CHEM 120: General Chemistry 1
Scott Hopkins
Estimated study time: 1 hr 6 min
Table of contents
CHEM 120 is a first-year general chemistry course at the University of Waterloo, covering stoichiometry review, quantum theory, atomic structure, chemical bonding (Lewis structures, VSEPR, valence bond theory, and molecular orbital theory), the behaviour of gases, and intermolecular forces. The quantum theory and atomic structure modules are taught by Prof Scott Hopkins; the bonding and gases modules are co-taught with Prof Sen. Course administration is coordinated by Dr Jake Fisher. Assignments and tests are delivered through the Mobius online system.
Module 0: Stoichiometry and Solution Chemistry Review
Limiting Reagents and Percent Yield
The limiting reagent is the reactant that is completely consumed first and thereby determines the maximum amount of product that can form. To identify it, convert the mass of each reactant to moles, then divide by its stoichiometric coefficient. The reactant with the smallest resulting ratio is limiting.
Percent yield relates the actual experimental result to the theoretical maximum:
\[\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%\]A worked example: suppose 14.0 g of CO reacts with 8.0 g of H₂ to form methanol (CH₃OH). Moles of CO = 14.0/28.01 = 0.4998 mol; moles of H₂ = 8.0/2.016 = 3.97 mol. The reaction is CO + 2 H₂ → CH₃OH, so dividing by stoichiometric coefficients gives 0.4998/1 = 0.50 for CO and 3.97/2 = 1.98 for H₂. CO is limiting. Theoretical yield of CH₃OH = 0.4998 mol × 32.04 g/mol = 16.0 g. If only 9.5 g is recovered, percent yield = (9.5/16.0) × 100% = 59%.
Simultaneous and Mixture Reaction Problems
When two reactions occur simultaneously, or when a mixture of compounds reacts, a system of algebraic equations is required. Define variables for the unknown quantities (typically moles of each component), set up one equation per measured experimental observable, and solve the system.
Example — mixture problem: A mixture of NaCl and KCl has total mass 3.595 g. When dissolved and treated with excess AgNO₃, 8.640 g of AgCl precipitate forms. Let \(x\) = moles of NaCl and \(y\) = moles of KCl. Two equations arise:
\[58.44x + 74.55y = 3.595\]\[143.32(x + y) = 8.640\]From the second equation, \(x + y = 0.06027\) mol. Substituting into the first gives a linear system solvable for \(x\) and \(y\), and hence the mass percent of each component.
Molarity, Dilution, and Solution Stoichiometry
Molarity (\(c\) or \(M\)) is defined as moles of solute per litre of solution:
\[c = \frac{n}{V}\]To prepare a solution from a liquid reagent, the following chain is used: density (g/mL) × volume (mL) → mass → moles → molarity.
Example: A bottle of H₂SO₄ is labelled 98.0% w/w and has density 1.84 g/mL. Moles of H₂SO₄ per litre = (1840 g/L × 0.980) / 98.08 g/mol = 18.4 mol/L.
For dilution problems, the amount of solute is conserved:
\[c_1 V_1 = c_2 V_2\]Precipitation Reactions: Molecular, Complete Ionic, and Net Ionic Equations
A precipitation reaction occurs when two aqueous solutions are mixed and an insoluble product forms. Writing equations requires three forms:
- Molecular equation — all species written as complete formulas.
- Complete ionic equation (CIE) — all strong electrolytes written as dissociated ions.
- Net ionic equation (NIE) — spectator ions (present on both sides unchanged) are cancelled; only species that change chemical form remain.
Solubility rules (abbreviated):
- All nitrates (NO₃⁻) and ammonium (NH₄⁺) salts are soluble.
- All alkali-metal salts are soluble.
- Chlorides, bromides, and iodides are soluble except with Ag⁺, Pb²⁺, and Hg₂²⁺.
- Sulfates are soluble except with Ba²⁺, Sr²⁺, Pb²⁺, and Ca²⁺.
- Hydroxides and sulfides are insoluble except with alkali metals and Ba²⁺.
- Carbonates and phosphates are insoluble except with alkali metals and NH₄⁺.
Example: Mixing BaCl₂(aq) with Na₂SO₄(aq):
Molecular: \(\text{BaCl}_2(aq) + \text{Na}_2\text{SO}_4(aq) \to \text{BaSO}_4(s) + 2\,\text{NaCl}(aq)\)
NIE: \(\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \to \text{BaSO}_4(s)\)
For purity calculations, if an impure sample of MgSO₄ is reacted with excess BaCl₂, the mass of BaSO₄ precipitate is used to back-calculate the moles of SO₄²⁻ and hence the mass of pure MgSO₄ in the sample.
Acid–Base Chemistry (Brønsted–Lowry)
In the Brønsted–Lowry framework, an acid is a proton (H⁺) donor and a base is a proton acceptor. Strong acids (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄) ionise completely in water; all others are weak. Strong bases (alkali metal hydroxides, Ba(OH)₂) fully dissociate; weak bases (NH₃, amines) only partially react.
An amphiprotic species can act as either acid or base. Water is the classic example: in an acid–base reaction it can accept a proton (acting as base) or donate one (acting as acid).
Oxidation States
Oxidation states (oxidation numbers) are formal charges assigned by convention to track electron transfer:
- Pure elements: oxidation state = 0.
- Monoatomic ions: oxidation state = charge.
- O is usually −2 (except in peroxides, −1, or with F).
- H is usually +1 (except in metal hydrides, −1).
- Sum of oxidation states = overall charge on the species.
In a redox reaction, the species that is oxidised loses electrons (oxidation state increases); the species that is reduced gains electrons (oxidation state decreases).
Module 1: The Periodic Table and Green Chemistry
Atomic Mass and Isotopic Abundance
Elements exist as mixtures of isotopes — atoms with the same atomic number (protons) but different mass numbers (protons + neutrons). The average atomic mass listed on the periodic table is the weighted mean of all stable isotope masses:
\[\bar{m} = \sum_i f_i \, m_i\]where \(f_i\) is the natural fractional abundance of isotope \(i\) and \(m_i\) is its mass.
Example — bromine: Bromine has two naturally occurring isotopes: ⁷⁹Br (mass 78.918 u) and ⁸¹Br (mass 80.916 u). The average atomic mass of 79.904 u implies:
\[78.918\,f + 80.916\,(1-f) = 79.904\]Solving: \(f = 0.506\), so ⁷⁹Br accounts for about 50.6% of natural bromine and ⁸¹Br about 49.4%. This near-equal split explains why bromine’s mass spectrum shows a characteristic pair of peaks of roughly equal intensity separated by two mass units.
Atomic Mass Intervals and the Periodic Table
Atomic masses across the periodic table do not increase in perfectly equal steps. Moving along a period, mass increments vary because different numbers of neutrons are added and nuclear binding energies change. The general pattern is that atomic mass increases across a period as nuclear charge increases.
Green Chemistry: Atom Economy and E-Factor
Traditional chemistry metrics (yield, purity) do not capture how much of the starting material ends up in the waste stream. Green chemistry uses two complementary metrics:
Atom economy measures the fraction of reactant atoms incorporated into the desired product:
\[\text{atom economy} = \frac{M_\text{product}}{M_\text{all reactants}} \times 100\%\]A reaction with 100% atom economy produces no by-products at all (e.g., addition reactions). Substitution and elimination reactions typically have lower atom economy because they generate by-products.
E-factor (environmental factor) measures waste generated per unit of product:
\[E = \frac{\text{mass of waste (kg)}}{\text{mass of product (kg)}}\]Low E-factors are desirable. The pharmaceutical industry historically has very high E-factors (25–100+) compared to bulk chemical production (~1–5), motivating redesign of synthesis routes to reduce waste, use benign solvents, and incorporate renewable feedstocks.
Module 2: Quantum Theory and Atomic Structure
The Nature of Light: Electromagnetic Radiation
Light is an electromagnetic wave characterised by its wavelength \(\lambda\) (metres), frequency \(\nu\) (hertz, s⁻¹), and amplitude. All electromagnetic radiation travels through vacuum at the speed of light:
\[c = \lambda \nu, \qquad c = 2.998 \times 10^8\,\text{m s}^{-1}\]The electromagnetic spectrum spans from radio waves (long \(\lambda\), low \(\nu\)) through microwaves, infrared, visible light (approximately 400–700 nm), ultraviolet, X-rays, to gamma rays (short \(\lambda\), high \(\nu\)).
Planck’s Quantum Hypothesis
Classical physics predicted that a hot object (a “blackbody”) should emit infinite energy at short wavelengths — the ultraviolet catastrophe. In 1900 Max Planck resolved this by proposing that electromagnetic energy is quantised: it can only be emitted or absorbed in discrete packets called quanta (photons). The energy of a single photon is:
\[E = h\nu = \frac{hc}{\lambda}\]where \(h = 6.626 \times 10^{-34}\,\text{J s}\) is Planck’s constant. This seemingly small adjustment — energy comes in chunks rather than a continuum — was revolutionary and marks the birth of quantum mechanics.
The Photoelectric Effect
In the photoelectric effect, light shining on a metal surface ejects electrons. Classical wave theory predicted that any frequency of light, given enough intensity, should eventually eject electrons. Experiment showed otherwise:
- Below a threshold frequency \(\nu_0\), no electrons are ejected regardless of intensity.
- Above \(\nu_0\), electrons are ejected immediately, with kinetic energy proportional to \(\nu\).
Einstein (1905) explained this by treating light as a stream of photons. Each photon carries energy \(h\nu\). To eject an electron, the photon must supply at least the work function \(\Phi\) (the binding energy of an electron in that metal). Excess energy becomes kinetic energy of the ejected electron:
\[E_k = h\nu - \Phi\]If \(h\nu < \Phi\), the electron cannot be freed. This established the particle nature of light.
Atomic Emission Spectra and the Rydberg Formula
When gaseous hydrogen is excited (by electrical discharge or heat), it emits light only at discrete wavelengths — a line spectrum rather than a continuous rainbow. Each series of lines corresponds to electrons falling to a fixed lower level:
| Series | Lower level (\(n_1\)) | Region |
|---|---|---|
| Lyman | 1 | Ultraviolet |
| Balmer | 2 | Visible |
| Paschen | 3 | Infrared |
The wavelengths are given exactly by the Rydberg formula:
\[\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \quad n_2 > n_1\]where \(R_H = 1.097 \times 10^7\,\text{m}^{-1}\) is the Rydberg constant and \(n_1, n_2\) are positive integers.
The Bohr Model
Niels Bohr (1913) proposed that electrons in hydrogen orbit the nucleus in fixed circular orbits of quantised radii. Each orbit has a definite energy:
\[E_n = -\frac{R_H h c}{n^2} = -\frac{13.6\,\text{eV}}{n^2}, \quad n = 1, 2, 3, \ldots\]The negative sign indicates a bound state; \(n = 1\) is the ground state (most stable). Emission occurs when an electron falls from \(n_2\) to \(n_1\):
\[\Delta E = E_{n_2} - E_{n_1} = hc R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\]The Bohr model successfully predicts hydrogen’s spectrum but fails for multi-electron atoms and offers no physical justification for why orbits are quantised. It was superseded by quantum mechanics.
Wave–Particle Duality and the de Broglie Hypothesis
If light can behave as a particle (photoelectric effect), perhaps particles can behave as waves. Louis de Broglie (1924) proposed that any particle with momentum \(p = mv\) has an associated de Broglie wavelength:
\[\lambda = \frac{h}{mv}\]This is significant only for very light particles (electrons, atoms) at the quantum scale; macroscopic objects have wavelengths far too small to observe. The hypothesis was confirmed by the Davisson–Germer experiment (1927): a beam of electrons directed at a nickel crystal produced a diffraction pattern — unmistakable wave behaviour.
The Heisenberg Uncertainty Principle
Classical physics assumed position and momentum could both be measured exactly. In quantum mechanics this is impossible. Heisenberg showed that the uncertainties in position (\(\Delta x\)) and momentum (\(\Delta p\)) are fundamentally linked:
\[\Delta x \cdot \Delta p \geq \frac{\hbar}{2}, \qquad \hbar = \frac{h}{2\pi}\]A smaller uncertainty in position requires a larger uncertainty in momentum, and vice versa. This is not a limitation of measurement instruments — it is a fundamental feature of nature. Electrons do not have definite trajectories; they have probability distributions.
The Schrödinger Equation and Wave Functions
Erwin Schrödinger (1926) formulated the full quantum mechanical description of the atom. The time-independent Schrödinger equation is:
\[\hat{H}\psi = E\psi\]where \(\hat{H}\) is the Hamiltonian operator (representing the total energy), \(\psi\) is the wave function, and \(E\) is the energy of the state.
The wave function \(\psi(r, \theta, \phi)\) is not directly observable, but \(|\psi|^2\) gives the probability density — the probability of finding an electron per unit volume at a given point in space.
Quantum Numbers
Solving the Schrödinger equation for hydrogen yields a set of quantised states, each characterised by four quantum numbers:
| Quantum Number | Symbol | Allowed Values | Physical Meaning |
|---|---|---|---|
| Principal | \(n\) | 1, 2, 3, … | Shell; governs energy and size |
| Angular momentum | \(l\) | 0, 1, 2, …, \(n-1\) | Subshell; governs shape |
| Magnetic | \(m_l\) | \(-l, \ldots, 0, \ldots, +l\) | Orientation of orbital |
| Spin | \(m_s\) | \(+\tfrac{1}{2},\, -\tfrac{1}{2}\) | Intrinsic spin of electron |
The subshell labels \(l = 0, 1, 2, 3\) correspond to the letters s, p, d, f respectively.
Atomic Orbitals: Shapes and Nodes
An atomic orbital is a one-electron wave function characterised by \((n, l, m_l)\). Its shape is determined by \(l\):
| Subshell | \(l\) | Shape | Angular nodes | Radial nodes |
|---|---|---|---|---|
| s | 0 | Spherical | 0 | \(n-1\) |
| p | 1 | Dumbbell (two lobes) | 1 | \(n-2\) |
| d | 2 | Four-lobed (mostly) | 2 | \(n-3\) |
| f | 3 | Complex | 3 | \(n-4\) |
Total nodes = \(n - 1\) = angular nodes + radial nodes.
The 1s orbital is a sphere centred on the nucleus. The 2p orbitals are dumbbell-shaped along the \(x\), \(y\), or \(z\) axis (for \(m_l = -1, 0, +1\)). The 3d orbitals have four-lobed “clover” shapes (four of them) plus one \(d_{z^2}\) with a unique doughnut-plus-lobe shape.
Radial distribution functions plot \(4\pi r^2 |\psi|^2\) versus \(r\) and show the probability of finding an electron in a thin shell at radius \(r\). The most probable radius for the hydrogen 1s orbital is exactly the Bohr radius \(a_0 = 52.9\) pm. The 2s orbital has a radial node — a spherical surface where \(\psi = 0\).
Boundary surface plots (90% probability surfaces) give the familiar orbital shapes drawn in chemistry courses.
Module 3: Multi-Electron Atoms and Periodic Trends
Orbital Energies in Multi-Electron Atoms
In hydrogen, orbitals with the same \(n\) but different \(l\) are degenerate (have the same energy). In multi-electron atoms this degeneracy is broken because each electron is repelled by all other electrons in addition to being attracted to the nucleus.
Shielding (screening): Inner electrons partially shield outer electrons from the full nuclear charge. An outer electron “sees” an effective nuclear charge \(Z_\text{eff} < Z\).
Penetration: s orbitals have appreciable electron density near the nucleus (they penetrate the inner shells), so they experience less shielding and are lower in energy than p orbitals with the same \(n\). Similarly, p orbitals are lower in energy than d orbitals of the same \(n\).
\[1s < 2s < 2p < 3s < 3p < 3d \approx 4s < 4p < \ldots\](The exact ordering of 3d and 4s depends on \(Z\).)
Pauli Exclusion Principle
No two electrons in the same atom can have identical sets of all four quantum numbers. Equivalently, each orbital can hold at most two electrons and they must have opposite spins (\(m_s = +\tfrac{1}{2}\) and \(m_s = -\tfrac{1}{2}\)).
Aufbau Principle and the (n + l) Rule
Electrons fill orbitals starting from the lowest energy. The approximate filling order follows the \(n + l\) rule: orbitals with smaller \(n + l\) fill first; if tied, smaller \(n\) fills first. This generates the familiar diagonal filling diagram and predicts the block structure of the periodic table.
Hund’s Rule
When filling orbitals of the same energy (degenerate orbitals, e.g., the three 2p orbitals), one electron enters each orbital with the same spin before any orbital receives a second electron. This minimises electron–electron repulsion and gives the lowest-energy configuration.
Example — oxygen (Z = 8): Configuration is 1s² 2s² 2p⁴. The four 2p electrons fill as: ↑↓ in one 2p, and ↑ in each of the other two (Hund’s rule requires parallel spins in half-filled orbitals).
Electron Configurations
| Element | \(Z\) | Configuration |
|---|---|---|
| H | 1 | 1s¹ |
| He | 2 | 1s² |
| Li | 3 | [He] 2s¹ |
| Be | 4 | [He] 2s² |
| B | 5 | [He] 2s² 2p¹ |
| C | 6 | [He] 2s² 2p² |
| N | 7 | [He] 2s² 2p³ |
| O | 8 | [He] 2s² 2p⁴ |
| F | 9 | [He] 2s² 2p⁵ |
| Ne | 10 | [He] 2s² 2p⁶ |
| Na | 11 | [Ne] 3s¹ |
Ions are formed by adding or removing electrons. Cations lose electrons starting from the highest-energy (outermost) orbital; anions gain electrons into the lowest available orbital.
Diamagnetic species have all electrons paired; paramagnetic species have one or more unpaired electrons and are attracted into a magnetic field.
Atomic Radius Trends
Atomic radius is defined as half the distance between adjacent nuclei in a homonuclear diatomic or crystal. Two primary trends govern atomic size:
- Across a period (left to right): Atomic radius decreases. The number of protons increases (rising \(Z_\text{eff}\)) while electrons are added to the same shell, pulling the electron cloud inward.
- Down a group: Atomic radius increases. Each successive period adds a new electron shell further from the nucleus, overwhelming the increase in nuclear charge.
Ionization Energy Trends
First ionization energy (IE₁) is the energy required to remove the outermost electron from a gaseous atom:
\[\text{M}(g) \to \text{M}^+(g) + e^-, \quad \Delta H = IE_1\]- Across a period: IE₁ generally increases (higher \(Z_\text{eff}\), smaller radius, stronger hold on electrons).
- Down a group: IE₁ decreases (outer electrons are further from the nucleus and more shielded).
Important exceptions:
- Be vs B (Group 2 vs 13): IE₁(B) < IE₁(Be) because boron’s outermost electron is in a 2p orbital (higher energy, easier to remove than Be’s 2s).
- N vs O (Group 15 vs 16): IE₁(O) < IE₁(N) because oxygen’s fourth 2p electron must pair in an already-occupied 2p orbital, experiencing extra electron–electron repulsion.
Successive ionisation energies (IE₂, IE₃, …) increase sharply when a core (noble-gas configuration) electron is removed.
Electron Affinity
Electron affinity (EA) is the energy change when a gaseous atom gains an electron:
\[\text{M}(g) + e^- \to \text{M}^-(g), \quad \Delta H = EA\]More negative EA means the anion is more stable (the element accepts an electron more readily). Halogens have the most negative EA values (F and Cl particularly). Noble gases have positive EA (the anion would be unstable). Exceptions arise for the same reasons as in IE trends (N has less negative EA than O due to half-filled 2p stability).
Module 4: Lewis Structures, Formal Charge, and VSEPR
Ionic vs Covalent Bonding
Ionic bonds form between atoms with large electronegativity differences (typically metal and non-metal). The more electronegative atom essentially takes one or more electrons completely, forming anions and cations held together by electrostatic attraction. Lattice energy is the energy released on forming one mole of the ionic solid from gaseous ions; it correlates with ionic charge and inversely with ionic radius.
Covalent bonds form when two atoms share one or more pairs of electrons. The shared electron density between the nuclei creates attraction that holds them together.
Lewis Structures
Lewis dot symbols represent an atom’s valence electrons as dots. A Lewis structure shows how valence electrons are arranged in a molecule: bonding pairs (shared between atoms) and lone pairs (non-bonding, on one atom).
Rules for drawing Lewis structures:
- Count total valence electrons.
- Connect atoms with single bonds (using 2 electrons each).
- Complete octets on terminal atoms with lone pairs.
- Assign remaining electrons to the central atom.
- If the central atom lacks an octet, convert lone pairs on adjacent atoms to multiple bonds.
- Check formal charges and minimise them.
Formal charge is the charge an atom would carry if all bonding electrons were shared equally:
\[FC = (\text{valence electrons}) - (\text{lone pair electrons}) - \frac{1}{2}(\text{bonding electrons})\]The best Lewis structure minimises formal charges and places any negative formal charge on the more electronegative atom.
Resonance
Some molecules cannot be described by a single Lewis structure. Resonance structures are multiple valid Lewis structures that differ only in the placement of electrons (not atoms). The actual molecule is a resonance hybrid — intermediate between all contributing structures. Resonance delocalises electron density and generally stabilises the molecule.
Example — ozone (O₃): Two resonance structures exist, each with a double bond to one terminal oxygen and a single bond to the other. The hybrid has bond order 1.5 for each O–O bond and equal bond lengths.
Example — benzene (C₆H₆): Six resonance structures (Kekulé structures) contribute equally. All C–C bonds are equivalent with bond order 1.5.
Expanded Octets
Third-period and heavier elements can accommodate more than eight electrons in their valence shell by using empty d orbitals. Examples: PCl₅ (10 electrons around P), SF₆ (12 electrons around S), XeF₄ (12 electrons around Xe). Boron and beryllium compounds (BF₃, BeCl₂) can be electron-deficient with fewer than eight electrons.
Bond Dissociation Energy and Bond Order
Bond dissociation energy (BDE) is the energy required to break one mole of a given bond in the gas phase. BDE increases with bond order and decreases with bond length:
| Bond | Bond order | Length (pm) | BDE (kJ/mol) |
|---|---|---|---|
| C–C | 1 | 154 | 347 |
| C=C | 2 | 134 | 614 |
| C≡C | 3 | 120 | 839 |
| N–N | 1 | 145 | 163 |
| N=N | 2 | 125 | 418 |
| N≡N | 3 | 110 | 945 |
Electronegativity and Bond Polarity
Electronegativity (\(\chi\)) measures an atom’s ability to attract bonding electrons toward itself. The Pauling scale runs from F (most electronegative, \(\chi = 4.0\)) down to Cs/Fr (least electronegative, \(\chi \approx 0.7\)).
A bond between atoms with different electronegativities is polar covalent — the electron density is skewed toward the more electronegative atom, creating partial charges \(\delta^+\) and \(\delta^-\). The bond dipole moment \(\mu\) points from the positive toward the negative end (by convention). For a diatomic: \(\mu = \delta \cdot r\).
Whether a molecule is polar depends on both bond polarity and molecular geometry. A symmetric arrangement (e.g., CO₂, CCl₄, BF₃) can have polar bonds but zero net dipole moment because the bond vectors cancel.
VSEPR Theory
Valence Shell Electron Pair Repulsion (VSEPR) theory predicts molecular geometry from the number of electron domains (bonding pairs + lone pairs) around the central atom. Electron domains repel each other and adopt arrangements that maximise angular separation.
| Electron domains | Arrangement | Bond angle |
|---|---|---|
| 2 | Linear | 180° |
| 3 | Trigonal planar | 120° |
| 4 | Tetrahedral | 109.5° |
| 5 | Trigonal bipyramidal | 90°/120° |
| 6 | Octahedral | 90° |
Lone pairs occupy more space than bonding pairs, compressing bond angles slightly. The molecular geometry (shape) describes only the positions of atoms, not lone pairs:
| Electron domains | Lone pairs | Molecular geometry | Example |
|---|---|---|---|
| 2 | 0 | Linear | CO₂ |
| 3 | 0 | Trigonal planar | BF₃ |
| 3 | 1 | Bent | SO₂ |
| 4 | 0 | Tetrahedral | CH₄ |
| 4 | 1 | Trigonal pyramidal | NH₃ |
| 4 | 2 | Bent | H₂O |
| 5 | 0 | Trigonal bipyramidal | PCl₅ |
| 5 | 1 | See-saw | SF₄ |
| 5 | 2 | T-shaped | ClF₃ |
| 5 | 3 | Linear | XeF₂ |
| 6 | 0 | Octahedral | SF₆ |
| 6 | 1 | Square pyramidal | BrF₅ |
| 6 | 2 | Square planar | XeF₄ |
Module 5: Valence Bond Theory and Molecular Orbital Theory
Valence Bond (VB) Theory: Overview
Valence bond theory is a localised picture of bonding. A bond forms when two atoms approach and their half-filled atomic orbitals overlap, allowing electrons to be shared. VB theory explains molecular geometry through the concept of orbital hybridisation.
Sigma and Pi Bonds
- Sigma (σ) bond: Head-on (on-axis) overlap of two orbitals. Electron density is concentrated along the internuclear axis. A sigma bond is present in every covalent bond.
- Pi (π) bond: Sideways overlap of two parallel p orbitals. Electron density is concentrated above and below the internuclear axis. Pi bonds form the “second” and “third” components of double and triple bonds.
A single bond = 1 σ bond. A double bond = 1 σ + 1 π. A triple bond = 1 σ + 2 π.
Pi bonds restrict rotation about the internuclear axis (rotating breaks the overlap), which is why double bonds give rise to geometric (cis/trans) isomers.
Hybridisation of Atomic Orbitals
Pure atomic orbitals (s, p, d) do not have the correct shapes or directional properties to explain observed molecular geometries. Hybridisation mathematically combines atomic orbitals on the same atom to form new hybrid orbitals with the correct geometry. The number of hybrid orbitals equals the number of atomic orbitals mixed.
| Hybridisation | Orbitals mixed | Number of hybrids | Geometry | Bond angle |
|---|---|---|---|---|
| sp | 1s + 1p | 2 | Linear | 180° |
| sp² | 1s + 2p | 3 | Trigonal planar | 120° |
| sp³ | 1s + 3p | 4 | Tetrahedral | 109.5° |
| sp³d | 1s + 3p + 1d | 5 | Trigonal bipyramidal | 90°/120° |
| sp³d² | 1s + 3p + 2d | 6 | Octahedral | 90° |
Energy consideration: Hybridisation costs energy (the s orbital is lower in energy than p, so mixing them raises the average energy). This cost is more than recovered by the improved overlap in stronger bonds.
Examples:
- BeCl₂: sp hybridised; two sp orbitals point 180° apart; linear.
- BF₃: sp² hybridised; three sp² orbitals in a plane + one unhybridised p perpendicular; trigonal planar.
- CH₄: sp³ hybridised; four equivalent sp³ orbitals in a tetrahedron.
- NH₃: sp³ hybridised; three sp³ orbitals form N–H sigma bonds, one sp³ holds the lone pair; trigonal pyramidal.
- H₂O: sp³ hybridised; two sp³ form O–H bonds, two sp³ hold lone pairs; bent.
Multiple Bonds in VB Theory
Ethene (C₂H₄): Each carbon is sp² hybridised. The three sp² orbitals form sigma bonds (two C–H and one C–C). The remaining unhybridised 2p orbital on each carbon, perpendicular to the molecular plane, overlaps sideways to form the π bond. The molecule is planar.
Ethyne (C₂H₂): Each carbon is sp hybridised. Two sp orbitals form sigma bonds (one C–H and one C–C). The two remaining unhybridised 2p orbitals on each carbon overlap in two perpendicular planes to form two π bonds. The molecule is linear.
Limitations of VB theory: VB theory cannot easily explain the paramagnetism of O₂ or the bonding in molecules where resonance is important (e.g., benzene). Molecular orbital theory provides a more complete picture.
Molecular Orbital (MO) Theory: Overview
MO theory is a delocalised approach. Electrons are not confined to individual bonds but are spread across the entire molecule in molecular orbitals (MOs) — wave functions that extend over all atoms.
MOs are constructed by the Linear Combination of Atomic Orbitals (LCAO) approximation: when two atomic orbitals \(\phi_a\) and \(\phi_b\) combine, two MOs result:
\[\psi_\text{bonding} = \phi_a + \phi_b\]\[\psi_\text{antibonding} = \phi_a - \phi_b\]The bonding MO has electron density between the nuclei, lowering the energy relative to the isolated atoms. The antibonding MO has a node between the nuclei (zero electron density), raising the energy above the isolated atoms. Antibonding orbitals are labelled with a star (e.g., σ*).
MO Diagrams for First-Period Diatomics
For H₂ and He₂, only 1s orbitals combine:
- H₂: Two electrons fill the σ(1s) bonding MO. Bond order = (2 − 0)/2 = 1. Stable.
- H₂⁺: One electron in σ(1s). Bond order = (1 − 0)/2 = ½. Stable (weakly).
- He₂: Four electrons fill both σ(1s) and σ*(1s). Bond order = (2 − 2)/2 = 0. Does not exist as a stable covalent molecule (only an extremely weak van der Waals dimer at very low temperature).
- He₂⁺: Three electrons: two in σ(1s), one in σ*(1s). Bond order = (2 − 1)/2 = ½. Stable.
Bond order formula:
\[\text{Bond order} = \frac{(\text{bonding electrons}) - (\text{antibonding electrons})}{2}\]A bond order > 0 indicates a stable molecule. Higher bond order = shorter, stronger bond.
MO Diagrams for Second-Period Diatomics
For Li₂ through Ne₂, both 2s and 2p orbitals contribute to MOs. The full MO diagram includes:
From 2s: σ(2s) and σ*(2s) From 2p: σ(2p), two degenerate π(2p), σ*(2p), two degenerate π*(2p)
Critically, for Li₂ through N₂, sp mixing (interaction between the σ(2s) and σ(2p) MOs) raises the energy of σ(2p) above the two degenerate π(2p) MOs. For O₂ and F₂, this sp mixing is weaker and the standard ordering [σ(2p) below π(2p)] is restored.
\[\sigma(2s) < \sigma^*(2s) < \pi(2p) = \pi(2p) < \sigma(2p) < \pi^*(2p) = \pi^*(2p) < \sigma^*(2p)\]\[\sigma(2s) < \sigma^*(2s) < \sigma(2p) < \pi(2p) = \pi(2p) < \pi^*(2p) = \pi^*(2p) < \sigma^*(2p)\]| Molecule | Valence electrons | Bond order | Magnetic behaviour |
|---|---|---|---|
| Li₂ | 2 | 1 | Diamagnetic |
| Be₂ | 4 | 0 | Does not exist |
| B₂ | 6 | 1 | Paramagnetic (2 unpaired in π) |
| C₂ | 8 | 2 | Diamagnetic |
| N₂ | 10 | 3 | Diamagnetic |
| O₂ | 12 | 2 | Paramagnetic (2 unpaired in π*) |
| F₂ | 14 | 1 | Diamagnetic |
| Ne₂ | 16 | 0 | Does not exist |
O₂ paramagnetism is a triumph of MO theory over VB theory. VB theory predicts all electrons paired in O₂; MO theory correctly predicts two unpaired electrons in degenerate π*(2p) orbitals (by Hund’s rule).
Module 6: Gases, Kinetic Theory, and Intermolecular Forces
Ideal Gas Law
An ideal gas is a theoretical model in which:
- Gas molecules have negligible volume.
- There are no intermolecular attractions or repulsions.
- Collisions between molecules are perfectly elastic.
The ideal gas law combines Boyle’s Law (\(P \propto 1/V\) at constant \(T, n\)), Charles’ Law (\(V \propto T\) at constant \(P, n\)), and Avogadro’s Law (\(V \propto n\) at constant \(P, T\)):
\[PV = nRT\]where \(P\) is pressure, \(V\) is volume, \(n\) is moles of gas, \(T\) is absolute temperature (Kelvin), and \(R\) is the universal gas constant:
\[R = 8.314\,\text{J mol}^{-1}\text{K}^{-1} = 0.08314\,\text{L bar mol}^{-1}\text{K}^{-1} = 0.08206\,\text{L atm mol}^{-1}\text{K}^{-1}\]Standard conditions:
- STP (IUPAC, current definition): 0°C (273.15 K), 1 bar. Molar volume = 22.71 L/mol.
- STP (older, common in many textbooks): 0°C, 1 atm. Molar volume = 22.41 L/mol.
Pressure unit conversions: 1 atm = 101.325 kPa = 1.01325 bar ≈ 760 mmHg = 760 torr.
Derived Gas Laws
| Law | Variables held constant | Relationship |
|---|---|---|
| Boyle’s | \(n, T\) | \(P_1 V_1 = P_2 V_2\) |
| Charles' | \(n, P\) | \(V_1/T_1 = V_2/T_2\) |
| Gay-Lussac’s | \(n, V\) | \(P_1/T_1 = P_2/T_2\) |
| Avogadro’s | \(P, T\) | \(V_1/n_1 = V_2/n_2\) |
| Combined | \(n\) | \(P_1 V_1/T_1 = P_2 V_2/T_2\) |
Gas Stoichiometry
When gases are involved in chemical reactions, volumes can be related to moles via the ideal gas law. At the same temperature and pressure, gas volumes are proportional to moles (Avogadro’s Law), so volume ratios equal mole ratios directly.
Example: 2 H₂(g) + O₂(g) → 2 H₂O(g). At STP, 10 L of H₂ reacts with 5 L of O₂ (1:2 ratio) to produce 10 L of H₂O vapour.
For mixed gas/solid/liquid problems, convert to moles using appropriate formulas, apply stoichiometry in moles, then convert the gas product back to volume using PV = nRT.
Dalton’s Law of Partial Pressures
In a mixture of ideal gases, each component exerts its own partial pressure independently of the others. The total pressure is the sum of all partial pressures:
\[P_\text{total} = P_1 + P_2 + P_3 + \cdots\]The partial pressure of component \(i\) is related to its mole fraction \(\chi_i\):
\[P_i = \chi_i \, P_\text{total}, \qquad \chi_i = \frac{n_i}{n_\text{total}}\]Collecting gas over water: When a gas is collected by displacement of water, the total pressure measured is the sum of the gas pressure and the vapour pressure of water at that temperature:
\[P_\text{gas} = P_\text{total} - P_\text{H_2O}(T)\]Kinetic Molecular Theory
The kinetic molecular theory (KMT) provides the microscopic justification for the ideal gas law. Key assumptions:
- Gas consists of point-mass particles in constant random motion.
- No intermolecular forces (between collisions).
- All collisions are elastic (kinetic energy is conserved).
- Average kinetic energy is proportional to absolute temperature.
From KMT, the Maxwell–Boltzmann speed distribution describes the probability of finding a molecule with a given speed \(v\):
\[f(v) = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 \exp\!\left(-\frac{mv^2}{2k_B T}\right)\]This distribution gives three characteristic speeds:
\[v_\text{mp} = \sqrt{\frac{2RT}{M}}, \qquad v_\text{avg} = \sqrt{\frac{8RT}{\pi M}}, \qquad v_\text{rms} = \sqrt{\frac{3RT}{M}}\]where \(M\) is the molar mass in kg/mol. The root-mean-square speed \(v_\text{rms}\) is largest; the most probable speed \(v_\text{mp}\) is smallest:
\[v_\text{mp} < v_\text{avg} < v_\text{rms}\]As temperature increases, the distribution broadens and shifts to higher speeds. As molar mass increases, the distribution narrows and shifts to lower speeds.
Effusion and Diffusion: Graham’s Law
Effusion is the escape of gas molecules through a tiny orifice into a vacuum. Diffusion is the spreading of one gas through another.
Graham’s Law states that effusion rate (and approximately diffusion rate) is inversely proportional to the square root of molar mass:
\[\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\]This follows directly from KMT: lighter gases move faster and therefore effuse more rapidly. Graham’s Law is used to separate isotopes (e.g., ²³⁵UF₆ from ²³⁸UF₆ in uranium enrichment) and to identify unknown gases by comparing effusion times.
Real Gases and Deviations from Ideality
Real gases deviate from ideal behaviour at high pressure (molecules are forced close together; their own volume and intermolecular forces matter) and low temperature (slow molecules spend more time near each other; attractions become significant).
The compressibility factor \(Z\) quantifies deviation from ideality:
\[Z = \frac{PV}{nRT}\]For an ideal gas, \(Z = 1\) always. For real gases, \(Z < 1\) when attractive forces dominate (e.g., CO₂ at moderate pressure) and \(Z > 1\) when repulsive forces (finite molecular volume) dominate (e.g., H₂ and He at high pressure).
Van der Waals Equation
The van der Waals equation corrects the ideal gas law for both molecular volume (\(b\)) and intermolecular attractions (\(a\)):
\[\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT\]In terms of molar volume \(\bar{V} = V/n\):
\[\left(P + \frac{a}{\bar{V}^2}\right)(\bar{V} - b) = RT\]- The \(a/\bar{V}^2\) term corrects pressure upward because intermolecular attractions reduce the force of collisions with the walls.
- The \(b\) term corrects volume downward because molecules have finite size and the free volume available is \(V - nb\).
Both \(a\) and \(b\) increase with molecular size and complexity. In the limit \(a = b = 0\), the ideal gas law is recovered.
Example: One mole of CO₂ (\(a = 3.640\,\text{L}^2\text{bar mol}^{-2}\), \(b = 0.04267\,\text{L mol}^{-1}\)) in 0.5 L at 300 K:
Ideal gas: \(P = nRT/V = (1)(0.083145)(300)/0.5 = 49.89\,\text{bar}\)
Van der Waals: \(P = \frac{RT}{\bar{V}-b} - \frac{a}{\bar{V}^2} = \frac{(0.083145)(300)}{0.5-0.04267} - \frac{3.640}{0.5^2} = 54.166 - 14.56 = \ldots\) (a lower value than ideal), consistent with CO₂ experiencing net attractive forces.
Virial Equation of State
An alternative to van der Waals is the virial equation, which expands \(Z\) as a power series in \(1/\bar{V}\):
\[Z = 1 + \frac{B}{\bar{V}} + \frac{C}{\bar{V}^2} + \cdots\]The coefficients \(B\), \(C\), … (virial coefficients) are temperature-dependent and account for pairwise, three-body, etc., interactions. Truncating at the second term gives a good approximation at moderate pressures.
Macroscopic Properties of Liquids and IMFs
When a gas is cooled or compressed sufficiently, intermolecular forces (IMFs) become important enough to condense it into a liquid or solid. Several macroscopic properties of liquids reflect IMF strength:
- Vapour pressure: Pressure of vapour above a liquid at equilibrium. Stronger IMFs → lower vapour pressure (molecules are harder to escape).
- Boiling point: Temperature at which vapour pressure = external pressure (usually 1 atm). Stronger IMFs → higher boiling point.
- Surface tension: Energy per unit area required to increase the liquid surface. Stronger IMFs → higher surface tension.
- Viscosity: Resistance to flow. Stronger IMFs → higher viscosity.
Dipole Moment and Polarizability
Dipole moment (\(\mu\)) measures charge separation. A molecule has \(\mu \neq 0\) when positive and negative charge centres do not coincide. For two charges \(\pm\delta\) separated by distance \(r\): \(\mu = \delta r\). Units: coulomb-metres (C·m) or debyes (D).
Polarizability (\(\alpha\)) measures how easily a molecule’s electron cloud can be distorted by an external electric field. An induced dipole moment \(\mu_\text{ind} = \alpha \cdot E\) results. Larger, heavier molecules have more diffuse electron clouds and greater polarizability.
Types of Intermolecular Forces
| Force | Present in | Relative strength | Origin |
|---|---|---|---|
| London dispersion (LDF) | All molecules and atoms | Weakest (but universal) | Instantaneous dipole – induced dipole |
| Dipole–dipole | Polar molecules | Moderate | Permanent dipole – permanent dipole |
| Hydrogen bond | Molecules with H bonded to N, O, or F | Strongest (of these three) | Strong dipole; lone-pair attraction |
London dispersion forces (LDFs): Even nonpolar molecules experience fleeting, instantaneous dipoles due to momentary asymmetry in electron distribution. This instantaneous dipole induces a dipole in a neighbouring molecule, producing a net attractive force. LDF strength increases with polarizability (∝ molecular size and number of electrons). LDFs always contribute to IMFs in every molecule.
Interaction energy ∝ \(\alpha^2 / r^6\).
Dipole–dipole interactions: Polar molecules orient so that their positive ends point toward neighbours’ negative ends. Interaction energy ∝ \(\mu^2 / r^6\).
Hydrogen bonding: A uniquely strong directional intermolecular force occurring when H is covalently bonded to N, O, or F, and the partially positive H is attracted to a lone pair on N, O, or F of a neighbouring molecule. Represented as X–H ··· Y, where X and Y are N, O, or F. Hydrogen bonds are 5–30 kJ/mol — much weaker than covalent bonds (150–1000 kJ/mol) but significantly stronger than ordinary dipole–dipole interactions.
Hydrogen bonding in H₂O explains:
- Very high boiling point (100°C vs. −60°C expected from periodic trend).
- High surface tension and viscosity.
- Anomalously low density of ice (open hexagonal lattice) relative to liquid water.
IMF Example Problems
Comparing ICl and Br₂: Both have 70 electrons, so comparable LDFs. But ICl is polar (I and Cl differ in electronegativity → permanent dipole), giving dipole–dipole interactions in addition to LDFs. Br₂ is homonuclear and therefore nonpolar — only LDFs. Result: ICl has stronger IMFs and a higher boiling point (97°C vs. 59°C for Br₂).
Comparing CH₄ and CCl₄: Both are nonpolar (tetrahedral symmetry) and have no hydrogen bonding. CH₄ has 10 electrons; CCl₄ has 74 electrons. More electrons → greater polarizability → stronger LDFs → higher boiling point for CCl₄.
Comparing acetone and 1-propanol: Both are polar (similar electron count: 32 vs. 34). But 1-propanol has an O–H group → hydrogen bonding, absent in acetone (C=O group, no O–H). Stronger IMFs in 1-propanol → higher boiling point (370 K vs. 329 K).
Comparing ethanol and ethylene glycol (vapor pressure): Both form hydrogen bonds. Ethanol has one O–H → one H-bond per molecule. Ethylene glycol has two O–H groups → two H-bonds per molecule. More H-bonds → stronger IMFs → lower vapor pressure for ethylene glycol.
Cis- vs. trans-dichloroethene: Both have identical electron counts (same LDFs) and no H–bonds. In the cis isomer, the two C–Cl bond dipoles point in similar directions and do not cancel → net dipole → dipole–dipole interactions. In the trans isomer, the bond dipoles cancel by symmetry → no net dipole → no dipole–dipole interactions. The cis isomer has higher boiling point.
The Hydrogen Chalcogenide Anomaly
Consider the hydrides of Group 16: H₂O, H₂S, H₂Se, H₂Te. Down the group:
- Dipole moment decreases (electronegativity of chalcogen decreases).
- Number of electrons increases (LDFs increase).
Without H₂O, the remaining three (H₂S, H₂Se, H₂Te) show the trend expected from increasing LDFs: boiling point increases from H₂S to H₂Te.
H₂O is a dramatic outlier — its boiling point (100°C) is far above what the periodic trend would predict (~−80°C extrapolated) because of extensive hydrogen bonding. H₂O is the only molecule in this group where H is bonded to an atom electronegative enough (O) to form strong H-bonds.
The same anomaly is seen in Group 15 (NH₃ is an outlier) and Group 17 (HF is an outlier), but not in Group 14 (CH₄ has no H-bonds and sits smoothly on the LDF trend).
Summary of Key Equations and Constants
| Quantity | Formula | Notes |
|---|---|---|
| Photon energy | \(E = h\nu = hc/\lambda\) | \(h = 6.626 \times 10^{-34}\) J s |
| Photoelectric effect | \(E_k = h\nu - \Phi\) | \(\Phi\) = work function |
| Rydberg formula | \(1/\lambda = R_H(1/n_1^2 - 1/n_2^2)\) | \(R_H = 1.097 \times 10^7\) m⁻¹ |
| Bohr energies | \(E_n = -13.6/n^2\) eV | Hydrogen only |
| de Broglie wavelength | \(\lambda = h/(mv)\) | Matter waves |
| Heisenberg uncertainty | \(\Delta x \cdot \Delta p \geq \hbar/2\) | \(\hbar = h/(2\pi)\) |
| Ideal gas law | \(PV = nRT\) | \(R = 8.314\) J mol⁻¹ K⁻¹ |
| RMS speed | \(v_\text{rms} = \sqrt{3RT/M}\) | \(M\) in kg/mol |
| Graham’s law | \(r_1/r_2 = \sqrt{M_2/M_1}\) | Effusion/diffusion |
| van der Waals | \((P + an^2/V^2)(V-nb) = nRT\) | Real gas correction |
| Compressibility | \(Z = PV/(nRT)\) | 1 for ideal gas |
| Bond order (MO) | \((n_b - n_a)/2\) | \(n_b\) bonding, \(n_a\) antibonding |
| Formal charge | \(FC = V - LP - \tfrac{1}{2}BP\) | \(V\) valence, LP lone, BP bonding |
| Percent yield | \((\text{actual}/\text{theoretical}) \times 100\%\) | |
| Average atomic mass | \(\bar{m} = \sum f_i m_i\) | |
| Atom economy | \(M_\text{product}/M_\text{all reactants} \times 100\%\) | |
| Molarity dilution | \(c_1 V_1 = c_2 V_2\) |
Appendix A: Worked Gas Law Problems
Problem 1 — Finding Pressure with van der Waals
One mole of CO₂ is confined to a 0.500 L container at 300 K. Compare the pressure predicted by the ideal gas law and the van der Waals equation. For CO₂: \(a = 3.640\,\text{L}^2\text{atm mol}^{-2}\), \(b = 0.04267\,\text{L mol}^{-1}\).
Ideal gas:
\[P = \frac{nRT}{V} = \frac{(1\,\text{mol})(0.08206\,\text{L atm mol}^{-1}\text{K}^{-1})(300\,\text{K})}{0.500\,\text{L}} = 49.2\,\text{atm}\]Van der Waals:
\[P = \frac{nRT}{V - nb} - \frac{an^2}{V^2} = \frac{(1)(0.08206)(300)}{0.500 - (1)(0.04267)} - \frac{(3.640)(1)^2}{(0.500)^2}\]\[P = \frac{24.618}{0.45733} - \frac{3.640}{0.250} = 53.82 - 14.56 = 39.3\,\text{atm}\]The van der Waals pressure (39.3 atm) is lower than ideal (49.2 atm) because CO₂ molecules exert attractive forces on each other, reducing the pressure exerted on the container walls. This means \(Z < 1\) for CO₂ under these conditions.
Problem 2 — Effusion Rate Ratio
How much faster does ⁴He effuse compared to ²³⁸UF₆?
\[\frac{r_\text{He}}{r_\text{UF_6}} = \sqrt{\frac{M_\text{UF_6}}{M_\text{He}}} = \sqrt{\frac{352.0\,\text{g/mol}}{4.003\,\text{g/mol}}} = \sqrt{87.9} \approx 9.38\]Helium effuses about 9.4 times faster than UF₆. This large ratio makes gas diffusion a viable — if slow — method for isotope separation (e.g., uranium enrichment uses ²³⁵UF₆ vs. ²³⁸UF₆ with a ratio of only 1.004 per stage, requiring thousands of stages).
Problem 3 — Partial Pressures
A 3.00 L flask at 25°C contains 0.100 mol N₂ and 0.200 mol O₂. What is the partial pressure of each gas and the total pressure?
\[P_{\text{N}_2} = \frac{n_{\text{N}_2}RT}{V} = \frac{(0.100)(0.08206)(298)}{3.00} = 0.815\,\text{atm}\]\[P_{\text{O}_2} = \frac{(0.200)(0.08206)(298)}{3.00} = 1.630\,\text{atm}\]\[P_\text{total} = 0.815 + 1.630 = 2.445\,\text{atm}\]Mole fraction of N₂: \(\chi_{\text{N}_2} = 0.100/0.300 = 0.333\). Check: \(0.333 \times 2.445 = 0.815\,\text{atm}\). ✓
Appendix B: Worked Quantum Mechanics Problems
Problem 1 — Photoelectric Effect
Light of wavelength 200 nm strikes a metal with work function \(\Phi = 4.61\,\text{eV}\). What is the kinetic energy of the ejected electron?
First, find the photon energy:
\[E = \frac{hc}{\lambda} = \frac{(6.626 \times 10^{-34})(2.998 \times 10^8)}{200 \times 10^{-9}} = 9.93 \times 10^{-19}\,\text{J}\]Convert work function: \(\Phi = 4.61\,\text{eV} \times 1.602 \times 10^{-19}\,\text{J/eV} = 7.39 \times 10^{-19}\,\text{J}\)
\[E_k = E - \Phi = 9.93 \times 10^{-19} - 7.39 \times 10^{-19} = 2.54 \times 10^{-19}\,\text{J} = 1.59\,\text{eV}\]Problem 2 — Hydrogen Energy Levels
What is the wavelength of light emitted when hydrogen’s electron drops from \(n = 4\) to \(n = 2\) (the Hα line of the Balmer series)?
\[\frac{1}{\lambda} = R_H \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = (1.097 \times 10^7)\left(\frac{1}{4} - \frac{1}{16}\right) = (1.097 \times 10^7)(0.1875) = 2.057 \times 10^6\,\text{m}^{-1}\]\[\lambda = \frac{1}{2.057 \times 10^6} = 4.86 \times 10^{-7}\,\text{m} = 486\,\text{nm}\]This falls in the blue-green region of the visible spectrum.
Problem 3 — de Broglie Wavelength
What is the de Broglie wavelength of an electron moving at \(1.00 \times 10^6\,\text{m/s}\)?
\[\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{(9.109 \times 10^{-31})(1.00 \times 10^6)} = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-25}} = 7.28 \times 10^{-10}\,\text{m} = 0.728\,\text{nm}\]This is on the order of atomic bond lengths (~0.1–0.3 nm), which is why electron diffraction by crystal lattices is observable and why electrons exhibit wave behaviour in atomic-scale experiments.
Appendix C: Bonding — Additional Notes
Lattice Energy and the Born–Haber Cycle
Lattice energy (\(U\)) is the energy released when gaseous cations and anions combine to form one mole of the ionic solid. It cannot be measured directly but is calculated via a Born–Haber cycle — a thermodynamic cycle applying Hess’s Law to the formation of an ionic solid from its constituent elements:
\[\Delta H^\circ_f = \Delta H_\text{sub}(\text{metal}) + IE_1(\text{metal}) + \tfrac{1}{2}D(\text{bond}) + EA(\text{nonmetal}) + U\]For NaCl, the large lattice energy (~787 kJ/mol) makes the overall formation process exothermic even though several individual steps are endothermic.
Lattice energy magnitude increases with:
- Higher ionic charges (compare NaCl vs. MgO: 787 vs. ~3795 kJ/mol).
- Smaller ionic radii (stronger electrostatic attraction at shorter distances).
This is captured qualitatively by Coulomb’s law: \(U \propto \frac{q^+ q^-}{r}\).
VSEPR and Molecular Polarity
Knowing molecular geometry from VSEPR, we can determine whether a molecule is polar:
- Identify all bond dipoles (each polar bond has a dipole pointing from \(\delta^+\) to \(\delta^-\)).
- Add the bond dipole vectors as a vector sum.
- If the vector sum is non-zero, the molecule is polar; if zero, nonpolar.
Examples:
- CO₂ (linear, O=C=O): Two bond dipoles point in opposite directions; sum = 0. Nonpolar.
- H₂O (bent): Two O–H bond dipoles do not cancel; net dipole points toward O. Polar.
- CCl₄ (tetrahedral): Four C–Cl bond dipoles in perfect tetrahedral symmetry; sum = 0. Nonpolar.
- CHCl₃ (tetrahedral): The C–H bond is less polar than C–Cl; the four bond dipoles do not cancel. Polar.
- NH₃ (trigonal pyramidal): Three N–H bond dipoles plus the lone pair on N all contribute in the same general direction. Polar.
Multiple Bond Structures: Practice Examples
Nitric acid (HNO₃): The central N is bonded to: one –OH (single bond), two =O (one with double bond), giving N a formal charge. A resonance structure delocalises the double bond between the two N=O positions. Total valence electrons = 5 + 3(6) + 1 = 24.
Sulfur dioxide (SO₂): Two resonance structures: S=O in one position, S–O in the other. The actual molecule has equal S–O bond lengths (~143 pm) intermediate between single (~160 pm) and double (~143 pm), confirming resonance. The bent geometry (VSEPR: 3 electron domains on S including one lone pair) gives a net dipole moment.
Ozone (O₃): Similar to SO₂ in structure: bent geometry, two resonance structures, bond order 1.5. Ozone is polar.
Exceptions to the Octet Rule: Summary
| Type | Example | Reason |
|---|---|---|
| Electron-deficient | BF₃, BeCl₂ | B (group 13) and Be (group 2) need only 6 or 4 electrons |
| Odd-electron (radical) | NO, NO₂ | Odd number of electrons; one must be unpaired |
| Expanded octet | PCl₅, SF₆, XeF₄ | Period 3+ elements use empty d orbitals |
For expanded octets, the best structure generally minimises formal charges — often the structure with double bonds (even if it gives 10 or 12 electrons on the central atom) is preferred over a structure with more single bonds and higher formal charges.
Appendix D: Course Administrative Notes
Mobius Online System
Assignments, participation challenges, tests, and the final exam are delivered through the Mobius system, accessed via links in the Learn course page. Key points:
- Always print the assignment before starting online work. Since some modules span two weeks, you can begin reviewing printed questions early and work on them offline.
- Entering answers in Mobius and pressing Quit and Save does not use up an attempt — you can save partial work as many times as needed.
- Submitting the assignment uses one attempt and is equivalent to handing in your work. If you do not press Submit, you receive no grade.
- Many questions require specific formats (e.g., exponential notation to three significant figures). Read all instructions on each question carefully.
- The Mobius Gradebook (found in the Online Assignments folder in your content tab) lets you review which questions you answered correctly or incorrectly and see instructor feedback after submission.
- Assignment grades are pushed to the Learn Gradebook automatically (with up to a 5-minute delay after submission or resubmission).
Communication
Post questions about the course and Mobius system to the Piazza discussion board. Course staff aim to respond as quickly as possible. Do not use Mobius for the first time during a test — familiarise yourself with the interface during Module 0 assignments.