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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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ChemistryClass 122 marksmedium

The d- and f-Block Elements

Why do transition elements show variable oxidation states?

Reveal model answer + marking points

Transition elements show variable oxidation states because the energies of the (n-1)d and the ns orbitals are very close to each other. As a result, after the ns electrons are removed, a variable number of the (n-1)d electrons can also take part in bonding. This allows the elements to exhibit several oxidation states that usually differ by one unit (for example, iron shows +2 and +3, and manganese shows +2 to +7).

Marking-scheme points

  • Energies of (n-1)d and ns orbitals are very close
  • A variable number of d electrons can participate in bonding
  • Example: Fe shows +2 and +3, Mn shows +2 to +7
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ChemistryClass 122 marksmedium

The d- and f-Block Elements

Why are most transition metal compounds coloured?

Reveal model answer + marking points

Most transition metal ions are coloured because they have partially filled d orbitals. In the presence of ligands or other ions, the five d orbitals split into two sets of slightly different energy. An electron can absorb a particular wavelength (colour) of visible light and jump from the lower to the higher set of d orbitals (a d-d transition). The colour we see is the complementary colour of the light absorbed. Ions with completely empty or completely filled d orbitals (such as Sc3+ or Zn2+) are colourless.

Marking-scheme points

  • Partially filled d orbitals split into two energy levels
  • Electron absorbs visible light and jumps (d-d transition)
  • Observed colour is complementary to the absorbed light; d0 and d10 ions are colourless
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ChemistryClass 122 marksmedium

The d- and f-Block Elements

Why do transition metals and their compounds act as good catalysts?

Reveal model answer + marking points

Transition metals and their compounds act as good catalysts mainly because: (1) they show variable oxidation states, so they can readily form intermediate compounds with the reactants and provide an alternative path of lower activation energy; and (2) they have the ability to adsorb reactant molecules on their surfaces (due to incomplete d orbitals), which brings the reactants close together and weakens their bonds. Examples: iron in the Haber process and vanadium pentoxide in the Contact process.

Marking-scheme points

  • Variable oxidation states allow formation of intermediates
  • Provide a lower activation energy path
  • Adsorb reactants on their surface (incomplete d orbitals)
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ChemistryClass 122 marksmedium

The d- and f-Block Elements

Write the oxidising action of potassium dichromate in acidic medium. What happens to the colour of the solution?

Reveal model answer + marking points

Potassium dichromate (K2Cr2O7) is a strong oxidising agent in acidic medium. Its oxidising half-reaction is Cr2O7(2-) + 14H+ + 6e- -> 2Cr3+ + 7H2O, in which chromium is reduced from the +6 to the +3 oxidation state. During this the colour of the solution changes from orange (dichromate) to green (Cr3+). It is used, for example, to oxidise ferrous ions to ferric ions and iodide to iodine.

Cr2O7(2-) + 14H+ + 6e- -> 2Cr3+ + 7H2O

Marking-scheme points

  • Cr2O7(2-) + 14H+ + 6e- -> 2Cr3+ + 7H2O
  • Chromium reduced from +6 to +3
  • Colour changes from orange to green
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ChemistryClass 122 marksmedium

The d- and f-Block Elements

What is lanthanoid contraction? State one of its consequences.

Reveal model answer + marking points

Lanthanoid contraction is the steady, regular decrease in the atomic and ionic radii of the lanthanoid elements with increasing atomic number, from lanthanum to lutetium. It occurs because as the atomic number increases, electrons are added to the inner 4f subshell, which shields the nuclear charge poorly; so the effective nuclear charge on the outer electrons increases and the radius decreases. A consequence is that the elements of the second and third transition series (such as zirconium and hafnium) have almost the same size and very similar properties, making them difficult to separate.

Marking-scheme points

  • Steady decrease in size of lanthanoids with increasing atomic number
  • Cause: poor shielding by 4f electrons -> higher effective nuclear charge
  • Consequence: Zr and Hf have similar sizes and properties
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ChemistryClass 122 marksmedium

The d- and f-Block Elements

Why do transition metals have high melting points, and why are many of their compounds paramagnetic?

Reveal model answer + marking points

Transition metals have high melting and boiling points because their atoms have a large number of unpaired d electrons that form strong metallic bonds (in addition to bonds from the s electrons), so a lot of energy is needed to break them. Many of their compounds are paramagnetic because their ions contain unpaired electrons in the d orbitals; the magnetic moment increases with the number of unpaired electrons and can be estimated by the spin-only formula, magnetic moment = sqrt(n(n+2)) Bohr magnetons, where n is the number of unpaired electrons.

magnetic moment = sqrt(n(n+2)) BM

Marking-scheme points

  • High melting points: strong metallic bonding from unpaired d electrons
  • Paramagnetism due to unpaired d electrons in ions
  • Spin-only magnetic moment = sqrt(n(n+2)) Bohr magnetons
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ChemistryClass 122 marksmedium

Coordination Compounds

Distinguish between the primary and secondary valency of a metal in Werner's theory.

Reveal model answer + marking points

According to Werner's theory, a metal in a coordination compound has two types of valency. The primary valency is ionisable and corresponds to the oxidation state of the metal; it is satisfied by negative ions and is non-directional. The secondary valency is non-ionisable and corresponds to the coordination number of the metal; it is satisfied by ligands, is directional, and determines the geometry of the complex.

Marking-scheme points

  • Primary valency: ionisable, equals oxidation state, satisfied by anions
  • Secondary valency: non-ionisable, equals coordination number, satisfied by ligands
  • Secondary valency is directional and fixes the geometry
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ChemistryClass 122 markseasy

Coordination Compounds

Define ligand and coordination number.

Reveal model answer + marking points

A ligand is an ion or a molecule that has at least one lone pair of electrons which it can donate to the central metal atom or ion, forming a coordinate bond (for example Cl-, NH3, H2O, CN-). The coordination number of the central metal ion is the total number of coordinate bonds it forms with the ligands, that is, the number of ligand donor atoms directly attached to it (for example, in [Cu(NH3)4]2+ the coordination number of copper is 4).

Marking-scheme points

  • Ligand: donates a lone pair to the metal (e.g. NH3, Cl-, CN-)
  • Coordination number: number of donor atoms bonded to the metal
  • Example: [Cu(NH3)4]2+ has coordination number 4
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ChemistryClass 122 marksmedium

Coordination Compounds

What is a chelate ligand and an ambidentate ligand? Give one example of each.

Reveal model answer + marking points

A chelate ligand (chelating ligand) is a polydentate ligand that binds to the same central metal ion through two or more donor atoms, forming a ring structure; this gives extra stability (the chelate effect). An example is ethylenediamine (en), which is bidentate. An ambidentate ligand is a monodentate ligand that has two different donor atoms and can attach to the metal through either one (but only one at a time); an example is the nitrite ion, which can bind through nitrogen (-NO2) or through oxygen (-ONO).

Marking-scheme points

  • Chelate ligand: polydentate, forms a ring (e.g. ethylenediamine)
  • Chelation gives extra stability (chelate effect)
  • Ambidentate ligand: two possible donor atoms, e.g. NO2 (via N or O)
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ChemistryClass 122 marksmedium

Coordination Compounds

What is the spectrochemical series? Distinguish between strong field and weak field ligands.

Reveal model answer + marking points

The spectrochemical series is an arrangement of ligands in order of their increasing crystal field splitting power (the magnitude of delta they produce). A part of it is: I- < Br- < Cl- < F- < H2O < NH3 < en < CN- < CO. Strong field ligands (such as CN- and CO) produce a large splitting (large delta), favouring low-spin complexes; weak field ligands (such as halides and water) produce a small splitting (small delta), favouring high-spin complexes.

Marking-scheme points

  • Ligands arranged by increasing crystal field splitting power
  • Weak field (I-, Br-, H2O): small delta, high-spin
  • Strong field (CN-, CO): large delta, low-spin
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