Chemical Bonding and Shapes of Molecules

Almost nothing in nature exists as free atoms. Atoms join because the combination has lower energy than the separate atoms, and how they join — by transferring electrons, sharing them, or pooling them — decides everything about the substance: whether it melts at 801 °C like NaCl or boils at −161 °C like CH₄, whether it conducts, and what it dissolves in.

5.1 Valence shell, valence electron and octet theory

The valence shell is the outermost shell of an atom; the electrons in it are the valence electrons, and only these take part in bonding. Sodium (2, 8, 1) has one; chlorine (2, 8, 7) has seven.

In 1916 Kössel and Lewis noted that the noble gases — with ns²np⁶ — are almost completely unreactive, which gives the octet rule.

The rule is a good first guess, not a law. It fails in four ways:

Failure	Examples
Incomplete octet	BeCl₂ (4 e⁻ on Be), BF₃ (6 e⁻ on B), AlCl₃
Expanded octet (elements of period 3 onwards, which have vacant d orbitals)	PCl₅ (10 e⁻), SF₆ (12 e⁻), IF₇
Odd-electron molecules	NO (11 e⁻), NO₂, ClO₂
Noble gases do react	XeF₂, XeF₄, XeO₃

It also says nothing about the shape or the strength of a bond — which is why the rest of this unit exists.

5.2 Ionic bond and its properties

Na (2, 8, 1) − e⁻ → Na⁺ (2, 8) and Cl (2, 8, 7) + e⁻ → Cl⁻ (2, 8, 8), so

2Na + Cl₂ → 2Na⁺Cl⁻

Three conditions favour it: low ionization energy of the metal, high electron affinity of the non-metal, and a high lattice energy — the energy released when one mole of the crystal forms from its gaseous ions. As a working rule ΔEN must exceed about 1.7 (Na–Cl is 3.0 − 0.9 = 2.1). Lattice energy is 788 kJ mol⁻¹ for NaCl but 3791 kJ mol⁻¹ for MgO, whose ions carry double charges — hence melting points of 801 °C and 2852 °C.

The electrostatic field of an ion is the same in every direction, so the ionic bond is non-directional: there is no "molecule" of NaCl, only a giant lattice in which every Na⁺ touches six Cl⁻.

5.3 Covalent bond and coordinate covalent bond; properties of covalent compounds

A covalent bond is formed by the mutual sharing of one or more electron pairs between two atoms, each contributing an equal number of electrons. Sharing one pair gives a single bond (H–H), two pairs a double bond (O=O), three pairs a triple bond (N≡N).

Two standard examples: NH₃ + H⁺ → NH₄⁺ (the lone pair on N is donated to a bare proton) and NH₃ + BF₃ → H₃N→BF₃ (the same lone pair completes boron's octet). Once formed, a coordinate bond is indistinguishable from an ordinary covalent bond — all four N–H bonds in NH₄⁺ are identical.

Property	Ionic compounds	Covalent compounds
Physical state	hard, brittle crystalline solids	gases, liquids or soft solids
Melting / boiling point	very high (NaCl 801 °C)	low (CCl₄ 77 °C)
Electrical conduction	conduct when molten or in water, not when solid	do not conduct (no free ions)
Solubility	soluble in polar solvents (water)	soluble in non-polar solvents (benzene, ether)
Nature of bond	non-directional	strongly directional
Isomerism	not shown	shown
Rate of reaction	fast (ionic reactions)	slow (bonds must break)

5.4 Lewis dot structure of common compounds of s and p block elements

A Lewis (electron-dot) structure shows every valence electron as a dot — shared pairs between the symbols, lone pairs around them. To draw one:

1. Add up the valence electrons of all atoms; add one for each negative charge and subtract one for each positive charge. Call this A.
2. Work out the electrons needed to give every atom a full shell (8 each, 2 for hydrogen). Call this N.
3. The number of shared electrons is S = N - A, i.e. S/2 bond pairs.
4. The remaining (A - S)/2 pairs are placed as lone pairs, oxygen and the halogens first.

5.5 Resonance

When a single Lewis structure cannot account for all the observed properties of a molecule, several structures are written that differ only in the positions of electrons, never of atoms.

Conditions: the canonical forms must have the same arrangement of atoms, the same number of paired electrons, and roughly comparable energies.

5.6 VSEPR theory and shapes of simple molecules

The Valence Shell Electron Pair Repulsion theory (Sidgwick and Powell, developed by Gillespie and Nyholm) predicts shape from electron-pair counting alone. Its postulates:

1. The shape of a molecule depends on the total number of electron pairs (bonding + lone) in the valence shell of the central atom.
2. These pairs repel one another and arrange themselves as far apart as possible.
3. Repulsion falls in the order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, so lone pairs squeeze the bond angles.
4. A multiple bond is treated as a single "super pair" for counting shape, but it repels more strongly than a single bond.

The number of electron pairs X around the central atom is

where V is the number of valence electrons of the central atom, M the number of monovalent atoms attached, c the positive charge and a the negative charge on the ion.

Molecule	Bond pairs	Lone pairs	X	Hybridisation	Arrangement of pairs	Shape	Bond angle
BeF₂	2	0	2	sp	linear	linear	180°
CO₂	2	0	2	sp	linear	linear	180°
BF₃	3	0	3	sp²	trigonal planar	trigonal planar	120°
CH₄	4	0	4	sp³	tetrahedral	tetrahedral	109.5°
CH₃Cl	4	0	4	sp³	tetrahedral	tetrahedral (distorted)	≈110°
NH₃	3	1	4	sp³	tetrahedral	trigonal pyramidal	107°
PH₃	3	1	4	sp³	tetrahedral	trigonal pyramidal	93.5°
H₂O	2	2	4	sp³	tetrahedral	bent (V-shaped)	104.5°
H₂S	2	2	4	sp³	tetrahedral	bent (V-shaped)	92°
PCl₅	5	0	5	sp³d	trigonal bipyramidal	trigonal bipyramidal	120° and 90°
SF₆	6	0	6	sp³d²	octahedral	octahedral	90°

5.7 Elementary idea of Valence Bond Theory

Valence Bond Theory (Heitler, London and Pauling) treats a covalent bond as the overlap of two half-filled atomic orbitals holding electrons of opposite spin. As two H atoms approach, the attraction of each nucleus for the other's electron lowers the potential energy; at 74 pm it reaches a minimum 436 kJ mol⁻¹ below the free atoms, and any closer the nuclei repel. That minimum is the bond: 74 pm is the bond length and 436 kJ mol⁻¹ the bond dissociation energy of H₂.

The greater the overlap, the stronger the bond. Two kinds occur:

Bond	Overlap	Strength	Rotation	Example
σ (sigma)	head-on, along the internuclear axis (s–s, s–p, p–p)	strong	free	H–H, C–H
π (pi)	sideways, above and below the axis (p–p)	weaker	not free	second bond of C=C

A double bond is therefore σ + π and a triple bond σ + 2π. Simple VBT cannot explain why CH₄ has four identical bonds at 109.5°, or why O₂ is paramagnetic; the first difficulty is removed by hybridisation.

5.8 Hybridization involving s and p orbitals only

The number of hybrids always equals the number of orbitals mixed, and hybrids overlap better than pure orbitals, so their bonds are stronger.

Hybridisation	Orbitals mixed	Hybrids formed	Geometry	Angle	Examples
sp	one s + one p	2	linear	180°	BeF₂, BeCl₂, C₂H₂, CO₂
sp²	one s + two p	3	trigonal planar	120°	BF₃, C₂H₄, SO₂
sp³	one s + three p	4	tetrahedral	109.5°	CH₄, NH₃, H₂O, NH₄⁺

In NH₃ and H₂O the central atom is still sp³, but one and two of the four hybrids respectively hold lone pairs, which is exactly why the observed angles (107° and 104.5°) fall short of 109.5°.

5.9 Bond characteristics: bond length, ionic character, dipole moment

Bond length is the equilibrium internuclear distance. It decreases as bond order rises, as the atoms get smaller, and as the s character of the hybrid increases (sp > sp² > sp³).

Bond	C–C	C=C	C≡C	H–H	O–H	C–Cl
Length / pm	154	134	120	74	96	177
Bond energy / kJ mol⁻¹	348	614	839	436	463	328

Ionic character. No bond is purely covalent unless the two atoms are identical. When they differ in electronegativity the shared pair is pulled to the more electronegative atom, giving partial charges δ⁺ and δ⁻ — a polar covalent bond. The bigger the ΔEN, the more ionic the bond; at ΔEN ≈ 1.7 it is half ionic.

Dipole moment measures that polarity:

where q is the separated charge and d the distance between the centres of positive and negative charge. The unit is the debye, 1 D = 3.336 × 10⁻³⁰ C m. Dipole moment is a vector drawn from δ⁺ to δ⁻, so the molecular dipole is the vector sum of the bond dipoles.

Molecule	μ / D	Why
HF, HCl, HBr, HI	1.91, 1.03, 0.79, 0.38	falls as ΔEN falls
H₂O	1.85	bent, so bond dipoles do not cancel
NH₃	1.47	pyramidal; lone pair adds to the resultant
CO₂, BF₃, CCl₄	0	symmetrical, bond dipoles cancel exactly
CHCl₃	1.04	symmetry broken by replacing one Cl with H

5.10 Van der Waals' force and molecular solids

Even molecules with no net charge attract one another weakly. These van der Waals forces are worth 1–40 kJ mol⁻¹ against ≈400 for a covalent bond.

Type	Acts between	Example
Dipole–dipole	two polar molecules	HCl, CH₃Cl
Dipole–induced dipole	a polar and a non-polar molecule	HCl and Ar
London dispersion	any two molecules; an instantaneous dipole induces one in its neighbour	Cl₂, I₂, noble gases

Dispersion forces grow with the number of electrons, so the halogens turn from gas to solid down the group: F₂ boils at −188 °C, Cl₂ at −34 °C, Br₂ at 59 °C, and I₂ melts at 114 °C. Molecular solids (I₂, dry ice, naphthalene, ice) are held by these forces alone, so they are soft, volatile, low-melting and non-conducting.

5.11 Hydrogen bonding and its application

It is intermolecular when it joins separate molecules (water, HF, alcohols, carboxylic acids) and intramolecular when it closes a ring inside one molecule. The difference is measurable: o-nitrophenol, bonded to itself, melts at 45 °C, while p-nitrophenol, which can only bond to its neighbours, melts at 114 °C.

Applications and consequences:

• High boiling points: water boils at 100 °C when H₂S, with a heavier molecule, boils at −60 °C.
• Ice floats. In ice every molecule is hydrogen-bonded tetrahedrally to four others, making an open cage. Ice has a density of 0.92 g cm⁻³ against 1.00 for water, so lakes freeze from the top and fish survive the Himalayan winter.
• Solubility: ethanol, sugar, glucose and ammonia dissolve freely in water because they can hydrogen-bond to it; hydrocarbons cannot and do not.
• Association: ethanoic acid exists as a dimer in benzene, so its measured molecular mass is 120 instead of 60.
• Biology: the two strands of DNA are held together by hydrogen bonds between base pairs, and proteins owe their α-helix to them.
• High surface tension and viscosity of water, and the stiffness of cellulose and nylon fibres.

5.12 Metallic bonding and properties of metallic solids

In a metal the valence electrons are so loosely held that they leave their parent atoms and move freely through the whole crystal.

Its strength rises with the number of valence electrons contributed and falls with cation size: Na (1 e⁻, m.p. 98 °C) < Mg (2 e⁻, 650 °C) < Al (3 e⁻, 660 °C and far harder).

Property of metallic solids	Explanation
Electrical conductivity	delocalised electrons drift under an applied field
Thermal conductivity	mobile electrons carry kinetic energy quickly
Metallic lustre	free electrons absorb and re-emit light of all frequencies
Malleability and ductility	layers slip without breaking a directional bond
High density, high m.p. and b.p.	close packing and strong non-directional attraction
Alloy formation	similar-sized cations substitute freely in the lattice

Chapter summary

• Atoms bond to reach a noble-gas octet (duplet for H). The octet rule fails for incomplete octets (BF₃), expanded octets (SF₆), odd-electron molecules (NO) and noble-gas compounds.
• Ionic bond = complete transfer (ΔEN > 1.7), giving a non-directional giant lattice: high melting point, conducts only when molten or aqueous.
• Covalent bond = shared pairs; a coordinate bond is a shared pair donated entirely by one atom (NH₄⁺, H₃N→BF₃).
• Lewis structures: shared pairs S = N - A; lone pairs = (A-S)/2; formal charge = valence − lone-pair electrons − ½(bonding electrons).
• Resonance: one hybrid structure, not a mixture; it equalises bond lengths and lowers the energy by the resonance energy.
• VSEPR: X = \frac{1}{2}(V + M - c + a) pairs; repulsion lp–lp > lp–bp > bp–bp, so CH₄ 109.5° > NH₃ 107° > H₂O 104.5°.
• sp, sp² and sp³ hybrids give linear (180°), trigonal planar (120°) and tetrahedral (109.5°) geometries; a double bond is σ + π, a triple bond σ + 2π.
• \mu = q \times d in debye; a symmetrical molecule has \mu = 0 even with polar bonds; % ionic character = (\mu_{obs}/\mu_{ionic}) \times 100.
• Van der Waals forces (weak, grow with size) < hydrogen bonds (10–40 kJ mol⁻¹, need F, O or N) < covalent/ionic/metallic bonds.

Practice questions

Group A — Multiple choice (1 mark each)

1. The shape of the PH₃ molecule is [1]atrigonal planarbtrigonal pyramidalctetrahedraldbent
2. Which of the following has zero dipole moment? [1]aH₂ObNH₃cBF₃dCHCl₃
3. The number of coordinate covalent bonds in the ammonium ion is [1]a0b1c2d4
4. The correct order of bond angle is [1]aH₂O > NH₃ > CH₄bCH₄ > NH₃ > H₂OcNH₃ > CH₄ > H₂OdCH₄ > H₂O > NH₃
5. Which property is not a consequence of hydrogen bonding? [1]aice floats on waterbHF is a liquid near room temperaturecmethane has a very low boiling pointdethanol mixes with water in all proportions

Answers to Group A

1. (b) — X = \frac{1}{2}(5+3) = 4: three bond pairs and one lone pair.

2. (c) — BF₃ is trigonal planar and symmetrical, so the three bond dipoles cancel.

3. (b) — one, formed when N donates its lone pair to H⁺; all four bonds then become identical.

4. (b) — lone pairs repel more strongly, so each lone pair closes the angle: 109.5°, 107°, 104.5°.

5. (c) — CH₄ has no F, O or N, so there is no hydrogen bonding at all; its low boiling point is due to weak dispersion forces.

Group B — Short answer (5 marks each)

1. What is a coordinate covalent bond? Explain its formation in NH₄⁺ and in the addition compound H₃N→BF₃. [5]
2. State the postulates of VSEPR theory and use them to account for the fall in bond angle from CH₄ (109.5°) to NH₃ (107°) to H₂O (104.5°). [5]
3. Define dipole moment. The observed dipole moment of HBr is 0.79 D and the H–Br bond length is 1.41 Å. Calculate the percentage ionic character of the bond. [5]
4. What is hydrogen bonding? Explain why water boils at 100 °C while H₂S boils at −60 °C, and why ice floats on water. [5]
5. Give any five points of difference between ionic and covalent compounds. [5]

Answers to Group B

2. All three have four electron pairs and sp³ hybridisation, so the ideal angle is 109.5°. CH₄ has no lone pair and keeps it. NH₃ has one lone pair; since lp–bp repulsion exceeds bp–bp repulsion, the three bonds are pushed closer to 107°. H₂O has two lone pairs, and the additional lp–lp repulsion closes the angle further to 104.5°.

3. \mu = q \times d is the product of the separated charge and its separation, measured in debye. For a fully ionic H–Br bond, \mu_{ionic} = (1.602\times10^{-19})(1.41\times10^{-10}) = 2.259\times10^{-29} C m = 2.259\times10^{-29}/3.336\times10^{-30} = 6.77 D. Hence

% ionic character = (0.79/6.77)\times100 = 11.7\%.

4. A hydrogen bond is the attraction between an H atom bonded to F, O or N and a lone pair on another such atom. Water molecules are extensively hydrogen-bonded, so a great deal of extra energy is needed to separate them; sulphur is too large and too weakly electronegative to hydrogen-bond, so H₂S is held only by weak van der Waals forces and boils 160 °C lower. In ice each molecule is hydrogen-bonded to four neighbours in an open tetrahedral cage containing empty space; this makes ice (0.92 g cm⁻³) less dense than liquid water (1.00 g cm⁻³), so it floats.

Group C — Long answer (8 marks each)

1. (a) State the postulates of VSEPR theory. Predict, with reasons, the shape, hybridisation and bond angle of BF₃, PCl₅, SF₆ and H₂O. [5](b) Both BeF₂ and H₂O are triatomic, yet BeF₂ is linear and H₂O is bent. Explain. [3]
2. (a) What is hybridisation? Describe sp, sp² and sp³ hybridisation with one example of each, stating the geometry and bond angle. [5](b) The dipole moment of H₂O is 1.85 D but that of CO₂ is zero, although both contain polar bonds. Explain, and calculate the resultant for water from an O–H bond moment of 1.5 D and a bond angle of 104.5°. [3]

Answers to Group C

1.(b) In BeF₂, X = \frac{1}{2}(2+2) = 2: beryllium has two bond pairs and no lone pair, so the pairs get as far apart as possible at 180° and the molecule is linear (sp). In H₂O, X = \frac{1}{2}(6+2) = 4: oxygen has two bond pairs and two lone pairs, arranged tetrahedrally (sp³). Only the atoms are seen, so the shape is bent, and the lone pairs squeeze the angle to 104.5°.

2.(b) CO₂ is linear, so its two equal C=O bond dipoles point in exactly opposite directions and cancel: \mu = 0. Water is bent, so its bond dipoles add vectorially to a non-zero resultant: \mu = 2(1.5)\cos(104.5^{\circ}/2) = 3.0 \times 0.6122 = 1.84 D, in agreement with the measured 1.85 D.