# Question f41c1

Oct 16, 2016

$\text{FeO }$ and ${\text{ Fe"_2"O}}_{3}$

#### Explanation:

The keys here are the Roman numerals used to name the two compounds.

As you know, transition metals can have multiple oxidation states. In order to distinguish between these possible oxidation states, we use Roman numerals.

The Roman numerals are meant to describe the oxidation state of the transition metal in a given compound.

In this case, you have

• iron(II) $\to {\text{Fe}}^{2 +}$
• iron(III) $\to {\text{Fe}}^{3 +}$

The (II) Roman numeral tells you that iron has a $+ 2$ oxidation state in iron(II) oxide, i.e. a $2 +$ charge. Similarly, the (III) Roman numeral tells you that iron has a $+ 3$ oxidation state in iron(III) oxide, i.e. a $3 +$ charge.

Now, both of these ionic compounds have oxide, ${\text{O}}^{2 -}$, as their anion. As you know, ionic compounds must be electrically neutral.

This implies that the overall positive charge coming from the cations must be balanced by the overall negative charge coming from the anions.

For iron(II) oxide, you have ${\text{Fe}}^{2 +}$ as the cation and ${\text{O}}^{2 -}$ as the anion, which means that the chemical formula will be

["Fe"^(2+)] + ["O"^(2-)] -> "FeO" -> iron(II) oxide

For iron(III) oxide, you have ${\text{Fe}}^{3 +}$ as the cation and ${\text{O}}^{2 -}$ as the anion. This time, you need to have $2$ iron(III) cations to get an overall positive charge of $6 +$ and $3$ oxide anions to get an overall negative charge of $6 -$.

The chemical formula will thus be

color(blue)(2) xx ["Fe"^color(red)(3+)] + color(red)(3) xx ["O"^color(blue)(2-)] -> "Fe"_2"O"_3 -># iron(III) oxide