In a neutral compound formed from Al^{3+} ions and S^{2-} ions, what is the ratio of Al^{3+} ions to S^{2-} ions?

• A. 1 Al^{3+} : 2 S^{2-}
• B. 2 Al^{3+} : 3 S^{2-}
• C. 1 Al^{3+} : 1 S^{2-}
• D. 3 Al^{3+} : 2 S^{2-}

Answer: (B)

Verified step by step guidance

1. Identify the charges on the ions involved: Aluminum ion is \(\mathrm{Al^{3+}}\) with a charge of +3, and sulfide ion is \(\mathrm{S^{2-}}\) with a charge of -2.
2. To form a neutral compound, the total positive charge must balance the total negative charge. Let the number of aluminum ions be \(x\) and the number of sulfide ions be \(y\).
3. Set up the charge balance equation based on the charges and quantities: \$3x = 2y$.
4. Solve the equation for the ratio \(\frac{x}{y}\): \(\frac{x}{y} = \frac{2}{3}\), meaning for every 2 aluminum ions, there are 3 sulfide ions.
5. Express the ratio of \(\mathrm{Al^{3+}}\) to \(\mathrm{S^{2-}}\) ions as \$2 : 3$ to ensure the compound is electrically neutral.