KMU MDCAT Paper 2025 -

1. Consider the chlorination of methane to methyl chloride. The attack of a chlorine free radical on methane occurs in which phase?
A. Before initiation
B. Initiation
C. Propagation
D. Termination

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Correct Option: C
Option A Explained: Incorrect. “Before initiation” represents the stage where no reactions have started and no reactive free radicals are present in the system yet.
Option B Explained: Incorrect. The initiation step only involves the homolytic cleavage of Cl₂ by sunlight to produce chlorine free radicals (Cl•), without any attack on methane.
Option C Explained: Correct. During propagation, a chlorine radical (Cl•) attacks a methane molecule (CH₄) to form HCl and a methyl radical (CH₃•), driving the chain reaction.
Option D Explained: Incorrect. The termination step involves two free radicals combining to form a stable molecule, which stops the reaction rather than continuing the attack.

2. A patient with pancreatic insufficiency shows reduced activity of an enzyme that hydrolyzes the peptide bond at the carboxyl end of proteins and peptides. Which enzyme is deficient?
A. Elastase
B. Pepsin
C. Carboxypeptidase
D. Collagenase

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Correct Option: C
Option A Explained: Incorrect. Elastase is an endopeptidase that breaks internal peptide bonds in elastin fibers, not at the carboxyl end of a polypeptide.
Option B Explained: Incorrect. Pepsin is a stomach enzyme, active at highly acidic pH, that breaks internal peptide bonds (endopeptidase), not the carboxyl end.
Option C Explained: Correct. Carboxypeptidase is a pancreatic exopeptidase that specifically cleaves peptide bonds starting from the C-terminal (carboxyl end) of proteins.
Option D Explained: Incorrect. Collagenase specifically degrades collagen fibers in connective tissue and does not perform general carboxyl-end peptide cleavage.

3. The force of attraction due to temporary dipoles is:
A. Dipole-dipole force
B. Dispersion force
C. Debye force
D. Ion-dipole force

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Correct Option: B
Option A Explained: Incorrect. Dipole-dipole forces occur between molecules that have permanent dipoles (polar molecules), not temporary ones.
Option B Explained: Correct. Dispersion forces (London dispersion forces) are temporary dipole-induced dipole attractions that occur in all molecules, especially non-polar ones, due to instantaneous electron shifts.
Option C Explained: Incorrect. Debye forces (dipole-induced dipole) occur between a permanent polar molecule and a non-polar molecule, not between two temporary dipoles.
Option D Explained: Incorrect. Ion-dipole forces represent the attraction between a fully charged ion and a polar molecule with a permanent dipole.

4. In which case does a chemical reaction go to completion?
A. When Kc is small but positive
B. When Kc is large but positive
C. When Kc is negative
D. When Kc is approximately equal to 1

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Correct Option: B
Option A Explained: Incorrect. A small positive Kc indicates that equilibrium favours the reactants greatly, meaning the reaction barely proceeds forward.
Option B Explained: Correct. A very large positive Kc (Kc ≫ 1) means products are heavily favoured at equilibrium, so the reactants are almost completely converted to products.
Option C Explained: Incorrect. The equilibrium constant Kc is a ratio of positive concentration terms, so it can never mathematically be a negative number.
Option D Explained: Incorrect. When Kc is approximately 1, reactants and products are present in roughly equal concentrations, indicating a reversible reaction that does not go to completion.

5. A student adds bromine water to ethene and observes rapid decolourisation. What makes ethene more reactive than ethane in this reaction?
A. Ethene has a higher molecular mass
B. Ethene contains a weak and exposed pi bond
C. Ethane contains fewer sigma bonds
D. Ethene undergoes substitution more readily

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Correct Option: B
Option A Explained: Incorrect. Molecular mass has no bearing on chemical reactivity in this rapid addition reaction.
Option B Explained: Correct. Ethene (CH₂=CH₂) contains a double bond with an exposed, electron-rich π (pi) bond, which is easily attacked by electrophiles such as bromine.
Option C Explained: Incorrect. Ethane actually contains more sigma bonds (7) than ethene (5), but these are strong and less accessible, making ethane less reactive.
Option D Explained: Incorrect. Ethene undergoes electrophilic addition reactions with bromine water, not substitution reactions.

6. What best describes the overall reaction in electrophilic aromatic substitution in benzene?
A. Addition of an electrophile across a double bond
B. Substitution of a halogen by a nucleophile
C. Substitution of a proton (H⁺) by an electrophile on the aromatic ring
D. Substitution of a methyl group by a nucleophile

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Correct Option: C
Option A Explained: Incorrect. Benzene does not readily undergo addition, since this would destroy its exceptionally stable, conjugated aromatic ring system.
Option B Explained: Incorrect. This describes nucleophilic aromatic substitution, which is unfavourable and not relevant for unsubstituted benzene.
Option C Explained: Correct. An electrophile attacks the electron-rich benzene ring, forming an intermediate, which then loses a proton (H⁺) to restore aromaticity.
Option D Explained: Incorrect. Unsubstituted benzene has no methyl group, and the reaction is driven by an electrophile, not a nucleophile.

7. Real gases do not reach absolute zero in practice because:
A. Molecular collisions become inelastic due to increased kinetic energy
B. Intermolecular forces become negligible and molecules disperse
C. Kinetic energy of molecules increases due to compression
D. Intermolecular forces exceed the kinetic energy of molecules

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Correct Option: D
Option A Explained: Incorrect. At very low temperatures, molecular kinetic energy decreases drastically, meaning molecules slow down rather than speed up.
Option B Explained: Incorrect. As gases are cooled, intermolecular forces become significant and dominant, rather than negligible.
Option C Explained: Incorrect. Kinetic energy is directly proportional to temperature, so cooling a gas toward absolute zero decreases kinetic energy.
Option D Explained: Correct. At extremely low temperatures, the low kinetic energy of molecules cannot overcome their intermolecular attractions, causing the gas to liquefy or solidify before reaching 0 K.

8. The energy of an orbital is determined by:
A. Hund’s Rule
B. Pauli Exclusion Principle
C. (n + l) Rule
D. Boyle’s Law

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Correct Option: C
Option A Explained: Incorrect. Hund’s rule describes how degenerate orbitals (orbitals of the same energy) are filled by electrons, not how orbital energy levels are determined.
Option B Explained: Incorrect. The Pauli exclusion principle states that no two electrons can have the same four quantum numbers; it does not govern orbital energies.
Option C Explained: Correct. The (n + l) rule (Madelung’s rule) determines the relative energy of subshells; subshells with lower (n + l) values have lower energy and fill first.
Option D Explained: Incorrect. Boyle’s Law relates the pressure and volume of a gas and is unrelated to atomic structure or orbital energy.

9. The Ksp value of salt AB (AB ⇌ A⁺ + B⁻) is 9 × 10⁻⁸. Its molar solubility will be:
A. 3 × 10⁻⁴ mol dm⁻³
B. 9 × 10⁻³ mol dm⁻³
C. 3 × 10⁻³ mol dm⁻³
D. 9 × 10⁻⁴ mol dm⁻³

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Correct Option: A
Option A Explained: Correct. For a binary salt AB, Ksp = s². Thus, molar solubility s = √Ksp = √(9 × 10⁻⁸) = 3 × 10⁻⁴ mol dm⁻³.
Option B Explained: Incorrect. This value does not correspond to the square root of the given Ksp value.
Option C Explained: Incorrect. This value results from an error in placing the decimal exponent and does not equal √(9 × 10⁻⁸).
Option D Explained: Incorrect. This value is not the correct square root of the given Ksp and does not satisfy Ksp = s².

10. Water changes from a liquid at 4°C to ice at 0°C. What is the change in volume?
A. 9% increase
B. 9% decrease
C. 19% increase
D. 19% decrease

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Correct Option: A
Option A Explained: Correct. Due to hydrogen bonding, water molecules form an open, cage-like crystal structure upon freezing, causing ice to occupy about 9% more volume than liquid water.
Option B Explained: Incorrect. Water shows anomalous expansion on freezing, meaning its volume increases rather than decreases.
Option C Explained: Incorrect. While volume does increase, a 19% increase is far too large; the actual expansion is approximately 9%.
Option D Explained: Incorrect. Water expands on freezing, so its volume cannot decrease — this is why ice floats on liquid water.

11. For an exothermic reaction, the energy level of the reactants is:
A. Less than that of the products
B. More than that of the products
C. Equal to that of the products
D. Zero

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Correct Option: B
Option A Explained: Incorrect. If reactants had lower energy than products, the reaction would absorb heat from the surroundings, making it endothermic.
Option B Explained: Correct. Exothermic reactions release heat energy, meaning the reactants must possess higher internal chemical energy than the products.
Option C Explained: Incorrect. If reactant and product energy levels were identical, there would be no net exchange of heat energy (ΔH = 0).
Option D Explained: Incorrect. All chemical compounds possess non-zero internal bond energy; reactant molecules can never have zero energy.

12. In an electrolytic cell, when current passes through a solution, the anode is:
A. A positive electrode where oxidation occurs
B. A negative electrode where oxidation occurs
C. A positive electrode where reduction occurs
D. A negative electrode where reduction occurs

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Correct Option: A
Option A Explained: Correct. In an electrolytic cell, the anode is connected to the positive terminal of the battery and is always the site where oxidation (loss of electrons) occurs.
Option B Explained: Incorrect. The anode in an electrolytic cell is positive, not negative, though oxidation does occur there.
Option C Explained: Incorrect. Oxidation, not reduction, always occurs at the anode, regardless of the type of cell.
Option D Explained: Incorrect. The electrode where reduction occurs is the cathode, which is negative in an electrolytic cell, not the anode.

13. An increase in the internal energy of a chemical system can lead to all of the following EXCEPT:
A. An increase in temperature due to a rise in the kinetic energy of molecules
B. A phase change such as melting or evaporation
C. A chemical reaction, if the energy supplied is sufficient to break bonds
D. An increase in temperature due to a drop in the kinetic energy of molecules

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Correct Option: D
Option A Explained: Incorrect. This is a valid outcome, since temperature is a direct measure of average molecular kinetic energy, which rises when internal energy increases.
Option B Explained: Incorrect. This is a valid outcome, since increasing internal energy can break intermolecular forces, causing a transition from solid to liquid or liquid to gas.
Option C Explained: Incorrect. This is a valid outcome, since added thermal energy can supply the activation energy needed to break reactant bonds and trigger a reaction.
Option D Explained: Correct. This statement is chemically impossible, since temperature cannot increase while the kinetic energy of the molecules is dropping.

14. In an ethene molecule, each carbon atom has three hybridized sp² orbitals, which are:
A. Coplanar
B. Tetrahedral
C. Linear
D. Pyramidal

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Correct Option: A
Option A Explained: Correct. The three sp² hybrid orbitals on each carbon atom lie in a single flat plane at angles of 120° to minimize electron repulsion, i.e., they are coplanar.
Option B Explained: Incorrect. Tetrahedral geometry is formed by four sp³ hybrid orbitals at angles of 109.5°, as seen in saturated hydrocarbons like methane.
Option C Explained: Incorrect. Linear geometry is formed by two sp hybrid orbitals at an angle of 180°, as seen in triple-bonded alkynes like ethyne.
Option D Explained: Incorrect. Pyramidal geometry is typical of molecules with three bond pairs and one lone pair, such as ammonia (NH₃), not hybridized carbon atoms in ethene.

15. Which of the following best explains the reaction between beryllium (Be) and oxygen (O)?
A. Be burns vigorously with oxygen, forming a layer of BeO that accelerates oxidation of the remaining metal
B. Be reacts with oxygen, forming a layer of BeO that protects the metal from further oxidation
C. Be reacts slowly with oxygen to form a volatile oxide, BeO, which evaporates quickly
D. Be is the only alkaline earth metal that does not react with oxygen

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Correct Option: B
Option A Explained: Incorrect. Beryllium does not burn vigorously in bulk form, since the oxide layer halts rather than accelerates further reaction.
Option B Explained: Correct. Beryllium reacts with atmospheric oxygen to form a tough, non-porous, protective surface layer of BeO, which shields the underlying metal from further oxidation.
Option C Explained: Incorrect. Beryllium oxide (BeO) is an ionic compound with a very high melting point and is non-volatile; it does not evaporate at normal temperatures.
Option D Explained: Incorrect. Beryllium does react with oxygen, particularly in powder form or when heated to high temperatures, so it is not completely inert.

16. The value of R in atm dm³ mol⁻¹ K⁻¹ is:
A. 0.0821
B. 0.821
C. 62.4
D. 8.314

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Correct Option: A
Option A Explained: Correct. When pressure is measured in atmospheres (atm) and volume in cubic decimeters (dm³), the ideal gas constant R equals 0.0821.
Option B Explained: Incorrect. This value is ten times larger than the correct value of R (0.0821) in these units.
Option C Explained: Incorrect. 62.4 is the value of R used when pressure is measured in torr or millimetres of mercury.
Option D Explained: Incorrect. 8.314 is the SI value of the gas constant, expressed in joules per mole per kelvin (J mol⁻¹ K⁻¹).

17. The rate of a chemical reaction changes with:
A. Concentration of reactant molecules
B. Concentration of product molecules
C. Concentration of both reactant and product molecules
D. The rate constant

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Correct Option: A
Option A Explained: Correct. According to the Law of Mass Action, the rate of a chemical reaction is directly proportional to the active mass (concentration) of the reacting substances.
Option B Explained: Incorrect. Reaction rate depends on collisions between reactants, so it does not depend on the concentration of products in a forward reaction.
Option C Explained: Incorrect. The concentration of products does not influence the rate of a simple, one-way forward chemical reaction.
Option D Explained: Incorrect. The specific rate constant (k) remains constant at a given temperature; it does not itself change to alter the reaction rate.

18. According to Planck’s quantum theory, if the frequency of a photon is doubled, the value of ‘h’ will be:
A. Doubled
B. Increased 3 times
C. Increased 4 times
D. Unchanged

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Correct Option: D
Option A Explained: Incorrect. Planck’s constant ‘h’ is a fundamental universal constant and never changes with changes in photon frequency.
Option B Explained: Incorrect. Fundamental physical constants are independent of experimental or wave variables and cannot increase.
Option C Explained: Incorrect. While the energy of the photon (E = hν) doubles when the frequency doubles, ‘h’ itself remains constant.
Option D Explained: Correct. Planck’s constant (6.626 × 10⁻³⁴ J s) is a fundamental constant and remains completely unchanged by frequency variations.

19. The specific rate constant (k) of a reaction is related to the concentration of reactants:
A. Directly
B. Inversely
C. Exponentially
D. It is independent of concentration

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Correct Option: D
Option A Explained: Incorrect. The rate of the reaction depends directly on concentration, but the specific rate constant (k) itself does not.
Option B Explained: Incorrect. The specific rate constant ‘k’ is unaffected by reactant concentration; it is not inversely proportional to it.
Option C Explained: Incorrect. There is no exponential dependence of the specific rate constant on the concentration of reacting species.
Option D Explained: Correct. The rate constant ‘k’ is independent of concentration; it is affected only by changes in temperature or the presence of a catalyst.

20. An experiment shows that heating a protein disrupts its alpha-helix structure. Which level of protein structure is mainly affected?
A. Primary
B. Secondary
C. Tertiary
D. Quaternary

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Correct Option: B
Option A Explained: Incorrect. Primary structure is held together by strong covalent peptide bonds, which are not easily broken by gentle heating.
Option B Explained: Correct. The α-helix is a secondary protein structure held together by weak hydrogen bonds, which are easily disrupted by thermal energy.
Option C Explained: Incorrect. While tertiary structure may also be lost during complete denaturation, the α-helix is specifically a feature of secondary structure.
Option D Explained: Incorrect. Quaternary structure refers to the spatial arrangement of multiple polypeptide subunits and is not defined by individual α-helices.

21. Which of the following is NOT a postulate of the kinetic molecular theory of gases?
A. Gas molecules undergo elastic collisions
B. Gas molecules are in continuous random motion
C. Gas molecules do not exert pressure when they collide with the walls of the container
D. Gas molecules are far apart from each other

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Correct Option: C
Option A Explained: Incorrect. This is a valid postulate; gas collisions are perfectly elastic, meaning total kinetic energy is conserved during collisions.
Option B Explained: Incorrect. This is a valid postulate; gas molecules move continuously, randomly, and in straight lines between collisions.
Option C Explained: Correct. This statement is false, since the pressure exerted by a gas is actually caused by the collisions of its molecules against the container walls.
Option D Explained: Incorrect. This is a valid postulate; the distance between gas molecules is very large compared to the actual size of the molecules.

22. The number of spectral series observed in the hydrogen spectrum is:
A. 2
B. 3
C. 5
D. 7

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Correct Option: C
Option A Explained: Incorrect. The hydrogen spectrum contains far more than two series of lines across the ultraviolet, visible, and infrared regions.
Option B Explained: Incorrect. Three series is too few; there are several named series corresponding to transitions to different energy levels.
Option C Explained: Correct. The standard curriculum defines five main observable spectral series for hydrogen: Lyman, Balmer, Paschen, Brackett, and Pfund.
Option D Explained: Incorrect. While minor higher-order series (such as Humphreys) exist, standard textbooks define only 5 main series.

23. The most electronegative element in the periodic table is:
A. Fluorine
B. Chlorine
C. Oxygen
D. Nitrogen

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Correct Option: A
Option A Explained: Correct. Fluorine has the highest electronegativity (4.0 on the Pauling scale) due to its small atomic radius and high effective nuclear charge.
Option B Explained: Incorrect. Chlorine is larger than fluorine, so its valence shell is farther from the nucleus, giving it a lower electronegativity of 3.0.
Option C Explained: Incorrect. Oxygen is highly electronegative (3.4), but it still ranks below fluorine due to its lower effective nuclear charge.
Option D Explained: Incorrect. Nitrogen has an electronegativity of 3.0, significantly lower than that of fluorine.

24. If ΔH for a reaction is positive, then decreasing the temperature will cause the reaction to:
A. Move forward
B. Move in reverse
C. Have no effect
D. Move in both the forward and reverse directions

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Correct Option: B
Option A Explained: Incorrect. Decreasing temperature removes heat, which shifts an endothermic (positive ΔH) reaction in the heat-producing, reverse direction.
Option B Explained: Correct. A positive ΔH indicates an endothermic reaction; cooling shifts the equilibrium in the exothermic (reverse) direction to help restore lost heat.
Option C Explained: Incorrect. Temperature changes always shift the equilibrium position of a chemical system with a non-zero enthalpy change.
Option D Explained: Incorrect. A chemical system can only shift net in one direction at a time, not both forward and reverse simultaneously.

25. The addition of water to propene in the presence of sulfuric acid produces:
A. Propan-1-ol
B. Propan-2-ol
C. Butan-1-ol
D. Ethanol

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Correct Option: B
Option A Explained: Incorrect. Propan-1-ol is only a minor product, since acid-catalyzed hydration proceeds mainly through the more stable secondary carbocation (Markovnikov’s rule).
Option B Explained: Correct. According to Markovnikov’s rule, H⁺ adds to the terminal CH₂ carbon and OH⁻ adds to the more substituted CH carbon, yielding propan-2-ol as the major product.
Option C Explained: Incorrect. Propene has only three carbon atoms, so it cannot form a four-carbon alcohol such as butanol.
Option D Explained: Incorrect. Hydration of the three-carbon propene molecule cannot yield the two-carbon compound ethanol.

26. In which of the following species does the central atom use sp² hybridization?
A. PH₃
B. NH₃
C. CH₃⁺
D. SbH₃

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Correct Option: C
Option A Explained: Incorrect. Phosphine (PH₃) has 3 bond pairs and 1 lone pair, giving it sp³ hybridization with a pyramidal shape.
Option B Explained: Incorrect. Ammonia (NH₃) has 3 bond pairs and 1 lone pair, giving it sp³ hybridization with a pyramidal shape.
Option C Explained: Correct. The methyl carbocation (CH₃⁺) has only 3 bond pairs and no lone pair, giving it sp² hybridization and a flat, trigonal planar shape.
Option D Explained: Incorrect. Stibine (SbH₃) has 3 bond pairs and 1 lone pair on antimony, so it does not adopt sp² hybridization.

27. In a molecule of phenol, the carbon atom that is attached to the –OH group is:
A. sp hybridized
B. sp² hybridized
C. sp³ hybridized
D. Unhybridized

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Correct Option: B
Option A Explained: Incorrect. Linear, triple-bonded carbons are sp hybridized, whereas benzene-ring carbons are double-bonded and trigonal planar.
Option B Explained: Correct. In phenol, all carbon atoms of the benzene ring, including the one bonded to –OH, are involved in conjugated double bonds and are therefore sp² hybridized.
Option C Explained: Incorrect. An sp³ carbon would require four single sigma bonds, which would disrupt the aromaticity of the benzene ring.
Option D Explained: Incorrect. Carbons in a stable aromatic ring undergo hybridization to form equivalent, coplanar bonds.

28. Which one of the following is a planar molecule?
A. NH₃
B. H₂O
C. BF₃
D. CH₄

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Correct Option: C
Option A Explained: Incorrect. Ammonia has a pyramidal shape due to the steric influence of its lone pair, making it three-dimensional rather than planar.
Option B Explained: Incorrect. Water has a bent, non-planar-in-3D shape; it is BF₃ that shows perfect trigonal planar symmetry.
Option C Explained: Correct. Boron trifluoride (BF₃) has 3 bond pairs and no lone pair around boron, giving it a perfectly trigonal planar geometry with 120° bond angles.
Option D Explained: Incorrect. Methane (CH₄) has 4 bond pairs and no lone pairs, resulting in a three-dimensional tetrahedral geometry.

29. Which of the following has the highest bond energy?
A. HCl
B. HI
C. HF
D. HBr

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Correct Option: C
Option A Explained: Incorrect. The H–Cl bond is longer and weaker than the H–F bond, so it requires less energy to break.
Option B Explained: Incorrect. H–I has the longest bond length and the lowest bond energy among the hydrogen halides.
Option C Explained: Correct. Fluorine is the smallest and most electronegative halogen, forming an exceptionally short, highly polar, and strong H–F bond with high bond energy.
Option D Explained: Incorrect. The H–Br bond is weaker than both H–Cl and H–F due to the larger atomic size of bromine.

30. Identify the correct electronic configuration for an element with atomic number 24.
A. 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁴ 4s²
B. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
C. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁴
D. 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 4p⁶

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Correct Option: B
Option A Explained: Incorrect. This is the configuration predicted by the simple Aufbau order, but it is not the actual, more stable configuration adopted by chromium.
Option B Explained: Correct. Chromium (Z = 24) is an exception to the Aufbau principle: one electron shifts from 4s to 3d, giving the more stable, half-filled [Ar] 4s¹ 3d⁵ configuration.
Option C Explained: Incorrect. This configuration ignores the well-known half-filled d-subshell stability exception shown by chromium.
Option D Explained: Incorrect. This configuration is not physically valid, since the 4p subshell lies at a higher energy and is not filled before 3d.

31. Balance the following equation using the oxidation number method: Cu + H₂SO₄ → CuSO₄ + SO₂ + H₂O
A. Cu + H₂SO₄ → CuSO₄ + SO₂ + H₂O
B. Cu + 2H₂SO₄ → CuSO₄ + SO₂ + H₂O
C. Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
D. 2Cu + 2H₂SO₄ → 2CuSO₄ + SO₂ + 2H₂O

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Option A Explained: Incorrect. This equation is unbalanced; the numbers of hydrogen, sulfur, and oxygen atoms do not match on both sides.
Option B Explained: Incorrect. The hydrogen atoms are unbalanced; there are 4 hydrogens on the left but only 2 on the product side.
Option C Explained: Correct. This is the fully balanced equation, with 1 Cu, 4 H, 2 S, and 8 O atoms on both sides.
Option D Explained: Incorrect. The atoms of copper and sulfur become unbalanced when copper is doubled without a matching change in the products.

32. The compound that reacts with phenylhydrazine to form a crystalline phenylhydrazone derivative is:
A. Butanal
B. 1,3-Butadiene
C. Ethyl acetate
D. Ethanol

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Correct Option: A
Option A Explained: Correct. Butanal is an aldehyde; aldehydes contain a reactive carbonyl group (C=O) that undergoes nucleophilic addition with phenylhydrazine to form crystalline phenylhydrazones.
Option B Explained: Incorrect. 1,3-Butadiene is a conjugated diene and lacks the carbonyl (C=O) group required to react with phenylhydrazine.
Option C Explained: Incorrect. Ethyl acetate is an ester; esters do not readily form phenylhydrazone derivatives under standard conditions.
Option D Explained: Incorrect. Ethanol is an alcohol and lacks the carbonyl (C=O) group necessary to react with phenylhydrazine.

33. If 4 g of H₂ reacts with 2 moles of O₂ to form water, which reactant is in excess?
A. H₂ only
B. O₂ only
C. H₂O only
D. Both O₂ and H₂

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Correct Option: B
Option A Explained: Incorrect. 4 g of H₂ is exactly 2 moles. Since 2 moles of H₂ require only 1 mole of O₂, all of the hydrogen is completely consumed, making it the limiting reactant.
Option B Explained: Correct. The reaction is 2H₂ + O₂ → 2H₂O. 2 moles of H₂ (4 g) consume only 1 mole of O₂; since 2 moles of O₂ are provided, 1 mole of O₂ remains unreacted, i.e., in excess.
Option C Explained: Incorrect. H₂O is a product of the reaction and cannot be classified as a reactant.
Option D Explained: Incorrect. Both reactants cannot be in excess simultaneously; one must be limiting unless they are present in exact stoichiometric proportions.

34. If 10 moles of magnesium react with excess oxygen, calculate the theoretical yield of magnesium oxide, MgO. (Molar mass: Mg = 24 g/mol, O = 16 g/mol)
A. 160 g
B. 240 g
C. 320 g
D. 400 g

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Correct Option: D
Option A Explained: Incorrect. 160 g corresponds mathematically to only 4 moles of MgO, not the 10 moles produced here.
Option B Explained: Incorrect. 240 g represents the mass of 10 moles of the reactant magnesium (24 g/mol), not the MgO product.
Option C Explained: Incorrect. 320 g represents 8 moles of MgO, which does not match the expected 1:1 molar yield from 10 moles of Mg.
Option D Explained: Correct. The reaction is 2Mg + O₂ → 2MgO, a 1:1 mole ratio of Mg to MgO. Thus, 10 moles of Mg yield 10 moles of MgO: mass = 10 mol × (24 + 16) g/mol = 400 g.

35. Acetaldehyde reacts with ethanol in the presence of an acid catalyst and initially produces:
A. An acetal
B. A hemiacetal
C. A diol
D. Diethoxyethane

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Correct Option: B
Option A Explained: Incorrect. An acetal is the final product, formed when the intermediate hemiacetal reacts with a second molecule of ethanol.
Option B Explained: Correct. Acid-catalyzed addition of one equivalent of alcohol to an aldehyde initially forms a hemiacetal, containing one –OH and one –OR group on the same carbon.
Option C Explained: Incorrect. Diols contain two hydroxyl (–OH) groups, which is not the functional group initially formed in this reaction.
Option D Explained: Incorrect. Diethoxyethane is the systematic name for the final stable acetal, not the initial hemiacetal intermediate.

36. What is the correct IUPAC name of the compound Br–CH₂–CH₂–CO–CH₃?
A. 1-Bromobutan-3-one
B. 4-Bromobutan-2-one
C. 2-Bromobutan-4-one
D. 3-Bromobutan-1-one

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Correct Option: B
Option A Explained: Incorrect. This numbering gives the carbonyl group (the principal characteristic group) a higher locant than necessary, which violates IUPAC lowest-locant rules for the suffix.
Option B Explained: Correct. IUPAC rules require the principal characteristic group (the ketone, –one) to receive the lowest possible locant. Numbering from the methyl end places the carbonyl at C-2 and the bromine substituent at C-4, giving 4-bromobutan-2-one.
Option C Explained: Incorrect. This name incorrectly assigns the carbonyl group locant 4, which is not the lowest possible locant for the principal characteristic group.
Option D Explained: Incorrect. This name is structurally inconsistent with the given compound, since it places the ketone group at the terminal carbon.

37. Consider the reaction: 2Na + Cl₂ → 2NaCl. If 4 moles of Na and 2 moles of Cl₂ are reacted, how much Cl₂ will remain unreacted?
A. 0 mol
B. 0.5 mol
C. 1 mol
D. 1.5 mol

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Correct Option: A
Option A Explained: Correct. The balanced ratio of Na to Cl₂ is 2:1. Thus, 4 moles of Na react completely with exactly 2 moles of Cl₂, leaving 0 moles of Cl₂ unreacted.
Option B Explained: Incorrect. Since all 2 moles of Cl₂ react completely, there is no fractional amount left over.
Option C Explained: Incorrect. 1 mole of Cl₂ would be left over only if just 2 moles of Na had reacted.
Option D Explained: Incorrect. This value would imply that almost none of the chlorine reacted, which is inconsistent with the given stoichiometry.

38. The electronic configuration of Fe³⁺ (Z = 26) is:
A. [Ar] 4s² 3d³
B. [Ar] 4s¹ 3d⁴
C. [Ar] 4s⁰ 3d⁵
D. [Ar] 4s⁰ 3d⁶

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Correct Option: C
Option A Explained: Incorrect. Iron loses its 4s electrons first upon ionization, so Fe³⁺ cannot retain a 4s² configuration.
Option B Explained: Incorrect. Both 4s electrons are removed before any 3d electron is lost, so retaining 4s¹ is incorrect for Fe³⁺.
Option C Explained: Correct. Neutral Fe is [Ar] 4s² 3d⁶. To form Fe³⁺, it loses both 4s electrons and one 3d electron, giving [Ar] 4s⁰ 3d⁵, a stable, half-filled d-subshell.
Option D Explained: Incorrect. This is the configuration of the Fe²⁺ ion, not the Fe³⁺ ion.

39. If the pressure and temperature of a fixed amount of gas are both doubled, the new volume will be:
A. Doubled
B. Halved
C. Tripled
D. Unchanged

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Correct Option: D
Option A Explained: Incorrect. Doubling pressure alone would halve the volume, and doubling temperature alone would double it; together, these effects cancel out.
Option B Explained: Incorrect. The volume does not halve, since the simultaneous rise in temperature expands the gas, offsetting the compression from increased pressure.
Option C Explained: Incorrect. There is no physical or mathematical basis for the volume to triple under these conditions.
Option D Explained: Correct. From the ideal gas law, V = nRT/P; doubling both T and P cancels out, so the volume remains unchanged.

40. The IUPAC name of CH₃–CH(CH₃)–C≡CH is:
A. Hex-1-yne
B. Pent-2-yne
C. Pent-4-yne
D. 3-Methylbut-1-yne

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Correct Option: D
Option A Explained: Incorrect. The longest carbon chain contains only 4 carbons, making it a derivative of butyne, not hexyne, which has 6 carbons.
Option B Explained: Incorrect. Pent-2-yne is a straight 5-carbon chain with a triple bond at carbon-2, which does not match this branched structure.
Option C Explained: Incorrect. Pent-4-yne is an invalid name, since the triple bond must be numbered from the end that gives it the lowest possible locant.
Option D Explained: Correct. The longest chain containing the triple bond has 4 carbons (butyne). Numbering from the triple-bond end places a methyl substituent on carbon-3, giving 3-methylbut-1-yne.

41. According to Le Chatelier’s Principle, when the pressure of a gaseous equilibrium system is increased, the equilibrium shifts towards:
A. No change in the equilibrium position
B. The side with greater volume
C. The side with fewer moles of gas (lower volume)
D. The side with more moles of gas

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Correct Option: C
Option A Explained: Incorrect. Increasing pressure does shift the equilibrium whenever the number of gas moles differs on the two sides of the reaction.
Option B Explained: Incorrect. Shifting toward the side with greater volume would raise the pressure further, which contradicts Le Chatelier’s principle.
Option C Explained: Correct. Increasing pressure forces the system to shift toward the side that occupies less volume, that is, the side with fewer gas moles, to relieve the added stress.
Option D Explained: Incorrect. Shifting toward more moles of gas would increase system volume and pressure, opposite to the required stress relief.

42. Consider the reaction 2A + B₂ → 2AB. Which of the following mixtures would make A the limiting reagent?
A. 300 atoms of A and 400 molecules of B₂
B. 100 atoms of A and 50 molecules of B₂
C. 2 mol of A and 1 mol of B₂
D. 5 mol of A and 2.5 mol of B₂

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Option A Explained: Correct. The mole ratio of A to B₂ is 2:1. 300 atoms of A require only 150 molecules of B₂; since 400 are present, B₂ is in excess and A is the limiting reagent.
Option B Explained: Incorrect. 100 atoms of A react exactly with 50 molecules of B₂ (a 2:1 ratio), so both reactants are consumed completely with no limiting reagent.
Option C Explained: Incorrect. 2 moles of A react exactly with 1 mole of B₂, a perfect stoichiometric ratio with no limiting reactant.
Option D Explained: Incorrect. 5 moles of A react exactly with 2.5 moles of B₂, so neither reactant is in excess or limiting.

43. On increasing the temperature, the rate of a reaction increases mainly because:
A. The activation energy of the reaction increases
B. The concentration of the reacting molecules increases
C. The collision frequency increases
D. The energy of the molecules decreases

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Correct Option: C
Option A Explained: Incorrect. Activation energy is a characteristic of the reaction pathway and remains constant regardless of temperature changes.
Option B Explained: Incorrect. A temperature increase does not affect the concentration (moles per unit volume) of the reacting species.
Option C Explained: Correct. Raising the temperature increases the kinetic energy and speed of the molecules, which increases both the collision frequency and the fraction of effective (successful) collisions.
Option D Explained: Incorrect. Heating increases the kinetic energy of molecules; it never decreases the energy of reacting molecules.

44. Given: heat of sublimation of Na = 108 kJ/mol, ionization energy of Na = 496 kJ/mol, bond dissociation energy of Cl₂ = 242 kJ/mol, electron affinity of Cl = –349 kJ/mol, and enthalpy of formation of NaCl = –411 kJ/mol, calculate the lattice energy of NaCl.
A. –678 kJ/mol
B. –787 kJ/mol
C. –819 kJ/mol
D. –832 kJ/mol

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Correct Option: B
Option A Explained: Incorrect. This value does not result from correctly applying the Born–Haber cycle to the given thermodynamic data.
Option B Explained: Correct. By the Born–Haber cycle: ΔHf = ΔHsub + IE + ½BDE + EA + LE. So –411 = 108 + 496 + 121 + (–349) + LE, giving LE = –411 – 376 = –787 kJ/mol.
Option C Explained: Incorrect. This value does not correspond to the correct summation of the Born–Haber cycle terms for NaCl.
Option D Explained: Incorrect. This value is not supported by the thermodynamic parameters given for the Born–Haber cycle of NaCl.

45. The molecular orbitals formed by the π-electrons in benzene are:
A. Localized
B. Delocalized
C. Hybridized
D. Polarized

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Correct Option: B
Option A Explained: Incorrect. Localized orbitals confine electrons between two specific nuclei, which does not represent the shared electron cloud found in benzene.
Option B Explained: Correct. The six π-electrons in benzene are completely delocalized across the hexagonal ring of carbon atoms, forming a stable, continuous molecular orbital system.
Option C Explained: Incorrect. Individual atomic orbitals (such as s and p) hybridize to form hybrid orbitals like sp², but molecular orbitals themselves are not described as “hybridized.”
Option D Explained: Incorrect. Benzene is a symmetric, non-polar hydrocarbon ring, so its molecular orbitals are highly symmetric and non-polarized.

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