, the results are not sufficient.

What can be said, judging from the molecular orbitals of $\ce{(ICl3)2}$, is that apart from the occupied core d-orbitals of iodine, no other d-orbitals are involved in any bonding - as it is expected. Shown here are the occupied non-core orbitals of $\ce{(ICl3)2}$. It is noteworthy, that the HOMO is fully antibonding.[3]

occupied non-core orbitals of (ICl3)2

Finally a short warning:
While the assumption made by VSEPR may lead to the approximately correct conclusion for the interhalogenide compounds $\ce{XF3}$, where $\ce{X$=$Cl, Br, I}$, that the molecules are T-shaped, it is doubtfully the correct conclusion for the much less electronegative chlorine. This can be mainly explained by Bent's rule. Even in the monomeric compound the bonding situation is somewhat more complicated than you would expect.

Notes and References

  1. The point group $D_\mathrm{2h}$ specifies that there is a main symmetry axis $C_2$ which passes through the centre of the ring. Additionally there are also two $C_2$ axis perpendicular to this and a horizontal mirror plane $\sigma_\mathrm{h}$. This means the molecule is planar. There are also non-planar molecules that can be described by this point group.)
  2. As Max pointed out (Thank you), the structure is taken from the solid state. K. H. Boswijk and E. H. Wiebenga, Acta Cryst. 1954, 7, 417-423. It mentions that it was purified by sublimation. This is no absolute proof of a gaseous dimer, but at least very suggestive. For our purposes we assume that the gas phase calculations are sufficient.
  3. Orbitals obtained at the DF-BP86/def2-SVP level of theory with Gaussian09 rev. D. The energy increases row-wise from left to right and from top to bottom. The last shown structure is the molecule itself.
  • 1
    $\begingroup$ In the first paragraph I'd add (1) the dimer is only in solid form. (2) $D_{2h}$ symmetry is planar for those who don't understand the point group symmetry notation. // "The bonding situation in the solid state of the compound...." and "... $D_{2h}$ (a planar symmetry) ..." $\endgroup$
    – MaxW
    Commented Nov 5, 2015 at 17:24
  • 3
    $\begingroup$ @Max There is no conclusive evidence, that the dimer only exists in solid state. During my literature query I did not come across an article which investigates other states of matter. It is probably highly unstable towards decomposition. The low level calculation, however, predicts a local minimum on the potential energy hypersurface, so it could exist in gas phase - if only kinetically stable. In any state of matter the bonding situation will be more complicated than it is suggested by the book. I need to think about this a bit more before I make any edits. $\endgroup$ Commented Nov 6, 2015 at 5:03

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