Structure and Bonding: Ionic and Covalent Bonds, VB and MO Theory, Shapes, Polarity, Weak Interactions, Lattice Energy and HSAB

This chapter answers Section XL0.2 of XL0 Chemistry, the part of the Life Sciences (XL) paper that every XL candidate answers: “Ionic and covalent bonding, MO and VB approaches for diatomic molecules, VSEPR theory and shape of molecules, hybridization, resonance, dipole moment, structure parameters such as bond length, bond angle and bond energy, hydrogen bonding and van der Waals interactions. Ionic solids, ionic radii and lattice energy (Born‐Haber cycle). HSAB principle.” It moves from the electron pair to the solid: Lewis structures and formal charge, valence-bond and molecular-orbital pictures of diatomic molecules, VSEPR shapes and hybridisation, resonance and its effect on bond length and energy, dipole moments, the hydrogen bonds and van der Waals forces that hold biomolecules in shape, ionic lattices and the Born–Haber cycle, and Pearson’s hard and soft acids and bases. The numericals are bond orders, formal charges, lone-pair counts, percentage ionic character from dipole moments and lattice enthalpies from Born–Haber data.

1. Ionic and covalent bonding, Lewis structures and formal charge

An ionic bond forms when one atom transfers electrons to another and the resulting ions attract electrostatically; it is favoured by a low ionisation energy on one side and a high electron affinity on the other, and the energy that pays for it is the lattice energy (Section 6). A covalent bond is a shared electron pair. Real bonds lie on a continuum: as a rule of thumb an electronegativity difference above about 1.7 gives a mainly ionic bond. Fajans’ rules say when an ionic bond gains covalent character: a small, highly charged cation polarises the anion strongly; a large, highly charged anion is easily polarised; and a cation with an 18-electron shell (Cu⁺, Ag⁺) polarises more than a noble-gas cation of the same size. So LiI is more covalent than LiF, AlCl₃ more covalent than NaCl, and CuCl more covalent than NaCl.

A Lewis structure places every valence electron as a bond pair or lone pair, giving each atom an octet where possible (H a duet). Among valid structures, the best usually minimises formal charge, FC = V − N − B/2, where V is the valence electrons of the free atom, N the non-bonding electrons on it and B the bonding electrons around it. Worked example, ozone O–O=O: the central O has V = 6, N = 2, B = 6, so FC = +1; the single-bonded terminal O has 6 − 6 − 1 = −1; the double-bonded one 6 − 4 − 2 = 0. The charges sum to zero, as they must for a neutral molecule. Formal charge is a bookkeeping device, not the real charge on an atom.

2. Valence-bond and molecular-orbital approaches for diatomic molecules

Valence-bond (VB) theory, from Heitler and London’s treatment of H₂ (1927), forms a bond by overlap of two half-filled atomic orbitals whose electrons pair with opposite spins; the bond stays localised between two atoms. End-on overlap gives a σ bond, sideways overlap of p orbitals a π bond, so N≡N is one σ and two π. VB keeps the chemist’s picture of lines between atoms, but applied to O₂ it pairs every electron and predicts a diamagnetic molecule.

Molecular-orbital (MO) theory combines atomic orbitals (LCAO) into orbitals spread over the whole molecule: in-phase combination gives a lower bonding MO, out-of-phase a higher antibonding (*) MO. Electrons fill MOs by Aufbau, Pauli and Hund, and bond order = (N_b − N_a)/2. For O₂ and F₂ the valence order is σ2s < σ2s < σ2p_z < π2p_x = π2p_y < π2p_x = π2p_y < σ2p_z. For B₂, C₂ and N₂, s–p mixing pushes σ2p_z above the π2p pair. O₂ has 12 valence electrons; the last two occupy the two degenerate π* orbitals singly with parallel spins, so O₂ has bond order (8 − 4)/2 = 2 and two unpaired electrons: liquid oxygen is drawn into a magnet, which MO theory explains and VB does not. B₂ (bond order 1) is paramagnetic only because of s–p mixing; N₂ (bond order 3) is diamagnetic. He₂ has bond order 0 and does not exist, while He₂⁺ (0.5) does.

Dioxygen species: adding electrons to π* weakens and lengthens the bond
SpeciesValence electronsBond orderUnpaired electrons
O₂⁺ (dioxygenyl)112.51
O₂1222
O₂⁻ (superoxide)131.51
O₂²⁻ (peroxide)1410
⚠️ Which MO order to use
Bond order does not depend on the order chosen for the 2p MOs, but the count of unpaired electrons can: B₂ is paramagnetic only with s–p mixing, and C₂ is diamagnetic only with it. Use the mixed order up to N₂ and the unmixed one from O₂ onwards. Superoxide, which the enzyme superoxide dismutase destroys, is a radical with one unpaired electron.

3. VSEPR theory, hybridisation and the shapes of molecules

VSEPR counts the electron domains (bond pairs plus lone pairs; a multiple bond counts as one domain) around the central atom and places them as far apart as possible. Lone pairs are held by one nucleus and spread wider, so repulsion falls in the order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, and lone pairs squeeze bond angles. Hybridisation is the VB description of the same geometry: mixing s and p (and d) orbitals gives equivalent hybrids pointing to the domains.

Electron domains, hybridisation and shapes
DomainsHybridisationNo lone pairWith lone pairs
2spLinear, 180° (BeCl₂ gas, CO₂)None
3sp²Trigonal planar, 120° (BF₃)1 lone pair: bent (SO₂)
4sp³Tetrahedral, 109.5° (CH₄)1: pyramidal NH₃, 107°; 2: bent H₂O, 104.5°
5sp³dTrigonal bipyramidal (PCl₅)1: see-saw SF₄; 2: T-shaped ClF₃; 3: linear XeF₂, I₃⁻ (lone pairs equatorial)
6sp³d²Octahedral, 90° (SF₆)1: square pyramidal BrF₅; 2: square planar XeF₄

Bond angles also track hybrid character. More s character means a wider angle and a shorter, stronger bond: sp (50 % s) 180°, sp² (33 %) 120°, sp³ (25 %) 109.5°. Down a group the angle closes because the larger central atom uses nearly pure p orbitals for bonding: NH₃ 107° against PH₃ about 93.5°, and H₂O 104.5° against H₂S about 92°. Electronegative substituents also narrow the angle (NF₃ about 102° against NH₃ 107°) by pulling bonding pairs away from the centre.

4. Resonance and structure parameters: bond length, bond angle and bond energy

When one Lewis structure cannot describe a molecule, the real structure is a resonance hybrid of contributors that differ only in where electrons sit, never in where atoms sit. The contributors do not interconvert; the hybrid is a single structure, lower in energy than any contributor, and the difference is the resonance energy (about 150 kJ mol⁻¹ for benzene). Major contributors have more bonds, complete octets, little charge separation and negative charge on the more electronegative atom. In carbonate, three equivalent structures spread one double bond over three C–O bonds, so each has bond order 4/3 and all three are equal in length. In the peptide group, the nitrogen lone pair delocalises into C=O, giving the C–N bond partial double-bond character: it is about 132 pm instead of the 147 pm of a C–N single bond, and the six atoms of the peptide unit are held planar and do not rotate about C–N.

Higher bond order means a shorter and stronger bond. For carbon–carbon bonds, single, double and triple bonds are about 154, 134 and 120 pm long, with average bond enthalpies of about 348, 614 and 839 kJ mol⁻¹; benzene’s six equal bonds (about 139 pm) sit between single and double. Along N₂, O₂ and F₂ the bond order falls 3, 2, 1 and the bond enthalpy falls about 945, 498 and 159 kJ mol⁻¹. A double bond is stronger than a single one but less than twice as strong, because a π overlap is weaker than a σ overlap. Bond angle follows from VSEPR and hybridisation (Section 3).

5. Dipole moment, hydrogen bonding and van der Waals interactions

A dipole moment is μ = q × d, measured in debye (1 D = 3.336 × 10⁻³⁰ C m). A molecule’s dipole is the vector sum of its bond dipoles and lone-pair contributions, so symmetric molecules have μ = 0 even with polar bonds: CO₂, BF₃, CCl₄, PCl₅, SF₆, XeF₄ and trans-1,2-dichloroethene. H₂O, NH₃, SO₂, CHCl₃, SF₄ and the cis isomer are polar. NH₃ (about 1.47 D) is far more polar than NF₃ (about 0.23 D): in NH₃ the N–H bond dipoles point towards N and add to the lone-pair moment, whereas in NF₃ the N–F dipoles point away from N and largely cancel it. Percentage ionic character = μ(observed)/μ(ionic) × 100, where μ(ionic) = e × d is the moment if a full electron charge sat at the bond length.

A hydrogen bond D–H···A forms when H is bonded to a small electronegative donor D (N, O or F) and points at a lone pair on an acceptor A (N, O or F). It is strongest when nearly linear, and at a few to a few tens of kJ mol⁻¹ it is roughly a tenth as strong as a covalent bond. It explains why H₂O boils at 100 °C while heavier H₂S boils near −60 °C, why ice is less dense than water (an open tetrahedral network), and why o-nitrophenol, with an intramolecular hydrogen bond, is more volatile than p-nitrophenol, whose intermolecular hydrogen bonds link molecules together. In biology it pairs bases (two in A·T, three in G·C), holds the α-helix (C=O of residue i to N–H of residue i + 4) and β-sheets, and gives enzymes their specificity.

Van der Waals interactions are the weakest (well under 5 kJ mol⁻¹ per contact) and act between all molecules: Keesom forces between permanent dipoles, Debye forces between a dipole and the dipole it induces, and London dispersion forces between instantaneous and induced dipoles, which exist even between non-polar atoms and grow with size and polarisability. The attraction falls off as 1/r⁶, and at contact a steep repulsion sets in, which defines the van der Waals radius. Individually tiny, thousands of them add up to stabilise the packed core of a folded protein and the stacking of DNA bases.

6. Ionic solids, ionic radii and lattice energy by the Born–Haber cycle

In an ionic solid each ion is surrounded by as many counter-ions as fit without the counter-ions touching each other, so the radius ratio r₊/r₋ predicts the coordination number: 0.225–0.414 gives 4 (tetrahedral, ZnS), 0.414–0.732 gives 6 (octahedral, NaCl) and above 0.732 gives 8 (cubic, CsCl). For NaCl, 102/181 = 0.56 predicts 6:6 coordination, as observed. Ionic radii are partitions of measured interionic distances, so they depend on coordination number and on the reference scale chosen.

Lattice enthalpy is the enthalpy needed to separate one mole of an ionic solid into gaseous ions. It grows with the product of the ionic charges and falls with the distance between them, roughly as z₊z₋/(r₊ + r₋), so MgO (2+, 2−) has a lattice enthalpy about four times that of NaF, and NaF > NaCl > KCl. It cannot be measured directly, so the Born–Haber cycle obtains it from Hess’s law. Worked example for NaCl, in kJ mol⁻¹: ΔfH(NaCl, s) = −411, sublimation of Na = +107, first IE of Na = +496, ½ bond enthalpy of Cl₂ = +122 and electron gain enthalpy of Cl = −349. The route via gaseous ions must equal the direct formation, so the energy of forming the lattice from gaseous ions is −411 − (107 + 496 + 122 − 349) = −787, and the lattice enthalpy is +787 kJ mol⁻¹.

⚠️ Signs and stoichiometry in Born–Haber cycles
Check whether the stem defines lattice enthalpy for breaking the lattice (positive) or forming it (negative). Use half the bond enthalpy for each X atom of a diatomic gas, and for MX₂ use the full bond enthalpy and twice the electron gain enthalpy. For M²⁺ include both the first and second ionisation energies.

7. The HSAB principle

Pearson sorted Lewis acids and bases into hard (small, highly charged, weakly polarisable) and soft (large, low charge, easily polarisable). The HSAB principle states that hard acids prefer hard bases and soft acids prefer soft bases. Hard–hard pairs bind mainly by electrostatic attraction, soft–soft pairs by covalent sharing. Thus AgF is soluble but AgI is highly insoluble (soft Ag⁺ with soft I⁻), LiF is the least soluble lithium halide (hard–hard), and hard Al³⁺ and Fe³⁺ occur in nature as oxides while soft Hg²⁺ occurs as the sulfide cinnabar, HgS.

Common hard, borderline and soft species
ClassAcidsBases
HardH⁺, Li⁺, Na⁺, K⁺, Mg²⁺, Ca²⁺, Al³⁺, Cr³⁺, Fe³⁺F⁻, OH⁻, H₂O, NH₃, Cl⁻, carboxylate, phosphate
BorderlineFe²⁺, Co²⁺, Ni²⁺, Cu²⁺, Zn²⁺, Pb²⁺Br⁻, N₃⁻, pyridine, imidazole
SoftCu⁺, Ag⁺, Au⁺, Hg²⁺, Cd²⁺, Pt²⁺, Pd²⁺I⁻, RS⁻, R₂S, CN⁻, CO, R₃P, H⁻
🎯 HSAB in living systems
Hard Mg²⁺ and Ca²⁺ bind the oxygen of phosphates and carboxylates (Mg²⁺–ATP, Ca²⁺-binding proteins). Borderline Zn²⁺ and Cu²⁺ bind cysteine sulfur and histidine nitrogen in enzymes and zinc fingers. Soft Hg²⁺ and Cd²⁺ are toxic largely because they seize the soft thiol groups of cysteine, and thiol-containing chelating agents are used against heavy-metal poisoning.

Key takeaways

  • Fajans’ rules give ionic bonds covalent character (small, highly charged cations; large anions; 18-electron cations). Formal charge FC = V − N − B/2 picks the best Lewis structure and sums to the overall charge.
  • Bond order = (N_b − N_a)/2. MO theory, unlike VB, explains the paramagnetism of O₂ (two unpaired π* electrons); O₂⁺, O₂, O₂⁻ and O₂²⁻ have bond orders 2.5, 2, 1.5 and 1. s–p mixing up to N₂ makes B₂ paramagnetic.
  • VSEPR counts domains (2 to 6: linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral; sp to sp³d²); lone pairs repel most, sit equatorial in a trigonal bipyramid and close bond angles.
  • Resonance spreads bond order (4/3 in carbonate, partial double bond in the planar peptide group); higher bond order means shorter, stronger bonds.
  • Dipole moments add as vectors; % ionic character = μ(obs)/(e·d) × 100. Hydrogen bonds (N, O, F donors and acceptors) and van der Waals forces (Keesom, Debye, London; 1/r⁶) shape water, DNA and proteins.
  • Radius ratio predicts coordination (0.414 and 0.732 are the boundaries); lattice enthalpy scales with z₊z₋/(r₊ + r₋) and comes from the Born–Haber cycle; HSAB pairs hard with hard and soft with soft.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Liquid oxygen is drawn into the gap between the poles of a strong magnet. Which account of this observation is correct?

    1. Valence-bond theory: the σ and π bonds of O=O each leave one electron unpaired on an oxygen
    2. Molecular-orbital theory: s–p mixing raises σ2p above π2p and leaves two σ electrons unpaired
    3. Molecular-orbital theory: the highest occupied orbital is σ*2p, holding one unpaired electron
    4. Molecular-orbital theory: two electrons occupy the degenerate π*2p orbitals singly, spins parallel
    Show answer

    Answer: D — Molecular-orbital theory: two electrons occupy the degenerate π*2p orbitals singly, spins parallel

    O₂ has 12 valence electrons: σ2s² σ2s² σ2p² π2p⁴ π2p². By Hund’s rule the two π* electrons occupy the two degenerate π* orbitals singly with parallel spins, so O₂ is paramagnetic with bond order (8 − 4)/2 = 2. The valence-bond (and Lewis) O=O structure pairs every electron and so predicts a diamagnetic molecule, which is the classic failure that MO theory repairs. σ*2p is empty in O₂, and s–p mixing matters for B₂, C₂ and N₂, not O₂; even where it applies it does not unpair σ electrons.

  2. Using molecular-orbital theory with bond order = (number of bonding electrons − number of antibonding electrons)/2, the bond order of the superoxide ion, O₂⁻, to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 1.5

    O₂⁻ has 13 valence electrons: σ2s² σ2s² σ2p² π2p⁴ π2p³. Bonding electrons = 2 + 2 + 4 = 8 and antibonding = 2 + 3 = 5, so bond order = (8 − 5)/2 = 1.5. Counting the 1s electrons as well adds two to each side and leaves the answer unchanged. Writing 2.5 confuses superoxide with O₂⁺, which has one electron fewer than O₂, not one more.

  3. Use the MO energy order with s–p mixing (σ2p above π2p) for B₂, C₂ and N₂, and the order without it for O₂ and F₂. Which of the following statements are correct?

    1. He₂⁺ has a bond order of zero and therefore cannot exist
    2. B₂ is paramagnetic, with one electron in each of the two π2p orbitals
    3. The highest occupied molecular orbital of F₂ is σ*2p
    4. N₂ has a bond order of 3 and no unpaired electrons
    Show answer

    Answer: B — B₂ is paramagnetic, with one electron in each of the two π2p orbitals; D — N₂ has a bond order of 3 and no unpaired electrons

    B₂ has 6 valence electrons: σ2s² σ2s² and then, because s–p mixing puts π2p below σ2p, one electron in each π2p orbital, so it is paramagnetic. N₂ has σ2s² σ2s² π2p⁴ σ2p², bond order (8 − 2)/2 = 3, all paired. He₂⁺ has σ1s² σ1s¹, bond order 0.5, and is a known species. In F₂ (14 valence electrons) the π orbitals are full and σ2p is empty, so the HOMO is π2p.

  4. According to VSEPR theory, what is the shape of the ClF₃ molecule?

    1. T-shaped, with two lone pairs in equatorial positions
    2. Trigonal planar, with three bond pairs at 120° as in BF₃
    3. Trigonal pyramidal, with one lone pair as in ammonia
    4. See-saw, with one lone pair in an equatorial position
    Show answer

    Answer: A — T-shaped, with two lone pairs in equatorial positions

    Chlorine has 7 valence electrons; three form Cl–F bonds and four make two lone pairs, giving five domains arranged as a trigonal bipyramid. The lone pairs take two equatorial positions, where they suffer fewer 90° repulsions, leaving a T-shaped molecule with F–Cl–F angles slightly under 90°. Trigonal planar ignores the lone pairs, and see-saw is the shape of SF₄, which has four bond pairs and one lone pair.

  5. Applying VSEPR theory, the total number of lone pairs on the central atoms of ClF₃, XeF₂ and SF₄ taken together (count only the lone pairs on Cl, Xe and S) is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 6

    Lone pairs = (valence electrons − electrons used in bonds)/2. ClF₃: (7 − 3)/2 = 2. XeF₂: (8 − 2)/2 = 3. SF₄: (6 − 4)/2 = 1. Total = 2 + 3 + 1 = 6. These give T-shaped, linear and see-saw molecules, all from five electron domains.

  6. A Lewis structure of the sulfate ion, SO₄²⁻, is drawn with four S–O single bonds and a complete octet on every atom. Using formal charge = V − N − B/2, the formal charge on the sulfur atom in this structure is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 2

    Sulfur has V = 6 valence electrons; in this structure it carries no lone pairs (N = 0) and has 8 bonding electrons (B = 8), so FC = 6 − 0 − 4 = +2. Each oxygen has V = 6, N = 6 and B = 2, so FC = −1, and 2 + 4(−1) = −2 matches the ion’s charge. The expanded-octet structure with two S=O bonds lowers the sulfur formal charge to 0.

  7. In which pair do the central atoms have the same hybridisation?

    1. [PCl₄]⁺ and [PCl₆]⁻
    2. SF₄ and CH₄
    3. XeF₄ and SF₆
    4. BF₃ and NH₃
    Show answer

    Answer: C — XeF₄ and SF₆

    XeF₄ has four bond pairs and two lone pairs and SF₆ six bond pairs: both have six domains and sp³d² hybridisation. [PCl₄]⁺ is sp³ but [PCl₆]⁻ sp³d²; SF₄ (four bonds plus one lone pair) is sp³d while CH₄ is sp³; BF₃ is sp² while NH₃, with its lone pair, is sp³. The common slip is counting only bonded atoms and forgetting lone pairs, which would wrongly match SF₄ with CH₄.

  8. Which order of carbon–oxygen bond lengths is correct?

    1. CO < CO₂ < CO₃²⁻ < CH₃OH
    2. CO₂ < CO < CO₃²⁻ < CH₃OH
    3. CO < CO₃²⁻ < CO₂ < CH₃OH
    4. CH₃OH < CO₃²⁻ < CO₂ < CO
    Show answer

    Answer: A — CO < CO₂ < CO₃²⁻ < CH₃OH

    Bond length falls as bond order rises. CO has a triple bond (order 3), each C=O of CO₂ is a double bond (2), resonance spreads one π bond over the three C–O bonds of carbonate (4/3 each), and methanol has a C–O single bond (1). So CO < CO₂ < CO₃²⁻ < CH₃OH. Placing carbonate before CO₂ forgets that resonance dilutes its single π bond over three bonds; the fully reversed order confuses length with strength.

  9. In proteins, the C–N bond of each peptide group is about 132 pm long, shorter than a typical C–N single bond of about 147 pm, and rotation about it is restricted. What is the reason?

    1. The amide nitrogen is sp³ hybridised, and its high s character shortens the bond
    2. A hydrogen bond between C=O and N–H of the same peptide group locks it rigidly
    3. Bulky side chains on the neighbouring α-carbons collide and block all rotation
    4. Resonance delocalises the N lone pair into C=O, giving partial double-bond character
    Show answer

    Answer: D — Resonance delocalises the N lone pair into C=O, giving partial double-bond character

    The nitrogen lone pair conjugates with the carbonyl group; the contributor O⁻–C=N⁺ gives the C–N bond partial double-bond character, so it is shorter than a single bond and the peptide unit is planar and rigid. The nitrogen is sp² (planar), not sp³, and sp³ has less s character, not more. Hydrogen bonds in helices and sheets link different peptide groups, and side-chain clashes restrict rotation about the N–Cα and Cα–C bonds, not about C–N.

  10. The dipole moment of HCl is 1.03 D and its bond length is 127 pm. Taking e = 1.602 × 10⁻¹⁹ C and 1 D = 3.336 × 10⁻³⁰ C m, the percentage ionic character of the H–Cl bond, to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 17

    A fully ionic H⁺Cl⁻ pair would have μ(ionic) = e × d = 1.602 × 10⁻¹⁹ × 127 × 10⁻¹² = 2.035 × 10⁻²⁹ C m = 2.035 × 10⁻²⁹/3.336 × 10⁻³⁰ = 6.10 D. Percentage ionic character = 1.03/6.10 × 100 = 16.9, which is 17. Forgetting to convert pm to m or C m to D gives answers off by powers of ten.

  11. o-Nitrophenol can be separated from p-nitrophenol by steam distillation because the ortho isomer is the more volatile. What makes it more volatile?

    1. Its larger dipole moment, which strengthens the attractions between its molecules
    2. Its lower molar mass, because the nitro group sits closer to the hydroxyl group
    3. An intramolecular hydrogen bond, which leaves no O–H free to bond its neighbours
    4. Stronger London dispersion forces, which act only between non-polar molecules
    Show answer

    Answer: C — An intramolecular hydrogen bond, which leaves no O–H free to bond its neighbours

    In the ortho isomer the O–H hydrogen bonds to an oxygen of the neighbouring nitro group within the same molecule, closing a six-membered ring. With its O–H used up, the molecule barely associates with its neighbours or with water, so it is volatile. The para isomer cannot do this, so its O–H forms intermolecular hydrogen bonds that raise its boiling point. Both isomers have the same molar mass, a larger dipole would lower volatility, and London forces act between all molecules, polar or not.

  12. Which of the following statements about hydrogen bonds and van der Waals interactions are correct?

    1. London dispersion forces act between all molecules, including non-polar ones such as CH₄
    2. Ice is denser than liquid water because hydrogen bonds pull the molecules closer in the solid
    3. The much higher boiling point of H₂O than of H₂S is due to hydrogen bonding
    4. A single van der Waals contact is typically stronger than a hydrogen bond
    Show answer

    Answer: A — London dispersion forces act between all molecules, including non-polar ones such as CH₄; C — The much higher boiling point of H₂O than of H₂S is due to hydrogen bonding

    Instantaneous dipoles arise in every electron cloud, so dispersion forces act between all molecules, and they are the only attraction between CH₄ molecules. Water’s O–H···O hydrogen bonds raise its boiling point to 100 °C, whereas H₂S, with a weakly electronegative S, boils near −60 °C despite its larger mass. Ice is less dense than water because hydrogen bonding locks it into an open tetrahedral network. A van der Waals contact is weaker than a hydrogen bond, usually by several times.

  13. Data for magnesium chloride, in kJ mol⁻¹: enthalpy of formation of MgCl₂(s) = −641; enthalpy of sublimation of Mg = +148; first and second ionisation energies of Mg = +738 and +1451; bond dissociation enthalpy of Cl₂ = +243; electron gain enthalpy of Cl = −349 per mole of Cl atoms. Using a Born–Haber cycle, the lattice enthalpy of MgCl₂, defined as the enthalpy change for MgCl₂(s) → Mg²⁺(g) + 2Cl⁻(g), in kJ mol⁻¹, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 2523

    By Hess’s law, ΔfH = ΔsubH + IE₁ + IE₂ + D(Cl₂) + 2ΔegH − ΔH(lattice). So ΔH(lattice) = 148 + 738 + 1451 + 243 + 2(−349) − (−641) = 148 + 738 + 1451 + 243 − 698 + 641 = 2523 kJ mol⁻¹. The full Cl₂ bond enthalpy is used because two Cl atoms are needed. Using one electron gain enthalpy gives 2872, omitting IE₂ gives 1072, and taking half the bond enthalpy gives 2401.5.

  14. Which order of lattice enthalpies is correct?

    1. KCl > NaCl > NaF > MgO, because larger ions hold one another more strongly
    2. MgO > NaF > NaCl > KCl, because of higher charges first and then smaller ions
    3. NaF > MgO > NaCl > KCl, because fluoride is the smallest anion in the set
    4. MgO > NaCl > NaF > KCl, because chloride is more polarisable than fluoride
    Show answer

    Answer: B — MgO > NaF > NaCl > KCl, because of higher charges first and then smaller ions

    Lattice enthalpy scales roughly as z₊z₋/(r₊ + r₋). MgO has charges 2+ and 2−, a fourfold larger product than the singly charged salts, and small ions, so it is far the largest. Among the 1:1 salts, the smaller F⁻ makes NaF exceed NaCl, and the larger K⁺ makes KCl the smallest. Fluoride being small cannot outweigh the fourfold charge product of MgO; the polarisability of chloride does not raise its lattice enthalpy above that of fluoride; and larger ions are held less strongly, not more.

  15. An ionic solid MX has r(M⁺) = 95 pm and r(X⁻) = 181 pm. Using the radius-ratio rule (0.225–0.414 tetrahedral, 0.414–0.732 octahedral, above 0.732 cubic), what coordination is predicted for M⁺?

    1. Six, octahedral, as in the rock-salt (NaCl) structure
    2. Four, tetrahedral, as in the zinc blende (ZnS) structure
    3. Eight, cubic, as in the caesium chloride structure
    4. Twelve, as in the close packing of equal-sized spheres
    Show answer

    Answer: A — Six, octahedral, as in the rock-salt (NaCl) structure

    r₊/r₋ = 95/181 = 0.525, which lies between 0.414 and 0.732, so each cation is expected to have six anion neighbours at the corners of an octahedron, the rock-salt arrangement. Eight-fold cubic coordination needs a ratio above 0.732, and four-fold needs one below 0.414. Twelve-fold coordination arises only for equal spheres, not for a cation much smaller than its anion.

  16. According to the HSAB principle, which metal ion is expected to bind most strongly to the thiolate (RS⁻) side chain of cysteine?

    1. Mg²⁺, because its high charge density attracts any anion strongly
    2. Hg²⁺, a large and polarisable soft acid matching the soft base
    3. Ca²⁺, because it binds the carboxylate side chains of proteins well
    4. Al³⁺, because a +3 charge gives the strongest electrostatic pull
    Show answer

    Answer: B — Hg²⁺, a large and polarisable soft acid matching the soft base

    Thiolate sulfur is large and polarisable, a soft base, so it binds best to a soft acid such as Hg²⁺, which is why mercury poisons thiol enzymes. Mg²⁺, Ca²⁺ and Al³⁺ are hard acids that prefer hard oxygen donors (water, carboxylate, phosphate). The high-charge arguments are tempting, but HSAB pairs by hardness, not by charge alone: hard–hard binding is electrostatic, soft–soft binding is covalent.

  17. Which of the following statements are consistent with the HSAB principle and the usual classification of acids and bases?

    1. Silver iodide is far less soluble in water than silver fluoride
    2. Fe³⁺ is classed as a soft acid because of its high charge
    3. Iodide is a harder base than fluoride because it is larger
    4. Mg²⁺ binding the phosphate oxygens of ATP is a hard acid–hard base pairing
    Show answer

    Answer: A — Silver iodide is far less soluble in water than silver fluoride; D — Mg²⁺ binding the phosphate oxygens of ATP is a hard acid–hard base pairing

    Soft Ag⁺ binds soft I⁻ covalently, so AgI is highly insoluble, while the mismatched AgF dissolves readily. Mg²⁺ is a small, hard cation and phosphate oxygen is a hard base, so Mg²⁺–ATP is a hard–hard pair. A small, highly charged cation such as Fe³⁺ is hard, not soft, and a large, polarisable anion such as I⁻ is softer than F⁻, not harder.

  18. Consider the solid-state exchange LiI + CsF → LiF + CsI. Which prediction, with its reason, is correct?

    1. It does not proceed, because the small Li⁺ and large I⁻ already form the strongest pair
    2. It proceeds, because hard Li⁺ pairs with hard F⁻ and the softer Cs⁺ with soft I⁻
    3. It does not proceed, because Cs⁺ is the harder cation and keeps hold of F⁻
    4. It proceeds, because the total ionic charge is larger on the product side
    Show answer

    Answer: B — It proceeds, because hard Li⁺ pairs with hard F⁻ and the softer Cs⁺ with soft I⁻

    Li⁺ is the smallest, hardest alkali cation and Cs⁺ the largest and most polarisable, so it is the softest; F⁻ is hard and I⁻ soft. HSAB favours the hard–hard LiF and soft–soft CsI. In lattice terms, the very large lattice enthalpy of LiF, made of two small ions, more than repays the change from CsF to CsI, so the exchange is exothermic. Cs⁺ is softer, not harder, than Li⁺, and every ion carries a charge of one on both sides, so charge cannot drive the reaction.