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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.
PhysicsClass 123 marksmedium
Nuclei
State the radioactive decay law. What fraction of a radioactive sample remains after 3 half-lives?
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The radioactive decay law states that the rate of decay of a radioactive sample is directly proportional to the number of undecayed nuclei present at that instant: N = N0 e^(-lambda t), where lambda is the decay constant. The half-life T is related to it by T = 0.693/lambda. After each half-life, half of the sample remains, so after 3 half-lives the fraction remaining = (1/2)^3 = 1/8 of the original sample.
N = N0 (1/2)^(t/T)
Marking-scheme points
- ✓Decay law: N = N0 e^(-lambda t); half-life T = 0.693/lambda
- ✓Fraction after n half-lives = (1/2)^n
- ✓After 3 half-lives: (1/2)^3 = 1/8 remains
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Semiconductor Electronics
Explain how a p-n junction diode works as a half-wave rectifier.
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A rectifier converts alternating current (AC) into direct current (DC). In a half-wave rectifier, a single diode is connected in series with the AC source and a load resistor. During the positive half-cycle of the AC input, the diode is forward biased and conducts, so current flows through the load. During the negative half-cycle, the diode is reverse biased and does not conduct, so no current flows. As a result, output is obtained only during one half of each cycle, giving a pulsating DC. Its efficiency is low because half the input is wasted.
Marking-scheme points
- ✓Rectifier converts AC into DC; uses one diode
- ✓Positive half-cycle: diode forward biased -> conducts
- ✓Negative half-cycle: diode reverse biased -> no output (pulsating DC)
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Semiconductor Electronics
Explain the working of a full-wave rectifier using two diodes.
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A full-wave rectifier converts both halves of the AC input into DC. It uses two diodes with a centre-tapped transformer and a load resistor. During the positive half-cycle, one diode is forward biased and conducts while the other is reverse biased; during the negative half-cycle, the second diode conducts while the first does not. In both half-cycles, the current through the load flows in the same direction, so output is obtained during the whole cycle. This gives a smoother, more efficient pulsating DC than a half-wave rectifier.
Marking-scheme points
- ✓Uses two diodes and a centre-tapped transformer
- ✓Each diode conducts during one half-cycle
- ✓Current through load is in the same direction for both halves (full-wave DC)
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Semiconductor Electronics
Write the truth tables of the OR, AND and NOT logic gates for inputs A and B.
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A logic gate is a digital circuit that gives an output based on its inputs following a logical rule (using binary 0 and 1). OR gate (output Y = A + B): Y = 1 if any input is 1; for inputs (0,0),(0,1),(1,0),(1,1) the outputs are 0,1,1,1. AND gate (Y = A.B): Y = 1 only if both inputs are 1; outputs are 0,0,0,1. NOT gate (Y = not A): it has a single input and inverts it, so input 0 gives output 1 and input 1 gives output 0.
OR: Y = A + B; AND: Y = A.B; NOT: Y = not A
Marking-scheme points
- ✓OR (Y = A + B): output 1 if any input is 1 -> 0,1,1,1
- ✓AND (Y = A.B): output 1 only if both inputs 1 -> 0,0,0,1
- ✓NOT: single input, inverts it (0 -> 1, 1 -> 0)
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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.
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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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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)
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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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Solutions
What are colligative properties? Name the four colligative properties.
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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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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)
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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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Solutions
Calculate the osmotic pressure of a 0.1 M glucose solution at 300 K. (R = 0.0821 L atm K^-1 mol^-1)
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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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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.
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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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Electrochemistry
Write the Nernst equation for a general electrode and explain the terms.
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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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Electrochemistry
State Kohlrausch's law of independent migration of ions and give one application.
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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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Chemical Kinetics
Distinguish between order and molecularity of a reaction.
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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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Chemical Kinetics
Derive the integrated rate equation for a first order reaction.
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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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The p-Block Elements
Describe the Haber process for the manufacture of ammonia.
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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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The p-Block Elements
Describe the Contact process for the manufacture of sulphuric acid.
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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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The d- and f-Block Elements
How is potassium permanganate prepared from pyrolusite? State its oxidising action in acidic medium.
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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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Coordination Compounds
State the main rules for the IUPAC nomenclature of coordination compounds and name [Cu(NH3)4]SO4.
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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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Coordination Compounds
Name the main types of isomerism shown by coordination compounds.
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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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Coordination Compounds
Explain the geometry of a complex using valence bond theory. Distinguish inner and outer orbital complexes.
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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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