Chapter 25, Problem 1P

Interpretation Introduction

Interpretation: The oxidation number of N in nitrate, nitrite, NO, N₂O and N₂ needs to be determined.

Concept introduction: The oxidation state of an atom is zero in a neutral substance that contains atoms of just one component. The oxidation state of straightforward ions is adequate for the charge on the ion.

Answer to Problem 1P

The oxidation numbers of N in nitrate, nitrite, NO, N₂O and N₂ is +5, +3, +2, +1 and 0 respectively.

Explanation of Solution

The most stable oxidation state of oxygen is zero.

The formula of the nitrate is ${\text{NO}}_3{}^{-}$

The oxidation state of N in nitrate can be calculated as follows:

  $\begin{array}{l}{\text{ -1 = X}}_{\text{N}}\text{ + 3}{\text{.X}}_{\text{o}}\\ -1={\text{ X}}_{\text{N}}+3(-2)\\ {\text{ X}}_{\text{N}}\text{= +5}\end{array}$

The formula of the nitrite is ${\text{NO}}_2{}^{-}$

The oxidation state of N in nitrite can be calculated as follows:

  $\begin{array}{l}{\text{ -1 = X}}_{\text{N}}{\text{ + 2X}}_{\text{o}}\\ -1={\text{ X}}_{\text{N}}+2(-2)\\ {\text{ X}}_{\text{N}}\text{= +3}\end{array}$

For NO, the charge on the overall molecule is zero thus,

  $\begin{array}{l}{\text{0 = X}}_{\text{N}}{\text{ + X}}_{\text{o}}\\ \text{0 }={\text{ X}}_{\text{N}}+(-2)\\ {\text{ X}}_{\text{N}}\text{= +2}\end{array}$

For N₂O , the charge is also zero.

  $\begin{array}{l}\text{0 = 2}{\text{.X}}_{\text{N}}{\text{ + X}}_{\text{o}}\\ \text{0 }=\text{ 2}{\text{.X}}_{\text{N}}+(-2)\\ {\text{ X}}_{\text{N}}\text{= +1}\end{array}$

N₂also has zero charge thus,

  $\begin{array}{l}\text{0 = 2 }{\text{. X}}_{\text{N}}\\ {\text{X}}_{\text{N}}=\text{ 0}\end{array}$