Chapter 8, Problem 54QP

Interpretation Introduction

Interpretation:

The wrong point in the given Lewis structures is to be stated, and the correct Lewis structure of the molecule is to be drawn.

Concept Introduction:

Lewis structure is a representation of the bonding and non-bonding 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 electrons pairs.

Answer to Problem 54QP

Solution:

a)

The double bond is present between the carbon and nitrogen atoms.

The correct Lewis structure is as follows:

b)

The double bond is present between the carbon and hydrogen atoms.

The correct Lewis structure is as follows:

c)

A single bond is present between the $\text{Sn}$ and $\text{O}$ atoms.

The correct Lewis structure is as follows:

d)

The lone pair of electrons on boron.

The correct Lewis structure is as follows:

e)

The double bond between the oxygen and fluorine atoms.

The correct Lewis structure is as follows:

f)

The single bond between the carbon and oxygen atoms.

The correct Lewis structure is as follows:

g)

The lone pair of electrons of the nitrogen atom is missing.

The correct Lewis structure is as follows:

Explanation of Solution

a)

The given Lewis structure is as follows:

The electronic configurations of hydrogen, carbon, and nitrogen in $\text{HCN}$ are as follows:

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

Carbon has four electrons in its outermost shell and requires four electrons to complete its outermost shell of electrons, while nitrogen requires three electrons to complete its octet and hydrogen requires one electron to obtain its fully-filled electronic configuration. Therefore, the Lewis structure of $\text{HCN}$ contains one $\text{C}-\text{H}$ bond and one $\text{C}-\text{N}$ triple bond.

The correct Lewis structure of $\text{HCN}$ is as follows:

Hence, the given Lewis structure of $\text{HCN}$ is incorrect because a double bond is present between the carbon and nitrogen atoms in place of a triple bond.

b)

The given Lewis structure is as follows:

The electronic configurations of carbon and hydrogen in ${\text{C}}_2{\text{H}}_2$ are as follows:

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

The carbon atom contains four electrons in its outermost shell and the hydrogen atom contains one valence electron in its $\text{1s}$ subshell. Therefore, carbon has a tendency to share four electrons and hydrogen has a tendency to share one electron for them to complete their outermost shells. The Lewis structure of ${\text{C}}_2{\text{H}}_2$ contains one $\text{C}-\text{C}$ triple bond and two $\text{C}-\text{H}$ single bonds.

The correct Lewis structure of ${\text{C}}_2{\text{H}}_2$ is as follows:

Hence, the given Lewis structure of ${\text{C}}_2{\text{H}}_2$ is incorrect because a double bond is present between carbon and hydrogen in place of a single bond.

c)

The given Lewis structure is as follows:

The electronic configurations of tin and oxygen in ${\text{SnO}}_2$ are as follows:

$\begin{array}{c}\text{Sn}=\left[\text{Kr}\right]{\text{4d}}^{10}{\text{5s}}^2{\text{5p}}^2\\ \text{O}=1{\text{s}}^2{\text{2s}}^2{\text{2p}}^4\end{array}$

A carbon atom has a tendency to form four bonds because of the presence of four valence electrons in its outermost shell, while oxygen has a tendency to form two bonds due to the presence of two electrons in its outermost shell. The Lewis structure of ${\text{SnO}}_2$ contains two $\text{Sn}-\text{O}$ double bonds and lone pairs on the oxygen atom.

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

Hence, the given Lewis structure of ${\text{SnO}}_2$ is wrong because it contains a single bond between the tin and oxygen atoms.

d)

The given Lewis structure is as follows:

The electronic configurations of boron and fluorine in ${\text{BF}}_3$ are as follows:

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

The boron atom contains three electrons in its outermost shell and the fluorine atom contains five electrons in its $\text{2p}$ subshell. Therefore, boron has a tendency to share three electrons and fluorine has a tendency to share one electron for them to complete their outermost shells. The Lewis structure of ${\text{BF}}_3$ contains three $\text{B}-\text{F}$ bonds and no lone pairs on the boron atom.

The Lewis structure of ${\text{BF}}_3$ is as follows:

Hence, the given Lewis structure of ${\text{BF}}_3$ is wrong because it contains one lone pair of electrons on the boron atom, which is already bonded to three fluorine atoms.

e)

The given Lewis structure is as follows:

The electronic configurations of hydrogen, oxygen, and fluorine in $\text{HOF}$ are as follows:

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

Hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, fluorine has a tendency to form one bond because of the presence of five electrons in its $2\text{p}$ subshell, and oxygen has a tendency to form two bonds due to the presence of four electrons in its $2\text{p}$ subshell. Therefore, the Lewis structure of $\text{HOF}$ contains a $\text{H}-\text{O}$ single bond and a $\text{O}-\text{F}$ single bond.

The correct Lewis structure of $\text{HOF}$ is as follows:

Hence, the given Lewis structure of $\text{HOF}$ is incorrect because it contains a double bond between the oxygen and fluorine atoms.

f)

The given Lewis structure is as follows:

The electronic configurations of fluorine, carbon, oxygen, and hydrogen in $\text{FCHO}$ are as follows:

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

The fluorine atom has a tendency to form one bond because of the presence of five valence electrons in its $\text{2p}$ subshell, hydrogen has a tendency to form one bond because of the presence of one electron in its outermost shell, the carbon atom has a tendency to form four bonds because it has four electrons in its valence shell, and oxygen has a tendency to form two bonds due to the presence of four electrons in its $2\text{p}$ subshell. On the basis of the bond-forming tendency of fluorine, carbon, hydrogen and oxygen atoms, the Lewis structure of $\text{FCHO}$ is drawn as:

Hence, the given Lewis structure of $\text{FCHO}$ is incorrect because it contains a single bond between carbon and oxygen, due to which the valency of the carbon atom is not satisfied.

g)

The given Lewis structure is,

The electronic configurations of nitrogen and chlorine in ${\text{NF}}_3$ are:

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

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 donate three electrons and fluorine has a tendency to accept one electron for them to complete their outermost shells. The Lewis structure of ${\text{NF}}_3$ contains three $\text{N}-\text{F}$ bonds and one lone pair on the nitrogen atom.

The correct Lewis structure of ${\text{NF}}_3$ is as follows:

Hence, the given Lewis structure of ${\text{NF}}_3$ is incorrect because it does not contain a lone pair of electrons on the nitrogen atom.