Chapter 1, Problem 41P

Write a Lewis formula for each of the following organic molecules:

${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$

(vinyl chloride: starting material for the preparation of PVC plastics)

${\text{C}}_{\text{2}}{\text{HBrClF}}_{\text{3}}$

(halothane: a nonflammable inhalation anesthetic; all three fluorines are

bonded to the same carbon)

${\text{C}}_{\text{2}}{\text{Cl}}_{\text{2}}{\text{F}}_{\text{4}}$

(Freon 114: formerly used as a refrigerant and as an aerosol propellant; each

carbon bears one chlorine)

Interpretation Introduction

Interpretation:

The Lewis formula for each of the given organic molecules is to be written.

Concept introduction:

When writing the Lewis formula for a molecule, the total number of valence electrons in the molecule is first determined based on its molecular formula.

The bonded atoms are connected by a shared pair of electrons which is shown by a dash.

The number of electrons in the bonds is subtracted from the total number of valence electrons. This gives the number of electrons that remain to be added in the formula.

These remaining electrons are added as unshared electron pairs in such a way that the most electronegative atoms have eight electrons around them.

Second-row elements cannot have more than eight valence electrons.

Answer to Problem 41P

Solution:

Explanation of Solution

a) ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$ (vinyl chloride: starting material for the preparation of PVC plastics)

The total number of valence electrons in ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$ (vinyl chloride) is $\text{18}$. Each carbon atom contributes $\text{4}$ valence electrons, and each hydrogen atom contributes $\text{1}$ valence electron. Each chlorine atom contributes $\text{7}$ valence electrons.

The basic framework of the given molecule can be drawn as follows.

The above figure shows the presence of $5$ bonds in the molecule. This implies that $10$ electrons are to be shown as bonded pair in the Lewis formula of the given molecule. $8$ electrons will then remain to be shown.

Three pairs of unshared electrons are assigned to the chlorine atom and one pair of unshared electrons is assigned on one of the carbon atoms as follows.

The octet of each atom in the molecule, except one carbon atom, is complete. Hence, to complete the octet of this one carbon atom, the unshared pair of electrons on the other carbon atom shifts towards the bond between the two carbon atoms resulting in the formation of a double bond.

Now the octet of each atom is complete. In this way, the Lewis structure of vinyl chloride was written.

b) ${\text{C}}_{\text{2}}{\text{HBrClF}}_{\text{3}}$ (halothane: a nonflammable inhalation anesthetic; all three fluorine atoms are

bonded to the same carbon)

${\text{C}}_{\text{2}}{\text{HBrClF}}_{\text{3}}$ has a total number of valence electrons equal to $44$. Each carbon atom contributes $4$ valence electrons and each hydrogen atom contributes $1$ valence electron to the molecule. Similarly, each halogen atom contributes $7$ valence electrons to the molecule. It is given that all three fluorine atoms are bonded to the same carbon atom.

The basic framework can therefore, be drawn as follows.

It is evident from the figure drawn for the given molecule that there are a total of $7$ bonds in the molecule. This implies that $14$ electrons are to be shown as bonding electrons in the required Lewis formula. $30$ electrons will then remain to be shown as unpaired electrons.

Of these $30$ unpaired electrons, three pairs are assigned on each halogen atom to complete their octets.

In all, $15$ pairs of unshared electrons are assigned on the halogen atoms. Thus, all the valence electrons are shown in the structure.

In this way, the Lewis structure of ${\text{C}}_{\text{2}}{\text{HBrClF}}_{\text{3}}$ was written.

c) ${\text{C}}_{\text{2}}{\text{Cl}}_{\text{2}}{\text{F}}_{\text{4}}$ (Freon $\text{114:}$ formerly used as a refrigerant and as an aerosol propellant; each

carbon bears one chlorine)

${\text{C}}_{\text{2}}{\text{Cl}}_{\text{2}}{\text{F}}_{\text{4}}$ has a total of $50$ number of valence electrons. Each carbon atom contributes $4$ valence electrons and each hydrogen atom contributes $1$ valence electron to the molecule. Similarly, each halogen atom contributes $7$ valence electrons to the molecule. It is given that each carbon atom is bonded to one chlorine atom in the given molecule.

The basic framework of the molecule can be drawn as follows.

The above figure shows that there are $7$ bonds in the given molecule, which means $14$ electrons are to be shown as bonding pair of electrons in the Lewis formula. Therefore, $36$ electrons will remain to be represented as unpaired electrons.

Three pairs of unshared electrons are assigned on each halogen atom in order to complete their octets.

In all, $18$ pairs of unshared electrons are assigned on the halogen atoms. Thus, all the valence electrons are shown in the structure.

In this way, the Lewis structure of ${\text{C}}_{\text{2}}{\text{Cl}}_{\text{2}}{\text{F}}_{\text{4}}$ was written.