Ionic compound formation table displaying combinations of metal cations and non-metal anions.

Problem Analysis:

The task involves filling in a table that represents the formation of ionic compounds. The table has cations (positively charged ions) on the left and anions (negatively charged ions) at the top. The goal is to determine the formula of the ionic compound formed by each pair of cation and anion.

#### Key Concepts:
1. Ionic Compounds: Formed when a cation and an anion combine in such a way that the total charge is neutral.
2. Charge Balance: The number of positive charges must equal the number of negative charges in the compound.
3. Subscripts: Indicate the number of each ion needed to balance the charges. If the subscript is 1, it is typically omitted.

#### Given Data:
- Cations:
- \( \text{Na}^+ \) (sodium ion, +1 charge)
- \( \text{K}^+ \) (potassium ion, +1 charge)
- \( \text{Mg}^{2+} \) (magnesium ion, +2 charge)
- \( \text{Al}^{3+} \) (aluminum ion, +3 charge)

- Anions:
- \( \text{Cl}^- \) (chloride ion, -1 charge)
- \( \text{I}^- \) (iodide ion, -1 charge)
- \( \text{S}^{2-} \) (sulfide ion, -2 charge)
- \( \text{O}^{2-} \) (oxide ion, -2 charge)
- \( \text{Br}^- \) (bromide ion, -1 charge)
- \( \text{N}^{3-} \) (nitride ion, -3 charge)

Solution Approach:

For each pair of cation and anion, we need to determine the formula of the ionic compound by balancing the charges. This involves finding the smallest whole-number ratio of cations to anions that results in a net charge of zero.

#### Step-by-Step Calculation:

1. Row 1: \( \text{Na}^+ \)
- \( \text{Na}^+ \) has a +1 charge.
- For each anion:
- \( \text{Cl}^- \): \( \text{Na}^+ \) and \( \text{Cl}^- \) have opposite charges of +1 and -1, so they balance as \( \text{NaCl} \).
- \( \text{I}^- \): Similarly, \( \text{NaI} \).
- \( \text{S}^{2-} \): To balance +1 and -2, we need 2 \( \text{Na}^+ \) ions, resulting in \( \text{Na}_2\text{S} \).
- \( \text{O}^{2-} \): Similarly, \( \text{Na}_2\text{O} \).
- \( \text{Br}^- \): \( \text{NaBr} \).
- \( \text{N}^{3-} \): To balance +1 and -3, we need 3 \( \text{Na}^+ \) ions, resulting in \( \text{Na}_3\text{N} \).

2. Row 2: \( \text{K}^+ \)
- \( \text{K}^+ \) has a +1 charge.
- The process is similar to \( \text{Na}^+ \):
- \( \text{KCl} \), \( \text{KI} \), \( \text{K}_2\text{S} \), \( \text{K}_2\text{O} \), \( \text{KBr} \), \( \text{K}_3\text{N} \).

3. Row 3: \( \text{Mg}^{2+} \)
- \( \text{Mg}^{2+} \) has a +2 charge.
- For each anion:
- \( \text{Cl}^- \): To balance +2 and -1, we need 2 \( \text{Cl}^- \) ions, resulting in \( \text{MgCl}_2 \).
- \( \text{I}^- \): Similarly, \( \text{MgI}_2 \).
- \( \text{S}^{2-} \): \( \text{Mg}^{2+} \) and \( \text{S}^{2-} \) have opposite charges of +2 and -2, so they balance as \( \text{MgS} \).
- \( \text{O}^{2-} \): Similarly, \( \text{MgO} \).
- \( \text{Br}^- \): \( \text{MgBr}_2 \).
- \( \text{N}^{3-} \): To balance +2 and -3, we need 3 \( \text{Mg}^{2+} \) ions and 2 \( \text{N}^{3-} \) ions, resulting in \( \text{Mg}_3\text{N}_2 \).

4. Row 4: \( \text{Al}^{3+} \)
- \( \text{Al}^{3+} \) has a +3 charge.
- For each anion:
- \( \text{Cl}^- \): To balance +3 and -1, we need 3 \( \text{Cl}^- \) ions, resulting in \( \text{AlCl}_3 \).
- \( \text{I}^- \): Similarly, \( \text{AlI}_3 \).
- \( \text{S}^{2-} \): To balance +3 and -2, we need 3 \( \text{Al}^{3+} \) ions and 2 \( \text{S}^{2-} \) ions, resulting in \( \text{Al}_2\text{S}_3 \).
- \( \text{O}^{2-} \): Similarly, \( \text{Al}_2\text{O}_3 \).
- \( \text{Br}^- \): \( \text{AlBr}_3 \).
- \( \text{N}^{3-} \): \( \text{AlN} \) (since +3 and -3 balance directly).

Final Table:

Using the above calculations, the completed table is:

| | \( \text{Cl}^- \) | \( \text{I}^- \) | \( \text{S}^{2-} \) | \( \text{O}^{2-} \) | \( \text{Br}^- \) | \( \text{N}^{3-} \) |
|---------|-------------------|------------------|---------------------|---------------------|-------------------|---------------------|
| \( \text{Na}^+ \) | \( \text{NaCl} \) | \( \text{NaI} \) | \( \text{Na}_2\text{S} \) | \( \text{Na}_2\text{O} \) | \( \text{NaBr} \) | \( \text{Na}_3\text{N} \) |
| \( \text{K}^+ \) | \( \text{KCl} \) | \( \text{KI} \) | \( \text{K}_2\text{S} \) | \( \text{K}_2\text{O} \) | \( \text{KBr} \) | \( \text{K}_3\text{N} \) |
| \( \text{Mg}^{2+} \) | \( \text{MgCl}_2 \) | \( \text{MgI}_2 \) | \( \text{MgS} \) | \( \text{MgO} \) | \( \text{MgBr}_2 \) | \( \text{Mg}_3\text{N}_2 \) |
| \( \text{Al}^{3+} \) | \( \text{AlCl}_3 \) | \( \text{AlI}_3 \) | \( \text{Al}_2\text{S}_3 \) | \( \text{Al}_2\text{O}_3 \) | \( \text{AlBr}_3 \) | \( \text{AlN} \) |

Final Answer:

\[
\boxed{
\begin{array}{|c|c|c|c|c|c|c|}
\hline
& \text{Cl}^- & \text{I}^- & \text{S}^{2-} & \text{O}^{2-} & \text{Br}^- & \text{N}^{3-} \\
\hline
\text{Na}^+ & \text{NaCl} & \text{NaI} & \text{Na}_2\text{S} & \text{Na}_2\text{O} & \text{NaBr} & \text{Na}_3\text{N} \\
\hline
\text{K}^+ & \text{KCl} & \text{KI} & \text{K}_2\text{S} & \text{K}_2\text{O} & \text{KBr} & \text{K}_3\text{N} \\
\hline
\text{Mg}^{2+} & \text{MgCl}_2 & \text{MgI}_2 & \text{MgS} & \text{MgO} & \text{MgBr}_2 & \text{Mg}_3\text{N}_2 \\
\hline
\text{Al}^{3+} & \text{AlCl}_3 & \text{AlI}_3 & \text{Al}_2\text{S}_3 & \text{Al}_2\text{O}_3 & \text{AlBr}_3 & \text{AlN} \\
\hline
\end{array}
}
\]