NCERT Solutions for Class 11 Chemistry Chapter 4: Chemical Bonding and Molecular Structure – Free PDF Download

Chapter 4, “Chemical Bonding and Molecular Structure,” explains why and how atoms combine to form molecules and ions — covering the Kossel-Lewis approach, ionic and covalent bonding, VSEPR theory, dipole moment, hybridisation, valence bond theory, resonance, and molecular orbital theory (MOT) with bond order and magnetic behaviour of diatomic species.

NCERT Exercise Solutions

Question 4.1

Question: Explain the formation of a chemical bond.

Solution: A chemical bond is the force of attraction that holds two or more atoms or ions together in a molecule or compound. According to the Kossel-Lewis approach, atoms combine with each other to achieve a stable, noble-gas-like electronic configuration (octet) in their valence shell. This can happen either by complete transfer of one or more electrons from one atom to another, producing oppositely charged ions that attract each other electrostatically (ionic/electrovalent bond), or by mutual sharing of one or more pairs of electrons between two atoms so that both atoms attain a stable outer shell (covalent bond). In both cases the resulting bonded state has lower energy (greater stability) than the separate atoms, and this decrease in energy on bond formation is what drives and explains bond formation.

Question 4.2

Question: Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.

Solution: The number of dots shown around the symbol equals the number of valence electrons of the atom.
Mg (Group 2, 2 valence e^–): Mg with 2 dots.
Na (Group 1, 1 valence e^–): Na with 1 dot.
B (Group 13, 3 valence e^–): B with 3 dots.
O (Group 16, 6 valence e^–): O with 6 dots (2 lone pairs and 2 single dots).
N (Group 15, 5 valence e^–): N with 5 dots (1 lone pair and 3 single dots).
Br (Group 17, 7 valence e^–): Br with 7 dots (3 lone pairs and 1 single dot).

Question 4.3

Question: Write Lewis symbols for the following atoms and ions: S and S^2-; Al and Al³⁺; H and H^–.

Solution: S has 6 valence electrons, shown as S with 6 dots; S^2- has gained 2 electrons, so it is shown in square brackets as [S with 8 dots]^2- (complete octet). Al has 3 valence electrons, shown as Al with 3 dots; Al³⁺ has lost all 3 valence electrons, so it is shown simply as [Al]³⁺ with no dots. H has 1 valence electron, shown as H with 1 dot; H^– has gained 1 electron, shown in brackets as [H with 2 dots]^–.

Question 4.4

Question: Draw the Lewis structures for the following molecules and ions: H₂S, SiCl₄, BeF₂, CO₃^2-, HCOOH.

Solution:
H₂S: H–S–H, with sulfur carrying two lone pairs (bent/angular arrangement of atoms, S completes its octet with 2 bond pairs + 2 lone pairs).
SiCl₄: Si is the central atom singly bonded to four Cl atoms (Si–Cl × 4); each Cl carries 3 lone pairs, Si has a complete octet with no lone pair (tetrahedral arrangement).
BeF₂: F–Be–F, a linear structure; each F carries 3 lone pairs; Be has only 4 electrons around it (incomplete octet, an exception).
CO₃^2-: Central C is bonded to three O atoms — one C=O double bond and two C–O single bonds, each singly bonded O carrying a negative charge and 3 lone pairs, the doubly bonded O carrying 2 lone pairs. The actual ion is a resonance hybrid of three such structures with the double bond delocalised equally over all three C–O bonds, giving overall charge -2.
HCOOH (formic acid): H–C(=O)–O–H; the carbon is bonded to one H, doubly bonded to one O, and singly bonded to another O which is further bonded to H.

Question 4.5

Question: Define octet rule. Write its significance and limitations.

Solution: The octet rule states that atoms tend to combine (by transfer or sharing of electrons) so as to attain eight electrons in their outermost (valence) shell, thereby acquiring the stable electronic configuration of the nearest noble gas.
Significance: It provides a simple and useful basis for understanding the structures and formulae of a very large number of ionic and covalent compounds (e.g., NaCl, MgCl₂, CH₄, NH₃, H₂O, CCl₄) and helps predict how atoms bond.
Limitations: (i) Incomplete octet of the central atom – some stable molecules have fewer than 8 electrons around the central atom, e.g., LiCl, BeH₂, BCl₃. (ii) Odd-electron molecules – species like NO and NO₂ have an odd number of valence electrons, so the octet rule cannot be satisfied for all atoms. (iii) Expanded octet – many molecules such as PF₅, SF₆, and H₂SO₄ have a central atom with more than 8 electrons around it. (iv) The rule is based only on electron counting and does not explain the relative stability/energy of a molecule or the shape of a molecule. (v) It does not account for the fact that noble gases themselves (e.g., Xe) form compounds such as XeF₂, XeF₄.

Question 4.6

Question: Write the favourable factors for the formation of ionic bond.

Solution: Ionic bond formation is favoured when: (i) the metal atom has a low ionisation enthalpy, so that it can easily lose an electron to form a cation; (ii) the non-metal atom has a highly negative electron gain enthalpy, so that it can easily gain an electron to form an anion; and (iii) the lattice enthalpy of the resulting crystalline ionic solid is high, i.e., a large amount of energy is released when the oppositely charged ions arrange themselves into a stable, three-dimensional crystal lattice; this high lattice enthalpy compensates for the energy consumed in ion formation and makes the overall process energetically favourable.

Question 4.7

Question: Discuss the shape of the following molecules using the VSEPR model: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃.

Solution:
BeCl₂: 2 bond pairs, 0 lone pairs on Be → linear shape (Cl–Be–Cl, 180°).
BCl₃: 3 bond pairs, 0 lone pairs on B → trigonal planar shape (120°).
SiCl₄: 4 bond pairs, 0 lone pairs on Si → tetrahedral shape (109.5°).
AsF₅: 5 bond pairs, 0 lone pairs on As → trigonal bipyramidal shape (3 equatorial at 120°, 2 axial at 90° to equatorial plane).
H₂S: 2 bond pairs + 2 lone pairs on S → bent/angular shape (bond angle close to 92°, less than tetrahedral due to lone pair-lone pair repulsion).
PH₃: 3 bond pairs + 1 lone pair on P → trigonal pyramidal shape (bond angle about 93.5°, less than tetrahedral due to lone pair-bond pair repulsion).

Question 4.8

Question: Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

Solution: Both N in NH₃ and O in H₂O are sp³ hybridised, so the parent geometry in both cases is tetrahedral (ideal angle 109.5°). However, NH₃ has 3 bond pairs and 1 lone pair, while H₂O has 2 bond pairs and 2 lone pairs. The order of repulsive strength is lone pair-lone pair (lp-lp) > lone pair-bond pair (lp-bp) > bond pair-bond pair (bp-bp). In H₂O, the two lone pairs repel each other strongly (lp-lp repulsion), squeezing the H–O–H bond angle down to about 104.5°. In NH₃, there is only one lone pair, so the repulsion is of the weaker lp-bp type, and the H–N–H bond angle is compressed less, to about 107°. Hence, because water experiences greater lone-pair repulsion (two lone pairs) than ammonia (one lone pair), its bond angle is smaller.

Question 4.9

Question: How do you express the bond strength in terms of bond order?

Solution: Bond order is defined as the number of bonds (single, double or triple) between two bonded atoms in a molecule; equivalently, in molecular orbital theory, bond order = ½(number of electrons in bonding MOs – number of electrons in antibonding MOs). Bond strength (bond dissociation energy) is directly proportional to bond order — a higher bond order means a stronger, shorter bond, while a lower bond order means a weaker, longer bond. For example, a triple bond (bond order 3) is stronger and shorter than a double bond (bond order 2), which in turn is stronger and shorter than a single bond (bond order 1).

Question 4.10

Question: Define the bond length.

Solution: Bond length is defined as the equilibrium distance between the centres of the nuclei of two covalently (or otherwise) bonded atoms in a molecule. It is experimentally determined using techniques such as X-ray diffraction, electron diffraction, and spectroscopic methods, and is usually expressed in picometres (pm) or angstroms (Å).

Question 4.11

Question: Explain the important aspects of resonance with reference to the CO₃^2- ion.

Solution: The carbonate ion, CO₃^2-, cannot be represented satisfactorily by a single Lewis structure, since experimentally all three C–O bonds are found to be of equal length (136 pm), intermediate between a C=O double bond (~121 pm) and a C–O single bond (~143 pm). It can be drawn as three equivalent canonical (resonance) structures, in each of which the carbon has one C=O double bond with one particular oxygen and single bonds (bearing the negative charges) with the other two oxygens; the three structures differ only in which oxygen carries the double bond. The actual carbonate ion is a resonance hybrid of these three canonical forms, and is not any one of them individually — it has a single delocalised structure with all C–O bonds identical in length (intermediate between single and double bond) and the -2 charge spread equally over the three oxygen atoms. This illustrates that resonance structures are hypothetical/imaginary, differ only in electron arrangement (not atomic positions), and the true molecule is a hybrid that is more stable (lower energy) than any single contributing structure.

Question 4.12

Question: 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.

Solution: Structure 1 shows phosphorus bonded to three –OH groups (three P–OH bonds) plus one P=O bond, while structure 2 shows phosphorus bonded to two –OH groups, one P=O bond, and one P–H bond directly attached to phosphorus. These two structures cannot be taken as canonical (resonance) forms of a single resonance hybrid, because true resonance structures must have the same relative positions of all atomic nuclei and differ only in the arrangement/distribution of electrons (as in CO₃^2-). Here, in structure 1 all three H atoms are attached through oxygen (three P–O–H), whereas in structure 2 one H atom is directly bonded to phosphorus (P–H) — the positions of the atoms themselves are different in the two structures, which is not permissible for canonical forms. In fact, only one of the two, structure 2 (with one direct P–H bond and two P–OH groups), is the experimentally correct structure of H₃PO₃; this is consistent with the observed fact that H₃PO₃ is a diprotic acid (only 2 ionisable/acidic hydrogens, since the P–H hydrogen is not acidic).

Question 4.13

Question: Write the resonance structures for SO₃, NO₂ and NO₃^–.

Solution:
SO₃: three equivalent canonical structures, each with sulfur singly bonded to two oxygens and doubly bonded to the third oxygen (trigonal planar S), the position of the S=O double bond rotating among the three oxygens across the three forms.
NO₂: two canonical (resonance) structures for this odd-electron, bent molecule, in which the N=O double bond and the N–O single bond (carrying the odd/unpaired electron on the singly bonded oxygen side) interchange positions between the two oxygen atoms.
NO₃^–: three equivalent canonical structures, each with nitrogen doubly bonded to one oxygen and singly bonded (with a formal negative charge) to the other two oxygens, the position of the N=O double bond rotating among the three oxygens (trigonal planar hybrid with equal N–O bond lengths).

Question 4.14

Question: 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.

Solution:
(a) K and S: Each K atom (1 valence e^–) transfers its single electron to S (6 valence e^–, needs 2 more). Two K atoms are needed to supply the 2 electrons S requires, giving 2K⁺ and S^2-, forming K₂S.
(b) Ca and O: Ca (2 valence e^–) transfers both its electrons to O (6 valence e^–, needs 2 more), giving Ca²⁺ and O^2- in a 1:1 ratio, forming CaO.
(c) Al and N: Al (3 valence e^–) transfers all 3 electrons to N (5 valence e^–, needs 3 more), giving Al³⁺ and N^3- in a 1:1 ratio, forming AlN.

Question 4.15

Question: 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.

Solution: CO₂ has an experimentally measured dipole moment of zero, even though each C=O bond is individually polar (O is more electronegative than C). This is only possible if the molecule is linear (O=C=O), so that the two equal and oppositely directed C=O bond dipoles cancel each other out exactly, giving a net dipole moment of zero. In contrast, H₂O has a non-zero dipole moment (1.84 D). This can only be explained if the molecule is bent/angular (not linear); in a bent H₂O molecule the two O–H bond dipoles do not point in exactly opposite directions, so they do not cancel and add up (along with the lone pair contribution) to give a net non-zero dipole moment directed along the bisector of the H–O–H angle. Thus, the observed (zero vs non-zero) dipole moments directly indicate the linear shape of CO₂ and the bent shape of H₂O.

Question 4.16

Question: Write the significance/applications of dipole moment.

Solution: Dipole moment measurements are useful to: (i) distinguish between polar and non-polar molecules (polar molecules have non-zero dipole moment); (ii) determine the shape/geometry of molecules, e.g., a zero dipole moment for a molecule like CO₂ or BF₃ indicates a symmetrical (linear/trigonal planar) shape, while a non-zero value indicates an unsymmetrical/bent/pyramidal shape; (iii) distinguish between geometrical isomers, such as cis- and trans- isomers (the cis isomer generally has a higher dipole moment than the corresponding trans isomer); (iv) calculate the percentage ionic character of a covalent bond, by comparing the experimentally observed dipole moment with the theoretical dipole moment calculated assuming 100% ionic character.

Question 4.17

Question: Define electronegativity. How does it differ from electron gain enthalpy?

Solution: Electronegativity is the tendency (relative measure) of an atom in a covalently bonded molecule to attract the shared pair of electrons of the bond towards itself. It differs from electron gain enthalpy in several ways: electronegativity is a relative, comparative property (measured on scales like the Pauling scale) applicable to an atom while it is part of a bonded molecule, has no fixed/absolute units, and cannot be measured directly by experiment. Electron gain enthalpy, on the other hand, is a fixed, experimentally measurable quantity (expressed in kJ mol^-1) defined as the enthalpy change when an isolated, neutral, gaseous atom gains an electron to form a gaseous anion; it is a property of the free/isolated atom and is independent of the atom being part of any particular bonded molecule.

Question 4.18

Question: Explain with the help of suitable example polar covalent bond.

Solution: A polar covalent bond is formed between two atoms of different electronegativities that share a pair of electrons unequally; the shared electron pair is displaced/pulled more towards the more electronegative atom, creating partial charges without complete transfer of electrons. For example, in the HCl molecule, chlorine (more electronegative) pulls the shared bonding electron pair towards itself, developing a partial negative charge (δ-) on Cl, while hydrogen develops a corresponding partial positive charge (δ+); this unequal charge distribution constitutes a permanent dipole and the H–Cl bond is thus a polar covalent bond (unlike a purely covalent bond, e.g., in Cl₂, where electron sharing is equal, and unlike a purely ionic bond where electrons are fully transferred).

Question 4.19

Question: Arrange the bonds in order of increasing ionic character in the molecules: LiF, K₂O, N₂, SO₂ and ClF₃.

Solution: Ionic character of a bond increases with increasing difference in electronegativity between the bonded atoms. N₂ is a homonuclear diatomic molecule with zero electronegativity difference, so it has purely covalent (non-polar, zero ionic character) bonds. In ClF₃ and SO₂, the electronegativity differences between the bonded atoms are small and comparable, giving these bonds a small but similar degree of ionic character (ClF₃ slightly less than SO₂). K₂O and LiF involve a metal bonded to a highly electronegative non-metal, giving large electronegativity differences and hence high ionic character, with LiF (electronegativity difference about 3.0) being more ionic than K₂O (difference about 2.6). Thus, the order of increasing ionic character is: N₂ < ClF₃ < SO₂ < K₂O < LiF.

Question 4.20

Question: 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.

Solution: The correct Lewis structure of acetic acid (CH₃COOH) is: the first (methyl) carbon is singly bonded to three hydrogen atoms and singly bonded to the second (carbonyl) carbon (4 single bonds total, complete octet, no lone pair on this C). The second carbon is singly bonded to the methyl carbon, doubly bonded to one oxygen atom (C=O, the carbonyl oxygen, which carries 2 lone pairs), and singly bonded to a second oxygen atom (which in turn is singly bonded to a hydrogen atom and carries 2 lone pairs, i.e., the hydroxyl –O–H group). So the correct structure is CH₃–C(=O)–O–H, with no double bonds to the methyl group and no C–H bonds on the carbonyl carbon.

Question 4.21

Question: 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?

Solution: According to the VSEPR model, four bond pairs of electrons around a central atom arrange themselves so as to minimise electron pair-electron pair repulsion, which is achieved at the maximum possible bond angle. In a tetrahedral arrangement, the four bond pairs are separated by an angle of 109.5°, whereas in a hypothetical square planar arrangement they would be separated by only 90° (adjacent) and 180° (opposite). The smaller 90° angle in the square planar arrangement causes much greater electron pair-electron pair (and H···H) repulsion than the 109.5° angle of the tetrahedral arrangement, making the square planar geometry much higher in energy and less stable. Consistent with this, CH₄ is experimentally found to have a dipole moment of zero and four equal C–H bond lengths/angles, confirming the symmetric tetrahedral (sp³ hybridised) geometry rather than square planar.

Question 4.22

Question: Explain why BeH₂ molecule has a zero dipole moment although the Be–H bonds are polar.

Solution: BeH₂ is a linear molecule (sp hybridised Be, H–Be–H, bond angle 180°, no lone pair on Be). Each individual Be–H bond is polar (since Be and H have different electronegativities), and each bond therefore has its own bond dipole moment vector. However, because the molecule is linear and symmetric, the two Be–H bond dipole vectors are equal in magnitude but point in exactly opposite directions (180° apart). Their vector sum is therefore zero, so despite having individually polar bonds, the overall/net dipole moment of the BeH₂ molecule is zero.

Question 4.23

Question: Which out of NH₃ and NF₃ has higher dipole moment and why?

Solution: NH₃ has a higher dipole moment (about 1.47 D) than NF₃ (about 0.24 D). In NH₃, hydrogen is less electronegative than nitrogen, so each N–H bond dipole points from H towards N (i.e., towards the nitrogen atom, in the same general direction as the lone pair on N); the bond dipoles and the lone pair dipole moment reinforce/add up, giving a relatively large net dipole moment. In NF₃, however, fluorine is more electronegative than nitrogen, so each N–F bond dipole points from N towards F (i.e., away from the nitrogen atom, opposite to the direction of the lone pair on N). Here the bond dipole moments and the lone pair dipole moment oppose/partly cancel each other, resulting in a much smaller net dipole moment for NF₃ compared to NH₃.

Question 4.24

Question: What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.

Solution: Hybridisation is the process of intermixing of atomic orbitals of slightly different energies (belonging to the same atom) so as to give rise to an equal number of new, equivalent orbitals (of the same shape and energy) called hybrid orbitals, which have definite geometrical orientations suited to forming bonds with maximum overlap and minimum repulsion.
sp hybridisation: one s and one p orbital mix to give 2 sp hybrid orbitals, oriented at 180° to each other, giving a linear shape/geometry (e.g., BeCl₂).
sp² hybridisation: one s and two p orbitals mix to give 3 sp² hybrid orbitals, oriented at 120° to each other in one plane, giving a trigonal planar shape/geometry (e.g., BCl₃).
sp³ hybridisation: one s and three p orbitals mix to give 4 sp³ hybrid orbitals, oriented at 109.5° to each other, giving a tetrahedral shape/geometry (e.g., CH₄).

Question 4.25

Question: Describe the change in hybridisation (if any) of the Al atom in the following reaction: AlCl₃ + Cl^– → AlCl₄^–.

Solution: In AlCl₃, aluminium is sp² hybridised (3 bond pairs, trigonal planar geometry, with an empty/vacant p orbital and an incomplete octet). When AlCl₃ accepts a chloride ion (Cl^–), the lone pair on Cl^– is donated into Al’s empty orbital, forming a coordinate (dative) bond and giving the tetrahedral AlCl₄^– ion, in which Al now has 4 bond pairs. Thus, the hybridisation of the Al atom changes from sp² (in AlCl₃) to sp³ (in AlCl₄^–).

Question 4.26

Question: Is there any change in the hybridisation of B and N atoms as a result of the following reaction? BF₃ + NH₃ → F₃B·NH₃.

Solution: In BF₃, boron is sp² hybridised (3 bond pairs, trigonal planar, with an empty p orbital/incomplete octet). In NH₃, nitrogen is sp³ hybridised (3 bond pairs + 1 lone pair, pyramidal). In the addition reaction, the lone pair of NH₃‘s nitrogen is donated into boron’s empty orbital, forming a coordinate bond. As a result, boron’s hybridisation changes from sp² to sp³ (it now has 4 bond pairs, roughly tetrahedral around B). Nitrogen’s hybridisation, however, remains sp³ — it still has 4 electron pairs around it (now all 4 used in bonding, including the coordinate bond, instead of 3 bond pairs + 1 lone pair), so there is no change in nitrogen’s hybridisation.

Question 4.27

Question: Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂ molecules.

Solution: In C₂H₄ (ethylene), each carbon is sp² hybridised. Each carbon uses its three sp² hybrid orbitals to form three sigma bonds (2 C–H sigma bonds + 1 C–C sigma bond) all lying in one plane at about 120° to each other. Each carbon also has one unhybridised p orbital, oriented perpendicular to this plane; the two parallel p orbitals (one from each carbon) overlap sideways (laterally) above and below the internuclear axis to form one pi bond. The C=C double bond is thus made up of one sigma bond + one pi bond.
In C₂H₂ (acetylene), each carbon is sp hybridised. Each carbon uses its two sp hybrid orbitals to form two sigma bonds (1 C–H sigma bond + 1 C–C sigma bond) in a linear arrangement. Each carbon also has two unhybridised, mutually perpendicular p orbitals; these two pairs of parallel p orbitals overlap sideways to form two mutually perpendicular pi bonds. The C≡C triple bond is thus made up of one sigma bond + two pi bonds.

Question 4.28

Question: What is the total number of sigma and pi bonds in the following molecules? (a) C₂H₂ (b) C₂H₄.

Solution:
(a) C₂H₂ (H–C≡C–H): 2 C–H sigma bonds + 1 C–C sigma bond = 3 sigma bonds; the C≡C triple bond additionally contains 2 pi bonds. Total: 3 sigma bonds and 2 pi bonds.
(b) C₂H₄ (H₂C=CH₂): 4 C–H sigma bonds + 1 C–C sigma bond = 5 sigma bonds; the C=C double bond additionally contains 1 pi bond. Total: 5 sigma bonds and 1 pi bond.

Question 4.29

Question: 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 2p_x; (c) 2p_y and 2p_y (d) 1s and 2s.

Solution: A sigma bond requires head-on (axial) overlap of orbitals along the internuclear axis (taken here as the x-axis). (a) 1s and 1s: both are spherically symmetric, so they can overlap head-on along the x-axis → forms a sigma bond. (b) 1s and 2p_x: the 2p_x orbital is oriented along the x-axis itself, so it can overlap head-on with the spherical 1s orbital → forms a sigma bond. (c) 2p_y and 2p_y: the 2p_y orbitals are oriented along the y-axis, i.e., perpendicular to the internuclear (x) axis; they can only overlap sideways (laterally), which would give a pi bond, not head-on overlap → does NOT form a sigma bond. (d) 1s and 2s: both are spherically symmetric and can overlap head-on along the x-axis → forms a sigma bond. Answer: (c) 2p_y and 2p_y will not form a sigma bond, because they are oriented perpendicular to the internuclear axis and can only undergo lateral overlap (giving a pi bond).

Question 4.30

Question: 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.

Solution:
(a) CH₃–CH₃ (ethane): both carbon atoms are sp³ hybridised.
(b) CH₃–CH=CH₂ (propene): the CH₃ carbon is sp³ hybridised; the two doubly-bonded carbons (–CH=CH₂) are sp² hybridised.
(c) CH₃–CH₂–OH (ethanol): both carbon atoms are sp³ hybridised.
(d) CH₃–CHO (acetaldehyde): the CH₃ carbon is sp³ hybridised; the carbonyl carbon (–CHO, C=O) is sp² hybridised.
(e) CH₃COOH (acetic acid): the CH₃ carbon is sp³ hybridised; the carboxyl carbon (–COOH, C=O) is sp² hybridised.

Question 4.31

Question: What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.

Solution: A bond pair is a pair of valence electrons that is shared between two bonded atoms and constitutes a covalent bond; for example, in the H₂ molecule, the single pair of electrons shared between the two hydrogen atoms is a bond pair. A lone pair is a pair of valence electrons present on an atom that is not shared with, or involved in bonding to, any other atom; for example, in the NH₃ molecule, nitrogen has one lone pair of electrons that is not used in bonding with the hydrogen atoms (and in H₂O, oxygen has two such lone pairs).

Question 4.32

Question: Distinguish between a sigma and a pi bond.

Solution: A sigma (σ) bond is formed by the head-on (axial) overlap of atomic orbitals along the internuclear axis (s-s, s-p, or p-p overlap along the axis); it is generally stronger, allows free rotation of atoms about the bond axis, and is always the first bond formed between two atoms. A pi (π) bond is formed by the sideways (lateral) overlap of two parallel, unhybridised p (or d) orbitals, with electron density concentrated above and below the plane containing the bonded nuclei (a nodal plane along the internuclear axis); it is generally weaker than a sigma bond, restricts free rotation about the bond axis, and can only form in addition to (never in place of) an already-existing sigma bond between the same two atoms, i.e., it occurs only in double and triple bonds.

Question 4.33

Question: Explain the formation of H₂ molecule on the basis of valence bond theory.

Solution: Consider two isolated hydrogen atoms, H_A and H_B, each with one electron in a 1s orbital, and with the two electrons having opposite spins. As the two atoms approach each other, their 1s orbitals begin to overlap; this overlap increases electron density in the region between the two nuclei. The nucleus of each atom now attracts the electron cloud of the other atom in addition to its own electron, and this increased nucleus-electron attractive force outweighs the electron-electron and nucleus-nucleus repulsive forces, causing the potential energy of the system to decrease as the atoms come closer together. The energy reaches a minimum at a particular internuclear distance (the bond length, 74 pm for H₂), beyond which further approach increases repulsion and raises the energy again. At this minimum-energy distance, a stable covalent (sigma) bond is formed between the two hydrogen atoms, and the energy released in the process (about 435 kJ mol^-1) is the bond formation energy (equal in magnitude to the bond dissociation energy).

Question 4.34

Question: Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

Solution: (i) The combining atomic orbitals must have the same or nearly the same energy. (ii) The combining atomic orbitals must have the same symmetry about the molecular (internuclear) axis. (iii) The combining atomic orbitals must overlap to the maximum possible extent, since a greater extent of overlap leads to a greater degree of bonding electron density between the nuclei and hence a more stable, stronger molecular orbital/bond.

Question 4.35

Question: Use molecular orbital theory to explain why the Be₂ molecule does not exist.

Solution: Beryllium has the electronic configuration 1s²2s², so each Be atom contributes 2 valence (2s) electrons, giving a total of 4 valence electrons for the Be₂ molecule. Filling these into molecular orbitals in order of increasing energy gives the configuration KK σ2s² σ*2s² (i.e., both the bonding σ2s and the antibonding σ*2s orbitals are completely filled with 2 electrons each). The bond order is calculated as ½(N_b – N_a) = ½(2 – 2) = 0. Since the bond order is zero, there is no net bonding between the two Be atoms, and hence the Be₂ molecule is not expected to be stable and does not exist under normal conditions, consistent with experimental observation.

Question 4.36

Question: Compare the relative stability of the following species and indicate their magnetic properties: O₂, O₂⁺, O₂^– (superoxide), O₂^2- (peroxide).

Solution: Using molecular orbital theory, the bond orders of the species are: O₂ = 2, O₂⁺ = 2.5, O₂^– = 1.5, and O₂^2- = 1. Since stability increases with increasing bond order, the order of relative stability is: O₂⁺ > O₂ > O₂^– > O₂^2-. Regarding magnetic properties: O₂ has 2 unpaired electrons (in the degenerate π*2p orbitals) and is paramagnetic; O₂⁺ has 1 unpaired electron and is paramagnetic; O₂^– has 1 unpaired electron and is paramagnetic; O₂^2- has all electrons paired (the π*2p orbitals are completely filled) and is diamagnetic.

Question 4.37

Question: Write the significance of a plus and a minus sign shown in representing the orbitals.

Solution: The plus (+) and minus (–) signs used in representing atomic and molecular orbitals denote the phase (algebraic sign of the amplitude) of the orbital wave function in a particular lobe or region — they do not represent any positive or negative electric charge. When two orbital lobes of the same sign (phase) overlap, they undergo constructive interference, reinforcing electron density between the nuclei and giving a bonding molecular orbital of lower energy. When two orbital lobes of opposite sign (phase) overlap, they undergo destructive interference, cancelling electron density between the nuclei (creating a node) and giving an antibonding molecular orbital of higher energy.

Question 4.38

Question: Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?

Solution: In PCl₅, phosphorus is sp³d hybridised, giving 5 equivalent sp³d hybrid orbitals directed towards the corners of a trigonal bipyramidal geometry: 3 orbitals lie in an equatorial plane at 120° to each other, and 2 orbitals lie axially (perpendicular to this plane) at 90° to the equatorial orbitals and 180° to each other. The two types of P–Cl bonds (axial and equatorial) are not equivalent: each axial bond pair experiences repulsion from all 3 equatorial bond pairs at a close 90° angle, whereas each equatorial bond pair experiences repulsion from only 2 axial bond pairs at 90° (the other equatorial bond pairs being farther away, at 120°). Because the axial bond pairs experience greater (90°) repulsive interactions overall than the equatorial bond pairs, the axial P–Cl bonds are pushed farther from phosphorus to minimise this repulsion, making the axial P–Cl bonds longer than the equatorial P–Cl bonds.

Question 4.39

Question: Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?

Solution: A hydrogen bond is a weak electrostatic (dipole-dipole type) attractive force formed between a hydrogen atom that is covalently bonded to a highly electronegative atom (such as F, O, or N) in one molecule (or part of a molecule), and another highly electronegative atom (F, O, or N, generally bearing a lone pair) present in a nearby molecule or in the same molecule. A hydrogen bond is stronger than ordinary van der Waals forces of attraction, although it is still considerably weaker than a typical covalent or ionic bond.

Question 4.40

Question: What is meant by the term bond order? Calculate the bond order of: N₂, O₂, O₂⁺ and O₂^–.

Solution: Bond order is defined, in molecular orbital theory, as one-half of the difference between the number of electrons present in bonding molecular orbitals (N_b) and the number of electrons present in antibonding molecular orbitals (N_a), i.e., Bond order = ½(N_b – N_a).
N₂ (14 electrons): MO configuration KK σ2s² σ*2s² σ2p_z² π2p_x² π2p_y². N_b = 8, N_a = 2. Bond order = ½(8 – 2) = 3.
O₂ (16 electrons): N_b = 10, N_a = 6. Bond order = ½(10 – 6) = 2.
O₂⁺ (15 electrons, one electron removed from an antibonding π* orbital of O₂): N_b = 10, N_a = 5. Bond order = ½(10 – 5) = 2.5.
O₂^– (17 electrons, one electron added to an antibonding π* orbital of O₂): N_b = 10, N_a = 7. Bond order = ½(10 – 7) = 1.5.

Notes and Extra Questions

This chapter builds the conceptual foundation for chemical bonding used throughout the rest of the syllabus, moving from the simple Kossel-Lewis electron-dot picture and octet rule, through VSEPR theory (used to predict and rationalise molecular shapes), to the quantum-mechanical valence bond theory (hybridisation, sigma and pi bonds, resonance) and finally molecular orbital theory, which explains bond order, bond stability, and the paramagnetism/diamagnetism of species like O₂ that simple Lewis structures cannot account for. Key recurring skills tested in the exercises include drawing correct Lewis structures, predicting shapes and hybridisation states, explaining dipole moment trends, and calculating bond order from molecular orbital electron configurations. Students should pay special attention to lone pair-lone pair vs lone pair-bond pair vs bond pair-bond pair repulsion trends (Q4.8) and to MOT-based bond order/stability/magnetism questions (Q4.35, Q4.36, Q4.40), since these integrate several concepts from the chapter and are frequently the basis of exam-style variations.

Frequently Asked Questions

How many exercise questions are there in NCERT Class 11 Chemistry Chapter 4, Chemical Bonding and Molecular Structure?
The current NCERT textbook (as published on ncert.nic.in) contains 40 end-of-chapter exercise questions in this chapter, numbered 4.1 through 4.40.

What is the difference between VSEPR theory and valence bond theory as used in this chapter?
VSEPR (Valence Shell Electron Pair Repulsion) theory is a simple model used to predict the shape/geometry of a molecule based on minimising repulsion between electron pairs (bond pairs and lone pairs) around a central atom. Valence bond theory goes further, using atomic orbital overlap and hybridisation to explain how and why covalent bonds (including sigma and pi bonds) form and to justify the geometries predicted by VSEPR in terms of orbital overlap.

Why do NH₃ and H₂O have different bond angles even though both are based on tetrahedral (sp³) electron geometry?
Because they have different numbers of lone pairs on the central atom. Water has 2 lone pairs (causing stronger lone pair-lone pair repulsion and a smaller H–O–H angle of about 104.5°), while ammonia has only 1 lone pair (causing weaker lone pair-bond pair repulsion and a slightly larger H–N–H angle of about 107°).

Why is O₂ paramagnetic even though its Lewis structure shows all electrons paired?
The simple Lewis dot structure of O₂ (with a double bond and paired lone pairs) cannot correctly predict its magnetic behaviour. Molecular orbital theory shows that O₂ has two unpaired electrons in its degenerate antibonding π*2p molecular orbitals, which correctly explains its experimentally observed paramagnetism (attraction to a magnetic field) — this is one of the key successes of molecular orbital theory over the simpler Lewis/valence bond picture.