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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.

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ChemistryClass 122 markseasy

Electrochemistry

What is a galvanic (voltaic) cell? Name its two electrodes and the reaction at each.

Reveal model answer + marking points

A galvanic cell is an electrochemical device that converts chemical energy into electrical energy through a spontaneous redox reaction (e.g. the Daniell cell). It has two electrodes: the anode, where oxidation (loss of electrons) takes place and which is the negative terminal; and the cathode, where reduction (gain of electrons) takes place and which is the positive terminal. A salt bridge connects the two half-cells and maintains electrical neutrality.

Marking-scheme points

  • Converts chemical energy into electrical energy (spontaneous redox)
  • Anode: oxidation, negative terminal
  • Cathode: reduction, positive terminal; salt bridge maintains neutrality
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ChemistryClass 122 marksmedium

Electrochemistry

What is standard electrode potential? What is taken as the reference electrode?

Reveal model answer + marking points

The standard electrode potential is the potential difference developed between an electrode and its solution of unit concentration (1 M) at 298 K and 1 bar pressure, measured with respect to a reference electrode. The standard hydrogen electrode (SHE) is taken as the reference electrode and is assigned a potential of exactly zero volt. Electrodes are compared with the SHE to obtain their standard potentials.

Marking-scheme points

  • Electrode potential under standard conditions (1 M, 298 K, 1 bar)
  • Reference: standard hydrogen electrode (SHE)
  • SHE assigned a potential of zero volt
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ChemistryClass 122 marksmedium

Electrochemistry

Write the relation between the standard Gibbs energy change and the EMF of a cell.

Reveal model answer + marking points

The standard Gibbs energy change of a cell reaction is related to the standard EMF by delta G(standard) = -n F E(standard), where n is the number of electrons transferred, F is the Faraday constant (96500 C/mol) and E(standard) is the standard cell EMF. A positive EMF gives a negative delta G, meaning the cell reaction is spontaneous. This relation also links electrochemistry with thermodynamics.

delta G(standard) = -n F E(standard)

Marking-scheme points

  • delta G(standard) = -n F E(standard)
  • n = electrons transferred; F = 96500 C/mol
  • Positive EMF gives negative delta G (spontaneous)
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ChemistryClass 122 marksmedium

Electrochemistry

State Faraday's two laws of electrolysis.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

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)

Reveal model answer + marking points

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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ChemistryClass 122 markseasy

Electrochemistry

Distinguish between primary and secondary cells with one example each.

Reveal model answer + marking points

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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ChemistryClass 122 markseasy

Chemical Kinetics

Define the rate of a chemical reaction. Distinguish between average and instantaneous rate.

Reveal model answer + marking points

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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ChemistryClass 122 markseasy

Chemical Kinetics

State the factors that affect the rate of a chemical reaction.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Chemical Kinetics

The rate constant of a first order reaction is 6.93 x 10^-3 s^-1. Calculate its half-life.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Chemical Kinetics

Write the Arrhenius equation and define activation energy.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Chemical Kinetics

How does a catalyst increase the rate of a reaction?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Chemical Kinetics

Write the units of the rate constant for a zero order and a first order reaction.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

The p-Block Elements

Name the elements of Group 15 and give their general valence shell electronic configuration and common oxidation states.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

The p-Block Elements

Why does ammonia act as a Lewis base and have a higher boiling point than phosphine (PH3)?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

The p-Block Elements

Describe the Ostwald process for the manufacture of nitric acid.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

The p-Block Elements

Distinguish between white phosphorus and red phosphorus.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

The p-Block Elements

How is ozone prepared? Why does it act as a powerful oxidising agent?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

The p-Block Elements

Why is fluorine the strongest oxidising agent among the halogens, and why does chlorine act as a bleaching agent?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

The p-Block Elements

Why are noble gases chemically inert? Name two compounds of xenon.

Reveal model answer + marking points

Noble gases (Group 18) are chemically inert because they have completely filled valence shells (ns2 np6, except helium which is 1s2), which is a very stable electronic configuration. As a result they have very high ionisation enthalpies and almost zero electron gain enthalpy, so they have little tendency to gain, lose or share electrons. Xenon, being the largest and most easily ionised, does form some compounds, for example xenon difluoride (XeF2) and xenon tetrafluoride (XeF4).

Marking-scheme points

  • Completely filled valence shell (ns2 np6) -> very stable
  • Very high ionisation enthalpy, no tendency to react
  • Xenon compounds: XeF2 and XeF4
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ChemistryClass 122 markseasy

The d- and f-Block Elements

What are transition elements? Why are they called so?

Reveal model answer + marking points

Transition elements are the d-block elements whose atoms or stable ions have partially filled d orbitals (their general configuration is (n-1)d(1-10) ns(1-2)). They are called transition elements because they are placed between the s-block metals and the p-block non-metals in the periodic table and show a gradual transition of properties from the highly electropositive s-block metals to the less electropositive p-block elements.

(n-1)d(1-10) ns(1-2)

Marking-scheme points

  • d-block elements with partially filled d orbitals in atoms/ions
  • General configuration (n-1)d(1-10) ns(1-2)
  • Lie between s-block and p-block, showing transitional properties
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