Question 3.10: Lewis Structures of Covalent Compounds State the number of v...
Lewis Structures of Covalent Compounds
State the number of valence electrons in each molecule and draw a Lewis structure for each:
(a) Hydrogen peroxide, \text{H}_{2}\text{O}_{2} (b) Methanol, \text{CH}_{3}\text{OH}
(c) Acetic acid, \text{CH}_{3}\text{COOH}
Strategy
To determine the number of valence electrons in a molecule, add the number of valence electrons contributed by each kind of atom in the molecule. To draw a Lewis structure, determine the connectivity of the atoms and connect bonded atoms by single bonds. Then arrange the remaining valence electrons so that each atom has a complete outer shell.
(a) A Lewis structure for hydrogen peroxide, \text{H}_{2}\text{O}_{2} , must show the 14 valence electron —six from each oxygen and the one from each hydrogen, for a total of 12 + 2 = 14 valence electrons. We know that hydrogen forms only one covalent bond, so the connectivity of atoms must be as follows:
\text{H}-\text{O}-\text{O}-\text{H}The three single bonds account for six valence electrons. The remaining eight valence electrons must be placed on the oxygen atoms to give each a complete octet:
\text{H}-\overset{\cdot \cdot }{\underset{\cdot \cdot }{\text{O}}}-\overset{\cdot \cdot }{\underset{\cdot \cdot }{\text{O}}}-\text{H} 
(b) A Lewis structure for methanol, \text{CH}_{3}\text{OH}, must show the four valence electrons from carbon, one from each hydrogen, and the six from oxygen for a total of 4 + 4 + 6 = 14 valence electrons. The connectivity of atoms in methanol is given on the left, below. The five single bonds in this partial structure account for ten valence electrons. The remaining four valence electrons must be placed on oxygen as two Lewis dot pairs to give it a complete octet.
\underset{ \quad \substack{\text{The order of}\\ {\text{attachment of atoms}}}}{\begin{matrix} &&\underset{|}{\text{H}} \\ \text{H}&-&\text{C}&-&\text{O}&-&\text{H}\\ &&\overset{|}{\text{H}} \end{matrix} } \underset{\substack{\text{Lewis dot}\\ {\text{structure}}}}{\begin{matrix} &&\underset{|}{\text{H}} \\ \text{H}&-&\text{C}&-& \overset{\cdot \cdot }{\underset{\cdot \cdot }{\text{O}}} &-&\text{H}\\ &&\overset{|}{\text{H}} \end{matrix} } 
(c) A molecule of acetic acid, \text{CH}_{3}\text{COOH}, must contain the four valence electrons from each carbon, the six from each oxygen, and the one from each hydrogen for a total of 8 + 12 + 4 = 24 valence electrons. The connectivity of atoms, shown on the left below, contains seven single bonds, which account for 14 valence electrons. The remaining ten electrons must be added in such a way that each carbon and oxygen atom has a complete outer shell of eight electrons. This can be done in only one way, which creates a double bond between carbon and one of the oxygens.
In this Lewis structure, each carbon has four bonds: one carbon has four single bonds, and the other carbon has two single bonds and one double bond. Each oxygen has two bonds and two unshared pairs of electrons: one oxygen has one double bond and two unshared pairs of electrons, and the other oxygen has two single bonds and two unshared pairs of electrons.
■ Quick Check 3.10
Draw a Lewis structure for each molecule. Each has only one possible order of attachment of its atoms, which is left for you to determine.
(a) Ethane, \text{C}_{2}\text{H}_{6} (b) Chloromethane, \text{CH}_{3}\text{Cl}
(c) Hydrogen cyanide, HCN