Which of the following correctly represents the Lewis dot structure for the strontium oxide ion pair (SrO)?

• A. Sr with one dot and O with seven dots, indicating Sr^{+} and O^{-}
• B. Sr with two dots and O with six dots (three pairs), indicating neutral atoms
• C. Sr with eight dots and O with no dots, indicating Sr^{8-} and O^{8+}
• D. Sr with no dots and O with eight dots (four pairs), indicating Sr^{2+} and O^{2-}

Answer: (D)

Verified step by step guidance

1. Step 1: Identify the elements involved and their typical valence electrons. Strontium (Sr) is an alkaline earth metal in Group 2, so it typically has 2 valence electrons. Oxygen (O) is a Group 16 element with 6 valence electrons.
2. Step 2: Determine the common ionic charges for these elements. Strontium tends to lose 2 electrons to form Sr^{2+}, and oxygen tends to gain 2 electrons to form O^{2-}, achieving noble gas electron configurations.
3. Step 3: Represent the Lewis dot structures for the neutral atoms first: Sr with 2 dots (representing its 2 valence electrons) and O with 6 dots (3 pairs) around it.
4. Step 4: Show the transfer of electrons from Sr to O to form ions. Sr loses its 2 valence electrons (no dots remain), becoming Sr^{2+}, and O gains 2 electrons to complete its octet, resulting in 8 dots (4 pairs) around O, indicating O^{2-}.
5. Step 5: Conclude that the correct Lewis structure for the SrO ion pair is Sr with no dots (Sr^{2+}) and O with eight dots (four pairs) (O^{2-}), reflecting the ionic nature of the compound.