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{:("List-I"," ""List-II"),((A)SiF(4),P."...

`{:("List-I"," ""List-II"),((A)SiF_(4),P."Lewis acid"),((B)CO_(2),Q."Lewis base"),((C) SO_(2),R."Bronstead acid"),((D)NH_(3),S."Bronstead base"):}`

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To solve the problem of matching compounds in List-I with their corresponding classifications in List-II, we will analyze each compound based on its properties as a Lewis acid, Lewis base, Brønsted acid, or Brønsted base. ### Step-by-Step Solution: 1. **Identify the Nature of Each Compound:** - **(A) SiF₄ (Silicon Tetrafluoride):** - SiF₄ can accept a pair of electrons due to the vacant d-orbitals in silicon. Therefore, it acts as a **Lewis acid**. - **(B) CO₂ (Carbon Dioxide):** - In CO₂, the carbon atom is electron-deficient because the electronegative oxygen atoms withdraw electron density. Thus, CO₂ can accept electrons and is also a **Lewis acid**. - **(C) SO₂ (Sulfur Dioxide):** - Similar to CO₂, SO₂ has a central sulfur atom that is electron-deficient due to the electronegative oxygen atoms. Hence, SO₂ is classified as a **Lewis acid**. - **(D) NH₃ (Ammonia):** - NH₃ has a lone pair of electrons on nitrogen, allowing it to donate this pair. Therefore, NH₃ is a **Lewis base**. Additionally, NH₃ can accept a proton (H⁺) to form NH₄⁺, making it a **Brønsted base** as well. 2. **Match Each Compound with the Correct Classification:** - (A) SiF₄ → P (Lewis acid) - (B) CO₂ → P (Lewis acid) - (C) SO₂ → P (Lewis acid) - (D) NH₃ → Q (Lewis base) and S (Brønsted base) 3. **Conclusion:** - There are no compounds in List-I that act as Brønsted acids since none of them can donate a proton (H⁺). ### Final Matching: - (A) SiF₄ → P (Lewis acid) - (B) CO₂ → P (Lewis acid) - (C) SO₂ → P (Lewis acid) - (D) NH₃ → Q (Lewis base) and S (Brønsted base)
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