# Question 832f0

Jan 22, 2016

Because uranium is in its $+ 4$ oxidation state.

#### Explanation:

First thing first, you're actually dealing with the chemical formula, not with the chemical equation.

Now, uranium(IV) oxide is an ionic compound that contains uranium cations and oxide anions.

As you can see, the name of the compound contains Roman numerals. As you know, Roman numerals are used when naming ionic compound in cases where the metal can have multiple oxidation states.

More specifically, the Roman numeral is used to indicate the oxidation state of the metal. In this case, the $\text{IV}$ Roman numeral tells you that uranium is in its $+ 4$ oxidation state.

For cations and anions, the oxidation state is equal to the overall charge they carry. So a $+ 4$ oxidation state for uranium will imply that the uranium cation looks like this

${\text{U}}^{\textcolor{red}{4 +}} \to$ the uranium(IV) cation

The oxide anion, on the other hand, will carry a $2 -$ charge.

${\text{O}}^{\textcolor{g r e e n}{2 -}} \to$ the oxide anion

Now, an ionic compound is always neutral. Notice that you have a $\textcolor{red}{4 +}$ positive charge on the cation, but only a $\textcolor{g r e e n}{2 -}$ negative charge on the anion.

This means that you will need two oxide anions to balance the positive charge of the uranium(IV) cation.

["U"]^color(red)(4+) 2["O"]^color(green)(2-) implies "UO"_2#

You'd get the same result by using the criss cross method for ionic compounds, which states that the charge of the cation becomes the subscript of the of the anion and vice versa

${\text{U"^color(red)(4+)"O"^color(green)(2-) <=> "U"_color(green)(2)"O}}_{\textcolor{red}{4}}$

This will be reduced to

${\text{U"_color(green)(2)"O"_color(red)(4) <=> "UO}}_{2}$

Hence, the chemical formula for uranium(IV) oxide, also called uranium dioxide, is ${\text{UO}}_{2}$.

The plate in the video was made using a glaze containing uranium oxide. The uranium made the plates radioactive!