Which of the following correctly describes the Lewis dot structure for the ionic compound barium sulfide (BaS)?

• A. Ba has no dots, S has eight dots, and there is no charge shown.
• B. Ba is shown as Ba^{2+} with no dots, S is shown as S^{2-} with eight dots.
• C. Ba has two dots, S has six dots, and both are neutral atoms.
• D. Ba is shown as Ba^{+} with one dot, S is shown as S^{-} with seven dots.

Answer: (B)

Verified step by step guidance

1. Step 1: Understand that barium sulfide (BaS) is an ionic compound formed between a metal (barium) and a nonmetal (sulfur). Metals tend to lose electrons to form cations, and nonmetals tend to gain electrons to form anions.
2. Step 2: Determine the charges on the ions. Barium (Ba) is in Group 2 of the periodic table, so it typically loses two electrons to form a Ba^{2+} ion. Sulfur (S) is in Group 16 and typically gains two electrons to form an S^{2-} ion.
3. Step 3: Draw the Lewis dot structure for the ions. For Ba^{2+}, since it has lost two electrons, it will have no dots around it (no valence electrons shown). For S^{2-}, it gains two electrons, so it will have eight dots representing a full octet around the sulfur symbol.
4. Step 4: Include the charges on the ions in the Lewis structure to indicate the ionic nature of the compound. Ba should be labeled as Ba^{2+} and S as S^{2-}.
5. Step 5: Verify that the overall compound is neutral by balancing the charges: Ba^{2+} and S^{2-} combine in a 1:1 ratio, resulting in a neutral ionic compound BaS.