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$\ce{I2Cl6}$ 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.
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.
(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]
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
