# Question eae8c

Aug 4, 2017

Lead(IV) oxide is ${\text{PbO}}_{2}$.

#### Explanation:

With lead(IV) oxide, we're given that $\text{Pb}$ is in the $4 +$ oxidation state. And $\text{O}$ is almost always going to be in the $2 -$ oxidation state.

If we switch these numbers to the opposite subscript, we would indeed obtain

overbrace("Pb")^(color(red)(bb(4+))) + overbrace("O")^(color(green)(bb(2-))) = ul("Pb"_(color(green)(bb(2)))"O"_(color(red)(bb(4)))

HOWEVER, ionic compounds will have their formula as the smallest whole number ratios of the subscripts. Each subscript here can be divided by $2$, leaving us with

color(blue)(ulbar(|stackrel(" ")(" ""PbO"_2" ")|)#