About this NEET Chemistry practice test

NEET Chemistry is famous for testing the textbook almost word for word — a slightly-off memory of an oxidation state or a named test is exactly as wrong as not knowing it at all. This set covers Physical Chemistry (mole concept, equilibrium, electrochemistry, kinetics), Organic Chemistry (reaction mechanisms, named tests like Lucas and Tollens') and Inorganic Chemistry (periodic trends, coordination compounds) in the same proportion the real paper uses. Every explanation spells out the exact reasoning — the calculation, the rule, or the mechanism — so you're reinforcing precise recall, not vague familiarity. Treat this the way you'd treat a final NCERT read-through: exact numbers, exact names, no approximations.

NEET Chemistry Practice Test sample questions

These starter questions help you launch a chemistry mock test quickly. Swap them with your own worksheet, notebook, or textbook questions any time.

1. 1. In the reaction MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O, how many moles of KMnO4 are required to completely oxidise 1 mole of FeSO4 in acidic medium?

2. 2. An organic compound contains carbon, hydrogen and oxygen in the mass ratio 40% : 6.7% : 53.3%. What is its empirical formula?

3. 3. How many unpaired electrons are present in the Cr³⁺ ion (Z = 24)?

4. 4. Using the Rydberg formula (R = 1.097 × 10⁷ m⁻¹), calculate the wavelength (in nm) of the spectral line emitted when an electron in a hydrogen atom falls from n = 4 to n = 2.

5. 5. What is the hybridisation of the central atom and the molecular shape of XeF4?

6. 6. According to molecular orbital theory, what is the bond order of O2, and is it paramagnetic or diamagnetic?

7. 7. For a reaction to be spontaneous at all temperatures, what must be true of the signs of ΔH and ΔS?

8. 8. If Kc for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) is 4 × 10⁻², what is Kc for the reaction ½N2(g) + 3/2H2(g) ⇌ NH3(g)?

9. 9. Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, calculate the standard EMF of the Daniell cell.

10. 10. A current of 2 A is passed through a CuSO4 solution for 965 seconds. Calculate the mass of copper deposited at the cathode.

11. 11. 0.1 mol of NaCl is dissolved in 1 kg of water. Taking Kf of water as 1.86 K kg mol⁻¹ and assuming complete dissociation (i = 2), calculate the depression in freezing point.

12. 12. A metal crystallises in a face-centred cubic (fcc) lattice with an edge length of 400 pm. Calculate the radius of the metal atom.

13. 13. Why does NH3 have a significantly higher boiling point than PH3?

14. 14. Among Sc³⁺, Ti³⁺, V³⁺ and Cr³⁺, which ion is colourless in aqueous solution, and why?

15. 15. What is the most common and most stable oxidation state shown by the lanthanide elements?

16. 16. Calculate the Effective Atomic Number (EAN) of iron in the complex ion [Fe(CN)6]⁴⁻.

17. 17. Determine the oxidation state of cobalt in [Co(NH3)5Cl]Cl2 and give its IUPAC name.

18. 18. How many sigma (σ) and pi (π) bonds are present in a molecule of benzene (C6H6)?

19. 19. What is the major product formed when HBr is added to propene in the absence of peroxide, and which rule explains it?

20. 20. Which simple test reagent distinguishes a terminal alkyne (such as propyne) from an internal alkyne, and what is observed?

21. 21. How does Lucas reagent (conc. HCl + anhydrous ZnCl2) help distinguish primary, secondary and tertiary alcohols?

22. 22. Why is phenol a stronger acid than ethanol?

23. 23. Which simple test can distinguish acetaldehyde from acetone, given that both give a positive iodoform test?

24. 24. Arrange trichloroacetic acid, chloroacetic acid, formic acid and acetic acid in decreasing order of acid strength.

25. 25. Arrange methylamine, dimethylamine, trimethylamine and ammonia in decreasing order of basicity in aqueous solution.

26. 26. What product is obtained when benzenediazonium chloride is treated with Cu2Cl2/HCl, and what is this reaction called?

27. 27. What type of glycosidic linkage joins the glucose units in the unbranched polysaccharide amylose?

28. 28. Which level of protein structure is stabilised mainly by hydrogen bonds between the backbone C=O and N–H groups, giving rise to the α-helix and β-pleated sheet?

29. 29. Identify the two monomers used to synthesise Nylon-6,6 and state the type of polymerisation involved.

30. 30. What are non-narcotic analgesics, and which commonly used drug is an example of one?

Answer key & quick explanations

Short answers for the sample questions above. Use this to self-check before generating a fresh AI-built mock test.

1. 1. In the reaction MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O, how many moles of KMnO4 are required to completely oxidise 1 mole of FeSO4 in acidic medium?

   0.2 mol (1/5 mol) of KMnO4

   MnO4⁻ is reduced to Mn²⁺, gaining 5 electrons per ion, while each Fe²⁺ loses only 1 electron on oxidation to Fe³⁺. So one mole of KMnO4 oxidises 5 moles of FeSO4, meaning only 0.2 mole of KMnO4 is needed for 1 mole of FeSO4.

2. 2. An organic compound contains carbon, hydrogen and oxygen in the mass ratio 40% : 6.7% : 53.3%. What is its empirical formula?

   CH2O

   Dividing each mass percentage by its atomic mass gives moles: C = 40/12 = 3.33, H = 6.7/1 = 6.7, O = 53.3/16 = 3.33. Dividing through by the smallest value (3.33) gives a simple ratio of C:H:O = 1:2:1, so the empirical formula is CH2O (this is also the empirical formula of glucose).

3. 3. How many unpaired electrons are present in the Cr³⁺ ion (Z = 24)?

   3 unpaired electrons

   Chromium's ground-state configuration is [Ar]3d5 4s1. Forming Cr³⁺ removes the single 4s electron first and then two more from the 3d subshell, leaving 3d3. By Hund's rule, all three d electrons occupy separate orbitals with parallel spins, giving 3 unpaired electrons.

4. 4. Using the Rydberg formula (R = 1.097 × 10⁷ m⁻¹), calculate the wavelength (in nm) of the spectral line emitted when an electron in a hydrogen atom falls from n = 4 to n = 2.

   486 nm

   Using 1/λ = R(1/2² − 1/4²) = 1.097×10⁷ × (1/4 − 1/16) = 1.097×10⁷ × 0.1875 ≈ 2.057×10⁶ m⁻¹. Taking the reciprocal gives λ ≈ 4.86×10⁻⁷ m, i.e. 486 nm, which is the well-known Hβ line of the Balmer series.

5. 5. What is the hybridisation of the central atom and the molecular shape of XeF4?

   sp3d2 hybridisation; square planar shape

   Xenon in XeF4 has 8 valence electrons, of which 4 form Xe–F bonds and 4 remain as two lone pairs, giving six electron domains and sp3d2 hybridisation. The two lone pairs occupy axial positions and cancel each other's effect, leaving the four fluorine atoms arranged in a square planar shape.

6. 6. According to molecular orbital theory, what is the bond order of O2, and is it paramagnetic or diamagnetic?

   Bond order = 2; paramagnetic

   The molecular orbital configuration of O2 places one electron each in the degenerate π*2px and π*2py antibonding orbitals, accounting for its paramagnetism. The bond order, calculated as ½(bonding electrons − antibonding electrons) = ½(10−6) = 2, matches the O=O double bond.

7. 7. For a reaction to be spontaneous at all temperatures, what must be true of the signs of ΔH and ΔS?

   ΔH must be negative (exothermic) and ΔS must be positive (entropy increases)

   Since ΔG = ΔH − TΔS, a negative ΔH and a positive ΔS make the −TΔS term negative at every temperature, so ΔG stays negative regardless of how high or low T is. If either sign were reversed or mixed, spontaneity would depend on temperature.

8. 8. If Kc for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) is 4 × 10⁻², what is Kc for the reaction ½N2(g) + 3/2H2(g) ⇌ NH3(g)?

   0.2 (2 × 10⁻¹)

   The second equation is exactly half of the first, so its equilibrium constant equals the square root of Kc for the original reaction. √(4×10⁻²) = 0.2.

9. 9. Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, calculate the standard EMF of the Daniell cell.

   +1.10 V

   In the Daniell cell, copper is reduced at the cathode and zinc is oxidised at the anode, so E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V. This positive value confirms the cell reaction is spontaneous as written.

10. 10. A current of 2 A is passed through a CuSO4 solution for 965 seconds. Calculate the mass of copper deposited at the cathode.

    0.635 g

    The total charge passed is Q = It = 2 × 965 = 1930 C, which corresponds to 1930/96500 = 0.02 mol of electrons. Since Cu²⁺ + 2e⁻ → Cu requires 2 moles of electrons per mole of copper, 0.01 mol of Cu is deposited, giving a mass of 0.01 × 63.5 = 0.635 g.

11. 11. 0.1 mol of NaCl is dissolved in 1 kg of water. Taking Kf of water as 1.86 K kg mol⁻¹ and assuming complete dissociation (i = 2), calculate the depression in freezing point.

    ΔTf = 0.372 K

    With complete dissociation, NaCl gives a van't Hoff factor i = 2, and the molality here is 0.1 mol/kg since 0.1 mol is dissolved in 1 kg of water. Applying ΔTf = i·Kf·m gives 2 × 1.86 × 0.1 = 0.372 K.

12. 12. A metal crystallises in a face-centred cubic (fcc) lattice with an edge length of 400 pm. Calculate the radius of the metal atom.

    ≈ 141.4 pm

    In an fcc unit cell, atoms touch along the face diagonal, so 4r = a√2, giving r = a√2/4. Substituting a = 400 pm gives r = 400 × 1.414/4 ≈ 141.4 pm.

13. 13. Why does NH3 have a significantly higher boiling point than PH3?

    Because NH3 molecules form strong hydrogen bonds with each other, while PH3 does not

    Nitrogen is small and highly electronegative, so the N–H bond in ammonia is polar enough to support extensive intermolecular hydrogen bonding. Phosphorus is larger and less electronegative, so P–H bonds in PH3 are far less polar and hydrogen bonding is negligible, leaving only weak van der Waals forces between molecules.

14. 14. Among Sc³⁺, Ti³⁺, V³⁺ and Cr³⁺, which ion is colourless in aqueous solution, and why?

    Sc³⁺ is colourless

    Sc³⁺ has the electronic configuration [Ar]3d0, so it has no d electrons available for d-d transitions. Since colour in transition-metal ions arises from electrons absorbing visible light while jumping between split d orbitals, an ion with an empty d subshell shows no such absorption and appears colourless.

15. 15. What is the most common and most stable oxidation state shown by the lanthanide elements?

    +3 oxidation state

    Lanthanides achieve a stable [Xe]4f^n configuration after losing the two 6s electrons and one 4f (or 5d) electron, so the +3 state is both the most common and generally the most stable across the series. Some elements do show +2 or +4 states, but only when this leads to a specially stable f0, f7 or f14 configuration.

16. 16. Calculate the Effective Atomic Number (EAN) of iron in the complex ion [Fe(CN)6]⁴⁻.

    36

    Since each CN⁻ carries a −1 charge and the complex ion has an overall charge of −4 with six ligands, iron must be in the +2 state, leaving Fe²⁺ with 24 electrons (26 − 2). Each CN⁻ donates one lone pair (2 electrons) through its carbon atom, so six ligands add 12 electrons, giving an EAN of 24 + 12 = 36, the electron count of krypton.

17. 17. Determine the oxidation state of cobalt in [Co(NH3)5Cl]Cl2 and give its IUPAC name.

    Co is in the +3 oxidation state; IUPAC name: pentaamminechloridocobalt(III) chloride

    The two chloride ions outside the coordination sphere each carry a −1 charge, so the complex cation [Co(NH3)5Cl]²⁺ must carry an overall +2 charge. Inside the complex, the five ammine ligands are neutral and the coordinated chloride contributes −1, so cobalt must be +3 to balance the charge to +2.

18. 18. How many sigma (σ) and pi (π) bonds are present in a molecule of benzene (C6H6)?

    12 sigma bonds and 3 pi bonds

    Benzene has 6 C–H sigma bonds and 6 C–C sigma bonds forming the ring, giving 12 sigma bonds in total. The remaining electron density is delocalised over the ring as three pi bonds, conventionally drawn as three alternating double bonds.

19. 19. What is the major product formed when HBr is added to propene in the absence of peroxide, and which rule explains it?

    2-bromopropane, explained by Markovnikov's rule

    Protonation of propene occurs at the terminal carbon because this generates the more stable secondary carbocation on the middle carbon. Bromide then attacks this carbocation, placing Br on the more substituted carbon, consistent with Markovnikov's rule that H adds to the carbon already bearing more hydrogens.

20. 20. Which simple test reagent distinguishes a terminal alkyne (such as propyne) from an internal alkyne, and what is observed?

    Ammoniacal AgNO3 (Tollens' reagent); a terminal alkyne gives a white precipitate of silver alkynide

    Terminal alkynes have an acidic ≡C–H hydrogen because the sp-hybridised carbon holds the bonding electrons closer to itself. This hydrogen is readily replaced by silver ion in ammoniacal AgNO3, forming an insoluble white silver acetylide precipitate, whereas internal alkynes have no such acidic hydrogen and show no reaction.

21. 21. How does Lucas reagent (conc. HCl + anhydrous ZnCl2) help distinguish primary, secondary and tertiary alcohols?

    3° alcohols give turbidity immediately, 2° alcohols within about 5-10 minutes, and 1° alcohols show no turbidity at room temperature

    The Lucas test converts the alcohol's –OH group into an alkyl chloride, and the reaction proceeds through a carbocation intermediate whose stability decides the rate. Tertiary carbocations form fastest, giving instant cloudiness from the insoluble chloride, while primary carbocations are too unstable to form readily at room temperature.

22. 22. Why is phenol a stronger acid than ethanol?

    Because the phenoxide ion formed after deprotonation is resonance-stabilised, unlike the ethoxide ion

    When phenol loses a proton, the resulting negative charge on oxygen can delocalise into the benzene ring through resonance, spreading the charge and lowering the energy of the phenoxide ion. Ethanol's conjugate base, the ethoxide ion, has no such delocalisation pathway, so it is far less stable and ethanol is a much weaker acid.

23. 23. Which simple test can distinguish acetaldehyde from acetone, given that both give a positive iodoform test?

    Tollens' reagent: acetaldehyde gives a silver mirror, acetone gives no reaction

    Acetaldehyde is an aldehyde with a hydrogen atom directly attached to the carbonyl carbon, which allows it to be readily oxidised and reduce Tollens' reagent to metallic silver. Acetone is a ketone with no such hydrogen on the carbonyl carbon, so it cannot be oxidised in this way and gives no silver mirror, even though both compounds are iodoform-positive.

24. 24. Arrange trichloroacetic acid, chloroacetic acid, formic acid and acetic acid in decreasing order of acid strength.

    Trichloroacetic acid > chloroacetic acid > formic acid > acetic acid

    Chlorine atoms withdraw electron density inductively, stabilising the conjugate base and increasing acidity, so trichloroacetic acid (three Cl atoms) is more acidic than chloroacetic acid (one Cl atom). Formic acid, having only a hydrogen atom attached to the carboxyl carbon, is more acidic than acetic acid, whose electron-donating methyl group destabilises the conjugate base.

25. 25. Arrange methylamine, dimethylamine, trimethylamine and ammonia in decreasing order of basicity in aqueous solution.

    (CH3)2NH > CH3NH2 > (CH3)3N > NH3

    In water, basicity depends on a balance between the electron-donating effect of alkyl groups, steric hindrance around nitrogen, and how well the resulting ammonium ion is stabilised by hydrogen bonding with solvent. Dimethylamine gets enough alkyl electron donation without too much steric or solvation penalty, while trimethylamine's three methyl groups hinder solvation of the ammonium ion so much that it falls below even methylamine, though all three amines remain more basic than ammonia itself.

26. 26. What product is obtained when benzenediazonium chloride is treated with Cu2Cl2/HCl, and what is this reaction called?

    Chlorobenzene, formed by the Sandmeyer reaction

    In the Sandmeyer reaction, the diazonium group (–N2+Cl⁻) is replaced by a chlorine atom using cuprous chloride (Cu2Cl2) in the presence of HCl, releasing nitrogen gas. This provides a reliable route to aryl chlorides that cannot easily be made by direct chlorination of benzene.

27. 27. What type of glycosidic linkage joins the glucose units in the unbranched polysaccharide amylose?

    α-1,4-glycosidic linkage

    Amylose is the unbranched component of starch in which successive glucose units are joined through α-1,4-glycosidic bonds between C1 of one glucose unit and C4 of the next. This linkage gives amylose its characteristic helical, unbranched structure, distinguishing it from the branched amylopectin, which also contains α-1,6 linkages.

28. 28. Which level of protein structure is stabilised mainly by hydrogen bonds between the backbone C=O and N–H groups, giving rise to the α-helix and β-pleated sheet?

    Secondary structure

    The secondary structure of a protein describes the local folding pattern of the polypeptide backbone, held in place by hydrogen bonds between the carbonyl oxygen and amide hydrogen of peptide bonds. These interactions produce regular, repeating structures such as the coiled α-helix and the extended, pleated β-sheet.

29. 29. Identify the two monomers used to synthesise Nylon-6,6 and state the type of polymerisation involved.

    Hexamethylenediamine and adipic acid, joined by condensation (step-growth) polymerisation

    Nylon-6,6 is formed when the amine groups of hexamethylenediamine react with the carboxylic acid groups of adipic acid, each condensation releasing a molecule of water and forming an amide linkage. Because the chain grows by repeated condensation reactions between two different bifunctional monomers, this is classified as step-growth (condensation) polymerisation.

30. 30. What are non-narcotic analgesics, and which commonly used drug is an example of one?

    Analgesics; aspirin (acetylsalicylic acid) is a common example

    Non-narcotic analgesics relieve pain by acting mainly at the site of injury or inflammation rather than depressing the central nervous system, so they do not impair consciousness or cause addiction like narcotic analgesics such as morphine. Aspirin is a widely used non-narcotic analgesic that also has antipyretic and anti-inflammatory properties.

Curriculum Mapping & Learning Guide

Use this breakdown to identify which skills each question tests and guide post-test review.

NEET Chemistry units covered

1. Chapter 1: Some Basic Concepts of Chemistry
2. Chapter 2: Atomic Structure
3. Chapter 3: Chemical Bonding
4. Chapter 4: Thermodynamics and Equilibrium
5. Chapter 5: Redox Reactions and Electrochemistry
6. Chapter 6: Solutions and Solid State
7. Chapter 7: p-Block, d-Block and f-Block Elements
8. Chapter 8: Coordination Compounds
9. Chapter 9: Hydrocarbons
10. Chapter 10: Alcohols, Phenols and Ethers
11. Chapter 11: Aldehydes, Ketones and Carboxylic Acids
12. Chapter 12: Organic Compounds Containing Nitrogen
13. Chapter 13: Biomolecules
14. Chapter 14: Polymers
15. Chapter 15: Chemistry in Everyday Life

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