Question

fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table:
some ionic compounds
cation anion empirical formula name of compound
cu+ f-
rb+ n3-
cu2+ i-
mg2+ p3-

Explanation:

Step1: Determine the empirical formula for \(Cu^{+}\) and \(F^{-}\)

The charges on \(Cu^{+}\) and \(F^{-}\) are \(+ 1\) and \(-1\) respectively. Using the criss - cross method (where the magnitude of the charge of one ion becomes the subscript of the other ion), the empirical formula is \(CuF\). The name of the compound is copper(I) fluoride (since copper has a \(+1\) oxidation state here).

Step2: Determine the empirical formula for \(Rb^{+}\) and \(N^{3 -}\)

The charge on \(Rb^{+}\) is \(+1\) and on \(N^{3 -}\) is \(-3\). Using the criss - cross method, the subscript of \(Rb\) is \(3\) and that of \(N\) is \(1\). So the empirical formula is \(Rb_{3}N\). The name of the compound is rubidium nitride.

Step3: Determine the empirical formula for \(Cu^{2+}\) and \(I^{-}\)

The charge on \(Cu^{2+}\) is \(+2\) and on \(I^{-}\) is \(-1\). Using the criss - cross method, the subscript of \(Cu\) is \(1\) and that of \(I\) is \(2\). So the empirical formula is \(CuI_{2}\). The name of the compound is copper(II) iodide (since copper has a \(+2\) oxidation state here).

Step4: Determine the empirical formula for \(Mg^{2+}\) and \(P^{3 -}\)

The charge on \(Mg^{2+}\) is \(+2\) and on \(P^{3 -}\) is \(-3\). Using the criss - cross method, we get \(Mg_{3}P_{2}\) (by taking the least common multiple of \(2\) and \(3\) which is \(6\), so \(Mg\) has a subscript of \(\frac{6}{2}=3\) and \(P\) has a subscript of \(\frac{6}{3} = 2\)). The name of the compound is magnesium phosphide.

Answer:

cation	anion	empirical formula	name of compound
\(Rb^{+}\)	\(N^{3 -}\)	\(Rb_{3}N\)	rubidium nitride
\(Cu^{2+}\)	\(I^{-}\)	\(CuI_{2}\)	copper(II) iodide
\(Mg^{2+}\)	\(P^{3 -}\)	\(Mg_{3}P_{2}\)	magnesium phosphide