Question

determine the formulas for these ionic compounds. copper(i) bromide: copper(i) oxide: copper(ii) bromide: copper(ii) oxide: iron(iii) bromide: iron(iii) oxide: lead(iv) bromide: lead(iv) oxide:

Explanation:

Step1: Recall ion charges

Copper(I) is $\text{Cu}^+$, bromide is $\text{Br}^-$, copper(II) is $\text{Cu}^{2 +}$, oxide is $\text{O}^{2-}$, iron(III) is $\text{Fe}^{3+}$, lead(IV) is $\text{Pb}^{4+}$.

Step2: Determine formula for copper(I) bromide

To balance charges, one $\text{Cu}^+$ and one $\text{Br}^-$ combine, formula is $\text{CuBr}$.

Step3: Determine formula for copper(I) oxide

Two $\text{Cu}^+$ are needed to balance one $\text{O}^{2-}$, formula is $\text{Cu}_2\text{O}$.

Step4: Determine formula for copper(II) bromide

One $\text{Cu}^{2+}$ combines with two $\text{Br}^-$, formula is $\text{CuBr}_2$.

Step5: Determine formula for copper(II) oxide

One $\text{Cu}^{2+}$ and one $\text{O}^{2-}$ combine, formula is $\text{CuO}$.

Step6: Determine formula for iron(III) bromide

One $\text{Fe}^{3+}$ combines with three $\text{Br}^-$, formula is $\text{FeBr}_3$.

Step7: Determine formula for iron(III) oxide

Two $\text{Fe}^{3+}$ are needed to balance three $\text{O}^{2-}$, formula is $\text{Fe}_2\text{O}_3$.

Step8: Determine formula for lead(IV) bromide

One $\text{Pb}^{4+}$ combines with four $\text{Br}^-$, formula is $\text{PbBr}_4$.

Step9: Determine formula for lead(IV) oxide

One $\text{Pb}^{4+}$ and two $\text{O}^{2-}$ combine, formula is $\text{PbO}_2$.

Answer:

copper(I) bromide: $\text{CuBr}$
copper(I) oxide: $\text{Cu}_2\text{O}$
copper(II) bromide: $\text{CuBr}_2$
copper(II) oxide: $\text{CuO}$
iron(III) bromide: $\text{FeBr}_3$
iron(III) oxide: $\text{Fe}_2\text{O}_3$
lead(IV) bromide: $\text{PbBr}_4$
lead(IV) oxide: $\text{PbO}_2$