Source: CHEM 2301 Learning Objectives, Ch. 2
Tags: molecular geometry, VSEPR, bond angles, hybridisation, sp3, sp2, sp, polarity, intermolecular forces, hydrogen bonding, dipole-dipole, London dispersion, van der Waals, boiling point, solubility, alkane nomenclature, cycloalkane, oxidation, reduction, functional groups
Difficulty: Introductory to Intermediate | Prerequisites: Chapter 1 (Lewis structures, electronegativity, bond polarity).
Chapter 2 moves from flat, 2D drawings to three-dimensional molecular shape. You will learn how to predict the geometry around any atom using VSEPR theory, connect that geometry to hybridisation (sp³, sp², sp), and use shape and polarity to predict physical properties like boiling point and solubility. The chapter also introduces alkane nomenclature, the concept of oxidation state in organic molecules, and the key hydrocarbon functional groups. This is where organic chemistry starts to feel three-dimensional.
Chapter 2 connects molecular structure to 3D shape (VSEPR and hybridisation), then connects shape to bulk properties through intermolecular forces. It also covers IUPAC naming of simple alkanes and cycloalkanes, organic oxidation and reduction, and the structural differences between alkanes, alkenes, alkynes and aromatics.
VSEPR (Valence Shell Electron Pair Repulsion)
A model for predicting molecular geometry. Electron groups (bonds and lone pairs) around a central atom spread out to minimise repulsion. Think of it as: electron clouds want to be as far apart as possible.
Molecular geometry
The 3D arrangement of atoms (not lone pairs) around a central atom. Common shapes: linear, trigonal planar, tetrahedral, bent, trigonal pyramidal.
Bond angle
The angle between two bonds originating from the same atom. Ideal angles: 180° (linear), 120° (trigonal planar), 109.5° (tetrahedral). Lone pairs compress bond angles slightly below the ideal.
Hybridisation
The mixing of atomic orbitals (s and p) to form new, equivalent hybrid orbitals. sp³ = four hybrid orbitals (tetrahedral), sp² = three (trigonal planar), sp = two (linear).
Polar molecule
A molecule with a net dipole moment. This requires polar bonds arranged asymmetrically so that individual bond dipoles do not cancel. In simple terms, the molecule has a "positive side" and a "negative side."
London dispersion forces (van der Waals forces)
Weak, temporary attractive forces between all molecules, caused by momentary fluctuations in electron density. Strength increases with molecular surface area and molar mass.
Dipole-dipole interactions
Attractive forces between the δ+ end of one polar molecule and the δ− end of another. Stronger than dispersion forces for molecules of similar size.
Hydrogen bonding
An especially strong type of dipole-dipole interaction. Occurs when H bonded to F, O, or N interacts with a lone pair on F, O, or N of a neighbouring molecule. This is why water has an unusually high boiling point for its size.
Functional group
A specific grouping of atoms within a molecule that determines its chemical reactivity. In Chapter 2, the focus is on hydrocarbon groups: alkane (C-C single bonds only), alkene (C=C double bond), alkyne (C≡C triple bond), and aromatic (benzene ring).
Oxidation (organic context)
A reaction in which a carbon atom gains bonds to oxygen or other electronegative atoms, or loses bonds to hydrogen. Think of it as: more bonds to electronegative atoms = more oxidised.
Reduction (organic context)
The reverse: a carbon gains bonds to hydrogen or loses bonds to electronegative atoms. More bonds to hydrogen = more reduced.
To predict geometry around any atom:
Count the number of electron groups (bonds + lone pairs) on the atom. A double or triple bond counts as one electron group.
Electron groups arrange to minimise repulsion. Two groups → linear (180°). Three → trigonal planar (120°). Four → tetrahedral (109.5°).
The molecular geometry describes only the positions of atoms, ignoring lone pairs.
Four electron groups, no lone pairs: tetrahedral (e.g. CH₄, sp³ carbon). Bond angle ~109.5°.
Four electron groups, one lone pair: trigonal pyramidal (e.g. NH₃). Bond angle ~107° because the lone pair compresses the bond angles.
Four electron groups, two lone pairs: bent (e.g. H₂O). Bond angle ~104.5°.
Three electron groups, no lone pairs: trigonal planar (e.g. BH₃, sp² carbon in ethylene). Bond angle ~120°.
Two electron groups: linear (e.g. CO₂, sp carbon in acetylene). Bond angle = 180°.
In a large molecule, each atom has its own local geometry. To estimate a bond angle, look at the atom at the vertex of the angle, count its electron groups, and assign the corresponding ideal angle. Lone pairs and electronegativity differences cause small deviations from ideal values, but the ideal is close enough for most exam purposes.
Hybridisation is a model that explains observed bond angles by mixing atomic orbitals into equivalent hybrid orbitals before bonding.
sp³: One s + three p orbitals → four sp³ hybrids, tetrahedral arrangement, 109.5°. Applies to carbon with four single bonds, nitrogen with three bonds and a lone pair, oxygen with two bonds and two lone pairs.
sp²: One s + two p orbitals → three sp² hybrids in a plane at 120°, plus one unhybridised p orbital perpendicular to the plane. The p orbital forms pi bonds. Applies to C=C double bonds, carbonyls (C=O), and carbocations.
sp: One s + one p orbital → two sp hybrids at 180°, plus two unhybridised p orbitals. Applies to C≡C triple bonds, C≡N (nitriles), and CO₂.
Count the number of sigma bonds plus lone pairs (called "steric number") on the atom:
Steric number 4 → sp³
Steric number 3 → sp²
Steric number 2 → sp
A double bond contains one sigma and one pi bond; a triple bond contains one sigma and two pi bonds. Only sigma bonds count for the steric number.
The more s character in a hybrid orbital, the closer the electrons are held to the nucleus. This means sp carbons are slightly more electronegative than sp² carbons, which are more electronegative than sp³ carbons. This is why terminal alkynes (sp C-H) are more acidic than alkene C-H (sp²) or alkane C-H (sp³).
Draw the molecule's 3D shape.
Add dipole arrows for each polar bond (pointing toward the more electronegative atom).
If the individual bond dipoles cancel by symmetry, the molecule is nonpolar (e.g. CO₂, CCl₄). If they do not cancel, the molecule has a net dipole and is polar (e.g. CHCl₃, H₂O).
Symmetric molecules with identical substituents tend to be nonpolar even if each bond is polar.
In order of increasing strength (for molecules of similar size):
London dispersion forces: present in all molecules. Strength scales with surface area and polarisability (number of electrons). This is why longer-chain hydrocarbons have higher boiling points than shorter ones.
Dipole-dipole interactions: present only in polar molecules. The δ+ end of one molecule attracts the δ− end of another.
Hydrogen bonding: a special, strong case of dipole-dipole. Requires H bonded to F, O, or N, interacting with a lone pair on F, O, or N nearby. Roughly 5 to 10 times stronger than ordinary dipole-dipole forces.
Boiling point: stronger IMFs = higher boiling point (more energy needed to pull molecules apart). For molecules of similar mass, H-bonding compounds boil highest, then dipole-dipole, then dispersion-only.
Melting point: similar trend, though crystal packing and symmetry also matter.
Solubility: "like dissolves like." Polar and H-bonding molecules dissolve well in polar solvents (water). Nonpolar molecules dissolve in nonpolar solvents (hexane). This is why oil and water do not mix.
Real-world note: Intermolecular forces determine whether a substance is a gas, liquid, or solid at room temperature. They also explain why ethanol (H-bonding) is miscible with water while diethyl ether (weak dipole, no H-bond donor) is not.
Find the longest continuous carbon chain. This gives the parent name (methane, ethane, propane, butane, pentane, hexane, heptane, octane, nonane, decane for 1 through 10 carbons).
Number the chain from the end that gives substituents the lowest set of numbers.
Name each substituent (methyl, ethyl, propyl, etc.) with its position number.
Alphabetise substituents in the final name. Use di-, tri-, tetra- for identical substituents, but these prefixes do not affect alphabetical order.
Combine: position numbers, then substituent name(s), then parent name with the suffix "-ane."
Example: 2-methylbutane = a four-carbon chain with a methyl group on C2.
Named by adding "cyclo-" before the parent name: cyclopropane (3C ring), cyclobutane (4C), cyclopentane (5C), cyclohexane (6C).
If the ring has substituents, number the ring to give the lowest set of numbers. If there is only one substituent, no number is needed.
When a ring is attached to a longer chain, the ring is named as a substituent (e.g. cyclohexylpentane, though if the ring has more carbons than the chain, the ring is the parent).
Organic chemists track oxidation differently from inorganic chemists. Rather than formal oxidation states, the practical rule is:
Oxidation: a carbon gains bonds to more electronegative atoms (O, N, halogen) or loses bonds to hydrogen.
Reduction: a carbon gains bonds to hydrogen or loses bonds to electronegative atoms.
Examples of increasing oxidation at carbon: alkane → alcohol → aldehyde → carboxylic acid → CO₂. Each step increases the number of C-O bonds.
A quick test: if the product has more C-H bonds than the reactant, reduction occurred. If it has fewer C-H bonds (or more C-O, C-N bonds), oxidation occurred.
Alkane (C-C, C-H only): all sp³ carbons, tetrahedral geometry, free rotation about C-C bonds. Relatively unreactive. Saturated (no pi bonds).
Alkene (C=C): the double-bond carbons are sp², trigonal planar, 120° bond angles. Contains one pi bond, which is the site of reactivity. Unsaturated.
Alkyne (C≡C): the triple-bond carbons are sp, linear, 180°. Contains two pi bonds. More reactive than alkenes in many contexts. Terminal alkynes have a mildly acidic C-H (pKa ~25).
Aromatic (benzene ring): six sp² carbons in a planar ring with delocalised pi electrons above and below the plane. Unusually stable due to aromaticity. Does not undergo the same addition reactions as simple alkenes.
The key structural differences: bond order increases from alkane (1) to alkene (2) to alkyne (3). As bond order rises, the C-C bond gets shorter and stronger, and the carbon atoms become more electronegative (more s character in the hybrid orbital).
Students often confuse electron geometry with molecular geometry. Electron geometry includes lone pairs in the count; molecular geometry describes only atom positions. Water has tetrahedral electron geometry but bent molecular geometry.
Students often think that any molecule with polar bonds is polar. Symmetry matters. CCl₄ has four polar C-Cl bonds, but they point toward the corners of a tetrahedron and cancel out. CCl₄ is nonpolar.
Students often assume hydrogen bonding can occur whenever hydrogen is present. Hydrogen bonding requires H directly bonded to F, O, or N (the donor), interacting with a lone pair on F, O, or N (the acceptor). A C-H bond does not participate in hydrogen bonding under normal circumstances.
Students often confuse oxidation with combustion. Oxidation in organic chemistry is a stepwise concept (gaining one C-O bond, for instance), not necessarily a reaction with O₂.
⚠️ Assigning hybridisation to every atom in a large molecule is a classic exam question. Use the steric number shortcut: count sigma bonds + lone pairs.
⚠️ Predicting whether a molecule is polar or nonpolar requires you to draw its 3D shape first. A flat, 2D drawing will mislead you.
⚠️ Ranking boiling points by IMF type and molecular weight is heavily tested. Remember: H-bonding > dipole-dipole > dispersion (for similar-sized molecules), and within a dispersion-only series, larger surface area = higher boiling point.
⚠️ Naming alkanes and cycloalkanes with multiple substituents trips students up. Practise the numbering rule: choose the direction that gives the lowest set of locants at the first point of difference.
⚠️ Organic oxidation/reduction questions may show a reaction and ask whether the substrate was oxidised or reduced. Count C-H bonds vs C-O (or C-halogen) bonds before and after.
True or false: An sp² carbon has bond angles of approximately 109.5°. (False, sp² gives ~120°; 109.5° is sp³.)
Fill in the blank: A molecule with polar bonds arranged symmetrically has a net dipole moment of _______. (Zero.)
True or false: Hydrogen bonding requires H bonded to C, N, or O. (False, it requires H bonded to F, O, or N, not C.)
Fill in the blank: The steric number of the oxygen atom in water is _______. (4: two bonds + two lone pairs → sp³.)
True or false: Converting an alcohol to an aldehyde is an oxidation. (True, the carbon gains a bond to oxygen and loses a bond to hydrogen.)
Q: What is the hybridisation of each carbon in propene (CH₂=CHCH₃)?
A: C1 and C2 (the double-bond carbons) are sp². C3 (the methyl carbon) is sp³.
Q: Predict whether CHCl₃ (chloroform) is polar or nonpolar. Explain.
A: Polar. The carbon is sp³ (tetrahedral), bonded to one H and three Cl atoms. The three C-Cl dipoles do not cancel because the molecule lacks the full symmetry of CCl₄. There is a net dipole pointing roughly from H toward the three Cl atoms.
Q: Rank the following in order of increasing boiling point: pentane, 1-butanol, diethyl ether. Explain.
A: Pentane < diethyl ether < 1-butanol. Pentane is nonpolar (dispersion only). Diethyl ether is polar but cannot donate hydrogen bonds (no O-H or N-H). 1-Butanol can both donate and accept hydrogen bonds, giving it the strongest IMFs and the highest boiling point.
Q: Give the IUPAC name for a five-carbon chain with methyl groups on C2 and C3.
A: 2,3-dimethylpentane.
Q: A reaction converts cyclohexanol (C₆H₁₁OH) to cyclohexanone (C₆H₁₀O). Has the substrate been oxidised or reduced?
A: Oxidised. The carbon bearing the OH lost a C-H bond and gained a second C-O bond (forming C=O). Fewer C-H bonds and more C-O bonds = oxidation.
Hybridisation and geometry from this chapter directly feed into Ch. 3 (conformational analysis), where you need to visualise tetrahedral carbons rotating and interacting in 3D. The IMF concepts return whenever you predict solvent choice or separation behaviour. Oxidation state tracking becomes critical when you study oxidation and reduction reactions of alcohols, aldehydes, and ketones later in the course. Hydrocarbon functional group recognition is the starting point for learning the reactivity patterns of alkenes (Ch. 8-9), alkynes (Ch. 10), and aromatics (Ch. 16-17).
VSEPR, molecular geometry, electron geometry, bond angle, hybridisation, sp3, sp2, sp, tetrahedral, trigonal planar, linear, bent, trigonal pyramidal, polar molecule, nonpolar, dipole moment, intermolecular forces, IMF, London dispersion, van der Waals, dipole-dipole, hydrogen bonding, H-bonding, boiling point, melting point, solubility, like dissolves like, IUPAC nomenclature, alkane naming, cycloalkane, methyl, ethyl, propyl, substituent, oxidation, reduction, functional group, alkane, alkene, alkyne, aromatic, benzene, saturated, unsaturated, CHEM 2301, organic chemistry I, UMN