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NEET Chemistry: Basic Concepts, Atomic Structure and Bonding · Chapter 3

Chemical Bonding and Molecular Structure

What to remember

  • Atoms bond to reach a stable noble-gas-like octet; ionic bonds form by electron transfer and covalent bonds by electron sharing.
  • VSEPR theory predicts shape from the number of electron pairs around the central atom; lone pairs repel more than bond pairs, so lp–lp > lp–bp > bp–bp.
  • Hybridisation: sp (linear, 180°), sp² (trigonal planar, 120°), sp³ (tetrahedral, 109.5°), sp³d (trigonal bipyramidal), sp³d² (octahedral).
  • Bond order = (Nb − Na)/2 from molecular orbital theory; unpaired electrons make a species paramagnetic, and O₂ is paramagnetic.

Kossel-Lewis approach, octet rule and Lewis structures

Atoms of noble gases are stable because they have eight electrons in the valence shell (two for helium). Kossel and Lewis proposed that other atoms gain, lose or share electrons to reach this arrangement. This is the octet rule.

A Lewis structure shows valence electrons as dots. Shared pairs form bonds and unshared pairs are lone pairs.

Steps to write a Lewis structure:

  • 1. Count the total valence electrons (add one for each negative charge, subtract one for each positive charge).
  • 2. Place the least electronegative atom in the centre (hydrogen is always terminal).
  • 3. Join atoms with single bonds and distribute the remaining electrons to give octets.
  • 4. If octets are incomplete, form multiple bonds.

Formal charge on an atom = (valence electrons in the free atom) − (non-bonding electrons) − ½ (bonding electrons). The sum of formal charges equals the charge on the species. The structure with the smallest formal charges is the most stable.

Example: in NH₄⁺, N has 5 − 0 − ½(8) = +1. In ozone, O₃, drawn with one double bond and one single bond, the central O has 6 − 2 − ½(6) = +1; the double-bonded O has 0; the single-bonded O has 6 − 6 − ½(2) = −1.

Limitations of the octet rule:

  • Incomplete octet: the central atom has fewer than eight electrons, as in BeH₂ (4), BF₃ (6) and AlCl₃ (6).
  • Odd-electron molecules, such as NO (11 electrons) and NO₂ (17 electrons).
  • Expanded octet: the central atom has more than eight electrons, as in PCl₅ (10) and SF₆ (12).
  • It does not explain the shapes of molecules, or the paramagnetism of O₂.

Ionic bond, covalent bond and bond parameters

An ionic bond forms by complete transfer of electrons from a metal to a non-metal. It is favoured by low ionisation enthalpy of the metal, high negative electron gain enthalpy of the non-metal and high lattice enthalpy.

Lattice enthalpy is the energy needed to separate one mole of a solid ionic compound into its gaseous ions. It is larger when the ions are small and highly charged. For example, NaCl is about 788 kJ mol⁻¹. Ionic compounds have high melting points, are soluble in water and conduct electricity in the molten state or in solution.

A covalent bond forms by mutual sharing of electrons. Single, double and triple bonds involve one, two and three shared pairs.

Bond parameterMeaningTrend
Bond lengthDistance between the nuclei of two bonded atomsTriple < double < single between the same atoms
Bond angleAngle between two bonds at the central atomCH₄ 109.5°, NH₃ about 107°, H₂O about 104.5°
Bond enthalpyEnergy to break one mole of bonds in the gas phaseTriple > double > single
Bond orderNumber of bonds between two atomsHigher bond order, shorter and stronger bond

Resonance: when a single Lewis structure cannot describe a molecule, two or more structures differing only in the position of electrons are written. The real molecule is a resonance hybrid, which is more stable than any one contributing structure. Examples: ozone (both O–O bonds are equal, bond order 1.5) and the carbonate ion (three equal C–O bonds).

Polarity, dipole moment and Fajans' rule

A bond between atoms of different electronegativity is polar. The dipole moment μ = charge (q) × distance (d). It is a vector quantity, pointing towards the more electronegative atom. The unit is the debye: 1 D = 3.336 × 10⁻³⁰ C m.

The overall dipole moment of a molecule is the vector sum of bond dipoles.

MoleculeShapeNet dipole moment
CO₂LinearZero (dipoles cancel)
BF₃Trigonal planarZero
CH₄TetrahedralZero
H₂OBentNon-zero (about 1.85 D)
NH₃PyramidalNon-zero (about 1.47 D)
NF₃PyramidalSmall (about 0.24 D)

NH₃ has a larger dipole moment than NF₃. In NH₃ the N–H bond dipoles and the lone pair dipole point in the same direction. In NF₃ the N–F bond dipoles point opposite to the lone pair dipole and partly cancel it.

Percent ionic character = (observed dipole moment / dipole moment for 100% ionic) × 100.

Worked example 1. A diatomic molecule has bond length 100 pm. For complete charge separation, μ = e × d = 1.602 × 10⁻¹⁹ C × 1 × 10⁻¹⁰ m = 1.602 × 10⁻²⁹ C m = 4.80 D. If the observed μ is 2.40 D, ionic character = 2.40/4.80 × 100 = 50%.

Fajans' rule: covalent character in an ionic bond increases with (a) a small cation, (b) a large anion and (c) a high charge on either ion. For the sodium halides, covalent character increases in the order NaF < NaCl < NaBr < NaI because the anion becomes larger.

VSEPR theory and shapes of molecules

Valence shell electron pair repulsion (VSEPR) theory: electron pairs around the central atom arrange themselves as far apart as possible. Lone pairs occupy more space than bond pairs. The repulsion order is lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. This is why H₂O (104.5°) has a smaller angle than NH₃ (107°), which has a smaller angle than CH₄ (109.5°).

Bond pairsLone pairsShapeExample
20LinearBeCl₂, CO₂
30Trigonal planarBF₃
21BentSO₂
40TetrahedralCH₄
31Trigonal pyramidalNH₃
22BentH₂O
50Trigonal bipyramidalPCl₅
41See-sawSF₄
32T-shapedClF₃
60OctahedralSF₆
51Square pyramidalBrF₅
42Square planarXeF₄

Valence bond theory and hybridisation

Valence bond theory: a covalent bond forms by the overlap of half-filled atomic orbitals. The greater the overlap, the stronger the bond.

  • A sigma (σ) bond forms by head-on (axial) overlap along the internuclear axis. It is strong.
  • A pi (π) bond forms by sidewise overlap of p orbitals. It is weaker and lies above and below the axis.
  • A single bond is one σ; a double bond is one σ + one π; a triple bond is one σ + two π.

For example, C₂H₄ has 5 σ and 1 π bonds; C₂H₂ has 3 σ and 2 π bonds.

Hybridisation is the mixing of atomic orbitals of similar energy on the same atom to form the same number of equivalent hybrid orbitals.

HybridisationOrbitals mixedGeometryAngleExample
spone s + one pLinear180°BeCl₂, C₂H₂
sp²one s + two pTrigonal planar120°BF₃, C₂H₄
sp³one s + three pTetrahedral109.5°CH₄, NH₃, H₂O
sp³done s + three p + one dTrigonal bipyramidal90° and 120°PCl₅
sp³d²one s + three p + two dOctahedral90°SF₆

In NH₃ and H₂O the central atom is sp³ hybridised, but lone pairs occupy some of the hybrid orbitals, so the shapes are pyramidal and bent.

Molecular orbital theory

Molecular orbital (MO) theory: atomic orbitals combine by the linear combination of atomic orbitals (LCAO) to form molecular orbitals. Two atomic orbitals give two MOs: a bonding MO (lower energy, constructive combination) and an antibonding MO (higher energy, destructive combination, marked with *). Electrons fill MOs by the aufbau principle, Pauli principle and Hund's rule.

Energy order of MOs:

  • Up to N₂ (with s–p mixing): σ1s < σ*1s < σ2s < σ*2s < (π2pₓ = π2p_y) < σ2p_z < (π*2pₓ = π*2p_y) < σ*2p_z.
  • For O₂ and F₂ (no s–p mixing): σ1s < σ*1s < σ2s < σ*2s < σ2p_z < (π2pₓ = π2p_y) < (π*2pₓ = π*2p_y) < σ*2p_z.

Bond order = ½ (Nb − Na), where Nb and Na are the electrons in bonding and antibonding MOs. A stable molecule needs bond order above zero. A species with unpaired electrons is paramagnetic; with all electrons paired it is diamagnetic.

SpeciesElectronsBond orderMagnetic behaviour
H₂21Diamagnetic
He₂40Does not exist
Li₂61Diamagnetic
B₂101Paramagnetic (two unpaired in π orbitals)
C₂122Diamagnetic
N₂143Diamagnetic
O₂162Paramagnetic (two unpaired in π*)
O₂⁺152.5Paramagnetic (one unpaired)
O₂⁻171.5Paramagnetic (one unpaired)
F₂181Diamagnetic

Worked example 2. N₂ has 14 electrons: σ1s² σ*1s² σ2s² σ*2s² π2p⁴ σ2p². Nb = 10, Na = 4, so bond order = (10 − 4)/2 = 3.

For O₂⁻ with 17 electrons, Nb = 10 and Na = 7, so bond order = 1.5.

Bond length increases as bond order falls: O₂⁺ < O₂ < O₂⁻.

Hydrogen bonding

A hydrogen bond is the attraction between a hydrogen atom covalently bonded to a highly electronegative atom (F, O or N) and a lone pair on another electronegative atom. It is written X–H···Y.

  • Intermolecular hydrogen bonding occurs between different molecules (H₂O, HF, NH₃, alcohols). It raises melting and boiling points and solubility in water.
  • Intramolecular hydrogen bonding occurs within one molecule (ortho-nitrophenol). It lowers boiling point, makes the compound volatile in steam and reduces solubility in water. Para-nitrophenol has intermolecular bonding and a higher boiling point.

Effects: water has a much higher boiling point than H₂S, H₂Se and H₂Te; ice has an open cage-like structure with empty spaces, so it is less dense than liquid water and floats.

NEET traps

  • 1. BF₃ is non-polar (symmetrical) but NH₃ and NF₃ are polar, and NH₃ has the larger dipole moment.
  • 2. H₂O has two lone pairs, so the angle is smaller than in NH₃; both are sp³.
  • 3. In O₂ the σ2p_z MO is below π2p; in N₂ the order is reversed.
  • 4. O₂ is paramagnetic; N₂, C₂ and F₂ are diamagnetic; B₂ is paramagnetic.
  • 5. Bond length rises as bond order falls: O₂⁺ < O₂ < O₂⁻.
  • 6. Higher lattice enthalpy means a more stable ionic solid.
  • 7. Intramolecular hydrogen bonding lowers boiling point; intermolecular raises it.
  • 8. Octet rule fails for BF₃, PCl₅, SF₆ and NO.
  • 9. Shape and electron-pair geometry differ when lone pairs are present: NH₃ has tetrahedral electron geometry but pyramidal shape.
  • 10. A π bond is formed by sidewise overlap and is weaker than a σ bond.

One-liners

  • 1. Formal charge = valence electrons − non-bonding electrons − ½ bonding electrons.
  • 2. Lattice enthalpy is the energy to separate one mole of a solid ionic compound into gaseous ions.
  • 3. μ = q × d; 1 D = 3.336 × 10⁻³⁰ C m.
  • 4. CO₂, BF₃ and CH₄ have zero dipole moment.
  • 5. Bond angles: CH₄ 109.5°, NH₃ about 107°, H₂O about 104.5°.
  • 6. SF₄ is see-saw; ClF₃ is T-shaped; XeF₄ is square planar; BrF₅ is square pyramidal.
  • 7. sp, sp², sp³ give 180°, 120° and 109.5°.
  • 8. SF₆ is sp³d² and octahedral; PCl₅ is sp³d and trigonal bipyramidal.
  • 9. A double bond is one σ and one π bond.
  • 10. Bond order = ½ (Nb − Na).
  • 11. Bond order of N₂ is 3 and of O₂ is 2.
  • 12. O₂⁺ has bond order 2.5; O₂⁻ has 1.5.
  • 13. He₂ does not exist because its bond order is zero.
  • 14. Hydrogen bonding is possible with F, O and N.
  • 15. Ortho-nitrophenol has intramolecular hydrogen bonding.

Practice questions

  1. The bond order of the N₂ molecule is

    1. 2
    2. 1
    3. 3
    4. 2.5
    Answer

    C. 3

    N₂: Nb = 10, Na = 4; (10 − 4)/2 = 3.

  2. The shape of the NH₃ molecule is

    1. trigonal planar
    2. trigonal pyramidal
    3. bent
    4. tetrahedral
    Answer

    B. trigonal pyramidal

    Three bond pairs and one lone pair give a pyramidal shape.

  3. The hybridisation of the central atom in BF₃ is

    1. sp²
    2. sp³
    3. sp³d
    4. sp
    Answer

    A. sp²

    Three bond pairs, no lone pair: sp², trigonal planar.

  4. Which of the following molecules has zero dipole moment?

    1. SO₂
    2. NH₃
    3. H₂O
    4. CO₂
    Answer

    D. CO₂

    CO₂ is linear and its two C=O dipoles cancel; the others are bent or pyramidal.

  5. Which of the following molecules does not obey the octet rule?

    1. CH₄
    2. NH₃
    3. BF₃
    4. H₂O
    Answer

    C. BF₃

    Boron in BF₃ has only six electrons.

  6. The formal charge on nitrogen in the NH₄⁺ ion is

    1. +1
    2. −1
    3. +2
    4. 0
    Answer

    A. +1

    5 − 0 − ½(8) = +1.

  7. The number of σ and π bonds in ethene, C₂H₄, is

    1. 4 σ and 2 π
    2. 3 σ and 3 π
    3. 4 σ and 1 π
    4. 5 σ and 1 π
    Answer

    D. 5 σ and 1 π

    Four C–H σ, one C–C σ and one C–C π.

  8. The shape of the SF₄ molecule is

    1. T-shaped
    2. see-saw
    3. tetrahedral
    4. square planar
    Answer

    B. see-saw

    Four bond pairs and one lone pair around S give a see-saw shape.

  9. The bond order of the O₂⁻ ion is

    1. 2
    2. 2.5
    3. 1
    4. 1.5
    Answer

    D. 1.5

    17 electrons: Nb = 10, Na = 7; (10 − 7)/2 = 1.5.

  10. Which of the following species is diamagnetic?

    1. O₂
    2. C₂
    3. B₂
    4. O₂⁺
    Answer

    B. C₂

    C₂ (12 electrons) has all electrons paired; B₂, O₂ and O₂⁺ have unpaired electrons.

  11. The correct order of increasing bond length is

    1. O₂ < O₂⁺ < O₂⁻
    2. O₂⁺ < O₂⁻ < O₂
    3. O₂⁺ < O₂ < O₂⁻
    4. O₂⁻ < O₂ < O₂⁺
    Answer

    C. O₂⁺ < O₂ < O₂⁻

    Bond orders are 2.5, 2 and 1.5; bond length increases as bond order falls.

  12. The hybridisation of sulphur in SF₆ is

    1. sp³d²
    2. sp²d
    3. sp³d
    4. sp³
    Answer

    A. sp³d²

    Six bond pairs give sp³d² and an octahedral shape.

  13. The number of lone pairs on the central atom of ClF₃ is

    1. 1
    2. 2
    3. 3
    4. 0
    Answer

    B. 2

    Cl has 7 valence electrons; 3 are used in bonds and 4 remain as two lone pairs.

  14. The correct order of increasing bond angle is

    1. H₂O < NH₃ < CH₄
    2. NH₃ < H₂O < CH₄
    3. CH₄ < NH₃ < H₂O
    4. H₂O < CH₄ < NH₃
    Answer

    A. H₂O < NH₃ < CH₄

    104.5° < 107° < 109.5°; more lone pairs mean more repulsion and a smaller angle.

  15. Which of the following compounds shows intramolecular hydrogen bonding?

    1. water
    2. hydrogen fluoride
    3. para-nitrophenol
    4. ortho-nitrophenol
    Answer

    D. ortho-nitrophenol

    The –OH and –NO₂ groups are close in the ortho isomer and bond within the same molecule.

  16. Given below are two statements: Statement I: Carbon dioxide is non-polar even though each C=O bond is polar. Statement II: The two bond dipoles in CO₂ are equal in magnitude and opposite in direction. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Equal and opposite dipoles in a linear molecule cancel.

  17. Given below are two statements: Statement I: N₂ has a bond order of 3, the highest among the diatomic molecules of the second period. Statement II: N₂ is paramagnetic. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    N₂ has all electrons paired, so it is diamagnetic.

  18. Given below are two statements: Statement I: In O₂, the π2p molecular orbitals have lower energy than the σ2p_z orbital. Statement II: In N₂, the π2p orbitals have lower energy than σ2p_z because of s–p mixing. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    In O₂ the σ2p_z is lower; in N₂ s–p mixing puts π2p lower.

  19. Given below are two statements: Statement I: An ionic compound with a higher lattice enthalpy is less stable. Statement II: Lattice enthalpy decreases as the charge on the ions increases. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Higher lattice enthalpy gives greater stability; it increases with ionic charge.

  20. Given below are two statements: Statement I: A small cation and a large anion favour covalent character in an ionic bond. Statement II: A greater charge on the cation increases its polarising power. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both follow from Fajans' rule.

  21. Which of the following statements are correct? 1. The dipole moment of NH₃ is greater than that of NF₃. 2. BF₃ is a polar molecule. 3. CH₄ is tetrahedral with a bond angle of 109.5°.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    B. 1 and 3 only

    BF₃ is symmetrical and non-polar, so 2 is false.

  22. Given below are two statements: Statement I: In sp² hybridisation, the three hybrid orbitals are directed at 120° to each other. Statement II: A π bond is formed by the head-on overlap of orbitals. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    π bonds form by sidewise overlap.

  23. Given below are two statements: Statement I: The He₂ molecule is stable because its bond order is 1. Statement II: The bond order of H₂ is 1. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    He₂ has bond order 0 and does not exist.

  24. Which of the following statements are correct? 1. Lone pair–lone pair repulsion is greater than lone pair–bond pair repulsion. 2. Bond pair–bond pair repulsion is the weakest. 3. Lone pairs occupy less space than bond pairs.

    1. 1 and 3 only
    2. 2 and 3 only
    3. 1, 2 and 3
    4. 1 and 2 only
    Answer

    D. 1 and 2 only

    Lone pairs occupy more space, so 3 is false.

  25. Given below are two statements: Statement I: Resonance structures differ only in the position of electrons and not in the position of atoms. Statement II: The resonance hybrid is more stable than any single contributing structure. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both are the basic features of resonance.

  26. Assertion (A): The dipole moment of BF₃ is zero. Reason (R): BF₃ is a symmetrical trigonal planar molecule in which the bond dipoles cancel. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Cancelling dipoles give zero net dipole moment.

  27. Assertion (A): The H–O–H bond angle in water is 109.5°. Reason (R): Lone pair–bond pair repulsion is greater than bond pair–bond pair repulsion. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    The angle is about 104.5° (A false); R is true.

  28. Assertion (A): The O₂ molecule is paramagnetic. Reason (R): All the electrons in O₂ are paired in molecular orbitals. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    O₂ has two unpaired electrons in π* orbitals, so R is false.

  29. Assertion (A): The bond order of C₂ is 2. Reason (R): C₂ is diamagnetic. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both are true but diamagnetism is not the reason for the bond order.

  30. Assertion (A): The NH₃ molecule is pyramidal. Reason (R): The nitrogen atom has three bond pairs and one lone pair. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    The lone pair pushes the bonds into a pyramidal shape.

  31. Assertion (A): The boiling point of HF is lower than that of HCl. Reason (R): HF molecules form hydrogen bonds with each other. Choose the correct option.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    HF has the higher boiling point because of hydrogen bonding, so A is false.

  32. Match List I with List II. List I: a. BeCl₂ b. BF₃ c. NH₃ d. SF₆. List II: i. octahedral ii. pyramidal iii. linear iv. trigonal planar

    1. a-iii, b-ii, c-iv, d-i
    2. a-iv, b-iii, c-ii, d-i
    3. a-iii, b-iv, c-ii, d-i
    4. a-iii, b-iv, c-i, d-ii
    Answer

    C. a-iii, b-iv, c-ii, d-i

    Linear, trigonal planar, pyramidal and octahedral.

  33. Match List I with List II. List I: a. C₂H₂ b. C₂H₄ c. NH₃ d. PCl₅. List II (hybridisation of the central atom or carbon): i. sp³d ii. sp iii. sp³ iv. sp²

    1. a-ii, b-iv, c-i, d-iii
    2. a-ii, b-iv, c-iii, d-i
    3. a-ii, b-iii, c-iv, d-i
    4. a-iv, b-ii, c-iii, d-i
    Answer

    B. a-ii, b-iv, c-iii, d-i

    C₂H₂ sp; C₂H₄ sp²; NH₃ sp³; PCl₅ sp³d.

  34. Match List I with List II. List I: a. N₂ b. O₂ c. O₂⁻ d. Li₂. List II (bond order): i. 1 ii. 2 iii. 3 iv. 1.5

    1. a-ii, b-iii, c-iv, d-i
    2. a-iii, b-ii, c-i, d-iv
    3. a-iii, b-ii, c-iv, d-i
    4. a-iii, b-iv, c-ii, d-i
    Answer

    C. a-iii, b-ii, c-iv, d-i

    N₂ 3; O₂ 2; O₂⁻ 1.5; Li₂ 1.

  35. Match List I with List II. List I: a. SF₄ b. ClF₃ c. BrF₅ d. XeF₄. List II: i. square planar ii. T-shaped iii. see-saw iv. square pyramidal

    1. a-iv, b-ii, c-iii, d-i
    2. a-ii, b-iii, c-iv, d-i
    3. a-iii, b-ii, c-iv, d-i
    4. a-iii, b-ii, c-i, d-iv
    Answer

    C. a-iii, b-ii, c-iv, d-i

    Lone-pair shapes from trigonal bipyramidal and octahedral arrangements.

  36. Match List I with List II. List I: a. B₂ b. O₂⁺ c. C₂ d. F₂. List II: i. diamagnetic, bond order 2 ii. paramagnetic, two unpaired electrons, bond order 1 iii. paramagnetic, one unpaired electron, bond order 2.5 iv. diamagnetic, bond order 1

    1. a-iii, b-ii, c-i, d-iv
    2. a-ii, b-iii, c-i, d-iv
    3. a-i, b-iii, c-ii, d-iv
    4. a-ii, b-iii, c-iv, d-i
    Answer

    B. a-ii, b-iii, c-i, d-iv

    B₂ BO 1 with two unpaired; O₂⁺ BO 2.5 with one unpaired; C₂ BO 2 diamagnetic; F₂ BO 1 diamagnetic.

  37. Match List I with List II. List I: a. Lattice enthalpy b. Bond enthalpy c. Formal charge d. Bond order. List II: i. valence electrons − non-bonding electrons − ½ bonding electrons ii. ½ (Nb − Na) iii. energy to separate a mole of solid ionic compound into gaseous ions iv. energy to break a mole of bonds in the gas phase

    1. a-i, b-iv, c-iii, d-ii
    2. a-iv, b-iii, c-i, d-ii
    3. a-iii, b-iv, c-i, d-ii
    4. a-iii, b-iv, c-ii, d-i
    Answer

    C. a-iii, b-iv, c-i, d-ii

    Each term matched to its definition.

  38. A diatomic molecule has a bond length of 100 pm and an observed dipole moment of 2.40 D. Given that the dipole moment for complete charge separation is 4.80 D, its percent ionic character is

    1. 25%
    2. 50%
    3. 100%
    4. 20%
    Answer

    B. 50%

    2.40/4.80 × 100 = 50%.

  39. The formal charge on the central oxygen atom of ozone, drawn with one O=O double bond and one O–O single bond, is

    1. 0
    2. +2
    3. −1
    4. +1
    Answer

    D. +1

    Central O: 6 − 2 − ½(6) = +1.

  40. The total number of electrons present in antibonding molecular orbitals of N₂ is

    1. 2
    2. 6
    3. 4
    4. 10
    Answer

    C. 4

    σ*1s² and σ*2s² give 4 electrons.

  41. A molecule has three bond pairs and two lone pairs on its central atom. Its shape is

    1. T-shaped
    2. trigonal bipyramidal
    3. see-saw
    4. trigonal planar
    Answer

    A. T-shaped

    Five electron pairs with two lone pairs gives a T-shape.

  42. Among NaF, NaCl, NaBr and NaI, the compound with the maximum covalent character is

    1. NaCl
    2. NaF
    3. NaBr
    4. NaI
    Answer

    D. NaI

    The largest anion, I⁻, is the most polarised by Na⁺ (Fajans' rule).

  43. Ice floats on liquid water because

    1. hydrogen bonding gives ice an open cage-like structure with empty spaces
    2. ice molecules are held by weaker covalent bonds
    3. ice contains molecules of lower mass
    4. ice has more water molecules per unit volume
    Answer

    A. hydrogen bonding gives ice an open cage-like structure with empty spaces

    The open hydrogen-bonded structure makes ice less dense than water.

  44. The hybridisation of sulphur in SF₄ is

    1. sp³d²
    2. sp³d
    3. sp³
    4. sp²
    Answer

    B. sp³d

    Four bond pairs and one lone pair make five electron pairs, so sp³d.

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