NEET Chemistry: Biomolecules and Practical Chemistry · Chapter 2

Principles Related to Practical Chemistry

What to remember

Lassaigne's test converts N, S and halogens into ionic NaCN, Na₂S and NaX. N gives Prussian blue with FeSO₄, S gives black PbS or violet nitroprusside colour, halogens give AgX precipitates (white, pale yellow, yellow).
Functional group tests: CAN and Lucas (alcohols), neutral FeCl₃ (phenol, violet), Tollens' and Fehling's (aldehydes), 2,4-DNP (all carbonyls), NaHCO₃ (carboxylic acids), carbylamine and azo dye tests (primary amines).
In permanganate titrations, KMnO₄ is a self-indicator, only dilute H₂SO₄ is used as the acid, and oxalic acid must be warmed to about 333–343 K (Mohr's salt is titrated at room temperature).
Cation groups are separated by controlling the concentration of S²⁻ or OH⁻ using the common ion effect so that only the sulphides or hydroxides with ionic product greater than K_sp precipitate.

Detection of N, S and halogens (Lassaigne's test)

The organic compound is fused with sodium metal. Covalent elements become ionic: Na + C + N → NaCN; 2Na + S → Na₂S; Na + X → NaX. The fused mass is boiled with water and filtered. The filtrate is the sodium fusion extract.

Element	Test on extract	Positive result
Nitrogen	Add FeSO₄, boil, add a little conc. H₂SO₄ (acid)	Prussian blue colour, Fe₄[Fe(CN)₆]₃
Sulphur	Add lead acetate (acetic acid)	Black precipitate of PbS
Sulphur	Add sodium nitroprusside	Violet colour
Nitrogen and sulphur together	Add acidified FeCl₃	Blood-red colour, Fe(SCN)²⁺ (NaSCN forms; no Prussian blue)
Halogens	Boil with dilute HNO₃, then add AgNO₃	White (Cl), pale yellow (Br), yellow (I) precipitate

Reactions for nitrogen: 6CN⁻ + Fe²⁺ → [Fe(CN)₆]⁴⁻; 3[Fe(CN)₆]⁴⁻ + 4Fe³⁺ → Fe₄[Fe(CN)₆]₃.
Boiling with HNO₃ before AgNO₃ removes CN⁻ and S²⁻ (as HCN and H₂S), which would also give silver precipitates.
AgCl dissolves in dilute NH₄OH; AgBr is sparingly soluble in conc. NH₄OH; AgI does not dissolve.

Tests for functional groups

Group	Reagent	Positive result
Alcohol	Ceric ammonium nitrate (yellow)	Yellow turns red
Alcohol class	Lucas reagent (ZnCl₂ + conc. HCl)	3°: turbid at once; 2°: turbid in about 5 minutes; 1°: no turbidity at room temperature
Phenol	Neutral FeCl₃	Violet (blue, green) colour
Aldehyde	Tollens' reagent [Ag(NH₃)₂]⁺	Silver mirror
Aliphatic aldehyde	Fehling's solution	Red-brown precipitate of Cu₂O
Aldehyde	Schiff's reagent	Pink or magenta colour restored
Aldehyde and ketone	2,4-DNP	Yellow-orange precipitate
Methyl ketone	Sodium nitroprusside and NaOH	Red colour
Carboxylic acid	NaHCO₃	Brisk effervescence of CO₂
Primary amine (1°)	CHCl₃ + alcoholic KOH, heat (carbylamine)	Foul-smelling isocyanide
Aromatic 1° amine	NaNO₂ + HCl at 273–278 K, then alkaline β-naphthol (azo dye)	Orange-red dye

Tollens': RCHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → RCOO⁻ + 2Ag + 4NH₃ + 2H₂O.
Fehling's A is aqueous CuSO₄; Fehling's B is alkaline sodium potassium tartrate. Aromatic aldehydes do not give the Fehling's test. Ketones do not respond to Tollens', Fehling's or Schiff's tests.
Phenols are weak acids and do not give effervescence with NaHCO₃ (they are weaker than carbonic acid).
Carbylamine: RNH₂ + CHCl₃ + 3KOH → RNC + 3KCl + 3H₂O (1° amines only; 2° and 3° amines do not).

Preparation of inorganic compounds

Mohr's salt, FeSO₄·(NH₄)₂SO₄·6H₂O. Equimolar amounts of FeSO₄ and (NH₄)₂SO₄ are dissolved in water containing dilute H₂SO₄, concentrated by heating and cooled to give pale green crystals. Dilute H₂SO₄ prevents hydrolysis of Fe²⁺ and its oxidation to Fe³⁺ (Fe³⁺ would give a yellow-brown colour). The solution must not be heated to dryness, so that oxidation is avoided.

Potash alum, K₂SO₄·Al₂(SO₄)₃·24H₂O. Aluminium scrap is dissolved in hot KOH, and dilute H₂SO₄ is then added.

2Al + 2KOH + 6H₂O → 2K[Al(OH)₄] + 3H₂
2K[Al(OH)₄] + H₂SO₄ → 2Al(OH)₃ + K₂SO₄ + 2H₂O
2Al(OH)₃ + 3H₂SO₄ → Al₂(SO₄)₃ + 6H₂O
On cooling, colourless octahedral crystals of alum separate. Extra H₂SO₄ prevents hydrolysis of Al³⁺ to Al(OH)₃ and so keeps the solution clear. Both Mohr's salt and alum are double salts.

Preparation of organic compounds

Compound	Reactants and conditions	Chemistry
Acetanilide	Aniline + acetic anhydride (or glacial acetic acid with Zn dust); recrystallise from hot water	Acetylation of –NH₂; Zn dust prevents oxidation of aniline. m.p. about 387 K
p-Nitroacetanilide	Acetanilide + conc. HNO₃ + conc. H₂SO₄ (nitrating mixture) below about 288 K	Electrophilic nitration; the acetyl group protects –NH₂ from oxidation and directs to para. m.p. about 487 K
Aniline yellow (p-aminoazobenzene)	Aniline + NaNO₂ + HCl at 273–278 K, then coupling with aniline	Diazotisation then azo coupling; yellow dye
Iodoform	Acetone or ethanol + I₂ + NaOH (or NaOCl), warm	Haloform reaction; yellow CHI₃, antiseptic smell. m.p. about 392 K

Iodoform: CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃COONa + 3NaI + 3H₂O. Compounds with CH₃CO– or CH₃CH(OH)– give the iodoform test; ethanal and ethanol do, but methanol and other primary alcohols do not.
Diazonium salts are made at low temperature (273–278 K) because they decompose above about 283 K.

Titrimetric exercises

Acid–base indicators

Indicator	pH range	Colour in acid	Colour in base
Methyl orange	3.1–4.4	Red	Yellow
Phenolphthalein	8.3–10.0	Colourless	Pink

Titration	Suitable indicator
Strong acid – strong base	Either
Weak acid – strong base (e.g. oxalic acid – NaOH)	Phenolphthalein
Strong acid – weak base (e.g. HCl – NH₃)	Methyl orange
Weak acid – weak base	No sharp end point; no common indicator

Oxalic acid vs KMnO₄

2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

The mixture is warmed to about 333–343 K because the reaction is slow at room temperature. Mn²⁺ formed acts as an autocatalyst.
KMnO₄ is a self-indicator: the end point is the first permanent faint pink colour. It is taken in the burette and the upper meniscus is read, because the solution is dark.
Dilute H₂SO₄ only. HCl is oxidised by KMnO₄ to Cl₂ (extra KMnO₄ is used). HNO₃ is itself an oxidising agent. Acetic acid is too weak to supply enough H⁺.

Mohr's salt vs KMnO₄

MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

The titration is done at room temperature (no heating). Dilute H₂SO₄ is again used. Molar mass of Mohr's salt is 392 g mol⁻¹.

Worked example 1. 25.0 mL of 0.10 M oxalic acid needs 20.0 mL of KMnO₄. Moles of oxalate = 2.5 mmol. Moles of MnO₄⁻ = 2.5 × 2/5 = 1.0 mmol. Molarity = 1.0/20.0 = 0.050 M. Check by normality: N(oxalic) = 0.20, N(KMnO₄) = 5M; 0.20 × 25 = N × 20 gives N = 0.25, so M = 0.050.

Worked example 2. 25.0 mL of Mohr's salt solution needs 20.0 mL of 0.050 M KMnO₄. Moles of MnO₄⁻ = 1.0 mmol; moles of Fe²⁺ = 5.0 mmol; molarity = 5.0/25.0 = 0.20 M.

Qualitative salt analysis: anions

Anion	Key test	Observation
CO₃²⁻	Dilute HCl	Effervescence of CO₂; lime water turns milky
S²⁻	Dilute H₂SO₄	H₂S, rotten-egg smell; lead acetate paper turns black
SO₃²⁻	Dilute HCl	SO₂, suffocating smell; acidified K₂Cr₂O₇ paper turns green
NO₂⁻	Dilute H₂SO₄	Light brown fumes of NO₂; also decolourises acidified KMnO₄
NO₃⁻	Conc. H₂SO₄ with Cu turnings; brown ring test (FeSO₄, then conc. H₂SO₄ down the side)	Brown fumes; brown ring at the junction, [Fe(H₂O)₅NO]²⁺
Cl⁻	AgNO₃; chromyl chloride test	White curdy precipitate soluble in NH₄OH; red vapours of CrO₂Cl₂
Br⁻	AgNO₃; layer test (CHCl₃ + chlorine water)	Pale yellow precipitate, sparingly soluble in NH₄OH; orange-brown layer
I⁻	AgNO₃; layer test	Yellow precipitate, insoluble in NH₄OH; violet layer
SO₄²⁻	BaCl₂	White precipitate insoluble in conc. HCl
PO₄³⁻	Ammonium molybdate + conc. HNO₃	Canary yellow precipitate
CH₃COO⁻	Warm with ethanol and conc. H₂SO₄; neutral FeCl₃	Fruity smell of ester; red colour
C₂O₄²⁻	Warm acidified KMnO₄; CaCl₂	Pink colour discharged, CO₂ evolved; white CaC₂O₄ insoluble in acetic acid

Chromyl chloride test: the red vapours are passed into NaOH to give a yellow solution (CrO₄²⁻), which gives a yellow precipitate of PbCrO₄ with lead acetate. Bromides and iodides do not give this test.
The brown ring test is for nitrate; nitrite also interferes and must be removed first.

Qualitative salt analysis: cations

Group	Group reagent	Ions	Precipitate
0	NaOH, warm (no precipitate)	NH₄⁺	NH₃ gas; brown with Nessler's reagent
II	H₂S with dilute HCl	Pb²⁺, Cu²⁺, As³⁺	PbS (black), CuS (black), As₂S₃ (yellow)
III	NH₄Cl + NH₄OH	Al³⁺, Fe³⁺	Al(OH)₃ (white, gelatinous), Fe(OH)₃ (reddish brown)
IV	H₂S with NH₄OH	Co²⁺, Ni²⁺, Mn²⁺, Zn²⁺	CoS, NiS (black), MnS (buff), ZnS (white)
V	(NH₄)₂CO₃ with NH₄Cl + NH₄OH	Ba²⁺, Sr²⁺, Ca²⁺	White carbonates
VI	No group reagent; test with (NH₄)₂HPO₄	Mg²⁺	White MgNH₄PO₄

Why the group reagents work (solubility-product reasoning)

Group II: in dilute HCl the H⁺ is high. H₂S ⇌ 2H⁺ + S²⁻ is pushed back (common ion effect), so S²⁻ is very low. Only sulphides of very low K_sp (CuS, PbS, As₂S₃) have ionic product greater than K_sp.
Group III: NH₄Cl suppresses the ionisation of NH₄OH, so OH⁻ is low. Only hydroxides of very low K_sp (Fe(OH)₃, Al(OH)₃) precipitate; Mg(OH)₂ and others stay in solution.
Group IV: in ammoniacal medium OH⁻ removes H⁺ from H₂S, so S²⁻ rises. Sulphides of higher K_sp (ZnS, MnS, CoS, NiS) now precipitate.
Group V: (NH₄)₂CO₃ is added in the presence of NH₄Cl, which keeps CO₃²⁻ low enough that MgCO₃ (higher K_sp) does not precipitate, while BaCO₃, SrCO₃ and CaCO₃ do.

Confirmatory tests

Ion	Test	Result
Fe³⁺	K₄[Fe(CN)₆]; KSCN	Prussian blue; blood-red colour
Cu²⁺	Excess NH₄OH	Deep blue [Cu(NH₃)₄]²⁺
Ni²⁺	Dimethylglyoxime, NH₄OH	Rosy red precipitate
Zn²⁺	K₄[Fe(CN)₆]	Bluish-white precipitate
Pb²⁺	K₂CrO₄ or KI	Yellow PbCrO₄ or PbI₂
Ba²⁺	K₂CrO₄; flame test	Yellow BaCrO₄; apple green flame
Sr²⁺	Flame test	Crimson red flame
Ca²⁺	(NH₄)₂C₂O₄; flame test	White CaC₂O₄; brick red flame

Chemistry of the experiments

Enthalpy of dissolution of CuSO₄. A known mass of CuSO₄ is dissolved in a known volume of water in an insulated calorimeter and the temperature change ΔT is measured. q = m c ΔT, and ΔH = –q/n (per mole). Anhydrous CuSO₄ dissolves with a rise in temperature (exothermic); CuSO₄·5H₂O dissolves with a fall in temperature (endothermic).

Enthalpy of neutralisation. For a strong acid and a strong base the net change is H⁺ + OH⁻ → H₂O, with ΔH about –57.1 kJ mol⁻¹. Weak acids or bases give a smaller (less negative) value because some energy is used in their ionisation.

Worked example 3. 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH are mixed and the temperature rises by 6.8 K. Take mass = 100 g and c = 4.18 J g⁻¹ K⁻¹. q = 100 × 4.18 × 6.8 = 2842 J. Moles of water formed = 0.050. ΔH = –2842/0.050 = –56.8 kJ mol⁻¹.

Preparation of sols

Lyophilic (solvent-loving, stable, reversible): starch sol (starch boiled in water), gum, gelatin, egg albumin.
Lyophobic (solvent-hating, need stabilisers, irreversible): Fe(OH)₃ sol, As₂S₃ sol, sulphur sol.
Fe(OH)₃ sol: FeCl₃ added drop by drop to boiling water: FeCl₃ + 3H₂O → Fe(OH)₃ + 3HCl. The sol is positive. It is purified by dialysis.
As₂S₃ sol: H₂S passed through a dilute As₂O₃ solution: As₂O₃ + 3H₂S → As₂S₃ + 3H₂O. The sol is negative.
Coagulation follows the Hardy–Schulze rule: for a positive sol, the order of power is PO₄³⁻ > SO₄²⁻ > Cl⁻; for a negative sol, Al³⁺ > Ba²⁺ > Na⁺.

Kinetics of iodide with hydrogen peroxide. H₂O₂ + 2I⁻ + 2H⁺ → 2H₂O + I₂. Rate = k[H₂O₂][I⁻], so the reaction is first order in each reactant. A fixed small amount of Na₂S₂O₃ and starch is added. The iodine formed is consumed at once: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. When the thiosulphate is used up, the blue starch–iodine colour appears. The time taken to the blue colour is inversely proportional to the rate. Doubling [I⁻] halves the time.

NEET traps

1. Lucas test: 3° alcohol reacts fastest; 1° gives no turbidity at room temperature.
2. Fehling's and Tollens' tests are given by aldehydes, not ketones; Fehling's does not work for aromatic aldehydes.
3. 2,4-DNP reacts with both aldehydes and ketones; Tollens' separates the two.
4. Phenol gives a violet colour with neutral FeCl₃ but no effervescence with NaHCO₃; carboxylic acids do the opposite.
5. The carbylamine test works with 1° amines only (aliphatic and aromatic), but the azo dye test needs an aromatic 1° amine.
6. Dilute HCl is used for group I; H₂S with dilute HCl for group II; H₂S with NH₄OH for group IV.
7. KMnO₄ is titrated with H₂SO₄ only, never with HCl or HNO₃.
8. Oxalic acid is warmed; Mohr's salt is not warmed.
9. Methyl orange suits strong acid–weak base titrations; phenolphthalein suits weak acid–strong base.
10. AgBr is sparingly soluble in NH₄OH, AgCl is soluble, AgI is insoluble.

One-liners

1. Sodium fusion extract contains CN⁻, S²⁻ and X⁻.
2. Prussian blue is Fe₄[Fe(CN)₆]₃.
3. N and S together give blood-red Fe(SCN)²⁺.
4. Neutral FeCl₃ gives a violet colour with phenol.
5. Ceric ammonium nitrate turns from yellow to red with alcohols.
6. Schiff's reagent turns pink with aldehydes.
7. Sodium nitroprusside gives a red colour with methyl ketones.
8. Mohr's salt is FeSO₄·(NH₄)₂SO₄·6H₂O.
9. Potash alum is K₂SO₄·Al₂(SO₄)₃·24H₂O.
10. Iodoform is a yellow solid with an antiseptic smell.
11. KMnO₄ is a self-indicator; its end point is permanent pink.
12. Methyl orange range is 3.1–4.4 and phenolphthalein range is 8.3–10.0.
13. Ammonium molybdate gives a canary yellow precipitate with phosphate.
14. Group V cations are precipitated as carbonates.
15. The enthalpy of neutralisation of a strong acid with a strong base is about –57.1 kJ mol⁻¹.

Practice questions

The Prussian blue colour obtained in the Lassaigne test for nitrogen is due to
1. Fe(SCN)₃
2. Fe₄[Fe(CN)₆]₃
3. Fe₃[Fe(CN)₆]₂
4. Na₄[Fe(CN)₅NOS]
Answer
B. Fe₄[Fe(CN)₆]₃
CN⁻ forms [Fe(CN)₆]⁴⁻ with Fe²⁺, which then reacts with Fe³⁺ to give Fe₄[Fe(CN)₆]₃.
A compound containing both nitrogen and sulphur gives a blood-red colour in the Lassaigne test with acidified FeCl₃. The coloured species is
1. Fe(SCN)²⁺
2. Fe₄[Fe(CN)₆]₃
3. PbS
4. Na₄[Fe(CN)₅NOS]
Answer
A. Fe(SCN)²⁺
With both N and S the fusion gives NaSCN, which forms the blood-red Fe(SCN)²⁺ with Fe³⁺.
In the test for halogens, the sodium fusion extract is boiled with dilute HNO₃ before adding AgNO₃ in order to
1. reduce the halide ions
2. dissolve the silver halide
3. decompose NaCN and Na₂S
4. convert the halide into a halogen
Answer
C. decompose NaCN and Na₂S
CN⁻ and S²⁻ would otherwise give silver precipitates; HNO₃ removes them as HCN and H₂S.
The silver halide that is pale yellow and only sparingly soluble in NH₄OH is
1. AgI
2. AgCl
3. Ag₂CO₃
4. AgBr
Answer
D. AgBr
AgCl is white and soluble; AgBr is pale yellow and sparingly soluble; AgI is yellow and insoluble.
Which alcohol gives turbidity immediately with Lucas reagent at room temperature?
1. Ethanol
2. 2-Methylpropan-2-ol
3. Butan-2-ol
4. Butan-1-ol
Answer
B. 2-Methylpropan-2-ol
Tertiary alcohols react fastest with ZnCl₂/conc. HCl; secondary alcohols take about 5 minutes; primary alcohols do not react at room temperature.
The brown ring test in qualitative analysis is used to detect
1. Cl⁻
2. PO₄³⁻
3. NO₃⁻
4. SO₄²⁻
Answer
C. NO₃⁻
Nitrate with FeSO₄ and conc. H₂SO₄ forms a brown ring of [Fe(H₂O)₅NO]²⁺.
During the determination of enthalpy of dissolution of anhydrous CuSO₄ in a calorimeter, the temperature of the water
1. rises, because the dissolution is exothermic
2. falls, because the dissolution is exothermic
3. rises, because the dissolution is endothermic
4. does not change
Answer
A. rises, because the dissolution is exothermic
Anhydrous CuSO₄ dissolves with release of heat; the hydrated salt dissolves with absorption of heat.
Which of the following does NOT give a red-brown precipitate with Fehling's solution?
1. Methanal
2. Ethanal
3. Propanal
4. Benzaldehyde
Answer
D. Benzaldehyde
Aromatic aldehydes do not respond to Fehling's solution; aliphatic aldehydes do.
An orange-yellow precipitate with 2,4-dinitrophenylhydrazine shows that the compound is
1. only an aldehyde
2. an aldehyde or a ketone
3. only a ketone
4. a primary amine
Answer
B. an aldehyde or a ketone
2,4-DNP reacts with the C=O group of both aldehydes and ketones.
Which of the following gives a red colour with sodium nitroprusside and NaOH?
1. Benzophenone
2. Pentan-3-one
3. Ethoxyethane
4. Propanone
Answer
D. Propanone
The test is given by methyl ketones (CH₃CO– with an active CH₃ group); propanone is CH₃COCH₃.
A sol of ferric hydroxide is prepared by adding a few drops of FeCl₃ solution to
1. dilute NaOH solution
2. concentrated HCl
3. boiling water
4. cold water
Answer
C. boiling water
Hydrolysis in boiling water: FeCl₃ + 3H₂O → Fe(OH)₃ (sol) + 3HCl.
Which of the following compounds gives the carbylamine test?
1. Ethylamine
2. Triethylamine
3. Diethylamine
4. N,N-Dimethylaniline
Answer
A. Ethylamine
Only primary amines give a foul-smelling isocyanide with CHCl₃ and alcoholic KOH.
Aniline is treated with NaNO₂ and HCl at 273–278 K and then with an alkaline solution of β-naphthol. The product is
1. a blood-red complex
2. a yellow precipitate of iodoform
3. a violet coloured complex
4. an orange-red azo dye
Answer
D. an orange-red azo dye
Diazotisation followed by coupling with β-naphthol gives an orange-red azo dye.
Dilute H₂SO₄ is added while preparing Mohr's salt in order to
1. reduce Fe²⁺ to iron metal
2. form potash alum
3. prevent hydrolysis of Fe²⁺ and its oxidation to Fe³⁺
4. increase the solubility of (NH₄)₂SO₄
Answer
C. prevent hydrolysis of Fe²⁺ and its oxidation to Fe³⁺
The acid suppresses hydrolysis of ferrous sulphate and keeps iron in the +2 state.
Zinc dust is added during the preparation of acetanilide from aniline and glacial acetic acid to
1. convert acetic acid to acetic anhydride
2. prevent oxidation of aniline during heating
3. increase the melting point
4. catalyse nitration
Answer
B. prevent oxidation of aniline during heating
Aniline is easily oxidised and darkens; zinc dust keeps it in the reduced form.
In the preparation of p-nitroacetanilide, aniline is first converted to acetanilide mainly because the acetyl group
1. protects the –NH₂ group from oxidation by the nitrating mixture
2. acts as a nitrating agent
3. makes the ring more acidic
4. converts the group into a meta director
Answer
A. protects the –NH₂ group from oxidation by the nitrating mixture
Direct nitration of aniline oxidises it; the acetamido group is protected and directs substitution to the para position.
Given below are two statements: Statement I: Tollens' reagent is reduced to silver by aldehydes. Statement II: Ketones do not give a silver mirror with Tollens' reagent. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both statements are correct.
Given below are two statements: Statement I: Phenol gives brisk effervescence with NaHCO₃. Statement II: Phenol gives a violet colour with neutral FeCl₃. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Phenol is a weaker acid than carbonic acid, so I is incorrect; II is correct.
Given below are two statements: Statement I: Primary amines give the carbylamine test. Statement II: Tertiary amines also give the carbylamine test. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Only primary amines give the test, so II is incorrect.
Given below are two statements: Statement I: Primary alcohols give turbidity immediately with Lucas reagent at room temperature. Statement II: Tertiary alcohols do not give turbidity with Lucas reagent at room temperature. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
The facts are reversed: tertiary alcohols react at once and primary alcohols do not react at room temperature.
Which of the following statements are correct? 1. Oxalic acid solution is warmed before titration with KMnO₄. 2. KMnO₄ titrations are done in the presence of dilute HCl. 3. KMnO₄ acts as its own indicator.
1. 1 and 2 only
2. 1 and 3 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
B. 1 and 3 only
Dilute H₂SO₄ is used, not HCl, because HCl is oxidised by KMnO₄; statements 1 and 3 are correct.
Which of the following statements are correct? 1. Chlorides give the chromyl chloride test. 2. The brown ring test detects nitrate. 3. Bromides give the chromyl chloride test.
1. 1 and 3 only
2. 2 and 3 only
3. 1, 2 and 3
4. 1 and 2 only
Answer
D. 1 and 2 only
Only chlorides give red CrO₂Cl₂ vapours; bromides and iodides do not.
Given below are two statements: Statement I: Group II cations are precipitated by H₂S in the presence of dilute HCl. Statement II: The high H⁺ concentration lowers the S²⁻ concentration, so only sulphides of very low K_sp precipitate. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Both are correct; the common ion effect of H⁺ suppresses ionisation of H₂S.
Given below are two statements: Statement I: In group III, NH₄Cl is added to decrease the OH⁻ concentration of NH₄OH. Statement II: Mg(OH)₂ is precipitated in group III. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
NH₄Cl keeps OH⁻ low so that only Fe(OH)₃ and Al(OH)₃ precipitate; Mg(OH)₂ stays in solution, so II is incorrect.
Given below are two statements: Statement I: Fe³⁺ gives a blue colour with KSCN. Statement II: Fe³⁺ gives Prussian blue with K₄[Fe(CN)₆]. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Fe³⁺ with KSCN gives a blood-red colour, so I is incorrect; II is correct.
Given below are two statements: Statement I: The solution of Mohr's salt is evaporated to dryness at the end of the preparation. Statement II: Dilute H₂SO₄ is not needed in the preparation of Mohr's salt. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Evaporating to dryness would oxidise Fe²⁺; dilute H₂SO₄ is needed. Both are incorrect.
Assertion (A): Dilute H₂SO₄ is added during the preparation of Mohr's salt. Reason (R): It prevents hydrolysis of Fe²⁺ and its oxidation to Fe³⁺.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
The acid suppresses hydrolysis and oxidation, so R explains A.
Assertion (A): Oxalic acid solution is warmed to about 333–343 K before titration with KMnO₄. Reason (R): The reaction between MnO₄⁻ and oxalate ions is instantaneous at room temperature.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
The reaction is slow at room temperature, which is why warming is needed; R is false.
Assertion (A): A sol of Fe(OH)₃ is positively charged. Reason (R): Fe(OH)₃ sol is prepared by adding FeCl₃ to boiling water.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
The sol adsorbs Fe³⁺ ions and is positive; the method of preparation does not explain the charge.
Assertion (A): Methyl orange is a suitable indicator for titrating oxalic acid against NaOH. Reason (R): Methyl orange changes colour in the pH range 3.1–4.4.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
Oxalic acid with NaOH needs phenolphthalein (end point in the basic range), so A is false; R is true.
Assertion (A): Sulphides of group II precipitate in dilute HCl, but ZnS does not. Reason (R): ZnS has a higher K_sp than CuS and needs a higher S²⁻ concentration to precipitate.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
In acid the S²⁻ concentration is low; only low-K_sp sulphides such as CuS precipitate.
Assertion (A): Diazonium salts are prepared at 273–278 K. Reason (R): Diazonium salts are stable at temperatures above 300 K.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
They decompose at higher temperature; R is false.
Match List I with List II. List I: a. Group II b. Group III c. Group IV d. Group V List II: i. NH₄Cl and NH₄OH ii. H₂S with dilute HCl iii. (NH₄)₂CO₃ with NH₄Cl and NH₄OH iv. H₂S with NH₄OH
1. a-ii, b-i, c-iii, d-iv
2. a-iv, b-i, c-ii, d-iii
3. a-i, b-ii, c-iv, d-iii
4. a-ii, b-i, c-iv, d-iii
Answer
D. a-ii, b-i, c-iv, d-iii
Group II: H₂S/dil. HCl; III: NH₄Cl/NH₄OH; IV: H₂S/NH₄OH; V: (NH₄)₂CO₃.
Match List I with List II. List I: a. CO₃²⁻ b. SO₃²⁻ c. S²⁻ d. PO₄³⁻ List II: i. Canary yellow precipitate with ammonium molybdate ii. Lead acetate paper turns black iii. Effervescence; lime water turns milky iv. Suffocating gas; acidified K₂Cr₂O₇ turns green
1. a-iv, b-iii, c-ii, d-i
2. a-iii, b-iv, c-i, d-ii
3. a-iii, b-iv, c-ii, d-i
4. a-ii, b-iv, c-iii, d-i
Answer
C. a-iii, b-iv, c-ii, d-i
CO₂ turns lime water milky; SO₂ turns dichromate green; H₂S blackens lead acetate paper; phosphate gives the yellow molybdate precipitate.
Match List I with List II. List I: a. Alcohol b. Phenol c. Aldehyde d. Methyl ketone List II: i. Schiff's reagent ii. Sodium nitroprusside iii. Neutral FeCl₃ iv. Ceric ammonium nitrate
1. a-iv, b-iii, c-i, d-ii
2. a-iii, b-iv, c-i, d-ii
3. a-iv, b-iii, c-ii, d-i
4. a-i, b-iii, c-iv, d-ii
Answer
A. a-iv, b-iii, c-i, d-ii
Alcohol: CAN; phenol: neutral FeCl₃; aldehyde: Schiff's; methyl ketone: nitroprusside.
Match List I with List II. List I: a. Fe³⁺ with K₄[Fe(CN)₆] b. Ni²⁺ with dimethylglyoxime c. Cu²⁺ with excess NH₄OH d. Ba²⁺ in a flame List II: i. Deep blue solution ii. Rosy red precipitate iii. Apple green flame iv. Prussian blue
1. a-iv, b-i, c-ii, d-iii
2. a-iv, b-ii, c-i, d-iii
3. a-ii, b-iv, c-i, d-iii
4. a-iv, b-ii, c-iii, d-i
Answer
B. a-iv, b-ii, c-i, d-iii
Fe³⁺: Prussian blue; Ni²⁺: rosy red; Cu²⁺: [Cu(NH₃)₄]²⁺ deep blue; Ba²⁺: green flame.
Match List I with List II. List I: a. Acetanilide b. Aniline yellow c. Iodoform d. Mohr's salt List II: i. Haloform reaction ii. Acetylation iii. Diazotisation and coupling iv. Equimolar FeSO₄ and (NH₄)₂SO₄ crystallised with dilute H₂SO₄
1. a-i, b-iii, c-ii, d-iv
2. a-ii, b-iii, c-i, d-iv
3. a-iii, b-ii, c-i, d-iv
4. a-ii, b-iii, c-iv, d-i
Answer
B. a-ii, b-iii, c-i, d-iv
Acetanilide by acetylation; aniline yellow by diazotisation and coupling; iodoform by the haloform reaction; Mohr's salt as a double salt.
Match List I with List II. List I: a. Cl⁻ b. Br⁻ c. I⁻ d. SO₄²⁻ List II: i. Violet layer with CHCl₃ and chlorine water ii. Orange-brown layer with CHCl₃ and chlorine water iii. Red vapours of CrO₂Cl₂ iv. White precipitate with BaCl₂ insoluble in conc. HCl
1. a-ii, b-iii, c-i, d-iv
2. a-iv, b-ii, c-i, d-iii
3. a-iii, b-i, c-ii, d-iv
4. a-iii, b-ii, c-i, d-iv
Answer
D. a-iii, b-ii, c-i, d-iv
Chloride: chromyl chloride test; bromide: orange-brown layer; iodide: violet layer; sulphate: BaSO₄.
25.0 mL of 0.10 M oxalic acid in warm dilute H₂SO₄ requires 20.0 mL of KMnO₄ solution for the end point. The molarity of KMnO₄ is
1. 0.20 M
2. 0.025 M
3. 0.050 M
4. 0.125 M
Answer
C. 0.050 M
Oxalate = 2.5 mmol; MnO₄⁻ = 2.5 × 2/5 = 1.0 mmol; M = 1.0/20.0 = 0.050 M. (Normality route: 0.20 × 25 = N × 20, N = 0.25, M = 0.25/5 = 0.050.)
25.0 mL of an acidified Fe²⁺ solution needs 20.0 mL of 0.020 M KMnO₄. The molarity of the Fe²⁺ solution is
1. 0.080 M
2. 0.016 M
3. 0.40 M
4. 0.032 M
Answer
A. 0.080 M
MnO₄⁻ = 0.40 mmol; Fe²⁺ = 5 × 0.40 = 2.0 mmol; M = 2.0/25.0 = 0.080 M. (Normality: KMnO₄ = 0.10 N; 0.10 × 20 = N × 25.)
The mass of Mohr's salt (molar mass 392 g mol⁻¹) needed to prepare 250 mL of 0.10 M solution is
1. 9.8 g
2. 98 g
3. 3.92 g
4. 39.2 g
Answer
A. 9.8 g
Moles = 0.10 × 0.250 = 0.025 mol; mass = 0.025 × 392 = 9.8 g.
100 mL of 0.50 M HCl is mixed with 100 mL of 0.50 M NaOH and the temperature rises by 3.4 K. Taking the mass of solution as 200 g and c = 4.18 J g⁻¹ K⁻¹, the enthalpy of neutralisation is about
1. –5.68 kJ mol⁻¹
2. –56.8 kJ mol⁻¹
3. –113.7 kJ mol⁻¹
4. –28.4 kJ mol⁻¹
Answer
B. –56.8 kJ mol⁻¹
q = 200 × 4.18 × 3.4 = 2842 J; moles of water = 0.050; ΔH = –2842/0.050 = –56.8 kJ mol⁻¹. (Check: 0.050 × 57.1 kJ = 2855 J gives ΔT = 3.4 K.)
In the iodide–hydrogen peroxide reaction, rate = k[H₂O₂][I⁻]. If the concentrations of both H₂O₂ and I⁻ are doubled, the time taken for the blue starch colour to appear (with a fixed amount of thiosulphate) becomes
1. four times
2. one-half
3. double
4. one-fourth
Answer
D. one-fourth
Rate becomes 2 × 2 = 4 times; the time for a fixed amount of iodine to form is inversely proportional to rate, so it becomes one-fourth.
The sodium fusion extract of an organic compound gives a blood-red colour with acidified FeCl₃ but no Prussian blue with FeSO₄. The compound contains
1. both nitrogen and sulphur
2. nitrogen only
3. sulphur only
4. nitrogen and a halogen
Answer
A. both nitrogen and sulphur
NaSCN forms when N and S are both present; it gives blood-red Fe(SCN)²⁺ and no free CN⁻ for the Prussian blue test.
A compound C₃H₆O gives an orange precipitate with 2,4-DNP and a yellow precipitate with I₂/NaOH, but no silver mirror with Tollens' reagent. The compound is
1. propan-1-ol
2. propanal
3. propanone
4. propan-2-ol
Answer
C. propanone
DNP shows a carbonyl group; no Tollens' result excludes propanal; propan-2-ol and propan-1-ol (C₃H₈O) do not fit the formula or the DNP test.

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1.Sunil made a paste of rice flour and added a drop of iodine solution to the paste. What will he see?
2.The process used for removing gangue from the ore works on the basis of difference of their chemical or physical properties and is called ______ of ores.
3.Which of the following is an example of a third-class lever?
4.Which of the following methods is NOT used for preventing iron from rusting?
5.Which of the following methods is most commonly used nowadays to determine nitrogen and protein content in foods?
6.Which of the following is the reverse of electroplating?
7.Which process is used to get petrol, petroleum gas, diesel, kerosene and fuel oil from petroleum?
8.Which technique is useful for the separation of colloidal particles?
9.Which of the following procedures would you use to separate ammonium chloride from sodium chloride?
10.Which of the following parts are cleaned using ultrasound?

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