Chemical Bonding and Molecular Structure

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XI, Chapter 4.

Two puzzles for this chapter

Water boils at 100 °C, but H₂S, a heavier molecule, boils at about −60 °C. And if you pour liquid oxygen between the poles of a strong magnet, it sticks there, even though the Lewis picture of O₂ shows all electrons paired. A chapter on bonding has to explain both, and the tools you build to do it also explain why water is bent, why CO₂ is linear and why BF₃ is flat. We start with the simplest idea, atoms want a full outer shell, and then add one better model at a time, each fixing what the one before could not.

Why atoms bond: the Kossel and Lewis idea

A chemical bond is the attractive force that holds atoms or ions together. Noble gases are chemically unreactive because their outer shell is complete: 8 electrons ($ns^2np^6$), or 2 for helium. In 1916 Kossel and Lewis suggested that other atoms react to reach the same arrangement, either by transferring electrons or by sharing them. This is the octet rule.

A Lewis dot symbol shows the valence electrons around the atomic symbol. The valence of an atom is the number of dots (for the first few groups) or 8 minus the dots (for groups 15 to 17).

Writing a Lewis structure. Add up all valence electrons (add one for each negative charge, subtract one for each positive charge). Put the least electronegative atom in the middle, join the atoms with single bonds, and place the remaining electrons as lone pairs on the outer atoms, then the centre. If the central atom is short of an octet, convert lone pairs into double or triple bonds.

Formal charge helps pick the best structure when several are possible. It is the charge an atom would have if all bonding pairs were shared equally.

$$\text{Formal charge} = V - N - \dfrac{B}{2}$$

where $V$ is the valence electrons in the free atom, $N$ the non bonding electrons on it and $B$ the bonding electrons around it. The best structure has the smallest formal charges, with negative charge on the most electronegative atoms. Formal charge is bookkeeping and is not a real charge or an oxidation state.

Worked example: carbonate ion. Draw the Lewis structure of CO₃²⁻ and assign formal charges.

The same method for ozone, O₃ (18 electrons, drawn as O=O–O), gives the central oxygen +1, the doubly bonded oxygen 0 and the singly bonded oxygen −1.

Where the octet rule breaks down

The octet rule also says nothing about the shape of a molecule or its energy, and it cannot explain why noble gases form compounds such as XeF₂ and KrF₂.

Pause and predict: BF₃ has only six electrons around boron. What would you expect it to do when an NH₃ molecule is nearby?

Check your answer: Boron is short of two electrons, so it accepts the lone pair of nitrogen from NH₃ and forms H₃N→BF₃, completing its octet. A species with an incomplete octet is hungry for electrons.

The ionic bond: why NaCl exists

Sodium loses an electron easily (low ionisation enthalpy) and chlorine gains one with a release of energy (negative electron gain enthalpy). Yet for gaseous atoms, forming Na⁺ and Cl⁻ costs a net $+147$ kJ mol⁻¹, because removing Na's electron takes more energy (496 kJ mol⁻¹) than Cl's gain releases (349 kJ mol⁻¹). The compound still forms because the ions pack into a crystal, and this releases lattice enthalpy, the energy needed to separate one mole of a solid ionic compound into gaseous ions. For NaCl this is 788 kJ mol⁻¹, so the overall change is about $147 - 788 = -641$ kJ mol⁻¹, which is strongly favourable. So the stability of an ionic compound comes from lattice enthalpy, not from completing an octet alone.

Ionic bond formation is favoured by a low ionisation enthalpy of the metal, a very negative electron gain enthalpy of the non metal, and a high lattice enthalpy. Lattice enthalpy rises when ions are small and highly charged. The Mg²⁺ and O²⁻ ions in MgO give a lattice enthalpy of roughly 3800 kJ mol⁻¹, nearly five times that of NaCl, which is why MgO melts above 2800 °C.

No bond is completely ionic. Fajans' rules say when a cation distorts (polarises) the electron cloud of an anion enough to give covalent character.

Bond parameters

Four numbers describe a covalent bond. Bond length is the equilibrium distance between the two nuclei. Between like atoms it is twice the covalent radius, and for Cl₂ the covalent radius is 99 pm. Bond angle is the angle between bonds around the central atom. Bond enthalpy is the energy needed to break one mole of a given bond in the gas phase. For molecules with several identical bonds we use the mean: in H₂O the first O–H bond breaks at 502 kJ mol⁻¹ and the second at 427, so the mean bond enthalpy is 464.5 kJ mol⁻¹. Bond order in the Lewis picture is the number of bonds between two atoms. As bond order rises, the bond becomes shorter and stronger.

Isoelectronic species have the same bond order: N₂, CO, CN⁻ and NO⁺ (14 electrons each) all have a triple bond, and F₂ and O₂²⁻ (18 electrons each) are single bonded.

Resonance. For ozone, any one Lewis structure has one long single and one short double O–O bond. Experiment shows two bonds of equal length, 128 pm, between a single bond (148 pm) and a double bond (121 pm). The real molecule is a resonance hybrid of the canonical forms. These forms have identical atom positions and differ only in where the electrons are drawn, and none of them exists on its own. The hybrid has lower energy than any contributing structure. Similar examples are CO₃²⁻ (three forms, each C–O bond equal) and benzene. In CO₃²⁻ the bond order is $\dfrac{1+1+2}{3} = \dfrac{4}{3}$ for each C–O bond.

Polarity and dipole moment. In a bond between two different atoms, the shared pair is pulled towards the more electronegative atom, which gains a partial negative charge $\delta^-$. The product of charge and distance is the dipole moment.

$$\mu = Q \times r \qquad 1\ \text{D} = 3.336 \times 10^{-30}\ \text{C m}$$

Dipole moment is a vector, so whether a molecule is polar depends on its shape as well as on its bonds. In a symmetrical molecule the bond dipoles cancel.

NH₃ and NF₃ have the same shape, but N–H bonds point their dipole towards N, in the same direction as the lone pair, while N–F bonds point towards F, against it. So NF₃, with more electronegative fluorine, is far less polar.

Worked example: dipole moments. (a) The H–F bond length is 91.7 pm and $\mu = 1.78$ D. What fraction of an electron charge sits on each atom? (b) The O–H bond dipole is about 1.5 D and the H–O–H angle is 104.5°. Estimate the dipole moment of water.

The common wrong step: concluding that a molecule with polar bonds is polar. CO₂ has two polar C=O bonds and $\mu = 0$, because they point in opposite directions. Water has $\mu \neq 0$ because it is bent.

Pause and predict: Both CO₂ and H₂O are AB₂ molecules with polar bonds. Why does only one of them have a dipole moment?

Check your answer: CO₂ is linear, so the two bond dipoles are equal and opposite and cancel. H₂O is bent, so the dipoles do not point in opposite directions and add up to a net dipole. Shape decides.

VSEPR: shapes from electron pair repulsion

VSEPR (valence shell electron pair repulsion) theory is the quickest way to the shape of a molecule. Electron pairs around the central atom, both bonding pairs and lone pairs, repel each other and spread as far apart as possible. A double or triple bond counts as one region of electron density. Lone pairs take more room than bonding pairs, because a lone pair is held by only one nucleus. So the repulsion order is

lone pair–lone pair > lone pair–bond pair > bond pair–bond pair

and each lone pair squeezes the bond angles. This explains why the angle drops from 109.5° in CH₄ (no lone pair) to 107° in NH₃ (one) and 104.5° in H₂O (two).

For a central atom bonded to monovalent atoms, find the total number of electron pairs.

$$\text{pairs} = \dfrac{V + X - C}{2}$$

where $V$ is the valence electrons of the central atom, $X$ the number of monovalent atoms attached, and $C$ the charge on the species (positive for a cation, negative for an anion). Then bond pairs $= X$ and lone pairs $=$ pairs $- X$.

Worked example: predict the shape. Find the shape and hybridisation of H₂O, NH₃, NH₄⁺, SF₄, ClF₃, XeF₂, BrF₅ and XeF₄.

The number of pairs fixes the basic geometry, and the lone pairs then decide which atoms you see.

Why lone pairs sit in the equatorial positions. A trigonal bipyramid has two axial and three equatorial positions. An axial position has three neighbours at 90°, an equatorial one only two. Lone pairs repel more strongly, so they take the roomier equatorial sites. That is why the lone pair in SF₄ is equatorial (giving a see saw) and the two lone pairs of ClF₃ are equatorial too, leaving a T. In PCl₅ itself the axial bonds are longer and weaker than the equatorial ones, which makes PCl₅ very reactive.

Pause and predict: XeF₂ has three lone pairs on xenon and two bond pairs. Why is it linear and not bent?

Check your answer: The three lone pairs sit in the equatorial plane at 120° to each other, which keeps their repulsions as small as possible. The two fluorine atoms are then forced to the axial positions on opposite sides.

Beyond the textbook. For the exam: lone pair repulsion lowers the bond angle from 109.5° in CH₄ to 107° in NH₃ and 104.5° in H₂O. In reality: down a group the angles fall much further than VSEPR predicts. H₂S is about 92°, and PH₃ about 93.5°. Larger central atoms bond almost purely with their p orbitals and barely mix in their s orbital. Why it matters: it shows that the ideal angles of a hybridisation scheme are a starting point. Use the CUET values (109.5°, 107°, 104.5°) for the first-row hydrides.

Valence bond theory and hybridisation

VSEPR gives shapes but not reasons for bonding. Valence bond theory (Heitler and London, Pauling) says a covalent bond forms when half filled atomic orbitals, with opposite spins, overlap. As two hydrogen atoms approach, the energy falls to a minimum at 74 pm, the bond length of H₂, with a depth of 435.8 kJ mol⁻¹, the bond enthalpy. Closer than that, repulsion between nuclei dominates. The greater the overlap, the stronger the bond.

A single bond is one σ bond. A double bond is one σ plus one π, and a triple bond is one σ plus two π.

Carbon has only two unpaired electrons ($2s^2 2p^2$) but forms four equal bonds in CH₄. Hybridisation (Pauling) explains this: orbitals of slightly different energy in the same atom mix to form the same number of equivalent hybrid orbitals, all pointing in set directions. The recipe is simple. Count the regions of electron density around the central atom (σ bonds plus lone pairs) and read off the hybridisation. Hybrid orbitals can hold lone pairs, as in H₂O, and the mixing does not require half filled orbitals.

More s character means a shorter, stronger bond. The s character is 50% in sp, 33% in sp² and 25% in sp³, which is why C≡C (120 pm) is shorter than C=C (134 pm), which is shorter than C–C (154 pm).

Worked example: bonds and hybridisation in an organic molecule. Count σ and π bonds in CH₃–CH=CH–C≡CH and state the hybridisation of each carbon.

The common wrong step: counting the shared pairs in a double bond as two σ bonds. Only the first bond between any two atoms is σ.

Beyond the textbook. For the exam: PCl₅ is sp³d and SF₆ is sp³d², which means using one and two d orbitals. In reality: modern calculations show that d orbitals contribute very little in these molecules. The bonding is better described with three centre, four electron bonds that do not need extra orbitals. The shape predictions are the same. Why it matters: hybridisation is a model for getting shapes right, not a physical event inside the atom, and that is why VSEPR and hybridisation always agree.

Molecular orbital theory

Valence bond theory cannot explain why O₂ is attracted by a magnet. Molecular orbital theory (Hund and Mulliken) treats the electrons as spread over the whole molecule. Atomic orbitals of similar energy and the same symmetry combine (LCAO) to give an equal number of molecular orbitals. Adding the waves (ψ_A + ψ_B) gives a bonding orbital with electron density between the nuclei and lower energy. Subtracting them (ψ_A − ψ_B) gives an antibonding orbital, written with a star (∗), which has a node between the nuclei and higher energy. Electrons fill these orbitals by the same Aufbau, Pauli and Hund rules as in atoms.

The order of energy depends on the molecule. For O₂ and F₂:

σ1s < σ∗1s < σ2s < σ∗2s < σ2p_z < (π2p_x = π2p_y) < (π∗2p_x = π∗2p_y) < σ∗2p_z

For Li₂ through N₂, σ2p_z sits above the π orbitals: σ2s < σ∗2s < (π2p_x = π2p_y) < σ2p_z < (π∗2p_x = π∗2p_y) < σ∗2p_z. The reason is that in the lighter atoms the 2s and 2p energies are close, so the σ2s and σ2p_z orbitals interact and push the σ2p_z level up.

$$\text{Bond order} = \tfrac{1}{2}\left(N_b - N_a\right)$$

where $N_b$ and $N_a$ are the numbers of electrons in bonding and antibonding orbitals. A positive value means a stable molecule. A higher bond order means a shorter, stronger bond. A molecule with unpaired electrons is paramagnetic (attracted to a magnet), and one with all electrons paired is diamagnetic.

Worked example: bond order and magnetism. Find the bond order and magnetic behaviour of O₂, O₂⁺, O₂⁻, N₂ and C₂.

Here are the other species the exam likes.

C₂ is unusual: its 12 electrons fill the π orbitals before σ2p_z, so both bonds are π and there is no σ bond from 2p_z. Bond length goes down as bond order goes up, so O₂⁺ (2.5) < O₂ (2) < O₂⁻ (1.5) < O₂²⁻ (1).

Pause and predict: Does N₂ gain or lose bond strength when it loses one electron to become N₂⁺?

Check your answer: It loses strength. The electron removed comes from the bonding σ2p_z orbital, so the bond order falls from 3 to 2.5. Compare with O₂ → O₂⁺, where the electron is removed from an antibonding orbital and the bond becomes stronger.

Beyond the textbook. For the exam: O₂ is paramagnetic with two unpaired electrons, a success of molecular orbital theory. In reality: you can see it. Liquid oxygen is pale blue and, when poured past a strong magnet, is held between the poles. The Lewis structure O=O has no unpaired electrons and cannot account for this. Why it matters: it is the clearest experimental reason for needing MOT.

Beyond the textbook. For the exam: bond order zero means He₂ does not exist. In reality: helium atoms do attract each other very weakly through van der Waals forces, and a He₂ pair has been detected at temperatures around a thousandth of a kelvin. Why it matters: it is not a chemical bond, and the pair separates at once at any ordinary temperature, so the exam answer holds.

Hydrogen bonding

When hydrogen is bonded to a very electronegative small atom (F, O or N), it carries a strong partial positive charge. It is attracted to a lone pair on a neighbouring F, O or N, and the dotted link H···X is the hydrogen bond. It is weaker than a covalent bond but stronger than ordinary van der Waals attractions, typically a tenth or less of the strength of a covalent bond.

Hydrogen bonding explains the puzzle we started with. Boiling points of the hydrides normally rise with molar mass, but the first member of groups 15, 16 and 17 breaks the pattern.

Ice is another consequence. In ice each water molecule is hydrogen bonded to four others in an open tetrahedral cage with empty space inside. When ice melts, the cages partly collapse, so water is denser than ice and ice floats. In o-nitrophenol the internal H bond locks the OH group, so the molecules do not link to each other. It has a lower melting point (318 K against 387 K for p-nitrophenol) and is volatile in steam. In p-nitrophenol the molecules are linked to each other.

Pause and predict: Ethanol (CH₃CH₂OH) and dimethyl ether (CH₃OCH₃) have the same molecular formula. Which boils higher?

Check your answer: Ethanol, at about 78 °C against −24 °C. It has an O–H group that forms hydrogen bonds between molecules, while the ether has no hydrogen on oxygen to donate.

Exam traps

Formula and reaction sheet

60-second revision

Where this chapter connects

• Before this: Classification of Elements and Periodicity in Properties gave you ionisation enthalpy, electron gain enthalpy and electronegativity, the quantities that decide whether a bond is ionic or covalent.
• Next: Thermodynamics uses bond enthalpies and lattice enthalpy to calculate energy changes in reactions.
• Forward link: Organic Chemistry Some Basic Principles and Techniques and Hydrocarbons apply sp³, sp² and sp carbon throughout, and Coordination Compounds extends hybridisation to metal complexes.