Question

How would you account for the following?
The oxidizing power of the following three oxo-ions in the series follows the order:
$\text{V}{{\text{O}}_{2}}^{+}<\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}^{2-}<\text{Mn}{{\text{O}}_{4}}^{-}$

Answer 1

Hint: For determining the oxidizing power of the oxo-ions, we need to first calculate the oxidation state of the central atom of every oxo-ions and then account for the question accordingly.

Complete step by step solution:
As given in the hint, first we will have to calculate the oxidation of the central atom of the given oxo-ions. The calculation of the given oxo-ions are given below-
For $\text{V}{{\text{O}}_{2}}^{+}$:
Let the oxidation state of Vanadium (V) is $x$.
$\begin{aligned} & \Rightarrow x-4=1 \\\ & \Rightarrow x=1+4=5 \\\ \end{aligned}$
For $\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}^{2-}$ :
Let the oxidation state of Chromium (Cr) is $x$.
$\begin{aligned} & \Rightarrow 2x-14=-2 \\\ & \Rightarrow 2x=-2+14=12 \\\ & \Rightarrow x=+6 \\\ \end{aligned}$
For $\text{Mn}{{\text{O}}_{4}}^{-}$
Let the oxidation state of Manganese (Mn) is $x$.
$\begin{aligned} & \Rightarrow x-8=-1 \\\ & \Rightarrow x=-1+8 \\\ & \Rightarrow x=+7 \\\ \end{aligned}$ $\begin{aligned} & \Rightarrow x-8=-1 \\\ & \Rightarrow x=-1+8 \\\ & \Rightarrow x=+7 \\\ \end{aligned}$
As per the concept, the ions in which the central atom is present in the highest oxidation state will have the highest oxidizing power. The reason why the compounds in which metal is in higher oxidation state are strong oxidizing agents is because higher oxidation state compounds have the ability to get reduced easily i.e. they change from a high to low oxidation state very easily. Here, the oxidation state of V in $\text{V}{{\text{O}}_{2}}^{+}$ is +5 and oxidation state of Cr in $\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}^{2-}$is +6, oxidation state of Mn is +7.
Therefore, the higher the oxidation state of oxo-ions, the higher is the oxidizing power.
Hence, the given oxo-ions in the series follow the order as-
$\text{V}{{\text{O}}_{2}}^{+}<\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}^{2-}<\text{Mn}{{\text{O}}_{4}}^{-}$

Note: We should also know that higher oxidation states in heavy metals of d-block elements are very stable due to the fact that only core electrons are present which is very stable due to the removal of valence electrons. Hence, it can lose one or more electrons from the unstable core giving higher oxidation state and stability. Due to this, covalent character also increases and so is the stability.