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Chemistry · Class 11 · Chapter 4

Chemical Bonding and Molecular Structure

Shapes, hybridisation, bond order and magnetic behaviour come up again and again in inorganic and organic chemistry. Practise drawing structures by hand until you can predict the shape of a molecule in a few seconds.

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

Formal charge = V − L − ½SV = valence electrons in the free atom, L = non-bonding (lone pair) electrons, S = bonding (shared) electrons

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.

Basic VSEPR shapes with bond pairs onlywww.iitmedicoguide.comBeBCPSLinearBeCl₂, 180°Trigonal planarBF₃, 120°TetrahedralCH₄, 109.5°TrigonalbipyramidalPCl₅, 90°, 120°OctahedralSF₆, 90°Hybridisation of the central atom, left to right: sp, sp², sp³, sp³d, sp³d²Wedge: bond towards you. Dashed: bond away from you.www.iitmedicoguide.com
The five geometries for two to six bond pairs with no lone pairs. With lone pairs present, the arrangement of electron pairs stays the same but the shape is described by the atoms alone.
Bond pairsLone pairsShapeExample
21BentSO2, O3
31Trigonal pyramidalNH3 (107°)
22BentH2O (104.5°)
41See-sawSF4
32T-shapedClF3
51Square pyramidalBrF5
42Square planarXeF4

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).

HybridisationOrbitals mixedGeometryExamples
spone s + one pLinear, 180°BeCl2, C in C2H2
sp2one s + two pTrigonal planar, 120°BCl3, C in C2H4
sp3one s + three pTetrahedral, 109°28′CH4, C2H6; also NH3, H2O with lone pairs
sp3ds + three p + one dTrigonal bipyramidalPCl5
sp3d2s + three p + two dOctahedralSF6

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.

Bond order = ½(Nb − Na)Nb, Na = electrons in bonding and antibonding orbitals. Bond order zero means the molecule does not exist (He2, Ne2).

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.

Molecular orbital diagram of O2 (valence electrons)www.iitmedicoguide.com↑↓↑↑↑↓↑↑↑↓↑↓2p2s2p2s↑↑↑↓↑↓↑↓↑↓↑↓σ*2pzπ*2px, π*2pyπ2px, π2pyσ2pzσ*2sσ2sO atomO atomO2 moleculeBond order = ½(8 − 4) = 2; two unpaired electrons in π* orbitals, so O2 is paramagneticEnergy increases upwards (not to scale)www.iitmedicoguide.com
Valence molecular orbitals of O2. The last two electrons go singly into the two π* orbitals with parallel spins (Hund's rule), which is why liquid oxygen is attracted by a magnet.
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:

  1. Tetrahedral
  2. See-saw
  3. Square planar
  4. Trigonal pyramidal
Show answer
B. Four bond pairs and one lone pair (sp3d); the lone pair sits in an equatorial position.

Which of these species is diamagnetic?

  1. O2
  2. O2−
  3. O2+
  4. O22−
Show answer
D. With 18 electrons the π* orbitals are full, so no electron is unpaired.

Which molecule has zero dipole moment?

  1. NH3
  2. H2O
  3. BF3
  4. CHCl3
Show answer
C. The three B−F bond moments cancel in the trigonal planar shape.

The hybridisation of each carbon atom in ethyne (C2H2) is:

  1. sp
  2. sp2
  3. sp3
  4. dsp2
Show answer
A. Each carbon forms two σ bonds and two π bonds; the molecule is linear.

The numbers of σ and π bonds in ethene (C2H4) are:

  1. 4 σ, 2 π
  2. 5 σ, 1 π
  3. 6 σ, 0 π
  4. 3 σ, 3 π
Show answer
B. Four C−H σ, one C−C σ and one C−C π.

The correct order of bond angles is:

  1. H2O > NH3 > CH4
  2. NH3 > CH4 > H2O
  3. CH4 > NH3 > H2O
  4. CH4 = NH3 = H2O
Show answer
C. 109.5° > 107° > 104.5°; each lone pair squeezes the angle further.

In the Lewis structure of ozone, the formal charge on the central oxygen atom is:

  1. −1
  2. 0
  3. +2
  4. +1
Show answer
D. 6 − 2 − ½(6) = +1.

Which species has the same bond order as N2?

  1. O2
  2. NO+
  3. O2−
  4. F2
Show answer
B. NO+ has 14 electrons, like N2, so its bond order is 3.
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