Chapter 8, Problem 43QP

Draw Lewis structures for the following molecules and ions: $\left(\text{a}\right){\text{ OF}}_{\text{2}}\text{, }\left(\text{b}\right){\text{ N}}_{\text{2}}{\text{F}}_{\text{2}}\text{, }\left(\text{c}\right){\text{ Si}}_{\text{2}}{\text{H}}_{\text{6}}\text{, }\left(\text{d}\right){\text{ OH}}^{\text{-}}\left(\text{e}\right){\text{ CH}}_{\text{2}}{\text{ClCOO}}^{\text{-}}\text{, }\left(\text{f}\right){\text{ CH}}_{\text{3}}{\text{NH}}_{\text{3}}{}^{\text{+}}$

Interpretation Introduction

Interpretation:

The Lewis structures of the given molecules and ions are to be drawn.

Concept introduction:

In Lewis dot symbol, for each element, dots are mentioned around the symbol of an atom.

In Lewis dot symbol, valence electrons are represented by dots.

Dots are placed above and below as well as to the left and right of symbol.

Number of dots is important in Lewis dot symbol but not the order in which the dots are placed around the symbol.

In writing symbol pairing is not done until absolutely necessary.

For metals, the number of dots represents the number of electrons that are lost when the atom forms a cation.

For second period nonmetals, the number of unpaired dots is the number of bonds the atom can form.

Atomic ions can also be represented by dot symbols, by simply adding (for anions) and subtracting (for cations) the appropriate number of dots from Lewis dot symbol.

Lewis structure is the representation of bonding and nonbonding electron pairs present in the outermost shell of all atoms present in the molecule.

The number of bonds formed by an atom in the molecule is determined by the valence electron pairs.

Answer to Problem 43QP

Solution:

a)

b)

c)

d)

e)

f)

Explanation of Solution

a) ${\text{OF}}_2$

The electronic configuration of oxygen and fluorine in ${\text{OF}}_2$ is as follows:

$\begin{array}{c}\text{O}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^4\\ \text{F}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^5\end{array}$

The oxygen atom contains fourvalence electrons in its $2\text{p}$ subshell and the fluorine atom contains five valence electrons in its $\text{2p}$ subshell. Therefore, oxygen has a tendency to share two electrons and fluorine has a tendency to share one electron, to complete their outermost shell. Therefore, the Lewis structure of ${\text{OF}}_2$ contains two $\text{O}-\text{F}$ bonds and alone pair on oxygen and fluorine atoms.

The Lewis structure of ${\text{OF}}_2$ is as follows:

b) ${\text{N}}_2{\text{F}}_2$

The electronic configuration of nitrogen and fluorine in ${\text{N}}_2{\text{F}}_2$ is as follows:

$\begin{array}{c}\text{N}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^3\\ \text{F}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^5\end{array}$

The nitrogen atom contains three valence electrons in its $2\text{p}$ subshell and the fluorine atom contains five valence electrons in its $\text{2p}$ subshell. Therefore, nitrogen has a tendency to share three electrons and fluorine has a tendency to share one electron to complete their outermost shell. Therefore, the Lewis structure of ${\text{N}}_2{\text{F}}_2$ contains two $\text{N}-\text{F}$ single bonds and one $\text{N}-\text{N}$ double bond.

The Lewis structure of ${\text{N}}_2{\text{F}}_2$ is as follows:

c) ${\text{Si}}_2{\text{H}}_6$

The electronic configuration of silicon and hydrogen in ${\text{Si}}_2{\text{H}}_6$ is as follows:

$\begin{array}{c}\text{Si}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^{6}3{\text{s}}^2{\text{3p}}^2\\ \text{H}=1{\text{s}}^1\end{array}$

The silicon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell and hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell.

The Lewis structure is as follows:

d) ${\text{OH}}^{-}$

The electronic configuration of oxygen and hydrogen in ${\text{OH}}^{-}$ is as follows:

$\begin{array}{c}\text{O}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^4\\ \text{H}=1{\text{s}}^1\end{array}$

The oxygen atom contains four valence electrons in its $2\text{p}$ subshell and the hydrogen atom contains one valance electrons in its $\text{1s}$ subshell. Therefore, oxygen has a tendency to accept two electrons and hydrogen has a tendency to donate one electron, to complete their outermost shell. Therefore, the Lewis structure of ${\text{OH}}^{-}$ contains one $\text{O}-\text{H}$ bond and lone pairs on the oxygen atom.

The Lewis structure of ${\text{OH}}^{-}$ is as follows:

e) ${\text{CH}}_2{\text{ClCOO}}^{-}$

The electronic configuration of oxygen, carbon, chlorine, and hydrogen in ${\text{CH}}_2{\text{ClCOO}}^{-}$ is as follows:

$\begin{array}{c}\text{O}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^4\\ \text{C}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^2\\ \text{Cl}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^{6}3{\text{s}}^2{\text{3p}}^5\\ \text{H}=1{\text{s}}^1\end{array}$

The carbon atom has a tendency to form four bonds because of the presence of four electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, chlorine has atendency to form one bond because of the presence of five electrons in its $\text{3p}$ subshell, and oxygen has a tendency to form two bonds due to the presence of two electrons in its outermost shell. On the basis of the bond-forming tendency of carbon, hydrogen, chlorine, and oxygen atoms, the Lewis structure of ${\text{CH}}_2{\text{ClCOO}}^{-}$ is drawn as

f) ${\text{CH}}_3{\text{NH}}_3^{+}$

The electronic configuration of carbon, nitrogen, and hydrogen in ${\text{CH}}_3{\text{NH}}_3^{+}$ is as follows:

$\begin{array}{c}\text{N}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^3\\ \text{C}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^2\\ \text{H}=1{\text{s}}^1\end{array}$

The carbon atom has a tendency to form four bonds because of the presence of four electrons in its outermost shell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, and nitrogen has a tendency to form four bonds due to the presence of three electrons in its $2\text{p}$ subshell and one lone pair in $2\text{s}$ subshell. On the basis of the bond-forming tendency of carbon, hydrogen, and nitrogen atoms, the Lewis structure of ${\text{CH}}_3{\text{NH}}_3^{+}$ is drawn as