# What is the formula for lead (IV) oxide?

Pb${O}_{2}$
For each $P {b}^{4 +}$ we need two negative oxide ions ${O}^{2 -}$ to balance out total positive and total negative charge.
The ratio of $P {b}^{4 +}$ and ${O}^{2 -}$ ios is 1:2 and hence the formula is Pb${O}_{2}$.