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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?

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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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ChemistryClass 122 marksmedium

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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ChemistryClass 123 marksmedium

The d- and f-Block Elements

How is potassium permanganate prepared from pyrolusite? State its oxidising action in acidic medium.

Reveal model answer + marking points

Potassium permanganate (KMnO4) is prepared from the mineral pyrolusite (MnO2). MnO2 is fused with potassium hydroxide (KOH) in the presence of air or an oxidising agent like KNO3 to give green potassium manganate (K2MnO4): 2MnO2 + 4KOH + O2 -> 2K2MnO4 + 2H2O. The manganate is then oxidised (electrolytically or by chlorine/ozone) to purple permanganate: 2K2MnO4 + Cl2 -> 2KMnO4 + 2KCl. In acidic medium KMnO4 is a strong oxidising agent: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O (Mn goes from +7 to +2).

MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

Marking-scheme points

  • 2MnO2 + 4KOH + O2 -> 2K2MnO4 (green manganate)
  • Manganate oxidised to KMnO4 (purple permanganate)
  • Acidic oxidation: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
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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 123 marksmedium

Coordination Compounds

State the main rules for the IUPAC nomenclature of coordination compounds and name [Cu(NH3)4]SO4.

Reveal model answer + marking points

Main rules: (1) the cation is named before the anion; (2) within the complex, the ligands are named in alphabetical order before the central metal; (3) the number of ligands is shown by prefixes di, tri, tetra, etc.; (4) the oxidation state of the metal is written in Roman numerals in brackets; and (5) if the complex ion is an anion, the metal name ends in -ate. For example, [Cu(NH3)4]SO4 is named tetraamminecopper(II) sulphate (copper is in the +2 state).

Marking-scheme points

  • Cation named first; ligands named alphabetically before the metal
  • Number of ligands by di, tri, tetra; oxidation state in Roman numerals
  • [Cu(NH3)4]SO4 = tetraamminecopper(II) sulphate
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ChemistryClass 123 marksmedium

Coordination Compounds

Name the main types of isomerism shown by coordination compounds.

Reveal model answer + marking points

Coordination compounds show two broad types of isomerism. Structural (constitutional) isomerism includes: (1) ionisation isomerism (different ions in solution, e.g. [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br); (2) linkage isomerism (ambidentate ligand attached through different atoms, e.g. -NO2 vs -ONO); (3) coordination isomerism; and (4) hydrate (solvate) isomerism. Stereoisomerism includes: (1) geometrical isomerism (cis and trans forms) and (2) optical isomerism (non-superimposable mirror images).

Marking-scheme points

  • Structural: ionisation, linkage, coordination, hydrate isomerism
  • Stereoisomerism: geometrical (cis-trans) and optical
  • Linkage isomerism needs an ambidentate ligand (e.g. NO2)
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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 123 markshard

Coordination Compounds

Explain the geometry of a complex using valence bond theory. Distinguish inner and outer orbital complexes.

Reveal model answer + marking points

According to valence bond theory, the central metal ion provides a number of empty hybrid orbitals equal to its coordination number, and each ligand donates a lone pair into these orbitals to form coordinate bonds; the type of hybridisation decides the geometry (e.g. sp3 tetrahedral, dsp2 square planar, sp3d2 or d2sp3 octahedral). An inner orbital complex uses inner (n-1)d orbitals for hybridisation (d2sp3), usually formed with strong field ligands and often low-spin. An outer orbital complex uses the outer nd orbitals (sp3d2), usually formed with weak field ligands and is high-spin.

Marking-scheme points

  • Metal provides empty hybrid orbitals; ligands donate lone pairs
  • Hybridisation decides geometry (sp3, dsp2, d2sp3, sp3d2)
  • Inner orbital (d2sp3, low-spin) vs outer orbital (sp3d2, high-spin)
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ChemistryClass 123 markshard

Coordination Compounds

Explain the splitting of d orbitals in an octahedral crystal field according to crystal field theory.

Reveal model answer + marking points

According to crystal field theory, the bonding between the metal and the ligands is purely electrostatic. In an isolated metal ion the five d orbitals have the same energy (degenerate). When six ligands approach along the axes in an octahedral field, they repel the d orbitals unequally: the two orbitals pointing along the axes (dx2-y2 and dz2, called eg) are repelled more and rise in energy, while the three orbitals pointing between the axes (dxy, dyz, dzx, called t2g) are repelled less and are lowered in energy. This energy gap between t2g and eg is the crystal field splitting energy (delta o).

Marking-scheme points

  • Metal-ligand bonding treated as electrostatic
  • Six ligands approach along the axes (octahedral)
  • d orbitals split into lower t2g and higher eg; gap = delta 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 123 marksmedium

Haloalkanes and Haloarenes

How are haloalkanes prepared from alcohols? Write the reactions.

Reveal model answer + marking points

Haloalkanes are commonly prepared from alcohols by replacing the -OH group with a halogen. (1) With hydrogen halides: R-OH + HX -> R-X + H2O (reactivity of alcohols is 3 degrees > 2 degrees > 1 degree). (2) With phosphorus halides: 3R-OH + PCl3 -> 3R-Cl + H3PO3, and R-OH + PCl5 -> R-Cl + POCl3 + HCl. (3) With thionyl chloride (the best method, as the by-products are gases): R-OH + SOCl2 -> R-Cl + SO2 + HCl.

R-OH + SOCl2 -> R-Cl + SO2 + HCl

Marking-scheme points

  • R-OH + HX -> R-X + H2O
  • 3R-OH + PCl3 -> 3R-Cl + H3PO3
  • R-OH + SOCl2 -> R-Cl + SO2 + HCl (best method)
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