Question 7.EP.4: Chlorofluorocarbons have been used as refrigerants for the p...
Chlorofluorocarbons have been used as refrigerants for the past 50 years, but their use is now being phased out worldwide because of concerns about atmospheric ozone depletion. Draw the Lewis structure of dichlorodifluoromethane, CF_{2}Cl_{2}, also known as DuPont’s Freon-12.
Strategy
Use the algorithm for drawing Lewis structures.
Step 1: Count the total number of valence electrons. Carbon has four, fluorine has seven, and chlorine has seven.
\begin{matrix} C & 1\times 4=4 \\ F & 2\times 7=14 \\ Cl & \underline{2\times 7=14} & \\ & Total= 32 \end{matrix}
Step 2: Draw the skeletal structure of the molecule. Carbon is listed first in the formula, and it is the least electronegative, so use it as the central atom.
\begin{matrix} & \underset{}{Cl} & \\ F & \ C \ & F \\ & \overset{\underset{}{} }{Cl} & \end{matrix}
Step 3: Place single bonds between all connected atoms.
\begin{matrix} & \underset{\mid }{Cl} & \\ & F— \ C \ — F & \\ & \overset{\mid }{Cl} & \end{matrix}
Step 4: Place the remaining valence electrons not accounted for in Step 3 on individual atoms until the octet rule is satisfied. Four single bonds have been drawn, and they require eight valence electrons. We have a total of 32 available electrons, so we are left with 32 − 8 = 24. This gives each of the four halogen atoms three lone pairs (6 electrons).
\begin{matrix}\underset{\mid}{:\overset{..}{Cl}: }\\ \overset{..}{:\underset{..}{F}}—C— \overset{..}{\underset{..}{F}:} \\\overset{\mid}{:\underset{..}{Cl}: } \end{matrix}
Step 5: Create multiple bonds for any atoms that do not have a full octet of valence electrons. Check to make sure each atom has an octet of eight electrons. Carbon has four bonding pairs, or eight electrons. Each halogen has one bonding pair and three lone pairs, so they also have octets. Multiple bonds are not needed, and the final structure is the one arrived at after Step 4.
Discussion
Because we are using an algorithm to draw these structures, it is important that each step be carried out correctly because any errors may be propagated through the rest of the process. The first step is the most critical. If we lose track of the number of valence electrons we need to include, we will have little chance of arriving at the correct structure. Many students have missed exam questions by adding a lone pair of electrons they “didn’t have” rather than making multiple bonds.