# Polar/ non-polar nature of HgI2

Is $\ce{HgI_2}$ polar or non-polar?

The electronegativity difference between Hg and I is about 0.66, so the compound should be almost covalent. However, I cannot find the structure of $\ce{HgI_2}$ as a single molecule, so, I cannot find if it is polar or not, as a whole.

• You seem to be using a false dichotomy. A great deal of covalent molecules are polar. Many are even more polar than $\ce{HgI2}$ (which is polar, too). – Ivan Neretin Feb 26 '17 at 10:01
• @IvanNeretin I did not say covalent molecules are non-polar. If a molecule is ionic then it is surely polar, but if it is covalent , we must have its stucture to tell, whether it is polar or not. For example, CCl4 is non-polar as a whole. – Shoubhik R Maiti Feb 28 '17 at 15:59
• Oh, you mean the overall dipole moment, Well, it's $0$, because the molecule is linear, much like CO2. – Ivan Neretin Feb 28 '17 at 16:10

The electronegativity difference using the Pauling scale is not accurate in heavy transition metals. This is because the Pauling electronegativity involves the measure of bond energies, while the bonding of several metals (including "noble" metals like gold, silver, platinum, mercury, etc.) with large halides are diffuse and do not have sigma character . This is because the frontier orbitals of these elements are in the $d$ block, so molecular orbital interactions, if any, are diffuse especially in larger 4d and 5d metals. In reality, compounds like $\ce{HgI_2}$, as well as ones like $\ce{PbI_2}$ and $\ce{BiBr_3}$ (which I work with) are ionic salts.
• The interesting thing about many of these solids is that that despite having ionic character, the polarizability of these compounds are so high (due to their large atomic sizes) that in solid form, they can have semiconductive or conductive properties. If you look at new solar cell materials, you will find that there is a lot of research on ionic materials known as perovskites, which have the formulas similar to $\ce{CH_3NH_3PbI_3}$ yet conduct electrons very efficiently. – Xahoc Feb 26 '17 at 9:58
• HSAB uses the elements' polarizability to make determinations on the type of bonding. But, you cannot use only HSAB. In the specific case of $\ce{HgI_2}$, iodine is a strong pi donor yet there are not orbitals on mercury to accept electron density, so the bonding becomes ionic. But, this does not mean that the ionicity is not weak or diffuse. Similarly, for $\ce{PbI_2}$ and $\ce{BiBr_3}$, the stabilizing energy by removing electrons from the 6p orbitals of lead results in a more ionic contribution than would be expected from HSAB alone. – Xahoc Feb 26 '17 at 20:11