Chapter 8, Problem 8.88QP

(a)

Interpretation Introduction

To determine: The lewis structure for ${\text{CH}}_{\text{3}}{\text{NO}}_{\text{2}}$.

Answer to Problem 8.88QP

Solution

The lewis structure for ${\text{CH}}_{\text{3}}{\text{NO}}_{\text{2}}$ is shown in Figure 1.

Explanation of Solution

Explanation

Number of valence electrons in hydrogen is $1$, number of valence electrons in nitrogen is $5$ and number of valence electrons in carbon is $4$. There is one carbon atom, three hydrogen atoms, one nitrogen atom and two oxygen atoms present in ${\text{CH}}_{\text{3}}{\text{NO}}_{\text{2}}$, therefore the total valence electrons are $\begin{array}{c}=4+\left(3\times 1\right)+5+\left(2\times 6\right)\\ =24\end{array}$

Carbon is bonded to three hydrogen atoms and nitrogen atom by single bond. With one oxygen, nitrogen is bonded by single bond while with other oxygen, it is bonded by double bond. Lone pairs of electrons present on oxygen atoms are delocalized which results in the formation of another lewis structure. Hence the lewis structure of ${\text{CH}}_{\text{3}}{\text{NO}}_{\text{2}}$ is,

Figure 1

(b)

Interpretation Introduction

To determine: The lewis structure for. ${\text{CNNO}}_{\text{2}}$; formal charge on each atom of ${\text{CNNO}}_{\text{2}}$ and the structure more likely to exist.

Answer to Problem 8.88QP

Solution

The lewis structures for. ${\text{CNNO}}_{\text{2}}$ is shown in Figure 4 and Figure 5. The formal charge on each atom of ${\text{CNNO}}_{\text{2}}$ is calculated. Both the resonating structures are preferred.

Explanation of Solution

Explanation

The two given possible skeletal for ${\text{CNNO}}_{\text{2}}$ are,

Figure 2

Figure 3

In the first skeletal, the nitrogen of $\text{C}\equiv \text{N}$ group is bonded with ${\text{NO}}_{\text{2}}$ group while in the second skeletal, carbon of $\text{C}\equiv \text{N}$ group is bonded with ${\text{NO}}_{\text{2}}$ group.

Number of valence electrons in carbon is $4$, number of valence electrons in nitrogen is $5$ and number of valence electrons in oxygen is $6$. There is one carbon atom, two nitrogen atom and two oxygen atoms present in first skeletal ( ${\text{CNNO}}_{\text{2}}$), therefore the total valence electrons are $\begin{array}{l}=4+\left(5\times 2\right)+\left(6\times 2\right)\\ =26\end{array}$

With one oxygen atom, nitrogen is bonded by single bond while with other oxygen; it is bonded by double bond. Lone pairs of electrons present on oxygen atoms are delocalized which results in the formation of another lewis structure. Hence the lewis structure of ${\text{CNNO}}_{\text{2}}$ is,

Figure 4

The formal charge on each atom of resonating structure (I) of ${\text{CNNO}}_{\text{2}}$ is calculated by the formula,

Formal charge is calculated as,

$\text{Formal}\text{charge}=\left(\text{Valence}\text{electrons}-\left(\begin{array}{l}\text{Lone}\text{pair}\text{electrons}\\ +\frac{1}{2}\text{Bond}\text{pair}\text{electrons}\end{array}\right)\right)$ (1)

Number of valence electrons in first nitrogen is $5$.

Number of lone pair electrons in first nitrogen is $0$.

Number of bond pair electrons in first nitrogen is $8$.

To calculate the formal charge on first nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

Number of valence electrons in carbon is $4$.

Number of lone pair electrons in carbon is $2$.

Number of bond pair electrons in carbon is $6$.

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

$\begin{array}{c}\text{Formal}\text{charge}=4-\left(2+\frac{1}{2}\left(6\right)\right)\\ =-1\end{array}$

Number of valence electrons in second nitrogen is $5$.

Number of lone pair electrons in second nitrogen is $0$.

Number of bond pair electrons in second nitrogen is $8$.

To calculate the formal charge on second nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

Number of valence electrons in oxygen (1) is $6$.

Number of lone pair electrons in oxygen (1) is $4$.

Number of bond pair electrons in oxygen (1) is $4$.

To calculate the formal charge on oxygen (1), substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

$\begin{array}{c}\text{Formal}\text{charge}=6-\left(4+\frac{1}{2}\left(4\right)\right)\\ =0\end{array}$

Number of valence electrons in oxygen (2) is $6$.

Number of lone pair electrons in oxygen (2) is $6$.

Number of bond pair electrons in oxygen (2) is $2$.

To calculate the formal charge on oxygen (2), substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

The formal charge on each atom of resonating structure (II) of ${\text{CNNO}}_{\text{2}}$ is calculated as,

Number of valence electrons in first nitrogen is $5$.

Number of lone pair electrons in first nitrogen is $0$.

Number of bond pair electrons in first nitrogen is $8$.

To calculate the formal charge on first nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

Number of valence electrons in carbon is $4$.

Number of lone pair electrons in carbon is $2$.

Number of bond pair electrons in carbon is $6$.

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

$\begin{array}{c}\text{Formal}\text{charge}=4-\left(2+\frac{1}{2}\left(6\right)\right)\\ =-1\end{array}$

Number of valence electrons in second nitrogen is $5$.

Number of lone pair electrons in second nitrogen is $0$.

Number of bond pair electrons in second nitrogen is $8$.

To calculate the formal charge on second nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

Number of valence electrons in oxygen (1) is $6$.

Number of lone pair electrons in oxygen (1) is $6$.

Number of bond pair electrons in oxygen (1) is $2$.

To calculate the formal charge on oxygen (1), substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

Number of valence electrons in oxygen (2) is $6$.

Number of lone pair electrons in oxygen (2) is $4$.

Number of bond pair electrons in oxygen (2) is $4$.

To calculate the formal charge on oxygen (2), substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

$\begin{array}{c}\text{Formal}\text{charge}=6-\left(4+\frac{1}{2}\left(4\right)\right)\\ =0\end{array}$

Number of valence electrons in carbon is $4$, number of valence electrons in nitrogen is $5$ and number of valence electrons in oxygen is $6$. There is one carbon atom, two nitrogen atom and two oxygen atoms present in first skeletal ( ${\text{NCNO}}_{\text{2}}$), therefore the total valence electrons are $4+5\times 2+6\times 2=26$. With one oxygen atom, nitrogen is bonded by single bond while with other oxygen; it is bonded by double bond. Lone pairs of electrons present on oxygen atoms are delocalized which results in the formation of another lewis structure. Hence the lewis structure of ${\text{NCNO}}_{\text{2}}$ is,

Figure 5

The formal charge on each atom of resonating structure (I) of ${\text{NCNO}}_{\text{2}}$ is calculated as,

Number of valence electrons in first nitrogen is $5$.

Number of lone pair electrons in first nitrogen is $2$.

Number of bond pair electrons in first nitrogen is $6$.

To calculate the formal charge on first nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

$\begin{array}{c}\text{Formal}\text{charge}=\text{5}-\left(2+\frac{1}{2}\left(6\right)\right)\\ =0\end{array}$

Number of valence electrons in carbon is $4$.

Number of lone pair electrons in carbon is $0$.

Number of bond pair electrons in carbon is $8$.

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

Number of valence electrons in second nitrogen is $5$.

Number of lone pair electrons in second nitrogen is $0$.

Number of bond pair electrons in second nitrogen is $8$.

To calculate the formal charge on second nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

Number of valence electrons in oxygen (1) is $6$.

Number of lone pair electrons in oxygen (1) is $4$.

Number of bond pair electrons in oxygen (1) is $4$.

To calculate the formal charge on oxygen (1), substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

$\begin{array}{c}\text{Formal}\text{charge}=6-\left(4+\frac{1}{2}\left(4\right)\right)\\ =0\end{array}$

Number of valence electrons in oxygen (2) is $6$.

Number of lone pair electrons in oxygen (2) is $6$.

Number of bond pair electrons in oxygen (2) is $2$.

To calculate the formal charge on oxygen (2), substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

The formal charge on each atom of resonating structure (II) of ${\text{NCNO}}_{\text{2}}$ is calculated as,

Number of valence electrons in first nitrogen is $5$.

Number of lone pair electrons in first nitrogen is $2$.

Number of bond pair electrons in first nitrogen is $6$.

To calculate the formal charge on first nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

$\begin{array}{c}\text{Formal}\text{charge}=\text{5}-\left(2+\frac{1}{2}\left(6\right)\right)\\ =0\end{array}$

Number of valence electrons in carbon is $4$.

Number of lone pair electrons in carbon is $0$.

Number of bond pair electrons in carbon is $8$.

To calculate the formal charge on carbon, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

Number of valence electrons in second nitrogen is $5$.

Number of lone pair electrons in second nitrogen is $0$.

Number of bond pair electrons in second nitrogen is $8$.

To calculate the formal charge on second nitrogen, substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

Number of valence electrons in oxygen (1) is $6$.

Number of lone pair electrons in oxygen (1) is $6$.

Number of bond pair electrons in oxygen (1) is $2$.

To calculate the formal charge on oxygen (1), substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

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

Number of valence electrons in oxygen (2) is $6$.

Number of lone pair electrons in oxygen (2) is $4$.

Number of bond pair electrons in oxygen (2) is $4$.

To calculate the formal charge on oxygen (2), substitute the value of valence electrons, lone pair electrons and bond pair electrons in the equation (1).

$\begin{array}{c}\text{Formal}\text{charge}=6-\left(4+\frac{1}{2}\left(4\right)\right)\\ =0\end{array}$

The resonating structure which possesses zero or minimum formal charge is preferred. The distribution of formal charges on both the molecules ${\text{CNNO}}_{\text{2}}$ and ${\text{NCNO}}_{\text{2}}$ is same. Hence both the resonating structures are preferred.

(c)

Interpretation Introduction

To determine: Whether the given two structures of ${\text{CNNO}}_{\text{2}}$ are resonance forms of each other.

Answer to Problem 8.88QP

Solution

The two structures of ${\text{CNNO}}_{\text{2}}$ are not resonance forms of each other.

Explanation of Solution

Explanation

In the resonating forms, the postion of atoms remains same. But in the two structures of ${\text{CNNO}}_{\text{2}}$, the postion of carbon and nitrogen is not same. Hence the two structures of ${\text{CNNO}}_{\text{2}}$ are not resonance forms of each other.

Conclusion

The two structures of ${\text{CNNO}}_{\text{2}}$ are not resonance forms of each other.