Question

which formula corresponds to lead (iv) oxide? pbo pb₂o₄ pb₂o pbo₂

Explanation:

Lead (IV) oxide means lead has an oxidation state of +4. In oxides, oxygen has an oxidation state of -2. Using the criss - cross method (the magnitude of the oxidation state of one ion becomes the sub - script of the other ion), for \(Pb^{4 +}\) and \(O^{2-}\), we get \(PbO_2\) (since \(\frac{4}{2}=2\) for the sub - script of \(O\) and \(1\) for \(Pb\) as \(\frac{2}{2} = 1\)). \(PbO\) has \(Pb\) in \(+2\) oxidation state. \(Pb_2O_4\) can be simplified to \(PbO_2\) (dividing sub - scripts by 2), but the non - simplified form is not the standard formula. \(Pb_2O\) has \(Pb\) in \(+1\) oxidation state.

Answer:

\(PbO_2\) (the fourth option)