Chapter 8, Problem 8.144AP

(a)

Interpretation Introduction

Interpretation: The Lewis structure of the given species, its resonating structure, assigned formal charge and the reason for the difference in the bond length of given species of nitrogen is to be stated.

Concept introduction: The Lewis symbols of the atoms indicate the total number of valence electrons surrounding the atom or the ions.

To determine: The Lewis structure of $\text{N}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}{}^{-}$, assigned formal charge and its resonance form.

Answer to Problem 8.144AP

Solution

The formal charge on nitrogen $1$, nitrogen $2$, nitrogen $3$, oxygen bonded to nitrogen with single bond and oxygen bonded to nitrogen with double bond is $+1,0,+1,-1$ and $0$.

Explanation of Solution

Explanation

The atomic number of nitrogen is $7$.

The electronic configuration of nitrogen is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^3$. Therefore, nitrogen has five valence electrons.

The atomic number of oxygen is $8$.

The electronic configuration of oxygen is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^4$. Therefore, oxygen has six valence electrons.

A species on gaining electron possesses a negative charge. The number of electrons gains or lost is equal to the charge raised on the structure of species.

Therefore, $\text{N}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}{}^{-}$ has an electron excess with that of the valence electrons of oxygen and nitrogen.

Thus, the total number of valence electrons in $\text{N}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}{}^{-}$ is $3\times 5+4\times 6+1=40$.

Therefore, the Lewis structure of $\text{N}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}{}^{-}$ is,

Figure 1

The above structure shows resonance as,

Figure 2

According to the above figure,

• Nitrogen $1$ has four bonds and therefore, eight bonding electrons.
• Nitrogen $2$ has six bonding electrons and two non bonding electrons.
• Nitrogen $3$ has eight bonding electrons.
• Oxygen bonded to nitrogen with single bond has two bonding electrons and six nonbonding electrons.
• Oxygen bonded to nitrogen with double bond has four bonding electrons and four nonbonding electrons.

The valence electrons of nitrogen are five and that of oxygen are six.

Formal charge is calculated by the formula,

$\text{Formal}\text{charge}=\left(\begin{array}{l}\text{Valence}\text{electrons}\\ -\left(\text{Number}\text{of}\text{nonbonding}\text{electrons}+\frac{\text{Number}\text{of}\text{bonding}\text{electrons}}{2}\right)\end{array}\right)$

Substitute the value of valence electrons and bonding and nonbonding electrons of nitrogen $1$ and $3$ in formal charge formula.

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{5}-\left(\text{0}+\frac{8}{2}\right)\right)\\ =5-4\\ =+1\end{array}$

Substitute the value of valence electrons and bonding and nonbonding electrons of nitrogen $2$ in formal charge formula.

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{5}-\left(2+\frac{6}{2}\right)\right)\\ =5-5\\ =0\end{array}$

Substitute the value of valence electrons and bonding and nonbonding electrons of oxygen bonded to nitrogen with single bond in formal charge formula.

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}-\left(6+\frac{2}{2}\right)\right)\\ =6-7\\ =-1\end{array}$

Substitute the value of valence electrons and bonding and nonbonding electrons of oxygen bonded to nitrogen with double bond in formal charge formula.

$\begin{array}{c}\text{Formal}\text{charge}=\left(\text{6}-\left(4+\frac{4}{2}\right)\right)\\ =6-6\\ =0\end{array}$

Therefore, the formal charge on nitrogen $1$, nitrogen $2$, nitrogen $3$, oxygen bonded to nitrogen with single bond and oxygen bonded to nitrogen with double bond is $+1,0,+1,-1$ and $0$.

(b)

Interpretation Introduction

To determine: The reason for the difference in the bond length of $\text{nitrogen}-\text{oxygen}$ in $\text{N}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}{}^{-}$ and in ${\text{N}}_{\text{2}}\text{O}$.

Answer to Problem 8.144AP

Solution

In the bond length of $\text{nitrogen}-\text{oxygen}$ in $\text{N}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}{}^{-}$ and in ${\text{N}}_{\text{2}}\text{O}$ is different due to different resonating patterns.

Explanation of Solution

Explanation

Due to the resonance effect, the $\text{nitrogen}-\text{oxygen}$ in $\text{N}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}{}^{-}$ resonates between double and single bond. Thus, the bond length of $\text{nitrogen}-\text{oxygen}$ in $\text{N}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}{}^{-}$ is between the $\text{nitrogen}-\text{oxygen}$ single bond and $\text{nitrogen}-\text{oxygen}$ double bond.

The Lewis structure of ${\text{N}}_{\text{2}}\text{O}$ is,

Figure 3

Therefore, $\text{nitrogen}-\text{oxygen}$ bond resonates between single, double and triple bond. Thus, $\text{nitrogen}-\text{oxygen}$ bond in ${\text{N}}_{\text{2}}\text{O}$ is smaller than that in $\text{N}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}{}^{-}$

(c)

Interpretation Introduction

To determine: The Lewis structure of ${\text{NH}}_{\text{4}}{}^{+}$.

Answer to Problem 8.144AP

Solution

The Lewis structure of ${\text{NH}}_{\text{4}}{}^{+}$ is shown below.

Explanation of Solution

Explanation

The atomic number of nitrogen is $7$.

The electronic configuration of nitrogen is $\left[\text{He}\right]{\text{2s}}^{\text{2}}{\text{2p}}^3$. Therefore, nitrogen has five valence electrons.

The atomic number of hydrogen is $1$.

The electronic configuration of hydrogen is ${\text{1s}}^1$. Therefore, hydrogen has one valence electron.

A species on losing electron possesses a positive charge. The number of electrons gains or lost is equal to the charge raised on the structure of species.

Therefore, ${\text{NH}}_{\text{4}}{}^{+}$ has an electron less with that of the valence electrons of hydrogen and nitrogen.

Thus, the total number of valence electrons in ${\text{NH}}_{\text{4}}{}^{+}$ is $5+4\times 1-1=8$.

Therefore, the Lewis structure of ${\text{NH}}_{\text{4}}{}^{+}$ is,

Figure 4

Conclusion

1. a. The formal charge on nitrogen $1$, nitrogen $2$, nitrogen $3$, oxygen bonded to nitrogen with single bond and oxygen bonded to nitrogen with double bond is $+1,0,+1,-1$ and $0$.
2. b. In the bond length of $\text{nitrogen}-\text{oxygen}$ in $\text{N}{\left({\text{NO}}_{\text{2}}\right)}_{\text{2}}{}^{-}$ and in ${\text{N}}_{\text{2}}\text{O}$ is different due to different resonating patterns.
3. c. The Lewis structure of ${\text{NH}}_{\text{4}}{}^{+}$ shows eight valence electrons.