Chapter 12, Problem 33E

Interpretation Introduction

Interpretation:

The reason as to why it is sometimes difficult to fit a nitrogen atom into a Lewis diagram that obeys the octet rule is to be stated. The species out of ${\text{N}}_{\text{2}}\text{O,}{\text{NO}}_{\text{2}},{\text{NF}}_{\text{3}},\text{NO,}{\text{N}}_{\text{2}}{\text{O}}_{\text{3}},{\text{N}}_{\text{2}}{\text{O}}_{\text{4}},\text{NOCl,}{\text{NO}}_2\text{Cl}$ whose Lewis diagram does not satisfy the octet rule are to be stated.

Concept introduction:

Lewis diagram is a representation of the chemical formula of substance with valance electrons of atoms. The Lewis structures are also called electron dot structures. In the Lewis structure, electrons are denoted by dots. Only the valence electrons are presented as dots in the Lewis structure.

Answer to Problem 33E

It is difficult to fit a nitrogen atom into a Lewis diagram that obeys the octet rule as nitrogen has odd number of valence electrons. The Lewis diagram of ${\text{NO}}_{\text{2}}$ and $\text{NO}$ does not satisfy the octet rule.

Explanation of Solution

To write the Lewis diagram for a compound, first the number of valence electrons is to be calculated.

The electrons which are present in the outermost shell of the atom and occupy the highest energy level of an atom present in its ground state are termed as valence electrons. The number of valence electrons present in the nitrogen atom is $5$. This means that the number of valence electrons is odd. Therefore, it is difficult to fit a nitrogen atom into a Lewis diagram that obeys the octet rule.

The valence electrons of nitrogen ($\text{N}$) are $5$.

The valence electrons of oxygen ($\text{O}$) are $6$.

The total number of valence electron for the molecule ${\text{N}}_{\text{2}}\text{O}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{number}\text{of}\text{valence}\text{electron}& =& \left(\left(2\times 5\right)+6\right){\text{e}}^{-}\\ & =& 16{\text{e}}^{-}\end{array}$

The total number of valence electron for the molecule ${\text{NO}}_{\text{2}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{number}\text{of}\text{valence}\text{electron}& =& \left(5+\left(2\times 6\right)\right){\text{e}}^{-}\\ & =& 17{\text{e}}^{-}\end{array}$

The valence electrons of fluorine ($\text{F}$) are $7$.

The total number of valence electron for the molecule ${\text{NF}}_{\text{3}}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{number}\text{of}\text{valence}\text{electron}& =& \left(5+\left(3\times 7\right)\right){\text{e}}^{-}\\ & =& 26{\text{e}}^{-}\end{array}$

The total number of valence electron for the molecule $\text{NO}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{number}\text{of}\text{valence}\text{electron}& =& \left(5+6\right){\text{e}}^{-}\\ & =& 11{\text{e}}^{-}\end{array}$

The total number of valence electron for the molecule ${\text{N}}_2{\text{O}}_3$ is calculated below.

$\begin{array}{rll}\text{Total}\text{number}\text{of}\text{valence}\text{electron}& =& \left(\left(2\times 5\right)+\left(3\times 6\right)\right){\text{e}}^{-}\\ & =& 28{\text{e}}^{-}\end{array}$

The total number of valence electron for the molecule ${\text{N}}_2{\text{O}}_4$ is calculated below.

$\begin{array}{rll}\text{Total}\text{number}\text{of}\text{valence}\text{electron}& =& \left(\left(2\times 5\right)+\left(4\times 6\right)\right){\text{e}}^{-}\\ & =& 34{\text{e}}^{-}\end{array}$

The valence electrons of chlorine ($\text{Cl}$) are $7$.

The total number of valence electron for the molecule $\text{NOCl}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{number}\text{of}\text{valence}\text{electron}& =& \left(5+6+7\right){\text{e}}^{-}\\ & =& 18{\text{e}}^{-}\end{array}$

The total number of valence electron for the molecule ${\text{NO}}_2\text{Cl}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{number}\text{of}\text{valence}\text{electron}& =& \left(5+\left(2\times 6\right)+7\right){\text{e}}^{-}\\ & =& 24{\text{e}}^{-}\end{array}$

Therefore, as ${\text{NO}}_{\text{2}}$ and $\text{NO}$ have an odd number of valence electrons, their Lewis diagrams will not satisfy the octet rule.

Conclusion

Nitrogen has an odd number of valence electrons and it is difficult to fit into a Lewis diagram that obeys the octet rule. Lewis diagrams of ${\text{NO}}_{\text{2}}$ and $\text{NO}$ do not satisfy the octet rule as they have an odd number of valence electrons.