Which Lewis electron-dot diagram best represents the bonding in neutral potassium iodide (KI)?

• A. K oxed{8} ext{ I} oxed{8}
• B. K^{+} oxed{} ext{ I}^{-} oxed{8}
• C. K^{+} oxed{8} ext{ I}^{-} oxed{1}
• D. K oxed{1} ext{ I} oxed{7}

Answer: (B)

Verified step by step guidance

1. Step 1: Understand the nature of bonding in potassium iodide (KI). KI is an ionic compound formed between potassium (K), a metal, and iodine (I), a nonmetal. Potassium tends to lose one electron to achieve a stable electron configuration, while iodine tends to gain one electron to complete its octet.
2. Step 2: Represent the electron transfer. Potassium (K) loses one electron to become a potassium ion (K^{+}), and iodine (I) gains one electron to become an iodide ion (I^{-}). This electron transfer results in the formation of ions with full octets.
3. Step 3: Draw the Lewis electron-dot diagram for K^{+}. Since potassium loses its one valence electron, it will have no dots around it, and it carries a positive charge. The symbol is written as \(K^{+}\) with no dots.
4. Step 4: Draw the Lewis electron-dot diagram for I^{-}. Iodine gains one electron, so it will have eight dots around it representing a full octet, and it carries a negative charge. The symbol is written as \(I^{-}\) with eight dots around it.
5. Step 5: Combine the two ions to represent the ionic bond in KI. The correct Lewis structure shows \(K^{+}\) without dots and \(I^{-}\) with eight dots, indicating the transfer of one electron from potassium to iodine and the resulting ionic bond.