Question 7.EP.6: Draw the Lewis structure of tetrafluoroethylene, C2F4, the m...

Step 1: Count the total number of valence electrons. Each carbon atom has four and each fluorine atom has seven, so there are 36 valence electrons.

Number of valence electrons = 2(4) + 4(7) = 36

Step 2: Draw the skeletal structure of the molecule. Carbon occurs first in the formula and it is the least electronegative, so we expect it to be the central atom. This is also consistent with our knowledge of organic chemistry and suggests that the carbon atoms will be in the center of the molecule and will be linked by a chemical bond. So we’ll place the carbon atoms in the middle, surrounded by fluorine atoms. We have some choices in arranging the fluorine atoms around the carbon atoms. It is a good general rule to start off with the most symmetrical arrangement, so we’ll place two fluorine atoms near each carbon, rather than three near one and one near the other.

F           F

F            C           C          F

Step 3: Place single bonds between all connected atoms. To hold the molecule together, we need at least one bond between the two carbon atoms and at least one bond between each carbon atom and each of the two adjacent fluorine atoms.

\overset{F}{\underset{F—C—}{\mid}}\overset{F}{\underset{—C—F}\mid \ {}}

Step 4: Place the remaining valence electrons not accounted for in Step 3 on individual atoms until the octet rule is satisfied. Making 5 bonds requires 10 electrons, so we still have 36 − 10 = 26 electrons available. Start by assigning three lone pairs to each of the outer fluorine atoms so that each has an octet. We use 24 electrons in this process, so there are two remaining valence electrons, which are placed for the moment on one of the carbon atoms. (Because the two carbon atoms look identical in the structure before we add this last pair, the choice of which carbon gets the electrons is arbitrary.)

\overset{:\overset{..}{F}: }{\underset{\overset{..}{:\underset{..}{F}} —\underset{..}{C}—}{\mid}}\overset{:\overset{..}{F}: }{\underset{C—\overset{..}{\underset{..}{F}:} }\mid \ {}}

Step 5: Create multiple bonds for any atoms that do not have a full octet of valence electrons. The carbon atom that did not receive a lone pair in the last step does not have an octet. So we’ll need to convert a lone pair from an adjacent atom into a bonding pair, creating a double bond. But should this double bond be formed between the two carbon atoms or between the carbon atom and one of the fluorine atoms? At least two factors point toward the carbon–carbon choice. First, we have already seen that carbon often forms multiple bonds, whereas fluorine does not. Second, placing the double bond between the two carbon atoms gives the molecule a much more symmetrical structure, and such arrangements are usually preferable.

\begin{matrix}\underset{ |}{:\overset{..}{F}:} \ \ \underset{ |}{:\overset{..}{F}:} \quad\quad\quad\quad\quad \ \underset{ |}{:\overset{..}{F}:} \ \underset{ |}{:\overset{..}{F}:}\\:\overset{..}{\underset{..}{F}}- \underset{.\underset{\Large{⭟}}{.}}{C} – C – \overset{..}{\underset{..}{F}}: \longrightarrow : \overset{..}{\underset{..}{F}} – C = C – \overset{..}{\underset{..}{F}} :\end{matrix}

Discussion
This example shows that using our algorithm for obtaining a Lewis structure still leads to the need to make an occasional decision. As you grow more experienced in chemistry, decisions such as where to place a double bond become easier. Often you can look for clues, as we did in this case. An element such as fluorine does not often form double bonds. Hydrogen atoms never make double bonds. As a general rule, it is not likely that an element with a valence of one (i.e., an element that forms one bond with hydrogen) will form a double bond. But what happens when the choice is less obvious? Such a situation is discussed in the next subsection.