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800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.

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ChemistryClass 112 marksmedium

Organic Chemistry: Some Basic Principles and Techniques

What is the inductive effect? Distinguish between +I and -I effects.

Reveal model answer + marking points

The inductive effect is the permanent displacement of the shared sigma-bond electron pair towards the more electronegative atom, transmitted through a chain of carbon atoms and decreasing with distance. Groups that push electrons away from themselves (electron-releasing, e.g. alkyl groups -CH3) show the +I effect, while groups that pull electrons towards themselves (electron-withdrawing, e.g. -NO2, -Cl) show the -I effect.

Marking-scheme points

  • Permanent shift of sigma electrons due to electronegativity difference
  • Transmitted through the chain, weakens with distance
  • +I: electron releasing (alkyl); -I: electron withdrawing (-NO2, -Cl)
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ChemistryClass 112 marksmedium

Organic Chemistry: Some Basic Principles and Techniques

What is resonance? What is resonance energy? Give one example.

Reveal model answer + marking points

Resonance is the representation of a molecule or ion by two or more structures (canonical/contributing structures) that differ only in the position of electrons, none of which alone describes it fully; the actual structure is a resonance hybrid. Resonance energy is the difference in energy between the resonance hybrid and the most stable contributing structure; the greater the resonance energy, the more stable the molecule. Example: benzene, which is a hybrid of two Kekule structures and is more stable than expected.

Marking-scheme points

  • Molecule described by several canonical structures; real one is the hybrid
  • Resonance energy = extra stability of hybrid over best single structure
  • Example: benzene (Kekule structures), carbonate ion
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ChemistryClass 112 markseasy

Organic Chemistry: Some Basic Principles and Techniques

Define electrophile and nucleophile with two examples each.

Reveal model answer + marking points

An electrophile (electron-loving) is an electron-deficient species that accepts a pair of electrons; examples: H+, NO2+ (also positively charged or neutral electron-deficient species like BF3). A nucleophile (nucleus-loving) is an electron-rich species that donates a pair of electrons; examples: OH-, CN- (also neutral species with lone pairs like NH3 and H2O).

Marking-scheme points

  • Electrophile: electron-deficient, accepts electron pair (H+, NO2+)
  • Nucleophile: electron-rich, donates electron pair (OH-, CN-)
  • Electrophiles are Lewis acids; nucleophiles are Lewis bases
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ChemistryClass 112 marksmedium

Organic Chemistry: Some Basic Principles and Techniques

Distinguish between homolytic and heterolytic bond fission. What species does each produce?

Reveal model answer + marking points

In homolytic fission, a covalent bond breaks so that each bonded atom takes one of the shared electrons, producing neutral free radicals (species with an unpaired electron); it usually occurs in the presence of heat or light in non-polar bonds. In heterolytic fission, the bond breaks so that one atom takes both shared electrons, producing oppositely charged ions (a cation and an anion, e.g. a carbocation and an anion).

Marking-scheme points

  • Homolytic: each atom keeps one electron -> free radicals
  • Heterolytic: one atom keeps both electrons -> ions (cation + anion)
  • Homolysis favoured by heat/light; heterolysis in polar bonds
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ChemistryClass 112 markseasy

Hydrocarbons

Describe the Wurtz reaction and decarboxylation as methods for preparing alkanes.

Reveal model answer + marking points

Wurtz reaction: two molecules of an alkyl halide react with sodium metal in dry ether to give a symmetrical alkane with double the number of carbon atoms, e.g. 2CH3Cl + 2Na -> CH3-CH3 + 2NaCl (ethane). Decarboxylation: the sodium salt of a carboxylic acid is heated with soda lime (NaOH + CaO) to give an alkane with one carbon less, e.g. CH3COONa + NaOH -> CH4 + Na2CO3 (methane).

2CH3Cl + 2Na -> C2H6 + 2NaCl

Marking-scheme points

  • Wurtz: 2 R-X + 2Na (dry ether) -> R-R + 2NaX (symmetrical alkane)
  • Decarboxylation: R-COONa + NaOH (soda lime) -> R-H + Na2CO3
  • Decarboxylation gives alkane with one carbon less
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ChemistryClass 112 marksmedium

Hydrocarbons

State Markovnikov's rule and illustrate it with the addition of HBr to propene.

Reveal model answer + marking points

Markovnikov's rule states that when an unsymmetrical reagent (HX) adds to an unsymmetrical alkene, the negative part of the reagent (X) attaches to the carbon bearing fewer hydrogen atoms, while hydrogen adds to the carbon bearing more hydrogen atoms. So HBr adds to propene (CH3-CH=CH2) to give mainly 2-bromopropane (CH3-CHBr-CH3), because the more stable secondary carbocation is formed. In the presence of peroxide, addition is anti-Markovnikov (peroxide/Kharasch effect), giving 1-bromopropane.

CH3-CH=CH2 + HBr -> CH3-CHBr-CH3

Marking-scheme points

  • Negative part (X) goes to carbon with fewer H atoms
  • HBr + CH3-CH=CH2 -> CH3-CHBr-CH3 (2-bromopropane)
  • Rule follows the more stable carbocation; peroxide reverses it
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ChemistryClass 112 marksmedium

Hydrocarbons

What is ozonolysis? What products are formed when propene undergoes ozonolysis?

Reveal model answer + marking points

Ozonolysis is the reaction of an alkene with ozone (O3) to form an unstable ozonide, which on reductive cleavage (with Zn and water) breaks the carbon-carbon double bond and gives carbonyl compounds (aldehydes and/or ketones). It is used to locate the position of the double bond. Propene (CH3-CH=CH2) on ozonolysis gives ethanal (CH3CHO) and methanal (HCHO).

CH3-CH=CH2 -> CH3CHO + HCHO

Marking-scheme points

  • Alkene + O3 -> ozonide -> (Zn/H2O) carbonyl compounds
  • Double bond is cleaved into two C=O fragments
  • Propene -> ethanal (CH3CHO) + methanal (HCHO)
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ChemistryClass 112 marksmedium

Hydrocarbons

Why does benzene undergo electrophilic substitution rather than addition? Name any two such reactions.

Reveal model answer + marking points

Benzene has a stable, delocalised aromatic pi-electron cloud (6 pi electrons). Addition reactions would destroy this aromatic stability, so benzene prefers substitution, which preserves the aromatic ring. The pi cloud attracts electrophiles, so it undergoes electrophilic substitution. Examples: nitration (with conc. HNO3 + conc. H2SO4, electrophile NO2+) and halogenation (with Cl2/FeCl3); others are sulphonation and Friedel-Crafts alkylation/acylation.

Marking-scheme points

  • Stable delocalised aromatic sextet; addition would destroy aromaticity
  • Substitution preserves the aromatic ring
  • Examples: nitration (NO2+), halogenation, sulphonation, Friedel-Crafts
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MathsClass 112 markseasy

Sets

If A = {1, 2, 3}, write the power set of A and state n(P(A)).

Reveal model answer + marking points

The power set is the set of all subsets of A. P(A) = { {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} }. The number of elements n(P(A)) = 2^n = 2^3 = 8.

n(P(A)) = 2^n

Marking-scheme points

  • Power set = set of all subsets (including empty set and A itself)
  • List all 8 subsets
  • n(P(A)) = 2^3 = 8
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MathsClass 112 markseasy

Sets

In a class of 35 students, 24 like cricket and 16 like football. If each student likes at least one of the two games, how many like both?

Reveal model answer + marking points

Let n(C) = 24, n(F) = 16, n(C union F) = 35 (each likes at least one). By the formula n(C union F) = n(C) + n(F) - n(C intersection F), we get 35 = 24 + 16 - n(C intersection F), so n(C intersection F) = 40 - 35 = 5. Hence 5 students like both games.

n(A union B) = n(A) + n(B) - n(A intersection B)

Marking-scheme points

  • n(C union F) = n(C) + n(F) - n(C intersection F)
  • 35 = 24 + 16 - x
  • x = 5 students like both
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MathsClass 112 markseasy

Sets

Define union, intersection and difference of two sets with an example each.

Reveal model answer + marking points

Union (A union B) is the set of all elements that are in A or in B or in both. Intersection (A intersection B) is the set of elements common to both A and B. Difference (A - B) is the set of elements that are in A but not in B. Example: if A = {1,2,3} and B = {2,3,4}, then A union B = {1,2,3,4}, A intersection B = {2,3}, and A - B = {1}.

Marking-scheme points

  • Union: elements in A or B (or both)
  • Intersection: elements common to A and B
  • Difference A - B: in A but not in B
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MathsClass 112 markseasy

Sets

Write the set {x : x is a real number, -3 <= x < 5} in interval notation, and write (2, 9] in set-builder form.

Reveal model answer + marking points

{x : -3 <= x < 5} includes -3 but not 5, so in interval notation it is [-3, 5). The interval (2, 9] excludes 2 and includes 9, so in set-builder form it is {x : x is a real number, 2 < x <= 9}.

Marking-scheme points

  • Square bracket = endpoint included; round bracket = excluded
  • {x : -3 <= x < 5} = [-3, 5)
  • (2, 9] = {x : 2 < x <= 9}
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MathsClass 112 markseasy

Relations and Functions

If A = {1, 2} and B = {3, 4}, find A x B and n(A x B).

Reveal model answer + marking points

The Cartesian product A x B is the set of all ordered pairs (a, b) with a in A and b in B. A x B = {(1,3), (1,4), (2,3), (2,4)}. The number of elements n(A x B) = n(A) x n(B) = 2 x 2 = 4.

n(A x B) = n(A) x n(B)

Marking-scheme points

  • A x B = set of ordered pairs (a, b)
  • A x B = {(1,3),(1,4),(2,3),(2,4)}
  • n(A x B) = n(A) x n(B) = 4
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MathsClass 112 markseasy

Relations and Functions

If the ordered pairs (x + 1, y - 2) = (3, 1), find the values of x and y.

Reveal model answer + marking points

Two ordered pairs are equal when their corresponding components are equal. So x + 1 = 3 gives x = 2, and y - 2 = 1 gives y = 3. Hence x = 2 and y = 3.

(a, b) = (c, d) => a = c and b = d

Marking-scheme points

  • Equal ordered pairs -> equate corresponding components
  • x + 1 = 3 -> x = 2
  • y - 2 = 1 -> y = 3
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MathsClass 112 markseasy

Relations and Functions

For the relation R = {(1, 2), (2, 4), (3, 6)}, write its domain and range. Define domain and range.

Reveal model answer + marking points

The domain of a relation is the set of all first components (inputs), and the range is the set of all second components (outputs). For R = {(1,2), (2,4), (3,6)}, the domain = {1, 2, 3} and the range = {2, 4, 6}.

Marking-scheme points

  • Domain = set of first components
  • Range = set of second components
  • Domain = {1,2,3}, Range = {2,4,6}
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MathsClass 112 marksmedium

Relations and Functions

Define a one-one (injective) function and an onto (surjective) function.

Reveal model answer + marking points

A function f from A to B is one-one (injective) if different elements of A have different images in B, i.e. f(x1) = f(x2) implies x1 = x2. A function f from A to B is onto (surjective) if every element of B is the image of at least one element of A, i.e. the range of f equals the codomain B.

Marking-scheme points

  • One-one: distinct inputs give distinct outputs
  • f(x1) = f(x2) => x1 = x2
  • Onto: range equals codomain (every output is attained)
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MathsClass 112 markseasy

Trigonometric Functions

Convert 5pi/6 radians into degrees and 240 degrees into radians.

Reveal model answer + marking points

Using pi radians = 180 degrees: 5pi/6 radians = (5/6) x 180 = 150 degrees. And 240 degrees = 240 x (pi/180) = 4pi/3 radians.

radians = degrees x (pi/180)

Marking-scheme points

  • pi radians = 180 degrees
  • 5pi/6 rad = 150 degrees
  • 240 degrees = 4pi/3 rad
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MathsClass 112 marksmedium

Trigonometric Functions

Find the value of sin 75 degrees.

Reveal model answer + marking points

Write 75 = 45 + 30. Using sin(A + B) = sin A cos B + cos A sin B: sin 75 = sin 45 cos 30 + cos 45 sin 30 = (1/sqrt2)(sqrt3/2) + (1/sqrt2)(1/2) = (sqrt3 + 1)/(2 sqrt2) = (sqrt6 + sqrt2)/4.

sin(A + B) = sin A cos B + cos A sin B

Marking-scheme points

  • 75 = 45 + 30
  • sin(A+B) = sinA cosB + cosA sinB
  • sin 75 = (sqrt6 + sqrt2)/4
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MathsClass 112 markseasy

Trigonometric Functions

Prove that 1 + tan^2 x = sec^2 x.

Reveal model answer + marking points

Start from the fundamental identity sin^2 x + cos^2 x = 1. Divide both sides by cos^2 x (cos x not equal to 0): sin^2 x/cos^2 x + cos^2 x/cos^2 x = 1/cos^2 x, which gives tan^2 x + 1 = sec^2 x. Hence 1 + tan^2 x = sec^2 x.

1 + tan^2 x = sec^2 x

Marking-scheme points

  • Start from sin^2 x + cos^2 x = 1
  • Divide both sides by cos^2 x
  • Get tan^2 x + 1 = sec^2 x
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MathsClass 112 marksmedium

Trigonometric Functions

Find the value of sin 15 degrees.

Reveal model answer + marking points

Write 15 = 45 - 30. Using sin(A - B) = sin A cos B - cos A sin B: sin 15 = sin 45 cos 30 - cos 45 sin 30 = (1/sqrt2)(sqrt3/2) - (1/sqrt2)(1/2) = (sqrt3 - 1)/(2 sqrt2) = (sqrt6 - sqrt2)/4.

sin(A - B) = sin A cos B - cos A sin B

Marking-scheme points

  • 15 = 45 - 30
  • sin(A-B) = sinA cosB - cosA sinB
  • sin 15 = (sqrt6 - sqrt2)/4
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