Chapter 20, Problem 72A

Interpretation Introduction

Interpretation: The ions that can act as a good oxidizing agent in the given list of compounds, $Mn{O}_4{}^{-},Mn{O}_4{}^{2-}$ , and $M{n}^{2+}$ are to be determined.

Concept introduction: The chemical species that undergo reduction and gain of electrons is found to be an oxidizing agent.

Answer to Problem 72A

  $Mn{O}_4{}^{-}$ is the species that is more likely acts as an oxidizing agent.

Explanation of Solution

In the given compounds, the one that has the highest oxidation number can act as a good oxidizing agent. So, the oxidation number of $Mn$ in each species is to be determined.

The oxidation number of $Mn$ in $Mn{O}_4{}^{-}$ is,

  $\begin{array}{c}x-8=-1\\ x=+7\end{array}$

The oxidation number of $Mn$ in $Mn{O}_4{}^{2-}$ is,

  $\begin{array}{c}x-8=-2\\ x=+6\end{array}$

The oxidation number of $Mn$ in $M{n}^{2+}$ is +2.

The high oxidation number is for $Mn{O}_4{}^{-}$ , Mn7+ so it is more likely to undergo reduction by the gain of electrons. So, $Mn{O}_4{}^{-}$ is the best oxidizing agent.