Question

write the empirical formula of at least four binary ionic compounds that could be formed from the following ions: fe²⁺, au³⁺, i⁻, s²⁻

Explanation:

Step1: Combine \(Fe^{2 +}\) with \(I^{-}\)

For \(Fe^{2+}\) and \(I^{-}\), using the criss - cross method (charge magnitudes as sub - scripts). The charge on \(Fe\) is \(+ 2\) and on \(I\) is \(-1\). So the formula is \(FeI_{2}\) (since \(|+2|\) and \(|-1|\), we have \(Fe_{1}I_{2}\)).

Step2: Combine \(Fe^{2 +}\) with \(S^{2 -}\)

For \(Fe^{2+}\) and \(S^{2-}\), the charge on \(Fe\) is \(+2\) and on \(S\) is \(-2\). Using the criss - cross method (dividing by the greatest common divisor of the charge magnitudes). The formula is \(FeS\) (since \(\frac{2}{2}=1\) for both ions).

Step3: Combine \(Au^{3 +}\) with \(I^{-}\)

For \(Au^{3+}\) and \(I^{-}\), the charge on \(Au\) is \(+3\) and on \(I\) is \(-1\). Using the criss - cross method, the formula is \(AuI_{3}\) (since \(|+3|\) and \(|-1|\), we have \(Au_{1}I_{3}\)).

Step4: Combine \(Au^{3 +}\) with \(S^{2 -}\)

For \(Au^{3+}\) and \(S^{2-}\), find the least common multiple of \(3\) and \(2\) (which is \(6\)). For \(Au\): \(\frac{6}{3}=2\), for \(S\): \(\frac{6}{2}=3\). The formula is \(Au_{2}S_{3}\).

Answer:

\(FeI_{2}\), \(FeS\), \(AuI_{3}\), \(Au_{2}S_{3}\)