Chapter 2, Problem 2.13E

(a)

Interpretation Introduction

Interpretation:

The most important Lewis structure of $\text{HONCO}$ along with equivalent resonance structures and formal charges has to be drawn.

Concept Introduction:

Lewis structures represent covalent bonds and describe valence electrons configuration of atoms. The covalent bonds are depicted by lines, and unshared electron pairs by pairs of dots. The sequence to write Lewis structure of some molecule is given as follows:

• The central atom is identified and various other atoms are arranged around it. This central atom so chosen is often the least electronegative.
• Total valence electrons is estimated.
• single bond is first placed between each atom pair.
• The electrons left can be allocated as unshared electron pairs or as multiple bonds around the right symbol of the element to satisfy the octet (or duplet) for each atom.
• Add charge on the overall structure in case of polyatomic cation or anion.

The formal charge on each atom in the Lewis structure can be calculated from the equation written as follows:

    $\text{Formal charge}=\left[V-L-\frac{1}{2}B\right]$

Here,

$V$ denotes valence electrons in free atom.

$L$ denotes electrons present as lone pairs.

$B$ denotes electrons present as bond pairs.

Explanation of Solution

Lewis structure possible for $\text{HONCO}$ is illustrated below:

Since the double and triple bonds can conjugate therefore delocalization occurs that results in various equivalent resonance structures as indicated below:

The formal charge on each atom in the Lewis structure is calculated from the equation as follows:

  $\text{Formal charge}=\left[V-L-\frac{1}{2}B\right]$        (1)

Substitute 5 for $V$, 2 for $L$ and 6 for $B$ in equation (1) to calculate formal charge on doubly bonded nitrogen in structure I.

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

Substitute 5 for $V$, 0 for $L$ and 8 for $B$ in equation (1) to calculate formal charge on triply bonded nitrogen in structure II.

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

Substitute 4 for $V$, 0 for $L$ and 8 for $B$ in equation (1) to calculate formal charge on central doubly- bonded carbon in structure I.

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

Substitute 5 for $V$, 2 for $L$ and 6 for $B$ in equation (1) to calculate formal charge on singly bonded nitrogen in structure II.

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

Substitute 4 for $V$, 2 for $L$ and 6 for $B$ in equation (1) to calculate formal charge on carbon in structure III.

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

Substitute 6 for $V$, 0 for $L$ and 8 for $B$ in equation (1) to calculate formal charge on terminal doubly- bonded oxygen.

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

Substitute 6 for $V$, 6 for $L$ and 2 for $B$ in equation (1) to calculate formal charge on terminal oxygen in structure II.

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

Substitute 6 for $V$, 6 for $L$ and 2 for $B$ in equation (1) to calculate formal charge on central oxygen in structure II.

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

Substitute 1 for $V$, 0 for $L$ and 2 for $B$ in equation (1) to calculate formal charge on terminal hydrogen.

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

Therefore the non-zero formal charges can be assigned as follows:

(b)

Interpretation Introduction

Interpretation:

The most important Lewis structure of ${\text{H}}_2\text{CSO}$ along with equivalent resonance structures and formal charges has to be drawn.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Lewis structure possible for ${\text{H}}_2\text{CSO}$ is illustrated below:

Since the double and triple bonds can conjugate therefore delocalization occurs that results in various equivalent resonance structures as indicated below:

The formal charge on each atom in the Lewis structure is calculated from the equation as follows:

  $\text{Formal charge}=\left[V-L-\frac{1}{2}B\right]$        (1)

Substitute 6 for $V$, 2 for $L$ and 8 for $B$ in equation (1) to calculate formal charge on sulphur in I.

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

Substitute 6 for $V$, 2 for $L$ and 6 for $B$ in equation (1) to calculate formal charge on sulphur in II.

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

Substitute 6 for $V$, 4 for $L$ and 4 for $B$ in equation (1) to calculate formal charge on oxygen in I.

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

Substitute 6 for $V$, 6 for $L$ and 2 for $B$ in equation (1) to calculate formal charge on oxygen in II.

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

Substitute 1 for $V$, 0 for $L$ and 2 for $B$ in equation (1) to calculate formal charge on terminal hydrogen.

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

Therefore the formal charges can be assigned as follows:

(c)

Interpretation Introduction

Interpretation:

The most important Lewis structure of ${\text{H}}_2\text{CNN}$ along with equivalent resonance structures and formal charges has to be drawn.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Lewis structure for $\text{ONCN}$ is illustrated below:

Since the double and triple bonds can conjugate therefore delocalization occurs that results in two equivalent resonance structures as indicated below:

The formal charge on each atom in the Lewis structure is calculated from the equation as follows:

  $\text{Formal charge}=\left[V-L-\frac{1}{2}B\right]$        (1)

Substitute 5 for $V$, 0 for $L$ and 8 for $B$ in equation (1) to calculate formal charge on doubly bonded central nitrogen in structure I.

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

Substitute 5 for $V$, 4 for $L$ and 4 for $B$ in equation (1) to calculate formal charge on doubly bonded terminal nitrogen in structure I.

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

Substitute 5 for $V$, 0 for $L$ and 8 for $B$ in equation (1) to calculate formal charge on triply bonded central nitrogen in structure II.

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

Substitute 5 for $V$, 2 for $L$ and 6 for $B$ in equation (1) to calculate formal charge on triply bonded terminal nitrogen in structure II.

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

Substitute 4 for $V$, 2 for $L$ and 6 for $B$ in equation (1) to calculate formal charge on carbon in structure II.

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

Substitute 1 for $V$, 0 for $L$ and 2 for $B$ in equation (1) to calculate formal charge on terminal hydrogen.

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

Therefore the formal charges in ${\text{H}}_2\text{CNN}$ can be assigned as follows:

(d)

Interpretation Introduction

Interpretation:

The most important Lewis structure of $\text{ONCN}$ along with equivalent resonance structures and formal charges has to be drawn.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Lewis structure for $\text{ONCN}$ is illustrated below:

Since the double and triple bonds can conjugate therefore delocalization occurs that results in two equivalent resonance structures as indicated below:

The formal charge on each atom in the Lewis structure is calculated from the equation as follows:

  $\text{Formal charge}=\left[V-L-\frac{1}{2}B\right]$        (1)

Substitute 5 for $V$, 4 for $L$ and 4 for $B$ in equation (1) to calculate formal charge on doubly bonded terminal nitrogen in structure II.

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

Substitute 5 for $V$, 0 for $L$ and 8 for $B$ in equation (1) to calculate formal charge on doubly bonded non-terminal nitrogen in structure II.

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

Therefore the non-zero formal charges in $\text{ONCN}$ can be assigned as follows: