Question

identify the charge of the ions in each compound, then type the formula below:
pb n
lead (iv) nitride
cu br
copper (iii) bromide
(type final formula)
(type final formula)
answer key

Explanation:

Step1: Determine charge of lead (IV) nitride

Lead (IV) has a charge of \(4+\). Nitride (\(N\)) has a charge of \(3-\).

Step2: Determine charge of copper (III) bromide

Copper (III) has a charge of \(3+\). Bromide (\(Br\)) has a charge of \(1-\).

Step3: Write formula for lead (IV) nitride

Using the criss - cross method (swap the magnitudes of the charges), for \(Pb^{4+}\) and \(N^{3 -}\), the formula is \(Pb_3N_4\).

Step4: Write formula for copper (III) bromide

For \(Cu^{3+}\) and \(Br^{1 -}\), the formula is \(CuBr_3\).

Answer:

Lead (IV) nitride: \(Pb^{4+}\), \(N^{3 -}\), \(Pb_3N_4\); Copper (III) bromide: \(Cu^{3+}\), \(Br^{1 -}\), \(CuBr_3\)