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 | cation | anion
nacl | na⁺ | cl⁻
cucl₂ | □ | □
nh₄f | □ | □
vi₃ | □ | □
cr(po₄)₂ | □ | □

Explanation:

Step1: Analyze \( \text{CuCl}_2 \)

In \( \text{CuCl}_2 \), the cation is \( \text{Cu}^{2+} \) (since the compound has 2 \( \text{Cl}^- \) ions, and to balance charge, Cu has +2) and the anion is \( \text{Cl}^- \).

Step2: Analyze \( \text{NH}_4\text{F} \)

For \( \text{NH}_4\text{F} \), the cation is \( \text{NH}_4^+ \) (ammonium ion) and the anion is \( \text{F}^- \) (fluoride ion).

Step3: Analyze \( \text{VI}_3 \)

In \( \text{VI}_3 \), the cation is \( \text{V}^{3+} \) (since there are 3 \( \text{I}^- \) ions, V has +3 charge) and the anion is \( \text{I}^- \) (iodide ion).

Step4: Analyze \( \text{Cr(PO}_4\text{)}_2 \)

For \( \text{Cr(PO}_4\text{)}_2 \), the cation is \( \text{Cr}^{4+} \) (since there are 2 \( \text{PO}_4^{3-} \) ions, total negative charge is \( 2\times(-3)= -6 \), so Cr has +4 to balance) and the anion is \( \text{PO}_4^{3-} \) (phosphate ion).

Answer:

Ionic Compound	Cation	Anion
\( \text{CuCl}_2 \)	\( \text{Cu}^{2+} \)	\( \text{Cl}^- \)
\( \text{NH}_4\text{F} \)	\( \text{NH}_4^+ \)	\( \text{F}^- \)
\( \text{VI}_3 \)	\( \text{V}^{3+} \)	\( \text{I}^- \)
\( \text{Cr(PO}_4\text{)}_2 \)	\( \text{Cr}^{4+} \)	\( \text{PO}_4^{3-} \)