In this chapter: Kössel-Lewis approach and the octet rule, Lewis structures and formal charge, limitations of the octet rule, ionic bond and lattice enthalpy, bond length, bond angle, bond enthalpy and bond order, resonance, dipole moment and Fajans' rules, VSEPR theory, valence bond theory, σ and π bonds, hybridisation, molecular orbital theory of homonuclear diatomic molecules, and hydrogen bonding.Kössel-Lewis approach and the octet rule
Atoms combine to reach a stable noble gas configuration, usually eight electrons in the valence shell (two for hydrogen and helium). Kössel explained ionic (electrovalent) bonds by transfer of electrons: Na gives one electron to Cl, and Na+ and Cl− are held by electrostatic attraction. Lewis and Langmuir explained covalent bonds by sharing: each shared pair counts towards the octet of both atoms. Sharing two pairs gives a double bond (O=O, CO2) and three pairs a triple bond (N≡N, HC≡CH).
Formal charge
When several Lewis structures are possible, the one with the lowest formal charges on the atoms is usually the most stable.
Worked example: Find the formal charges on the three oxygen atoms in the Lewis structure of ozone, O=O−O (central O has one lone pair; the double-bonded end O has two lone pairs; the single-bonded end O has three).Solution: Central O: 6 − 2 − ½(6) = +1. Double-bonded end O: 6 − 4 − ½(4) = 0. Single-bonded end O: 6 − 6 − ½(2) = −1. The charges add up to zero, as they must for a neutral molecule.
Limitations of the octet rule
- Incomplete octet: the central atom has fewer than eight electrons, as in LiCl, BeH2 and BCl3.
- Odd-electron molecules: NO and NO2 cannot give every atom an octet.
- Expanded octet: elements of period 3 onwards can use d orbitals and have more than eight electrons: PF5 (10), SF6 (12), H2SO4 (12 around S).
- Noble gases also form compounds such as XeF2, XeF4 and XeOF2. The octet rule says nothing about the shape of a molecule or the energy of its bonds.
Ionic bond and lattice enthalpy
Ionic bonds form easily between a metal of low ionisation enthalpy and a non-metal with a large negative electron gain enthalpy. The crystal is held together by lattice enthalpy, the energy needed to completely separate one mole of a solid ionic compound into gaseous ions. For NaCl it is 788 kJ mol−1.
Bond parameters
- Bond length: equilibrium distance between the nuclei of two bonded atoms. The covalent radius is half the distance between two identical bonded atoms.
- Bond angle: angle between the orbitals containing bonding electron pairs around the central atom, for example 104.5° in water.
- Bond enthalpy: energy needed to break one mole of bonds of a particular type in the gaseous state: H−H 435.8, O=O 498, N≡N 946.0 kJ mol−1.
- Bond order: number of bonds between two atoms (H2 1, O2 2, N2 3). Higher bond order means greater bond enthalpy and shorter bond length. Isoelectronic species have the same bond order: F2 and O22− have 1; N2, CO and NO+ have 3.
Resonance
When one Lewis structure cannot describe a molecule, it is shown as a resonance hybrid of two or more canonical structures. In O3 both O−O bonds are 128 pm, between the single bond (148 pm) and the double bond (121 pm). In CO32− all three C−O bonds are identical. The canonical forms have no real existence; the molecule does not flip between them. Resonance lowers the energy of the molecule and so stabilises it.
Polarity and dipole moment
In a bond between atoms of different electronegativity the shared pair shifts towards the more electronegative atom, giving a polar covalent bond. Dipole moment μ = charge × distance; its unit is the debye, 1 D = 3.33564 × 10−30 C m. It is a vector, and the molecular dipole moment is the vector sum of bond moments.
- CO2 (linear), BF3 (trigonal planar) and CH4 (tetrahedral) have zero dipole moment because the bond moments cancel. H2O is bent and has μ = 1.85 D.
- NH3 has a larger dipole moment than NF3. In NH3 the lone pair moment is in the same direction as the resultant of the N−H bond moments; in NF3 it opposes the N−F bond moments.
Fajans' rules: an ionic bond gains covalent character when the cation is small, the anion is large and the charges on the ions are high. A cation with an (n−1)dnns0 configuration polarises an anion more than one of similar size and charge with a noble gas configuration.
VSEPR theory
The shape of a molecule depends on the number of electron pairs, bonded and lone, around the central atom. The pairs repel each other and take up positions as far apart as possible. A multiple bond is treated as a single super pair. Repulsion falls in the order lone pair-lone pair > lone pair-bond pair > bond pair-bond pair, so lone pairs squeeze bond angles.
| Bond pairs | Lone pairs | Shape | Example |
|---|---|---|---|
| 2 | 1 | Bent | SO2, O3 |
| 3 | 1 | Trigonal pyramidal | NH3 (107°) |
| 2 | 2 | Bent | H2O (104.5°) |
| 4 | 1 | See-saw | SF4 |
| 3 | 2 | T-shaped | ClF3 |
| 5 | 1 | Square pyramidal | BrF5 |
| 4 | 2 | Square planar | XeF4 |
To find the number of lone pairs quickly: lone pairs on the central atom = (valence electrons of central atom − electrons used in bonds)/2. For NH3, (5 − 3)/2 = 1.
Valence bond theory and hybridisation
When two hydrogen atoms approach, attraction between each nucleus and the other electron competes with nucleus-nucleus and electron-electron repulsion. Energy falls to a minimum at 74 pm, the bond length of H2, and 435.8 kJ mol−1 is released. A covalent bond forms by overlap of half-filled atomic orbitals; the greater the overlap, the stronger the bond.
- σ bond: end-to-end (head-on) overlap along the internuclear axis: s-s, s-p or p-p.
- π bond: sideways overlap of p orbitals, above and below the internuclear axis. It is weaker than a σ bond. A double bond is one σ + one π; a triple bond is one σ + two π.
Hybridisation is the mixing of atomic orbitals of slightly different energies on the same atom to give an equal number of identical hybrid orbitals, which then form σ bonds (or hold lone pairs).
| Hybridisation | Orbitals mixed | Geometry | Examples |
|---|---|---|---|
| sp | one s + one p | Linear, 180° | BeCl2, C in C2H2 |
| sp2 | one s + two p | Trigonal planar, 120° | BCl3, C in C2H4 |
| sp3 | one s + three p | Tetrahedral, 109°28′ | CH4, C2H6; also NH3, H2O with lone pairs |
| sp3d | s + three p + one d | Trigonal bipyramidal | PCl5 |
| sp3d2 | s + three p + two d | Octahedral | SF6 |
In PCl5 the two axial bonds are longer than the three equatorial bonds, because axial pairs suffer more repulsion. This is why PCl5 is reactive. In ethene each carbon is sp2; the C=C bond is one sp2-sp2 σ bond and one π bond from the unhybridised 2p orbitals. In ethyne each carbon is sp, and the C≡C bond is one σ and two π bonds.
Molecular orbital theory
Atomic orbitals of comparable energy and the same symmetry about the molecular axis combine (linear combination of atomic orbitals) to form the same number of molecular orbitals. Addition gives a bonding MO (lower energy, electron density between the nuclei); subtraction gives an antibonding MO (higher energy, a node between the nuclei, marked *). Electrons fill MOs by the Aufbau principle, Pauli principle and Hund's rule.
The order of MO energies differs for light and heavy second-period molecules:
- Up to N2 (B2, C2, N2): σ1s < σ*1s < σ2s < σ*2s < (π2px = π2py) < σ2pz < (π*2px = π*2py) < σ*2pz
- O2 and F2: σ2pz lies below the π2p pair; the rest of the order is the same.
N2 (14 electrons) has bond order ½(10 − 4) = 3 and all electrons paired, so it is diamagnetic. O2 (16 electrons) has bond order ½(10 − 6) = 2 and two unpaired electrons in the π* orbitals, so it is paramagnetic. Valence bond theory cannot explain this.
Worked example: Compare the bond order and magnetic behaviour of O2+, O2, O2− and O22−.Solution: Start from O2 (Nb = 10, Na = 6) and change only the π* electrons. O2+: ½(10 − 5) = 2.5, one unpaired electron. O2: 2, two unpaired. O2−: ½(10 − 7) = 1.5, one unpaired. O22−: ½(10 − 8) = 1, no unpaired electron, so only the peroxide ion is diamagnetic. Bond length increases as bond order falls: O2+ < O2 < O2− < O22−.
Hydrogen bonding
When hydrogen is bonded to a highly electronegative atom (F, O or N), it carries a partial positive charge and is attracted to a lone pair on an electronegative atom of another molecule or of the same molecule. This attraction is much weaker than a covalent bond.
- Intermolecular: between different molecules, as in HF and H2O. It raises boiling points, which is why water is a liquid while H2S is a gas at room temperature.
- Intramolecular: within the same molecule, as in o-nitrophenol, where the −OH hydrogen bonds to an oxygen of the neighbouring −NO2 group.
Common mistakes: (1) Using the O2 order of MO energies for N2, or the other way round. (2) Forgetting lone pairs when naming the shape: NH3 is pyramidal and H2O is bent, even though both are sp3. (3) Assuming every molecule with polar bonds has a dipole moment; symmetrical CO2 and BF3 have none. (4) Counting a double bond as two σ bonds. (5) Treating resonance structures as real forms that interconvert.JEE and NEET focus
- Shapes and bond angles from VSEPR, including molecules with lone pairs (SF4, ClF3, XeF4, BrF5).
- Hybridisation of the central atom and counting σ and π bonds in organic and inorganic molecules.
- MO configuration, bond order and magnetic character of H2, He2, N2, O2 and their ions; bond length order.
- Dipole moment comparisons (NH3 vs NF3, cis vs trans isomers, symmetrical molecules).
- Formal charge, resonance structures and Fajans' rules.
- Effect of hydrogen bonding on boiling point and solubility.
Practice questions
The shape of SF4 is:
- Tetrahedral
- See-saw
- Square planar
- Trigonal pyramidal
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Which of these species is diamagnetic?
- O2
- O2−
- O2+
- O22−
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Which molecule has zero dipole moment?
- NH3
- H2O
- BF3
- CHCl3
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The hybridisation of each carbon atom in ethyne (C2H2) is:
- sp
- sp2
- sp3
- dsp2
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The numbers of σ and π bonds in ethene (C2H4) are:
- 4 σ, 2 π
- 5 σ, 1 π
- 6 σ, 0 π
- 3 σ, 3 π
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The correct order of bond angles is:
- H2O > NH3 > CH4
- NH3 > CH4 > H2O
- CH4 > NH3 > H2O
- CH4 = NH3 = H2O
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In the Lewis structure of ozone, the formal charge on the central oxygen atom is:
- −1
- 0
- +2
- +1
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Which species has the same bond order as N2?
- O2
- NO+
- O2−
- F2





