Answered

Fill in the name and empirical formula of each ionic compound that could be formed from the ions in this table:



cation		anionempirical formulaname of compound

$Sr ^{2+}$		$S ^{2-}$□□

$Sr ^{2+}$		$O ^{2-}$□□

$Sr ^{2+}$		$P ^{3-}$□□

$Sr ^{2+}$		$N ^{3-}$□□

Verified Answer :

  • Identify ion charges for \(Sr^{2+}\) (\(+2\)), \(S^{2-}\) (\(-2\)), \(O^{2-}\) (\(-2\)), \(P^{3-}\) (\(-3\)), and \(N^{3-}\) (\(-3\)).
  • Apply the neutrality principle, finding the smallest whole-number ratio \(x:y\) such that \(x \times (\text{cation charge}) + y \times (\text{anion charge}) = 0\).
  • Determine formulas: \(SrS\), \(SrO\), \(Sr_3P_2\), and \(Sr_3N_2\).
  • Name compounds as Strontium sulfide, Strontium oxide, Strontium phosphide, and Strontium nitride. \(\boxed{SrS, \text{Strontium sulfide}; SrO, \text{Strontium oxide}; Sr_3P_2, \text{Strontium phosphide}; Sr_3N_2, \text{Strontium nitride}}\)

Explanation

  1. Analysis of Ion Charges and Neutrality Rule

To determine the empirical formula and the name of the ionic compounds, we analyze the charges of the given ions. An ionic compound must be electrically neutral, meaning the total positive charge from the cations must balance the total negative charge from the anions. If the cation has a charge of \(+m\) and the anion has a charge of \(-n\), the smallest whole-number ratio \(x:y\) such that \(x(+m) + y(-n) = 0\) gives the empirical formula \(M_x A_y\). In this problem, the cation is always strontium, \(Sr^{2+}\), with a charge of \(+2\). We apply this to four different anions: sulfide (\(S^{2-}\)), oxide (\(O^{2-}\)), phosphide (\(P^{3-}\)), and nitride (\(N^{3-}\)).

  1. Compounds with Sulfide and Oxide

For \(Sr^{2+}\) and \(S^{2-}\):
The charges are \(+2\) and \(-2\). The smallest ratio to make the total charge zero is \(1:1\). Thus, the empirical formula is \(Sr_{1}S_{1}\) or \(SrS\). The compound is named by stating the metal followed by the non-metal with an '-ide' suffix: Strontium sulfide.

For \(Sr^{2+}\) and \(O^{2-}\):
The charges are \(+2\) and \(-2\). Similar to the first case, the smallest ratio is \(1:1\). The empirical formula is \(SrO\). The compound is named Strontium oxide.

  1. Compounds with Phosphide and Nitride

For \(Sr^{2+}\) and \(P^{3-}\):
The charges are \(+2\) and \(-3\). To balance these, we find the least common multiple of 2 and 3, which is 6. We need three \(Sr^{2+}\) ions (\(3 \times +2 = +6\)) and two \(P^{3-}\) ions (\(2 \times -3 = -6\)). The empirical formula is \(Sr_3P_2\). The compound is named Strontium phosphide.

For \(Sr^{2+}\) and \(N^{3-}\):
The charges are \(+2\) and \(-3\). Using the same logic as the phosphide, we need three \(Sr^{2+}\) ions and two \(N^{3-}\) ions to achieve neutrality. The empirical formula is \(Sr_3N_2\). The compound is named Strontium nitride.

  1. Conclusion of Results

We have determined the empirical formulas and names for all four combinations:

  • \(Sr^{2+}\) and \(S^{2-}\): \(SrS\), Strontium sulfide.
  • \(Sr^{2+}\) and \(O^{2-}\): \(SrO\), Strontium oxide.
  • \(Sr^{2+}\) and \(P^{3-}\): \(Sr_3P_2\), Strontium phosphide.
  • \(Sr^{2+}\) and \(N^{3-}\): \(Sr_3N_2\), Strontium nitride.

The final results are presented below.

Examples

In chemical manufacturing, it is essential to determine the correct proportions of reactants to form stable products. For instance, knowing that strontium forms specific ratios with oxygen or nitrogen allows engineers to predict the chemical makeup of ceramics or specialized electronics materials. By balancing charges as shown in this exercise, chemists ensure that the final ionic compounds created are electrically neutral and stable, which is a foundational requirement for laboratory research and industrial synthesis.

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