Question

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
cr²⁺ | cn⁻ | cr(cn)₂ | chromium(ii) cyanide
ca²⁺ | so₄²⁻ | caso₄ | calcium sulfate
pb⁴⁺ | c₂h₃o₂⁻ | pb(c₂h₃o₂)₄ | lead(iv) acetate

Explanation:

Step1: Identify the ions for the third row (Pb⁴⁺ and C₂H₃O₂⁻)

The cation is \( \text{Pb}^{4+} \) and the anion is \( \text{C}_2\text{H}_3\text{O}_2^- \) (acetate ion). To form a neutral ionic compound, the total positive charge must equal the total negative charge. Let the number of \( \text{Pb}^{4+} \) ions be \( x \) and the number of \( \text{C}_2\text{H}_3\text{O}_2^- \) ions be \( y \). So, \( 4x = 1y \) (since the charge of \( \text{Pb}^{4+} \) is +4 and \( \text{C}_2\text{H}_3\text{O}_2^- \) is -1). To balance the charges, we need \( x = 1 \) and \( y = 4 \) (because \( 4\times1 = 1\times4 \)).

Step2: Write the empirical formula

Using the number of ions from Step 1, the empirical formula is \( \text{Pb}(\text{C}_2\text{H}_3\text{O}_2)_4 \).

Step3: Name the compound

The cation is lead with a +4 charge (so we use Roman numeral IV), and the anion is acetate. So the name is Lead(IV) acetate.

Answer:

The empirical formula for the compound formed from \( \text{Pb}^{4+} \) and \( \text{C}_2\text{H}_3\text{O}_2^- \) is \( \text{Pb}(\text{C}_2\text{H}_3\text{O}_2)_4 \), and its name is Lead(IV) acetate.