# Question #1e11d

Aug 21, 2017

Here's how I see it.

#### Explanation:

We can think of the $\text{HgI"_3^"-}$ as being formed by the reaction

$\text{Hg"^"2+" + 3"I"^"-" → "HgI"_3^"-}$

The electron configuration of a neutral $\text{Hg}$ atom is

${\text{Hg"^0 = "[Xe] 6s"^2 "4f"^14 "5d}}^{10}$

For $\text{Hg"^"2+}$, the electron configuration is

${\text{Hg"^"2+" = "[Xe] 4f"^14 "5d}}^{10}$

The $\text{Hg"^"2+}$ ion, however, has vacant $\text{6s}$ and $\text{6p}$ orbitals.

It can hybridize a $\text{6s}$ and two of the $\text{6p}$ orbitals to form three new vacant ${\text{sp}}^{3}$ hybrid orbitals.

These can each overlap with a filled $\text{5p}$ orbital of an iodide ion.

Thus, the new ion is trigonal planar, and the $\text{Hg-I}$ bonds are coordinate covalent bonds in which the iodide ions are contributing both electrons of the covalent bond.