Question

write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: zn^{2+}, al^{3+}, br^{-}, o^{2-}

Explanation:

Step1: Combine \(Zn^{2+}\) with \(Br^{-}\)

For \(Zn^{2+}\) and \(Br^{-}\), using the criss - cross method (charge of cation becomes subscript of anion and vice - versa). The charge of \(Zn^{2+}\) is \(+ 2\) and of \(Br^{-}\) is \(-1\). So the formula is \(ZnBr_{2}\)

Step2: Combine \(Zn^{2+}\) with \(O^{2 -}\)

For \(Zn^{2+}\) and \(O^{2-}\), the charge of \(Zn^{2+}\) is \(+2\) and of \(O^{2-}\) is \(-2\). Using the criss - cross method (simplify the ratio), the formula is \(ZnO\)

Step3: Combine \(Al^{3+}\) with \(Br^{-}\)

For \(Al^{3+}\) and \(Br^{-}\), the charge of \(Al^{3+}\) is \(+3\) and of \(Br^{-}\) is \(-1\). Using the criss - cross method, the formula is \(AlBr_{3}\)

Step4: Combine \(Al^{3+}\) with \(O^{2 -}\)

For \(Al^{3+}\) and \(O^{2-}\), the charge of \(Al^{3+}\) is \(+3\) and of \(O^{2-}\) is \(-2\). Using the criss - cross method (find the least common multiple of 2 and 3 which is 6. So \(Al\) subscript is \(2\) and \(O\) subscript is \(3\)), the formula is \(Al_{2}O_{3}\)

Answer:

\(ZnBr_{2}\), \(ZnO\), \(AlBr_{3}\), \(Al_{2}O_{3}\)