What is the formula for titanium(III) oxide?

Jul 17, 2016

${\text{Ti"_2"O}}_{3}$

Explanation:

The name of an ionic compound starts with the name of the cation, which is the positively charged ion, and ends with the name of the anion, which is the negatively charged ion.

Now, when a transition metal forms the cation, a Roman numeral is added to its name to let you know about its charge in the compound.

In this case, the name of the cation is

titanium(III)

This means that you're dealing with a cation of titanium, $\text{Ti}$, that carries a $\textcolor{red}{3 +}$ charge, hence the (III) Roman numeral used in its name.

The name of the anion is

oxide

As you know, oxygen, $\text{O}$, is located in group 16, which means that it accepts two electrons to complete its octet. This tells you that the oxide anion will carry a $\textcolor{b l u e}{2 -}$ charge.

Now, an ionic compound must be electrically neutral. To balance the overall positive charge coming from the cation with the overall negative charge coming from the anion, a formula unit of titanium(III) oxide must contain

$\textcolor{b l u e}{2} \times \text{Ti"^(color(red)(3+))" }$ and ${\text{ "color(red)(3) xx "O}}^{\textcolor{b l u e}{2 -}}$

You can thus say that the chemical formula for titanium(III) oxide is

2 xx ["Ti"^(3+)] + 3 xx ["O"^(2-)] = color(green)(|bar(ul(color(white)(a/a)color(black)("Ti"_2"O"_3)color(white)(a/a)|)))