Question #eae8c

1 Answer
Aug 4, 2017

Answer:

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" ")|)#