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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.

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800 questions

ChemistryClass 113 marksmedium

Classification of Elements and Periodicity

How does atomic radius vary across a period and down a group? Give reasons.

Reveal model answer + marking points

Across a period (left to right), atomic radius decreases because nuclear charge increases while electrons are added to the same shell, so the increased effective nuclear charge pulls the electron cloud closer. Down a group, atomic radius increases because a new shell is added at each step and the number of inner shielding electrons increases, outweighing the rise in nuclear charge.

Marking-scheme points

  • Across period: radius decreases (rising effective nuclear charge, same shell)
  • Down group: radius increases (new shells added)
  • Shielding by inner electrons increases down a group
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ChemistryClass 112 marksmedium

Classification of Elements and Periodicity

Why is the first ionization enthalpy of nitrogen greater than that of oxygen?

Reveal model answer + marking points

Nitrogen has the configuration 1s2 2s2 2p3 with a half-filled 2p subshell, which is extra stable due to symmetry and exchange energy, so removing an electron requires more energy. Oxygen (1s2 2s2 2p4) has one paired electron in 2p; removing it relieves electron-electron repulsion and gives a stable half-filled configuration, so less energy is needed. Hence IE1 of nitrogen > oxygen.

Marking-scheme points

  • N has stable half-filled 2p3 configuration
  • O 2p4 has electron-pair repulsion, easier to remove
  • So IE1(N) > IE1(O)
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ChemistryClass 113 marksmedium

Classification of Elements and Periodicity

Define ionization enthalpy. Explain its trend across a period and down a group.

Reveal model answer + marking points

Ionization enthalpy is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state. Across a period it increases because nuclear charge increases and atomic size decreases, so the electron is held more tightly. Down a group it decreases because atomic size increases and shielding by inner electrons increases, so the outer electron is more easily removed.

M(g) -> M+(g) + e-

Marking-scheme points

  • Energy to remove electron from gaseous atom
  • Increases across a period (size down, nuclear charge up)
  • Decreases down a group (size and shielding up)
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ChemistryClass 112 markshard

Classification of Elements and Periodicity

Why is the electron gain enthalpy of chlorine more negative than that of fluorine?

Reveal model answer + marking points

Although fluorine is smaller, its 2p subshell is very compact, so the incoming electron experiences strong inter-electronic repulsion from the already crowded 2p electrons. In chlorine, the larger 3p subshell accommodates the incoming electron with less repulsion, so more energy is released. Hence chlorine has a more negative (more exothermic) electron gain enthalpy than fluorine.

Marking-scheme points

  • F is small, compact 2p -> high electron-electron repulsion
  • Cl larger 3p -> less repulsion for incoming electron
  • So electron gain enthalpy of Cl is more negative
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ChemistryClass 112 markseasy

Classification of Elements and Periodicity

Define electronegativity. How does it vary across a period and down a group?

Reveal model answer + marking points

Electronegativity is the tendency of an atom in a molecule to attract the shared pair of electrons in a covalent bond towards itself. It increases across a period (as size decreases and nuclear charge increases) and decreases down a group (as size increases). Fluorine is the most electronegative element. The Pauling scale is commonly used.

Marking-scheme points

  • Tendency to attract bonded (shared) electrons
  • Increases across a period, decreases down a group
  • F is most electronegative; Pauling scale
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ChemistryClass 112 marksmedium

Classification of Elements and Periodicity

Arrange the isoelectronic species O2-, F-, Na+ and Mg2+ in increasing order of ionic radius. Justify.

Reveal model answer + marking points

All four species have 10 electrons (isoelectronic). For isoelectronic species, the greater the nuclear charge (number of protons), the smaller the radius. Nuclear charges: O2- (8), F- (9), Na+ (11), Mg2+ (12). Increasing order of radius: Mg2+ < Na+ < F- < O2-.

Marking-scheme points

  • All are isoelectronic (10 electrons)
  • More protons -> smaller radius
  • Order: Mg2+ < Na+ < F- < O2-
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ChemistryClass 113 marksmedium

Classification of Elements and Periodicity

What are s-, p-, d- and f-block elements? Why are d-block elements called transition elements?

Reveal model answer + marking points

Elements are classified by the subshell into which the last (differentiating) electron enters: s-block (last electron in s), p-block (in p), d-block (in d) and f-block (in f). d-block elements are called transition elements because they lie between the s-block (metals) and p-block (non-metals) and show a gradual transition in properties; their atoms or common ions have partially filled d orbitals.

Marking-scheme points

  • Block = subshell receiving the last electron (s, p, d, f)
  • d-block lies between s- and p-blocks
  • Transition: atoms/ions have partially filled d orbitals
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ChemistryClass 112 markseasy

Chemical Bonding and Molecular Structure

Distinguish between an ionic bond and a covalent bond with one example each.

Reveal model answer + marking points

An ionic (electrovalent) bond is formed by the complete transfer of one or more electrons from one atom to another, producing oppositely charged ions held by electrostatic attraction (e.g. NaCl). A covalent bond is formed by the mutual sharing of electron pairs between atoms (e.g. H2 or Cl2). Ionic bonds form between metals and non-metals; covalent bonds form between non-metals.

Marking-scheme points

  • Ionic: complete transfer of electrons, e.g. NaCl
  • Covalent: sharing of electron pairs, e.g. H2
  • Ionic = metal + non-metal; covalent = non-metals
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ChemistryClass 113 marksmedium

Chemical Bonding and Molecular Structure

Using VSEPR theory, predict the shapes and bond angles of BF3, NH3 and H2O.

Reveal model answer + marking points

BF3: central B has 3 bond pairs and no lone pair -> trigonal planar, bond angle 120 deg. NH3: central N has 3 bond pairs and 1 lone pair -> pyramidal (trigonal pyramidal), bond angle about 107 deg. H2O: central O has 2 bond pairs and 2 lone pairs -> bent/angular, bond angle about 104.5 deg. Lone pairs repel more than bond pairs, reducing the angle from the ideal 109.5 deg in NH3 and H2O.

Marking-scheme points

  • BF3: 3 bp, 0 lp -> trigonal planar, 120 deg
  • NH3: 3 bp, 1 lp -> pyramidal, 107 deg
  • H2O: 2 bp, 2 lp -> bent, 104.5 deg
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ChemistryClass 112 marksmedium

Chemical Bonding and Molecular Structure

Why is the bond angle of H2O (104.5 deg) less than that of NH3 (107 deg)?

Reveal model answer + marking points

Both molecules are based on sp3 hybridisation with an ideal angle of 109.5 deg. NH3 has one lone pair, while H2O has two lone pairs. Lone pair-lone pair repulsion is stronger than lone pair-bond pair repulsion, and water has an extra lone pair, so its bond pairs are pushed closer together. Hence the bond angle of H2O (104.5 deg) is smaller than that of NH3 (107 deg).

Marking-scheme points

  • Both sp3, ideal 109.5 deg
  • NH3 has 1 lone pair; H2O has 2 lone pairs
  • More lone pairs -> greater repulsion -> smaller angle
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ChemistryClass 113 marksmedium

Chemical Bonding and Molecular Structure

Describe the hybridization, shape and bonding in an ethyne (C2H2) molecule.

Reveal model answer + marking points

In ethyne each carbon is sp hybridised. The two sp hybrid orbitals on each carbon form sigma bonds: one C-C sigma bond and one C-H sigma bond, giving a linear molecule with a bond angle of 180 deg. The remaining two unhybridised p orbitals on each carbon overlap sideways to form two pi bonds, so there is a triple bond (1 sigma + 2 pi) between the carbon atoms.

H-C(triple bond)C-H

Marking-scheme points

  • Each C is sp hybridised, molecule linear (180 deg)
  • sp orbitals form C-C and C-H sigma bonds
  • Two unhybridised p orbitals form 2 pi bonds (triple bond = 1 sigma + 2 pi)
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ChemistryClass 113 markshard

Chemical Bonding and Molecular Structure

Using molecular orbital theory, write the molecular orbital configuration of O2, calculate its bond order and explain its magnetic nature.

Reveal model answer + marking points

O2 has 16 electrons. Configuration: sigma(1s)2 sigma*(1s)2 sigma(2s)2 sigma*(2s)2 sigma(2pz)2 pi(2px)2 pi(2py)2 pi*(2px)1 pi*(2py)1. Bonding electrons Nb = 10, antibonding Na = 6. Bond order = (Nb - Na)/2 = (10 - 6)/2 = 2. The two unpaired electrons in the pi* antibonding orbitals make O2 paramagnetic.

Bond order = (Nb - Na)/2

Marking-scheme points

  • O2 has 16 electrons; two unpaired in pi* orbitals
  • Bond order = (Nb - Na)/2 = (10 - 6)/2 = 2
  • Unpaired electrons -> O2 is paramagnetic
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ChemistryClass 112 marksmedium

Chemical Bonding and Molecular Structure

Calculate the bond order of the nitrogen molecule (N2) using molecular orbital theory.

Reveal model answer + marking points

N2 has 14 electrons. Number of bonding electrons Nb = 10 and antibonding electrons Na = 4. Bond order = (Nb - Na)/2 = (10 - 4)/2 = 3. This corresponds to a nitrogen-nitrogen triple bond, which explains the very high stability and bond dissociation energy of N2.

Bond order = (Nb - Na)/2

Marking-scheme points

  • N2 has 14 electrons; Nb = 10, Na = 4
  • Bond order = (10 - 4)/2 = 3
  • Triple bond -> very stable molecule
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ChemistryClass 112 markseasy

Chemical Bonding and Molecular Structure

What is a coordinate (dative) bond? Explain with the example of the ammonium ion.

Reveal model answer + marking points

A coordinate bond is a covalent bond in which the shared pair of electrons is contributed by only one of the two bonded atoms (the donor). In the ammonium ion (NH4+), nitrogen in NH3 has a lone pair which it donates to a proton (H+) that has no electrons, forming the fourth N-H bond as a coordinate bond. Once formed, all four N-H bonds are identical.

NH3 + H+ -> NH4+

Marking-scheme points

  • Both shared electrons come from one atom (donor)
  • NH3 nitrogen lone pair donated to H+
  • All four N-H bonds become equivalent in NH4+
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ChemistryClass 112 marksmedium

Chemical Bonding and Molecular Structure

What is a hydrogen bond? State the conditions for its formation and its two types.

Reveal model answer + marking points

A hydrogen bond is a weak electrostatic attraction between a hydrogen atom covalently bonded to a highly electronegative atom (F, O or N) and the lone pair of another electronegative atom. Conditions: hydrogen must be attached to a small, highly electronegative atom. Types: intermolecular hydrogen bonding (between different molecules, e.g. in water and HF) and intramolecular hydrogen bonding (within the same molecule, e.g. o-nitrophenol).

Marking-scheme points

  • Attraction of H (bonded to F/O/N) with lone pair on another electronegative atom
  • Needs H on small, highly electronegative atom
  • Types: intermolecular and intramolecular
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ChemistryClass 112 marksmedium

Chemical Bonding and Molecular Structure

Why does water (H2O) have a much higher boiling point than hydrogen sulphide (H2S)?

Reveal model answer + marking points

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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ChemistryClass 113 marksmedium

Chemical Bonding and Molecular Structure

State Fajans' rules governing the covalent character of an ionic bond.

Reveal model answer + marking points

Fajans' rules: covalent character in an ionic compound increases when (1) the cation is small, (2) the anion is large, and (3) the cation has a high charge (all three increase the polarising power/polarisability). Also, cations with a pseudo-noble gas (18-electron) configuration cause greater polarisation than those with a noble gas configuration. Greater polarisation of the anion by the cation increases covalent character.

Marking-scheme points

  • Small cation -> more covalent character
  • Large anion -> more covalent character
  • High charge on ions and 18-electron cation -> more covalent character
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ChemistryClass 113 marksmedium

Chemical Bonding and Molecular Structure

What is dipole moment? Why is the dipole moment of CO2 zero while that of H2O is not?

Reveal model answer + marking points

Dipole moment (mu) is the product of the magnitude of charge and the distance between the centres of positive and negative charge; it is a vector quantity (unit: debye). CO2 is linear (O=C=O) and its two C=O bond dipoles are equal and opposite, so they cancel and the net dipole moment is zero. H2O is bent/angular, so its two O-H bond dipoles do not cancel and add up to give a net dipole moment (1.85 D). Hence CO2 is non-polar but H2O is polar.

mu = q x d

Marking-scheme points

  • mu = charge x distance (vector, unit debye)
  • CO2 linear: equal opposite dipoles cancel -> mu = 0
  • H2O bent: dipoles do not cancel -> net dipole (polar)
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ChemistryClass 112 marksmedium

Chemical Bonding and Molecular Structure

Distinguish between a sigma bond and a pi bond. Which is stronger and why?

Reveal model answer + marking points

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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ChemistryClass 112 markseasy

States of Matter

State Boyle's law and Charles's law with their mathematical expressions.

Reveal model answer + marking points

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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