The ionic compound containing ${F e}^{3 +}$ and $(S O_{4})^{2 -}$ would have the formula

${F e}_{2} S O_{4}$

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${F e}_{3} (S O_{4})_{2}$

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${F e}_{2} (S O_{4})_{3}$

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${F e}_{3} S O_{4}$

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Solution

The correct option is C
${F e}_{2} (S O_{4})_{3}$

The valency of ${F e}^{3 +}$ ion is 3 and that of $(S O_{4})^{2 -}$ ion is 2. So, on cross-multiplication of valencies, the formula of the ionic compound containing ${F e}^{3 +}$ and $(S O_{4})^{2 -}$ is $F e_{2} (S O_{4})_{3}$ . It is also called as iron(III) sulphate or Ferric sulphate.

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Similar questions

The ionic compound containing ${F e}^{3 +}$ and $(S O_{4})^{2 -}$ would have the formula

{F e}^{3 +}

(S O_{4})^{2 -}

The ionic compound containing ${F e}^{3 +}$ and $(S O_{4})^{2 -}$ would have the formula

{F e}^{3 +}

(S O_{4})^{2 -}

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