Animated Solution for Chemistry - Redox Reactions: The oxidation states of transition metal atoms in ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$, ${\text{KMnO}}_4$ and ${\text{K}}_2{\text{FeO}}_4$, respectively, are $x,y$ and $z$. The sum of $x,y$ and $z$ is ……… .

Have you ever wondered what makes certain chemicals such powerful oxidizing agents? The secret often lies in the oxidation states of their central transition metals.

In this problem, we are tasked with finding the oxidation states of chromium, manganese, and iron in three classic compounds: potassium dichromate (${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$), potassium permanganate (${\text{KMnO}}_4$), and potassium ferrate (${\text{K}}_2{\text{FeO}}_4$).

Let's break down the rules and solve this step-by-step!

Before we dive into the algebra, we need to establish our ground rules.

Potassium ($\text{K}$) is an alkali metal residing in Group 1 of the periodic table. Because it has one valence electron that it readily gives up, its oxidation state is always $+1$ in its compounds.

Oxygen ($\text{O}$), on the other hand, is highly electronegative. In typical oxides like the ones we are dealing with, oxygen almost always takes an oxidation state of $-2$.

Finally, the golden rule of neutral molecules: the algebraic sum of the oxidation states of all atoms in a neutral compound must be exactly zero.

Let's start with our first compound, potassium dichromate (${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$). We need to find the oxidation state of chromium, which the problem calls $x$.

We have two potassium atoms, two chromium atoms, and seven oxygen atoms. Setting up our equation based on the golden rule, we get:

Now, it's just simple algebra. Let's simplify the terms:

Combining the constants gives us $2x-12=0$. Moving the $12$ to the other side, we find $2x=12$, which means:

Chromium is sitting at a hefty $+6$ oxidation state!

Next up is the famous purple compound, potassium permanganate (${\text{KMnO}}_4$). We need to find the oxidation state of manganese, denoted as $y$.

This molecule contains one potassium, one manganese, and four oxygens. Our equation looks like this:

Let's simplify the numbers:

This simplifies to $y-7=0$. Solving for $y$, we get:

Manganese is maxed out at its highest possible oxidation state of $+7$. No wonder it's such a brilliant oxidizing agent!

Our final compound is potassium ferrate (${\text{K}}_2{\text{FeO}}_4$). We are looking for the oxidation state of iron, which is $z$.

With two potassiums, one iron, and four oxygens, we set up our final equation:

Simplifying the expression:

This leaves us with $z-6=0$. Solving for $z$ gives:

Iron in a $+6$ state is quite rare and highly unstable, making ferrate an exceptionally strong oxidizer.

We've successfully unlocked all three oxidation states: $x=6$, $y=7$, and $z=6$.

The question asks for the sum of these three values. Let's add them up:

And there we have it! The sum of the oxidation states is 19.

By systematically applying the rules of oxidation states, even the most intimidating transition metal complexes can be easily decoded.