Chapter 2, Problem 2B.19E

(a)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{NO}}^{+}$ has to be drawn and the formal charge has to be determined.

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 are estimated for each atoms.
• single bond is first placed between each atom pair.
• The electrons left can be allocated as unshared electron pairs or as multiple bonds around symbol of element to satisfy the octet (or duplet) for each atom.
• Add charge on overall structure in case of polytatomic 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

${\text{NO}}^{+}$ consists of one $\text{N}$ and one $\text{O}$ atom.

$\text{O}$ has atomic number 8; its electronic configuration is $\left[\text{He}\right]2{\text{s}}^{2}2{\text{p}}^{\text{4}}$. Thus, it possesses 6 valence electrons.

$\text{N}$ with atomic number 7, its electronic configuration is $\left[\text{He}\right]2{\text{s}}^{2}2{\text{p}}^3$. Thus, it possesses 5 valence electrons.

Thus total valence electrons is sum of the valence electrons for each atom along with uni-positive in ${\text{NO}}^{+}$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=6+5-1\\ =10\left(\text{5 pairs}\right)\end{array}$

Hence, 5 pairs are allocated to form Lewis structure of ${\text{NO}}^{+}$ is drawn as follows:

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

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

Substitute 5for $V$, 2 for $L$ and 6 for $B$ in equation (1).

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

Substitute 6for $V$, 2 for $L$ and 6 for $B$ in equation (1).

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

Hence $\text{N}$ has formal charge of 0 while $\text{O}$ has formal charge of $+1$. This is illustrated below:

(b)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{N}}_2$ has to be drawn and the formal charge has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

${\text{N}}_2$ consists of two $\text{N}$ atoms.

$\text{N}$ with atomic number 7; its electronic configuration is $\left[\text{He}\right]2{\text{s}}^{2}2{\text{p}}^3$. Thus, it possesses 5 valence electrons.

Thus total valence electrons is sum of the valence electrons on each atom in ${\text{N}}_2$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=5\left(2\right)\\ =10\end{array}$

Hence,  of these five pairs two are allocated as lone pairs and three pairs are added so as to form triple bond to obtain Lewis structure of ${\text{N}}_2$ as follows:

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

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

Substitute 5for $V$, 2 for $L$ and 6 for $B$ in equation (1).

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

Hence each $\text{N}$ has zero formal charge. This is illustrated below:

(c)

Interpretation Introduction

Interpretation:

The Lewis structure of $\text{CO}$ has to be drawn and the formal charge has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

$\text{CO}$ consists of one $\text{C}$ and one $\text{O}$ atom.

$\text{O}$ has atomic number 8; its electronic configuration is $\left[\text{He}\right]2{\text{s}}^{2}2{\text{p}}^{\text{4}}$. Thus, it possesses 6 valence electrons.

$\text{C}$ has atomic number 4; its electronic configuration is $\left[\text{He}\right]2{\text{s}}^{2}2{\text{p}}^2$. Thus, it possesses 4 valence electrons.

Thus total valence electron is sum of the valence electrons for each atom in $\text{CO}$. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=6+4\\ =10\left(\text{5 pairs}\right)\end{array}$

Hence, 5 pairs are allocated to form Lewis structure of $\text{CO}$ as follows:

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

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

Substitute 4for $V$, 2 for $L$ and 6 for $B$ in equation (1).

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

Substitute 6 for $V$, 2 for $L$ and 6 for $B$ in equation (1).

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

Hence $\text{C}$ has formal charge of $-1$ while $\text{O}$ has formal charge of $+1$. This is illustrated below:

(d)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{C}}_2{}^{2-}$ has to be drawn and the formal charge has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

${\text{C}}_2{}^{2-}$ consists of two $\text{C}$.

$\text{C}$ has atomic number 4; its electronic configuration is $\left[\text{He}\right]2{\text{s}}^{2}2{\text{p}}^2$. Thus, it possesses 4 valence electrons.

Thus total valence electron is sum of the valence electrons for each atom in ${\text{C}}_2{}^{2-}$ along with two uni-negative charge. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=4\left(2\right)+2\\ =10\left(\text{5 pairs}\right)\end{array}$

Hence, 5 pairs are arranged to form Lewis structure of ${\text{C}}_2{}^{2-}$ as follows:

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

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

Substitute 4 for $V$, 2 for $L$ and 6 for $B$ in equation (1).

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

Hence each $\text{C}$ has formal charge of $-1$. This is illustrated below:

(e)

Interpretation Introduction

Interpretation:

The Lewis structure of ${\text{CN}}^{-}$ has to be drawn and the formal charge has to be determined.

Concept Introduction:

Refer to part (a).

Explanation of Solution

${\text{CN}}^{-}$ consists of two $\text{C}$.

$\text{C}$ has atomic number 4; its electronic configuration is $\left[\text{He}\right]2{\text{s}}^{2}2{\text{p}}^2$. Thus, it possesses 4 valence electrons.

$\text{N}$ with atomic number 7, its electronic configuration is $\left[\text{He}\right]2{\text{s}}^{2}2{\text{p}}^3$. Thus, it possesses 5 valence electrons.

Thus total valence electron is sum of the valence electrons for each atom in ${\text{CN}}^{-}$ along with one uni-negative charge. It is calculated as follows:

  $\begin{array}{c}\text{Total valence electrons}=4+5+1\\ =10\left(\text{5 pairs}\right)\end{array}$

Hence, 5 pairs are arranged to form Lewis structure of ${\text{CN}}^{-}$ as follows:

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

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

Substitute 4 for $V$, 2 for $L$ and 6 for $B$ in equation (1).

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

Substitute 5 for $V$, 2 for $L$ and 6 for $B$ in equation (1).

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

Hence formal charge on $\text{C}$ is $-1$  and on $\text{N}$ is 0. This is illustrated below: