Questions: Fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table: Some ionic compounds cation anion empirical formula name of compound Pb^2+ I^- Pb^4+ I^- K^+ I^- Ca^2+ I^-

Fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table: Some ionic compounds cation anion empirical formula name of compound Pb^2+ I^- Pb^4+ I^- K^+ I^- Ca^2+ I^-
Transcript text: Fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table: \begin{tabular}{|c|c|c|c|} \hline \multicolumn{4}{|c|}{ Some ionic compounds } \\ \hline cation & anion & empirical formula & name of compound \\ \hline $\mathrm{Pb}^{2+}$ & $\mathrm{I}^{-}$ & $\square$ & $\square$ \\ \hline $\mathrm{Pb}^{4+}$ & $\mathrm{I}^{-}$ & $\square$ & $\square$ \\ \hline $\mathrm{K}^{+}$ & $\mathrm{I}^{-}$ & $\square$ & $\square$ \\ \hline $\mathrm{Ca}^{2+}$ & $\mathrm{I}^{-}$ & $\square$ & $\square$ \\ \hline \end{tabular}
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Solution

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Solution Steps

Step 1: Determine the Empirical Formula for Each Compound

To find the empirical formula of an ionic compound, balance the charges of the cations and anions so that the total charge is zero.

  1. For \(\mathrm{Pb}^{2+}\) and \(\mathrm{I}^{-}\):

    • \(\mathrm{Pb}^{2+}\) has a charge of +2, and \(\mathrm{I}^{-}\) has a charge of -1.
    • To balance the charges, two \(\mathrm{I}^{-}\) ions are needed for each \(\mathrm{Pb}^{2+}\) ion.
    • Empirical formula: \(\mathrm{PbI}_2\)

  2. For \(\mathrm{Pb}^{4+}\) and \(\mathrm{I}^{-}\):

    • \(\mathrm{Pb}^{4+}\) has a charge of +4, and \(\mathrm{I}^{-}\) has a charge of -1.
    • To balance the charges, four \(\mathrm{I}^{-}\) ions are needed for each \(\mathrm{Pb}^{4+}\) ion.
    • Empirical formula: \(\mathrm{PbI}_4\)

  3. For \(\mathrm{K}^{+}\) and \(\mathrm{I}^{-}\):

    • \(\mathrm{K}^{+}\) has a charge of +1, and \(\mathrm{I}^{-}\) has a charge of -1.
    • The charges are already balanced with one \(\mathrm{K}^{+}\) and one \(\mathrm{I}^{-}\).
    • Empirical formula: \(\mathrm{KI}\)

  4. For \(\mathrm{Ca}^{2+}\) and \(\mathrm{I}^{-}\):

    • \(\mathrm{Ca}^{2+}\) has a charge of +2, and \(\mathrm{I}^{-}\) has a charge of -1.
    • To balance the charges, two \(\mathrm{I}^{-}\) ions are needed for each \(\mathrm{Ca}^{2+}\) ion.
    • Empirical formula: \(\mathrm{CaI}_2\)
Step 2: Determine the Name of Each Compound

The name of an ionic compound is typically the name of the cation followed by the name of the anion. For metals with variable charges, the charge is indicated in Roman numerals.

  1. For \(\mathrm{PbI}_2\):

    • Cation: Lead (II)
    • Anion: Iodide
    • Name: Lead (II) iodide

  2. For \(\mathrm{PbI}_4\):

    • Cation: Lead (IV)
    • Anion: Iodide
    • Name: Lead (IV) iodide

  3. For \(\mathrm{KI}\):

    • Cation: Potassium
    • Anion: Iodide
    • Name: Potassium iodide

  4. For \(\mathrm{CaI}_2\):

    • Cation: Calcium
    • Anion: Iodide
    • Name: Calcium iodide

Final Answer

\[ \begin{array}{|c|c|c|c|} \hline \text{cation} & \text{anion} & \text{empirical formula} & \text{name of compound} \\ \hline \mathrm{Pb}^{2+} & \mathrm{I}^{-} & \boxed{\mathrm{PbI}_2} & \boxed{\text{Lead (II) iodide}} \\ \hline \mathrm{Pb}^{4+} & \mathrm{I}^{-} & \boxed{\mathrm{PbI}_4} & \boxed{\text{Lead (IV) iodide}} \\ \hline \mathrm{K}^{+} & \mathrm{I}^{-} & \boxed{\mathrm{KI}} & \boxed{\text{Potassium iodide}} \\ \hline \mathrm{Ca}^{2+} & \mathrm{I}^{-} & \boxed{\mathrm{CaI}_2} & \boxed{\text{Calcium iodide}} \\ \hline \end{array} \]

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