Chemical Bonding and Molecular Structure | ISC Class 11 Chemistry Notes

Electron sharing and transfer, Lewis structures, bond properties, molecular geometry, orbital theories, coordinate bonding and hydrogen bonding.

How do valence electrons explain chemical bonding?

A chemical bond is the attraction holding atoms, ions or other constituents together. Bonding lowers energy and increases stability. An ion is an atom or group carrying an electric charge.

Valence electrons occupy the outer shell and participate in combination. Inner electrons are generally not involved. A Lewis symbol shows one dot for each valence electron.

What does the octet rule mean?

The octet rule describes loss, gain or sharing to obtain eight outer electrons. Hydrogen attains a duplet, two outer electrons, like helium.

In the Kössel-Lewis approach, electron transfer produces oppositely charged ions; sharing produces covalent bonds. The octet rule helps explain these processes but does not completely explain bonding or stability.

How is a Lewis structure constructed?

1. Add the valence electrons supplied by all atoms. For methane, CH₄, carbon supplies four and the four hydrogen atoms supply one each.
2. Add one electron for each negative charge on an ion; subtract one for each positive charge. Charge changes the total electron count, not the number of nuclei.
3. Arrange the atoms in a skeletal structure. Generally, the least electronegative atom occupies the centre. Electronegativity is an atom's tendency in a bond to attract the shared electrons.
4. Place shared pairs between linked atoms, then distribute the remaining electrons as lone pairs or use them to form multiple bonds where needed. Check hydrogen's duplet and applicable octets.

A bond pair is shared; a lone pair is not involved in bonding. Lewis structures show electron accounting, not generally the actual molecular shape.

Worked example 1. Carbon supplies four valence electrons and oxygen supplies six. Construct the Lewis structure of carbon monoxide, CO.

Answer: The total is 4 + 6 = 10 electrons. A triple bond uses six electrons; the remaining four form one lone pair on each atom. Both atoms have octets. The structure is :C≡O:, where each colon denotes a lone pair and ≡ denotes three shared pairs.

How are ionic bonds formed and stabilised?

An ionic or electrovalent bond is electrostatic attraction between positive ions, called cations, and negative ions, called anions. Electrovalency counts their unit charges, including the sign.

For sodium chloride, NaCl, sodium loses an electron, written e⁻, and chlorine gains it: Na → Na⁺ + e⁻; Cl + e⁻ → Cl⁻. Opposite charges attract.

How are the required ionic Lewis structures shown?

In this compact Lewis notation, (··)₄ means four electron pairs surrounding the preceding atom. Brackets enclose each ion; superscripts specify charge.

Cation symbols omit completed inner shells; displayed anions have outer octets. Group 1 metals commonly form singly charged cations; Group 2 metals form doubly charged cations. Groups 16 and 17 provide the illustrated oxide-type and halide-type anions.

Which conditions favour ionic bonding?

• Low ionisation enthalpy: relatively little energy is required to remove an electron from a gaseous atom.
• Highly negative electron gain enthalpy: considerable energy is released when a gaseous atom accepts an electron. Electron gain can otherwise be endothermic, meaning energy is absorbed.
• Large lattice stabilisation: arranging gaseous ions into a solid releases sufficient energy to favour the crystal.
• A large electronegativity difference favours ionic character, although completely ionic and completely covalent bonds are ideal limits.

Most ionic compounds have metal-derived cations and non-metal-derived anions. Ammonium, NH₄⁺, is an exception. Ionic crystals have three-dimensional arrays of ions, called a crystal lattice, rather than separate NaCl molecules.

How does a Born-Haber cycle determine lattice enthalpy?

Lattice dissociation enthalpy is the energy needed to separate one mole of ionic solid into gaseous ions. The mole, mol, measures amount of substance; kJ mol⁻¹ means kilojoules per mole.

For NaCl, lattice dissociation requires +788 kJ mol⁻¹; reverse lattice formation releases this energy and has negative enthalpy change. The labels (s) and (g) mean solid and gas.

Hess's law allows enthalpy changes for steps to be added, provided the initial and final states are the same. A Born-Haber cycle applies this to ionic-solid formation through gaseous atoms and ions.

Derivation: the lattice dissociation expression

Let S be sodium's sublimation enthalpy, I its ionisation enthalpy, D the Cl₂ bond dissociation enthalpy, A chlorine's electron gain enthalpy, F the enthalpy for forming solid NaCl from its elements, and L its lattice dissociation enthalpy. All refer to the stated molar processes.

1. Convert Na(s) into Na(g), requiring S, and ½Cl₂(g) into Cl(g), requiring D/2. Sublimation means conversion directly from solid to gas.
2. Ionise Na(g), contributing I, and add its electron to Cl(g), contributing A. The result is Na⁺(g) and Cl⁻(g).
3. Form NaCl(s) from the gaseous ions. This contributes −L. The sum S + I + D/2 + A − L equals F.

L = S + I + D/2 + A − F

Worked example 2. For NaCl, use $S=108.4$, $I=496$, $D/2=121$, $A=-348.6$ and $F=-411.2$, all in kJ mol⁻¹. Find lattice dissociation and formation enthalpies.

Formula: $L=S+I+D/2+A-F$; lattice formation enthalpy $H=-L$. Substitute: $L=[108.4+496+121-348.6-(-411.2)]\,\mathrm{kJ\,mol^{-1}}$.

Answer: $L=+788\,\mathrm{kJ\,mol^{-1}}$, equivalent to +788000 J mol⁻¹. Lattice formation enthalpy is $-788\,\mathrm{kJ\,mol^{-1}}$, equivalent to −788000 J mol⁻¹. Reversing the process changes the sign, while retaining the magnitude.

The energy-level diagram links solid sodium and chlorine gas to gaseous atoms, gaseous ions and solid NaCl. Upward steps show sublimation, ionisation and chlorine dissociation; downward steps show electron gain and lattice formation.

See Fig. 5.9 in your NCERT textbook

Octet attainment alone does not establish stability: forming gaseous ions can require energy overall, compensated by energy released during lattice formation.

How do covalent bonds, variable valency and octet exceptions compare?

A covalent bond shares an electron pair. One, two or three shared pairs form single, double or triple bonds. Sharing is commonly favoured between non-metals when complete electron transfer is less favourable.

What do simple covalent Lewis structures contain?

Each bond line replaces two dots; add the listed lone pairs. Covalency counts shared pairs, whereas electrovalency counts ionic charges.

Where does the octet rule fail?

• Incomplete octets occur in species such as BeH₂ and BCl₃, with fewer than eight electrons around the central atom.
• Odd-electron molecules, including nitric oxide, NO, and nitrogen dioxide, NO₂, cannot give every atom an octet.
• Expanded octets occur in representations of PF₅ and SF₆, containing more than eight electrons around phosphorus and sulphur respectively.

The rule applies mainly to second-period elements and is useful for most organic compounds. It does not explain molecular shapes or relative energies.

Why can valency vary?

Phosphorus shows covalencies 3 and 5; sulphur 2, 4 and 6; chlorine 1, 3, 5 and 7. An orbital describes an electron's distribution. The traditional account uses different unpaired-electron counts, with promotion into available orbitals for expanded valency.

Variable electrovalency can arise from an unstable core: an arrangement left after electron loss that permits further loss. Fe²⁺ and Fe³⁺ illustrate loss of different numbers of outer and inner-subshell electrons. A subshell groups orbitals of the same type within a shell.

The inert pair effect is the tendency towards non-participation of the outer s-electron pair in heavier main-group elements. Here s names an orbital type. It helps explain lead's lower valency in Pb²⁺ compared with Pb⁴⁺.

Ionic solids generally have high melting points and conduct when molten or dissolved if ions can move. Simple molecular covalent substances generally melt lower and conduct poorly; giant covalent networks are exceptions to this simple comparison.

How do formal charge and resonance improve Lewis structures?

Formal charge compares free-atom valence electrons with electrons assigned in a Lewis structure. It is accounting, not actual charge separation.

Derivation: assigning formal charge

Let V be the free atom's valence-electron count, N the number of non-bonding electrons, B the number of bonding electrons around it, and C the formal charge in elementary-charge units, each the magnitude of an electron's charge.

1. Assign both electrons of each lone pair to the atom: this gives N assigned electrons.
2. Divide every shared pair equally between the bonded atoms: this gives B/2 more assigned electrons.
3. Subtract the total assigned electrons, N + B/2, from the free atom's count V.

C = V − N − B/2

Generally, the lowest-energy Lewis structure has the smallest formal charges. Their sum equals the species' total charge.

Worked example 3. In one ozone structure, O₃, the central oxygen has two non-bonding and six bonding electrons. Its double-bonded terminal oxygen has four non-bonding and four bonding electrons; its single-bonded terminal oxygen has six non-bonding and two bonding electrons. Each free oxygen has six valence electrons. Find the formal charges.

Answer: Central oxygen: 6 − 2 − 6/2 = +1. Double-bonded terminal oxygen: 6 − 4 − 4/2 = 0. Single-bonded terminal oxygen: 6 − 6 − 2/2 = −1. Their sum is zero.

Worked example 4. The carbon to oxygen bond length in CO₂ is 115 pm. Normal C=O and C≡O bond lengths are 121 pm and 110 pm respectively. Compare the measured length with both reference values and explain the need for resonance.

Calculate the differences: $121-115=6$ pm and $115-110=5$ pm.

Answer: The measured bond is shorter than the normal double bond by 6 pm, equivalent to 0.000000000006 m, and longer than the normal triple bond by 5 pm, equivalent to 0.000000000005 m. Its intermediate length is represented by a resonance hybrid of the contributing Lewis structures.

What is a resonance hybrid?

Resonance uses alternative electron arrangements, or canonical structures, when one Lewis structure is inadequate. Nuclear positions stay fixed. The actual structure is the resonance hybrid.

The hybrid has lower energy than any canonical form. Forms do not interconvert or exist in equilibrium. A double-headed arrow, ↔, connects their representations.

Structures I and II interchange single and double oxygen-oxygen bonds. Their labelled lengths are 148 pm and 121 pm. Structure III shows equal bonds of 128 pm. Here pm means picometre, equal to 10⁻¹² metre.

See Fig. 4.3 in your NCERT textbook

In the carbonate ion, CO₃²⁻, three equivalent canonical forms place the carbon-oxygen double bond on different oxygen atoms. All three actual carbon-oxygen bonds are equivalent. The nitrate ion, NO₃⁻, likewise has three equivalent forms with the double bond on different oxygen atoms.

CO₂ has contributions from O=C=O and charge-separated forms ⁻O-C≡O⁺ and ⁺O≡C-O⁻. Add lone pairs to complete oxygen octets. Resonance averages the overall bond characteristics.

What do bond length, bond angle, bond enthalpy and bond order measure?

Bond length is the equilibrium internuclear distance. Each atom contributes a covalent radius, approximately its core size in a bonded situation. For identical bonded atoms, this radius is half the distance.

Let R be bond length and r₁ and r₂ be the two atomic covalent radii, all in the same length unit.

R = r₁ + r₂

Bond angle is the angle between orbitals containing bonding pairs around the central atom, expressed in degrees, °. It helps describe molecular shape.

Worked example 5. For chlorine, a bonded pair of atoms has an internuclear separation of 198 pm, while the separation between neighbouring chlorine atoms in different molecules is 360 pm. Calculate the covalent and van der Waals radii.

Each radius is half its corresponding separation. Thus $r_{\mathrm{c}}=(198/2)\,\mathrm{pm}=99\,\mathrm{pm}$ and $r_{\mathrm{vdW}}=(360/2)\,\mathrm{pm}=180\,\mathrm{pm}$.

Answer: The covalent radius is 99 pm, equivalent to 0.000000000099 m. The van der Waals radius is 180 pm, equivalent to 0.000000000180 m. The larger value describes the atom in a nonbonded situation.

How does bond enthalpy describe strength?

Bond enthalpy is the energy required to break one mole of a particular bond between gaseous atoms. Higher bond dissociation enthalpy means stronger bonding; the H-H value is 435.8 kJ mol⁻¹.

Successive O-H bond breaking in water requires 502 and 427 kJ mol⁻¹. Removing the first hydrogen changes the remaining bond's chemical environment.

Let T be the sum of successive molar bond dissociation enthalpies, n the number of bonds per molecule being averaged, and E the mean bond enthalpy. The symbol n here counts bonds and has no unit.

E = T/n

Worked example 6. Water's two successive O-H bond dissociation enthalpies are 502 and 427 kJ mol⁻¹. Calculate the total and mean values.

Formula: $T=\Delta H_1+\Delta H_2$; $E=T/n$. Substitute: $T=(502+427)\,\mathrm{kJ\,mol^{-1}}$; $n=2$.

Answer: $T=929\,\mathrm{kJ\,mol^{-1}}$, equivalent to 929000 J mol⁻¹, for complete dissociation of gaseous water into atoms. $E=(929/2)\,\mathrm{kJ\,mol^{-1}}=464.5\,\mathrm{kJ\,mol^{-1}}$, equivalent to 464500 J mol⁻¹ per mole of O-H bonds, the mean value.

How does bond order relate to bond length?

In the Lewis description, bond order counts bonds between two atoms. Generally, increasing bond order increases bond enthalpy and decreases bond length.

Note: An average bond length is not necessarily the exact length in every molecule. Compare like bond types and retain the distinction between averages and molecule-specific measurements.

How do electronegativity, dipole moment and Fajans' rules describe polarity?

A non-polar covalent bond has equally attracted shared electrons, as in H₂. A polar covalent bond has unequal sharing, as in HF, where fluorine attracts electrons more strongly.

The symbols δ⁺ and δ⁻ mean partial positive and negative charges. Hydrogen is δ⁺ and fluorine δ⁻ in HF. Bond polarity differs from overall molecular polarity.

How is dipole moment calculated?

Dipole moment, μ, is the product of the magnitude of separated charge, Q, and the distance between the charge centres, r. When Q is measured in coulombs, C, and r in metres, m, μ is measured in C m.

μ = Q × r

The debye, D, satisfies 1 D = 3.33564 × 10⁻³⁰ C m. Dipole moment is a vector, so direction matters. The chemistry polarity arrow has a crossed positive end and points towards the negative end.

Worked example 7. Water has a dipole moment of 1.85 D. Using 1 D = 3.33564 × 10⁻³⁰ C m, convert this to C m.

Answer: μ = 1.85 × 3.33564 × 10⁻³⁰ C m = 6.17 × 10⁻³⁰ C m, rounded to three significant figures.

Linear BeF₂ has equal and opposite bond dipoles, giving zero resultant dipole moment. Trigonal planar BF₃ also has zero resultant. Water's bent arrangement prevents cancellation of its O-H bond dipoles.

Ammonia, NH₃, and nitrogen trifluoride, NF₃, are both pyramidal. In NH₃, the lone-pair contribution reinforces the resultant bond dipole; in NF₃ it opposes it. Their dipole moments are 1.47 D and 0.23 D respectively.

What do Fajans' rules predict?

Polarisation is distortion of an anion's electron cloud by a cation. Polarising power describes the cation's ability to cause distortion; polarisability describes how readily the anion's electron cloud distorts. Greater distortion increases covalent character.

• A smaller cation and a larger anion favour greater covalent character.
• A greater cation charge favours greater covalent character.
• For cations of the same size and charge, a transition-metal-type electronic configuration is more polarising than a noble-gas configuration.

Fajans' rules explain partial covalent character in ionic compounds and complement electronegativity differences. Purely ionic and purely covalent bonds are ideal limits.

How does VSEPR theory predict molecular shapes?

Valence Shell Electron Pair Repulsion theory, abbreviated VSEPR, predicts shapes by arranging electron pairs around a central atom to minimise their mutual repulsion. Count both bonding and lone pairs. Treat a multiple bond as one region of electron density for predicting geometry.

What shapes result without lone pairs?

Linear means straight; trigonal planar means triangular and flat; tetrahedral places four atoms at a tetrahedron's corners. A trigonal bipyramid adds opposite axial positions to a triangle; an octahedron has six surrounding positions.

How do lone pairs distort these arrangements?

Lone pairs occupy more space around the central atom than bond pairs. Repulsion decreases in the order lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. Here > means greater repulsion than.

CH₄ has a 109.5° bond angle, NH₃ has 107°, and H₂O has 104.5°. The additional lone-pair repulsions compress the bond angles. In a trigonal bipyramid, lone pairs favour equatorial positions, where they encounter fewer interactions at 90°.

The ammonia drawing shows nitrogen bonded to three hydrogen atoms with a lone-pair region above it. The water drawing shows oxygen bonded to two hydrogen atoms with two lone-pair regions. The label lp means lone pair.

See Figs. 4.13 and 4.14 in your NCERT textbook

How does orbital overlap explain covalent bond formation?

Valence bond theory, abbreviated VB theory, describes covalent bonding through the overlap of atomic orbitals containing electrons with opposite spins. An atomic orbital describes an electron's distribution around an atom; spin is an intrinsic electron property with two possible orientations.

Each hydrogen contributes a singly occupied 1s orbital: 1 names the shell and s the orbital type. Approach lowers energy to a minimum; closer approach makes repulsion raise it sharply.

Energy is plotted vertically against internuclear distance horizontally. The curve falls to a minimum at a bond length of 74 pm, then approaches the separated-atom energy at large distance. The bond-energy magnitude is labelled 435.8 kJ mol⁻¹.

See Fig. 4.8 in your NCERT textbook

Worked example 8. Dissociating one mole of gaseous H₂ into separate gaseous hydrogen atoms requires 435.8 kJ. Find the energy change when the same amount of H₂ forms from the separated atoms.

Reversing the process reverses the sign: $\Delta H_{\mathrm{form}}=-\Delta H_{\mathrm{diss}}=-435.8\,\mathrm{kJ\,mol^{-1}}$. For one mole, $\Delta H=(-435.8\,\mathrm{kJ\,mol^{-1}})(1\,\mathrm{mol})=-435.8\,\mathrm{kJ}$.

Answer: The energy change is −435800 J. Formation releases this energy, so the bonded molecule is more stable than the separated atoms.

Worked example 9. H₂ and F₂ each have one covalent bond. Their bond dissociation enthalpies are 435.8 and 155 kJ mol⁻¹ respectively. Calculate the difference and identify the stronger bond.

Subtract the smaller dissociation enthalpy from the larger: $\Delta=(435.8-155)\,\mathrm{kJ\,mol^{-1}}=280.8\,\mathrm{kJ\,mol^{-1}}$.

Answer: The difference is 280800 J mol⁻¹. The H-H bond requires more energy to break and is stronger. Equal numbers of shared electron pairs do not imply equal bond strengths.

How do sigma and pi bonds differ?

A sigma bond, symbol σ, forms by head-on overlap along the internuclear axis, the line joining the nuclei. It can involve s-s, s-p or p-p overlap. The p orbitals are directional orbitals with lobes on opposite sides of the nucleus.

A pi bond, symbol π, forms by sideways overlap of parallel orbitals perpendicular to the internuclear axis. Its electron density lies on either side of that axis. Head-on overlap is greater, making a sigma bond stronger than the corresponding pi bond.

In the localised VB description, H₂ has one σ bond; O₂ has one σ and one π bond; N₂ has one σ and two π bonds. In general, greater orbital overlap gives a stronger covalent bond.

Orbital signs denote phase, the wave-function sign, not electric charge. Suitable orientation and matching phase permit bonding overlap.

How does hybridisation explain simple molecular structures?

Hybridisation is the mixing of atomic orbitals of slightly different energies on one atom to produce a set of equivalent hybrid orbitals. The number of hybrid orbitals equals the number of atomic orbitals mixed. The labels s, p and d identify atomic orbital types.

Mixing requires nearly equal energies. Electron promotion is not essential, and filled orbitals can participate. Lone pairs still affect molecular shape.

What are the principal hybridisation patterns?

How are the bonds in ethane, ethene and ethyne formed?

• Ethane: each carbon is sp³ hybridised. Carbon-carbon overlap forms one σ bond, and six carbon-hydrogen overlaps form six more σ bonds.
• Ethene: each carbon is sp² hybridised. The molecule has five σ bonds. One unhybridised p orbital on each carbon overlaps sideways to form one π bond.
• Ethyne: each carbon is sp hybridised. The molecule has three σ bonds and two π bonds, formed by two pairs of unhybridised p orbitals.

In CO₂, two bonding regions around carbon give a linear arrangement with sp hybridisation. Each carbon-oxygen double bond is represented by one σ and one π bond.

What distinguishes axial and equatorial bonds in PCl₅?

The three equatorial bonds lie in one plane at 120° to one another. The two axial bonds lie above and below that plane, at 90° to it. Axial bond pairs experience greater repulsion, so axial bonds are slightly longer and slightly weaker.

SF₆ has six sp³d² hybrids directed towards a regular octahedron's corners, forming six sulphur-fluorine sigma bonds.

How does molecular orbital theory explain bond order and magnetism?

Molecular orbital theory, MO theory, places electrons in orbitals of the whole molecule. Homonuclear diatomic means two identical atoms. Two atomic orbitals produce two molecular orbitals.

A bonding molecular orbital has lower energy and increased electron density between nuclei. An antibonding molecular orbital has higher energy and a node between nuclei, a region of zero electron density. An asterisk, *, marks an antibonding orbital.

How are molecular orbitals formed and filled?

A wave function mathematically describes an electron wave. Let ψₐ and ψᵦ represent atomic wave functions, and ψbond and ψanti the resulting bonding and antibonding combinations. The schematic combinations are:

$\psi_{\mathrm{bond}} = \psi_a + \psi_b$

$\psi_{\mathrm{anti}} = \psi_a - \psi_b$

The linear combination of atomic orbitals, LCAO, adds or subtracts atomic wave functions. Reinforcement produces bonding; cancellation produces antibonding. Combination requires comparable energies, suitable symmetry and effective overlap.

1. Count all electrons, adding electrons for a negative charge and subtracting electrons for a positive charge.
2. Fill lower-energy orbitals first. This is the aufbau principle.
3. Place at most two electrons with opposite spins in each orbital. This is the Pauli exclusion principle.
4. Fill equal-energy orbitals singly with parallel spins before pairing. This is Hund's rule.

Take z as the internuclear direction, with x and y perpendicular to it. In σ2p_z, 2 names the shell and p_z names the contributing orbital; π2p_x and π2p_y are the corresponding sideways combinations. The core sequence is σ1s, σ*1s, σ2s, σ*2s.

For B₂, C₂ and N₂, the remaining increasing energy order is (π2p_x = π2p_y), σ2p_z, (π*2p_x = π*2p_y), σ*2p_z. For O₂ and F₂, σ2p_z lies below the two bonding π orbitals. The equals signs indicate equal energies.

Three panels show combinations of 1s, 2p_z and 2p_x orbitals. In each energy diagram, the bonding level lies below the parent atomic levels and the antibonding level lies above them. The accompanying orbital pictures distinguish reinforcement from cancellation.

See Fig. 4.20 in your NCERT textbook

How is bond order calculated?

Let b be bond order, Nᵦ the number of electrons in bonding orbitals, and Nₐ the number in antibonding orbitals. Bond order is dimensionless.

b = (Nᵦ − Nₐ)/2

Positive bond order indicates a stable bond; zero or negative bond order indicates no stable molecule. Increasing bond order generally strengthens and shortens bonding.

Diamagnetic species have all electrons paired and are repelled by a magnetic field. Paramagnetic species have unpaired electrons and are attracted by a magnetic field.

O₂ contains two unpaired electrons in separate antibonding π orbitals. This explains its observed paramagnetism. C₂ is an important qualification to the usual double-bond picture: its two bonds in this MO description arise from occupied bonding π orbitals.

How do electrons change the relative stability of molecular ions?

For oxygen, electron removal takes an electron from an antibonding orbital, strengthening the bond. Added electrons enter antibonding orbitals and weaken it. In nitrogen, removing a bonding electron or adding an antibonding electron lowers bond order by one-half. Equal bond orders predict comparable bond strength, not identical chemical behaviour.

Worked example 10. O₂ has ten bonding electrons and six antibonding electrons. Formation of O₂⁺ removes one antibonding electron. Calculate both bond orders.

Answer: For O₂, b = (10 − 6)/2 = 2. For O₂⁺, b = (10 − 5)/2 = 2.5. The ion has the higher bond order and therefore the stronger bond in this comparison.

How are coordinate bonds represented in ions and oxy-acids?

A coordinate or dative covalent bond shares a pair initially supplied by one atom, the donor. The acceptor accepts it; an arrow points from donor to acceptor.

After donation the pair participates in covalent bonding. The arrow records its origin without making otherwise equivalent bonds permanently distinguishable.

How do ammonium and hydronium form?

Ammonia donates nitrogen's lone pair to a hydrogen ion, H⁺, forming ammonium: NH₃ + H⁺ → NH₄⁺. The donor-to-acceptor step is N→H. The product has four equivalent N-H bonds and no nitrogen lone pair.

Water similarly donates an oxygen lone pair to H⁺: H₂O + H⁺ → H₃O⁺. This product is the hydronium ion. Oxygen has three O-H bonds and one remaining lone pair. Both products carry an overall positive charge.

How are the chlorine oxy-acids drawn?

An oxy-acid contains oxygen and ionisable hydrogen. Chlorous acid, HClO₂, chloric acid, HClO₃, and perchloric acid, HClO₄, each have an H-O-Cl group.

In the traditional coordinate-bond representations, add one Cl→O arrow for chlorous acid, two for chloric acid and three for perchloric acid. Each arrow terminates on an oxygen not bonded to hydrogen. These are localised representations of electron-pair donation.

If each coordinate link is counted as a single bond in a Lewis structure, the chlorine formal charges are respectively +1, +2 and +3, balanced by −1 on each terminal oxygen. The H-O oxygen has two lone pairs; each arrow-terminal oxygen has three.

How do nitric acid and ozone connect donation with resonance?

One octet-preserving Lewis form of nitric acid, HNO₃, is H-O-N⁺(=O)-O⁻. The terminal single N-O link can be represented by N→O. Nitrogen has no lone pair; the terminal single-bonded oxygen has three and each other oxygen has two.

One ozone form, O=O⁺-O⁻, can show central-to-terminal donation. Resonance interchanges terminal bonding descriptions; actual oxygen-oxygen bonds remain equal.

How does hydrogen bonding affect water, ice and alcohols?

A hydrogen bond attracts hydrogen already bonded to an electronegative atom towards another electronegative atom, commonly F, O or N. It is weaker than a covalent bond; the hydrogen is partially positive.

Dotted lines show hydrogen bonding; solid lines show covalent bonding. Thus O-H···O distinguishes the two interactions.

What is intermolecular hydrogen bonding?

Intermolecular hydrogen bonding occurs between separate molecules of the same or different substances. HF molecules associate through F-H···F attractions. Water molecules associate through O-H···O interactions, with oxygen lone pairs accepting hydrogen bonds.

In ice, hydrogen bonding creates an open tetrahedral arrangement, helping explain its lower density than liquid water. Melting permits closer packing.

Ethanol, C₂H₅OH, has an O-H group that donates hydrogen bonds and an oxygen that accepts them. It hydrogen-bonds with ethanol and water molecules.

What is intramolecular hydrogen bonding?

Intramolecular hydrogen bonding occurs within one molecule when a hydrogen lies suitably between electronegative atoms. In ortho-nitrophenol, the hydroxyl (O-H) and nitro (NO₂) groups occupy adjacent positions on a benzene ring, allowing an internal oxygen-hydrogen-oxygen interaction.

The drawing shows adjacent hydroxyl and nitro groups on a six-membered ring. A dotted line links the hydroxyl hydrogen to an oxygen of the neighbouring nitro group, showing an interaction within the same molecule.

See Fig. 4.22 in your NCERT textbook

Hydrogen bonding is maximum in the solid state and minimum in the gaseous state. It influences molecular association, structure and properties.

Glossary

• Octet rule — The tendency to gain, lose or share electrons to obtain eight outer-shell electrons.
• Ionic bond — Electrostatic attraction between oppositely charged ions in an ionic substance.
• Lattice enthalpy — Energy required to separate one mole of ionic solid into its gaseous constituent ions.
• Formal charge — Electron-accounting charge calculated by assigning lone-pair electrons fully and shared electrons equally to bonded atoms.
• Resonance hybrid — The actual structure represented collectively by canonical forms with the same nuclear arrangement.
• Dipole moment — Product of separated charge magnitude and the distance between positive and negative charge centres.
• Polarisation — Distortion of an anion's electron cloud by the attractive influence of a cation.
• Hybridisation — Mixing of atomic orbitals of slightly different energies to form equivalent hybrid orbitals.
• Antibonding orbital — A molecular orbital whose occupation opposes bonding and which has a node between the nuclei.
• Paramagnetism — Attraction towards a magnetic field associated with the presence of unpaired electrons.
• Coordinate bond — A covalent bond whose shared electron pair initially comes from one donor atom.
• Hydrogen bond — Attraction involving hydrogen bonded to an electronegative atom and another suitable electronegative atom.

Common errors and misconceptions

• Misconception: Every bonded atom must have an octet. Correct: Hydrogen attains a duplet; incomplete octets, odd-electron molecules and expanded-octet representations also occur.
• Misconception: Lattice formation and dissociation have the same sign. Correct: Formation releases energy; reversing it requires energy and reverses the enthalpy sign.
• Misconception: Formal charge measures actual charge separation. Correct: It is based on assigning shared electrons equally and is an electron-accounting device.
• Misconception: Resonance means rapid switching between structures. Correct: Canonical forms are representations; the species has a single resonance-hybrid structure.
• Misconception: Polar bonds guarantee a polar molecule. Correct: Molecular shape can make bond dipoles cancel, as in linear BeF₂.
• Misconception: Four electron-pair regions always mean a tetrahedral molecular shape. Correct: Lone pairs give ammonia a pyramidal shape and water a bent shape.

Exam-style questions with model answers

Q1. Define an ionic bond and explain its formation in NaCl, given that sodium loses one electron and chlorine gains that electron. [2 marks]

1. An ionic bond is the electrostatic attraction between oppositely charged ions.
2. Sodium becomes Na⁺ and chlorine becomes Cl⁻ through electron transfer; attraction between these ions holds the ionic solid together.

Q2. In O₃, each oxygen supplies six valence electrons. A Lewis form has one double and one single bond. Central oxygen has one lone pair, the double-bonded terminal oxygen has two, and the single-bonded terminal oxygen has three. Calculate all three formal charges. [3 marks]

1. Using formal charge = valence electrons − non-bonding electrons − half the bonding electrons, the central oxygen has 6 − 2 − 6/2 = +1.
2. The double-bonded terminal oxygen has four non-bonding electrons and four bonding electrons, so its formal charge is 6 − 4 − 4/2 = 0.
3. The single-bonded terminal oxygen has six non-bonding electrons and two bonding electrons, so its formal charge is 6 − 6 − 2/2 = −1.

Q3. For NaCl formation, the molar enthalpies in kJ mol⁻¹ are: sodium sublimation +108.4; sodium ionisation +496; dissociation of ½Cl₂ into Cl +121; chlorine electron gain −348.6; formation of NaCl(s) from its elements −411.2. Use a Born-Haber cycle to find lattice dissociation enthalpy and explain its significance. [5 marks]

1. Convert the elements into gaseous atoms. Sodium sublimation contributes +108.4 kJ mol⁻¹ and production of one mole of chlorine atoms contributes +121 kJ mol⁻¹.
2. Produce the gaseous ions. Sodium ionisation contributes +496 kJ mol⁻¹, while chlorine electron gain contributes −348.6 kJ mol⁻¹.
3. Let L denote lattice dissociation enthalpy in kJ mol⁻¹. Lattice formation contributes −L. Hess's law therefore gives −411.2 = 108.4 + 121 + 496 − 348.6 − L.
4. Rearrangement gives L = 108.4 + 121 + 496 − 348.6 + 411.2 = +788 kJ mol⁻¹.
5. The positive value is the energy required to separate the solid into gaseous ions. The reverse release of 788 kJ mol⁻¹ stabilises the crystal.

Q4. CH₄ has four bond pairs and no central lone pairs; NH₃ has three bond pairs and one lone pair; H₂O has two bond pairs and two lone pairs. Their bond angles are respectively 109.5°, 107° and 104.5°. Use VSEPR theory to explain their shapes and angle order. [4 marks]

1. All three have four electron-pair regions, which take a tetrahedral arrangement to minimise repulsion. With no lone pairs, methane also has a tetrahedral molecular shape and its angle is 109.5°.
2. Ammonia's one lone pair leaves three bonded atoms, giving a trigonal pyramidal shape. Stronger lone pair-bond pair repulsion compresses the angle to 107°.
3. Water's two lone pairs leave two bonded atoms, giving a bent shape. Additional lone-pair repulsions compress the angle further to 104.5°.
4. The repulsion order is lone pair-lone pair greater than lone pair-bond pair greater than bond pair-bond pair, explaining the stated decreasing angle order.

Q5. O₂ has ten bonding and six antibonding electrons, including two unpaired electrons in separate equal-energy antibonding π orbitals. O₂⁺ forms by removing one antibonding electron; O₂⁻ forms by adding one there, pairing one of the unpaired electrons. Calculate all three bond orders, rank bond strength and state the magnetic behaviour of O₂. [5 marks]

1. Bond order is half the difference between bonding and antibonding electron counts. For neutral oxygen, the given counts give (10 − 6)/2 = 2.
2. O₂⁺ retains ten bonding electrons but has five antibonding electrons. Its bond order is therefore (10 − 5)/2 = 2.5.
3. O₂⁻ retains ten bonding electrons and has seven antibonding electrons. Its bond order is therefore (10 − 7)/2 = 1.5.
4. Greater bond order indicates stronger bonding in this comparison, so bond strength decreases in the order O₂⁺, O₂, O₂⁻.
5. O₂ is paramagnetic because the given orbital arrangement contains two unpaired electrons. Paramagnetic substances are attracted by a magnetic field.

Q6. Water's successive O-H bond dissociation enthalpies are 502 and 427 kJ mol⁻¹. Calculate the mean O-H bond enthalpy and explain why a mean is used. [3 marks]

1. Complete separation of one mole of gaseous water into gaseous atoms requires the sum of the two steps: 502 + 427 = 929 kJ mol⁻¹.
2. There are two O-H bonds per molecule, so the mean bond enthalpy is 929/2 = 464.5 kJ mol⁻¹.
3. The two successive bond-breaking steps have different enthalpies because removing the first hydrogen changes the chemical environment of the remaining O-H bond. A mean represents the two values together.

Key takeaways

• Lattice formation can compensate for the energy needed to form gaseous ions, explaining why octet attainment alone is insufficient.
• Lewis structures account for valence electrons, while formal charge helps compare alternative representations without measuring actual charge separation.
• Resonance describes one actual hybrid structure, with electron distribution and bond characteristics represented collectively by its canonical forms.
• Molecular shape depends on bonding regions and lone pairs; shape also determines whether individual bond dipoles cancel.
• Molecular orbital populations determine bond order and magnetic behaviour; oxygen's two unpaired electrons explain its paramagnetism.
• Coordinate bonding identifies the initial donor of a shared pair; hydrogen bonding describes an additional attraction involving suitably bonded hydrogen. The origin of the pair does not make otherwise equivalent bonds permanently distinguishable.

Test yourself

In linear BeF₂, why is the resultant dipole moment zero despite two polar bonds?

The equal bond dipoles point in opposite directions and cancel, leaving no resultant molecular dipole moment.

What separates a resonance hybrid from its canonical forms?

The hybrid is the actual structure. Canonical forms are alternative electron representations, not separate molecules that interconvert.

What are the shapes of NH₃ and H₂O when their central atoms have respectively one and two lone pairs?

Ammonia is trigonal pyramidal, while water is bent. Their lone pairs occupy positions in tetrahedral electron-pair arrangements.

A species has ten bonding and eight antibonding electrons. What is its bond order?

Its bond order is (10 − 8)/2 = 1, calculated by halving the difference in the two electron populations.

Why are the axial bonds of PCl₅ slightly longer than its equatorial bonds?

Axial bond pairs experience more repulsive interactions with equatorial bond pairs, making axial bonds slightly longer and slightly weaker.

How does hydrogen bonding in ortho-nitrophenol differ from hydrogen bonding between water molecules?

Ortho-nitrophenol can form a hydrogen bond within one molecule; the specified water interaction links separate molecules.