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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.

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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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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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ChemistryClass 122 markseasy

Coordination Compounds

State any four applications of coordination compounds.

Reveal model answer + marking points

(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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ChemistryClass 122 markseasy

Haloalkanes and Haloarenes

Distinguish between haloalkanes and haloarenes with one example each.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Haloalkanes and Haloarenes

Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution?

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Haloalkanes and Haloarenes

What is a dehydrohalogenation (elimination) reaction of a haloalkane? State Saytzeff's rule.

Reveal model answer + marking points

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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ChemistryClass 122 marksmedium

Haloalkanes and Haloarenes

What is a chiral molecule? What is meant by optical activity?

Reveal model answer + marking points

A chiral molecule is one that is non-superimposable on its mirror image, just as the left and right hands are not superimposable; it usually contains at least one carbon atom bonded to four different groups (an asymmetric or chiral carbon). Optical activity is the property of such a chiral substance to rotate the plane of plane-polarised light; the two non-superimposable mirror-image forms are called enantiomers, one rotating the light to the right (dextrorotatory) and the other to the left (laevorotatory).

Marking-scheme points

  • Chiral molecule: non-superimposable on its mirror image (chiral carbon)
  • Optical activity: rotates the plane of plane-polarised light
  • Mirror-image forms = enantiomers (dextro- and laevorotatory)
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ChemistryClass 122 marksmedium

Haloalkanes and Haloarenes

What is a Grignard reagent? How is it prepared?

Reveal model answer + marking points

A Grignard reagent is an alkyl or aryl magnesium halide, R-Mg-X, which is a very important and reactive organometallic compound used in organic synthesis. It is prepared by the reaction of a haloalkane (or haloarene) with magnesium metal in the presence of dry ether: R-X + Mg -> R-Mg-X (in dry ether). Grignard reagents are highly reactive and must be prepared under anhydrous conditions, because even traces of water or moisture decompose them to alkanes.

R-X + Mg -> R-Mg-X (dry ether)

Marking-scheme points

  • Grignard reagent = alkyl/aryl magnesium halide (R-Mg-X)
  • Prepared: R-X + Mg -> R-Mg-X in dry ether
  • Very reactive; must be kept anhydrous (water decomposes it)
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ChemistryClass 122 markseasy

Haloalkanes and Haloarenes

Why is chloroform stored in dark coloured bottles filled up to the brim?

Reveal model answer + marking points

Chloroform (CHCl3) is stored in dark coloured bottles filled completely up to the brim because in the presence of air (oxygen) and sunlight it undergoes slow oxidation to form a highly poisonous gas, phosgene (carbonyl chloride, COCl2): 2CHCl3 + O2 -> 2COCl2 + 2HCl. Filling the bottle to the brim leaves no air space, and the dark bottle keeps out light; together these prevent the oxidation of chloroform to phosgene.

2CHCl3 + O2 -> 2COCl2 + 2HCl

Marking-scheme points

  • Chloroform is oxidised in air and light to poisonous phosgene (COCl2)
  • 2CHCl3 + O2 -> 2COCl2 + 2HCl
  • Dark bottle filled to the brim excludes light and air
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ChemistryClass 122 markseasy

Alcohols, Phenols and Ethers

Classify alcohols as primary, secondary and tertiary with one example each.

Reveal model answer + marking points

Alcohols are classified according to the type of carbon atom to which the -OH group is attached. In a primary (1 degree) alcohol the -OH is on a carbon attached to only one other carbon, e.g. ethanol (CH3CH2OH). In a secondary (2 degree) alcohol the -OH is on a carbon attached to two other carbons, e.g. propan-2-ol ((CH3)2CHOH). In a tertiary (3 degree) alcohol the -OH is on a carbon attached to three other carbons, e.g. 2-methylpropan-2-ol ((CH3)3COH).

Marking-scheme points

  • Primary: -OH carbon attached to one carbon (ethanol)
  • Secondary: -OH carbon attached to two carbons (propan-2-ol)
  • Tertiary: -OH carbon attached to three carbons (2-methylpropan-2-ol)
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ChemistryClass 122 marksmedium

Alcohols, Phenols and Ethers

Why is phenol more acidic than ethanol?

Reveal model answer + marking points

Phenol is more acidic than ethanol because the phenoxide ion formed after the loss of the proton is stabilised by resonance (the negative charge is delocalised into the benzene ring), which makes phenol lose its proton more easily. In contrast, the ethoxide ion from ethanol is not resonance-stabilised, and the alkyl group has an electron-releasing (+I) effect that further destabilises the ethoxide ion. Hence phenol ionises more readily and is a stronger acid than ethanol.

Marking-scheme points

  • Phenoxide ion is stabilised by resonance (charge delocalised into ring)
  • Ethoxide ion is not resonance-stabilised
  • Alkyl (+I) effect destabilises ethoxide, so ethanol is weaker acid
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ChemistryClass 122 marksmedium

Alcohols, Phenols and Ethers

Write the reaction of ethanol with sodium metal and its dehydration to ethene.

Reveal model answer + marking points

(1) Reaction with sodium: alcohols react with active metals like sodium to liberate hydrogen gas and form sodium alkoxide, 2C2H5OH + 2Na -> 2C2H5ONa + H2. This shows the acidic nature of the -OH group. (2) Dehydration: when ethanol is heated with concentrated sulphuric acid at about 443 K (170 degrees C), it loses a molecule of water to form ethene, C2H5OH -> CH2=CH2 + H2O. Concentrated H2SO4 acts as a dehydrating agent.

C2H5OH -> CH2=CH2 + H2O

Marking-scheme points

  • With sodium: 2C2H5OH + 2Na -> 2C2H5ONa + H2 (shows acidic -OH)
  • Dehydration with conc. H2SO4 at 443 K gives ethene
  • C2H5OH -> CH2=CH2 + H2O
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