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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.
ChemistryClass 123 marksmedium
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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Electrochemistry
State Faraday's two laws of electrolysis.
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Faraday's first law states that the mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte (m is proportional to Q = I t). Faraday's second law states that when the same quantity of charge is passed through different electrolytes, the masses of substances deposited or liberated are directly proportional to their chemical equivalent weights.
m proportional to Q = I t
Marking-scheme points
- ✓First law: mass deposited is proportional to charge passed (m proportional to It)
- ✓Second law: for same charge, mass is proportional to equivalent weight
- ✓One Faraday (96500 C) deposits one gram equivalent
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Electrochemistry
A current of 5 A is passed through a silver nitrate solution for 30 minutes. Calculate the mass of silver deposited. (Atomic mass of Ag = 108, F = 96500 C/mol)
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Charge passed Q = I t = 5 x (30 x 60) = 5 x 1800 = 9000 C. Silver is deposited by Ag+ + e- -> Ag, so 1 mole of electrons (96500 C) deposits 108 g of silver. Mass of Ag = (108/96500) x 9000 = (108 x 9000)/96500 = 972000/96500 = 10.07 g.
mass = (equivalent mass x Q)/96500
Marking-scheme points
- ✓Q = I t = 5 x 1800 = 9000 C
- ✓96500 C deposits 108 g of Ag (Ag+ + e- -> Ag)
- ✓Mass = (108 x 9000)/96500 = 10.07 g
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Electrochemistry
Distinguish between primary and secondary cells with one example each.
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A primary cell is one in which the redox reaction occurs only once and cannot be reversed, so it cannot be recharged and is discarded after use; e.g. the dry cell (Leclanche cell) and the mercury cell. A secondary cell is one that can be recharged by passing current through it in the opposite direction, so it can be used again and again; e.g. the lead storage battery and the nickel-cadmium cell.
Marking-scheme points
- ✓Primary cell: cannot be recharged, used once (dry cell)
- ✓Secondary cell: rechargeable, reusable (lead storage battery)
- ✓Secondary cells are recharged by passing current in reverse
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Chemical Kinetics
Define the rate of a chemical reaction. Distinguish between average and instantaneous rate.
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The rate of a chemical reaction is the change in the concentration of a reactant or product per unit time. The average rate is the change in concentration over a measurable time interval (delta concentration/delta time). The instantaneous rate is the rate of the reaction at a particular instant of time, obtained by making the time interval very small (the derivative d[concentration]/dt). Its units are usually mol L^-1 s^-1.
rate = -d[R]/dt = +d[P]/dt
Marking-scheme points
- ✓Rate = change in concentration per unit time
- ✓Average rate = delta concentration/delta time over an interval
- ✓Instantaneous rate = rate at a particular instant (d[c]/dt)
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Chemical Kinetics
State the factors that affect the rate of a chemical reaction.
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The rate of a chemical reaction is affected by: (1) the nature and concentration of the reactants (rate usually increases with concentration); (2) temperature (rate generally increases with a rise in temperature); (3) the presence of a catalyst (which increases the rate by providing an alternative path of lower activation energy); (4) the surface area of solid reactants (greater surface area gives a faster rate); and (5) for photochemical reactions, the intensity of light.
Marking-scheme points
- ✓Concentration of reactants and their nature
- ✓Temperature (rate increases with temperature)
- ✓Catalyst, surface area and (for photochemical reactions) light
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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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Chemical Kinetics
The rate constant of a first order reaction is 6.93 x 10^-3 s^-1. Calculate its half-life.
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For a first order reaction, the half-life is independent of the initial concentration and is given by t(1/2) = 0.693/k. Substituting k = 6.93 x 10^-3 s^-1: t(1/2) = 0.693/(6.93 x 10^-3) = 100 s. Thus the half-life of the reaction is 100 seconds.
t(1/2) = 0.693/k
Marking-scheme points
- ✓First order half-life t(1/2) = 0.693/k (independent of concentration)
- ✓= 0.693/(6.93 x 10^-3)
- ✓t(1/2) = 100 s
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Chemical Kinetics
Write the Arrhenius equation and define activation energy.
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The Arrhenius equation relates the rate constant to temperature: k = A e^(-Ea/RT), where k is the rate constant, A is the frequency (pre-exponential) factor, Ea is the activation energy, R is the gas constant and T is the absolute temperature. Activation energy (Ea) is the minimum extra energy that the reactant molecules must possess (above their average energy) for a collision to be effective and lead to a reaction. A higher Ea means a slower reaction.
k = A e^(-Ea/RT)
Marking-scheme points
- ✓k = A e^(-Ea/RT)
- ✓Ea = minimum extra energy needed for an effective collision
- ✓Higher Ea -> slower reaction
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Chemical Kinetics
How does a catalyst increase the rate of a reaction?
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A catalyst increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy. As a result, a larger fraction of the reactant molecules have enough energy to cross the (lowered) energy barrier, so more effective collisions occur and the reaction proceeds faster. The catalyst does not change the enthalpy or the equilibrium position of the reaction, and it is regenerated at the end of the reaction.
Marking-scheme points
- ✓Provides an alternative path of lower activation energy
- ✓More molecules can cross the lower energy barrier
- ✓Does not change the equilibrium; is regenerated
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Chemical Kinetics
Write the units of the rate constant for a zero order and a first order reaction.
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The units of the rate constant depend on the order of the reaction. For a zero order reaction, the rate = k, so the units of k are the same as the rate: mol L^-1 s^-1 (or mol L^-1 time^-1). For a first order reaction, rate = k[R], so k has units of s^-1 (or time^-1), which are independent of concentration. In general, the units of k are (mol L^-1)^(1-n) time^-1 for an nth order reaction.
units of k = (mol L^-1)^(1-n) time^-1
Marking-scheme points
- ✓Zero order: units of k are mol L^-1 s^-1
- ✓First order: units of k are s^-1 (time^-1)
- ✓General: (mol L^-1)^(1-n) time^-1 for order n
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The p-Block Elements
Name the elements of Group 15 and give their general valence shell electronic configuration and common oxidation states.
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The Group 15 elements (the nitrogen family) are nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb) and bismuth (Bi). Their general valence shell electronic configuration is ns2 np3 (a half-filled p subshell, which gives extra stability). Their common oxidation states are -3, +3 and +5; the stability of the +5 state decreases and that of the +3 state increases down the group due to the inert pair effect.
ns2 np3
Marking-scheme points
- ✓Group 15: N, P, As, Sb, Bi
- ✓General configuration ns2 np3 (half-filled p)
- ✓Oxidation states -3, +3, +5; +3 more stable down the group
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The p-Block Elements
Why does ammonia act as a Lewis base and have a higher boiling point than phosphine (PH3)?
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Ammonia (NH3) has a lone pair of electrons on the nitrogen atom which it can donate to an electron-deficient species, so it acts as a Lewis base. It has a higher boiling point than phosphine because nitrogen is small and highly electronegative, so NH3 molecules form strong intermolecular hydrogen bonds, whereas PH3 molecules are held only by weak van der Waals forces; more energy is needed to separate the hydrogen-bonded NH3 molecules.
Marking-scheme points
- ✓NH3 has a lone pair on N -> donates it -> Lewis base
- ✓N is small and electronegative -> NH3 forms hydrogen bonds
- ✓PH3 has only weak van der Waals forces -> lower boiling point
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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 Ostwald process for the manufacture of nitric acid.
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Nitric acid is manufactured by the Ostwald process, which uses ammonia as the starting material. The steps are: (1) catalytic oxidation of ammonia, 4NH3 + 5O2 -> 4NO + 6H2O, using a platinum-rhodium catalyst at about 500 K; (2) oxidation of nitric oxide, 2NO + O2 -> 2NO2; and (3) absorption of nitrogen dioxide in water, 3NO2 + H2O -> 2HNO3 + NO, where the NO produced is recycled. The dilute acid is then concentrated by distillation.
3NO2 + H2O -> 2HNO3 + NO
Marking-scheme points
- ✓4NH3 + 5O2 -> 4NO + 6H2O (Pt-Rh catalyst)
- ✓2NO + O2 -> 2NO2
- ✓3NO2 + H2O -> 2HNO3 + NO
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The p-Block Elements
Distinguish between white phosphorus and red phosphorus.
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White phosphorus consists of discrete tetrahedral P4 molecules; it is soft, poisonous, very reactive, glows in the dark (chemiluminescence), catches fire in air spontaneously and is stored under water. Red phosphorus has a polymeric chain structure of linked P4 units; it is comparatively hard, non-poisonous, much less reactive, does not glow in the dark and does not catch fire spontaneously. Red phosphorus is more stable than white phosphorus.
Marking-scheme points
- ✓White P: discrete P4 molecules, poisonous, very reactive, glows, stored under water
- ✓Red P: polymeric, non-poisonous, less reactive, stable
- ✓White P is converted to red P on heating in the absence of air
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The p-Block Elements
How is ozone prepared? Why does it act as a powerful oxidising agent?
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Ozone (O3) is prepared by passing a silent electric discharge through pure, dry oxygen: 3O2 -> 2O3 (the reaction is endothermic). Ozone acts as a powerful oxidising agent because it is unstable and readily decomposes to give nascent oxygen: O3 -> O2 + [O]. This nascent oxygen is very reactive and readily oxidises other substances (for example, it turns moist starch-iodide paper blue by liberating iodine).
3O2 -> 2O3; O3 -> O2 + [O]
Marking-scheme points
- ✓Silent electric discharge through dry O2: 3O2 -> 2O3
- ✓Ozone is unstable and gives nascent oxygen: O3 -> O2 + [O]
- ✓Nascent oxygen makes it a strong oxidising agent
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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 p-Block Elements
Why is fluorine the strongest oxidising agent among the halogens, and why does chlorine act as a bleaching agent?
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Fluorine is the strongest oxidising agent among the halogens because of its low bond dissociation energy (weak F-F bond), small atomic size and high hydration energy of the fluoride ion, which together make it accept electrons most readily. Chlorine acts as a bleaching agent because in the presence of moisture it produces nascent oxygen, Cl2 + H2O -> 2HCl + [O], and this nascent oxygen oxidises the coloured substance to a colourless one. The bleaching action of chlorine is permanent.
Cl2 + H2O -> 2HCl + [O]
Marking-scheme points
- ✓Fluorine: low F-F bond energy, small size, high hydration energy -> strongest oxidiser
- ✓Cl2 + H2O -> 2HCl + [O] (nascent oxygen)
- ✓Nascent oxygen bleaches (oxidises) coloured matter permanently
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