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800 questions

ChemistryClass 112 marksmedium

The p-Block Elements

What is the inert pair effect? Illustrate with an example from group 14.

Reveal model answer + marking points

The inert pair effect is the reluctance of the two outermost s-electrons (ns2) to take part in bonding, which becomes more pronounced down a group. As a result the lower oxidation state (2 less than the group valency) becomes more stable for heavier elements. For example, in group 14 the +2 state becomes more stable than +4 down the group, so Pb2+ is more stable than Pb4+, while carbon and silicon prefer +4.

Marking-scheme points

  • ns2 electron pair resists participating in bonding
  • Effect increases down a group
  • Group 14: Pb2+ more stable than Pb4+
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ChemistryClass 112 marksmedium

The p-Block Elements

Why does boron trifluoride (BF3) behave as a Lewis acid?

Reveal model answer + marking points

In BF3, boron has only six electrons in its valence shell after forming three B-F bonds, so it has an incomplete octet and an empty p-orbital. This makes BF3 electron-deficient, so it can accept a lone pair of electrons from a donor (Lewis base) such as NH3 to complete its octet (forming F3B<-NH3). A species that accepts an electron pair is a Lewis acid, hence BF3 is a Lewis acid.

BF3 + :NH3 -> F3B-NH3

Marking-scheme points

  • Boron has incomplete octet (only 6 electrons) and empty p-orbital
  • Electron deficient -> accepts a lone pair
  • Electron-pair acceptor = Lewis acid
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ChemistryClass 113 markshard

The p-Block Elements

Describe the structure and bonding in diborane (B2H6).

Reveal model answer + marking points

Diborane (B2H6) has two boron atoms and six hydrogen atoms. Four hydrogen atoms (two on each boron) are terminal and are bonded by normal two-centre two-electron (2c-2e) B-H bonds. The remaining two hydrogen atoms are bridging: each forms a three-centre two-electron (3c-2e) B-H-B bond, often called a banana bond. Each boron is sp3 hybridised. The molecule is electron-deficient because it does not have enough valence electrons for normal two-electron bonds throughout.

B2H6 (2 bridging + 4 terminal H)

Marking-scheme points

  • 4 terminal B-H bonds (normal 2c-2e bonds)
  • 2 bridging B-H-B bonds are 3c-2e (banana) bonds
  • Boron is sp3; molecule is electron-deficient
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ChemistryClass 112 marksmedium

The p-Block Elements

What is catenation? Why does carbon show it to a maximum extent? Name three allotropes of carbon.

Reveal model answer + marking points

Catenation is the self-linking of atoms of the same element to form long chains or rings. Carbon shows catenation to the maximum extent because the C-C bond is very strong (high bond energy) due to the small size of the carbon atom, allowing effective overlap. Three crystalline allotropes of carbon are diamond, graphite and fullerene (C60).

Marking-scheme points

  • Catenation = self-linking of like atoms into chains/rings
  • Carbon: small size gives very strong C-C bonds
  • Allotropes: diamond, graphite, fullerene
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ChemistryClass 112 marksmedium

The p-Block Elements

Why is diamond very hard while graphite is soft and a good conductor of electricity?

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In diamond each carbon is sp3 hybridised and bonded to four others in a rigid three-dimensional tetrahedral network, so it is extremely hard and a non-conductor. In graphite each carbon is sp2 hybridised and bonded to three others in flat hexagonal layers held together by weak van der Waals forces, so the layers slide easily (soft/lubricating). The fourth (unhybridised) electron of each carbon is delocalised over the layer, allowing graphite to conduct electricity.

Marking-scheme points

  • Diamond: sp3, rigid 3D tetrahedral network -> hard, non-conductor
  • Graphite: sp2 layers with weak forces between them -> soft
  • Delocalised fourth electron in graphite -> conducts electricity
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ChemistryClass 112 markseasy

Organic Chemistry: Some Basic Principles and Techniques

Give the IUPAC names of (CH3)2CHCH2CH3 and CH3COOH.

Reveal model answer + marking points

(CH3)2CHCH2CH3 has a four-carbon main chain (butane) with a methyl group on the second carbon, so its IUPAC name is 2-methylbutane. CH3COOH is a two-carbon carboxylic acid, so its IUPAC name is ethanoic acid (common name acetic acid).

Marking-scheme points

  • (CH3)2CHCH2CH3: longest chain butane, methyl at C-2
  • IUPAC name = 2-methylbutane
  • CH3COOH = ethanoic acid
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ChemistryClass 113 marksmedium

Organic Chemistry: Some Basic Principles and Techniques

What is structural isomerism? Name and briefly explain any three types.

Reveal model answer + marking points

Structural (constitutional) isomerism occurs when compounds have the same molecular formula but different arrangements of atoms. Three types: (1) Chain isomerism - different carbon skeletons, e.g. n-butane and isobutane (C4H10). (2) Position isomerism - same skeleton but a substituent or functional group in different positions, e.g. 1-propanol and 2-propanol. (3) Functional isomerism - same molecular formula but different functional groups, e.g. ethanol (alcohol) and dimethyl ether (C2H6O).

Marking-scheme points

  • Same molecular formula, different arrangement of atoms
  • Chain: different carbon skeleton (n-butane/isobutane)
  • Position and functional isomerism with examples
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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 113 marksmedium

Organic Chemistry: Some Basic Principles and Techniques

What is hyperconjugation? How does it explain the stability of carbocations?

Reveal model answer + marking points

Hyperconjugation is the delocalisation of the sigma electrons of a C-H bond (adjacent to a positively charged carbon or a double bond) into the empty p-orbital or pi-system; it is also called no-bond resonance. In carbocations, the more alkyl groups attached to the positive carbon, the more C-H bonds are available for hyperconjugation, so more delocalisation of charge occurs. Hence stability order is tertiary > secondary > primary > methyl carbocation.

Marking-scheme points

  • Delocalisation of adjacent C-H sigma electrons (no-bond resonance)
  • More alpha C-H bonds -> more hyperconjugation -> more stable
  • Carbocation stability: 3 deg > 2 deg > 1 deg > methyl
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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 113 marksmedium

Organic Chemistry: Some Basic Principles and Techniques

Briefly describe crystallisation, simple distillation and steam distillation as methods of purifying organic compounds.

Reveal model answer + marking points

Crystallisation: used to purify solids; the impure solid is dissolved in a suitable hot solvent, filtered, and cooled so that the pure compound crystallises out while soluble impurities remain in solution. Simple distillation: used to separate a volatile liquid from a non-volatile impurity or two liquids with a large difference in boiling points; the liquid is boiled and the vapour is condensed and collected. Steam distillation: used to purify liquids that are steam-volatile and immiscible with water; steam is passed through the mixture so the compound distils over below its normal boiling point (e.g. aniline).

Marking-scheme points

  • Crystallisation: dissolve in hot solvent, cool -> pure crystals
  • Simple distillation: separate liquids differing widely in boiling point
  • Steam distillation: for steam-volatile, water-immiscible liquids (e.g. aniline)
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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 113 marksmedium

Hydrocarbons

How is ethene prepared by dehydration of ethanol? How can you test for unsaturation in an alkene?

Reveal model answer + marking points

Ethene is prepared by heating ethanol with concentrated sulphuric acid at about 443 K (170 deg C), which removes a molecule of water (dehydration): CH3CH2OH -> CH2=CH2 + H2O. Test for unsaturation: (1) alkenes decolourise reddish-brown bromine water (Br2/H2O). (2) alkenes decolourise cold dilute alkaline KMnO4 (Baeyer's reagent), turning the purple colour colourless and forming a diol. These tests confirm a carbon-carbon double bond.

CH3CH2OH -> CH2=CH2 + H2O

Marking-scheme points

  • Ethanol + conc. H2SO4 at 443 K -> CH2=CH2 + H2O
  • Bromine water test: alkene decolourises reddish-brown Br2 water
  • Baeyer's test: alkene decolourises cold dilute alkaline KMnO4
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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 113 marksmedium

Hydrocarbons

State Huckel's rule of aromaticity. Why is benzene aromatic?

Reveal model answer + marking points

Huckel's rule states that a planar, cyclic, fully conjugated ring is aromatic if it contains (4n + 2) pi electrons, where n = 0, 1, 2, 3... Benzene is aromatic because it is planar and cyclic, every carbon is sp2 hybridised with continuous conjugation (a delocalised pi system), and it has 6 pi electrons, which fits (4n + 2) with n = 1. This delocalisation gives benzene extra stability (resonance/aromatic stabilisation).

(4n + 2) pi electrons

Marking-scheme points

  • Aromatic: planar, cyclic, conjugated with (4n + 2) pi electrons
  • Benzene: planar, sp2, fully conjugated ring
  • 6 pi electrons (n = 1) -> aromatic and extra stable
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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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ChemistryClass 113 marksmedium

Hydrocarbons

How would you chemically distinguish between ethane, ethene and ethyne?

Reveal model answer + marking points

Ethane (alkane) does not react with bromine water or Baeyer's reagent (no decolourisation) as it is saturated. Ethene (alkene) decolourises both bromine water and cold dilute alkaline KMnO4 (Baeyer's reagent) but gives no precipitate with ammoniacal silver nitrate. Ethyne (terminal alkyne) also decolourises bromine water and Baeyer's reagent, and in addition forms a white precipitate of silver acetylide with ammoniacal silver nitrate (and a red precipitate with ammoniacal cuprous chloride), because of its acidic terminal hydrogen.

Marking-scheme points

  • Ethane: no reaction with bromine water / Baeyer's (saturated)
  • Ethene: decolourises bromine water and Baeyer's reagent
  • Ethyne: also gives white precipitate with ammoniacal AgNO3 (terminal C-H)
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