Chapter 8, Problem 8.92QP

Interpretation Introduction

Interpretation: The preferred resonance form among the given ions is to be stated on the basis of formal charges.

Concept introduction: Formal charges play an important role in choosing between the possible molecular structures. The preferred structure is the one in which formal charges are zero.

To determine: The preferred resonance form among the given ions on the basis of formal charges.

Answer to Problem 8.92QP

Solution

The preferred resonance form in the ${\text{CNO}}^{-}$ ion is resonating structure (a).

The preferred resonance form in the ${\text{NCO}}^{-}$ ion is resonating structure (b).

The preferred resonance form in the ${\text{CON}}^{-}$ ion is resonating structure (b).

Explanation of Solution

Explanation

The given ion is ${\text{CNO}}^{-}$. Number of valence electrons in oxygen is $6$ , number of valence electrons in carbon is $4$ and number of valence electrons in nitrogen is $5$. There is one nitrogen atom, one carbon atom and one oxygen atom present in ${\text{CNO}}^{-}$. There is an additional valence electron due to $-1$ charge on ${\text{CNO}}^{-}$.

Therefore the total valence electrons are $\begin{array}{l}=6+4+5+1\\ =16\end{array}$

The lone pair of electrons on oxygen and bonding electrons is delocalized which results in the formation of lewis structures. The lewis structures for ${\text{CNO}}^{-}$ are,

Figure 1

The formal charge on each atom of resonating structure (a) of ${\text{CNO}}^{-}$ is calculated as,

Formal charge is calculated as,

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

Number of valence electrons in nitrogen is $5$.

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

Number of bond pair electrons in nitrogen is $10$.

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

$\begin{array}{c}\text{Formal}\text{charge}=5-\left(0+\frac{1}{2}\left(10\right)\right)\\ =0\end{array}$

Number of valence electrons in oxygen is $6$.

Number of lone pair electrons in oxygen is $4$.

Number of bond pair electrons in oxygen is $4$.

To calculate the formal charge on oxygen, 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 lone pair electrons in carbon is $0$.

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).

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

The formal charge on each atom of resonating structure (b) of ${\text{CNO}}^{-}$ is calculated as,

Number of valence electrons in nitrogen is $5$.

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

Number of bond pair electrons in nitrogen is $10$.

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

$\begin{array}{c}\text{Formal}\text{charge}=5-\left(0+\frac{1}{2}\left(10\right)\right)\\ =0\end{array}$

Number of valence electrons in oxygen is $6$.

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

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

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

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

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 $4$.

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

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

According to formal charge calculations, there is a formal charge of $+1$ on oxygen atom in the resonanting structure (b). Oxygen is more electronegative than nitrogen. Formal positive charge is not favorable on most electronegative atom. Hence the resonating structure (a) is preffered.

The given ion is ${\text{NCO}}^{-}$. Number of valence electrons in oxygen is $6$ , number of valence electrons in carbon is $4$ and number of valence electrons in nitrogen is $5$. There is one nitrogen atom, one carbon atom and one oxygen atom present in ${\text{NCO}}^{-}$. There is an additional valence electron due to $-1$ charge on ${\text{NCO}}^{-}$.

Therefore the total valence electrons are $\begin{array}{l}=6+4+5+1\\ =16\end{array}$

Since carbon is least electronegative, it will act as central atom. The lone pair of electrons on oxygen and bonding electrons is delocalized which results in the formation of lewis structures. The lewis structures for ${\text{NCO}}^{-}$ are,

Figure 2

The formal charge on each atom of resonating structure (a) of ${\text{NCO}}^{-}$ is calculated as,

Number of valence electrons in nitrogen is $5$.

Number of lone pair electrons in nitrogen is $4$.

Number of bond pair electrons in nitrogen is $4$.

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

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

Number of valence electrons in oxygen is $6$.

Number of lone pair electrons in oxygen is $4$.

Number of bond pair electrons in oxygen is $4$.

To calculate the formal charge on oxygen, 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 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).

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

The formal charge on each atom of resonating structure (b) of ${\text{NCO}}^{-}$ is calculated as,

Number of valence electrons in nitrogen is $5$.

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

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

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

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

Number of valence electrons in oxygen is $6$.

Number of lone pair electrons in oxygen is $6$.

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

To calculate the formal charge on oxygen, 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 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).

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

The formal charge on each atom of resonating structure (c) of ${\text{NCO}}^{-}$ is calculated as,

Number of valence electrons in nitrogen is $5$.

Number of lone pair electrons in nitrogen is $6$.

Number of bond pair electrons in nitrogen is $2$.

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

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

Number of valence electrons in oxygen is $6$.

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

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

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

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

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).

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

According to formal charge calculations, there is a formal charge of $-1$ on oxygen atom in the resonanting structure (b). Oxygen is more electronegative than nitrogen. Formal negative charge is favorable on most electronegative atom. Hence the resonating structure (b) is preferred.

The given ion is ${\text{CON}}^{-}$. Number of valence electrons in oxygen is $6$ , number of valence electrons in carbon is $4$ and number of valence electrons in nitrogen is $5$. There is one nitrogen atom, one carbon atom and one oxygen atom present in ${\text{CON}}^{-}$. There is an additional valence electron due to $-1$ charge on ${\text{CON}}^{-}$. Therefore the total valence electrons are $6+4+5+1=16$. The lone pair of electrons on oxygen and bonding electrons is delocalized which results in the formation of lewis structuresThe lewis structures for ${\text{CON}}^{-}$ are,

Figure 3

The formal charge on each atom of resonating structure (a) of ${\text{CON}}^{-}$ is calculated as,

Number of valence electrons in nitrogen is $5$.

Number of lone pair electrons in nitrogen is $4$.

Number of bond pair electrons in nitrogen is $2$.

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

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

Number of valence electrons in oxygen is $6$.

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

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

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

$\begin{array}{c}\text{Formal}\text{charge}=6-\left(2+\frac{1}{2}\left(8\right)\right)\\ =0\end{array}$

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).

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

The formal charge on each atom of resonating structure (b) of ${\text{CON}}^{-}$ is calculated as,

Number of valence electrons in nitrogen is $5$.

Number of lone pair electrons in nitrogen is $4$.

Number of bond pair electrons in nitrogen is $4$.

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

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

Number of valence electrons in oxygen is $6$.

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

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

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

$\begin{array}{c}\text{Formal}\text{charge}=6-\left(2+\frac{1}{2}\left(8\right)\right)\\ =0\end{array}$

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 $4$.

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

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

The formal charge on each atom of resonating structure (c) of ${\text{CON}}^{-}$ is calculated as,

Number of valence electrons in nitrogen is $5$.

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

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

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

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

Number of valence electrons in oxygen is $6$.

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

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

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

$\begin{array}{c}\text{Formal}\text{charge}=6-\left(2+\frac{1}{2}\left(8\right)\right)\\ =0\end{array}$

Number of valence electrons in carbon is $4$.

Number of lone pair electrons in carbon is $4$.

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

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

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

According to formal charge calculations, resonating structure (a) and (c) least electronegative atom carbon possess $-1$ formal charge. In resonating structure (b), more electronegative atom nitrogen carries $-1$ formal charge. Hence resonating structure (b) is preffered for ${\text{CON}}^{-}$ ion.

Conclusion

The preferred resonance form in the ${\text{CNO}}^{-}$ ion is resonating structure (a).

The preferred resonance form in the ${\text{NCO}}^{-}$ ion is resonating structure (b).

The preferred resonance form in the ${\text{CON}}^{-}$ ion is resonating structure (b).