Question

Question #eae8c

Answer 1

Lead(IV) oxide is `"PbO"_2`.

Explanation:

With lead(IV) oxide, we're given that `"Pb"` is in the `4+` oxidation state. And `"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" ")|)`