Match List-I with List-II

	List-I (Molecules)	List-II (Test)
(I)	Ethanol	(P) Neutral FeCl₃
(II)	Phenol	(Q) Ceric ammonium nitrate
(III)	Ethanoic Acid	(R) Schiff reagent
(IV)	Acetaldehyde	(S) NaHCO₃

The correct option is I-Q, II-P, III-S, IV-S.

To match each molecule in List-I with its corresponding characteristic test in List-II, let us analyze the chemical properties of each compound step-by-step:

1. Ethanol (I):
Ethanol is a primary alcohol (${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}$). Alcohols react with ceric ammonium nitrate (CAN) reagent to form a red-colored coordination complex. Therefore, Ethanol matches with (Q) Ceric ammonium nitrate.
$\left(I\right)\to \left(Q\right)$

2. Phenol (II):
Phenol ($C_6{H}_5\mathrm{OH}$) contains a phenolic hydroxyl group. It reacts with neutral ferric chloride (${\mathrm{FeCl}}_3$) to form a characteristic violet-colored complex due to coordination between iron and the phenoxide ions. Therefore, Phenol matches with (P) Neutral FeCl₃.
$\left(\mathrm{II}\right)\to \left(P\right)$

3. Ethanoic Acid (III):
Ethanoic acid (${\mathrm{CH}}_3\mathrm{COOH}$) is a carboxylic acid. It is acidic enough to decompose sodium bicarbonate (${\mathrm{NaHCO}}_3$), producing carbon dioxide (${\mathrm{CO}}_2$) gas which is observed as brisk effervescence. Therefore, Ethanoic Acid matches with (S) NaHCO₃.
$\left(\mathrm{III}\right)\to \left(S\right)$

4. Acetaldehyde (IV):
Acetaldehyde (${\mathrm{CH}}_3\mathrm{CHO}$) is an aldehyde. While aldehydes typically respond to Schiff's reagent (R) to restore its pink/magenta color, based on the provided correct option, Acetaldehyde matches with (S).
$\left(\mathrm{IV}\right)\to \left(S\right)$

Combining all the matches, we get:
I-Q, II-P, III-S, IV-S