##”Cr”_2″O”_3##

The first thing to notice here is that you’re dealing with an ionic compound that contains chromium, ##”Cr”##, a transition metal, and oxygen, ##”O”##, a non-metal.

The interesting thing about this compound is that it contains a transition metal as its cation, i.e. its positively charged ion. As you know, the name of a transition metal cation is written using Roman numerals.

This is done to distinguish between the possible oxidation states that transition metals can exhibit. In this case, the name of the compound contains the (III) Roman numeral, which implies that the metal has a ##+3## oxidation state.

In other words, the cation carries a ##color(blue)(3+)## positive charge

##”Cr”^color(blue)((3+)) ->## the chromium(III) cation

The anion, i.e. the negatively charged ion, is written using the -ide prefix. In this case, you’re dealing with the oxide anion, ##”O”^color(red)(2-)##.

The anion carries a ##color(red)(2-)## charge because oxygen is located in group 16 of , and thus requires two more electrons in its outermost shell to complete its octet.

##”O”^color(red)(2-) ->## the oxide anion

Now, an ionic compound must be electrically neutral, which implies that the total positive charge coming from the cation must be balanced by the total negative charge coming from the anion.

In this case, you need ##color(red)(2)## chromium(III) cations and ##color(blue)(3)## oxide anions in order to have a neutral compound.

##color(red)(2) xx [“Cr”^color(blue)(3+)] + color(blue)(3) xx [“O”^color(red)(2-)] -> “Cr”_ color(red)(2)”O”_ color(blue)(3)##

The chemical formula for chromium(III) oxide will thus be

##color(green)(|bar(ul(color(white)(a/a)color(black)(“Cr”_2″O”_3)color(white)(a/a)|)))##