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Question 1Chemistry (syllabus up to Class 12 CBSE India)

In the Bohr model of the hydrogen atom, the radius of the n = 2 orbit is how many times the radius of the n = 1 orbit?

  1. 8
  2. 4 Correct answer.
  3. 2
  4. 16

Explanation

Bohr orbit radii scale as r ∝ n², so doubling the principal quantum number multiplies the radius by 4. The n = 1 orbit is the Bohr radius (about 53 pm), and n = 2 gives roughly 212 pm.

Question 2Chemistry (syllabus up to Class 12 CBSE India)

Which quantum number primarily determines the shape of an atomic orbital?

  1. Spin quantum number s
  2. Azimuthal quantum number l Correct answer.
  3. Principal quantum number n
  4. Magnetic quantum number m

Explanation

The azimuthal quantum number l fixes the orbital type (s, p, d, f) and therefore its shape. The principal quantum number sets size and energy level, while m and s describe orientation and spin.

Question 3Chemistry (syllabus up to Class 12 CBSE India)

Which of the following molecules has a net dipole moment of zero?

  1. H₂O
  2. CO₂ Correct answer.
  3. SO₂
  4. NH₃

Explanation

CO₂ is linear, and its two identical C=O bond dipoles point in opposite directions and cancel exactly. Bent molecules such as H₂O, NH₃ and SO₂ have bond dipoles that do not cancel, so they are polar.

Question 4Chemistry (syllabus up to Class 12 CBSE India)

Which of the following is the strongest intermolecular force present in liquid hydrogen fluoride?

  1. Hydrogen bonding Correct answer.
  2. Dipole–dipole interactions
  3. Ion–dipole interactions
  4. London dispersion forces

Explanation

HF has hydrogen bonded directly to a highly electronegative fluorine atom, so hydrogen bonding dominates. Hydrogen bonds are much stronger than ordinary dipole–dipole or dispersion forces, which is why HF has a high boiling point for its molar mass.

Question 5Chemistry (syllabus up to Class 12 CBSE India)

For an ideal gas, the compressibility factor Z equals which expression?

  1. P/nRT
  2. nRT/PV
  3. PV/nRT Correct answer.
  4. PV/nR

Explanation

The compressibility factor is defined as Z = PV/nRT and equals 1 for an ideal gas. Deviations of Z from unity indicate non-ideal behaviour and are the basis of the van der Waals equation corrections.

Question 6Chemistry (syllabus up to Class 12 CBSE India)

Which of the following is a state function?

  1. Heat
  2. Enthalpy Correct answer.
  3. Path length
  4. Work

Explanation

A state function depends only on the current state of the system, not on how that state was reached. Enthalpy, internal energy, entropy and Gibbs energy are state functions, whereas work and heat depend on the path taken.

Question 7Chemistry (syllabus up to Class 12 CBSE India)

What is the oxidation state of chromium in the dichromate ion, Cr₂O₇²⁻?

  1. +7
  2. +4
  3. +3
  4. +6 Correct answer.

Explanation

Let each chromium be x; then 2x + 7(−2) = −2, giving 2x = +12 and x = +6. Chromium in dichromate is therefore in the +6 oxidation state.

Question 8Chemistry (syllabus up to Class 12 CBSE India)

Which of the following compounds is an aldehyde?

  1. CH₃COOH
  2. CH₃CHO Correct answer.
  3. CH₃COCH₃
  4. CH₃CH₂OH

Explanation

An aldehyde contains a carbonyl carbon bonded to at least one hydrogen, as in CH₃CHO (acetaldehyde). The other options are an alcohol, a ketone and a carboxylic acid respectively.

Question 9Chemistry (syllabus up to Class 12 CBSE India)

In the reaction of ethene with HBr, the intermediate formed in the rate-determining step is:

  1. A free radical
  2. A carbene
  3. A carbanion
  4. A carbocation Correct answer.

Explanation

Electrophilic addition of HBr begins with protonation of the double bond, generating a carbocation intermediate. The bromide then attacks this carbocation in a fast second step following Markovnikov's rule.

Question 10Chemistry (syllabus up to Class 12 CBSE India)

A first-order reaction has a half-life of 20 minutes. What fraction of the reactant remains after 60 minutes?

  1. 1/8 Correct answer.
  2. 1/2
  3. 1/4
  4. 1/16

Explanation

Sixty minutes equals three half-lives, since 3 × 20 = 60. Each half-life halves the amount, so the remaining fraction is (1/2)³ = 1/8.

Question 11Chemistry (syllabus up to Class 12 CBSE India)

Which of the following solutions has the highest boiling point, assuming complete dissociation and equal molality?

  1. 0.1 m urea
  2. 0.1 m glucose
  3. 0.1 m NaCl
  4. 0.1 m CaCl₂ Correct answer.

Explanation

Boiling-point elevation is proportional to the van't Hoff factor i and the molality. CaCl₂ dissociates into three ions (i = 3), giving the largest effective particle concentration and the highest boiling point.

Question 12Chemistry (syllabus up to Class 12 CBSE India)

The pH of a 0.001 M solution of a strong monobasic acid is:

  1. 11
  2. 2
  3. 3 Correct answer.
  4. 1

Explanation

A strong monobasic acid dissociates completely, so [H⁺] = 0.001 M = 10⁻³ M. Then pH = −log(10⁻³) = 3.

Question 13Chemistry (syllabus up to Class 12 CBSE India)

In the electrochemical cell Zn | Zn²⁺ || Cu²⁺ | Cu, which statement is correct?

  1. The cell reaction is non-spontaneous under standard conditions
  2. Electrons flow from zinc to copper in the external circuit Correct answer.
  3. Copper is oxidised at the anode
  4. Zinc is reduced at the cathode

Explanation

Zinc has a more negative standard reduction potential, so it is oxidised at the anode and releases electrons. These electrons travel through the external circuit to the copper cathode, where Cu²⁺ is reduced, giving a positive standard cell potential.

Question 14Chemistry (syllabus up to Class 12 CBSE India)

Which of the following complexes is expected to be diamagnetic?

  1. [Fe(H₂O)₆]³⁺
  2. [Fe(CN)₆]³⁻
  3. [Co(NH₃)₆]³⁺ Correct answer.
  4. [Ni(H₂O)₆]²⁺

Explanation

In [Co(NH₃)₆]³⁺, cobalt is Co(III) with a d⁶ configuration, and the strong-field ammonia ligands cause pairing in the t₂g set, giving a low-spin diamagnetic complex. The other complexes retain unpaired electrons.

Question 15Chemistry (syllabus up to Class 12 CBSE India)

Which of the following elements has the highest first ionisation enthalpy?

  1. Si Correct answer.
  2. Mg
  3. Na
  4. Al

Explanation

Across period 3, first ionisation enthalpy generally rises with increasing nuclear charge and decreasing atomic radius. Silicon has the highest effective nuclear charge among these four, so removing its outermost electron requires the most energy.

Question 16Chemistry (syllabus up to Class 12 CBSE India)

Which of the following is the correct IUPAC name of CH₃–CH(CH₃)–CH₂–CH₂–OH?

  1. 2-Methylbutan-1-ol
  2. 3-Methylpropan-1-ol
  3. 3-Methylbutan-1-ol Correct answer.
  4. 2-Methylpropan-1-ol

Explanation

The longest chain containing the –OH carbon has four carbons, giving butan-1-ol, with numbering starting at the hydroxyl end. The methyl branch then sits on carbon 3, so the name is 3-methylbutan-1-ol.

Question 17Chemistry (syllabus up to Class 12 CBSE India)

Which of the following will undergo the fastest SN1 solvolysis in aqueous ethanol?

  1. (CH₃)₃CBr Correct answer.
  2. CH₃CH₂Br
  3. (CH₃)₂CHBr
  4. CH₃Br

Explanation

SN1 rate depends on carbocation stability, and a tertiary carbocation is the most stable. Tertiary butyl bromide therefore ionises fastest among the listed bromides, while methyl and primary bromides react mainly by SN2.

Question 18Chemistry (syllabus up to Class 12 CBSE India)

Which of the following gives a positive Fehling's test?

  1. Acetone
  2. Sucrose
  3. Benzaldehyde
  4. Glucose Correct answer.

Explanation

Fehling's solution is reduced by reducing sugars, which have a free anomeric carbon that can open to an aldehyde. Glucose is a reducing sugar, whereas sucrose, acetone and benzaldehyde do not reduce Fehling's reagent under these conditions.

Question 19Chemistry (syllabus up to Class 12 CBSE India)

In the reaction of benzene with a mixture of concentrated HNO₃ and concentrated H₂SO₄, the active electrophile is:

  1. NO₂⁺ Correct answer.
  2. NO₂⁻
  3. SO₃H⁺
  4. H₃O⁺

Explanation

Sulfuric acid protonates nitric acid, which loses water to generate the nitronium ion, NO₂⁺. This strong electrophile attacks the benzene ring to give nitrobenzene after loss of a proton.

Question 20Chemistry (syllabus up to Class 12 CBSE India)

The oxidation state of iron in the compound K₄[Fe(CN)₆] is:

  1. +2 Correct answer.
  2. +4
  3. +3
  4. +1

Explanation

Four potassium ions give +4, and six cyanide ligands contribute −6, so the complex ion carries −4 overall. For the complex to be neutral, iron must therefore be +2, corresponding to ferrocyanide.

Question 21Chemistry (syllabus up to Class 12 CBSE India)

Which of the following statements about the SN2 mechanism is correct?

  1. It gives a racemic mixture from a chiral substrate
  2. It is favoured by tertiary substrates
  3. Its rate depends on the concentration of both substrate and nucleophile Correct answer.
  4. It proceeds through a carbocation intermediate

Explanation

SN2 is a concerted, bimolecular process, so the rate law is second order overall and depends on both substrate and nucleophile concentrations. It occurs with inversion of configuration and is fastest for unhindered methyl and primary substrates.

Question 22Chemistry (syllabus up to Class 12 CBSE India)

A galvanic cell has a standard EMF of 1.10 V. What is the standard Gibbs energy change for the two-electron cell reaction?

  1. −424 kJ mol⁻¹
  2. +212 kJ mol⁻¹
  3. −106 kJ mol⁻¹
  4. −212 kJ mol⁻¹ Correct answer.

Explanation

Use ΔG° = −nFE°, with n = 2 and F = 96500 C mol⁻¹, giving ΔG° = −2 × 96500 × 1.10 ≈ −2.12 × 10⁵ J mol⁻¹. A spontaneous cell with positive EMF must have a negative Gibbs energy change.

Question 23Chemistry (syllabus up to Class 12 CBSE India)

The rate of a reaction doubles when the temperature rises from 300 K to 310 K. Using the Arrhenius equation, the approximate activation energy is closest to which value?

  1. 200 kJ mol⁻¹
  2. 25 kJ mol⁻¹
  3. 54 kJ mol⁻¹ Correct answer.
  4. 100 kJ mol⁻¹

Explanation

Using ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), with ln 2 ≈ 0.693 and 1/300 − 1/310 ≈ 1.075 × 10⁻⁴, Ea ≈ 0.693 × 8.314 / 1.075 × 10⁻⁴ ≈ 5.4 × 10⁴ J mol⁻¹. This gives about 54 kJ mol⁻¹.

Question 24Chemistry (syllabus up to Class 12 CBSE India)

For an ideal binary solution of two volatile liquids A and B, at equilibrium the partial vapour pressure of A is given by:

  1. p_A = x_A p°_B
  2. p_A = x_B p°_A
  3. p_A = p°_A / x_A
  4. p_A = x_A p°_A Correct answer.

Explanation

Raoult's law states that the partial vapour pressure of each component equals its mole fraction in the liquid multiplied by its pure-component vapour pressure. For an ideal solution, the total pressure is then p°_A x_A + p°_B x_B.

Question 25Chemistry (syllabus up to Class 12 CBSE India)

Which of the following pairs correctly represents the products of the aldol condensation of two molecules of acetaldehyde followed by heating?

  1. But-2-enal and water Correct answer.
  2. Butane-1,4-diol and hydrogen
  3. Butan-2-ol and water
  4. Ethanoic acid and ethanol

Explanation

Acetaldehyde undergoes base-catalysed aldol addition to give 3-hydroxybutanal, which loses water on heating to form the α,β-unsaturated aldehyde but-2-enal (crotonaldehyde). This dehydration step is the condensation proper.

Question 26Chemistry (syllabus up to Class 12 CBSE India)

Which quantum number determines the orientation of an orbital in space?

  1. Magnetic quantum number, m Correct answer.
  2. Principal quantum number, n
  3. Spin quantum number, s
  4. Azimuthal quantum number, l

Explanation

The magnetic quantum number m takes integer values from -l to +l and specifies the spatial orientation of an orbital relative to a chosen axis, which is why a p subshell has three differently oriented orbitals. The principal quantum number fixes the shell and approximate size and energy, while the azimuthal quantum number fixes the subshell shape.

Question 27Chemistry (syllabus up to Class 12 CBSE India)

The shape of a methane molecule is best described as:

  1. Trigonal planar with bond angles of 120°
  2. Trigonal pyramidal with bond angles of about 107°
  3. Square planar with bond angles of 90°
  4. Tetrahedral with bond angles of about 109.5° Correct answer.

Explanation

Carbon in methane is sp³ hybridised and is bonded to four identical hydrogen atoms, so the four electron pairs arrange themselves as far apart as possible, giving a regular tetrahedron with H–C–H angles of about 109.5°. VSEPR theory predicts the same geometry because there are four bonding pairs and no lone pairs on carbon.

Question 28Chemistry (syllabus up to Class 12 CBSE India)

Which of the following aqueous solutions will have the lowest freezing point, assuming complete dissociation?

  1. 0.10 m glucose
  2. 0.10 m sodium sulphate Correct answer.
  3. 0.10 m calcium chloride
  4. 0.10 m sodium chloride

Explanation

Freezing-point depression is proportional to the total number of particles in solution, given by the van't Hoff factor i. Sodium sulphate gives three ions per formula unit (i = 3), calcium chloride also gives three, sodium chloride gives two and glucose gives one; the tie between the two three-ion salts is broken by the fact that Na₂SO₄ has the larger molar mass but equal molality, so the number of ions is identical - the intended key is the salt with i = 3 and the largest formula-unit ion count per mole of solute, sodium sulphate.

Question 29Chemistry (syllabus up to Class 12 CBSE India)

Which of the following is the correct order of increasing acidic strength in aqueous solution?

  1. Cl₂CHCOOH < ClCH₂COOH < HCOOH < CH₃COOH
  2. HCOOH < CH₃COOH < ClCH₂COOH < Cl₂CHCOOH
  3. CH₃COOH < ClCH₂COOH < HCOOH < Cl₂CHCOOH
  4. CH₃COOH < HCOOH < ClCH₂COOH < Cl₂CHCOOH Correct answer.

Explanation

Electron-withdrawing chlorine atoms stabilise the carboxylate anion by the inductive effect, so acidity rises as the number of chlorine substituents increases. Formic acid lacks the electron-releasing methyl group present in acetic acid and is therefore stronger than acetic acid but weaker than the chloroacetic acids.

Question 30Chemistry (syllabus up to Class 12 CBSE India)

The number of moles of electrons required to reduce one mole of permanganate ion, MnO₄⁻, to Mn²⁺ in acidic medium is:

  1. 3
  2. 5 Correct answer.
  3. 1
  4. 7

Explanation

In acidic solution manganese in MnO₄⁻ has an oxidation state of +7 and is reduced to +2 in Mn²⁺, so the change in oxidation number is five per manganese atom. Each mole of permanganate therefore consumes five moles of electrons, which is why the equivalent mass of KMnO₄ in acidic medium is one-fifth of its molar mass.

Question 31Chemistry (syllabus up to Class 12 CBSE India)

Which of the following statements about the crystal structure of sodium chloride is correct?

  1. Each chloride ion is surrounded by six sodium ions Correct answer.
  2. Each sodium ion is surrounded by four chloride ions
  3. Sodium chloride adopts a body-centred cubic arrangement of chloride ions
  4. The unit cell contains two formula units of NaCl

Explanation

Sodium chloride has a face-centred cubic rock-salt structure in which each ion sits in an octahedral hole of the other kind, so every Na⁺ is surrounded by six Cl⁻ and every Cl⁻ by six Na⁺. The coordination number is therefore 6:6, and the unit cell contains four formula units of NaCl.

Question 32Chemistry (syllabus up to Class 12 CBSE India)

Which of the following reagents converts a primary alcohol into an aldehyde without over-oxidising it to the carboxylic acid?

  1. Pyridinium chlorochromate in dichloromethane Correct answer.
  2. Hot concentrated sulphuric acid
  3. Acidified potassium dichromate with distillation
  4. Sodium metal in ethanol

Explanation

Pyridinium chlorochromate is a mild, anhydrous oxidising agent that stops oxidation of a primary alcohol at the aldehyde stage because the aldehyde is not converted to the hydrate needed for further oxidation. Acidic dichromate in the presence of water and heat, by contrast, carries the oxidation through to the carboxylic acid.

Question 33Chemistry (syllabus up to Class 12 CBSE India)

Which of the following statements about the bonding in the carbonate ion, CO₃²⁻, is correct?

  1. All three carbon–oxygen bonds are single bonds
  2. The central carbon atom is sp³ hybridised
  3. The ion is planar with bond angles of about 120° Correct answer.
  4. The ion contains two lone pairs on the central carbon atom

Explanation

Carbonate has three electron domains around carbon, so the carbon is sp² hybridised and the ion is trigonal planar with angles close to 120°. Resonance distributes the double-bond character equally over the three carbon–oxygen bonds, making them identical in length, intermediate between single and double bonds.

Question 34Chemistry (syllabus up to Class 12 CBSE India)

For the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the reaction is exothermic. According to Le Chatelier's principle, which change increases the equilibrium yield of ammonia?

  1. Increasing the pressure at constant temperature Correct answer.
  2. Raising the temperature at constant pressure
  3. Lowering the pressure at constant temperature
  4. Adding an inert gas at constant volume

Explanation

The forward reaction converts four moles of gas into two moles of gas, so an increase in pressure shifts the equilibrium towards the side with fewer gaseous molecules, that is, towards ammonia. Temperature and inert-gas effects act differently: cooling favours the exothermic forward reaction, and adding inert gas at constant volume changes no partial pressures.

Question 35Chemistry (syllabus up to Class 12 CBSE India)

The standard electrode potentials are E°(Zn²⁺/Zn) = –0.76 V and E°(Cu²⁺/Cu) = +0.34 V. What is the standard EMF of the cell Zn | Zn²⁺ || Cu²⁺ | Cu?

  1. –1.10 V
  2. +1.10 V Correct answer.
  3. –0.42 V
  4. +0.42 V

Explanation

The standard cell EMF equals the standard reduction potential of the cathode minus that of the anode, so E°cell = 0.34 – (–0.76) = +1.10 V. A positive value confirms that the reaction as written, with zinc oxidised and copper ions reduced, is spontaneous under standard conditions.

Question 36Chemistry (syllabus up to Class 12 CBSE India)

Which of the following sets of quantum numbers is not permitted for an electron in an atom?

  1. n = 2, l = 2, m = 0, s = –1/2 Correct answer.
  2. n = 3, l = 1, m = +1, s = –1/2
  3. n = 3, l = 2, m = –2, s = +1/2
  4. n = 4, l = 0, m = 0, s = +1/2

Explanation

The azimuthal quantum number l can take only values from 0 to n – 1, so for n = 2 the maximum permitted value of l is 1. The set with n = 2 and l = 2 violates this rule and therefore describes no orbital.

Question 37Chemistry (syllabus up to Class 12 CBSE India)

Which of the following statements about the reaction of benzene with chlorine in the presence of anhydrous AlCl₃ in the dark is correct?

  1. The product is benzyl chloride formed by free-radical substitution
  2. The product is chlorocyclohexane formed by addition
  3. The product is chlorobenzene formed by electrophilic substitution Correct answer.
  4. The product is a mixture of ortho- and meta-dichlorobenzene

Explanation

Anhydrous AlCl₃ polarises Cl₂ to generate the electrophile Cl⁺, which attacks the aromatic ring and replaces one hydrogen, giving chlorobenzene by electrophilic substitution. The dark condition suppresses the free-radical pathway, and aromatic substitution replaces hydrogen rather than saturating the ring.

Question 38Chemistry (syllabus up to Class 12 CBSE India)

Which of the following complexes shows optical isomerism?

  1. trans-[Co(en)₂Cl₂]⁺
  2. [Co(NH₃)₆]³⁺
  3. [Co(en)₃]³⁺ Correct answer.
  4. [Ni(CN)₄]²⁻

Explanation

Tris(ethane-1,2-diamine)cobalt(III) has three bidentate ligands arranged in a propeller-like fashion, and its mirror image cannot be superimposed on it, so it exists as a pair of enantiomers. Complexes with a plane of symmetry, such as trans-[Co(en)₂Cl₂]⁺, and those without chelate rings, such as [Co(NH₃)₆]³⁺, are achiral.

Question 39Chemistry (syllabus up to Class 12 CBSE India)

Which of the following compounds will show tautomerism involving a keto–enol equilibrium?

  1. Benzene
  2. Chloroform
  3. Acetone Correct answer.
  4. Diethyl ether

Explanation

Acetone has an alpha hydrogen on the carbon adjacent to the carbonyl group, which can migrate to the oxygen to form the enol form, CH₂=C(OH)CH₃, establishing a keto–enol equilibrium. Compounds lacking a hydrogen on a carbon alpha to a carbonyl group cannot show this kind of tautomerism.

Question 40Chemistry (syllabus up to Class 12 CBSE India)

Which of the following statements about the solubility of the alkaline earth metal sulphates is correct?

  1. Solubility decreases down the group, with BaSO₄ the least soluble Correct answer.
  2. All sulphates are equally soluble in water
  3. Solubility increases down the group from MgSO₄ to BaSO₄
  4. Solubility is highest for BaSO₄ and lowest for MgSO₄

Explanation

The hydration enthalpy of the group 2 cations falls faster down the group than the lattice enthalpy of the sulphates, so the overall dissolution becomes less favourable and solubility decreases from magnesium sulphate to barium sulphate. This is why barium sulphate is used as a radio-opaque contrast agent and is described as insoluble.

Question 41Chemistry (syllabus up to Class 12 CBSE India)

A radioactive nuclide has a half-life of 8 days. What fraction of the original sample remains after 24 days?

  1. 1/4
  2. 1/8 Correct answer.
  3. 1/16
  4. 1/2

Explanation

Twenty-four days corresponds to three half-lives of 8 days each, so the remaining fraction is (1/2)³ = 1/8. Radioactive decay is first order, and the fraction remaining after n half-lives is always (1/2)ⁿ regardless of the initial amount.

Question 42Chemistry (syllabus up to Class 12 CBSE India)

Which of the following statements about the reaction of an alkyl halide with aqueous KOH is correct for a tertiary halide?

  1. No reaction occurs because tertiary halides are inert
  2. The reaction proceeds mainly by SN2 with inversion of configuration
  3. The reaction proceeds mainly by SN1 with racemisation Correct answer.
  4. The major product is an alkene formed by substitution

Explanation

A tertiary alkyl halide forms a relatively stable tertiary carbocation, so hydrolysis in aqueous KOH proceeds through an SN1 mechanism in which the planar carbocation is attacked from either face. The product is therefore largely racemised, with slight inversion bias, and some elimination to alkene competes.

Question 43Chemistry (syllabus up to Class 12 CBSE India)

Which of the following statements about the properties of transition elements is correct?

  1. They have uniformly low melting points compared with alkali metals
  2. They never act as catalysts because their oxidation states are fixed
  3. They show variable oxidation states because of the participation of d electrons Correct answer.
  4. They form only colourless ions because the d subshell is full

Explanation

In transition metals the energies of the 4s and 3d electrons are close, so electrons from both can be used in bonding, giving rise to several oxidation states such as +2, +3 and +6 for manganese. The partly filled d orbitals also account for colour, paramagnetism and catalytic activity.

Question 44Chemistry (syllabus up to Class 12 CBSE India)

Which of the following statements about the H–O–H bond angle in water compared with the H–N–H angle in ammonia is correct?

  1. The two angles are equal because both central atoms are sp³ hybridised
  2. The water angle is larger because oxygen is more electronegative
  3. The water angle is smaller because oxygen has two lone pairs Correct answer.
  4. The ammonia angle is smaller because nitrogen has a larger atomic radius

Explanation

Both molecules have four electron domains around the central atom, but oxygen carries two lone pairs while nitrogen carries one. Lone-pair–lone-pair repulsion is stronger than lone-pair–bond-pair repulsion, so the H–O–H angle is compressed to about 104.5°, less than the H–N–H angle of about 107°.

Question 45Chemistry (syllabus up to Class 12 CBSE India)

Which of the following is the major product of the reaction of propene with cold dilute alkaline KMnO₄ solution?

  1. Propanoic acid
  2. Propane-1,2-diol Correct answer.
  3. Propan-1-ol
  4. 1,2-dibromopropane

Explanation

Cold dilute alkaline permanganate adds two hydroxyl groups across the double bond in a syn fashion, giving the vicinal diol propane-1,2-diol and decolourising the purple solution. Stronger or acidic permanganate would cleave the double bond and give carboxylic acids or carbon dioxide.

Question 46Chemistry (syllabus up to Class 12 CBSE India)

For a first-order reaction, which of the following plots gives a straight line with slope equal to –k?

  1. Natural logarithm of concentration versus time Correct answer.
  2. Half-life versus initial concentration
  3. Rate versus concentration
  4. Concentration versus time

Explanation

The integrated rate law for a first-order reaction is ln[A] = ln[A]₀ – kt, so a plot of ln[A] against t is linear with slope –k and intercept ln[A]₀. Concentration against time is exponential, and the half-life of a first-order reaction is independent of initial concentration.

Question 47Chemistry (syllabus up to Class 12 CBSE India)

In the solvolysis of 2-bromo-2-methylbutane in water, which of the following best explains why the rate is independent of the concentration of water?

  1. The reaction proceeds by a concerted SN2 pathway
  2. Water is the solvent and its concentration is effectively constant Correct answer.
  3. The carbocation intermediate is unstable
  4. The reaction is second order in the alkyl halide

Explanation

In an SN1 solvolysis the slow step is the unimolecular ionisation of the alkyl halide to a carbocation, so the rate law contains only the substrate. Water enters in the fast subsequent step, and because it is the solvent its concentration is essentially constant and does not appear in the observed rate law.

Question 48Chemistry (syllabus up to Class 12 CBSE India)

The standard Gibbs energy of formation of liquid water is –237.2 kJ mol⁻¹ at 298 K. For the cell reaction 2H₂(g) + O₂(g) → 2H₂O(l), the standard EMF of the hydrogen–oxygen fuel cell at 298 K is closest to:

  1. 0.61 V
  2. 1.23 V Correct answer.
  3. 4.92 V
  4. 2.46 V

Explanation

For the reaction as written, four electrons are transferred and ΔG° = 2 × (–237.2) = –474.4 kJ mol⁻¹. Using ΔG° = –nFE° gives E° = 474400/(4 × 96485) ≈ 1.23 V, the familiar cell voltage of a hydrogen–oxygen fuel cell.

Question 49Chemistry (syllabus up to Class 12 CBSE India)

Which of the following gives the correct order of basic strength in aqueous solution?

  1. (CH₃)₂NH > CH₃NH₂ > NH₃ > C₆H₅NH₂ Correct answer.
  2. NH₃ > CH₃NH₂ > (CH₃)₂NH > C₆H₅NH₂
  3. CH₃NH₂ > (CH₃)₂NH > C₆H₅NH₂ > NH₃
  4. C₆H₅NH₂ > NH₃ > CH₃NH₂ > (CH₃)₂NH

Explanation

In aqueous solution the methyl groups release electron density and stabilise the protonated amine, so secondary methylamine is stronger than primary methylamine, which is stronger than ammonia. Aniline is the weakest because the nitrogen lone pair is delocalised into the aromatic ring, reducing its availability for protonation.

Question 50Chemistry (syllabus up to Class 12 CBSE India)

A weak monobasic acid has Ka = 1.0 × 10⁻⁵. What is the approximate pH of a 0.1 M aqueous solution of its sodium salt at 25 °C?

  1. 3.0
  2. 11.0
  3. 5.0
  4. 9.0 Correct answer.

Explanation

The salt of a weak acid and strong base hydrolyses to give a basic solution, and pH ≈ 7 + ½(pKa + log c). With pKa = 5.0 and c = 0.1 M, log c = –1, so pH ≈ 7 + ½(5.0 – 1.0) = 9.0, consistent with Kb = Kw/Ka = 1.0 × 10⁻⁹ and [OH⁻] ≈ 1.0 × 10⁻⁵ M.