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{:(,"Column I (Solvent)",,"Column II (Na...

`{:(,"Column I (Solvent)",,"Column II (Nature)",),((A),"Methyl alcohol"(CH_(3)OH),(p),"Protophilic",),((B),"Benzene" (C_(6)H_(6)),(q),"Protogenic",),((C),"Ammonia" (NH_(3)),(r),"Amphiprotic",),((D),"Acetic acid" (CH_(3)CO OH),(s),"Aprotic",):}`

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The correct Answer is:
To solve the question of matching solvents in Column I with their respective nature in Column II, we will analyze each solvent and identify its nature step by step. ### Step-by-Step Solution: 1. **Identify Methyl Alcohol (A)**: - Methyl alcohol (CH₃OH) is known to act as a protophilic solvent. This means it can accept protons (H⁺ ions) due to the presence of the hydroxyl (-OH) group. - Therefore, we match A with P (Protophilic). 2. **Identify Benzene (B)**: - Benzene (C₆H₆) is a non-polar organic solvent and does not donate protons. Hence, it is classified as aprotic. - Thus, we match B with S (Aprotic). 3. **Identify Ammonia (C)**: - Ammonia (NH₃) can act as both a proton donor and a proton acceptor, making it amphiprotic. It can donate a proton to form NH₂⁻ or accept a proton to form NH₄⁺. - Therefore, we match C with R (Amphiprotic). 4. **Identify Acetic Acid (D)**: - Acetic acid (CH₃COOH) is a weak acid that can donate protons, which classifies it as protogenic. - Thus, we match D with Q (Protogenic). ### Final Matches: - A (Methyl alcohol) → P (Protophilic) - B (Benzene) → S (Aprotic) - C (Ammonia) → R (Amphiprotic) - D (Acetic acid) → Q (Protogenic) ### Summary of Matches: - A - P - B - S - C - R - D - Q
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Knowledge Check

  • {:(,"Column I",,"Column II"),((A),"Inorganic benzene",(p),"An allotrope of carbon"),((B),"Fulerene",(q),"Orthosilicate"),((C),"Phenacite",(r),"An ore of boron"),((D),"Colemanite",(s),"Borazine" (B_(3)N_(3)H_(6))):}

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