Question

complete the table below by writing the symbols for the cation and anion that make up each ionic compound. the first row has been completed for you.

ionic compound	nacl	fepo₄	(nh₄)₂s	mn₂(co₃)₃	nii₂
anion	cl⁻

Explanation:

Step1: Analyze \( \text{FePO}_4 \)

In \( \text{FePO}_4 \), the phosphate ion is \( \text{PO}_4^{3-} \). To balance the charge, iron must have a \( +3 \) charge (since \( (+3) + (-3) = 0 \)). So the cation is \( \text{Fe}^{3+} \) and the anion is \( \text{PO}_4^{3-} \).

Step2: Analyze \( (\text{NH}_4)_2\text{S} \)

Ammonium ion is \( \text{NH}_4^+ \), and sulfide ion is \( \text{S}^{2-} \). The formula has 2 ammonium ions to balance the \( -2 \) charge of sulfide. So cation is \( \text{NH}_4^+ \), anion is \( \text{S}^{2-} \).

Step3: Analyze \( \text{Mn}_2(\text{CO}_3)_3 \)

Carbonate ion is \( \text{CO}_3^{2-} \). There are 3 carbonate ions (total charge \( 3\times(-2) = -6 \)), so 2 Mn ions must have total charge \( +6 \), meaning each Mn has \( +3 \) ( \( \text{Mn}^{3+} \) ). Anion is \( \text{CO}_3^{2-} \).

Step4: Analyze \( \text{NiI}_2 \)

Iodide ion is \( \text{I}^- \). There are 2 iodide ions (total charge \( -2 \)), so Ni has \( +2 \) charge ( \( \text{Ni}^{2+} \) ). Anion is \( \text{I}^- \).

Answer:

Ionic Compound	Cation	Anion
\( \text{FePO}_4 \)	\( \text{Fe}^{3+} \)	\( \text{PO}_4^{3-} \)
\( (\text{NH}_4)_2\text{S} \)	\( \text{NH}_4^+ \)	\( \text{S}^{2-} \)
\( \text{Mn}_2(\text{CO}_3)_3 \)	\( \text{Mn}^{3+} \)	\( \text{CO}_3^{2-} \)
\( \text{NiI}_2 \)	\( \text{Ni}^{2+} \)	\( \text{I}^- \)