Previous-year papers
Original PYQ-style practice with full worked solutions, modeled on the real exam pattern — plus where to find the official past papers, free.
Previous-year questions (PYQs) are the single best signal for what an exam will ask next. Themes repeat, question styles recur, and the difficulty band stays remarkably stable year to year.
ChemVidya gives you original PYQ-style practice — questions written in our own words, modeled on the topics, pattern and difficulty of real papers, each with a full step-by-step worked solution. The official past papers themselves are published free by the exam bodies; practise those, then use our solutions and the full question bank to master the concepts behind them.
Where to get the official past papers (free)
The actual previous-year papers are released free of charge by the bodies that conduct each exam — CSIR-HRDG and the NTA for CSIR-NET, the organising IITs for GATE Chemistry and IIT-JAM, and the state agencies for SET/SLET. We do not republish those copyrighted papers here. Download them from the official sources, then use ChemVidya's worked solutions and the full question bank to understand the reasoning behind every answer.
PYQ-style practice (modeled on past CSIR-NET papers)
Here is a set of 24 original practice questions written by ChemVidya. They are not reproductions of any official paper — each is reworded in our own words and modeled on the topics, pattern and difficulty band that recur in CSIR-NET Chemical Sciences (Part B/C). Every question comes with a full worked solution — expand each one to check your reasoning.
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Which reagent best converts a terminal alkyne (R–C≡C–H) directly into a methyl ketone (R–CO–CH3)?
- A. B2H6 then H2O2/NaOH (anti-Markovnikov)
- B. HgSO4 / dilute H2SO4 (Markovnikov hydration)
- C. Na / liquid NH3
- D. O3 then Zn/H2O
Show worked solution
Answer: B
Acid-catalysed hydration of a terminal alkyne follows Markovnikov addition: water adds so that OH goes to the more substituted carbon, giving an enol that tautomerises to the methyl ketone R–CO–CH3. Hydroboration–oxidation gives the anti-Markovnikov aldehyde instead; dissolving-metal reduction gives a trans-alkene; ozonolysis cleaves the triple bond.
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The major product of monobromination of toluene with Br2 in the presence of FeBr3 is:
- A. ortho/para-bromotoluene
- B. benzyl bromide
- C. meta-bromotoluene
- D. benzoic acid
Show worked solution
Answer: A
With a Lewis acid (FeBr3) the reaction is electrophilic aromatic substitution on the ring. The methyl group is an activating, ortho/para-directing substituent, so the major products are o- and p-bromotoluene. Benzyl bromide forms only under radical/side-chain conditions (light or peroxides, no Lewis acid).
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Which pair of compounds are enantiomers?
- A. D-glucose and D-fructose
- B. cis- and trans-2-butene
- C. (R)-2-butanol and (S)-2-butanol
- D. n-butane and isobutane
Show worked solution
Answer: C
Enantiomers are non-superimposable mirror images differing in configuration at every stereocentre. (R)- and (S)-2-butanol fit this exactly. cis/trans-2-butene are geometric (diastereomeric) isomers; glucose/fructose are constitutional isomers (aldose vs ketose); butane/isobutane are constitutional (chain) isomers.
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In an SN1 solvolysis, the rate is greatest for which substrate?
- A. methyl bromide
- B. n-butyl bromide
- C. isobutyl bromide
- D. tert-butyl bromide
Show worked solution
Answer: D
SN1 rate depends on carbocation stability. tert-Butyl bromide ionises to a tertiary carbocation, which is the most stabilised by hyperconjugation and induction, so it solvolyses fastest. Primary and methyl halides give very unstable cations and react through SN2 instead.
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The Diels–Alder reaction between 1,3-butadiene and ethene gives which product?
- A. benzene
- B. cyclohexene
- C. cyclohexane
- D. 1,3-cyclohexadiene
Show worked solution
Answer: B
The Diels–Alder is a [4+2] cycloaddition: a conjugated diene (4 π electrons) reacts with a dienophile (2 π electrons) to form a six-membered ring with one remaining double bond. Butadiene + ethene therefore gives cyclohexene, not the fully saturated or aromatic ring.
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Which compound is the most acidic?
- A. cyclohexanol
- B. ethanol
- C. phenol
- D. water
Show worked solution
Answer: C
Acidity tracks the stability of the conjugate base. The phenoxide ion delocalises its negative charge into the aromatic ring, so phenol (pKa about 10) is far more acidic than alcohols (pKa 16–18) or water (pKa 15.7), whose alkoxide/hydroxide anions are not resonance-stabilised.
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Treatment of an aldehyde (RCHO, R ≠ H) with a Grignard reagent (R′MgX) followed by aqueous workup gives:
- A. a secondary alcohol
- B. a primary alcohol
- C. a tertiary alcohol
- D. a carboxylic acid
Show worked solution
Answer: A
A Grignard adds its carbanion to the carbonyl carbon. With formaldehyde you get a primary alcohol, with a ketone a tertiary alcohol, and with any other aldehyde (RCHO, R not H) the product after protonation is a secondary alcohol bearing two carbon substituents on the carbinol carbon.
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Applying Huckel's rule to the cyclopentadienyl system, which statement is correct?
- A. The cyclopentadienyl anion is antiaromatic
- B. The cyclopentadienyl cation is aromatic
- C. Neutral cyclopentadiene is aromatic
- D. The cyclopentadienyl anion is aromatic
Show worked solution
Answer: D
Huckel's rule requires a planar, fully conjugated cyclic system with (4n+2) π electrons. The cyclopentadienyl anion has 6 π electrons (n=1) in a planar ring and is aromatic; the cation has 4 π electrons and is antiaromatic. Neutral cyclopentadiene has an sp3 CH2 that breaks conjugation, so it is non-aromatic.
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Considering only the t2g/eg occupancy, which low-spin octahedral d-configuration gives the largest crystal field stabilisation energy?
- A. d10
- B. d5
- C. d6
- D. d0
Show worked solution
Answer: C
For a low-spin octahedral d6 ion all six electrons occupy the lower t2g set, giving CFSE = −2.4 Δo (plus pairing terms) — the maximum among the options. d10 and d0 give zero CFSE from the simple formula, and d5 is smaller.
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Which molecule has a trigonal bipyramidal electron geometry and a see-saw molecular shape?
- A. CH4
- B. SF4
- C. BF3
- D. PCl5
Show worked solution
Answer: B
SF4 has five electron domains (four bonding + one lone pair) around sulfur, i.e. trigonal bipyramidal electron geometry. The lone pair occupies an equatorial position to minimise repulsion, distorting the shape to see-saw. PCl5 (no lone pair) stays trigonal bipyramidal; CH4 is tetrahedral; BF3 is trigonal planar.
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The spin-only magnetic moment of a high-spin d5 octahedral complex (e.g. Mn(II)) is approximately:
- A. 3.87 BM
- B. 1.73 BM
- C. 2.83 BM
- D. 5.92 BM
Show worked solution
Answer: D
Spin-only moment μ = √(n(n+2)) BM, where n is the number of unpaired electrons. High-spin d5 has 5 unpaired electrons, so μ = √(5×7) = √35 = 5.92 BM. The values 1.73, 3.87 and 2.83 BM correspond to n = 1, 3 and 2 respectively.
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Which of the following is the strongest field ligand in the spectrochemical series?
- A. CO
- B. H2O
- C. F−
- D. Cl−
Show worked solution
Answer: A
The spectrochemical series orders ligands by the size of the crystal field splitting they produce. CO is a strong π-acceptor and sits at the high-field end, producing a large Δ and typically low-spin complexes. The order here is CO > H2O > F− > Cl−.
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In a blast furnace, the species that mainly reduces iron(III) oxide in the upper, cooler zone is:
- A. hydrogen
- B. coke (C) directly
- C. limestone (CaCO3)
- D. carbon monoxide (CO)
Show worked solution
Answer: D
In the moderate-temperature region of the blast furnace, gaseous CO is the principal reducing agent: Fe2O3 + 3CO → 2Fe + 3CO2. Direct reduction by solid carbon becomes significant only in the hottest lower zone; limestone acts as a flux to remove silica as slag, not as a reductant.
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Which species is isoelectronic with the carbonate ion, CO32−?
- A. SO3
- B. ClO3−
- C. NO3−
- D. PO43−
Show worked solution
Answer: C
Isoelectronic species have the same number of atoms and the same total number of valence electrons. CO32− has 4 atoms and 24 valence electrons and is trigonal planar. NO3− (N + 3O + 1 charge) also has 4 atoms and 24 valence electrons with the same trigonal-planar D3h structure.
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The lanthanide contraction is primarily responsible for which observation?
- A. The high reactivity of alkali metals
- B. The very similar atomic radii of Zr and Hf
- C. The colour of transition-metal complexes
- D. The diagonal relationship of Li and Mg
Show worked solution
Answer: B
Across the lanthanides the poorly shielding 4f electrons cause a steady decrease in radius (the lanthanide contraction). This offsets the expected increase down to the third transition series, so Hf ends up almost the same size as Zr, making the two elements chemically very similar and hard to separate.
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Which oxide is amphoteric (reacts with both acids and bases)?
- A. Al2O3
- B. Na2O
- C. SO3
- D. MgO
Show worked solution
Answer: A
Amphoteric oxides react with both acids and bases. Al2O3 dissolves in acid to give Al3+ salts and in alkali to give aluminate, [Al(OH)4]−. Na2O and MgO are basic; SO3 is acidic (gives H2SO4 with water).
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For a first-order reaction, the half-life is:
- A. inversely proportional to initial concentration
- B. directly proportional to initial concentration
- C. independent of the initial concentration
- D. proportional to the square of initial concentration
Show worked solution
Answer: C
For a first-order reaction t1/2 = ln2 / k, which contains no concentration term, so the half-life is constant regardless of starting concentration. (Zero-order half-life is proportional to [A]0; second-order is inversely proportional to [A]0.)
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A reaction has ΔH = −40 kJ/mol and ΔS = −120 J/(mol·K). Above what temperature does it become non-spontaneous (ΔG > 0)?
- A. below about 333 K
- B. never; it is always spontaneous
- C. at all temperatures
- D. above about 333 K
Show worked solution
Answer: D
ΔG = ΔH − TΔS. With both terms negative, ΔG turns positive when the TΔS magnitude exceeds the ΔH magnitude: T > ΔH/ΔS = 40000 J / 120 J·K−1 = 333 K. So above ~333 K the reaction is non-spontaneous.
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The de Broglie wavelength of a particle is:
- A. inversely proportional to its momentum
- B. directly proportional to its momentum
- C. independent of its mass
- D. proportional to its kinetic energy
Show worked solution
Answer: A
The de Broglie relation is λ = h / p, where p is momentum. Wavelength is therefore inversely proportional to momentum (and hence to mass and speed): heavier or faster particles have shorter wavelengths.
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For an ideal gas undergoing a reversible isothermal expansion, which statement is true?
- A. ΔH is greater than zero
- B. ΔU = 0 and q = −w
- C. w = 0 and q = ΔU
- D. ΔU > 0 and q = 0
Show worked solution
Answer: B
For an ideal gas, internal energy depends only on temperature, so an isothermal process has ΔU = 0 (and ΔH = 0). By the first law, ΔU = q + w = 0, hence q = −w: the heat absorbed exactly equals the work done by the gas during expansion.
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Raising the temperature increases the rate of a reaction mainly because it:
- A. always shifts the equilibrium toward products
- B. lowers the activation energy of the reaction
- C. increases the enthalpy change of the reaction
- D. increases the fraction of molecules with energy ≥ activation energy
Show worked solution
Answer: D
The Arrhenius equation k = A·exp(−Ea/RT) shows that raising T increases the fraction of collisions with energy at or above Ea, raising the rate constant. Temperature does not change Ea itself, nor the reaction's ΔH; equilibrium position depends on the sign of ΔH, not simply on heating.
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A buffer is most effective at resisting pH change when:
- A. pH = pKa of the weak acid
- B. pH is two units below pKa
- C. the acid is fully ionised
- D. only the conjugate base is present
Show worked solution
Answer: A
From the Henderson–Hasselbalch equation, pH = pKa + log([A−]/[HA]). Buffer capacity is maximal when [A−] = [HA], i.e. pH = pKa, because equal reservoirs of acid and conjugate base can neutralise added base or acid most effectively.
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For a spontaneous galvanic cell, the standard cell potential E°cell and Gibbs energy satisfy:
- A. E°cell > 0 and ΔG > 0
- B. E°cell > 0 and ΔG < 0
- C. E°cell < 0 and ΔG < 0
- D. E°cell = 0 at all times
Show worked solution
Answer: B
The relation is ΔG = −nFE°cell. For a spontaneous galvanic cell ΔG must be negative, which requires E°cell to be positive (n and F are positive). A negative E°cell would correspond to a non-spontaneous (electrolytic) process.
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Which quantum numbers correctly describe a 3d orbital?
- A. n = 3, l = 3
- B. n = 3, l = 1
- C. n = 3, l = 2
- D. n = 2, l = 2
Show worked solution
Answer: C
The principal quantum number n gives the shell (3 here) and the azimuthal quantum number l gives the subshell, where l = 0,1,2,3 correspond to s,p,d,f. A d orbital has l = 2, so a 3d orbital is n = 3, l = 2. Note l = 2 is not allowed when n = 2.