Practical Chemistry Previous Year Questions (JEE Main 2026)

Read each test in turn:

Flame test — apple green: An apple/grass-green flame is characteristic of $Ba^{2+}$. This eliminates $Ca^{2+}$ (brick-red flame) and $Mn^{2+}$.

Group precipitate + acetic acid + $K_2CrO_4$ → yellow precipitate: $Ba^{2+}$ falls in Group V and its precipitate dissolves in acetic acid. Treatment with chromate gives yellow barium chromate:

Note that $BaSO_4$ would not dissolve in acetic acid, so the salt’s anion is not sulphate.

Sodium carbonate extract + conc. $HNO_3$ + ammonium molybdate → canary yellow precipitate: This is the confirmatory test for the phosphate ion, giving ammonium phosphomolybdate:

So the cation is $Ba^{2+}$ and the anion is $PO_4^{3-}$.

Answer: B

Passing $H_2S$ in the presence of $NH_4OH$ (alkaline medium) is the Group IV reagent, precipitating cations as sulphides.

Classify each cation by its analytical group:

• $Pb^{2+}$ and $Cu^{2+}$ → Group II (precipitated by $H_2S$ in acidic medium).
• $Fe^{3+}$ → Group III (precipitated by $NH_4OH$/$NH_4Cl$ as hydroxide, not sulphide).
• $Mn^{2+}$ → Group IV, precipitated by $H_2S$ in the presence of $NH_4OH$ as buff-coloured $MnS$: $$Mn^{2+} + H_2S \xrightarrow{\ NH_4OH\ } MnS\!\downarrow + 2H^+$$

So the ion precipitated here is $Mn^{2+}$. Manganese has electron configuration $[Ar]\,3d^5\,4s^2$, giving a highest oxidation state of $+7$ (as in $KMnO_4$).

Answer: D

Match each compound to the test that identifies its functional group:

• A. Cyclohexanol is a secondary alcohol → the Lucas test (III) distinguishes 1°, 2°, 3° alcohols by rate of turbidity with conc. HCl/$ZnCl_2$.
• B. Cyclohexylamine is a primary amine → Hinsberg’s reagent test (I) with benzenesulphonyl chloride gives a product soluble in alkali.
• C. Cyclohexanecarbaldehyde is an aldehyde → Tollen’s test (IV) gives a silver mirror with ammoniacal $AgNO_3$.
• D. Phenol → Phthalein dye test (II); phenol with phthalic anhydride gives phenolphthalein.

Therefore: A-III, B-I, C-IV, D-II.

Answer: A

A primary standard must be a stable, high-purity solid of exactly known composition that is non-hygroscopic, so it can be weighed directly to make a standard solution.

Statement I: Only potassium dichromate ($K_2Cr_2O_7$) meets this test — it is anhydrous and non-hygroscopic, so it is a primary standard. Sodium dichromate, however, is hygroscopic (deliquescent) and is normally obtained as the dihydrate $Na_2Cr_2O_7\cdot 2H_2O$; it absorbs moisture and cannot be weighed to an exactly known composition, so it is not a primary standard. Because the statement lumps both dichromates together as primary standards, Statement I is false.

Statement II: Phenolphthalein is a weak organic acid (not a base). It stays colourless and essentially un-ionised in acidic medium, and dissociates (turning pink) only in basic medium, pH $\approx 8.3$–$10$. So calling it a weak base that dissociates in acidic medium is false on both counts. Statement II is false.

Hence both Statement I and Statement II are false.

Answer: B

$SO_2$ is a reducing gas. A paper that turns green on exposure to $SO_2$ contains acidified dichromate: the orange $Cr_2O_7^{2-}$ (Cr in $+6$) is reduced to green $Cr^{3+}$:

The orange-to-green colour change is characteristic of $K_2Cr_2O_7$.

Checking the others: KI-starch turns blue-black with oxidising gases (not $SO_2$); acidified $KMnO_4$ is decolourised (purple → colourless), not green; $Pb(CH_3COO)_2$ paper detects $H_2S$ (turns black), not $SO_2$.

Answer: D