# Question a242a

Nov 4, 2016

Yes $I C {l}_{4}$ has four resonance structures.

#### Explanation:

Iodine has 7valance electrons. It can form one set of a paired sigma bonds. It also forms 3 sets of double bonds. (coordinate bonds)

As all four Chlorine atoms are actually identical the location of the double bonds and single bond are variable. The double bonds and single bond can shift from one chlorine atom to the other. This creates four possible resonance structures.

In reality none of these resonance structures exists for very long. What actually happens is that the electrons shift with such rapidity that the bonds create an average bond of electrons between the four chlorines and the Iodine.

Nov 5, 2016

Most chemists would say that $\text{ICl"_4^"-}$ has no resonance contributors.

#### Explanation:

The Lewis structure of $\text{ICl"_4^"-}$ is

This is the most stable structure because it has the minimum number of formal charges.

The formal charge of $\text{I}$ is -1 and the formal charges on each $\text{Cl}$ are zero.

Theoretically, you could form more Lewis structures by moving a lone pair from a $\text{Cl}$ to form an $\text{I=Cl}$ double bond.

This would give a structure like "Cl"_3stackrelcolor(blue)("-2")("I")"="stackrelcolor(blue)("+1")("Cl")#.

$\text{I}$ would then have 14 valence electrons, but this is probably OK because $\text{I}$ can expand its valence shell.

However, the structure is highly unlikely for two reasons:

1. It puts a positive charge on $\text{Cl}$, which is the second most electronegative atom in the Periodic Table.
2. It puts two positive charges adjacent to a negative charge, and this requires a large amount of energy.

Conclusion: $\text{ICl"_4^"-}$ has no low-energy resonance contributors.