NCERT Solutions for Class 11 Chemistry Chapter 4: Chemical Bonding and Molecular Structure (2026–27)
These Class 11 Chemistry Chapter 4 solutions cover Chemical Bonding and Molecular Structure in complete detail. Every one of the 40 NCERT exercise questions is reproduced verbatim and solved step by step — Lewis structures, octet rule, ionic and covalent bonding, bond parameters, resonance, dipole moment, VSEPR shapes, valence bond theory, hybridisation, molecular orbital theory and hydrogen bonding — all updated for the session 2026–27.
Class: 11Subject: ChemistryChapter: 4Unit Name: Chemical Bonding and Molecular StructureExercises: 40 questions solvedSession: 2026–27
A chemical bond is the attractive force that holds atoms, ions or other constituents together in a chemical species. Chapter 4 explains why atoms combine and what shapes the resulting molecules take, building up four major models in order. The Kössel–Lewis approach introduces Lewis symbols, the octet rule, ionic (electrovalent) and covalent bonds, formal charge and the limitations of the octet rule. Bond parameters — bond length, bond angle, bond enthalpy, bond order, resonance and bond polarity (dipole moment) — describe real molecules quantitatively. The VSEPR theory predicts molecular geometry from the repulsion of valence-shell electron pairs. Valence Bond (VB) theory explains covalent bonding through orbital overlap and hybridisation (sp, sp2, sp3, sp3d, sp3d2), while Molecular Orbital (MO) theory accounts for bond order and magnetic behaviour of diatomic molecules. The chapter closes with hydrogen bonding. This is one of the highest-weightage chapters in the Class 11 board and NEET/JEE syllabus.
Key Concepts & Definitions
Chemical bond: the attractive force that holds constituent atoms or ions together in a molecule or crystal; bonding lowers the energy of the system and increases stability.
Octet rule: atoms tend to gain, lose or share valence electrons so as to acquire eight electrons (a stable noble-gas octet) in their outermost shell (a duplet for H).
Ionic (electrovalent) bond: the electrostatic attraction between oppositely charged ions formed by complete transfer of electrons from a metal to a non-metal; its stability is measured by lattice enthalpy.
Covalent bond: a bond formed by the mutual sharing of one or more electron pairs between atoms (single, double or triple bond).
Formal charge: the difference between the number of valence electrons of a free atom and the electrons assigned to it in a Lewis structure (lone-pair electrons + half the shared electrons).
Resonance: the representation of a molecule by two or more equivalent-energy Lewis (canonical) structures; the true structure is the resonance hybrid, which is more stable than any single canonical form.
Dipole moment (μ): the product of the magnitude of charge and the distance of separation between the centres of positive and negative charge; a vector, measured in Debye (D).
VSEPR theory: the shape of a molecule depends on the number of bonding and lone pairs around the central atom, arranged to minimise repulsion (lp–lp > lp–bp > bp–bp).
Hybridisation: the intermixing of atomic orbitals of nearly equal energy to form an equal number of equivalent hybrid orbitals with definite geometry.
Bond order: the number of bonds between two atoms; in MO theory, ½ (Nb − Na).
Hydrogen bond: the attractive force between an H atom bonded to a highly electronegative atom (F, O, N) and another electronegative atom; weaker than a covalent bond but stronger than van der Waals forces.
Important Formulas
Formal charge = (valence electrons in free atom) − (lone-pair electrons) − ½ (bonding electrons)
Dipole moment μ = Q × r (1 D = 3.33564 × 10−30 C·m)
Bond order (MO theory) = ½ (Nb − Na), where Nb = electrons in bonding MOs, Na = electrons in antibonding MOs
% ionic character ≈ (observed μ / μ for 100% ionic bond) × 100
Relation: higher bond order → greater bond enthalpy → shorter bond length
NCERT Exercise Solutions — Chapter 4
All questions are reproduced verbatim from the NCERT textbook (Reprint 2026–27). Answers are original and exam-ready.
4.1 Explain the formation of a chemical bond.
ANSWERA chemical bond is the attractive force that holds two or more atoms together in a molecule or ion. Atoms combine because, except for the noble gases, an isolated atom is unstable; by bonding, the system lowers its energy and becomes more stable.According to the Kössel–Lewis (electronic) theory, atoms combine to attain a stable noble-gas (octet) configuration in their valence shell by one of two ways:(i) Electrovalent (ionic) bond: by complete transfer of one or more valence electrons from a metal to a non-metal, forming oppositely charged ions held together by electrostatic attraction, e.g. Na+Cl−.(ii) Covalent bond: by mutual sharing of electron pairs between atoms, so that each atom attains an octet, e.g. the Cl–Cl bond in Cl2. Sharing of one, two or three pairs gives single, double or triple bonds respectively.
4.2 Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.
ANSWERA Lewis symbol shows the valence electrons as dots around the element symbol.Mg (2 valence e−): Mg with 2 dots → Mg··Na (1 valence e−): Na with 1 dot → Na·B (3 valence e−): B with 3 dotsO (6 valence e−): O with 6 dots (two lone pairs + two single electrons)N (5 valence e−): N with 5 dots (one lone pair + three single electrons)Br (7 valence e−): Br with 7 dots (three lone pairs + one single electron)
4.3 Write Lewis symbols for the following atoms and ions: S and S2−; Al and Al3+; H and H−
ANSWERS (valence shell 3s23p4, 6 valence e−): S with 6 dots.S2−: sulphur gains 2 electrons → 8 valence electrons (octet), shown with 8 dots enclosed in brackets with a 2− charge: [··S··]2− (all four sides paired).Al (3s23p1, 3 valence e−): Al with 3 dots.Al3+: aluminium loses 3 electrons → no valence electrons left, shown as [Al]3+ with no dots.H (1 valence e−): H with 1 dot → H·H−: hydrogen gains 1 electron → duplet (2 electrons), shown as [H··]−, the helium configuration.
4.4 Draw the Lewis structures for the following molecules and ions: H2S, SiCl4, BeF2, CO32−, HCOOH
ANSWERH2S (8 valence e−): S is central, bonded by single bonds to two H atoms; S carries two lone pairs. H–S–H (S has 2 lp). Each H attains a duplet, S an octet.SiCl4 (32 valence e−): Si is central, four single Si–Cl bonds; each Cl carries three lone pairs. Si attains an octet; tetrahedral.BeF2 (16 valence e−): Be is central with two single Be–F bonds; each F has three lone pairs. Be has only 4 electrons (incomplete octet, an exception to the octet rule). F–Be–F, linear.CO32− (24 valence e−): C is central; one C=O double bond and two C–O single bonds; the two singly-bonded O atoms carry the negative charges. The ion is a resonance hybrid of three equivalent structures (all C–O bonds equal). C attains an octet.HCOOH (formic acid, 18 valence e−): C is central, bonded to one H, one =O (double bond) and one –O–H (single bond, O carries 2 lone pairs and is bonded to H). Structure: H–C(=O)–O–H.
4.5 Define octet rule. Write its significance and limitations.
ANSWEROctet rule: atoms combine by transferring or sharing valence electrons in such a way that each atom attains a stable octet (eight electrons) in its valence shell, like the nearest noble gas (H attains a duplet).Significance: it successfully explains the formation of ionic and covalent bonds and the stoichiometry of most molecules and ions, especially of second-period (organic) elements.Limitations: (i) Incomplete octet — central atoms in LiCl, BeH2, BCl3, AlCl3 have fewer than 8 electrons. (ii) Odd-electron molecules — NO and NO2 cannot satisfy the octet on all atoms. (iii) Expanded octet — elements of period 3 and beyond (PF5, SF6, H2SO4) have more than 8 electrons. (iv) It does not explain molecular shape, relative stability or the bonding of noble gases (XeF2, KrF2).
4.6 Write the favourable factors for the formation of ionic bond.
ANSWERIonic bonds form readily when ion formation and lattice formation are energetically easy:(i) Low ionization enthalpy of the metal, so the cation forms easily.(ii) High negative electron gain enthalpy of the non-metal, so the anion forms easily.(iii) High lattice enthalpy — a large amount of energy released on packing the ions into a crystal lattice. This is the most decisive factor; it is favoured by small, highly charged ions.
4.7 Discuss the shape of the following molecules using the VSEPR model: BeCl2, BCl3, SiCl4, AsF5, H2S, PH3
4.8 Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
ANSWERBoth N and O are sp3 hybridised, so the ideal angle is 109.5°. In NH3 the central N has 3 bond pairs and 1 lone pair, while in H2O the central O has 2 bond pairs and 2 lone pairs.Lone pair–lone pair repulsion > lone pair–bond pair > bond pair–bond pair. The extra lone pair in water produces stronger repulsion that compresses the bond angle further. Hence the bond angle decreases from 109.5° to 107° in NH3 (one lone pair) and to 104.5° in H2O (two lone pairs). Therefore the bond angle in water is less than that in ammonia.
4.9 How do you express the bond strength in terms of bond order?
ANSWERBond order is the number of bonds between two atoms. Bond strength is directly related to bond order: the higher the bond order, the greater the bond dissociation enthalpy and the shorter (stronger) the bond.For example, the bond order of N2 (triple bond) is 3 and its bond enthalpy is 946 kJ mol−1; O2 (bond order 2) has 498 kJ mol−1; and H2 (bond order 1) has 435.8 kJ mol−1. Thus bond strength increases as bond order increases.
4.10 Define the bond length.
ANSWERBond length is the equilibrium distance between the nuclei of two bonded atoms in a molecule. It is measured by spectroscopic, X-ray diffraction and electron-diffraction techniques and is usually expressed in picometres (pm) or angstrom (Å). For a covalent bond it equals the sum of the covalent radii of the two atoms, R = rA + rB.
4.11 Explain the important aspects of resonance with reference to the CO32− ion.
ANSWERA single Lewis structure of CO32− (one C=O and two C–O bonds) wrongly suggests unequal bonds. Experimentally, all three carbon–oxygen bonds in CO32− are identical in length.To explain this, the ion is described as a resonance hybrid of three equivalent canonical structures (I, II, III) in which the double bond is shown to each oxygen in turn. Important aspects: (i) the canonical forms have no real existence; (ii) the molecule does not flip between forms; (iii) the actual structure is the single resonance hybrid; (iv) resonance averages out the bond characteristics so all bonds become equal; and (v) the hybrid is more stable (lower energy) than any single canonical form (resonance energy).
4.12 H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3? If not, give reasons for the same.
ANSWERNo, the two structures cannot be taken as canonical (resonance) forms.A basic condition for resonance is that the canonical structures must have the same relative positions of all atoms — only the positions of electrons (bonds and lone pairs) may differ. In the two structures of H3PO3, the position of one hydrogen atom is different (in one it is on oxygen, in the other on phosphorus). Since the atomic skeletons differ, they are not resonance structures but two different molecules (tautomers/isomers).
4.13 Write the resonance structures for SO3, NO2 and NO3−.
ANSWERSO3: three equivalent canonical structures in which the S=O double bond is placed on each of the three oxygen atoms in turn, the other two being S–O single bonds (each terminal O then carries the appropriate charges so all S–O bonds become equivalent in the hybrid).NO2: two resonance structures in which the N=O double bond and the N–O single bond interchange between the two oxygen atoms, so both N–O bonds are equivalent.NO3−: three equivalent canonical structures in which the N=O double bond is shown to each of the three oxygen atoms in turn (the other two being N–O− single bonds), making all three N–O bonds identical in the resonance hybrid.
4.14 Use Lewis symbols to show electron transfer between the following atoms to form cations and anions: (a) K and S (b) Ca and O (c) Al and N.
ANSWER(a) K and S: Each K (1 valence e−) loses 1 electron; S (6 valence e−) gains 2 electrons. So 2 K atoms transfer to 1 S atom: 2K + S → 2K+ + S2−, giving K2S.(b) Ca and O: Ca (2 valence e−) loses 2 electrons; O (6 valence e−) gains 2 electrons. Ca + O → Ca2+ + O2−, giving CaO.(c) Al and N: Al (3 valence e−) loses 3 electrons; N (5 valence e−) gains 3 electrons. Al + N → Al3+ + N3−, giving AlN.
4.15 Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment.
ANSWERCO2: the central C atom has no lone pair (2 double bonds, AB2), so the molecule is linear (O=C=O). The two equal C=O bond dipoles point in exactly opposite directions and cancel out, giving a net dipole moment of zero.H2O: the central O atom has two lone pairs (AB2E2), so the molecule is bent (angle 104.5°). The two O–H bond dipoles do not cancel; their vector sum gives a net dipole moment of 1.85 D.Thus the observed zero dipole moment of CO2 proves it is linear, while the non-zero dipole moment of water proves it is bent.
4.16 Write the significance/applications of dipole moment.
ANSWER(i) It helps decide the polarity of a molecule (zero μ = non-polar; non-zero μ = polar).(ii) It is used to predict the geometry/shape of molecules (e.g. zero μ of CO2 → linear; finite μ of H2O → bent).(iii) It gives the percentage ionic character of a bond (ratio of observed to theoretical 100% ionic dipole moment).(iv) It helps distinguish between cis and trans (and ortho/meta/para) isomers and determines symmetry.
4.17 Define electronegativity. How does it differ from electron gain enthalpy?
ANSWERElectronegativity is the tendency of an atom in a chemical bond (a combined state) to attract the shared pair of electrons towards itself.
Electronegativity
Electron gain enthalpy
Tendency of a bonded atom to attract the shared electron pair.
Energy change when an isolated gaseous atom gains an electron.
A relative number (no units); cannot be measured directly.
An absolute, measurable quantity expressed in kJ mol−1.
Property of an atom in a molecule.
Property of an isolated atom.
4.18 Explain with the help of suitable example polar covalent bond.
ANSWERA polar covalent bond is a covalent bond between two atoms of different electronegativities, in which the shared electron pair is displaced towards the more electronegative atom. This creates partial charges (δ+ on the less electronegative, δ− on the more electronegative atom).Example: in HF, fluorine is far more electronegative than hydrogen, so the bonding pair lies closer to F. The bond becomes Hδ+–Fδ−, a polar covalent bond having a dipole moment of 1.78 D.
4.19 Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3.
ANSWERIonic character increases with increasing electronegativity difference between the bonded atoms. N2 (identical atoms) is purely covalent; ClF3 and SO2 are covalent with small differences; K2O and LiF have large differences (metal–non-metal).Increasing ionic character: N2 < SO2 < ClF3 < K2O < LiF.
4.20 The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.
ANSWERAcetic acid is CH3COOH (24 valence electrons). The correct Lewis structure has:• A methyl carbon bonded to three H atoms by single C–H bonds and to the carboxyl carbon by a single C–C bond.• The carboxyl carbon double-bonded to one oxygen (C=O) and single-bonded to a second oxygen (C–O), which is in turn single-bonded to an H (O–H).• The carbonyl O carries two lone pairs; the hydroxyl O carries two lone pairs. Every atom satisfies its octet (duplet for H). Structure: CH3–C(=O)–O–H.
4.21 Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar?
ANSWERIn CH4 the carbon is sp3 hybridised and has four bond pairs. According to VSEPR, the four electron pairs arrange themselves to be as far apart as possible to minimise repulsion.In a tetrahedral shape the H–C–H angle is 109.5°, whereas in a square planar shape it would be only 90°. The larger 109.5° angle keeps the bond pairs farther apart, giving minimum repulsion and maximum stability. Hence CH4 adopts the tetrahedral geometry, not square planar.
4.22 Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar.
ANSWERBeH2 is a linear molecule (H–Be–H, sp hybridised Be, bond angle 180°). Each Be–H bond is polar and has a definite bond dipole.However, the two equal Be–H bond dipoles point in exactly opposite directions and so cancel each other out. The vector sum (resultant) is zero, so the net dipole moment of BeH2 is zero, even though the individual bonds are polar.
4.23 Which out of NH3 and NF3 has higher dipole moment and why?
ANSWERNH3 has the higher dipole moment (1.47 D) compared with NF3 (0.23 D), even though F is more electronegative than H.In NH3, the orbital dipole due to the lone pair on N points in the same direction as the resultant of the three N–H bond dipoles, so they add up → large net dipole moment.In NF3, the bond dipoles point from N towards F (away from N), opposite to the lone-pair orbital dipole. The two partly cancel → small net dipole moment.
4.24 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals.
ANSWERHybridisation is the intermixing of atomic orbitals of slightly different energies belonging to the same atom to form an equal number of new, equivalent orbitals (hybrid orbitals) of identical shape and energy, oriented to give minimum repulsion.sp: one s + one p orbital → two sp hybrids at 180° → linear (e.g. BeCl2, C2H2).sp2: one s + two p orbitals → three sp2 hybrids at 120° in a plane → trigonal planar (e.g. BCl3, C2H4).sp3: one s + three p orbitals → four sp3 hybrids at 109.5° → tetrahedral (e.g. CH4, SiCl4).
4.25 Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl3 + Cl− → AlCl4−
ANSWERIn AlCl3, Al has 3 bond pairs and no lone pair, so it is sp2 hybridised and trigonal planar.When AlCl3 accepts a chloride ion to form AlCl4−, the empty p orbital of Al accepts the lone pair of Cl−, giving four bonds. Al now becomes sp3 hybridised and the ion is tetrahedral. Thus the hybridisation changes from sp2 to sp3.
4.26 Is there any change in the hybridisation of B and N atoms as a result of the following reaction? BF3 + NH3 → F3B·NH3
ANSWERBoron: in BF3, B has 3 bond pairs, no lone pair → sp2 (trigonal planar). After forming the adduct it has four bonds → hybridisation changes to sp3 (tetrahedral). Yes, B changes from sp2 to sp3.Nitrogen: in NH3, N is already sp3 (3 bond pairs + 1 lone pair). It donates its lone pair to B but remains tetrahedrally sp3. No change for N (it stays sp3).
4.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules.
ANSWERC2H4 (ethene – double bond): each carbon is sp2 hybridised. The two carbons form one C–C sigma (σ) bond by head-on overlap of sp2 orbitals; each C also forms two σ C–H bonds. The unhybridised 2pz orbitals on the two carbons overlap sideways to form one pi (π) bond. So a double bond = 1 σ + 1 π.C2H2 (ethyne – triple bond): each carbon is sp hybridised. One C–C σ bond forms by sp–sp overlap, and each C forms one σ C–H bond. The two unhybridised p orbitals on each carbon overlap sideways to form two π bonds. So a triple bond = 1 σ + 2 π.(Diagrams cannot be shown; the bonding is described above.)
4.28 What is the total number of sigma and pi bonds in the following molecules? (a) C2H2 (b) C2H4
ANSWER(a) C2H2 (H–C≡C–H): 2 C–H σ bonds + 1 C–C σ bond = 3 sigma bonds; the triple bond contributes 2 pi bonds.(b) C2H4 (H2C=CH2): 4 C–H σ bonds + 1 C–C σ bond = 5 sigma bonds; the double bond contributes 1 pi bond.
4.29 Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px; (c) 2py and 2py (d) 1s and 2s.
ANSWERA sigma bond forms by end-on (axial) overlap along the internuclear (x-) axis. Orbitals lying perpendicular to this axis can only overlap sideways and give a π bond.(c) 2py and 2py will NOT form a sigma bond. These orbitals are oriented along the y-axis, perpendicular to the internuclear x-axis, so they can only overlap laterally to form a π bond, not a σ bond.(a), (b) and (d) all involve s orbitals or the px orbital, which overlap along the x-axis and do form σ bonds.
4.30 Which hybrid orbitals are used by carbon atoms in the following molecules? (a) CH3–CH3; (b) CH3–CH=CH2; (c) CH3–CH2–OH; (d) CH3–CHO (e) CH3COOH
ANSWER(a) CH3–CH3: both carbons sp3.(b) CH3–CH=CH2: the CH3 carbon is sp3; the two doubly-bonded carbons (CH= and =CH2) are sp2.(c) CH3–CH2–OH: both carbons sp3.(d) CH3–CHO: the CH3 carbon is sp3; the carbonyl carbon (–CHO) is sp2.(e) CH3COOH: the CH3 carbon is sp3; the carboxyl carbon (–COOH) is sp2.
4.31 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.
ANSWERBond pair: the pair of electrons shared between two atoms and involved in bonding. Example: in H2O each O–H bond is a shared bond pair (water has 2 bond pairs).Lone pair (non-bonding pair): a pair of valence electrons that belongs to one atom only and is not shared. Example: the oxygen atom in H2O has 2 lone pairs; the nitrogen in NH3 has 1 lone pair.
4.32 Distinguish between a sigma and a pi bond.
ANSWER
Sigma (σ) bond
Pi (π) bond
Formed by end-on (axial/head-on) overlap of orbitals.
Formed by sideways (lateral) overlap of orbitals.
Stronger; greater extent of overlap.
Weaker; smaller extent of overlap.
Can exist independently (single bond).
Exists only along with a σ bond (in double/triple bonds).
Free rotation possible about the bond.
No free rotation; restricts rotation.
Electron cloud is symmetrical about the internuclear axis.
Electron cloud lies above and below the internuclear axis.
4.33 Explain the formation of H2 molecule on the basis of valence bond theory.
ANSWERConsider two hydrogen atoms A and B with nuclei NA, NB and electrons eA, eB. When they are far apart there is no interaction. As they approach, new attractive forces (each nucleus with the other atom’s electron: NA–eB, NB–eA) and repulsive forces (eA–eB and NA–NB) begin to operate.At a particular distance the attractive forces dominate, the potential energy reaches a minimum, and the system becomes most stable. At this point the two atoms are held together, forming an H2 molecule with a bond length of 74 pm. The energy released equals the bond enthalpy (435.8 kJ mol−1), so H2 is more stable than two isolated H atoms. The bond results from the overlap of the two half-filled 1s orbitals with opposite electron spins.
4.34 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.
ANSWER(i) The combining atomic orbitals must have nearly the same energy.(ii) They must have the same symmetry about the molecular (internuclear) axis.(iii) They must overlap to the maximum extent; the greater the overlap, the stronger the bond and the lower the energy of the bonding molecular orbital.
4.35 Use molecular orbital theory to explain why the Be2 molecule does not exist.
ANSWERBe2 has 8 electrons. Its MO configuration is σ1s2 σ*1s2 σ2s2 σ*2s2.Number of bonding electrons Nb = 4; antibonding electrons Na = 4.Bond order = ½(Nb − Na) = ½(4 − 4) = 0. Since the bond order is zero, no net bond is formed, so the Be2 molecule does not exist.
4.36 Compare the relative stability of the following species and indicate their magnetic properties: O2, O2+, O2− (superoxide), O22− (peroxide)
ANSWERBond order = ½(Nb − Na). Total electrons: O2+ = 15, O2 = 16, O2− = 17, O22− = 18.
Species
Electrons
Bond order
Magnetic nature
O2+
15
2.5
Paramagnetic (1 unpaired e−)
O2
16
2.0
Paramagnetic (2 unpaired e−)
O2− (superoxide)
17
1.5
Paramagnetic (1 unpaired e−)
O22− (peroxide)
18
1.0
Diamagnetic (0 unpaired e−)
Relative stability (higher bond order = more stable): O2+ > O2 > O2− > O22−.
4.37 Write the significance of a plus and a minus sign shown in representing the orbitals.
ANSWERThe plus (+) and minus (−) signs of an orbital do not represent positive or negative charge. They represent the sign (phase) of the wave function (ψ) in different lobes of the orbital.For bonding, lobes of the same sign must overlap (positive overlap), which gives a bonding molecular orbital. Overlap of lobes of opposite sign (positive–negative) cancels out (zero/negative overlap) and gives an antibonding interaction. So the signs indicate which overlaps are constructive (bond-forming).
4.38 Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds?
ANSWERIn PCl5, phosphorus is sp3d hybridised, forming five sp3d hybrid orbitals directed towards the corners of a trigonal bipyramid. Three equatorial bonds lie in a plane at 120°, and two axial bonds lie perpendicular to that plane at 90°.The axial bonds are longer (and weaker) than the equatorial bonds because each axial bond pair suffers repulsion from three equatorial bond pairs at 90°, whereas each equatorial bond pair faces only two bond pairs at 90°. The greater repulsion on the axial bonds lengthens them.
4.39 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
ANSWERHydrogen bond: the electrostatic attraction between a hydrogen atom covalently bonded to a highly electronegative atom (F, O or N) and a lone pair on another electronegative atom of the same or a different molecule. It is represented by a dotted line, e.g. in HF, H2O and NH3.A hydrogen bond is stronger than van der Waals forces (its energy, about 10–40 kJ mol−1, is much larger), but it is much weaker than a covalent bond.
4.40 What is meant by the term bond order? Calculate the bond order of: N2, O2, O2+ and O2−.
ANSWERBond order is half the difference between the number of electrons in bonding and antibonding molecular orbitals: B.O. = ½(Nb − Na).N2 (14 e−): Nb = 10, Na = 4 → B.O. = ½(10 − 4) = 3.O2 (16 e−): Nb = 10, Na = 6 → B.O. = ½(10 − 6) = 2.O2+ (15 e−): Nb = 10, Na = 5 → B.O. = ½(10 − 5) = 2.5.O2− (17 e−): Nb = 10, Na = 7 → B.O. = ½(10 − 7) = 1.5.
Extra Practice Questions
Short Answer Type Questions
Q1. Why is the dipole moment of CCl4 zero?
ANSWERCCl4 is tetrahedral and symmetrical. The four equal C–Cl bond dipoles point towards the corners of a tetrahedron and their vector sum is zero, so the net dipole moment is zero.
Q2. Define lattice enthalpy.
ANSWERLattice enthalpy is the energy required to completely separate one mole of a solid ionic compound into its gaseous constituent ions. For NaCl it is 788 kJ mol−1; a high value indicates a stable ionic crystal.
Q3. Why is the bond angle in H2S less than in H2O?
ANSWERSulphur is less electronegative and larger than oxygen, so the bonding pairs lie farther from the central atom and repel each other less. S also uses nearly pure p orbitals. Hence the H–S–H angle (~92°) is smaller than the H–O–H angle (104.5°).
Q4. What is meant by a coordinate (dative) bond? Give an example.
ANSWERA coordinate bond is a covalent bond in which the shared electron pair is contributed by only one of the two atoms (donor). Example: in the ammonium ion NH4+, N donates its lone pair to H+.
Q5. Why does the energy of the resonance hybrid lie lower than that of any canonical form?
ANSWERDelocalisation of electrons over more than one position spreads the electron density and lowers the energy. The difference between the energy of the hybrid and the most stable canonical form is the resonance energy, which makes the molecule more stable.
Long Answer Type Questions
Q1. Explain the salient features of molecular orbital theory.
ANSWERAccording to MO theory: (i) atomic orbitals of comparable energy and proper symmetry combine to form molecular orbitals; (ii) the number of MOs formed equals the number of combining atomic orbitals — a lower-energy bonding MO and a higher-energy antibonding MO; (iii) bonding MOs (σ, π) increase electron density between nuclei and stabilise the molecule, while antibonding MOs (σ*, π*) decrease it; (iv) MOs are filled following the Aufbau principle, Pauli exclusion principle and Hund’s rule; (v) bond order = ½(Nb − Na) decides stability — a positive value means the molecule exists; and (vi) the presence of unpaired electrons makes a species paramagnetic, explaining, for example, the paramagnetism of O2 which the Lewis/VB models could not.
Q2. State the postulates of VSEPR theory and use them to explain the shapes of CH4, NH3 and H2O.
ANSWERPostulates: (i) molecular shape depends on the number of valence-shell electron pairs (bonded + lone) around the central atom; (ii) electron pairs repel one another and arrange themselves to be as far apart as possible; (iii) lone pairs occupy more space than bond pairs, so repulsion order is lp–lp > lp–bp > bp–bp; (iv) a multiple bond is treated as a single super-pair. Application: all three central atoms are sp3 (4 electron pairs, ideal 109.5°). CH4 has 4 bond pairs and no lone pair → perfect tetrahedron, 109.5°. NH3 has 3 bp + 1 lp → trigonal pyramidal, angle reduced to 107°. H2O has 2 bp + 2 lp → bent, angle further reduced to 104.5°. The increasing number of lone pairs progressively compresses the bond angle.
Q3. Compare ionic and covalent compounds with respect to their physical properties.
ANSWERIonic compounds are crystalline solids with high melting and boiling points because strong electrostatic forces (lattice energy) hold the ions together; they are usually soluble in polar solvents like water; they conduct electricity in the molten or aqueous state (mobile ions) but not as solids; and they are generally hard and brittle. Covalent compounds are often gases, liquids or soft solids with low melting and boiling points (weak intermolecular forces); they are generally soluble in non-polar solvents; they are mostly poor conductors of electricity because they have no free ions or electrons; and they consist of discrete molecules. These differences arise because ionic bonding is non-directional and very strong, while covalent bonding is directional and the molecules are held by weaker intermolecular forces.
For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.
A-R 1. Assertion: The dipole moment of CO2 is zero.
Reason: CO2 is a linear molecule in which the two equal bond dipoles cancel out.
A-R 2. Assertion: O2 is paramagnetic.
Reason: O2 has two unpaired electrons in its antibonding π* molecular orbitals.
A-R 3. Assertion: The bond angle in NH3 is greater than in H2O.
Reason: Nitrogen in NH3 has one lone pair whereas oxygen in H2O has two lone pairs.
A-R 4. Assertion: BeH2 has a non-zero dipole moment.
Reason: The Be–H bonds in BeH2 are polar.
A-R 5. Assertion: A pi bond is weaker than a sigma bond.
Reason: A pi bond is formed by sideways overlap, which is less effective than the end-on overlap of a sigma bond.
Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).
Common Mistakes to Avoid
Watch out for these
Confusing formal charge with real charge — formal charge is only a bookkeeping device, not actual charge separation.
Forgetting that the molecule with more lone pairs has the smaller bond angle (H2O < NH3 < CH4).
Assuming a molecule with polar bonds must be polar — symmetrical molecules (CO2, BeH2, BF3, CCl4) have zero net dipole moment.
Treating canonical (resonance) structures as if the molecule keeps switching between them — only the single resonance hybrid exists.
Mistaking the + and − signs on orbitals for charges — they are wave-function phases.
Writing wrong bond order — always use ½(Nb − Na) and count electrons carefully.
Forgetting that NH3 > NF3 in dipole moment because of lone-pair direction, not just electronegativity.
How to score full marks in this chapter
Always state the number of bond pairs and lone pairs before predicting a shape, and quote the exact bond angles (180°, 120°, 109.5°, 107°, 104.5°). For dipole-moment questions, mention both the bond polarity and the vector cancellation due to symmetry. For MO questions, write the full electronic configuration, compute bond order with ½(Nb − Na), and state the magnetic nature from unpaired electrons. Use subscripts and superscripts correctly when writing formulae and charges, and remember the rule: higher bond order → higher bond enthalpy → shorter bond length.
Frequently Asked Questions
What is Class 11 Chemistry Chapter 4 about?
Chapter 4, Chemical Bonding and Molecular Structure, explains why and how atoms combine. It covers the Kössel–Lewis approach, octet rule, ionic and covalent bonds, formal charge, bond parameters, resonance, dipole moment, VSEPR theory, valence bond theory, hybridisation, molecular orbital theory and hydrogen bonding.
How many exercise questions are there in Chapter 4?
There are 40 numbered exercise questions (4.1 to 4.40). All of them are reproduced verbatim from the NCERT textbook and solved step by step on this page.
Why is O2 paramagnetic but N2 is not?
Molecular orbital theory shows that O2 has two unpaired electrons in its antibonding π* orbitals, making it paramagnetic, whereas all electrons in N2 are paired, making it diamagnetic.
Are these Class 11 Chemistry Chapter 4 solutions free?
Yes. All ClearStudy NCERT Solutions for Class 11 Chemistry are free and follow the official NCERT textbook for the session 2026–27.
