Chapter 4: Chemical Bonding and Molecular Structure (Chemical Bonding)

A chemical bond is the attractive force holding atoms together. Atoms (other than noble gases) are not stable alone; by bonding they lower their energy, usually by attaining a noble-gas configuration (octet).

(i) Transfer of electrons from an electropositive atom to an electronegative one gives ions held by electrostatic attraction: an ionic bond (Na + Cl → Na⁺Cl⁻).

(ii) Sharing of electron pairs gives a covalent bond (H–H, Cl–Cl); if both shared electrons come from one atom, it is a coordinate bond.

In orbital terms, half-filled orbitals overlap and the energy falls to a minimum at the bond length.

Atoms bond (by transfer or sharing of electrons) to lower their energy, usually reaching an octet.

The number of dots equals the number of valence electrons.

Mg: 2 dots; Na: 1; B: 3; O: 6 (two pairs + two singles); N: 5 (one pair + three singles); Br: 7 (three pairs + one single)

S: 6 dots; S²⁻: 8 dots (four pairs) round S in brackets with charge 2−. Al: 3 dots; Al³⁺: no dots, [Al]³⁺. H: one dot, H·; H⁻: one pair, [H:]⁻.

S 6 dots, S²⁻ 8 dots; Al 3 dots, Al³⁺ none; H 1 dot, H⁻ 2 dots

H₂S (8 valence e⁻): H–S–H with two lone pairs on S.

SiCl₄ (32 e⁻): Si in the centre with four Si–Cl single bonds; three lone pairs on each Cl.

BeF₂ (16 e⁻): F–Be–F; three lone pairs on each F; Be has only 4 electrons (incomplete octet).

CO₃²⁻ (24 e⁻): C in the centre with one C=O and two C–O bonds; the double-bonded O has 2 lone pairs, each single-bonded O has 3 lone pairs and a −1 formal charge; the whole in brackets with charge 2− (three resonance forms).

HCOOH (18 e⁻): H–C(=O)–O–H; the C=O oxygen has 2 lone pairs and the –OH oxygen has 2 lone pairs.

As described: H₂S bent with 2 lp on S; SiCl₄ four single bonds; BeF₂ incomplete octet on Be; CO₃²⁻ with resonance; HCOOH H–C(=O)–OH.

Octet rule: atoms combine by losing, gaining or sharing electrons so as to have eight electrons in their valence shell.

Significance: it explains why atoms combine, and the formation and valencies of many ionic and covalent compounds.

Limitations: (i) incomplete octet of the central atom: LiCl, BeH₂, BCl₃; (ii) odd-electron molecules: NO, NO₂; (iii) expanded octet: PF₅, SF₆, H₂SO₄; (iv) noble gases do form compounds (XeF₂, XeF₄, XeOF₂); (v) it does not explain the shapes of molecules; (vi) it says nothing about the energy (stability) of molecules.

Eight valence electrons; explains bonding but fails for incomplete/expanded octets, odd-electron molecules, noble-gas compounds and shapes.

(i) Low ionisation enthalpy of the metal atom (easy cation formation). (ii) High negative electron gain enthalpy of the non-metal (easy anion formation). (iii) High lattice enthalpy: small ions with high charges attract strongly and release much energy on forming the crystal, outweighing the energy spent on ionisation.

Low ionisation enthalpy, high negative electron gain enthalpy, high lattice enthalpy.

BeCl₂: 2 bp, 0 lp: linear, 180°.

BCl₃: 3 bp, 0 lp: trigonal planar, 120°.

SiCl₄: 4 bp, 0 lp: tetrahedral, 109.5°.

AsF₅: 5 bp, 0 lp: trigonal bipyramidal (120° and 90°).

H₂S: 2 bp + 2 lp (tetrahedral arrangement): bent (V-shaped), angle less than 109.5° (about 92°).

PH₃: 3 bp + 1 lp: trigonal pyramidal, angle less than 109.5° (about 93°).

Linear; trigonal planar; tetrahedral; trigonal bipyramidal; bent; trigonal pyramidal

Both central atoms have four electron pairs (sp³). N in NH₃ has 3 bp + 1 lp; O in H₂O has 2 bp + 2 lp. Since lp–lp > lp–bp > bp–bp repulsion, the two lone pairs in water push the O–H bonds closer (104.5°) than the single lone pair does in ammonia (107°).

Two lone pairs in H₂O compress the angle more (104.5° vs 107°).

Bond strength (bond enthalpy) increases as bond order increases, and bond length decreases. E.g. H₂ (order 1), O₂ (2), N₂ (3): 435.8, 498 and 946 kJ mol⁻¹. In MO theory, bond order $=\frac{1}{2}(N_b-N_a)$; a larger value means a stronger, more stable bond.

Higher bond order ⇒ stronger and shorter bond.

The equilibrium distance between the nuclei of two bonded atoms in a molecule. In a covalent molecule it equals the sum of the covalent radii of the two atoms ($r_A+r_B$); e.g. H–H 74 pm, C–C 154 pm.

Equilibrium internuclear distance of two bonded atoms.

(i) No single Lewis structure fits: one C=O and two C–O bonds would give unequal bond lengths, but experiment shows all three C–O bonds are equal (intermediate between single and double).

(ii) CO₃²⁻ is a resonance hybrid of three canonical forms, with the double bond on each O in turn; they differ only in the position of electrons, not of atoms.

(iii) In the hybrid each C–O has bond order 4/3 and each O carries −2/3 charge.

(iv) The hybrid is more stable than any canonical form (resonance energy).

(v) The canonical forms are imaginary; only the hybrid exists.

Three equivalent canonical forms; the real ion is their hybrid with three equal C–O bonds.

No. Canonical forms must have the same positions of atoms and differ only in the arrangement of electrons. In structure 1 one H is bonded to P; in structure 2 that H is on O. The atoms are in different places, so these are two different molecules (tautomers), not resonance forms. (Actual H₃PO₃ has structure 1, which is why it is dibasic.)

No; the positions of atoms differ, which is not allowed in resonance.

SO₃: S in the centre bonded to three O; three equivalent forms, each with one S=O and two S⁺–O⁻ bonds (the double bond on each O in turn).

NO₂ (17 e⁻, odd): O=N–O ↔ O–N=O, with one unpaired electron on N, formal charge +1 on N and −1 on the single-bonded O.

NO₃⁻ (24 e⁻): three forms, each with one N=O and two N–O⁻; N carries +1; the double bond on each O in turn (like CO₃²⁻).

SO₃: 3 forms; NO₂: 2 forms; NO₃⁻: 3 forms (double bond moving round).

(a) 2K· + S (6 dots) → 2K⁺ + [S, 8 dots]²⁻: K₂S.

(b) Ca (2 dots) + O (6 dots) → Ca²⁺ + [O, 8 dots]²⁻: CaO.

(c) Al (3 dots) + N (5 dots) → Al³⁺ + [N, 8 dots]³⁻: AlN.

K₂S, CaO, AlN

μ(CO₂) = 0 although each C=O bond is polar: the two equal bond moments must point in exactly opposite directions and cancel, so O=C=O is linear (180°).

μ(H₂O) = 1.85 D: the O–H bond moments do not cancel, so they must be at an angle; the molecule is bent (104.5°).

μ(CO₂) = 0 ⇒ linear; μ(H₂O) ≠ 0 ⇒ bent.

(i) Polarity: μ = 0 means non-polar, μ ≠ 0 polar. (ii) Shape: for AB₂, μ = 0 means linear (CO₂), μ ≠ 0 bent (H₂O); for AB₃, planar (BF₃) vs pyramidal (NH₃). (iii) Per cent ionic character $=\frac{{\mu }_{\text{observed}}}{{\mu }_{\text{ionic}}}\times 100$. (iv) Distinguishing isomers: cis and trans (cis has larger μ), and o-, m-, p-disubstituted benzenes (p-dichlorobenzene has μ = 0).

Polarity, shape, ionic character and distinguishing isomers.

Electronegativity: the ability of an atom in a chemical compound to attract the shared electron pair towards itself.

Electron gain enthalpy is the energy change when an isolated gaseous atom gains an electron; it is measurable (kJ mol⁻¹) and fixed for an element. Electronegativity is a relative, dimensionless number (Pauling scale, F = 4.0), not directly measurable, and it varies with the bonding situation.

Electronegativity: pull on a shared pair in a molecule; electron gain enthalpy: energy change for an isolated atom.

When two bonded atoms differ in electronegativity, the shared pair is displaced towards the more electronegative atom, which gets a partial negative charge (δ−), the other a partial positive charge (δ+). Example: HF, where F (4.0) attracts the pair more than H (2.1): Hᵟ⁺–Fᵟ⁻; the bond has a dipole moment. Other examples: H–Cl, O–H in water.

A covalent bond with unequal sharing, e.g. Hᵟ⁺–Fᵟ⁻.

Ionic character grows with electronegativity difference: N₂ (0) is purely covalent; S–O and Cl–F are polar covalent (≈ 1.0 each); K–O (3.5 − 0.8 = 2.7) and Li–F (4.0 − 1.0 = 3.0) are ionic.

N₂ < SO₂ < ClF₃ < K₂O < LiF

Hydrogen can form only one bond. Correct structure: H₃C–C(=O)–O–H. The methyl carbon has three C–H bonds and one C–C bond; the second carbon has a double bond to one O (2 lone pairs) and a single bond to the O of the OH group (2 lone pairs, bonded to H). Every C and O has an octet (24 valence electrons).

H₃C–C(=O)–O–H, each H with one bond and every C, O with an octet.

C in CH₄ is sp³ hybridised: four hybrid orbitals point to the corners of a regular tetrahedron (109.5°). A square planar shape would need dsp² hybridisation, but carbon has no d orbitals in its valence shell. Also, four bond pairs are farthest apart at 109.5°; at 90° (square planar) repulsions would be greater and the molecule less stable.

sp³ carbon gives a tetrahedron; square planar needs dsp² and more repulsion.

BeH₂ is linear (Be is sp, H–Be–H 180°). The two equal Be–H bond moments point in opposite directions and cancel.

Linear molecule: equal bond moments cancel, μ = 0.

NH₃ (1.47 D, against 0.23 D for NF₃). Both are pyramidal with a lone pair on N. In NH₃, N is more electronegative than H, so the N–H bond moments point towards N, in the same direction as the lone-pair moment: they add. In NF₃, F is more electronegative, so the bond moments point towards F, opposite to the lone-pair moment: they largely cancel.

NH₃: bond moments and lone-pair moment add (in NF₃ they oppose).

Hybridisation: mixing of atomic orbitals of slightly different energies of the same atom to give an equal number of new orbitals of identical shape and energy, which then form bonds.

sp: one s + one p → 2 orbitals at 180°, linear (BeCl₂); 50% s character.

sp²: one s + two p → 3 orbitals at 120° in a plane, trigonal planar (BCl₃); 33% s.

sp³: one s + three p → 4 orbitals at 109.5°, tetrahedral (CH₄); 25% s.

Each hybrid orbital has a large lobe (used in bonding) and a small one.

Mixing of orbitals; sp linear (180°), sp² trigonal planar (120°), sp³ tetrahedral (109.5°).

In AlCl₃, Al uses three orbitals (sp²): trigonal planar, with one empty 3p orbital. Cl⁻ donates a lone pair into this empty orbital, so four orbitals are now used (sp³): tetrahedral AlCl₄⁻.

sp² → sp³

B changes from sp² (planar BF₃ with an empty 2p orbital) to sp³ when it accepts the lone pair of N. N does not change: it is sp³ in NH₃ (3 bp + 1 lp) and stays sp³ in the adduct (4 bp).

B: sp² → sp³; N: remains sp³

C₂H₄: each C is sp². One sp² of each C overlaps head-on to form the C–C σ bond; the other two sp² of each C overlap with H 1s (four C–H σ bonds, all in one plane, 120°). The unhybridised 2p orbitals, perpendicular to the plane, overlap sideways to form one π bond above and below the plane. C=C = 1σ + 1π.

C₂H₂: each C is sp. sp–sp overlap gives the C–C σ bond and sp–1s the two C–H σ bonds (linear, 180°). The two unhybridised p orbitals on each C overlap sideways in two perpendicular planes: two π bonds. C≡C = 1σ + 2π.

C=C: 1σ (sp²–sp²) + 1π; C≡C: 1σ (sp–sp) + 2π.

(a) H–C≡C–H: 2 C–H σ + C–C (1σ + 2π). (b) H₂C=CH₂: 4 C–H σ + C–C (1σ + 1π).

(a) 3σ, 2π (b) 5σ, 1π

2p_y lobes are perpendicular to the x-axis, so two 2p_y orbitals can only overlap sideways, giving a π bond. All the others overlap along the axis (s orbitals are spherical; 2pₓ lies along x) and give σ bonds.

(c) 2p_y and 2p_y

Only single bonds: sp³; one double bond: sp²; triple bond: sp.

(a) sp³, sp³ (b) sp³, sp², sp² (c) sp³, sp³ (d) sp³, sp² (e) sp³, sp²

Bond pair: a pair of electrons shared between two atoms. Lone pair: a valence-shell pair on an atom that is not shared.

Examples: CH₄: 4 bp, 0 lp on C; NH₃: 3 bp, 1 lp on N; H₂O: 2 bp, 2 lp on O.

Shared vs unshared pairs; e.g. H₂O has 2 bond pairs and 2 lone pairs.

σ: axial overlap, stronger, free rotation; π: sideways overlap, weaker, no rotation, only with σ.

Two H atoms A and B approach each other. New attractive forces (nucleus A–electron B, nucleus B–electron A) and repulsive forces (electron–electron, nucleus–nucleus) arise. Attraction exceeds repulsion, so the potential energy falls, reaching a minimum at 74 pm: the 1s orbitals overlap, the electrons pair with opposite spins, and a stable H₂ molecule forms, releasing 435.8 kJ mol⁻¹. If they come closer, repulsion dominates and the energy rises steeply.

Overlap of 1s orbitals with paired spins at the energy minimum (74 pm).

(i) The combining atomic orbitals must have the same or nearly the same energy (1s with 1s, not with 2s). (ii) They must have the same symmetry about the molecular axis (with z as the axis, 2p_z combines with 2p_z but not with 2pₓ). (iii) They must overlap to the maximum extent.

Similar energy, same symmetry, maximum overlap.

Be₂ has 8 electrons: σ1s² σ*1s² σ2s² σ*2s². $N_b=4$, $N_a=4$, so bond order $=\frac{1}{2}(4-4)=0$. With no net bonding, Be₂ is not stable.

Bond order = 0

O₂: KK σ2s² σ*2s² σ2p_z² π2p⁴ π*2p²; in the valence MOs $N_b=8$, $N_a=4$.

Stability: O₂⁺ > O₂ > O₂⁻ > O₂²⁻; O₂⁺, O₂, O₂⁻ paramagnetic, O₂²⁻ diamagnetic

They are the signs (phases) of the wave function ψ in the lobes, not electric charges. Lobes of the same sign overlap constructively (${\psi }_A+{\psi }_B$): electron density builds up between the nuclei, giving a bonding MO. Lobes of opposite sign overlap destructively (${\psi }_A-{\psi }_B$): a node between the nuclei, an antibonding MO.

Phase of the wave function; same sign → bonding, opposite → antibonding.

P: [Ne]3s² 3p³ → excited 3s¹ 3p³ 3d¹. One 3s, three 3p and one 3d mix to give five sp³d orbitals pointing to the corners of a trigonal bipyramid, each overlapping a Cl 3p orbital.

Three equatorial P–Cl bonds lie in a plane at 120°; two axial bonds are above and below at 90° to it. Each axial bond pair is at 90° to three equatorial pairs, so it suffers more repulsion; the axial bonds stretch and are longer (240 pm vs 202 pm) and weaker.

sp³d, trigonal bipyramidal; axial bonds face more bond-pair repulsion.

A hydrogen bond is the attraction between an H atom bonded to a highly electronegative atom (F, O, N) and another electronegative atom (F, O, N) of the same or another molecule, e.g. H–F···H–F. It is stronger than van der Waals forces (about 10–40 kJ mol⁻¹) but much weaker than a covalent bond.

Stronger than van der Waals forces.

Bond order $=\frac{1}{2}(N_b-N_a)$ (in Lewis terms, the number of bonds between two atoms).

N₂ (14 e⁻): σ1s² σ*1s² σ2s² σ*2s² π2p⁴ σ2p_z²: $\frac{1}{2}(10-4)=3$. O₂ (16 e⁻): $\frac{1}{2}(10-6)=2$. O₂⁺ (15 e⁻): $\frac{1}{2}(10-5)=2.5$. O₂⁻ (17 e⁻): $\frac{1}{2}(10-7)=1.5$.

N₂ 3, O₂ 2, O₂⁺ 2.5, O₂⁻ 1.5