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ChemistryClass 112 marksmedium

Redox Reactions

What is a disproportionation reaction? Give one example.

Reveal model answer + marking points

A disproportionation reaction is a redox reaction in which the same element in a single oxidation state is simultaneously oxidised and reduced (to a higher and a lower oxidation state). Example: 2H2O2 -> 2H2O + O2, where oxygen in the -1 state is both oxidised (to 0 in O2) and reduced (to -2 in H2O). Another example is Cl2 + 2NaOH -> NaCl + NaOCl + H2O.

Marking-scheme points

  • Same element in one oxidation state is both oxidised and reduced
  • Example: 2H2O2 -> 2H2O + O2 (O goes -1 to -2 and 0)
  • Also Cl2 + 2NaOH -> NaCl + NaOCl + H2O
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ChemistryClass 112 marksmedium

Redox Reactions

In the reaction Zn + CuSO4 -> ZnSO4 + Cu, identify the species oxidised, the species reduced, the oxidising agent and the reducing agent.

Reveal model answer + marking points

Zinc goes from 0 to +2 (loses electrons), so Zn is oxidised and acts as the reducing agent. Copper goes from +2 (in CuSO4) to 0 (in Cu) by gaining electrons, so Cu2+ is reduced and CuSO4 acts as the oxidising agent. Thus Zn is the reducing agent and CuSO4 is the oxidising agent.

Zn + Cu2+ -> Zn2+ + Cu

Marking-scheme points

  • Zn: 0 -> +2, oxidised, reducing agent
  • Cu2+: +2 -> 0, reduced, oxidising agent (CuSO4)
  • Electrons transfer from Zn to Cu2+
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ChemistryClass 112 marksmedium

Hydrogen

Why is the position of hydrogen anomalous in the periodic table?

Reveal model answer + marking points

Hydrogen resembles both alkali metals (group 1) and halogens (group 17), so its position is anomalous. Like alkali metals, it has one valence electron (1s1), forms H+ and shows +1 oxidation state. Like halogens, it is one electron short of a noble gas configuration, is diatomic (H2), and can gain an electron to form the hydride ion (H-). Because it fits neither group perfectly, its placement is debated.

Marking-scheme points

  • 1s1: resembles alkali metals (forms H+, +1 state)
  • One electron short of He: resembles halogens (forms H-, diatomic)
  • Fits neither group fully -> anomalous position
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ChemistryClass 112 marksmedium

Hydrogen

Explain why hydrogen peroxide (H2O2) can act both as an oxidising agent and as a reducing agent.

Reveal model answer + marking points

In H2O2 the oxidation number of oxygen is -1, which is intermediate between 0 (in O2) and -2 (in H2O). Therefore it can be reduced to -2 (acting as an oxidising agent) or oxidised to 0 (acting as a reducing agent), depending on the other reactant. For example, it oxidises PbS to PbSO4 (oxidising agent) and reduces acidified KMnO4 (reducing agent).

Marking-scheme points

  • Oxygen in H2O2 is in intermediate -1 state
  • Can be reduced to -2 -> oxidising agent
  • Can be oxidised to 0 -> reducing agent
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ChemistryClass 112 marksmedium

The s-Block Elements

Why are alkali metals strong reducing agents?

Reveal model answer + marking points

Alkali metals have low ionization enthalpies because of their large size and a single loosely held valence electron. They readily lose this electron to form unipositive ions, i.e. they are easily oxidised. Since a substance that is easily oxidised is a good reducing agent, alkali metals are strong reducing agents. Their reducing power generally increases down the group.

M -> M+ + e-

Marking-scheme points

  • Low ionization enthalpy, large size, single valence electron
  • Easily lose electron (easily oxidised)
  • Easily oxidised -> strong reducing agents
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ChemistryClass 112 marksmedium

The s-Block Elements

What is a diagonal relationship? Why do lithium and magnesium show similar properties?

Reveal model answer + marking points

A diagonal relationship is the similarity in properties between an element and the element placed diagonally to its lower right in the periodic table (e.g. Li and Mg, Be and Al, B and Si). Lithium and magnesium resemble each other because they have similar atomic and ionic sizes and nearly the same charge-to-size (polarising power) ratio. For example, both form nitrides directly with nitrogen and both form covalent, water-soluble compounds unlike the rest of their groups.

Marking-scheme points

  • Diagonal similarity: element and one to its lower-right
  • Li-Mg have similar size and charge/size ratio
  • Both form nitrides and show covalent character
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ChemistryClass 112 marksmedium

The s-Block Elements

Why does lithium show anomalous behaviour compared with the other alkali metals?

Reveal model answer + marking points

Lithium differs from the other alkali metals because of its very small atomic and ionic size, high charge density (high polarising power) and the absence of d-orbitals in its valence shell. As a result, its compounds have appreciable covalent character; for example, LiCl is soluble in organic solvents, Li forms a nitride (Li3N) and its carbonate and hydroxide decompose on heating, unlike those of Na and K.

Marking-scheme points

  • Very small size and high polarising power (high charge density)
  • Compounds show covalent character (LiCl soluble in organic solvents)
  • Forms nitride; Li2CO3 and LiOH decompose on heating
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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 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?

Reveal model answer + marking points

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 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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