Chemical Bonding and Molecular Structure · Lesson 12 of 12
Chapter Summary and Practice
“Connect electron counting, shape, polarity, orbital models, and intermolecular attraction through whole-chapter reasoning and mixed practice.”
• Choose an appropriate bonding model for electron counts, structure, energetics, polarity, or magnetism. • Connect Lewis structures, formal charge, resonance, molecular geometry, and dipole cancellation. • Apply lattice enthalpy, bond enthalpy, dipole moment, and molecular-orbital bond order with correct meanings and units. • Relate hybridisation and orbital overlap to sigma and pi bonding. • Solve mixed problems spanning the complete chapter and explain why each method applies.
Choose a model that answers the question
Chemical bonding has several connected layers. Atoms and ions are held together through stabilising interactions, but no single drawing answers every question about a substance. A complete explanation might need electron accounting, a three-dimensional arrangement, energy measurements, and an orbital occupancy picture.
The central habit is to identify what must be predicted. A Lewis structure accounts for electrons and connectivity. VSEPR predicts shape from domains. Bond-dipole reasoning uses that shape to discuss polarity. Valence bond and hybridisation descriptions organise local overlap, while molecular orbitals connect whole-molecule occupancy with bond order and magnetic properties.
| Question | Useful starting model | Essential check |
|---|---|---|
| How many bonding and lone pairs are shown? | Lewis structures | Total valence-electron count and ionic charge |
| Is a proposed electron drawing plausible? | Formal charge and octet limits | Charge sum, permitted atom counts, and chemical context |
| Why are several bonds equivalent? | Resonance | Same nuclear skeleton in every contributor |
| What is the shape? | VSEPR | All domains, including lone pairs |
| Is the whole molecule polar? | Dipole vectors and geometry | Possible cancellation of bond dipoles |
| How is a local covalent bond described? | Valence bond overlap and hybridisation | Orbitals, phase, direction, and spin |
| Why is a diatomic species magnetic? | MO filling | Correct energy order and unpaired electrons |
| What stabilises an ionic crystal? | Lattice interactions and enthalpy | Complete stated energy processes |
| What joins suitable groups or molecules? | Hydrogen bonding | A donor, an acceptor, and suitable proximity |
Start with electron accounting
Count outer-shell electrons, add electrons for negative charge, and subtract for positive charge. Make a sensible skeleton, distribute pairs, and use multiple bonds when appropriate. A bond line represents two shared electrons. Count the pair around both atoms for an octet check, but only once when checking the molecule’s total electrons.
The octet rule helps many second-period structures but has incomplete-octet, odd-electron, and expanded-octet exceptions. Hydrogen generally counts a duplet. Formal charge helps compare drawings and checks that charges sum correctly, but it is not a measurement of the actual partial charge in a polar bond.
Problem
Compare NH₃ and the ammonium ion using the same sequence of questions.
- 1.NH₃ has valence electrons. Three N–H bonds use six and one lone pair uses two.
- 2.Four domains give a tetrahedral electron arrangement, but only three positions contain H atoms, so the shape is trigonal pyramidal. The bond-dipole contributions do not cancel, giving a polar molecule.
- 3.The ammonium ion has valence electrons. All four pairs are used in N–H bonds, with no central lone pair.
- 4.The four bonding directions are tetrahedral. Its positive ionic charge should not be confused with the kind of neutral-molecule dipole comparison made for NH₃; charge and charge separation are distinct ideas.
Use measurements to test electron drawings
Bond length is the equilibrium nuclear separation and bond angle is the angle between bonded directions. Covalent and van der Waals radii refer to different contact situations. Bond enthalpy measures the energy needed to break a specified gaseous bond, while average bond enthalpy combines several dissociation steps or related environments.
For comparable atom pairs, higher bond order generally means a shorter, stronger bond. Equivalent experimental bonds may require resonance. Ozone’s two O–O bonds and carbonate’s three C–O bonds are not permanently selected single and double bonds. The real species is one hybrid, not a collection of rapidly interconverting canonical molecules.
Problem
Use carbonate to connect electron counting, resonance, and shape.
- 1.The electron total is . An octet drawing has one C=O bond and two C–O bonds.
- 2.The three equivalent choices of double-bond position are canonical structures with the same nuclei. Their resonance hybrid explains three equivalent C–O bonds, with average order in this model.
- 3.Around carbon there are three bonding directions and no lone pair. The central electron arrangement and molecular shape are trigonal planar.
- 4.Each multiple-bond region counts as one domain for this shape prediction. It would be wrong to count the double bond as two independent VSEPR directions.
Keep energetic processes separate
Ionisation of an isolated gaseous atom requires energy. Electron gain can release or require energy, depending on the stated process. Lattice formation can release enough energy to stabilise an ionic crystal even when making separated gaseous ions is energetically uphill.
Lattice enthalpy in this chapter is the positive energy required to separate one mole of solid into gaseous ions. The reverse formation process has the opposite sign. This is different from bond enthalpy for a specified covalent bond and from a complete enthalpy of formation starting with elements in their standard states.
Problem
A gaseous molecule requires 500 and 430 kJ per mole in two successive bond-cleavage steps. An ionic solid has lattice dissociation enthalpy 780 kJ per mole. Interpret the numbers.
- 1.For the gaseous molecule, add the two cleavage steps: for breaking both bonds.
- 2.The average for those two bonds is . The separate measured steps remain different.
- 3.The ionic value refers to pulling an extended solid into gaseous ions. Forming one mole of that lattice from its gaseous ions instead has enthalpy .
- 4.These values refer to different processes and cannot be treated as the energy of one identical kind of “bond”.
Shape and polarity require directions
The domain arrangements for two through six regions are linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. Lone pairs remain part of the electron arrangement but are excluded when naming the positions of atoms. Their stronger repulsions help explain bent water, pyramidal ammonia, see-saw SF₄, T-shaped ClF₃, square pyramidal BrF₅, and square planar XeF₄.
Once shape is known, combine bond dipoles as vectors. Linear BeF₂ and trigonal planar BF₃ have polar bonds whose effects cancel. Bent water has a nonzero net dipole. NH₃ and NF₃ show that even two molecules of the same shape cannot be ranked just by the electronegativity difference in one bond: directions of the contributions matter.
Fajans’ rules connect the two bonding limits. A small highly charged cation strongly polarises an anion, while a large anion is more easily distorted. Greater density drawn between the nuclei increases covalent character. Ionic and covalent labels therefore describe useful limits rather than a universal sharp division.
Overlap, hybridisation, and multiple bonds
Valence bond theory explains the H₂ bond through orbital overlap and a lower-energy paired-electron distribution. Its potential-energy curve has a minimum at the equilibrium length: too-close approach raises repulsion. Phase signs on orbitals refer to wavefunctions, not positive and negative electric charges.
Head-on overlap produces sigma bonding; sideways p-orbital overlap produces pi bonding. In the ordinary localised framework, single, double, and triple bonds contain one sigma bond with zero, one, or two additional pi bonds. Do not count the two density regions of one pi bond as two pi bonds.
| Orbital description | Directions or example | Unhybridised p orbitals per central atom in the basic s/p set |
|---|---|---|
| sp | Linear BeCl₂; carbon in ethyne | 2 |
| sp² | Trigonal planar BCl₃; carbon in ethene | 1 |
| sp³ | Tetrahedral framework of CH₄, NH₃, H₂O; carbon in ethane | 0 |
| Conventional sp³d | Trigonal-bipyramidal PCl₅ | Not obtained by the preceding simple leftover-p count |
| Conventional sp³d² | Octahedral SF₆ | Not obtained by the preceding simple leftover-p count |
Hybridisation changes the orbital description on one atom, and the number of hybrids equals the number mixed. Lone pairs may occupy hybrids. The expanded geometrical schemes also include square planar dsp² and octahedral d²sp³ examples in the source. Conventional labels are models of orbital construction, not direct measurements of how much each atomic orbital participates in a real molecule.
Problem
Relate carbon hybridisation to the complete sigma and pi counts.
- 1.Ethane has sp³ carbon, six C–H sigma bonds, and one C–C sigma bond: seven sigma and zero pi.
- 2.Ethene has sp² carbon and retains one p orbital per carbon. Four C–H sigma bonds plus one C–C sigma bond give five sigma; sideways p overlap gives one pi.
- 3.Ethyne has sp carbon and retains two p orbitals per carbon. Two C–H sigma bonds plus one C–C sigma bond give three sigma; the two perpendicular sideways overlaps give two pi.
- 4.Each C–C multiple bond contains only one sigma component. The remaining bond multiplicity is supplied by pi interactions.
Molecular orbitals explain occupancy and magnetism
LCAO creates as many MOs as the atomic orbitals supplied. Constructive overlap gives a bonding combination and destructive overlap an antibonding one. Atomic orbitals combine effectively only when their energies, symmetry, and spatial overlap are suitable. The resulting MOs are filled using Aufbau, Pauli, and Hund rules.
Select the correct order: the bonding pi-2p pair lies below sigma-2p for B₂, C₂, and N₂, but above it for O₂ and F₂. A cancelling filled core may be omitted consistently. Count electrons in bonding and antibonding orbitals, rather than just the number of occupied boxes.
Problem
Compare O₂ with O₂⁺ and O₂⁻.
- 1.O₂ has two unpaired electrons in the antibonding pi pair and bond order 2.
- 2.Removing one antibonding electron gives O₂⁺ with order 2.5 and one unpaired electron. Its bond is expected to be stronger and shorter than O₂ in this related comparison.
- 3.Adding one antibonding electron gives O₂⁻ with order 1.5 and one unpaired electron. Its bond is expected to be weaker and longer.
- 4.All three are paramagnetic, although their bond orders differ. Magnetic nature and bond strength are connected through occupancy but are not the same property.
Hydrogen bonding adds another level of structure
A hydrogen attached to F, O, or N in the chapter’s principal examples can be partially positive and interact with a suitable electron-rich acceptor. The solid donor bond and dotted hydrogen-bond contact are different links. Hydrogen bonding between water or HF molecules is intermolecular, while the contact between neighbouring hydroxyl and nitro groups in o-nitrophenol is intramolecular.
These additional interactions help explain molecular association and internal arrangements. They are generally weaker than the donor covalent bonds, yet many contacts together can strongly affect a substance. Do not confuse hydrogen bonding with making an H–H molecule or assume that the presence of any H atom guarantees a suitable contact.
Formal charge is not actual partial charge. An electron-domain arrangement is not always the molecular shape. A polar bond need not make a polar molecule. A multiple bond is not several sigma bonds. An electron in an antibonding orbital does not strengthen the bond simply because it is an extra electron.
Mixed chapter check
These questions move between models. For each, identify the property being tested before using a formula or memorised shape.
Quiz
Which model is the most direct starting point for predicting whether O₂ is paramagnetic?
Which structure has a tetrahedral electron arrangement but a bent molecular shape?
What must remain unchanged between ordinary resonance contributors?
Which statement about lattice enthalpy uses the definition in this chapter?
A neutral linear molecule with two identical polar bonds can be nonpolar because:
In the ordinary localised picture, how many sigma and pi components form one triple bond?
Adding one electron to an antibonding MO changes bond order by:
Which comparison predicts greater covalent character using anion polarisability?
Which contact is intramolecular?
A formal-charge sum of −1 is appropriate for:
Practice Problems
- Define a chemical bond and explain why lowering energy is more fundamental than saying atoms “want” an octet.
- Draw Lewis symbols for Mg, Na, B, O, and Br. Then explain electron transfer in potassium–sulfur, calcium–oxygen, and aluminium–nitrogen compounds.
- Draw complete Lewis structures for H₂S, SiCl₄, BeF₂, CO, nitrite, carbonate, and ammonium. Check the electron totals.
- Calculate formal charges in CO and one nitrite contributor. Distinguish these charges from real partial charges.
- Give an incomplete-octet example, an odd-electron example, and an expanded-octet example. Explain a different limitation of the octet rule beyond electron-count exceptions.
- Draw resonance contributors for ozone, carbonate, nitrate, SO₃, and NO₂, using an unpaired electron where necessary. Explain what stays fixed and what changes.
- Two proposed drawings of phosphorous acid attach an H nucleus to different atoms. Explain why they cannot be treated simply as resonance contributors.
- Define bond length, bond angle, bond order, and bond enthalpy, then explain which comparisons are valid for the C–C, C=C, and C≡C series.
- Use successive dissociation enthalpies 510 and 430 kJ per mole to calculate an average bond enthalpy and explain why averaging does not make the two steps identical.
- Explain the favourable factors for ionic-solid formation. Write a lattice-dissociation equation and reverse its enthalpy sign for lattice formation.
- Predict the shapes of BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃, SF₄, ClF₃, BrF₅, and XeF₄. Show central-domain and lone-pair counts.
- Explain the angle sequence methane, ammonia, water. Compare a tetrahedral methane model with a proposed square planar one.
- Calculate a dipole moment for and . Convert to debye.
- Use dipole vectors to compare H₂O with CO₂ and BeH₂. Explain why NH₃ has a larger molecular moment than NF₃.
- Distinguish electronegativity from electron-gain enthalpy. Use qualitative polarity reasoning to compare N₂, ClF, SO₂, K₂O, and LiF without assuming a precise numerical ionic percentage.
- Sketch H₂ formation on a potential-energy curve. Mark the equilibrium distance and explain why attraction alone cannot determine it.
- Explain positive-phase, negative-phase, and zero overlap. With x as the axis, decide which of 1s–1s, 1s–2pₓ, 2pᵧ–2pᵧ, and 1s–2s can give sigma overlap by symmetry.
- Describe sp, sp², and sp³ sets; show sigma and pi formation in ethene and ethyne; explain the conventional PCl₅ and SF₆ schemes and axial-bond inequality in PCl₅.
- Predict the change in conventional hybridisation in AlCl₃ + Cl⁻ → AlCl₄⁻ and in BF₃ + NH₃ → F₃B–NH₃. Explain why the latter donor–acceptor bond does not create extra electrons.
- Identify the hybridisation of every carbon in CH₃–CH₃, CH₃–CH=CH₂, CH₃–CH₂–OH, CH₃–CHO, and CH₃COOH.
- State LCAO conditions and explain, within the elementary MO model, the zero-order predictions for He₂, Be₂, and Ne₂.
- Compare B₂, C₂, N₂, O₂, and F₂ by bond order and magnetism, using the correct two energy orders.
- Calculate and compare the bond orders of N₂, O₂, O₂⁺, O₂⁻, and O₂²⁻. For the oxygen family, predict relative lengths and count unpaired electrons.
- Draw intermolecular hydrogen bonding in HF or water and intramolecular hydrogen bonding in o-nitrophenol. Compare a hydrogen bond with the donor covalent bond and with a weak van der Waals contact.
Key Takeaways
• Begin with the question: electron count, shape, energy, polarity, orbital overlap, magnetism, or molecular association. • Lewis structures, formal charge, and resonance describe electron distribution with important model limits. • VSEPR and dipole vectors connect three-dimensional arrangement with molecular polarity. • Valence bond theory and hybridisation describe local overlap; molecular orbital theory predicts occupancy, bond order, and magnetism. • Lattice formation stabilises ionic crystals, while hydrogen bonding adds interactions within or between suitable molecules. • Always check electrons, charges, units, orbital counts, and the physical meaning of the process before accepting an answer.
Previous · Lesson 11
Hydrogen Bonding and Its Effects
Next
End of chapter