Question 13.8: Lewis Structures for Molecules That Violate the Octet Rule I...

We can follow the same stepwise procedure we used previously for sulfur hexafluoride.

Step 1: Sum the valence electrons.

5 + 5 ( 7 ) =  40 electrons
↑       ↑
P      Cl

Step 2: Indicate single bonds between bound atoms.

Step 3: Distribute the remaining electrons. In this case, 30 electrons (40 - 10) remain. These are used to satisfy the octet rule for each chlorine atom. The final Lewis structure is

Note that phosphorus, a third-row element, exceeds the octet rule by two electrons.
In the PC l_{5} and S F_{6} molecules, the third-row central atoms (P and S, respectively) are assigned the extra electrons. However, in molecules having more than one atom that can exceed the octet rule, it is not always clear which atom should have the extra electrons. Consider the Lewis structure for the triiodide ion ( I_{3} ^{-} ), which has

3  (7) + 1 =  22  valence electrons
↑           ↑
I           – 1 charge

Indicating the single bonds gives I—I—I. At this point 18 electrons (22 - 4) remain. Trial and error will convince you that one of the iodine atoms must exceed the octet rule, but which one?
The rule we will follow is that when it is necessary to exceed the octet rule for one of several third-row (or higher) elements, assume that the extra electrons are placed on the central atom.
Thus for I_{3} ^{-} the Lewis structure is

where the central iodine exceeds the octet rule. This structure agrees with known properties of I_{3} ^{-} .