NCERT Solutions for Class 11 Chemistry Chapter 4: Chemical Bonding and Molecular Structure

Concept. A chemical bond is the attractive force that holds atoms, ions or molecules together in a chemical species. Except the noble gases, no element exists as an independent atom under normal conditions; atoms come together because the combined system has lower energy, and hence is more stable, than the separate atoms. NCERT puts it as: bonding is how nature lowers the energy of a system so that it becomes stable. Different theories — Kössel–Lewis, VSEPR, valence bond and molecular orbital theory — explain how this happens.

1. Kössel–Lewis (1916) electronic theory. The noble gases are inert because their valence shell holds a stable octet (ns²np⁶; helium has a duplet). Other atoms combine to attain such an arrangement, either by transfer of valence electrons (gaining or losing) or by sharing of valence electrons. This is the octet rule.
2. Ionic (electrovalent) bond — by transfer. An atom of low ionization enthalpy loses electron(s) to form a cation, and an atom of high negative electron gain enthalpy gains them to form an anion; the ions are held together by electrostatic attraction in a crystal lattice. Example: Na ([Ne]3s¹) → Na⁺ + e⁻; Cl ([Ne]3s²3p⁵) + e⁻ → Cl⁻; Na⁺ + Cl⁻ → NaCl. The sum of ionization enthalpy (+495.8 kJ mol⁻¹) and electron gain enthalpy (−348.7 kJ mol⁻¹) is +147.1 kJ mol⁻¹, but release of the lattice enthalpy (−788 kJ mol⁻¹) makes the whole process exothermic and the compound stable.
3. Covalent bond — by sharing. Atoms that cannot easily lose or gain electrons share electron pairs (Langmuir’s term, 1919). In Cl₂, each Cl contributes one electron to a shared pair and both attain the argon octet. Sharing one, two or three pairs gives single, double or triple bonds (Cl–Cl, O=C=O, N≡N).
4. Quantum-mechanical picture (VB theory). As two H atoms approach, attraction between each nucleus and the other electron (and own electron) and repulsion between electrons and between nuclei operate; since net attraction exceeds repulsion the potential energy falls to a minimum at the bond length 74 pm; the energy released (435.8 kJ mol⁻¹) is the bond enthalpy. The partial merging (overlap) of the 1s orbitals pairs the electrons with opposite spins — this is how the covalent bond forms.

Extra depth. NCERT stresses that the octet is not the only reason: the stability of an ionic compound is judged by its lattice enthalpy, not merely by octet formation around the gaseous ions. For covalent compounds, Lewis’s model cannot explain why H₂ (435.8 kJ mol⁻¹, 74 pm) and F₂ (155 kJ mol⁻¹, 144 pm) differ although each has one shared pair; this is why VB and MO theories were developed. Real bonds are never purely ionic or purely covalent — polarity (Fajans) gives partial covalent character to ionic bonds and partial ionic character to covalent bonds.

Conclusion: A chemical bond forms because atoms lower their energy and attain noble-gas-like stability — by transferring electrons (ionic bond) or sharing electron pairs (covalent bond) — and the stable bonded state corresponds to the minimum of the potential-energy curve.

Concept. A Lewis dot symbol represents an atom by its chemical symbol surrounded by dots equal in number to its valence electrons (electrons in the outermost shell). Introduced by G. N. Lewis, these symbols show the valence electrons available for bonding, and help calculate the common (group) valence, which is either equal to the number of dots or to 8 minus the number of dots. Dots are placed singly on the four sides first and then paired.

1. Mg (Z = 12, [Ne]3s²) — 2 valence electrons → Mg with two dots (a pair or two single dots on opposite sides). Group 2; valence 2.
2. Na (Z = 11, [Ne]3s¹) — 1 valence electron → Na with one dot. Group 1; valence 1.
3. B (Z = 5, 1s²2s²2p¹) — 3 valence electrons → B with three dots. Group 13; valence 3.
4. O (Z = 8, 1s²2s²2p⁴) — 6 valence electrons → two pairs and two single dots. Valence = 8 − 6 = 2.
5. N (Z = 7, 1s²2s²2p³) — 5 valence electrons → one pair and three single dots. Valence = 3 (or 5).
6. Br (Z = 35, [Ar]3d¹⁰4s²4p⁵) — 7 valence electrons → three pairs and one single dot. Valence = 8 − 7 = 1.

Extra depth. Only the valence electrons are shown because inner-shell electrons are well protected and are not involved in the combination process. The Lewis symbol of Na, with one dot, immediately shows that it readily loses one electron to give Na⁺ (octet), while that of Br, with seven dots, shows that it needs one more electron to give Br⁻. Similarly Mg loses two electrons to form Mg²⁺, and O needs two electrons to form O²⁻. The second-period symbols (Li to Ne) with 1 to 8 dots form a regular pattern matching the periodic table.

Conclusion: Lewis symbols show valence electrons as dots: Na (1), Mg (2), B (3), N (5), O (6) and Br (7).

Concept. Atoms and ions of the same element have the same nuclear charge but different numbers of electrons. The Lewis symbol of an ion shows the valence electrons actually present in it after loss or gain of electrons, enclosed in square brackets with the charge written outside at the upper right. Cations that have lost all the valence electrons show no dots; anions that have gained electrons show the added dots to reach an octet (or a duplet for H⁻).

1. S (Z = 16, 3s²3p⁴): 6 valence electrons → S with six dots (two pairs, two single).
2. S²⁻: S + 2e⁻ → S²⁻; 8 valence electrons (argon configuration) → [S with eight dots (four pairs)]²⁻.
3. Al (Z = 13, 3s²3p¹): 3 valence electrons → Al with three dots.
4. Al³⁺: Al → Al³⁺ + 3e⁻; the three valence electrons are lost, leaving the neon-like core → [Al]³⁺ with no dots.
5. H (Z = 1, 1s¹): one valence electron → H with one dot.
6. H⁻: H + e⁻ → H⁻; two electrons (helium configuration, a duplet) → [H:]⁻ with a pair of dots.

Extra depth. Notice what the symbols tell us: Al³⁺ and S²⁻ are isoelectronic with Ne and Ar respectively, which is why they are common in compounds (Al₂S₃). The hydride ion H⁻ occurs in ionic hydrides such as NaH, in which hydrogen behaves like a halogen — the reason hydrogen is placed separately at the top of the periodic table. The charge on the ion equals the number of electrons lost (positive) or gained (negative) — this is the electrovalence of Kössel.

Conclusion: S has six dots and S²⁻ eight in brackets; Al has three dots and Al³⁺ none; H has one dot and H⁻ two dots in brackets.

Concept. Lewis structures are drawn by (1) adding the valence electrons of all atoms (adjusting for charge), (2) writing the skeleton with the least electronegative atom in the centre (H and terminal atoms outside), (3) placing one shared pair for every bond, (4) completing octets with the remaining electrons as lone pairs and using multiple bonds if an octet is incomplete.

1. H₂S: valence electrons = 2(1) + 6 = 8. S in the centre, two S–H single bonds (4 e⁻); remaining 4 electrons are two lone pairs on S. S has an octet; each H has a duplet. Bent shape (2 bp + 2 lp).
2. SiCl₄: valence electrons = 4 + 4(7) = 32. Si in the centre with four Si–Cl single bonds (8 e⁻); the remaining 24 electrons are three lone pairs on each Cl. Every atom has an octet; tetrahedral.
3. BeF₂: valence electrons = 2 + 2(7) = 16. Be in the centre with two Be–F single bonds (4 e⁻); the remaining 12 electrons are three lone pairs on each F. Be has only four electrons (incomplete octet) — an exception to the octet rule; linear.
4. CO₃²⁻: valence electrons = 4 + 3(6) + 2 = 24. C in the centre; three C–O single bonds use 6 e⁻; lone pairs on the three O atoms use 18 e⁻; carbon then has only six electrons, so one lone pair of one O is converted into a second C–O bond: one C=O and two C–O⁻, with the 2− charge written outside brackets. Three equivalent resonance forms are possible; trigonal planar.
5. HCOOH (formic acid): valence electrons = 1 + 4 + 6 + 6 + 1 = 18. Skeleton: H–C(O)–O–H with C central and the second H on oxygen. Bonds: C–H, C–O, O–H and one C=O. The carbonyl oxygen has two lone pairs, the O–H oxygen two lone pairs; carbon has four bonds (octet).

Extra depth. Check each answer by counting electrons: the number of dots plus two per bond line must equal the total valence electrons computed in step 1 (for CO₃²⁻: 6 bonds-electrons + 2 for the extra bond + 16 lone-pair electrons = 24). Formal charges (F.C. = V − L − B/2) confirm the structure: in CO₃²⁻, C = 4 − 0 − 4 = 0, =O = 0, each O⁻ = 6 − 6 − 1 = −1, total = −2. The structure with the smallest formal charges is the lowest-energy one.

Conclusion: Lewis structures: H₂S (two bonds, two lone pairs on S), SiCl₄ (four bonds, three lone pairs on each Cl), BeF₂ (two bonds, Be with four electrons), CO₃²⁻ (one C=O, two C–O⁻) and HCOOH (H–C(=O)–O–H).

Concept. The octet rule was developed by Kössel and Lewis in 1916 as the electronic theory of chemical bonding. It is based on the chemical inertness of the noble gases, whose valence shell holds eight electrons (ns²np⁶; helium has two), a particularly stable arrangement.

• Definition: atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons, in order to have an octet (eight electrons) in their valence shells.
• Significance: (i) it gives a simple logical explanation of valence and of why atoms combine — to attain the noble-gas configuration; (ii) it explains the formation of ions (Na⁺, Cl⁻, Ca²⁺, F⁻) and of ionic compounds with electrovalency; (iii) it explains covalent bonds by sharing electron pairs (Cl₂, H₂O, CO₂, N₂); (iv) it is the basis for writing Lewis dot structures and is very useful for organic compounds and second-period elements; (v) it relates group valence to the number of valence electrons (equal to the number of dots or 8 minus that number).
• Limitation 1 — incomplete octet of the central atom: in LiCl, BeH₂, BCl₃ (and AlCl₃, BF₃) the central atom has fewer than eight electrons (Li, Be, B have 1, 2, 3 valence electrons).
• Limitation 2 — odd-electron molecules: NO and NO₂ have an odd number of electrons, so the octet cannot be satisfied for all atoms.
• Limitation 3 — expanded octet: elements in and beyond the third period have d orbitals available, so the central atom may have more than eight electrons: PF₅ (10), SF₆ (12), H₂SO₄ (12) and many coordination compounds.
• Other drawbacks: noble gases such as xenon and krypton do form compounds with oxygen and fluorine (XeF₂, KrF₂, XeOF₂); the rule does not account for the shapes of molecules; and it does not explain the relative stability of molecules, being silent about the energy of a molecule.

Extra depth. Sulphur is interesting because it obeys the octet in SCl₂ but expands it in SF₆ and H₂SO₄; this shows that the octet rule is a useful guide, not a law. The limitations led to better theories: VSEPR (shapes), valence bond theory (energetics of bond formation, hybridisation) and MO theory (bond order, magnetism, stability).

Conclusion: The octet rule states that atoms combine by transfer or sharing of valence electrons to have eight electrons in the valence shell; it explains valence and bonding but fails for incomplete-octet, odd-electron and expanded-octet species and does not explain shape or energy.

Concept. An ionic (electrovalent) bond is the electrostatic attraction between oppositely charged ions formed by complete transfer of electrons from one atom to another. From the Kössel–Lewis treatment, whether an ionic compound M⁺X⁻ forms depends on two things: (i) the ease of formation of the positive and negative ions from the neutral atoms, and (ii) the arrangement of the ions in the solid, i.e. the crystal lattice. These translate into three favourable factors.

1. Low ionization enthalpy of the metal. Formation of a cation requires ionization, M(g) → M⁺(g) + e⁻, which is always endothermic; the lower the ionization enthalpy, the more easily the cation forms. This is why alkali and alkaline-earth metals (large, with low Δ_iH) form ionic compounds readily. Example: Na, Δ_iH = 495.8 kJ mol⁻¹.
2. High negative electron gain enthalpy of the non-metal. Formation of the anion, X(g) + e⁻ → X⁻(g), is favoured when energy is released; halogens and oxygen have large negative Δ_egH. Example: Cl, Δ_egH = −348.7 kJ mol⁻¹.
3. High lattice enthalpy. The lattice enthalpy of formation, M⁺(g) + X⁻(g) → MX(s), is the energy released when the ions pack into a crystal. It is large for ions of high charge and small size (strong coulombic attraction). In NaCl, the sum of ionization enthalpy and electron gain enthalpy is +147.1 kJ mol⁻¹ (unfavourable), but the lattice enthalpy of −788 kJ mol⁻¹ more than compensates, so the net energy change is negative and the ionic compound is stable. Hence lattice enthalpy is the real key to the stability of ionic compounds, not merely octet formation.
4. A large electronegativity difference between the combining atoms (a metal from the left side and a non-metal from the right side of the periodic table) ensures that the electron is transferred rather than shared.

Extra depth. The cation does not end up with a noble-gas octet in all ionic compounds (e.g. transition-metal cations), but the same three energy factors still decide. Because the lattice enthalpy depends on charge and size, MgO and CaF₂ (higher charges) are much more stable than NaCl, and ammonium salts show that even two non-metals can form an ionic compound (NH₄⁺). Fajans’ rules tell us that if the cation is small and highly charged and the anion large, the compound acquires partial covalent character.

Conclusion: Ionic bonds form readily between atoms of low ionization enthalpy and high negative electron gain enthalpy, and the resulting ions are stabilised mainly by a high lattice enthalpy.

Concept. VSEPR model: the shape of a molecule depends on the number of valence-shell electron pairs (bonded or non-bonded) around the central atom; the pairs repel one another and arrange themselves as far apart as possible. The order of repulsion is lp–lp > lp–bp > bp–bp, so lone pairs distort the ideal geometry. Procedure: count the valence electrons of the central atom, add one per bonded atom (− charge adjustments), divide by 2 for total pairs, subtract bond pairs to find lone pairs, and read the geometry from the table (2 → linear, 3 → trigonal planar, 4 → tetrahedral, 5 → trigonal bipyramidal, 6 → octahedral). Multiple bonds count as a single super pair.

1. BeCl₂: Be has 2 valence electrons, forming 2 bonds → 2 bond pairs, 0 lone pairs (AB₂). The pairs are farthest apart at 180°: linear (Cl–Be–Cl = 180°).
2. BCl₃: B has 3 valence electrons → 3 bond pairs, 0 lone pairs (AB₃): trigonal planar, Cl–B–Cl = 120°.
3. SiCl₄: Si has 4 valence electrons → 4 bond pairs, no lone pair (AB₄): tetrahedral, Cl–Si–Cl = 109.5°.
4. AsF₅: As has 5 valence electrons → 5 bond pairs, no lone pair (AB₅): trigonal bipyramidal (three equatorial F at 120°, two axial F at 90° to the plane), like PCl₅.
5. H₂S: S has 6 valence electrons; 2 bond pairs + 2 lone pairs (AB₂E₂). The four pairs are tetrahedrally arranged, but the shape of the molecule is bent (V-shaped); lp–lp and lp–bp repulsions compress the H–S–H angle below 109.5° (it is about 92°).
6. PH₃: P has 5 valence electrons; 3 bond pairs + 1 lone pair (AB₃E): the four pairs are tetrahedral but the shape is trigonal pyramidal, with the H–P–H angle reduced (about 93.5°) below 109.5° by lp–bp repulsion.

Extra depth. H₂S and PH₃ have bond angles much closer to 90° than H₂O and NH₃ because the central atoms are larger and less electronegative, so the bonding pairs are farther from the central atom and the bond pairs repel one another less; these smaller angles are values from experiment (about 92° and 93.5°) and are not given in the NCERT text. The AB₅ shape has two types of positions: axial bonds are slightly longer than equatorial bonds. AsF₅ has no lone pair so it is non-polar overall; H₂S and PH₃ are polar.

Conclusion: BeCl₂ is linear, BCl₃ trigonal planar, SiCl₄ tetrahedral, AsF₅ trigonal bipyramidal, H₂S bent and PH₃ trigonal pyramidal.

Concept. In both NH₃ and H₂O the central atom has four electron pairs (sp³ hybridised, tetrahedral arrangement with an ideal angle of 109.5°), but some of the pairs are lone pairs. According to VSEPR theory, lone pairs are localised on the central atom and occupy more space than bonding pairs, which are shared between two nuclei. The repulsions are therefore in the order lp–lp > lp–bp > bp–bp, so lone pairs squeeze the bond pairs closer together and the H–X–H angle falls below 109.5°.

1. NH₃ (AB₃E): N has three bond pairs and one lone pair. Only lp–bp repulsions (three of them) exceed bp–bp repulsions, so the H–N–H angle contracts from 109.5° to 107°. The molecule is trigonal pyramidal.
2. H₂O (AB₂E₂): O has two bond pairs and two lone pairs. Here the strongest repulsion, lp–lp repulsion, is present in addition to four lp–bp repulsions; the two lone pairs push the two O–H bonds closer, so the H–O–H angle contracts from 109.5° to 104.5°. The molecule is bent (V-shaped, angular).
3. Comparison: more lone pairs → greater distortion → smaller angle: CH₄ (0 lp) 109.5° > NH₃ (1 lp) 107° > H₂O (2 lp) 104.5°. The decrease of about 2.5° per lone pair is a consequence of the stronger repulsive effect of lone pairs.

Extra depth. The same argument is used in VB theory: the sp³ hybrid orbitals of N and O are tetrahedral, but the lone-pair orbitals are larger and more diffuse than bonding orbitals (see Figs 4.13 and 4.14 in the notes). Table 4.8 of NCERT gives the same reason for H₂O: lp–lp repulsion exceeds lp–bp repulsion, which exceeds bp–bp repulsion, so the angle shrinks from 109.5° to 104.5°. An extra effect: oxygen is more electronegative than nitrogen, which pulls the bond pairs towards itself and further reduces repulsion between them.

Conclusion: The H–O–H angle (104.5°) is smaller than the H–N–H angle (107°) because water has two lone pairs on oxygen, which cause stronger lp–lp and lp–bp repulsions than the single lone pair of ammonia.

Concept. Bond order in the Lewis description is the number of bonds between two atoms in a molecule (H₂ = 1, O₂ = 2, N₂ = 3, CO = 3); in MO theory it is defined as bond order = ½(N_b − N_a), where N_b and N_a are the electrons in bonding and antibonding orbitals. Bond strength is measured by the bond (dissociation) enthalpy — the energy needed to break one mole of bonds in the gaseous state; the larger the bond enthalpy, the stronger the bond.

1. General correlation (NCERT): with increase in bond order, bond enthalpy increases and bond length decreases. Thus bond strength increases with bond order: the more shared electron pairs (or the greater the excess of bonding over antibonding electrons), the stronger the attraction between the nuclei and the shorter and stronger the bond.
2. Illustration with diatomic molecules: H₂ (b.o. 1) 435.8 kJ mol⁻¹, 74 pm; O₂ (b.o. 2) 498 kJ mol⁻¹, 121 pm; N₂ (b.o. 3) 946.0 kJ mol⁻¹, 109 pm — one of the highest bond enthalpies of any diatomic molecule.
3. Illustration with carbon bonds: C–C (b.o. 1) 154 pm > C=C (b.o. 2) 133 pm > C≡C (b.o. 3) 120 pm; the bond length falls as the bond order rises.
4. A bond order of zero (e.g. He₂, Be₂) means no net bonding and no stable molecule; a positive bond order means a stable molecule. In general bond strength: b.o. 3 > b.o. 2.5 > b.o. 2 > b.o. 1.5 > b.o. 1 (compare O₂⁺ 2.5, O₂ 2, O₂⁻ 1.5).

Extra depth. Isoelectronic molecules and ions have identical bond orders: F₂ and O₂²⁻ (b.o. 1), N₂, CO and NO⁺ (b.o. 3). Resonance gives fractional bond orders: in O₃ each O–O bond has order 1.5 and a bond length (128 pm) between O–O (148 pm) and O=O (121 pm). The correlation is approximate: bond enthalpy also depends on atomic size and lone-pair repulsion (e.g. F₂ has a weak single bond, 155 kJ mol⁻¹, despite b.o. 1).

Conclusion: Bond strength increases with bond order: a higher bond order means a larger bond enthalpy and a shorter bond length (N₂ with bond order 3 is stronger than O₂ with 2 and H₂ with 1).

Concept. The distance between two bonded atoms is one of the most important bond parameters that determine molecular size and shape (along with bond angle, bond enthalpy, bond order and bond polarity).

• Definition: Bond length is the equilibrium distance between the nuclei of two bonded atoms in a molecule. It is the distance at which the attractive and repulsive forces balance and the potential energy of the system is a minimum (for H₂, 74 pm).
• How measured: by spectroscopic, X-ray diffraction and electron-diffraction techniques.
• Contribution of each atom: each atom of the bonded pair contributes to the bond length: R = r_A + r_B, where r_A and r_B are the covalent radii of atoms A and B. The covalent radius is half the distance between two similar atoms joined by a covalent bond in the same molecule (Cl: 198/2 = 99 pm); the van der Waals radius is half the distance between two similar atoms in separate molecules in a solid (Cl: 360/2 = 180 pm).
• Factors: bond length decreases with increase in bond order (C–C 154, C=C 133, C≡C 120 pm) and increases with the size of the atoms (HF 92, HCl 127, HBr 141, HI 160 pm; F₂ 144, Cl₂ 199, Br₂ 228, I₂ 267 pm).

Extra depth. Bond lengths are additive: for a bond A–B the length is roughly the sum of the covalent radii (e.g. C–Cl ≈ 77 + 99 = 176 pm). In ionic crystals the corresponding quantity is the sum of the ionic radii. In resonance hybrids the bond length is intermediate (O₃ 128 pm; CO₃²⁻ all equal; CO₂ 115 pm). Shorter bonds are stronger bonds (higher bond enthalpy).

More to know. Bond length is a property of the pair of atoms and of the bond order, not of the atom alone: the same two atoms give different lengths in single, double and triple bonds (C–C 154 pm, C=C 133 pm, C≡C 120 pm; C–N 143, C=N 128, C≡N 116 pm; C–O 143, C=O 121 pm). For heteronuclear bonds the length is roughly the sum of the two covalent radii (Table 4.4): H–Cl = 37 + 99 = 136 pm calculated against 127 pm measured — the small difference arises because the shared pair is pulled towards the more electronegative atom, shortening the bond. In a polar bond the observed length is therefore usually a little shorter than the sum of covalent radii. Bond lengths are used, together with bond angles, to describe the geometry of a molecule completely (e.g. H₂O: O–H 96 pm, H–O–H 104.5°); they also reflect resonance — the O–O bonds of ozone (128 pm) are intermediate between a single (148 pm) and a double bond (121 pm).

Conclusion: Bond length is the equilibrium distance between the nuclei of two bonded atoms, equal to the sum of their covalent radii (R = r_A + r_B), and it decreases as bond order increases.

Concept. Resonance is the concept introduced to deal with molecules and ions whose experimentally determined parameters (bond lengths, bond angles) cannot be explained by any one Lewis structure. Whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding and non-bonding pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule accurately. The carbonate ion is the classic example (Problem 4.3, Fig. 4.4 of the notes).

1. The problem with one structure. A single Lewis structure of CO₃²⁻ (24 valence electrons) needs one C=O double bond and two C–O⁻ single bonds. That predicts unequal bonds (double bond shorter, single bonds longer), but experiment shows that all three carbon–oxygen bonds are equal in length and equivalent (about 129 pm, between C=O and C–O).
2. Canonical forms. Three structures I, II and III can be drawn in which the double bond is placed between carbon and a different oxygen each time and the negative charges sit on the other two oxygens. They differ only in the positions of electrons, not of the nuclei, and they have the same energy.
3. Resonance hybrid. The actual ion is the resonance hybrid of I, II and III, represented by double-headed arrows between the canonical forms. In the hybrid each C–O bond has an average bond order of 4/3 ≈ 1.33 (one σ bond plus one-third of a π bond), and each oxygen carries two-thirds of a negative charge.
4. Stabilisation. The energy of the resonance hybrid is lower than that of any single canonical structure, so the ion is more stable than predicted from any one structure.
5. Averaging of bond characteristics. Resonance averages the bond characteristics as a whole: all three C–O bonds are identical in length and strength, and the three O atoms are equivalent.

Extra depth. Misconceptions to avoid (NCERT): (1) the canonical forms have no real existence; (2) the molecule does not exist for a fraction of time in one canonical form and for other fractions in the others; (3) there is no equilibrium between canonical forms such as exists between tautomers (keto and enol); (4) the molecule has a single structure — the resonance hybrid — which cannot be depicted by one Lewis structure. Other examples of resonance: O₃ (both O–O 128 pm), CO₂ (115 pm), NO₃⁻, SO₃.

Conclusion: In CO₃²⁻ the three canonical forms (one C=O and two C–O⁻ each) combine into a resonance hybrid in which all three C–O bonds are equal (bond order 1.33), and the hybrid is more stable than any single form.

Concept. For two structures to be canonical forms (resonance structures) of one resonance hybrid, they must have the same positions of the nuclei (the same skeleton, the same arrangement of atoms) and differ only in the distribution of bonding and non-bonding electrons, while having similar energies. Resonance involves a movement of electrons, never of atoms.

1. Structure 1 has one hydrogen bonded directly to phosphorus (a P–H bond), two O–H groups and one P=O: H–P(=O)(OH)₂.
2. Structure 2 has all three hydrogens attached to oxygen atoms (three O–H groups) with a lone pair on phosphorus: P(OH)₃.
3. The two structures therefore differ in the position of a hydrogen atom (and hence of a nucleus): H is on P in structure 1 but on O in structure 2. They also differ in the bonding arrangement around phosphorus (P=O and P–H versus three P–O single bonds and a lone pair).
4. Conclusion: No — structures 1 and 2 cannot be taken as canonical forms of a resonance hybrid because the positions of the atoms differ. They are different compounds / isomeric forms of the same molecular formula (related like the keto and enol forms in tautomerism), and the real structure of phosphorous acid is structure 1 (it is a dibasic acid — only the two O–H hydrogens are acidic).

Extra depth. NCERT explicitly says that canonical forms “with similar energy, positions of nuclei, bonding and non-bonding pairs of electrons” are the ones that resonate. Structures that differ in the position of an atom are related by tautomerism, not resonance; between tautomers there may be an equilibrium (a real shift of H), while between canonical forms there is none (the forms have no real existence). In P(OH)₃ phosphorus would have a lone pair; actual H₃PO₃ is H–P(=O)(OH)₂ because P=O and P–H are more stable than a lone pair on P with P(OH)₃. The 3d-orbital participation of P allows this expanded-octet-type bonding.

Conclusion: No. Structures 1 and 2 of H₃PO₃ differ in the position of a hydrogen atom (and the bonding around P), so they are not resonance forms of one hybrid.

Concept. The method is the same for all three: write a valid Lewis structure, find the atoms on which a double bond can be placed in more than one equivalent way, and write all such structures joined by double-headed arrows. Each structure has the same nuclear positions and the same energy; the true species is their resonance hybrid, with all equivalent bonds identical in length.

1. SO₃ (24 valence electrons; S central): there are three equivalent canonical structures; in each, sulphur is bonded to two oxygens by double bonds and to the third by a single bond, with the negative charge on the singly bonded oxygen (formal charge of S = +1). The double bond can be placed on any two of the three S–O bonds in three ways. Hybrid: three equivalent S–O bonds, each of bond order 1.67 (5/3), trigonal planar, O–S–O = 120°.
2. NO₂ (17 valence electrons, an odd-electron molecule): there are two equivalent canonical structures, O=N–O⁻ and ⁻O–N=O, in which the unpaired electron resides on nitrogen (formal charge of N = +1). Hybrid: both N–O bonds are equal, with bond order 1.5; the molecule is bent.
3. NO₃⁻ (24 valence electrons): three equivalent canonical structures, each with one N=O double bond and two N–O⁻ single bonds (N has formal charge +1, each O⁻ has −1, total −1). Hybrid: three identical N–O bonds, bond order 1.33, trigonal planar, O–N–O = 120°.

Extra depth. Formal charges (V − L − B/2) validate the structures: in NO₃⁻ N = 5 − 0 − 4 = +1, =O = 6 − 4 − 2 = 0, each O⁻ = 6 − 6 − 1 = −1, so the total is +1 + 0 − 2 = −1. Fully modern descriptions of SO₃ also show three S=O double bonds (an expanded octet using 3d orbitals), but the octet-obeying forms drawn here are the ones used at this level; both predict three equal bonds. Resonance explains why all N–O bonds in NO₃⁻ are equal and why the nitrate ion is so stable.

Conclusion: SO₃ and NO₃⁻ each have three equivalent resonance structures and NO₂ has two; in each hybrid the equivalent bonds are identical.

Concept. Electron transfer between atoms is the basis of the Kössel model of ionic bonding: a metal atom loses its valence electron(s) to form a cation with a noble-gas configuration, and a non-metal atom gains them to form an anion with an octet; the ions are held by electrostatic attraction. The Lewis symbol shows the valence electrons as dots, and the number of electrons transferred must balance.

1. (a) K and S: K (Group 1, 1 valence electron) and S (Group 16, 6 valence electrons). S needs two electrons for an octet; each K atom gives one. So 2 K· + ·S· (6 dots) → 2 K⁺ + [S (8 dots)]²⁻, giving K₂S. K⁺ has the [Ar] configuration, S²⁻ also [Ar].
2. (b) Ca and O: Ca (Group 2, 2 valence electrons) loses two; O (6 valence electrons) gains two. Ca (2 dots) + O (6 dots) → Ca²⁺ + [O (8 dots)]²⁻, giving CaO. Ca²⁺ is [Ar]; O²⁻ is [Ne].
3. (c) Al and N: Al (Group 13, 3 valence electrons) loses three; N (5 valence electrons) gains three. Al (3 dots) + N (5 dots) → Al³⁺ + [N (8 dots)]³⁻, giving AlN. Both ions are isoelectronic with neon.

Extra depth. The total positive charge always equals the total negative charge — the principle of electroneutrality: 2(+1) + (−2) = 0 for K₂S, (+2) + (−2) = 0 for CaO, (+3) + (−3) = 0 for AlN. The electrovalence of an element equals the number of charges on its ion (K = 1, Ca = 2, Al = 3, S = 2, O = 2, N = 3). Lattice enthalpy (high for the highly charged ions in CaO and AlN) provides the energy that makes the transfer favourable.

Conclusion: (a) 2 K + S → 2 K⁺ + S²⁻; (b) Ca + O → Ca²⁺ + O²⁻; (c) Al + N → Al³⁺ + N³⁻.

Concept. Dipole moment (μ) is the product of the charge and the distance between the centres of positive and negative charge (μ = Q × r); it is a vector quantity. For a polyatomic molecule the dipole moment is the vector sum of the individual bond dipoles, and so it depends on both the polarity of the bonds and on how they are arranged in space (the molecular geometry). Thus a measured dipole moment reveals the shape.

1. CO₂: each C=O bond is polar (O is more electronegative), but the experimentally measured dipole moment of CO₂ is zero (0 D). This is possible only if the two bond dipoles are equal and point in exactly opposite directions so that they cancel — i.e. only if the molecule is linear (O=C=O, 180°). A bent CO₂ would have a non-zero resultant.
2. H₂O: each O–H bond is polar, and the measured dipole moment of water is 1.85 D (6.17 × 10⁻³⁰ C m), which is not zero. A non-zero resultant arises only if the two O–H bond dipoles are not collinear — i.e. the molecule must be bent. Since the bond angle is 104.5°, the vector sum of the two bond dipoles points along the bisector of the angle: μ = 2 × μ_O–H cos(52.25°), a resultant of 1.85 D.
3. Hence the dipole-moment data show that CO₂ is linear and H₂O is bent, in agreement with VSEPR theory: two bond pairs and no lone pair at carbon (AB₂, linear); two bond pairs and two lone pairs at oxygen (AB₂E₂, bent).

Extra depth. The same reasoning explains BeF₂ (linear, μ = 0), BF₃ (trigonal planar, μ = 0), CH₄ (tetrahedral, μ = 0) and CCl₄ (μ = 0) versus H₂S (bent, 0.95 D), NH₃ (1.47 D), CHCl₃ (1.04 D). The dipole moment of H₂O has an important consequence: it makes water a polar solvent that dissolves ionic compounds and forms hydrogen bonds. Refer to the dipole figure of the notes (section 4.3.6) for the vector pictures of H₂O and BeF₂.

Conclusion: CO₂ has zero dipole moment because its two equal C=O bond dipoles cancel in a linear molecule, whereas H₂O has a net dipole moment of 1.85 D because its two O–H bond dipoles at 104.5° add up — hence H₂O is bent.

Concept. The dipole moment (μ) is the product of the magnitude of the charge (Q) and the distance (r) between the centres of positive and negative charge: μ = Q × r. It is expressed in debye (1 D = 3.33564 × 10⁻³⁰ C m), is a vector quantity, and is represented in chemistry by a crossed arrow (cross at the positive end, head at the negative end). It is an experimentally measurable quantity that connects the electronic charge distribution of a molecule with its properties.

1. Polarity of bonds and molecules. A non-zero dipole moment shows a polar molecule (HF 1.78 D, HCl 1.07 D, H₂O 1.85 D); μ = 0 shows a non-polar molecule (H₂, CH₄, CCl₄, CO₂). The larger the dipole moment, the more polar the bond: HF (1.78 D) > HCl (1.07 D) > HBr (0.79 D) > HI (0.38 D).
2. Shape (geometry) of molecules. Since the molecular dipole moment is the vector sum of bond dipoles, it reveals the geometry: CO₂ (0 D) is linear while H₂O (1.85 D) is bent; BF₃ (0 D) is planar while NH₃ (1.47 D) is pyramidal; BeF₂ (0 D) is linear; CH₄ (0 D) is symmetrical tetrahedral while CHCl₃ (1.04 D) is not.
3. Comparing similar molecules. NH₃ (4.90 × 10⁻³⁰ C m) has a larger dipole moment than NF₃ (0.80 × 10⁻³⁰ C m) because the lone-pair dipole reinforces the N–H resultant but opposes the N–F resultant.
4. Percentage ionic character. Ionic character = (observed μ / μ for 100 % ionic) × 100. For HCl: bond length 127 pm, so the 100 % ionic value would be 1.602 × 10⁻¹⁹ C × 127 × 10⁻¹² m = 2.03 × 10⁻²⁹ C m = 6.1 D; observed 1.07 D → about 17.5 % ionic. This confirms that no bond is wholly ionic or covalent.
5. Isomers and physical properties. Dipole moments help distinguish isomers (e.g. cis and trans forms, ortho/meta/para isomers) and explain solubility (“like dissolves like”: polar solutes in polar solvents), boiling points and intermolecular forces.

Extra depth. Experimental dipole moments were historically the first evidence for the non-linear structure of water and the pyramidal structure of ammonia. The dipole moment also underlies the idea of partial ionic character of covalent bonds and partial covalent character of ionic bonds (Fajans); molecules with high dipole moments (H₂O, HF) form strong hydrogen bonds and strong dipole–dipole attractions, which raise their boiling points. (The last three uses are standard chemistry; NCERT itself gives the first three.)

Conclusion: Dipole moment (μ = Q × r) indicates the polarity of bonds and molecules, helps to deduce molecular shape (CO₂ linear, H₂O bent, BF₃ planar, NH₃ pyramidal), gives the percentage ionic character, and helps distinguish isomers and explain solubility.

Concept. Electronegativity is a qualitative measure of the ability of an atom in a chemical compound to attract the shared electrons to itself (Unit 3). It is not a measurable quantity, but numerical scales have been developed: Pauling (most widely used; arbitrarily F = 4.0), Mulliken–Jaffe and Allred–Rochow. It helps predict the nature of the force holding two atoms together — ionic, polar covalent or non-polar covalent — which is exactly the issue of this chapter (bond polarity, dipole moments, Fajans’ rules).

Extra depth. The two are related but not identical: F has the highest electronegativity (4.0) but Cl has the most negative electron gain enthalpy (−349 vs −328 kJ mol⁻¹) because the added electron in F enters the small n = 2 shell and suffers repulsion. Electronegativity difference between bonded atoms measures bond polarity: H–H (0) non-polar; H–F (4.0 − 2.1 = 1.9) strongly polar; Na–Cl (3.0 − 0.9 = 2.1) ionic. Non-metals have high electronegativity and high negative electron gain enthalpy; metals have low values of both.

More to know. Electronegativity also helps to classify bonds: when the electronegativity difference between two atoms is zero the bond is purely covalent and non-polar; a small difference gives a polar covalent bond; a very large difference gives an essentially ionic bond. It is related to other periodic properties: electronegativity increases with decreasing atomic radius, so across a period it rises from Li (1.0) to F (4.0) and down a group it falls from F (4.0) to At (2.2); non-metals have high electronegativity, metals low. Because it varies with the oxidation state and hybridisation of the atom, the tabulated Pauling values are only approximate averages. Electron gain enthalpy, in contrast, is a precise thermochemical quantity used in Born–Haber-type energy calculations of ionic compounds such as NaCl (−348.7 kJ mol⁻¹ for Cl).

Conclusion: Electronegativity is the relative, non-measurable tendency of a bonded atom to attract shared electrons, whereas electron gain enthalpy is the measurable energy change when an isolated gaseous atom gains an electron.

Concept. In a covalent bond between two identical atoms (H₂, O₂, Cl₂, N₂, F₂) the shared electron pair is equally attracted by both nuclei and sits exactly between them — a non-polar covalent bond. When the bonded atoms are different, with different electronegativities, the shared pair is displaced towards the more electronegative atom, and the bond is called a polar covalent bond.

1. Example HF: fluorine (electronegativity 4.0) is far more electronegative than hydrogen (2.1), so the shared electron pair in H–F is displaced towards fluorine.
2. Partial charges: the displacement gives fluorine a fractional negative charge (δ−) and hydrogen a fractional positive charge (δ+): H^δ+–F^δ−. The ends of the bond have opposite partial charges separated by the bond length (92 pm).
3. Dipole moment: because of polarisation the molecule has a dipole moment μ = Q × r; for HF μ = 1.78 D; it is shown by a crossed arrow H → F pointing from the positive towards the negative end.
4. Other examples: HCl (1.07 D), HBr (0.79 D), HI (0.38 D), H₂O (O–H bonds), NH₃ (N–H bonds). The polarity decreases as the electronegativity difference decreases down Group 17: HF > HCl > HBr > HI.

Extra depth. NCERT notes that a bond that is purely ionic or purely covalent is an ideal case; in reality every bond has some character of the other kind, and even the H–H bond has a little ionic character. A polar covalent bond is therefore the general case, lying between pure covalent and pure ionic. In a polyatomic molecule the net dipole moment is the vector sum of the bond dipoles: in BF₃ and CO₂ the bonds are polar but the molecules are non-polar due to symmetry.

Conclusion: A polar covalent bond is a covalent bond between atoms of different electronegativity in which the shared pair is displaced towards the more electronegative atom, as in H^δ+–F^δ− (μ = 1.78 D).

Concept. The ionic character of a bond increases with the difference in electronegativity between the bonded atoms. A bond between identical atoms is purely covalent (0 % ionic); a bond between a highly electropositive metal and a highly electronegative non-metal is largely ionic. We use the Pauling electronegativities of Unit 3 (F 4.0, O 3.5, Cl 3.0, S 2.5, N 3.0, Li 1.0, K 0.8).

1. N₂ (N≡N): identical atoms, ΔEN = 0: purely covalent, non-polar — the lowest ionic character.
2. SO₂ (S–O): ΔEN = 3.5 − 2.5 = 1.0 — polar covalent, a bond between two non-metals.
3. ClF₃ (Cl–F): ΔEN = 4.0 − 3.0 = 1.0 — polar covalent, also between two non-metals; it is placed after SO₂ in the standard answer because the Cl–F bond is between the most electronegative element and a large halogen atom (the two values are very close, both about 1.0 on the NCERT scale).
4. K₂O (K–O): ΔEN = 3.5 − 0.8 = 2.7 — a metal with a very low ionization enthalpy and a non-metal: largely ionic.
5. LiF (Li–F): ΔEN = 4.0 − 1.0 = 3.0 — the largest difference of all, so the most ionic (note: the Li⁺ ion is very small, which gives LiF some covalent character by Fajans’ rules, but its electronegativity difference still makes it the most ionic of the five).

Extra depth. Order of increasing ionic character: N₂ < SO₂ < ClF₃ < K₂O < LiF. The values for SO₂ and ClF₃ are almost the same on the scale of Table 3.8 (each 1.0), so their relative order is a matter of fine distinction (with more precise Pauling values, 0.86 for S–O and 0.82 for Cl–F, the ordering of these two covalent compounds is marginal); the clearly separated groups are non-polar (N₂), polar covalent (SO₂, ClF₃) and ionic (K₂O, LiF). Fajans’ rules supplement the electronegativity argument: the greater the charge and the smaller the cation, the more covalent the compound.

Conclusion: The order of increasing ionic character is N₂ < SO₂ < ClF₃ < K₂O < LiF, following the increasing electronegativity difference between the bonded atoms.

Concept. The skeleton of acetic acid is H₃C–C(O)–O–H: a methyl carbon bonded to the carboxyl carbon, which carries one carbonyl oxygen and one hydroxyl oxygen. We must write the correct bonds and lone pairs by counting valence electrons: 2 C (2 × 4 = 8) + 4 H (4 × 1 = 4) + 2 O (2 × 6 = 12) = 24 valence electrons.

1. Errors in the given drawing: (i) one H is shown joined to carbon by a double bond (“H=C”) — but hydrogen can form only one bond (it attains a duplet), so all C–H bonds must be single; (ii) the carbonyl C–O is shown as a single bond with an incomplete representation — it must be a C=O double bond so that carbon has an octet; (iii) lone pairs are shown on carbon — carbon has four bonds and no lone pair; (iv) the hydroxyl oxygen is missing its second lone pair.
2. Correct structure: CH₃ carbon forms three C–H single bonds (6 e⁻) and one C–C single bond (2 e⁻). The carboxyl carbon forms the C–C bond, a C=O double bond (4 e⁻) and a C–O single bond (2 e⁻); the hydroxyl O forms the O–H single bond (2 e⁻). Bonding electrons: 3 × 2 + 2 + 4 + 2 + 2 = 16 (i.e. 8 bonds counted with the double bond as two). Lone pairs: two on the carbonyl oxygen and two on the hydroxyl oxygen (8 e⁻). Total = 16 + 8 = 24 ✓.
3. Octet check: each carbon has four bonds (8 e⁻); the carbonyl oxygen has 4 shared + 4 lone-pair electrons = 8 and the hydroxyl oxygen has 4 shared + 4 lone-pair electrons = 8; each hydrogen has a duplet.

Extra depth. Formal charges in the correct structure are all zero (C: 4 − 0 − 4 = 0; =O: 6 − 4 − 2 = 0; –OH oxygen: 6 − 4 − 2 = 0; H: 1 − 0 − 1 = 0), which confirms it is the lowest-energy structure (NCERT: the lowest energy structure has the smallest formal charges). In the acetate ion CH₃COO⁻, formed by loss of the O–H hydrogen as H⁺, the two C–O bonds become equivalent by resonance (each bond order 1.5).

Conclusion: The correct Lewis structure of acetic acid is H₃C–C(=O)–O–H with single C–H bonds, a C=O double bond, two lone pairs on each oxygen and no lone pair on carbon.

Concept. Both arrangements put four bond pairs around carbon, so the choice between them is decided by the repulsion between those four electron pairs (VSEPR theory) and by the way carbon’s orbitals can combine (hybridisation). The molecule adopts the geometry of minimum energy, i.e. the one that keeps the four pairs as far apart as possible.

1. VSEPR argument. In a tetrahedral CH₄ the H–C–H angle is 109.5°, the maximum possible separation of four pairs on a sphere. In the hypothetical square planar arrangement the angle between adjacent C–H bonds would be only 90°. Bond pairs at 90° repel much more strongly than at 109.5°; the square planar structure would therefore be of higher energy and less stable.
2. Counting repulsions. In the tetrahedron all six bond–bond angles are 109.5°. In the square, four adjacent pairs are at 90° (strong repulsion) and only two (diagonally opposite) at 180°. The total repulsion is therefore greater in the square planar arrangement.
3. Hybridisation (VB theory). Carbon in CH₄ uses one 2s and three 2p orbitals (after promotion of an electron) to give four equivalent sp³ hybrid orbitals directed to the corners of a tetrahedron, at 109.5° to one another (25 % s-character, 75 % p-character). This hybridisation gives maximum overlap with the H 1s orbitals and minimum repulsion; no hybridisation scheme of s and p orbitals of carbon gives a square planar array (square planar geometry needs d orbitals, i.e. dsp², as in [Ni(CN)₄]²⁻, which carbon does not have in its valence shell).
4. Experimental evidence. If CH₄ were square planar, CH₂Cl₂ would exist as two isomers (the two Cl atoms either adjacent, at 90°, or opposite, at 180°); in reality only one CH₂Cl₂ is known, which agrees with the tetrahedron where all positions are equivalent. Measured H–C–H angles are 109.5°.

Extra depth. Both shapes have a zero dipole moment, so the dipole moment alone cannot decide between them (NCERT Table 4.5 lists CH₄ = 0), which is why the energy and isomer arguments are decisive. A similar argument shows why BF₃ is trigonal planar (3 pairs at 120°) and why AB₄ molecules with no lone pair are tetrahedral. The existence of square planar XeF₄ (AB₄E₂) is explained because the two lone pairs sit at 180° to each other above and below the plane.

Conclusion: CH₄ is tetrahedral and not square planar because the tetrahedral arrangement (109.5°, sp³ hybridisation) minimises repulsion between the four bond pairs, whereas the square planar arrangement would force them to 90°, raising the energy.

Concept. The dipole moment of a molecule is the vector sum of its bond dipoles. Whether polar bonds give a polar molecule therefore depends on the molecular geometry. The Be–H bonds are polar (H is more electronegative than Be, 2.1 vs 1.5), but BeH₂ is a symmetrical linear molecule, so the individual dipoles cancel.

1. Shape of BeH₂. Be (1s²2s²) has two valence electrons and promotes one 2s electron to 2p; one 2s and one 2p orbital hybridise to two sp hybrid orbitals at 180°, each overlapping with the 1s orbital of an H atom. By VSEPR, two bond pairs and no lone pair give a linear molecule, H–Be–H = 180°. (BeH₂ is also an incomplete-octet molecule, with only four electrons around Be.)
2. Bond dipoles. Each Be–H bond has the same dipole moment, directed from Be (positive end) towards H (negative end). In a linear molecule these two equal dipoles point in opposite directions along the same axis.
3. Vector addition. The resultant is μ = μ₁ + μ₂ = (+μ) + (−μ) = 0. Hence BeH₂ is non-polar overall (like BeF₂ in NCERT, equal bond dipoles pointing in opposite directions cancel each other).

Extra depth. A zero dipole moment is characteristic of symmetrical molecules: linear AB₂ (BeH₂, BeF₂, CO₂), trigonal planar AB₃ (BF₃), tetrahedral AB₄ (CH₄, CCl₄). Conversely, a measured dipole moment of zero for an AB₂ molecule proves it is linear, while a non-zero moment (H₂O, H₂S) proves it is bent.

More to know. The dipole moment of BeH₂ is just one member of a family of results: linear molecules whose bond dipoles are equal and opposite — BeCl₂, BeF₂, CO₂, CS₂, HgCl₂ — are non-polar although every bond is polar. If the molecule were bent (like H₂O or H₂S) the same bond dipoles would not cancel. Therefore dipole moment is used to confirm the shape predicted by VSEPR or hybridisation: μ = 0 for an AB₂ molecule confirms the linear structure (sp hybridised central atom). Also note that BeH₂ exists as a polymeric solid with bridging hydrogens, but the gaseous monomer is linear.

Conclusion: BeH₂ is linear (sp hybridised Be, H–Be–H = 180°), so its two equal Be–H bond dipoles are equal and opposite and cancel, giving a net dipole moment of zero.

Concept. Both NH₃ and NF₃ are pyramidal (sp³ nitrogen, three bond pairs and one lone pair; 107° in NH₃). The dipole moment of each is the vector sum of the three N–X bond dipoles and the dipole due to the lone pair (the orbital dipole). Their directions decide the result, not just the electronegativity of X.

1. NH₃: N (electronegativity 3.0) is more electronegative than H (2.1), so each N–H bond dipole points towards N. The resultant of the three N–H dipoles therefore points towards nitrogen, along the same direction as the lone-pair orbital dipole. The two effects add: μ = 4.90 × 10⁻³⁰ C m (1.47 D).
2. NF₃: F (4.0) is more electronegative than N (3.0), so each N–F bond dipole points away from N, towards F. The resultant of the three N–F dipoles is therefore opposite in direction to the lone-pair dipole, which points the other way. The orbital dipole partly cancels the bond-dipole resultant: μ = 0.80 × 10⁻³⁰ C m (0.23 D), much smaller.
3. Conclusion: NH₃ has the higher dipole moment (4.90 × 10⁻³⁰ C m vs 0.80 × 10⁻³⁰ C m), although fluorine is more electronegative than nitrogen.

Extra depth. This is the explanation given by NCERT in section 4.3.6: in NH₃ the lone-pair orbital dipole acts in the same direction as the resultant of the N–H bond dipoles, while in NF₃ it acts opposite to the resultant of the N–F bond dipoles and weakens it. The case is a good example of why electronegativity alone cannot predict dipole moments of polyatomic molecules — geometry and lone pairs matter. (Table 4.5 lists the values in debye: NH₃ 1.47 D, NF₃ 0.23 D.)

Conclusion: NH₃ has the higher dipole moment because its lone-pair dipole reinforces the resultant of the N–H bond dipoles, whereas in NF₃ the lone-pair dipole opposes the resultant of the N–F bond dipoles.

Concept. Simple overlap of pure atomic orbitals cannot explain the shapes of molecules like CH₄ (tetrahedral, 109.5°), NH₃ (pyramidal, 107°) and H₂O (bent, 104.5°); for this Pauling introduced the concept of hybridisation.

• Definition: hybridisation is the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of a new set of orbitals of equivalent energies and shape. The new orbitals are called hybrid orbitals. E.g. one 2s and three 2p orbitals of carbon give four sp³ hybrid orbitals.
• Features: number of hybrid orbitals = number of atomic orbitals mixed; they are equivalent in energy and shape; they form stronger bonds than pure orbitals because of better overlap; they are directed in space to minimise repulsion, so the type of hybridisation indicates the geometry. Promotion of an electron is not essential, and even filled orbitals (lone pairs) may hybridise.
• sp hybrid orbitals: one s + one p (p_z) → two sp orbitals, each with 50 % s- and 50 % p-character, pointing in opposite directions along the z-axis (180°, linear, ‘diagonal’). Each hybrid has a large positive lobe and a very small negative lobe. Example: BeCl₂, C₂H₂.
• sp² hybrid orbitals: one s + two p → three sp² orbitals (33 % s, 67 % p) lying in one plane at 120° (trigonal planar). Example: BCl₃, C₂H₄ (the unhybridised p orbital is perpendicular to the plane and forms the π bond).
• sp³ hybrid orbitals: one s + three p → four sp³ orbitals (25 % s, 75 % p) directed to the corners of a regular tetrahedron, 109.5° apart. Example: CH₄, C₂H₆; also NH₃ and H₂O (with lone pairs occupying one and two hybrids; angles 107° and 104.5°).

Extra depth. Each hybrid orbital is lopsided: one large lobe (used for overlap with another atom, giving a strong σ bond) and one very small lobe on the opposite side — see Figs 4.10–4.12 in the notes. Increasing s-character makes the hybrid orbital shorter, closer to the nucleus and more electronegative, which explains why C–H bonds shorten from sp³ (109 pm in ethane) to sp² (108 pm in ethene) and the bond angle widens. d orbitals extend hybridisation to sp³d (trigonal bipyramidal, PCl₅) and sp³d² (octahedral, SF₆).

Conclusion: Hybridisation is the intermixing of atomic orbitals of slightly different energies to give equivalent hybrid orbitals: sp gives two orbitals at 180° (linear), sp² three at 120° (trigonal planar) and sp³ four at 109.5° (tetrahedral).

Concept. Hybridisation depends on the number of regions of electron density (bond pairs + lone pairs) around the central atom. Aluminium (Group 13, 3s²3p¹) in AlCl₃ and AlCl₄⁻ has a different number of bonds, so its hybridisation differs.

1. In AlCl₃: in the excited state Al is 3s¹3p_x¹3p_y¹ with an empty 3p_z orbital. One 3s and two 3p orbitals hybridise to give three sp² hybrid orbitals which bond to three Cl atoms: a trigonal planar molecule with Cl–Al–Cl = 120°. Aluminium has only six electrons (an incomplete octet) and a vacant 3p orbital, so AlCl₃ is an electron-deficient Lewis acid.
2. In AlCl₄⁻: the Cl⁻ ion donates a lone pair into the vacant 3p orbital of Al, forming a fourth Al–Cl bond (a coordinate bond). Now Al uses one 3s and three 3p orbitals: four sp³ hybrid orbitals directed to the corners of a tetrahedron (Cl–Al–Cl = 109.5°). Aluminium now has an octet.
3. Change: hybridisation of Al changes from sp² to sp³, and the geometry from trigonal planar to tetrahedral; the vacant p orbital is used.

Extra depth. This is a typical Lewis acid–base reaction: AlCl₃ is the acid (electron-pair acceptor) and Cl⁻ the base (donor). The same change occurs for boron in BF₃ + F⁻ → BF₄⁻ and in BF₃ + NH₃ (next exercise). Note that the change in the number of electron pairs around the central atom (3 → 4) — not the number of electrons given by the incoming ligand — decides the new hybridisation. In solid AlCl₃ and in non-polar solvents, Al achieves the octet by forming the dimer Al₂Cl₆ with sp³ Al as well.

Conclusion: On forming AlCl₄⁻ the Al atom changes from sp² hybridisation (trigonal planar AlCl₃) to sp³ hybridisation (tetrahedral AlCl₄⁻) as its vacant 3p orbital accepts the lone pair of Cl⁻.

Concept. In BF₃ the boron atom is electron-deficient (six electrons, one vacant 2p orbital) and is a Lewis acid; in NH₃ the nitrogen atom carries a lone pair and is a Lewis base. They react to form an adduct (coordinate or dative bond) F₃B←NH₃ in which the lone pair of N is donated into the vacant orbital of B.

1. Boron in BF₃: ground state 2s²2p¹; excited 2s¹2p_x¹2p_y¹; one 2s and two 2p orbitals form three sp² hybrids (trigonal planar, F–B–F = 120°) with an empty unhybridised 2p_z orbital.
2. Boron in F₃B·NH₃: boron now forms four bonds (three B–F and one B–N); the empty 2p orbital takes part, so one 2s and three 2p orbitals form four sp³ hybrid orbitals directed to the corners of a tetrahedron. Hybridisation of B changes from sp² to sp³ (geometry from trigonal planar to tetrahedral; boron attains an octet).
3. Nitrogen in NH₃: three N–H bond pairs and one lone pair → sp³ hybridised (pyramidal, 107°).
4. Nitrogen in F₃B·NH₃: the lone pair is now used in the coordinate bond to boron, giving four bond pairs (three N–H, one N–B) and no lone pair; nitrogen is still surrounded by four electron pairs and remains sp³ hybridised (now tetrahedral). No change in the hybridisation of N (only the lone pair becomes a bond pair).

Extra depth. The H–N–H angle in the adduct widens from 107° towards 109.5° because the lone pair (which exerted a greater repulsion) has become a bond pair; the F–B–F angle in the adduct narrows from 120° to about 109.5°. The reaction is the same type as AlCl₃ + Cl⁻ → AlCl₄⁻ in the previous exercise. The B–N bond formed is a coordinate (dative) covalent bond: both electrons come from nitrogen, but once formed it is indistinguishable from other covalent bonds.

Conclusion: B changes from sp² (in BF₃) to sp³ (in F₃B·NH₃), whereas N remains sp³ hybridised before and after the reaction.

Concept. Multiple bonds between carbon atoms are explained by VB theory through hybridisation plus sigma and pi overlap: a σ bond is formed by head-on overlap of hybrid orbitals along the internuclear axis, and each additional bond between the same two atoms is a π bond formed by sidewise overlap of unhybridised p orbitals.

1. Double bond in C₂H₄ (ethene) — sp² hybridisation. Each carbon mixes 2s + 2p_x + 2p_y to give three sp² hybrids in a plane at 120°, leaving 2p_z unhybridised and perpendicular to the plane. One sp² hybrid of each carbon overlaps head-on to form the C–C σ bond; the other two form C–H σ bonds with H 1s orbitals (total 5 σ bonds, all in one plane). The two parallel 2p_z orbitals overlap sidewise to give the π bond, with electron clouds above and below the molecular plane. Result: C=C = 1 σ + 1 π; C=C 134 pm; H–C–H 117.6°.
2. Triple bond in C₂H₂ (ethyne) — sp hybridisation. Each carbon mixes 2s + 2p_z to give two sp hybrids at 180°, leaving 2p_x and 2p_y unhybridised and perpendicular to the molecular axis and to each other. One sp hybrid of each carbon overlaps head-on forming the C–C σ bond; the other forms a C–H σ bond with H 1s. The two pairs of parallel p orbitals (2p_x–2p_x and 2p_y–2p_y) overlap sidewise giving two π bonds in mutually perpendicular planes. Result: C≡C = 1 σ + 2 π; C≡C 120 pm; the molecule is linear.

Extra depth. A π bond is weaker than a σ bond because sideways overlap is less extensive than head-on overlap, which is why alkenes and alkynes are chemically reactive at the π bond (addition reactions). The π electron cloud in ethene lies above and below the molecular plane, which also makes the molecule rigid (no free rotation about the C=C bond). In ethyne the two π bonds together form a cylindrical electron cloud around the C–C axis. For comparison, ethane has an sp³–sp³ C–C σ bond of 154 pm.

Conclusion: In C₂H₄ each carbon is sp² and the C=C bond is one σ plus one π bond; in C₂H₂ each carbon is sp and the C≡C bond is one σ plus two π bonds.

Concept. To count σ and π bonds, draw the Lewis structure and remember: every single bond is one σ bond; every double bond is one σ + one π; every triple bond is one σ + two π (π bonds are formed in addition to a σ bond between the same two atoms).

1. (a) C₂H₂ (H–C≡C–H): two C–H single bonds = 2 σ; one C≡C triple bond = 1 σ + 2 π. Total 3 σ bonds and 2 π bonds.
2. (b) C₂H₄ (H₂C=CH₂): four C–H single bonds = 4 σ; one C=C double bond = 1 σ + 1 π. Total 5 σ bonds and 1 π bond.

Extra depth. Check using hybridisation: in C₂H₂ the sp hybrids give three σ bonds (C–C, 2 C–H) and the four unhybridised p orbitals give two π bonds; in C₂H₄ sp² hybrids give five σ bonds (C–C, 4 C–H) and the two unhybridised p orbitals give one π bond. A useful general rule: total bonds in a molecule = number of σ bonds + number of π bonds; the number of σ bonds is always equal to (number of atoms − 1) for an acyclic molecule: C₂H₂ has 4 atoms → 3 σ; C₂H₄ has 6 atoms → 5 σ.

More to know. Worked checks on other molecules. CH₄: four C–H bonds, all single → 4 σ, 0 π. H₂O: two O–H bonds → 2 σ. N₂: N≡N → 1 σ + 2 π. CO₂ (O=C=O): two C=O double bonds → 2 σ + 2 π. HCN (H–C≡N): C–H (1 σ) + C≡N (1 σ + 2 π) → 2 σ + 2 π. Ethane C₂H₆: 7 σ, 0 π. Ethene: 5 σ, 1 π. Ethyne: 3 σ, 2 π. The pattern shows that increasing the bond order between the carbons from 1 to 2 to 3 does not change the number of σ bonds between them (always one) — it only adds π bonds, and that the number of C–H σ bonds falls (6 → 4 → 2) as each H is replaced by an extra C–C bond.

Conclusion: (a) C₂H₂ has 3 σ and 2 π bonds; (b) C₂H₄ has 5 σ and 1 π bond.

Concept. A sigma (σ) bond is formed by head-on (end-to-end, axial) overlap of orbitals along the internuclear axis; the overlap region is symmetrical about this axis. A pi (π) bond is formed by sidewise overlap of parallel p orbitals whose axes are perpendicular to the internuclear axis. The question therefore asks which pair cannot overlap head-on along the x-axis.

1. (a) 1s and 1s: two spherical orbitals overlap along any line, including the x-axis: head-on s–s overlap → σ bond (as in H₂).
2. (b) 1s and 2p_x: the 2p_x lobes point along the x-axis (the internuclear axis), so the s orbital can overlap with one lobe head-on: s–p axial overlap → σ bond (as in HF).
3. (c) 2p_y and 2p_y: the 2p_y orbitals are perpendicular to the x-axis; with the nuclei on the x-axis the two p_y orbitals are parallel to each other and perpendicular to the internuclear axis, so they can only overlap sidewise, forming a π bond, not a σ bond. This pair does not form a sigma bond.
4. (d) 1s and 2s: both spherical; head-on s–s overlap (the 1s and 2s of different atoms) → σ bond.
5. Answer: (c) 2p_y and 2p_y — they overlap laterally to give a π bond.

Extra depth. In general: s–s, s–p_axis and p_axis–p_axis overlaps give σ bonds; p–p overlaps with orbitals perpendicular to the axis give π bonds. A p_y orbital cannot overlap with an s orbital along the x-axis at all, because the positive and negative lobes of p_y overlap the s orbital equally and the net overlap is zero (Fig. 4.9 (i) in the notes). Since the extent of overlap is less in π bonds, they are weaker than σ bonds.

Conclusion: Out of the four pairs, 2p_y and 2p_y will not form a sigma bond along the x-axis, because their lobes are perpendicular to the internuclear axis and can overlap only sidewise to form a π bond.

Concept. The hybridisation of a carbon atom is determined by its number of σ bonds (bonds to other atoms) plus lone pairs: 4 σ bonds → sp³ (tetrahedral); 3 σ bonds (and one π bond) → sp² (trigonal planar); 2 σ bonds (and two π bonds) → sp (linear). A double bond is one σ + one π, so a carbon with a C=C or C=O double bond has three σ bonds and is sp².

1. (a) CH₃–CH₃ (ethane): each carbon has four σ bonds (3 C–H + 1 C–C) → both carbons sp³.
2. (b) CH₃–CH=CH₂ (propene): CH₃ carbon: four σ bonds → sp³; the =CH– carbon: three σ bonds (C–CH₃, C–H, C=C σ) + one π → sp²; the =CH₂ carbon: three σ bonds (2 C–H, C=C σ) + one π → sp².
3. (c) CH₃–CH₂–OH (ethanol): CH₃ carbon: four σ → sp³; CH₂ carbon (C–CH₃, 2 C–H, C–O): four σ → sp³. Both carbons sp³.
4. (d) CH₃–CHO (acetaldehyde): CH₃ carbon sp³; carbonyl carbon (C–CH₃, C–H, C=O σ) + one π → three σ bonds → sp².
5. (e) CH₃COOH (acetic acid): CH₃ carbon sp³; carboxyl carbon (C–CH₃, C–OH, C=O σ) + one π → three σ bonds → sp².

Extra depth. The hybridisation of carbon controls the geometry and bond lengths: sp³ carbon is tetrahedral (109.5°), sp² carbon is trigonal planar (about 120°), and a C–H bond becomes shorter and stronger as s-character increases (sp³–s 109 pm in ethane, sp²–s 108 pm in ethene). C=C (134 pm) and C=O (121 pm) double bonds involve sp² carbons. An sp-hybridised carbon would occur in alkynes (HC≡CH) and nitriles (C≡N) — none in this list.

Conclusion: (a) both sp³; (b) CH₃ sp³, both alkene carbons sp²; (c) both sp³; (d) CH₃ sp³, CHO carbon sp²; (e) CH₃ sp³, COOH carbon sp².

Concept. In a Lewis structure the valence electrons around an atom are of two kinds: those that are shared between two atoms to form a covalent bond, and those that stay on one atom and are not involved in bonding. This distinction is basic to Lewis structures, to VSEPR theory (lone pairs repel more strongly than bond pairs) and to the idea of coordinate bonds.

• Bond pair (bonding pair or shared pair): a pair of electrons that is shared between two atoms and holds them together as a covalent bond. It is attracted by two nuclei and is therefore confined to the region between them. A single bond has one bond pair, a double bond two, a triple bond three. Example: in H₂O (H–O–H) there are two bond pairs (one in each O–H bond); in Cl₂ there is one bond pair.
• Lone pair (non-bonding pair, unshared pair): a pair of valence electrons on an atom that is not shared with any other atom and is not involved in bonding. It is localised on one atom only (attracted by one nucleus), so it occupies more space than a bond pair. Example: in H₂O the oxygen atom has two lone pairs; in NH₃ nitrogen has one lone pair; in Cl₂ each Cl has three lone pairs.
• Illustration with H₂O: total valence electrons = 2(1) + 6 = 8 = 4 pairs. Two pairs are bond pairs (O–H); the other two pairs on O are lone pairs (written as pairs of dots). The oxygen atom attains an octet (2 bond pairs + 2 lone pairs) and each H a duplet.
• Consequences: lone pairs repel more than bond pairs (order lp–lp > lp–bp > bp–bp), so molecules with lone pairs have distorted shapes (NH₃ 107°, H₂O 104.5°); a lone pair can be donated to form a coordinate bond (NH₃ in F₃B·NH₃).

Extra depth. Lone pairs and bond pairs together make up the electron pairs that determine the geometry in VSEPR theory; a double or triple bond counts as a single ‘super pair’. In VB theory lone pairs occupy hybrid orbitals (sp³ hybrid with an electron pair) — e.g. in NH₃ one of the four sp³ orbitals of N holds the lone pair and three form σ bonds with H. The number of lone pairs on the central atom = (valence electrons − electrons used in bonds)/2.

Conclusion: A bond pair is an electron pair shared between two atoms (e.g. each O–H bond in H₂O), while a lone pair is a pair of valence electrons that belongs to one atom and is not shared (e.g. the two pairs on oxygen in H₂O).

Concept. Covalent bonds are classified by the way atomic orbitals overlap: σ bond — overlap is end-to-end (head-on) along the internuclear axis; π bond — overlap is sideways (lateral) by parallel orbitals perpendicular to the internuclear axis. In multiple bonds, a σ bond is always present and the additional bonds are π bonds.

Extra depth. Counting: C₂H₄ has five σ and one π bond; C₂H₂ has three σ and two π bonds; N₂ has one σ and two π bonds. The π bonds are the weak and reactive part of multiple bonds, which is why C=C undergoes addition across the double bond; in C₂ molecule (MO theory) the double bond consists of both π bonds. Because a π bond requires parallel p orbitals, a double bond locks the molecule in a plane (ethene is planar, with no free rotation, giving rise to cis–trans isomerism).

More to know. Where each type occurs. All single bonds are σ bonds (C–H in CH₄, O–H in H₂O, F–F in F₂). In ethene the C=C bond is sp²–sp² σ plus p–p π (the π cloud above and below the plane makes alkenes planar and rigid); in ethyne, C≡C is sp–sp σ plus two perpendicular p–p π bonds (a cylindrical cloud). In MO language, a σ MO is symmetrical about the bond axis (σ1s, σ2p_z), while a π MO (π2p_x, π2p_y) has a nodal plane through the nuclei; this is why the MO order differs for B₂, C₂, N₂ (where the two π MOs lie below σ2p_z) and in C₂ the double bond consists of two π bonds and no σ bond from the 2p orbitals.

Conclusion: A σ bond is formed by head-on axial overlap, is stronger and cylindrically symmetrical, while a π bond is formed by sidewise overlap of parallel p orbitals, is weaker and lies above and below the molecular plane; a π bond exists only along with a σ bond.

Concept. The H₂ molecule is the simplest molecule and the best illustration of valence bond (VB) theory, introduced by Heitler and London (1927) and developed by Pauling. VB theory describes bond formation by overlap of half-filled atomic orbitals of the two atoms, with the electrons paired (opposite spins).

1. Starting point: two hydrogen atoms A and B, with nuclei N_A, N_B and electrons e_A, e_B in their 1s orbitals. At a large distance there is no interaction.
2. Forces on approach: attractive forces arise between N_A–e_A and N_B–e_B (old forces) and, newly, between N_A–e_B and N_B–e_A (new attractive forces). Repulsive forces arise between e_A–e_B and N_A–N_B (new repulsive forces). Attractive forces tend to bring the atoms together, repulsive forces to push them apart.
3. Net effect: experiments show the new attractive forces are greater than the new repulsive forces; so as the atoms approach, the potential energy decreases. At a certain internuclear distance the net force of attraction balances the force of repulsion and the system has minimum energy; the two atoms are then bonded as a stable H₂ molecule with bond length 74 pm. Bringing the nuclei still closer raises the energy sharply (repulsion).
4. Orbital overlap: at this distance the two 1s orbitals undergo partial interpenetration (overlap), which results in the pairing of the two electrons with opposite spins and an increase of electron density between the nuclei. The electrons now belong to both nuclei, which are held together: this is the covalent (σ, s–s overlap) bond.
5. Energy: energy is released during bond formation and equals the bond enthalpy, 435.8 kJ mol⁻¹ (the depth of the minimum in the potential energy curve); the same energy must be supplied to dissociate one mole of H₂. H₂(g) + 435.8 kJ mol⁻¹ → 2H(g). The greater the overlap, the stronger the bond.