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ChemistryClass 123 marksmedium

The Solid State

Calculate the number of atoms present per unit cell in simple cubic, body-centred cubic (bcc) and face-centred cubic (fcc) structures.

Reveal model answer + marking points

A corner atom is shared by 8 unit cells (contributes 1/8), a face atom by 2 (contributes 1/2) and a body-centre atom belongs entirely to one cell (contributes 1). Simple cubic: 8 corners x 1/8 = 1 atom. Body-centred cubic (bcc): (8 x 1/8) + 1 (centre) = 2 atoms. Face-centred cubic (fcc): (8 x 1/8) + (6 x 1/2) = 1 + 3 = 4 atoms.

Marking-scheme points

  • Corner atom contributes 1/8, face atom 1/2, body-centre 1
  • Simple cubic: 1 atom; bcc: 2 atoms
  • fcc: 1 + 3 = 4 atoms
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ChemistryClass 123 markshard

The Solid State

An element with atomic mass 56 g/mol has a bcc structure with edge length 288 pm. Calculate its density. (NA = 6.022 x 10^23)

Reveal model answer + marking points

For bcc, Z = 2. Edge a = 288 pm = 2.88 x 10^-8 cm, so a^3 = (2.88 x 10^-8)^3 = 2.39 x 10^-23 cm^3. Density d = Z M/(a^3 NA) = (2 x 56)/(2.39 x 10^-23 x 6.022 x 10^23) = 112/14.39 = 7.79 g/cm^3.

d = Z M/(a^3 NA)

Marking-scheme points

  • d = Z M/(a^3 NA); bcc Z = 2
  • a^3 = (2.88e-8)^3 = 2.39 x 10^-23 cm^3
  • d = 112/14.39 = 7.79 g/cm^3
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ChemistryClass 123 marksmedium

Solutions

What are colligative properties? Name the four colligative properties.

Reveal model answer + marking points

Colligative properties are those properties of a solution that depend only on the number of solute particles present and not on their chemical nature. The four colligative properties are: (1) relative lowering of vapour pressure; (2) elevation of boiling point; (3) depression of freezing point; and (4) osmotic pressure. They are used to determine the molar mass of a solute.

Marking-scheme points

  • Depend only on the number of solute particles, not their nature
  • Relative lowering of vapour pressure and elevation of boiling point
  • Depression of freezing point and osmotic pressure
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ChemistryClass 123 markshard

Solutions

When 3 g of a non-volatile solute is dissolved in 100 g of water, the freezing point is depressed by 0.93 K. Calculate the molar mass of the solute. (Kf for water = 1.86 K kg/mol)

Reveal model answer + marking points

The depression of freezing point is given by delta Tf = Kf x molality = Kf x (w2 x 1000)/(M2 x w1), where w2 = mass of solute, w1 = mass of solvent, M2 = molar mass of solute. So 0.93 = 1.86 x (3 x 1000)/(M2 x 100) = 55.8/M2. Therefore M2 = 55.8/0.93 = 60 g/mol.

delta Tf = Kf x (w2 x 1000)/(M2 x w1)

Marking-scheme points

  • delta Tf = Kf x (w2 x 1000)/(M2 x w1)
  • 0.93 = 1.86 x (3 x 1000)/(M2 x 100) = 55.8/M2
  • M2 = 60 g/mol
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ChemistryClass 123 marksmedium

Solutions

Calculate the osmotic pressure of a 0.1 M glucose solution at 300 K. (R = 0.0821 L atm K^-1 mol^-1)

Reveal model answer + marking points

Osmotic pressure is given by pi = C R T, where C is the molar concentration. Here C = 0.1 mol/L, R = 0.0821 L atm K^-1 mol^-1, T = 300 K. So pi = 0.1 x 0.0821 x 300 = 2.463 atm. Osmotic pressure is a colligative property used to find the molar mass of macromolecules.

pi = C R T

Marking-scheme points

  • pi = C R T
  • = 0.1 x 0.0821 x 300
  • pi = 2.463 atm
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ChemistryClass 123 marksmedium

Electrochemistry

Calculate the standard EMF of a Daniell cell given the standard electrode potentials E(Cu2+/Cu) = +0.34 V and E(Zn2+/Zn) = -0.76 V.

Reveal model answer + marking points

In the Daniell cell, zinc (lower/more negative potential) is the anode and copper is the cathode. The standard EMF = E(cathode) - E(anode) = E(Cu2+/Cu) - E(Zn2+/Zn) = (+0.34) - (-0.76) = 0.34 + 0.76 = 1.10 V. Since the EMF is positive, the cell reaction is spontaneous.

E(cell) = E(cathode) - E(anode)

Marking-scheme points

  • Zn = anode (more negative), Cu = cathode
  • EMF = E(cathode) - E(anode)
  • = 0.34 - (-0.76) = 1.10 V (spontaneous)
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ChemistryClass 123 marksmedium

Electrochemistry

Write the Nernst equation for a general electrode and explain the terms.

Reveal model answer + marking points

The Nernst equation gives the electrode (or cell) potential under non-standard conditions. For the reaction M(n+) + n e- -> M, the electrode potential E = E(standard) - (2.303 RT/nF) log(1/[M(n+)]). At 298 K, substituting the constants, E = E(standard) - (0.0591/n) log(1/[M(n+)]), where E(standard) is the standard electrode potential, n is the number of electrons transferred, F is the Faraday constant, and the term in brackets is the reaction quotient. It shows how potential varies with concentration.

E = E(standard) - (0.0591/n) log Q

Marking-scheme points

  • Gives potential under non-standard conditions
  • E = E(standard) - (0.0591/n) log Q at 298 K
  • n = electrons transferred; depends on ion concentration
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ChemistryClass 123 marksmedium

Electrochemistry

State Kohlrausch's law of independent migration of ions and give one application.

Reveal model answer + marking points

Kohlrausch's law states that at infinite dilution, the molar conductivity of an electrolyte is the sum of the individual contributions of its cations and anions, each migrating independently. Mathematically, the limiting molar conductivity = (number of cations x limiting molar conductivity of cation) + (number of anions x limiting molar conductivity of anion). Applications: it is used to calculate the limiting molar conductivity of weak electrolytes (which cannot be found by extrapolation) and the degree of dissociation of a weak electrolyte.

Marking-scheme points

  • At infinite dilution, molar conductivity = sum of ionic contributions
  • Ions migrate independently
  • Used to find limiting molar conductivity of weak electrolytes and degree of dissociation
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ChemistryClass 123 marksmedium

Chemical Kinetics

Distinguish between order and molecularity of a reaction.

Reveal model answer + marking points

The order of a reaction is the sum of the powers of the concentration terms in the experimentally determined rate law; it can be zero, fractional or a whole number and is an experimental quantity. Molecularity is the number of reacting species (atoms, ions or molecules) that collide simultaneously in an elementary reaction; it is always a whole number (1, 2 or 3) and is a theoretical concept. Order is defined for overall reactions, whereas molecularity is defined only for elementary reactions.

Marking-scheme points

  • Order: sum of powers in the rate law (experimental, can be fractional/zero)
  • Molecularity: number of species in an elementary step (whole number)
  • Order applies to overall reactions; molecularity only to elementary steps
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ChemistryClass 123 marksmedium

Chemical Kinetics

Derive the integrated rate equation for a first order reaction.

Reveal model answer + marking points

For a first order reaction R -> P, the rate = -d[R]/dt = k[R]. Rearranging, d[R]/[R] = -k dt. Integrating between limits [R0] at t = 0 and [R] at time t: ln([R]/[R0]) = -k t, so [R] = [R0] e^(-k t). Converting to base 10 logarithms, k = (2.303/t) log([R0]/[R]). This shows that for a first order reaction, a plot of log[R] against t is a straight line.

k = (2.303/t) log([R0]/[R])

Marking-scheme points

  • Rate = k[R]; integrate d[R]/[R] = -k dt
  • ln([R]/[R0]) = -k t
  • k = (2.303/t) log([R0]/[R])
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ChemistryClass 123 marksmedium

The p-Block Elements

Describe the Haber process for the manufacture of ammonia.

Reveal model answer + marking points

Ammonia is manufactured industrially by the Haber process, in which nitrogen and hydrogen combine directly: N2(g) + 3H2(g) <=> 2NH3(g), and the forward reaction is exothermic and proceeds with a decrease in the number of moles. According to Le Chatelier's principle, the optimum conditions are a high pressure (about 200 atmospheres), a moderately low temperature (about 700 K), and a catalyst of finely divided iron with molybdenum (or K2O and Al2O3) as a promoter. The ammonia formed is removed by liquefaction to shift the equilibrium forward.

N2 + 3H2 <=> 2NH3

Marking-scheme points

  • N2 + 3H2 <=> 2NH3 (exothermic, fewer moles on product side)
  • Optimum: about 200 atm pressure, about 700 K temperature
  • Catalyst: finely divided iron with a promoter (e.g. molybdenum)
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ChemistryClass 123 marksmedium

The p-Block Elements

Describe the Contact process for the manufacture of sulphuric acid.

Reveal model answer + marking points

Sulphuric acid is manufactured by the Contact process. The steps are: (1) sulphur or sulphide ore is burnt to form sulphur dioxide, S + O2 -> SO2; (2) sulphur dioxide is catalytically oxidised to sulphur trioxide, 2SO2 + O2 <=> 2SO3, using vanadium pentoxide (V2O5) as catalyst at about 720 K and about 2 atm; (3) sulphur trioxide is absorbed in concentrated sulphuric acid to form oleum, SO3 + H2SO4 -> H2S2O7; and (4) the oleum is diluted with water to give sulphuric acid, H2S2O7 + H2O -> 2H2SO4.

2SO2 + O2 -> 2SO3 (V2O5)

Marking-scheme points

  • S + O2 -> SO2
  • 2SO2 + O2 <=> 2SO3 (V2O5 catalyst)
  • SO3 + H2SO4 -> oleum (H2S2O7); then diluted to H2SO4
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ChemistryClass 123 marksmedium

The d- and f-Block Elements

How is potassium permanganate prepared from pyrolusite? State its oxidising action in acidic medium.

Reveal model answer + marking points

Potassium permanganate (KMnO4) is prepared from the mineral pyrolusite (MnO2). MnO2 is fused with potassium hydroxide (KOH) in the presence of air or an oxidising agent like KNO3 to give green potassium manganate (K2MnO4): 2MnO2 + 4KOH + O2 -> 2K2MnO4 + 2H2O. The manganate is then oxidised (electrolytically or by chlorine/ozone) to purple permanganate: 2K2MnO4 + Cl2 -> 2KMnO4 + 2KCl. In acidic medium KMnO4 is a strong oxidising agent: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O (Mn goes from +7 to +2).

MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

Marking-scheme points

  • 2MnO2 + 4KOH + O2 -> 2K2MnO4 (green manganate)
  • Manganate oxidised to KMnO4 (purple permanganate)
  • Acidic oxidation: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
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ChemistryClass 123 marksmedium

Coordination Compounds

State the main rules for the IUPAC nomenclature of coordination compounds and name [Cu(NH3)4]SO4.

Reveal model answer + marking points

Main rules: (1) the cation is named before the anion; (2) within the complex, the ligands are named in alphabetical order before the central metal; (3) the number of ligands is shown by prefixes di, tri, tetra, etc.; (4) the oxidation state of the metal is written in Roman numerals in brackets; and (5) if the complex ion is an anion, the metal name ends in -ate. For example, [Cu(NH3)4]SO4 is named tetraamminecopper(II) sulphate (copper is in the +2 state).

Marking-scheme points

  • Cation named first; ligands named alphabetically before the metal
  • Number of ligands by di, tri, tetra; oxidation state in Roman numerals
  • [Cu(NH3)4]SO4 = tetraamminecopper(II) sulphate
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ChemistryClass 123 marksmedium

Coordination Compounds

Name the main types of isomerism shown by coordination compounds.

Reveal model answer + marking points

Coordination compounds show two broad types of isomerism. Structural (constitutional) isomerism includes: (1) ionisation isomerism (different ions in solution, e.g. [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br); (2) linkage isomerism (ambidentate ligand attached through different atoms, e.g. -NO2 vs -ONO); (3) coordination isomerism; and (4) hydrate (solvate) isomerism. Stereoisomerism includes: (1) geometrical isomerism (cis and trans forms) and (2) optical isomerism (non-superimposable mirror images).

Marking-scheme points

  • Structural: ionisation, linkage, coordination, hydrate isomerism
  • Stereoisomerism: geometrical (cis-trans) and optical
  • Linkage isomerism needs an ambidentate ligand (e.g. NO2)
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ChemistryClass 123 markshard

Coordination Compounds

Explain the geometry of a complex using valence bond theory. Distinguish inner and outer orbital complexes.

Reveal model answer + marking points

According to valence bond theory, the central metal ion provides a number of empty hybrid orbitals equal to its coordination number, and each ligand donates a lone pair into these orbitals to form coordinate bonds; the type of hybridisation decides the geometry (e.g. sp3 tetrahedral, dsp2 square planar, sp3d2 or d2sp3 octahedral). An inner orbital complex uses inner (n-1)d orbitals for hybridisation (d2sp3), usually formed with strong field ligands and often low-spin. An outer orbital complex uses the outer nd orbitals (sp3d2), usually formed with weak field ligands and is high-spin.

Marking-scheme points

  • Metal provides empty hybrid orbitals; ligands donate lone pairs
  • Hybridisation decides geometry (sp3, dsp2, d2sp3, sp3d2)
  • Inner orbital (d2sp3, low-spin) vs outer orbital (sp3d2, high-spin)
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ChemistryClass 123 markshard

Coordination Compounds

Explain the splitting of d orbitals in an octahedral crystal field according to crystal field theory.

Reveal model answer + marking points

According to crystal field theory, the bonding between the metal and the ligands is purely electrostatic. In an isolated metal ion the five d orbitals have the same energy (degenerate). When six ligands approach along the axes in an octahedral field, they repel the d orbitals unequally: the two orbitals pointing along the axes (dx2-y2 and dz2, called eg) are repelled more and rise in energy, while the three orbitals pointing between the axes (dxy, dyz, dzx, called t2g) are repelled less and are lowered in energy. This energy gap between t2g and eg is the crystal field splitting energy (delta o).

Marking-scheme points

  • Metal-ligand bonding treated as electrostatic
  • Six ligands approach along the axes (octahedral)
  • d orbitals split into lower t2g and higher eg; gap = delta o
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ChemistryClass 123 marksmedium

Haloalkanes and Haloarenes

How are haloalkanes prepared from alcohols? Write the reactions.

Reveal model answer + marking points

Haloalkanes are commonly prepared from alcohols by replacing the -OH group with a halogen. (1) With hydrogen halides: R-OH + HX -> R-X + H2O (reactivity of alcohols is 3 degrees > 2 degrees > 1 degree). (2) With phosphorus halides: 3R-OH + PCl3 -> 3R-Cl + H3PO3, and R-OH + PCl5 -> R-Cl + POCl3 + HCl. (3) With thionyl chloride (the best method, as the by-products are gases): R-OH + SOCl2 -> R-Cl + SO2 + HCl.

R-OH + SOCl2 -> R-Cl + SO2 + HCl

Marking-scheme points

  • R-OH + HX -> R-X + H2O
  • 3R-OH + PCl3 -> 3R-Cl + H3PO3
  • R-OH + SOCl2 -> R-Cl + SO2 + HCl (best method)
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ChemistryClass 123 markshard

Haloalkanes and Haloarenes

Distinguish between the SN1 and SN2 mechanisms of nucleophilic substitution.

Reveal model answer + marking points

In the SN2 (substitution nucleophilic bimolecular) mechanism, the reaction occurs in a single step; the nucleophile attacks the carbon from the side opposite the leaving group, and bond breaking and bond making occur simultaneously. Its rate depends on both the substrate and the nucleophile, and it gives inversion of configuration (Walden inversion); it is favoured by primary halides. In the SN1 (substitution nucleophilic unimolecular) mechanism, the reaction occurs in two steps through a carbocation intermediate; its rate depends only on the substrate, and it usually gives a racemic mixture. It is favoured by tertiary halides and polar protic solvents.

Marking-scheme points

  • SN2: one step, backside attack, inversion; rate depends on both reactants; favoured by primary halides
  • SN1: two steps via carbocation; rate depends only on substrate; racemisation
  • SN1 favoured by tertiary halides and polar protic solvents
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ChemistryClass 123 marksmedium

Alcohols, Phenols and Ethers

How are alcohols prepared by the hydration of alkenes and by the reduction of aldehydes and ketones?

Reveal model answer + marking points

(1) Acid-catalysed hydration of alkenes: an alkene adds water in the presence of dilute sulphuric acid following Markovnikov's rule to give an alcohol, e.g. CH2=CH2 + H2O -> CH3CH2OH. (2) Reduction of carbonyl compounds: aldehydes on reduction (with H2/Ni or NaBH4 or LiAlH4) give primary alcohols (R-CHO -> R-CH2OH), and ketones on reduction give secondary alcohols (R-CO-R' -> R-CH(OH)-R'). Alcohols can also be prepared from Grignard reagents reacting with carbonyl compounds.

R-CHO + 2[H] -> R-CH2OH

Marking-scheme points

  • Hydration of alkenes (Markovnikov): CH2=CH2 + H2O -> CH3CH2OH
  • Reduction of aldehyde -> primary alcohol
  • Reduction of ketone -> secondary alcohol
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