About this IIT JEE Chemistry practice test

JEE Chemistry splits almost exactly one-third each into Physical, Organic and Inorganic, and most students have a favourite third they over-prepare while quietly under-revising the other two. This set forces balance: real numerical problems in Physical Chemistry (Nernst equation, kinetics, equilibrium), genuine mechanism-based questions in Organic (regiochemistry, named reactions like Cannizzaro), and precise factual recall in Inorganic (coordination chemistry, periodic trends). Every explanation walks through the actual reasoning — the mechanism, the formula, or the rule — instead of just stating the answer. Work through all three sections evenly here, the same way the real JEE paper will test you.

IIT JEE 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. A 4.4 g sample of propane (C3H8) is burnt in the presence of 6.4 g of oxygen gas. Identify the limiting reagent and calculate the mass of CO2 formed, assuming the reaction goes to completion for the limiting reagent.

2. 2. An electron, initially at rest, is accelerated through a potential difference of 100 V. Calculate its de Broglie wavelength (in picometres).

3. 3. A real gas obeys Z ≈ 1 + (b − a/RT)P/(RT) at low pressure. At 300 K, its van der Waals constants are a = 4.17 atm L² mol⁻² and b = 0.0371 L mol⁻¹. Determine whether Z is greater than or less than 1 at low pressure and justify your answer using the sign of (b − a/RT).

4. 4. Given the following standard enthalpies of reaction: C(graphite) + O2(g) → CO2(g), ΔH° = −393.5 kJ/mol; H2(g) + ½O2(g) → H2O(l), ΔH° = −285.8 kJ/mol; C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH° = −1367 kJ/mol. Calculate the standard enthalpy of formation of liquid ethanol.

5. 5. A reaction has ΔH = +30.5 kJ/mol and ΔS = +100 J K⁻¹ mol⁻¹. Calculate the minimum temperature above which the reaction becomes spontaneous.

6. 6. At 500 K, Kc for N2(g) + 3H2(g) ⇌ 2NH3(g) is 6.0 × 10⁻² mol⁻² L². At equilibrium, [N2] = 3.0 mol/L and [H2] = 2.0 mol/L. Calculate the equilibrium concentration of NH3.

7. 7. Calculate the pH of a buffer solution prepared by mixing 0.10 mol of CH3COOH and 0.15 mol of CH3COONa in enough water to make 1 L of solution. (Ka of CH3COOH = 1.8 × 10⁻⁵)

8. 8. For the cell Zn(s) | Zn²⁺(0.01 M) || Cu²⁺(1.0 M) | Cu(s), E°cell = 1.10 V. Calculate the cell potential at 298 K.

9. 9. A first-order reaction has a rate constant of 1.15 × 10⁻³ s⁻¹. Calculate the time required for the amount of reactant to decrease from 5 g to 3 g.

10. 10. 0.6 g of a non-volatile, non-electrolyte solute is dissolved in 100 g of water. The resulting solution freezes at −0.186°C. Given Kf of water = 1.86 K kg mol⁻¹, calculate the molar mass of the solute.

11. 11. Using VSEPR theory, predict the hybridization of the central atom and the molecular shape of ClF3.

12. 12. Arrange NH3, PH3 and H2O in increasing order of bond angle, and explain the trend.

13. 13. Using molecular orbital theory, determine the bond order of the superoxide ion, O2⁻, and predict its magnetic behaviour.

14. 14. PCl5 is a well-known stable compound, but NCl5 does not exist. Explain this difference in terms of the atomic orbitals available to the central atom.

15. 15. White phosphorus reacts with excess chlorine gas. Write the balanced equation for the product formed, and state the hybridization and shape of phosphorus in that product.

16. 16. Calculate the spin-only magnetic moment of the complex ion [Fe(H2O)6]²⁺. (Atomic number of Fe = 26)

17. 17. Identify the lanthanide that commonly shows a stable +2 oxidation state in addition to +3, and explain why. Also identify a lanthanide that shows a stable +4 oxidation state.

18. 18. Give the IUPAC name of the coordination compound [Co(NH3)5Cl]Cl2, and state the oxidation state of cobalt in it.

19. 19. Calculate the crystal field stabilization energy (in terms of Δo) for the complex ion [Ti(H2O)6]³⁺.

20. 20. [Co(NH3)5(NO2)]Cl2 and [Co(NH3)5(ONO)]Cl2 are isomers of each other. Name the type of isomerism shown, and identify each form.

21. 21. 1-Methylcyclohexene is treated with HBr in the presence of peroxides. Identify the major product and explain why its regiochemistry differs from the reaction carried out without peroxides.

22. 22. Arrange ethene, propene and 2-methylpropene (isobutylene) in increasing order of their rate of electrophilic addition with HBr, and justify the order.

23. 23. 2-Bromobutane is heated with alcoholic KOH. Identify the major organic product and name the mechanism and rule that determine it.

24. 24. Chlorobenzene requires much harsher conditions (623 K, 300 atm) than an alkyl halide to undergo nucleophilic substitution with NaOH. Explain why aryl halides are far less reactive than alkyl halides in nucleophilic substitution.

25. 25. Explain, with reference to the stability of the conjugate base, why phenol is a stronger acid than ethanol and reacts with NaOH while ethanol does not.

26. 26. Anisole (methoxybenzene) is heated with excess concentrated HI. Identify the major products and explain which bond is cleaved and why.

27. 27. Arrange acetaldehyde, acetone and benzaldehyde in decreasing order of their reactivity toward nucleophilic addition, and justify the order.

28. 28. Benzaldehyde is treated with concentrated NaOH. Identify the two organic products formed and name the reaction and its mechanism.

29. 29. Arrange aniline, methylamine, ammonia and N,N-dimethylaniline in decreasing order of basicity in aqueous solution, and justify the order.

30. 30. Glucose contains an aldehyde group in its open-chain form but does not give a positive Schiff's test. Explain why, identifying the predominant structural form of glucose in aqueous solution.

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. A 4.4 g sample of propane (C3H8) is burnt in the presence of 6.4 g of oxygen gas. Identify the limiting reagent and calculate the mass of CO2 formed, assuming the reaction goes to completion for the limiting reagent.

   O2 is limiting; 5.28 g of CO2 is formed.

   C3H8 + 5O2 → 3CO2 + 4H2O needs a 1:5 mole ratio. Moles taken are 0.1 mol propane and 0.2 mol O2, but 0.1 mol propane would require 0.5 mol O2, so oxygen runs out first. The 0.2 mol O2 available yields (3/5)×0.2 = 0.12 mol CO2, i.e. 0.12 × 44 = 5.28 g.

2. 2. An electron, initially at rest, is accelerated through a potential difference of 100 V. Calculate its de Broglie wavelength (in picometres).

   ≈ 122.7 pm

   For an electron accelerated from rest, kinetic energy eV equals p²/2m, so λ = h/√(2meV). Substituting the standard constants gives the shortcut formula λ(Å) = 12.27/√V, and with V = 100 this gives 12.27/10 = 1.227 Å, i.e. 122.7 pm.

3. 3. A real gas obeys Z ≈ 1 + (b − a/RT)P/(RT) at low pressure. At 300 K, its van der Waals constants are a = 4.17 atm L² mol⁻² and b = 0.0371 L mol⁻¹. Determine whether Z is greater than or less than 1 at low pressure and justify your answer using the sign of (b − a/RT).

   Z < 1 at low pressure.

   Compute a/RT = 4.17/(0.0821×300) = 4.17/24.63 ≈ 0.169 L mol⁻¹, which is larger than b = 0.0371 L mol⁻¹. Since (b − a/RT) is negative, the coefficient of P in the Z-expression is negative, so Z falls below 1. Physically this means attractive forces (the a-term) dominate over the finite-size (b-term) correction for this gas at 300 K, making it more compressible than an ideal gas.

4. 4. Given the following standard enthalpies of reaction: C(graphite) + O2(g) → CO2(g), ΔH° = −393.5 kJ/mol; H2(g) + ½O2(g) → H2O(l), ΔH° = −285.8 kJ/mol; C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l), ΔH° = −1367 kJ/mol. Calculate the standard enthalpy of formation of liquid ethanol.

   ΔHf°(C2H5OH, l) ≈ −277.4 kJ/mol

   By Hess's law, the formation reaction 2C(graphite) + 3H2(g) + ½O2(g) → C2H5OH(l) can be built from 2×(combustion of C) + 3×(formation of water) − (combustion of ethanol). That gives 2(−393.5) + 3(−285.8) − (−1367) = −787 − 857.4 + 1367 = −277.4 kJ/mol.

5. 5. A reaction has ΔH = +30.5 kJ/mol and ΔS = +100 J K⁻¹ mol⁻¹. Calculate the minimum temperature above which the reaction becomes spontaneous.

   T > 305 K

   Spontaneity requires ΔG = ΔH − TΔS < 0. Setting ΔG = 0 to find the crossover temperature gives T = ΔH/ΔS = 30500 J/(100 J K⁻¹) = 305 K. Since both ΔH and ΔS are positive, the reaction is entropy-driven and becomes spontaneous only above this temperature.

6. 6. At 500 K, Kc for N2(g) + 3H2(g) ⇌ 2NH3(g) is 6.0 × 10⁻² mol⁻² L². At equilibrium, [N2] = 3.0 mol/L and [H2] = 2.0 mol/L. Calculate the equilibrium concentration of NH3.

   [NH3] = 1.2 mol/L

   Kc = [NH3]²/([N2][H2]³), so [NH3]² = Kc × [N2][H2]³ = 0.06 × 3.0 × (2.0)³ = 0.06 × 24 = 1.44. Taking the square root gives [NH3] = 1.2 mol/L.

7. 7. Calculate the pH of a buffer solution prepared by mixing 0.10 mol of CH3COOH and 0.15 mol of CH3COONa in enough water to make 1 L of solution. (Ka of CH3COOH = 1.8 × 10⁻⁵)

   pH ≈ 4.92

   This is a Henderson-Hasselbalch problem: pH = pKa + log([salt]/[acid]). pKa = −log(1.8×10⁻⁵) ≈ 4.74, and log(0.15/0.10) = log(1.5) ≈ 0.176, so pH ≈ 4.74 + 0.18 = 4.92.

8. 8. For the cell Zn(s) | Zn²⁺(0.01 M) || Cu²⁺(1.0 M) | Cu(s), E°cell = 1.10 V. Calculate the cell potential at 298 K.

   Ecell ≈ 1.16 V

   For Zn + Cu²⁺ → Zn²⁺ + Cu (n = 2), the Nernst equation gives Ecell = E° − (0.0591/n) log([Zn²⁺]/[Cu²⁺]) = 1.10 − (0.0591/2) log(0.01/1.0). Since log(0.01) = −2, this becomes 1.10 − (0.02955)(−2) = 1.10 + 0.059 = 1.159 V. The low Zn²⁺ concentration pushes the potential above the standard value.

9. 9. A first-order reaction has a rate constant of 1.15 × 10⁻³ s⁻¹. Calculate the time required for the amount of reactant to decrease from 5 g to 3 g.

   t ≈ 444 s (about 7.4 minutes)

   For a first-order reaction, t = (2.303/k) log(A0/At). Here A0/At = 5/3 ≈ 1.667, and log(1.667) ≈ 0.222. So t = (2.303/1.15×10⁻³) × 0.222 ≈ 2003 × 0.222 ≈ 444 s.

10. 10. 0.6 g of a non-volatile, non-electrolyte solute is dissolved in 100 g of water. The resulting solution freezes at −0.186°C. Given Kf of water = 1.86 K kg mol⁻¹, calculate the molar mass of the solute.

    M ≈ 60 g/mol

    ΔTf = Kf × m gives molality m = 0.186/1.86 = 0.1 mol/kg. In 100 g (0.1 kg) of water this corresponds to 0.1 × 0.1 = 0.01 mol of solute. Molar mass = mass/moles = 0.6 g/0.01 mol = 60 g/mol.

11. 11. Using VSEPR theory, predict the hybridization of the central atom and the molecular shape of ClF3.

    sp3d hybridization; T-shaped geometry.

    Chlorine contributes 7 valence electrons, 3 of which form bonds to fluorine, leaving 2 lone pairs on Cl. Five electron domains around the central atom correspond to sp3d hybridization and a trigonal bipyramidal electron geometry, but with both lone pairs occupying equatorial positions the observed molecular shape is T-shaped.

12. 12. Arrange NH3, PH3 and H2O in increasing order of bond angle, and explain the trend.

    PH3 (93.5°) < H2O (104.5°) < NH3 (107°)

    PH3 has the smallest bond angle because phosphorus is large and only weakly electronegative, so its bonding pairs stay close to the almost-pure p orbitals with little s-character, keeping the angle near 90°. Nitrogen, being more electronegative than phosphorus, pulls bonding electron density toward itself, giving NH3 the largest angle among these three. Water's angle sits between the two group-15 hydrides here because its two lone pairs compress the H-O-H angle below the tetrahedral value but not as severely as in PH3.

13. 13. Using molecular orbital theory, determine the bond order of the superoxide ion, O2⁻, and predict its magnetic behaviour.

    Bond order = 1.5; paramagnetic (one unpaired electron).

    O2⁻ has 13 valence electrons, one more than neutral O2's 12. Filling the standard MO diagram gives the configuration ...σ2p2z π2p2x π2p2y π*2p2 π*2p1, so bonding electrons total 8 and antibonding electrons total 5. Bond order = (8−5)/2 = 1.5, and because one π* orbital carries a single unpaired electron, the ion is paramagnetic.

14. 14. PCl5 is a well-known stable compound, but NCl5 does not exist. Explain this difference in terms of the atomic orbitals available to the central atom.

    Nitrogen has no accessible d orbitals to expand its octet beyond four bonds, so NCl5 cannot form; phosphorus can use 3d orbitals to accommodate five bond pairs.

    Nitrogen is a second-period element restricted to 2s and 2p orbitals, giving it a maximum of four orbitals for bonding or lone pairs, so its covalency is capped at 4 (as in NH4+). Phosphorus, in the third period, has energetically accessible 3d orbitals that allow sp3d hybridization, letting it expand its valence shell to five bonds and form PCl5.

15. 15. White phosphorus reacts with excess chlorine gas. Write the balanced equation for the product formed, and state the hybridization and shape of phosphorus in that product.

    P4 + 10Cl2 → 4PCl5; sp3d hybridization, trigonal bipyramidal shape.

    With limited chlorine, P4 first gives PCl3 (P4 + 6Cl2 → 4PCl3), but excess Cl2 chlorinates further to PCl5. In PCl5, phosphorus is sp3d hybridized, giving a trigonal bipyramidal shape with three equatorial and two axial P-Cl bonds of unequal length.

16. 16. Calculate the spin-only magnetic moment of the complex ion [Fe(H2O)6]²⁺. (Atomic number of Fe = 26)

    μ ≈ 4.90 BM (4 unpaired electrons)

    Fe²⁺ has a d6 configuration, and H2O is a weak-field ligand, so the octahedral complex is high spin with configuration t2g4eg2. Filling the five d orbitals by Hund's rule leaves 4 unpaired electrons. Using μ = √(n(n+2)) BM with n = 4 gives √24 ≈ 4.90 BM.

17. 17. Identify the lanthanide that commonly shows a stable +2 oxidation state in addition to +3, and explain why. Also identify a lanthanide that shows a stable +4 oxidation state.

    Europium (Eu) commonly shows +2; cerium (Ce) commonly shows +4.

    Eu²⁺ has the configuration [Xe]4f7, a half-filled 4f subshell that is extra stable, so europium readily retains the +2 state alongside its usual +3. On the other side, Ce4+ has an empty [Xe]4f0 configuration, and this closed-subshell stability is why cerium is the classic example of a lanthanide favouring +4.

18. 18. Give the IUPAC name of the coordination compound [Co(NH3)5Cl]Cl2, and state the oxidation state of cobalt in it.

    Pentaamminechloridocobalt(III) chloride; Co is in the +3 oxidation state.

    The two chloride ions outside the bracket carry a total charge of −2, so the complex ion [Co(NH3)5Cl]²⁺ must carry +2 overall. Since NH3 is neutral and the coordinated Cl⁻ contributes −1, cobalt must be +3 to balance the charge. Naming the ligands alphabetically (ammine before chlorido) and adding the oxidation state gives pentaamminechloridocobalt(III) chloride.

19. 19. Calculate the crystal field stabilization energy (in terms of Δo) for the complex ion [Ti(H2O)6]³⁺.

    CFSE = −0.4Δo (i.e., −2Δo/5)

    Ti³⁺ has a single d electron (d1 configuration). In an octahedral field this electron occupies one of the three lower-energy t2g orbitals, each of which contributes −0.4Δo to the stabilization energy. With no electrons in the eg set and no pairing involved, the total CFSE is simply −0.4Δo.

20. 20. [Co(NH3)5(NO2)]Cl2 and [Co(NH3)5(ONO)]Cl2 are isomers of each other. Name the type of isomerism shown, and identify each form.

    Linkage isomerism; the N-bonded form is the nitro isomer and the O-bonded form is the nitrito isomer.

    The ambidentate nitrite ligand can coordinate to cobalt either through nitrogen or through oxygen, giving two distinct complexes with the same formula but different donor atoms attached to the metal — the defining feature of linkage isomerism. When NO2⁻ binds through N it is called the nitro isomer, and when it binds through O (as ONO⁻) it is called the nitrito isomer.

21. 21. 1-Methylcyclohexene is treated with HBr in the presence of peroxides. Identify the major product and explain why its regiochemistry differs from the reaction carried out without peroxides.

    2-Bromo-1-methylcyclohexane (Br on the less-substituted alkene carbon), via the peroxide (anti-Markovnikov) radical mechanism.

    Peroxides trigger a free-radical chain mechanism unique to HBr addition. The Br• radical adds first to whichever carbon gives the more stable carbon radical, which here is the carbon bearing the methyl group (C1); this places Br on the other alkene carbon (C2), and the resulting tertiary radical then abstracts H from another HBr molecule. Without peroxide, the ionic mechanism instead forms the more stable tertiary carbocation at C1 and delivers Br there, giving 1-bromo-1-methylcyclohexane instead — so the peroxide effect reverses where the bromine ends up.

22. 22. Arrange ethene, propene and 2-methylpropene (isobutylene) in increasing order of their rate of electrophilic addition with HBr, and justify the order.

    ethene < propene < 2-methylpropene

    Electrophilic addition of HBr proceeds through a carbocation intermediate, and its rate tracks how stable that intermediate is. Ethene can only form a primary carbocation, propene forms a secondary one, and 2-methylpropene forms a tertiary carbocation stabilized by hyperconjugation and the +I effect of three methyl groups, so it reacts fastest of the three.

23. 23. 2-Bromobutane is heated with alcoholic KOH. Identify the major organic product and name the mechanism and rule that determine it.

    But-2-ene (predominantly) is the major product, formed by E2 elimination following Zaitsev's rule.

    Alcoholic KOH is a strong base/poor nucleophile in a non-aqueous medium, conditions that favour bimolecular elimination (E2) over substitution. Elimination can remove a β-hydrogen from either side of the C-Br bond, giving either but-1-ene or but-2-ene, and Zaitsev's rule predicts the more substituted, more stable alkene — but-2-ene — as the major product.

24. 24. Chlorobenzene requires much harsher conditions (623 K, 300 atm) than an alkyl halide to undergo nucleophilic substitution with NaOH. Explain why aryl halides are far less reactive than alkyl halides in nucleophilic substitution.

    Resonance delocalization of the chlorine lone pair into the ring gives the C-Cl bond partial double-bond character, making it shorter and stronger, and the sp2 carbon holds the bonding electrons more tightly than an sp3 carbon does.

    In chlorobenzene, one lone pair on chlorine conjugates with the aromatic ring, so the C-Cl bond has some double-bond character and is both shorter and stronger than the C-Cl bond in an alkyl halide. The carbon bearing the halogen is also sp2 hybridized, which is more electronegative than an sp3 carbon and grips the shared electrons more tightly, further discouraging nucleophilic attack. Consequently, only forcing conditions of high temperature and pressure can drive the substitution.

25. 25. Explain, with reference to the stability of the conjugate base, why phenol is a stronger acid than ethanol and reacts with NaOH while ethanol does not.

    The phenoxide ion is resonance-stabilized, delocalizing the negative charge into the ring, whereas the ethoxide ion has no such stabilization; hence phenol (pKa ≈ 10) is far more acidic than ethanol (pKa ≈ 16).

    Once phenol loses its proton, the resulting phenoxide ion spreads its negative charge over the oxygen and three ring carbons through resonance, which lowers its energy and favours proton loss. Ethoxide has only the inductive effect of the ethyl group to disperse charge, with no resonance option, leaving it far less stable and ethanol correspondingly less acidic. This is why phenol reacts with NaOH to form sodium phenoxide, while ethanol — a weaker acid than water itself — does not.

26. 26. Anisole (methoxybenzene) is heated with excess concentrated HI. Identify the major products and explain which bond is cleaved and why.

    Phenol and methyl iodide (CH3I); the alkyl-oxygen bond is cleaved, not the aryl-oxygen bond.

    Protonation of the ether oxygen by HI gives an oxocarbenium-like intermediate, and the iodide ion then attacks the methyl carbon in an SN2 step because it is unhindered and primary. The aryl-oxygen bond is never broken because it has partial double-bond character from conjugation with the ring and because backside attack on the sp2 ring carbon is not feasible, so cleavage occurs exclusively on the alkyl side, releasing phenol and methyl iodide.

27. 27. Arrange acetaldehyde, acetone and benzaldehyde in decreasing order of their reactivity toward nucleophilic addition, and justify the order.

    acetaldehyde > benzaldehyde > acetone

    Nucleophilic addition to a carbonyl is favoured by a more electrophilic, less hindered carbonyl carbon. Acetaldehyde has only one small alkyl group, so its carbonyl carbon is both the least hindered and the least electron-rich among the three. Benzaldehyde's phenyl ring donates some electron density into the carbonyl by resonance, making it less reactive than acetaldehyde but still more reactive than acetone, which bears two electron-donating, bulkier alkyl groups.

28. 28. Benzaldehyde is treated with concentrated NaOH. Identify the two organic products formed and name the reaction and its mechanism.

    Sodium benzoate and benzyl alcohol, formed by the Cannizzaro reaction.

    Benzaldehyde has no alpha-hydrogen, so it cannot undergo aldol-type chemistry and instead undergoes the Cannizzaro reaction under concentrated base. Hydroxide first adds to the carbonyl of one benzaldehyde molecule to give a tetrahedral alkoxide intermediate, which then transfers a hydride ion to the carbonyl carbon of a second benzaldehyde molecule. This disproportionation oxidizes one molecule to sodium benzoate and reduces the other to benzyl alcohol.

29. 29. Arrange aniline, methylamine, ammonia and N,N-dimethylaniline in decreasing order of basicity in aqueous solution, and justify the order.

    methylamine > ammonia > N,N-dimethylaniline > aniline

    Methylamine is more basic than ammonia because the methyl group's +I effect pushes electron density onto nitrogen, making its lone pair more available for protonation. In aniline, however, the nitrogen lone pair conjugates extensively with the benzene ring, so it is far less available for protonation, making aniline the weakest base of the four. N,N-dimethylaniline sits above aniline because the +I effect of its two methyl groups partly compensates for this same ring delocalization, but it still falls short of ammonia since the aromatic resonance loss dominates over the alkyl electron donation.

30. 30. Glucose contains an aldehyde group in its open-chain form but does not give a positive Schiff's test. Explain why, identifying the predominant structural form of glucose in aqueous solution.

    Glucose exists almost entirely as cyclic hemiacetals (α- and β-D-glucopyranose) in water, leaving essentially no free aldehyde available to react with Schiff's reagent.

    In aqueous solution, the C1 aldehyde of glucose reacts intramolecularly with the C5 hydroxyl group to form a six-membered pyranose ring, and this cyclic hemiacetal equilibrium accounts for well over 99% of the molecules present at any instant. Because Schiff's reagent needs a free aldehyde in appreciable concentration to regenerate its magenta colour, and only a vanishingly small fraction of glucose exists in the open-chain aldehyde form at equilibrium, the test comes out negative despite the aldehyde being present in principle.

Curriculum Mapping & Learning Guide

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

IIT JEE Chemistry units covered

1. Chapter 1: Some Basic Concepts of Chemistry
2. Chapter 2: Atomic Structure
3. Chapter 3: Chemical Bonding and Molecular Structure
4. Chapter 4: States of Matter
5. Chapter 5: Thermodynamics
6. Chapter 6: Equilibrium (Chemical & Ionic)
7. Chapter 7: Redox Reactions and Electrochemistry
8. Chapter 8: Chemical Kinetics
9. Chapter 9: Solutions
10. Chapter 10: p-Block, d-Block and f-Block Elements
11. Chapter 11: Coordination Compounds
12. Chapter 12: Hydrocarbons (Alkanes, Alkenes, Alkynes)
13. Chapter 13: Haloalkanes and Haloarenes
14. Chapter 14: Alcohols, Phenols and Ethers
15. Chapter 15: Aldehydes, Ketones and Carboxylic Acids
16. Chapter 16: Amines and Biomolecules

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