Chapter SH, Problem 18.7APP

To determine

To Establish:The formula for lead (IV) oxide.

Answer to Problem 18.7APP

For lead (IV) oxide the formula is ${\text{PbO}}_2$ .

Explanation of Solution

Introduction:

The symbol and oxidation number of the positive ion:

Lead (IV): ${\text{Pb}}^{4+}$

The symbol and oxidation number of the negative ion:

Oxygen: ${\text{O}}^{2-}$

First write the symbol of the element having positive charge followed by the symbol of the element having negative charge.

  ${\text{Pb}}^{4+}{\text{O}}^{2-}$

Without the sign, one ion’s charge forms the subscript of the other.

  ${\text{Pb}}_2{\text{O}}_4$

The subscripts should be the lowest whole numbers that maintain the ratio of the ions:

  ${\text{PbO}}_2$

By determining whether the compound is neutral the formula obtained can be checked. The charge on one lead (IV) ions is $4+$ , and the charge on two oxygen is $4-$ . The compound is neutral, so the formula is correct.

Conclusion:

Forlead (IV) oxidethe formula is ${\text{PbO}}_2$ .