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Chapter 20: Problem 28

Draw Lewis structures for \(\mathrm{HCFC}-123\left(\mathrm{CF}_{3} \mathrm{CHCl}_{2}\right)\) and \(\mathrm{CF}_{3} \mathrm{CFH}_{2}\)

Short Answer

Expert verified
The Lewis structures for CF3CHCl2 and CF3CFH2 can be obtained by first determining the central atom (which is carbon), then counting the total number of valence electrons for each molecule. Afterward, distribute these electrons across the molecule aiming to fill each atom's octet. CF3CFH2 will be a free radical due to the uneven number of electrons.

Step by step solution

01

Identify the central atom

For CF3CHCl2, since carbon is less electronegative, it is the central atom. In CF3CFH2, the central atom is also carbon.
02

Count the total number of valence electrons

For CF3CHCl2, there are 7 valence electrons from each F atom (total 21), 4 from C atom, one from H atom and seven from each Cl atom (total 14), this sums up to 40 valence electrons. For CF3CFH2, there are seven valence electrons from each F atom (total 21), four from each C atom (total 8), and one from each H atom (total 2), this sums up to 31 valence electrons.
03

Distribute the electrons

Try to distribute so that each atom is surrounded by eight electrons (Octet rule).
04

Lewis structure for CF3CHCl2

To follow the octet rule, place the three F atoms, one H atom, and two Cl atoms around the C atom. Connect them with single bonds. This uses up 12 electrons. Now distribute the remaining 28 electrons around F and Cl atoms to complete their octets.
05

Lewis structure for CF3CFH2

First, connect two C atoms and two H atoms around one C. This uses up 8 electrons. Place three F atoms around the other C atom. This uses up 6 more electrons. Now, distribute the remaining 17 electrons around F and C atoms to complete their octets. Since we have an odd number of electrons, one electron will remain unpaired, indicating a free radical molecule.

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