Question

write the empirical formula for at least four ionic compounds that could be formed from the following ions: no₃⁻, fe²⁺, cn⁻, pb⁴⁺

Explanation:

Step1: Combine Fe²⁺ with NO₃⁻

For \(Fe^{2 +}\) and \(NO_{3}^{-}\), using the criss - cross method (charge of \(Fe^{2+}\) is \(+ 2\), charge of \(NO_{3}^{-}\) is \(-1\)). The formula is \(Fe(NO_{3})_{2}\)

Step2: Combine Fe²⁺ with CN⁻

For \(Fe^{2 +}\) and \(CN^{-}\), charge of \(Fe^{2+}\) is \(+ 2\), charge of \(CN^{-}\) is \(-1\). The formula is \(Fe(CN)_{2}\)

Step3: Combine Pb⁴⁺ with NO₃⁻

For \(Pb^{4 +}\) and \(NO_{3}^{-}\), charge of \(Pb^{4+}\) is \(+ 4\), charge of \(NO_{3}^{-}\) is \(-1\). The formula is \(Pb(NO_{3})_{4}\)

Step4: Combine Pb⁴⁺ with CN⁻

For \(Pb^{4 +}\) and \(CN^{-}\), charge of \(Pb^{4+}\) is \(+ 4\), charge of \(CN^{-}\) is \(-1\). The formula is \(Pb(CN)_{4}\)

Answer:

\(Fe(NO_{3})_{2}\), \(Fe(CN)_{2}\), \(Pb(NO_{3})_{4}\), \(Pb(CN)_{4}\)