Untitled - Kelly Walsh High School

140 CHEMISTRY FOR THE UTTERLY CONFUSED
Quick Tip
Quick Tip
N 8 4(8) 40. The determination of A uses 8 for xenon and 7 for each fluorine
atom. This gives A 8 4(7) 36. Using the values for N and A, we
find S 40 36 4. If S 4, then there are S/2 2 bonds. This is a problem
since we already know we need at least 4 bonds. This means that we have a
compound that is an exception to the octet rule.
In a compound that is an exception to the octet rule, there is usually only one
atom, other than hydrogen, that is an exception. There are few nonhydrogen
compounds with more than one exception present.
Even though we have an exception, we can still complete the Lewis structure.
We need to draw a bond from each of the fluorine atoms to the central xenon.
This gives us 4 bonds and uses 8 electrons. Each fluorine atom needs to complete
its octet. The bond accounts for 2 electrons, so we need 6 more electrons
(3 pairs) for each. Therefore, we add 3 separate pairs to each of the fluorine
atoms. Six electrons per fluorine times 4 fluorine atoms accounts for 24 electrons.
Our Lewis structure now contains 8 24 32 electrons. The number of
available electrons (A) is 36, so we still need to add 36 32 4 electrons.
These 4 electrons will give us 2 pairs. The xenon atom will get these pairs and
become an exception to the octet rule. The actual placement of the pairs is not
important as long as it is obvious that they are with the central atoms and not
one of the fluorine atoms. The final Lewis structure is:
F
F Xe F
F
When you are not certain where to put electrons, remember that the most electronegative
element will get its octet. This element will not exceed an octet.