Animated Solution for Chemistry - States of Matter: Experimentally, it was found that a metal oxide has formula $M_{0.98}O$. Metal $M$, present as $M^{2+}$ and $M^{3+}$ in its oxide. Fraction of the metal which exists as $M^{3+}$ would be

Imagine you are holding a perfectly formed crystal of a metal oxide. In an ideal world, for every single oxygen atom, there would be exactly one metal atom. This perfect $1:1$ ratio is what we call stoichiometry.

But nature is rarely perfect.

In the real world, crystals often have defects. The problem we are tackling today introduces us to a fascinating imperfection known as a metal deficiency defect. The formula given is $M_{0.98}O$. This tells us a story: for every $1.00$ mole of oxygen ions ($O^{2-}$), there are only $0.98$ moles of metal ions ($M$). Some metal atoms have simply vanished from the lattice!

You might wonder, if positive metal ions are missing, wouldn't the crystal become negatively charged?

Here is where the Principle of Electrical Neutrality steps in. A stable crystal must always have a net charge of zero. The total positive charge must perfectly balance the total negative charge.

Let's break down the charges. We know that one mole of oxide ions ($O^{2-}$) carries a total charge of $-2$. Therefore, to keep the crystal neutral, the $0.98$ moles of metal ions must collectively carry a charge of $+2$.

But how can fewer atoms carry the same amount of charge?

The only way $0.98$ moles of metal can carry a $+2$ charge is if some of the metal ions step up and take on a higher charge. The problem states that the metal exists in two oxidation states: $M^{2+}$ and $M^{3+}$.

To solve this, we need to set up a simple algebraic equation. Let's assume the number of moles of the higher-charged $M^{3+}$ ions is $x$.

Since the total moles of metal is $0.98$, the remaining moles must belong to the $M^{2+}$ ions. So, the moles of $M^{2+}$ will be $(0.98-x)$.

Now, let's calculate the total positive charge contributed by both types of ions.

Each $M^{2+}$ ion contributes a charge of $+2$, and each $M^{3+}$ ion contributes a charge of $+3$. We can write the total positive charge as:

We already established that this total positive charge must equal $+2$ to balance the oxygen. So, we equate them:

Now, we just need to solve for $x$. Let's expand the bracket:

Simplifying the $x$ terms:

Moving $1.96$ to the other side:

This tells us that there are $0.04$ moles of $M^{3+}$ ions in our $1$ mole sample of the oxide.

The final step is to find the percentage of the metal that exists as $M^{3+}$. We divide the moles of $M^{3+}$ by the total moles of metal and multiply by $100$:

And there we have it! Exactly $4.08\%$ of the metal ions have upgraded to a $+3$ state to keep the crystal perfectly balanced.