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The questions your board exam loves to ask
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 112 marksmedium
Chemical Bonding and Molecular Structure
Why does water (H2O) have a much higher boiling point than hydrogen sulphide (H2S)?
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Oxygen is much more electronegative and smaller than sulphur, so water molecules form strong intermolecular hydrogen bonds, whereas H2S molecules are held only by weak van der Waals (dipole) forces. Extra energy is needed to break the hydrogen bonds in water, so water has a much higher boiling point than H2S even though H2S has a higher molar mass.
Marking-scheme points
- ✓Water forms strong intermolecular H-bonds (O is small, electronegative)
- ✓H2S has only weak van der Waals forces
- ✓Breaking H-bonds needs more energy -> higher boiling point
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Chemical Bonding and Molecular Structure
Distinguish between a sigma bond and a pi bond. Which is stronger and why?
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A sigma bond is formed by the head-on (axial) overlap of orbitals along the internuclear axis, while a pi bond is formed by the sidewise (lateral) overlap of parallel p orbitals. A sigma bond is stronger because axial overlap is more effective and greater, giving a larger region of electron density between the nuclei; pi bonds have smaller lateral overlap and are weaker and more reactive.
Marking-scheme points
- ✓Sigma: head-on/axial overlap; pi: sidewise overlap of p orbitals
- ✓Sigma has greater, more effective overlap
- ✓Sigma bond is stronger than pi bond
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States of Matter
State Boyle's law and Charles's law with their mathematical expressions.
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Boyle's law: at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure, i.e. V is proportional to 1/P, so PV = constant. Charles's law: at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute (Kelvin) temperature, i.e. V is proportional to T, so V/T = constant.
PV = constant (Boyle); V/T = constant (Charles)
Marking-scheme points
- ✓Boyle: V proportional to 1/P at constant T -> PV = constant
- ✓Charles: V proportional to T at constant P -> V/T = constant
- ✓Temperature must be in Kelvin
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States of Matter
Write the ideal gas equation and give the value of the gas constant R in two units.
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The ideal gas equation is PV = nRT, where P = pressure, V = volume, n = number of moles, T = absolute temperature and R = universal gas constant. R = 0.0821 L atm K^-1 mol^-1 = 8.314 J K^-1 mol^-1.
PV = nRT
Marking-scheme points
- ✓PV = nRT
- ✓R = 0.0821 L atm K^-1 mol^-1
- ✓R = 8.314 J K^-1 mol^-1
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States of Matter
Write the van der Waals equation for n moles of a real gas and explain the significance of the constants a and b.
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The van der Waals equation is (P + a n^2/V^2)(V - nb) = nRT. The constant 'a' corrects for the intermolecular forces of attraction (it accounts for the pressure being lower than ideal), and the constant 'b' corrects for the finite volume actually occupied by the gas molecules (excluded volume). Real gases deviate from ideal behaviour at high pressure and low temperature.
(P + a n^2/V^2)(V - nb) = nRT
Marking-scheme points
- ✓(P + a n^2/V^2)(V - nb) = nRT
- ✓a: correction for intermolecular attraction
- ✓b: correction for finite molecular volume
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Thermodynamics
Define system and surroundings. Name the three types of thermodynamic systems.
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A system is the specified part of the universe under study; the surroundings are the rest of the universe outside the system that can interact with it. The three types are: open system (exchanges both matter and energy with surroundings), closed system (exchanges only energy, not matter) and isolated system (exchanges neither matter nor energy).
Marking-scheme points
- ✓System = part under study; surroundings = rest of universe
- ✓Open: exchanges matter and energy
- ✓Closed: only energy; Isolated: neither
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Thermodynamics
State the first law of thermodynamics and give its mathematical expression with sign convention.
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The first law of thermodynamics states that energy can neither be created nor destroyed, only transformed from one form to another; the total energy of an isolated system remains constant. Mathematically, delta U = q + w, where delta U is the change in internal energy, q is the heat added to the system (positive when absorbed) and w is the work done on the system (positive when done on the system).
delta U = q + w
Marking-scheme points
- ✓Energy is conserved (cannot be created or destroyed)
- ✓delta U = q + w
- ✓q positive if heat absorbed; w positive if work done on system
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Thermodynamics
Define enthalpy. Derive the relation between delta H and delta U for a reaction involving gases.
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Enthalpy (H) is the total heat content of a system at constant pressure, defined as H = U + PV. For a reaction at constant pressure and temperature, delta H = delta U + P delta V. For ideal gases, P delta V = delta ng RT, where delta ng is the change in the number of moles of gaseous species. Hence delta H = delta U + delta ng RT.
delta H = delta U + delta ng RT
Marking-scheme points
- ✓H = U + PV (heat content at constant pressure)
- ✓delta H = delta U + P delta V
- ✓For gases: delta H = delta U + delta ng RT
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Thermodynamics
State Hess's law of constant heat summation and mention one of its applications.
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Hess's law states that the total enthalpy change of a reaction is the same whether the reaction takes place in one step or in several steps, provided the initial and final conditions are the same. It follows from the fact that enthalpy is a state function. Applications: it is used to calculate enthalpies of formation, bond enthalpies and reaction enthalpies that cannot be measured directly.
Marking-scheme points
- ✓Total enthalpy change is path independent
- ✓Consequence of enthalpy being a state function
- ✓Used to find delta H that cannot be measured directly
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Thermodynamics
What is entropy? Predict the sign of delta S when ice melts into water.
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Entropy (S) is a thermodynamic state function that measures the degree of randomness or disorder of a system. When ice (a highly ordered solid) melts into water (a more disordered liquid), disorder increases, so the entropy increases and delta S is positive.
delta S = q(rev) / T
Marking-scheme points
- ✓Entropy = measure of randomness/disorder
- ✓Melting increases disorder (solid -> liquid)
- ✓delta S is positive for melting of ice
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Thermodynamics
Define standard enthalpy of formation and standard enthalpy of combustion.
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Standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states (at 298 K and 1 bar); e.g. for CO2 it is -393.5 kJ/mol. Standard enthalpy of combustion is the enthalpy change when one mole of a substance is completely burnt in excess oxygen under standard conditions; it is always negative (exothermic).
Marking-scheme points
- ✓Formation: 1 mol compound from elements in standard states
- ✓Combustion: 1 mol substance completely burnt in oxygen
- ✓Enthalpy of combustion is always negative
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Equilibrium
What is a reversible reaction? State two characteristics of chemical equilibrium.
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A reversible reaction is one that proceeds in both forward and backward directions under the same conditions. Characteristics of chemical equilibrium: (1) it is dynamic in nature, i.e. the forward and backward reactions continue at equal rates; (2) the observable properties (concentration, pressure, colour) remain constant with time; (3) it can be attained from either direction and is disturbed by changing conditions.
rate(forward) = rate(backward)
Marking-scheme points
- ✓Reversible: proceeds in both directions
- ✓Equilibrium is dynamic: forward rate = backward rate
- ✓Measurable properties stay constant with time
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Equilibrium
State Le Chatelier's principle. What is the effect of increasing pressure on the equilibrium N2(g) + 3H2(g) <=> 2NH3(g)?
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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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Equilibrium
Define an acid and a base according to the Bronsted-Lowry concept. What is a conjugate acid-base pair?
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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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Equilibrium
Calculate the pH of a 0.001 M HCl solution.
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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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Equilibrium
What is the ionic product of water (Kw)? State its value at 298 K and its relation with pH.
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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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Equilibrium
What is the common ion effect? Explain with a suitable example.
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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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Redox Reactions
Define oxidation and reduction in terms of electron transfer and oxidation number.
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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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Redox Reactions
Calculate the oxidation number of manganese in KMnO4 and of chromium in K2Cr2O7.
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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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Redox Reactions
Define an oxidising agent and a reducing agent with one example each.
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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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