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-
cr3(po4)4
vcl3
fe(oh)3
cuf2

Explanation:

Step1: Analyze \(Cr_3(PO_4)_4\)

• For \(Cr_3(PO_4)_4\), the phosphate ion \((PO_4)\) has a charge of \(-3\). Let the charge of \(Cr\) be \(x\). Using the formula \(3x+4\times(- 3)=0\) (since the compound is neutral), \(3x = 12\), so \(x = +4\). The cation is \(Cr^{4+}\) and the anion is \(PO_4^{3-}\).

Step2: Analyze \(VCl_3\)

• Chloride ion \((Cl)\) has a charge of \(-1\). Let the charge of \(V\) be \(y\). Using \(y+3\times(-1)=0\), \(y=+3\). The cation is \(V^{3+}\) and the anion is \(Cl^{-}\).

Step3: Analyze \(Fe(OH)_3\)

• Hydroxide ion \((OH)\) has a charge of \(-1\). Let the charge of \(Fe\) be \(z\). Using \(z + 3\times(-1)=0\), \(z=+3\). The cation is \(Fe^{3+}\) and the anion is \(OH^{-}\).

Step4: Analyze \(CuF_2\)

• Fluoride ion \((F)\) has a charge of \(-1\). Let the charge of \(Cu\) be \(w\). Using \(w+2\times(-1)=0\), \(w = +2\). The cation is \(Cu^{2+}\) and the anion is \(F^{-}\).

Answer:

ionic compound	cation	anion
\(VCl_3\)	\(V^{3+}\)	\(Cl^{-}\)
\(Fe(OH)_3\)	\(Fe^{3+}\)	\(OH^{-}\)
\(CuF_2\)	\(Cu^{2+}\)	\(F^{-}\)