Comprehensive A-Level Chemistry Revision Notes

Atomic Structure and Electron Configuration

Subatomic Particles

An atom consists of three fundamental subatomic particles: protons, neutrons, and electrons.

An atom of an element is represented as ${_Z^A\text{X}}$, where:

• $\text{X}$ is the chemical symbol.
• $\text{A}$ is the mass number (total number of protons and neutrons in the nucleus).
• $\text{Z}$ is the atomic number (number of protons in the nucleus).
• The number of neutrons is calculated as $\text{A} - \text{Z}$.

Isotopes

• Definition: Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.
• Chemical Properties: Isotopes have identical chemical properties because they possess the exact same electronic configuration.
• Physical Properties: Isotopes may display slightly varying physical properties (e.g., density, rate of diffusion) because they have different masses.

Historical Development of Atomic Models

• Early Sphere Model: Before the discovery of subatomic particles, atoms were thought to be tiny, indivisible spheres.
• Plum-Pudding Model: The discovery of the electron led J.J. Thomson to propose the plum-pudding model, which suggested that the atom was a sphere of positive charge with negative electrons embedded throughout it.

• Rutherford and Marsden Alpha Scattering Experiment: Alpha particles ($^4_2\text{He}^{2+}$) were directed at thin gold foil:
  • Observation: Most alpha particles passed straight through undeflected.
  • Deduction: Most of the atom is empty space, and most of its mass is concentrated in a tiny central region called the nucleus.
  • Observation: A small fraction of alpha particles were deflected at large angles or reflected straight back.
  • Deduction: The positive charge of the atom is concentrated at the center in the nucleus.

• Bohr Model: Niels Bohr adapted the nuclear model by proposing that electrons orbit the nucleus at specific, fixed distances in energy levels or shells. Theoretical calculations based on Bohr's model aligned closely with experimental spectral observations.

Principles of Electronic Structure

Electrons are organized into principal energy levels, sub-levels, and orbitals:

• Principal Energy Levels ($n$): Numbered $1, 2, 3, 4\text{...}$ where level $1$ is closest to the nucleus.
• Sub-levels: Principal levels are split into sub-levels labeled $s$, $p$, $d$, and $f$:
  • $s$ sub-level: Holds up to $2$ electrons ($1$ orbital).
  • $p$ sub-level: Holds up to $6$ electrons ($3$ orbitals).
  • $d$ sub-level: Holds up to $10$ electrons ($5$ orbitals).
  • $f$ sub-level: Holds up to $14$ electrons ($7$ orbitals).

• Atomic Orbitals: An orbital represents a three-dimensional region of space where there is a high mathematical probability of finding an electron. Each orbital can hold a maximum of $2$ electrons with opposite spins.
  • Shapes of Orbitals:
  • $s$ orbitals are spherical.
  • $p$ orbitals are dumbbell-shaped along three perpendicular axes ($p_x, p_y, p_z$).

• Order of Subshell Filling: Sub-shells fill in order of increasing energy:   $1s \rightarrow 2s \rightarrow 2p \rightarrow 3s \rightarrow 3p \rightarrow 4s \rightarrow 3d \rightarrow 4p \rightarrow 5s \rightarrow 4d \rightarrow 5p$Note: The $3d$ sub-level is higher in energy than the $4s$ sub-level, so $4s$ is filled before $3d$
• Hund's Rule: When filling orbitals of equal energy (degenerate orbitals in a sub-level), electrons occupy each orbital singly with parallel spins before pairing up to minimize electron-electron repulsion.

Electronic Configuration of Transition Metals and Ions

• Chromium ($\text{Cr}$, $Z=24$): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^5$
• Copper ($\text{Cu}$, $Z=29$): $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}$
• Formation of Ions: When d-block elements form positive ions, electrons are lost from the $4s$ subshell before the $3d$ subshell (e.g., $\text{Fe}$ is $[Ar]4s^2 3d^6$, while $\text{Fe}^{3+}$ is $[Ar]3d^5$).

Periodic Table Blocks

• $s$ block element: Outer electron occupies an $s$ sub-shell (e.g., Sodium: $1s^2 2s^2 2p^6 3s^1$).
• $p$ block element: Outer electron occupies a $p$ sub-shell (e.g., Chlorine: $1s^2 2s^2 2p^6 3s^2 3p^5$).
• $d$ block element: Outer electron occupies a $d$ sub-shell (e.g., Vanadium: $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3$).

Ionisation Energy

• First Ionisation Energy Definition: The first ionisation energy is the enthalpy change when one mole of gaseous atoms forms one mole of gaseous ions with a single positive charge.   $\text{H}_{(g)} \rightarrow \text{H}^+_{(g)} + e^-$
• Second Ionisation Energy Definition: The second ionisation energy is the enthalpy change when one mole of gaseous $1+$ ions forms one mole of gaseous $2+$ ions.   $\text{Ti}^+_{(g)} \rightarrow \text{Ti}^{2+}_{(g)} + e^-$

Factors Affecting Ionisation Energy

1. Nuclear Charge: Greater number of protons creates a stronger attraction for outer electrons.
2. Distance from Nucleus: Outer electrons further from the nucleus experience a weaker electrostatic force of attraction.
3. Shielding: Repulsion by filled inner electron shells reduces the net attraction felt by outer electrons.

Successive Ionisation Energies

• Successive ionisation energies for an element always increase because each subsequent electron is removed from a progressively more positive ion, strengthening the attraction on remaining electrons.
• A large jump between successive ionisation energies indicates the removal of an electron from an inner main shell closer to the nucleus with significantly less shielding.
  • Example: An element with ionisation energies of $590, 1150, 4940, 6480, 8120\text{\text{ kJ}}\text{ mol}^{-1}$ shows a large jump between the 2nd and 3rd ionisation energies. Therefore, it has 2 valence electrons and belongs to Group 2.

Periodicity in First Ionisation Energy

• Helium: Has the highest first ionisation energy because its electron is in the first shell closest to the nucleus, experiences no inner shell shielding, and has more protons than hydrogen.
• Trend Down a Group: First ionisation energy decreases down a group because outer electrons occupy shells further from the nucleus with increased shielding, overriding the increase in nuclear charge.
• Trend Across a Period: First ionisation energy generally increases across a period because nuclear charge increases while electrons are added to the same main shell (similar distance and shielding).
• Exceptions / Drops Across Periods 2 and 3:
  1. Drop between Group 2 and Group 3 ($\text{Mg} \rightarrow \text{Al}$): Magnesium has its outer electron in a $3s$ sub-level ($3s^2$), whereas Aluminum has its outer electron in a $3p$ sub-level ($3s^2 3p^1$). The $3p$ electron is higher in energy and partially shielded by the $3s$ electrons, making it easier to remove.
  2. Drop between Group 5 and Group 6 ($\text{P} \rightarrow \text{S}$): Phosphorus has the configuration $3s^2 3p^3$ (singly occupied $3p$ orbitals). Sulfur has the configuration $3s^2 3p^4$, containing a paired set of electrons in one $3p$ orbital. Inter-electron repulsion between the two paired electrons in the same orbital makes the outer electron easier to remove.

Mass Spectrometry and Time of Flight (TOF) Analysis

Principles of Mass Spectrometry

Mass spectrometers operate under a high vacuum to prevent gaseous sample ions from colliding with air particles, which would cause ionisation of air or deflection of the sample beam.

Steps in Time of Flight (TOF) Mass Spectrometry

1. Ionisation:
   • Electron Impact Ionisation:
     • The sample is vaporized and injected at low pressure.
     • An electron gun fires high-energy electrons at the sample, knocking off an outer electron to form positive ions: $\text{X}_{(g)} \rightarrow \text{X}^+_{(g)} + e^-$.
     • Used for elements and substances with low formula mass; causes fragmentation in larger organic molecules.
   • Electrospray Ionisation:
     • The sample is dissolved in a volatile, polar solvent and injected through a fine hypodermic needle connected to a high-voltage supply.
     • The sample molecules ($\text{M}$) gain a proton ($\text{H}^+$) from the solvent to form $\text{MH}^+$ ions: $\text{M}_{(g)} + \text{H}^+ \rightarrow \text{MH}^+_{(g)}$.
     • The solvent evaporates, leaving a fine aerosol of $\text{MH}^+$ ions.
     • Preferred for large organic molecules (e.g., proteins) because the 'softer' condition prevents fragmentation.
2. Acceleration:
   • Positive ions are accelerated by an electric field toward a negatively charged plate, giving all ions equal kinetic energy ($KE$).
   • $KE = \frac{1}{2} m v^2$
   • Where $KE$ is kinetic energy ($\text{J}$), $m$ is particle mass ($\text{kg}$), and $v$ is velocity ($\text{m s}^{-1}$).
3. Flight Tube (Ion Drift):
   • Ions pass through a hole in the negative plate into a field-free drift region of length $d$.
   • Because all ions have the same kinetic energy, their velocity depends on mass: lighter ions (smaller $m/z$) travel faster than heavier ions (larger $m/z$).
   • Time of flight ($t$) formula:      $t = \frac{d}{v} = d \times \frac{1}{\frac{2 KE}{m}} = d \times \frac{m}{2 KE}$$t = d \times \frac{m}{2 KE}$
4. Detection:
   • Positive ions strike a detector plate, acquiring electrons from the detector to generate a current.
   • The current magnitude is directly proportional to the abundance of that specific ion.

TOF Mass Spectrometer Calculation Example

• Problem: A sample of nickel contains the isotope $^{59}\text{Ni}$. The ions are accelerated to a kinetic energy of $1.000 \times 10^{-16}\text{ J}$ and travel through a flight tube $0.8000\text{ m}$ long. Calculate the flight time for a single $^{59}\text{Ni}^+$ ion ($L = 6.022 \times 10^{23}\text{ mol}^{-1}$).
• Step 1: Calculate mass of one ion in kilograms:   $\text{Mass in grams} = \frac{59}{6.022 \times 10^{23}} = 9.797 \times 10^{-23}\text{ g}$$\text{Mass in kg} = 9.797 \times 10^{-26}\text{ kg}$
• Step 2: Calculate flight time ($t$):   $t = d \times \frac{m}{2 KE}$$t = 0.8000 \times \frac{9.797 \times 10^{-26}}{2 \times 1.000 \times 10^{-16}}$$t = 1.771 \times 10^{-5}\text{ s}$

Relative Atomic Mass Calculations

Relative Atomic Mass ($A_r$) is the weighted average mass of an atom of an element relative to $\frac{1}{12}$th the mass of an atom of carbon-12.

• Formula for Percentage Abundances:   $A_r = \frac{\text{isotopic mass} \times \text{\text{\%}} \text{ abundance}}{100}$
• Formula for Relative Abundances:   $A_r = \frac{\text{isotopic mass} \times \text{relative abundance}}{\text{total relative abundance}}$

Worked Example 1: Magnesium Spectrum

• Data: $78.70\text{\%}$ of $^{24}\text{Mg}^+$, $10.13\text{\%}$ of $^{25}\text{Mg}^+$, $11.17\text{\%}$ of $^{26}\text{Mg}^+$.   $A_r = \frac{(78.70 \times 24) + (10.13 \times 25) + (11.17 \times 26)}{100} = 24.3$

Worked Example 2: Isotope Percentage Abundances

• Data: Copper has two isotopes $^{63}\text{Cu}$ and $^{65}\text{Cu}$. The $A_r$ is $63.55$. Calculate the percentage abundances.
• Calculation:   $63.55 = (y \times 63) + ((1 - y) \times 65)$$63.55 = 63 y + 65 - 65 y$$2 y = 1.45 \rightarrow y = 0.725$
  • Abundance of $^{63}\text{Cu} = 72.5\text{\%}$
  • Abundance of $^{65}\text{Cu} = 27.5\text{\%}$

Molecular Mass Spectra and Fragmentation

• Electron Impact: Molecular species fragment into smaller positive ions. The peak with the highest $m/z$ value corresponds to the unfragmented radical cation, called the molecular ion peak ($M^+$), which equals the relative molecular mass ($M_r$) of the compound.

• Electrospray Ionisation: No fragmentation occurs. The main peak represents the $\text{MH}^+$ ion. Subtract $1$ from the $m/z$ of the $\text{MH}^+$ peak to obtain the true $M_r$ of the molecule.
• Diatomic Halogen Spectra:
  • Chlorine ($^{35}\text{Cl} : ^{37}\text{Cl} = 3:1$) gives three diatomic $\text{Cl}_2^+$ peaks at $m/z = 70, 72, 74$ in a $9:6:1$ height ratio ($^{35}\text{Cl}^{35}\text{Cl}^+$, $^{35}\text{Cl}^{37}\text{Cl}^+$, $^{37}\text{Cl}^{37}\text{Cl}^+$).
  • Bromine ($^{79}\text{Br} : ^{81}\text{Br} = 1:1$) gives three diatomic $\text{Br}_2^+$ peaks at $m/z = 158, 160, 162$ in a $1:2:1$ height ratio. The $160$ peak is twice as high because $\text{Br}^{79}\text{-}\text{Br}^{81}$ can be formed in two combinations ($^{79}\text{Br}\text{-}^{81}\text{Br}$ and $^{81}\text{Br}\text{-}^{79}\text{Br}$).

Chemical Calculations, Mole Concept, and Solution Stoichiometry

Core Definitions

• The Mole: The amount of substance in grams that contains the same number of particles as there are atoms in 12 grams of carbon-12.
• Avogadro's Constant ($L$ or $N_A$): $6.022 \times 10^{23}\text{ mol}^{-1}$, representing the number of specified entities in one mole.   $\text{Number of particles} = \text{moles of substance} \times 6.022 \times 10^{23}$
• Relative Molecular Mass ($M_r$): Average mass of a molecule compared to $\frac{1}{12}$th of the mass of one atom of carbon-12.

Essential Equations for Calculations

1. Pure Solids, Liquids, and Gases:    $\text{moles } n = \frac{\text{mass (g)}}{M_r}$
2. Solutions:    $\text{concentration (mol dm}^{-3}\text{)} = \frac{\text{moles } n}{\text{volume (dm}^3\text{)}}$
3. Ideal Gas Equation:    $P V = n R T$
   • $P$ = Pressure in Pascals ($\text{Pa}$).
   • $V$ = Volume in cubic meters ($\text{m}^3$).
   • $n$ = Total moles of gas.
   • $R$ = Gas constant = $8.31\text{ J K}^{-1}\text{ mol}^{-1}$.
   • $T$ = Temperature in Kelvin ($\text{K}$).

Conversion Factors

• Mass: $1000\text{ mg} = 1\text{ g}$, $1000\text{ g} = 1\text{ kg}$, $1000\text{ kg} = 1\text{ tonne}$.
• Volume:   $\text{cm}^3 \rightarrow \text{dm}^3 : \text{divide by } 1000$$\text{cm}^3 \rightarrow \text{m}^3 : \text{divide by } 1{,}000{,}000$$\text{dm}^3 \rightarrow \text{m}^3 : \text{divide by } 1000$
• Temperature: {^\text{o}}\text{C} \rightarrow \text{K} : \text{add } 273$.\n- **Pressure**: ext{kPa} ightarrow ext{Pa} : ext{multiply by } 1000$.

Density Calculations

$\text{density} = \frac{\text{mass}}{\text{volume}}$

• Example: Calculate the number of molecules of ethanol in $0.500\text{ dm}^3$ of liquid ethanol ($\text{CH}_3\text{CH}_2\text{OH}$, density $= 0.789\text{ g cm}^{-3}$):   $\text{Volume} = 0.500\text{ dm}^3 = 500\text{ cm}^3$$\text{Mass} = \text{density} \times \text{volume} = 0.789 \times 500 = 394.5\text{ g}$$\text{Moles} = \frac{394.5}{46.0} = 8.576\text{ mol}$$\text{Number of molecules} = 8.576 \times 6.022 \times 10^{23} = 5.16 \times 10^{24}$

Empirical and Molecular Formulae

• Empirical Formula: The simplest whole-number ratio of atoms of each element in a compound.
• Molecular Formula: The actual number of atoms of each element in a molecule.

Determining Empirical Formula

1. Divide the mass or percentage mass of each element by its atomic mass ($A_r$) to obtain moles.
2. Divide each resulting mole value by the smallest calculated mole value.
3. Multiply ratios to obtain the simplest whole-number integers.

• Worked Example: A compound contains $1.82\text{ g}$ $\text{K}$, $5.93\text{ g}$ $\text{I}$, and $2.24\text{ g}$ $\text{O}$:
  • Moles $\text{K} = \frac{1.82}{39.1} = 0.0465$
  • Moles $\text{I} = \frac{5.93}{126.9} = 0.0467$
  • Moles $\text{O} = \frac{2.24}{16.0} = 0.140$
  • Divide by smallest ($0.0465$): $\text{K} = 1$, $\text{I} = 1$, $\text{O} = 3$.
  • Empirical Formula: $\text{KIO}_3$

Hydrated Salts and Water of Crystallisation

• Experiment - Heating in a Crucible:
  • Weigh an empty, clean, dry crucible and lid.
  • Add hydrated salt (e.g., $\text{CaSO}_4 \times x\text{H}_2\text{O}$) and reweigh.
  • Heat strongly with a Bunsen burner, cool, and reweigh.
  • Repeat heating and reweighing until constant mass is achieved (ensures complete loss of water).
  • Precautions: Using too small a mass ($< 0.100\text{ g}$) yields high percentage weighing uncertainties. Using too large a mass ($> 50\text{ g}$) leads to incomplete decomposition. A wet crucible causes an artificially high mass loss.
• Worked Example: $3.51\text{ g}$ of hydrated zinc sulfate ($\text{ZnSO}_4 \times x\text{H}_2\text{O}$) yields $1.97\text{ g}$ of anhydrous $\text{ZnSO}_4$ upon heating:   $\text{Mass of H}_2\text{O} = 3.51 - 1.97 = 1.54\text{ g}$$\text{Moles of ZnSO}_4 = \frac{1.97}{161.5} = 0.0122\text{ mol}$$\text{Moles of H}_2\text{O} = \frac{1.54}{18.0} = 0.0855\text{ mol}$$\text{Ratio } \frac{\text{H}_2\text{O}}{\text{ZnSO}_4} = \frac{0.0855}{0.0122} = 7$$x = 7 \rightarrow \text{ZnSO}_4 \times 7\text{H}_2\text{O}$

Solutions and Dilutions

• Mass Concentration: Measured in $\text{g dm}^{-3}$.   $\text{concentration (g dm}^{-3}\text{)} = \text{concentration (mol dm}^{-3}\text{)} \times M_r$
• Dissociation of Soluble Ionic Compounds: Soluble ionic solids dissociate completely into ions when dissolved, changing effective ion concentrations:   $\text{MgCl}_{2(s)} \rightarrow \text{Mg}^{2+}_{(aq)} + 2\text{Cl}^-_{(aq)}$   A $0.1\text{ mol dm}^{-3}$ $\text{MgCl}_2$ solution has a $0.1\text{ mol dm}^{-3}$ $\text{Mg}^{2+}$ concentration and a $0.2\text{ mol dm}^{-3}$ $\text{Cl}^-$ concentration.
• Standard Solution Preparation:
  • Weigh sample bottle containing solid on a $2$ d.p. balance.
  • Transfer solid to a beaker and reweigh sample bottle (weighing by difference).
  • Add $100\text{ cm}^3$ distilled water and stir with a glass rod until dissolved.
  • Transfer solution to a $250\text{ cm}^3$ volumetric flask using a funnel.
  • Rinse beaker, rod, and funnel with distilled water and add washings to the flask.
  • Make up to the mark with distilled water using a teat pipette so the bottom of the meniscus sits on the line.
  • Invert flask multiple times to ensure a uniform solution.
• Dilution Calculations:   $\text{New diluted concentration} = \text{original concentration} \times \frac{\text{original volume}}{\text{new diluted volume}}$

Gas Calculations

• Avogadro's Law: Equal volumes of gases under identical temperature and pressure conditions contain equal numbers of molecules.
• Molar Gas Volume: $1\text{ mole}$ of any gas occupies $24\text{ dm}^3$ at room temperature ($25^\text{o}\text{C}$) and pressure ($1\text{ atm}$).
• Changing Gas Conditions ($n$ remains constant):   $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Stoichiometry, Percentage Yield, and Atom Economy

$\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

$\text{percentage atom economy} = \frac{\text{mass of useful products}}{\text{mass of all reactants}} \times 100$

Note: Stoichiometric balancing numbers must be included when calculating atom economy.

Worked Example: Atom Economy

• Reaction: $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ (Desired product: $\text{Fe}$)   $\text{\% atom economy} = \frac{2 \times 55.8}{(2 \times 55.8 + 3 \times 16) + 3 \times (12 + 16)} \times 100 = 45.8\text{\%}$

Titrations and Practical Volumetric Analysis

Standard Titration Method

1. Rinse equipment: burette with titrant acid, volumetric pipette with alkali solution, conical flask with distilled water.
2. Pipette $25.0\text{ cm}^3$ of alkali into the conical flask. Touch the tip of the pipette to the liquid surface.
3. Fill the burette with acid, ensuring the jet space below the tap is completely filled and free of air bubbles.
4. Add a few drops ($2-3$) of indicator to the conical flask:
   • Phenolphthalein: Pink in alkali to colourless in acid (End point: pink just disappears).
   • Methyl Orange: Yellow in alkali to red in acid (End point: orange).
5. Place the flask on a white tile to observe color changes clearly.
6. Add acid while swirling constantly; add dropwise near the end point.
7. Record initial and final burette readings to $0.05\text{ cm}^3$.
8. Repeat titrations until at least two concordant results (titres within $0.10\text{ cm}^3$ of each other) are obtained.

Practical Points

• Conical Flask vs. Beaker: A conical flask is preferred because its sloping walls prevent splashing and liquid loss during swirling.
• Adding Distilled Water: Rinsing the sides of the conical flask with distilled water during titration ensures all reagents react. It does not alter the titre volume because water does not change the number of moles of acid/alkali present.
• Unfilled Jet Space: An air bubble in the burette jet space that fills during titration causes an artificially high titre reading.

Back Titrations

Used for insoluble solids (e.g., $\text{CaCO}_3$ in indigestion tablets) or reactions where end points are difficult to detect directly.

Worked Back Titration Example

• Data: A $950\text{ mg}$ impure $\text{CaCO}_3$ tablet reacted with $50.0\text{ cm}^3$ of $1.00\text{ mol dm}^{-3}$ $\text{HCl}$ (excess). The mixture was diluted to $100\text{ cm}^3$. A $10.0\text{ cm}^3$ sample required $11.1\text{ cm}^3$ of $0.300\text{ mol dm}^{-3}$ $\text{NaOH}$ for neutralisation. Calculate $\text{\%}$ mass of $\text{CaCO}_3$.
• Calculation:
  1. Moles $\text{NaOH}$ in titration $= 0.300 \times 0.0111 = 0.00333\text{ mol}$.
  2. Moles $\text{HCl}$ in $10.0\text{ cm}^3$ sample $= 0.00333\text{ mol}$.
  3. Moles $\text{HCl}$ remaining in $100\text{ cm}^3$ solution $= 0.00333 \times 10 = 0.0333\text{ mol}$.
  4. Initial moles $\text{HCl}$ added $= 1.00 \times 0.0500 = 0.0500\text{ mol}$.
  5. Moles $\text{HCl}$ reacted with $\text{CaCO}_3 = 0.0500 - 0.0333 = 0.0167\text{ mol}$.
  6. Equation: $\text{CaCO}_{3(s)} + 2\text{HCl}_{(aq)} \rightarrow \text{CaCl}_{2(aq)} + \text{CO}_{2(g)} + \text{H}_2\text{O}_{(l)}$      Moles $\text{CaCO}_3 = \frac{0.0167}{2} = 0.00835\text{ mol}$.
  7. Mass $\text{CaCO}_3 = 0.00835 \times 100.1 = 0.835\text{ g}$.
  8. $\text{\% purity} = \frac{0.835}{0.950} \times 100 = 87.9\text{\%}$.

Apparatus Uncertainties

$\text{\% uncertainty} = \frac{\text{uncertainty of apparatus}}{\text{measurement made}} \times 100$

• Readings vs. Measurements:
  • Reading: Single judgement value (e.g., thermometer, balance). Uncertainty is at least $\pm 0.5$ of the smallest division.
  • Measurement: Difference between two judgements (e.g., burette initial/final readings). Uncertainty is at least $\pm 1.0$ of the smallest division (e.g., two burette readings at $\pm 0.05\text{ cm}^3$ give an overall volume uncertainty of $\pm 0.10\text{ cm}^3$).
• Reducing Percentage Uncertainty: Use larger sample masses or larger titre volumes; use a 3 decimal place balance or equipment with higher resolution.

Chemical Bonding, Structure, and Molecular Shapes

Ionic Bonding

• Definition: The electrostatic force of attraction between oppositely charged ions formed by electron transfer.
• Factors Affecting Strength: Ionic bonds are stronger and melting points are higher when ions are smaller and have higher charges (e.g., $\text{MgO}$ has a higher melting point than $\text{NaCl}$ because $\text{Mg}^{2+}$ and $\text{O}^{2-}$ are smaller and more highly charged than $\text{Na}^+$ and $\text{Cl}^-$).
• Ionic Radii Trends in Isoelectronic Species:
  • Ions $\text{N}^{3-}, \text{O}^{2-}, \text{F}^-, \text{Na}^+, \text{Mg}^{2+}, \text{Al}^{3+}$ all possess the same electronic structure ($1s^2 2s^2 2p^6$, identical to $\text{Ne}$).
  • Proton count increases from $N (7)$ to $Al (13)$, increasing nuclear charge while electron shielding remains constant.
  • Effective nuclear pull per electron increases, pulling electrons closer and decreasing ionic radius.

Covalent Bonding

• Definition: A shared pair of electrons between non-metal atoms.
• Dative Covalent (Co-ordinate) Bonding: A covalent bond where both shared electrons originate from the same atom.
  • The donor atom must possess an available lone pair of electrons; the acceptor atom must be electron-deficient.
  • Represented diagrammatically by an arrow pointing from the donor atom to the acceptor atom.
  • Once formed, a dative covalent bond has identical properties and strength to an ordinary covalent bond.
  • Examples:
  • Ammonium ion ($\text{NH}_4^+$): Nitrogen donates its lone pair to an $\text{H}^+$ ion.
  • Hydronium ion ($\text{H}_3\text{O}^+$): Oxygen donates its lone pair to an $\text{H}^+$ ion.
  • Ammonia-boron trifluoride complex ($\text{H}_3\text{N}\rightarrow\text{BF}_3$): Nitrogen donates its lone pair to electron-deficient boron.

Metallic Bonding

• Definition: The electrostatic force of attraction between positive metal cations and delocalized outer-shell electrons.
• Factors Affecting Strength:
  1. Number of Protons: Greater nuclear charge creates stronger attraction for delocalized electrons.
  2. Number of Delocalized Electrons: More electrons per atom released into the sea of electrons increases bonding strength.
  3. Size of Cation: Smaller cations allow delocalized electrons to get closer to the positive nucleus, strengthening the bond (e.g., $\text{Mg}$ is stronger than $\text{Na}$).

Four Types of Crystal Structure

Detailed Crystal Descriptions

• Iodine ($\text{I}_2$): Molecular crystal consisting of $\text{I}_2$ molecules arranged regularly in a lattice, held together by weak Van der Waals forces.
• Ice ($\text{H}_2\text{O}$): Water molecules are held in a rigid 3D open network by hydrogen bonds, positioning molecules further apart than in liquid water, making ice less dense than water.
• Diamond: Each carbon atom forms four single covalent bonds to adjacent carbon atoms in a rigid tetrahedral arrangement. Bond angles are $109.5^\text{o}$.
• Graphite: Carbon atoms are arranged in planar hexagonal layers, with each carbon forming three covalent bonds. The fourth valence electron is delocalized along the layers, enabling conductivity. Layers are held together by weak Van der Waals forces, allowing them to slide easily.

Valence Shell Electron Pair Repulsion (VSEPR) Theory

Explanation Steps for Molecular Shapes

1. State the total number of electron pairs (bonding pairs and lone pairs) surrounding the central atom.
2. State that electron pairs repel each other and adopt positions as far apart as possible to minimize repulsion.
3. If no lone pairs are present, state that all electron pairs repel equally.
4. If lone pairs are present, state that lone pairs repel more than bonding pairs.
5. State the shape name and corresponding bond angle. Each lone pair reduces bond angles by approximately $2.5^\text{o}$.

Molecular Geometry Table

Electronegativity, Polarity, and Intermolecular Forces

Electronegativity

• Definition: Electronegativity is the relative power of an atom in a covalent bond to attract the electron density in that covalent bond toward itself.
• Pauling Scale: Ranges from $0$ to $4.0$. Fluorine is the most electronegative element with a value of $4.0$.
• Most Electronegative Elements: Fluorine ($\text{F}$), Oxygen ($\text{O}$), Nitrogen ($\text{N}$), and Chlorine ($\text{Cl}$).
• Periodic Trends:
  • Across a Period: Electronegativity increases due to increasing nuclear charge and decreasing atomic radius.
  • Down a Group: Electronegativity decreases due to increasing atomic radius and inner shell electron shielding.

Intermediate Bonding and Polarity

• Bond Types:
  • Pure Covalent: Elements have identical or near-identical electronegativities (\text{\Delta EN} \rightarrow 0).
  • Polar Covalent: Elements have moderate electronegativity differences ($0.3 \rightarrow 1.7$), causing uneven electron distribution ($\delta+$ and $\delta-$ ends).
  • Ionic: Elements have a large electronegativity difference ($> 1.7$).
• Polar Molecules and Symmetry:
  • A polar bond creates a permanent dipole.
  • If a molecule contains polar bonds but is completely symmetrical (e.g., $\text{CCl}_4, \text{CO}_2$), the individual bond dipoles cancel out, resulting in no net molecular dipole moment (non-polar molecule).
  • If the molecule is asymmetrical (e.g., $\text{CH}_3\text{Cl}$), the dipoles do not cancel, giving a net permanent dipole moment (polar molecule).

Intermolecular Forces

Intermolecular forces act between simple covalent molecules. They do not exist in giant ionic, metallic, or giant covalent structures.

1. Van der Waals (Induced Dipole-Dipole) Forces

• Occur between all simple covalent molecules and noble gas atoms.
• Mechanism: Electrons move constantly and randomly within orbitals. At any instant, electron density fluctuates, creating a temporary dipole. This temporary dipole induces an opposite dipole in an adjacent molecule, resulting in a weak attraction.
• Factors Affecting Magnitude:
  • Number of Electrons: Larger molecules with more electrons experience higher electron fluctuations, forming stronger induced dipoles and higher boiling points.
  • Surface Area / Shape: Unbranched, straight-chain alkanes have a larger surface contact area for Van der Waals interactions compared to compact, spherical branched alkanes, leading to higher boiling points.

2. Permanent Dipole-Dipole Forces

• Occur between polar molecules containing permanent dipoles (e.g., $\text{H-Cl}, \text{C=O}$).
• Permanent dipole-dipole attractions act in addition to Van der Waals forces, giving polar compounds higher boiling points than non-polar compounds of similar mass.

3. Hydrogen Bonding

• The strongest form of intermolecular force.
• Conditions Required: Occurs when a hydrogen atom is covalently bonded directly to one of the three most electronegative elements: Fluorine, Oxygen, or Nitrogen, each possessing an available lone pair of electrons.
• Diagram Requirements: Must explicitly show all lone pairs, partial charges ($\delta+$, $\delta-$), and the hydrogen bond line extending from a lone pair to the $\delta+$ hydrogen atom at an angle of $180^\text{o}$.

Boiling Point Trends of Hydrides

• Anomalously High Boiling Points ($\text{H}_2\text{O}, \text{HF}, \text{NH}_3$): Abnormally high compared to other group hydrides due to the presence of hydrogen bonding.
• Trend from $\text{H}_2\text{S} \rightarrow \text{H}_2\text{Te}$: Boiling points increase due to increasing molecular size and electron count, strengthening Van der Waals forces.

Energetics and Thermochemistry

Core Definitions

• Enthalpy Change ($\Delta H$): The amount of heat energy taken in or given out during any chemical change at constant pressure.
• Exothermic Change: Energy is transferred from the chemical system to the surroundings. The products have less energy than the reactants ($\Delta H$ is negative).
  • Examples: Combustion of fuels, oxidation of carbohydrates in respiration.
• Endothermic Change: Energy is transferred from the surroundings to the chemical system. The products have more energy than the reactants ($\Delta H$ is positive).
  • Examples: Thermal decomposition of calcium carbonate ($\text{CaCO}_3$).

Standard Conditions and Definitions

Standard conditions are:

• Pressure: $100\text{ kPa}$

• Temperature: $298\text{ K}$ ($25^\text{o}\text{C}$)

• Concentration of solutions: $1\text{ mol dm}^{-3}$

• Standard state: Physical state of a substance under $100\text{ kPa}$ and $298\text{ K}$.

• **Standard Enthalpy Change of Formation ($\Delta_f H^

\n\n)**: The enthalpy change when 1 mole of a compound is formed from its constituent elements under standard conditions, all reactants and products being in their standard states.\n   ext{Mg}{(s)} + ext{Cl}{2(g)} ightarrow ext{MgCl}_{2(s)}\n  *Note*: The enthalpy of formation of any element in its standard state is 0 ext{ kJ mol}^{-1}.\n- **Standard Enthalpy Change of Combustion ($\Delta_c H^\n\n

)*: The enthalpy change when 1 mole of a substance is burned completely in excess oxygen under standard conditions, all reactants and products being in their standard states.   $\text{CH}_{4(g)} + 2\text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)}$Note*: Incomplete combustion produces carbon monoxide or soot (carbon) and releases less energy.

Calorimetry and Experimental Enthalpy Determination

Calorimetric Equations

Energy change in calorimeter: Q = m \times c_p \times \text{\Delta} T

• $Q$ = Heat energy change ($\text{J}$).
• $m$ = Mass of solution or water ($\text{g}$) (assuming density $= 1\text{ g cm}^{-3}$, so $25\text{ cm}^3 = 25\text{ g}$).
• $c_p$ = Specific heat capacity ($\text{J g}^{-1}\text{ K}^{-1}$) (assumed equal to pure water, $4.18\text{ J g}^{-1}\text{ K}^{-1}$).
• $\Delta T$ = Temperature change ($\text{K}$ or ${^\text{o}}\text{C}$).

Enthalpy change per mole ($\Delta H$): $\Delta H = \frac{Q}{\text{moles of limiting reactant} \times 1000} \text{ (kJ mol}^{-1}\text{)}$

Sign Rule: If temperature increases, the reaction is exothermic ($\Delta H$ is negative).

Extrapolation Method for Temperature Corrections

For slow reactions where heat loss occurs simultaneously, temperature readings are taken at regular time intervals and extrapolated on a graph back to the time of reactant mixing.

• Procedure:
  1. Measure reagent initial temperature every minute for $3$ minutes.
  2. Add reactants at minute $4$ (do not record temperature at minute $4$).
  3. Record temperature every minute from minute $5$ for several minutes.
  4. Plot temperature versus time, draw a line of best fit through cooling points, and extrapolate back to minute $4$ to find the theoretical maximum temperature change ($\Delta T$).

Errors in Solution Calorimetry

• Heat loss to surroundings.
• Assuming solution heat capacity equals pure water ($4.18\text{ J g}^{-1}\text{ K}^{-1}$).
• Neglecting heat energy absorbed by the calorimeter apparatus.
• Incomplete reaction or slow dissolving rates.

Hess's Law

Hess's Law: The total enthalpy change for a chemical reaction is independent of the route by which the chemical change takes place.

• Enthalpy of Reaction from Enthalpies of Formation:   \Delta H_{\text{reaction}} = \text{\Sigma} \Delta_f H(\text{products}) - \text{\Sigma} \Delta_f H(\text{reactants})
• Enthalpy of Reaction from Enthalpies of Combustion:   \Delta H_{\text{reaction}} = \text{\Sigma} \Delta_c H(\text{reactants}) - \text{\Sigma} \Delta_c H(\text{products})

Bond Enthalpies

• Mean Bond Energy Definition: The enthalpy change required to break one mole of a specified covalent bond in gaseous molecules, averaged over a range of different compounds.   \Delta H = \text{\Sigma} (\text{bond energies broken}) - \text{\Sigma} (\text{bond energies formed})
• Values calculated using mean bond enthalpies are less accurate than those from formation/combustion cycles because mean bond values are averaged over many different environments.

Reaction Kinetics and Collision Theory

Collision Theory and Activation Energy

• Collision Theory: Chemical reactions occur only when colliding particles possess energy equal to or greater than the activation energy.
• Activation Energy ($E_A$): The minimum energy required by colliding particles to start a reaction by breaking bonds.

Maxwell-Boltzmann Energy Distribution

Shows the distribution of molecular kinetic energies present in a gas or liquid at a specific temperature.

• Key Features:
  • Passes through the origin $(0,0)$ because no molecules have zero energy.
  • The peak represents the most probable energy ($E_{mp}$).
  • The mean energy lies to the right of the peak.
  • The curve never touches the x-axis at high energy because there is no theoretical upper limit to molecular energy.
  • The total area under the curve represents the total number of particles present.

Effect of Temperature

• As temperature increases, average kinetic energy increases, shifting the peak to the right and lowering its height.
• Total area under the curve remains constant.
• A significantly larger fraction of molecules possess energy $\ge E_A$, increasing successful collision frequency and reaction rate.

Effect of Concentration and Pressure

• Increasing concentration increases the number of particles per unit volume.
• The shape and peak ($E_{mp}$) of the distribution remain unchanged, but the curve height increases, enlarging the total area.
• Collision frequency increases, increasing reaction rate.

Effect of Catalysts

• Definition: A catalyst increases the rate of a reaction without being chemically consumed by providing an alternative reaction pathway with a lower activation energy ($E_{A,\text{cat}}$).