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ChemistryClass 112 marksmedium

Equilibrium

State Le Chatelier's principle. What is the effect of increasing pressure on the equilibrium N2(g) + 3H2(g) <=> 2NH3(g)?

Reveal model answer + marking points

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to counteract (reduce) the effect of that change. Increasing the pressure shifts this equilibrium in the forward direction (towards NH3), because the forward reaction reduces the number of gas moles from 4 (1 + 3) to 2, thereby lowering the pressure.

Marking-scheme points

  • System shifts to oppose the imposed change
  • Higher pressure favours the side with fewer gas moles
  • Here forward reaction (4 -> 2 moles) is favoured -> more NH3
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ChemistryClass 112 markseasy

Equilibrium

Define an acid and a base according to the Bronsted-Lowry concept. What is a conjugate acid-base pair?

Reveal model answer + marking points

According to the Bronsted-Lowry concept, an acid is a substance that donates a proton (H+) and a base is a substance that accepts a proton. A conjugate acid-base pair is a pair of species that differ by a single proton; for example, in HCl + H2O -> H3O+ + Cl-, HCl/Cl- and H2O/H3O+ are conjugate acid-base pairs.

acid <=> base + H+

Marking-scheme points

  • Bronsted acid = proton donor; base = proton acceptor
  • Conjugate pair differs by one proton (H+)
  • Example: HCl/Cl- and H2O/H3O+
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ChemistryClass 112 markseasy

Equilibrium

Calculate the pH of a 0.001 M HCl solution.

Reveal model answer + marking points

HCl is a strong acid and dissociates completely, so [H+] = 0.001 M = 1 x 10^-3 M. pH = -log[H+] = -log(10^-3) = 3. The solution is acidic (pH < 7).

pH = -log[H+]

Marking-scheme points

  • Strong acid: [H+] = 10^-3 M
  • pH = -log[H+]
  • pH = 3 (acidic)
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ChemistryClass 112 marksmedium

Equilibrium

What is the ionic product of water (Kw)? State its value at 298 K and its relation with pH.

Reveal model answer + marking points

The ionic product of water Kw is the product of the molar concentrations of hydrogen and hydroxide ions in water: Kw = [H+][OH-]. At 298 K, Kw = 1.0 x 10^-14 mol^2 L^-2. Taking negative logarithm gives pKw = pH + pOH = 14 at 298 K. For pure (neutral) water, [H+] = [OH-] = 10^-7 M, so pH = 7.

Kw = [H+][OH-] = 1.0 x 10^-14

Marking-scheme points

  • Kw = [H+][OH-]
  • Kw = 1.0 x 10^-14 at 298 K
  • pH + pOH = 14; neutral water pH = 7
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ChemistryClass 112 marksmedium

Equilibrium

What is the common ion effect? Explain with a suitable example.

Reveal model answer + marking points

The common ion effect is the suppression of the degree of dissociation (ionisation) of a weak electrolyte by the addition of a strong electrolyte that provides an ion common to the weak electrolyte. For example, adding NH4Cl (which provides NH4+) to a solution of NH4OH suppresses the ionisation of NH4OH, decreasing the OH- concentration. This is used in qualitative analysis and to control pH.

Marking-scheme points

  • Adding a common ion suppresses ionisation of a weak electrolyte
  • Example: NH4Cl added to NH4OH suppresses OH-
  • Application: salt analysis and pH control
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ChemistryClass 112 markseasy

Redox Reactions

Define oxidation and reduction in terms of electron transfer and oxidation number.

Reveal model answer + marking points

Oxidation is the loss of electrons or an increase in oxidation number of an element; reduction is the gain of electrons or a decrease in oxidation number. Both occur simultaneously in a redox reaction. For example, in Zn -> Zn2+ + 2e-, zinc is oxidised (oxidation number 0 -> +2).

Marking-scheme points

  • Oxidation: loss of electrons / increase in oxidation number
  • Reduction: gain of electrons / decrease in oxidation number
  • Oxidation and reduction always occur together
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ChemistryClass 112 markseasy

Redox Reactions

Calculate the oxidation number of manganese in KMnO4 and of chromium in K2Cr2O7.

Reveal model answer + marking points

In KMnO4: K = +1, O = -2 (four O = -8). Let Mn = x. Then +1 + x + (-8) = 0, so x = +7. In K2Cr2O7: 2 K = +2, 7 O = -14. Let each Cr = y. Then +2 + 2y - 14 = 0, so 2y = 12, y = +6. Thus Mn is +7 and Cr is +6.

sum of oxidation numbers = charge on species

Marking-scheme points

  • Sum of oxidation numbers of a neutral compound = 0
  • KMnO4: +1 + x - 8 = 0 -> Mn = +7
  • K2Cr2O7: +2 + 2y - 14 = 0 -> Cr = +6
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ChemistryClass 112 markseasy

Redox Reactions

Define an oxidising agent and a reducing agent with one example each.

Reveal model answer + marking points

An oxidising agent is a substance that oxidises another substance by accepting electrons and is itself reduced; e.g. KMnO4, O2. A reducing agent is a substance that reduces another substance by donating electrons and is itself oxidised; e.g. H2, C (carbon). In a redox reaction the oxidising agent gains electrons and the reducing agent loses electrons.

Marking-scheme points

  • Oxidising agent: accepts electrons, gets reduced (e.g. KMnO4)
  • Reducing agent: donates electrons, gets oxidised (e.g. H2)
  • Oxidising agent oxidises the other species
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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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