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ChemistryClass 122 marksmedium
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
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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The p-Block Elements
Why are noble gases chemically inert? Name two compounds of xenon.
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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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The d- and f-Block Elements
What are transition elements? Why are they called so?
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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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The d- and f-Block Elements
Why do transition elements show variable oxidation states?
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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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The d- and f-Block Elements
Why are most transition metal compounds coloured?
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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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The d- and f-Block Elements
Why do transition metals and their compounds act as good catalysts?
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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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The d- and f-Block Elements
Write the oxidising action of potassium dichromate in acidic medium. What happens to the colour of the solution?
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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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The d- and f-Block Elements
What is lanthanoid contraction? State one of its consequences.
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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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The d- and f-Block Elements
Why do transition metals have high melting points, and why are many of their compounds paramagnetic?
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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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Coordination Compounds
Distinguish between the primary and secondary valency of a metal in Werner's theory.
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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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Coordination Compounds
Define ligand and coordination number.
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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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Coordination Compounds
What is a chelate ligand and an ambidentate ligand? Give one example of each.
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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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Coordination Compounds
What is the spectrochemical series? Distinguish between strong field and weak field ligands.
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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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Coordination Compounds
State any four applications of coordination compounds.
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(1) In biological systems: haemoglobin (a complex of iron) carries oxygen, and chlorophyll (a complex of magnesium) is essential for photosynthesis. (2) In metallurgy: metals like silver and gold are extracted and purified using complex formation (cyanide process). (3) In medicine: cisplatin is used in cancer treatment and EDTA complexes are used to treat lead poisoning. (4) In analytical chemistry and electroplating: complexes are used in the estimation of metal ions and in electroplating of metals.
Marking-scheme points
- ✓Biological: haemoglobin (Fe), chlorophyll (Mg)
- ✓Metallurgy: extraction of silver and gold (cyanide process)
- ✓Medicine (cisplatin), analysis and electroplating
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Haloalkanes and Haloarenes
Distinguish between haloalkanes and haloarenes with one example each.
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Haloalkanes (alkyl halides) are compounds in which one or more hydrogen atoms of an aliphatic hydrocarbon (alkane) are replaced by halogen atoms; the halogen is attached to an sp3 carbon (e.g. CH3Cl, chloromethane). Haloarenes (aryl halides) are compounds in which the halogen atom is directly attached to an sp2 carbon of an aromatic ring (e.g. C6H5Cl, chlorobenzene). Haloarenes are much less reactive than haloalkanes towards nucleophilic substitution.
Marking-scheme points
- ✓Haloalkane: halogen on sp3 (aliphatic) carbon (CH3Cl)
- ✓Haloarene: halogen on sp2 carbon of an aromatic ring (C6H5Cl)
- ✓Haloarenes are less reactive to nucleophilic substitution
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Haloalkanes and Haloarenes
Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution?
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Haloarenes are less reactive than haloalkanes towards nucleophilic substitution mainly because of: (1) resonance - the lone pair of the halogen delocalises into the ring, giving the carbon-halogen bond a partial double-bond character, so it is shorter and stronger and harder to break; (2) the halogen is attached to an sp2 hybridised carbon which is more electronegative and holds the shared electrons more tightly than the sp3 carbon in haloalkanes; and (3) repulsion between the electron-rich aromatic ring and the approaching nucleophile.
Marking-scheme points
- ✓Resonance gives the C-X bond partial double-bond character (stronger)
- ✓Halogen on sp2 carbon (more electronegative, holds electrons tightly)
- ✓Electron-rich ring repels the incoming nucleophile
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Haloalkanes and Haloarenes
What is a dehydrohalogenation (elimination) reaction of a haloalkane? State Saytzeff's rule.
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Dehydrohalogenation is a beta-elimination reaction in which a haloalkane loses a hydrogen halide (HX) when heated with an alcoholic solution of potassium hydroxide, forming an alkene. The hydrogen is removed from the beta-carbon (the carbon next to the one bearing the halogen). Saytzeff's rule states that in such an elimination the preferred (major) product is the more highly substituted (more stable) alkene, that is, the alkene formed by removal of the hydrogen from the beta-carbon having the fewer hydrogen atoms.
Marking-scheme points
- ✓Beta-elimination of HX with alcoholic KOH gives an alkene
- ✓Hydrogen removed from the beta-carbon
- ✓Saytzeff's rule: the more substituted (more stable) alkene is the major product
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