Compound Q has -OH group and is oxidizable to Cr³⁺ (green) Steps:

• PCl₅ treatment evolves HCl fumes, indicating labile H from -OH (alcohols) or -COOH (acids); carbonyls like aldehydes/ketones form gem-dichlorides without HCl.
• Acidified K₂Cr₂O₇ warming turns solution green (Cr⁶⁺ → Cr³⁺), confirming Q undergoes oxidation; ketones and carboxylic acids resist this.
• Eliminate A (CH₃CHO): aldehyde oxidizes (green) but no HCl with PCl₅ (RCHO + PCl₅ → RCHCl₂ + POCl₃).
• Select B (CH₃CH₂OH): primary alcohol gives HCl (ROH + PCl₅ → RCl + POCl₃ + HCl) and oxidizes to CH₃COOH (green Cr³⁺). Why B is correct:
• Ethanol reacts with PCl₅ to evolve HCl (due to -OH) and oxidizes via [O] from Cr₂O₇²⁻/H⁺ to acetate, reducing Cr⁶⁺ to green Cr³⁺ (standard alcohol oxidation test). Why the others are wrong:
• A: No HCl evolution with PCl₅, as aldehyde lacks replaceable -OH hydrogen.
• C: No reaction with PCl₅ (no HCl); ketone not oxidized, no green color.
• D: HCl with PCl₅ (from -COOH), but carboxylic acid not oxidized, no green color.

Final answer: B