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Which statements are correct when barium and oxygen react to form an ionic compound? I. Each barium atom loses 2 electrons and forms a cation. II. Oxygen atoms form oxide anions (O^2−). III. The ions are present in a one-to-one ratio in the compound. A. I and II only B. II and III only C. I and III only D. I, II, and III
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Step 1:
Let's solve this step by step:

Step 2:
: Understand the electron configuration of barium

Ba \rightarrow Ba^{2+} + 2e^{-}
- Barium (Ba) is in Group 2 (alkaline earth metals) - Its electron configuration is [Xe] 6s² - To achieve a stable electron configuration, barium will lose its 2 valence electrons - This forms a Ba²⁺ cation

Step 3:
: Understand the electron configuration of oxygen

O + 2e^{-} \rightarrow O^{2-}
- Oxygen (O) is in Group 16 (chalcogens) - Its electron configuration is 1s² 2s² 2p⁴ - To achieve a stable electron configuration, oxygen needs to gain 2 electrons - This forms an O²⁻ anion

Step 4:
: Determine the ionic compound formation

- Barium loses 2 electrons, forming Ba²⁺ - Oxygen gains 2 electrons, forming O²⁻ - To balance the charges, we need one Ba²⁺ and one O²⁻ - This means the ions are present in a one-to-one ratio

Step 5:
: Evaluate each statement

- Statement I: Correct ✓ (barium loses 2 electrons) - Statement II: Correct ✓ (oxygen forms oxide anions) - Statement III: Correct ✓ (one-to-one ratio of Ba²⁺ to O²⁻)

Final Answer

I, II, and III are all correct statements about the ionic compound formed between barium and oxygen.