A-Level Chemistry Revision Guide: Physical and Inorganic Chemistry

Redox and Standard Electrode Potential

• Definitions in Redox Chemistry:

  • Oxidation: The loss of electrons or an increase in oxidation number.
  • Reduction: The gain of electrons or a decrease in oxidation number.
  • Oxidising Agent: A species that accepts electrons, thereby facilitating oxidation in another species. It becomes reduced itself in the process.
  • Reducing Agent: A species that donates electrons, thereby facilitating reduction in another species. It becomes oxidised itself in the process.

• Rules for Assigning Oxidation Numbers:

  1. All uncombined elements have an oxidation number of zero (including diatomic molecules such as $\text{H}_2$ or $\text{Cl}_2$).
  2. Hydrogen in compounds has an oxidation number of $+1$, except when combined with a Group 1 metal in metal hydrides, where it is $-1$.
  3. Oxygen in compounds has an oxidation number of $-2$, except in peroxides where it is $-1$, or when bonded to fluorine, where it is $+2$
  4. Group 1 metals always have an oxidation number of $+1$, and Group 2 metals always have an oxidation number of $+2$
  5. The sum of the oxidation numbers of all atoms in a neutral compound must equal zero. In a polyatomic ion, the sum must equal the overall charge of the ion.
  6. In covalent bonds between different elements, the negative oxidation number is assigned to the more electronegative element.

• Determining Redox Reactions via Oxidation States:

  • An element is oxidised if its oxidation number increases.
  • An element is reduced if its oxidation number decreases.
  • Notation Distinction: Ions carry their charge sign at the end (e.g., $2-$), whereas oxidation numbers carry their sign at the beginning (e.g., $-2$).

• Worked Example: Oxidation of Iron(II) by Dichromate Ions:

  • Equation:     $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$
  • Oxidation state analysis:
  • Reactants: $\text{Cr} = +6$, $\text{O} = -2$, $\text{H} = +1$, $\text{Fe} = +2$
  • Products: $\text{Cr} = +3$, $\text{O} = -2$, $\text{H} = +1$, $\text{Fe} = +3$
  • Conclusion: Chromium is reduced from $+6$ down to $+3$. Iron is oxidised up from $+2$ to $+3$

• Constructing and Balancing Half-Equations:

  • Redox processes can be decomposed into an oxidation half-equation and a reduction half-equation.
  • Example reaction: Zinc strip immersed in copper(II) sulfate solution:     $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$
  • Oxidation half-equation:     $\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-$
  • Reduction half-equation:     $\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)$
  • Balancing Check: The net charge on both sides of a half-equation must balance.

• Electrochemical Cells and Half-Cells:

  • A half-cell contains the oxidized and reduced forms of an element or ion species.

  • Joining two distinct half-cells creates an electrochemical cell capable of producing an electromotive force (EMF).      

  • Components:

  • Electrodes: Conductive media allowing electron flow through an external circuit.

  • Salt Bridge: A tube containing an electrolyte (e.g., $\text{KCl}$) that completes the circuit by permitting ion migration without direct mixing of solutions.

  • High-Resistance Voltmeter: Measures the EMF of the cell without drawing significant current.

• Three Primary Types of Half-Cells:

  1. Metal / Metal Ion Half-Cell: Consists of a metal strip immersed in an aqueous solution of its ions at $1.0\,mol\,dm^{-3}$ (e.g., $\text{Zn}(s)/\text{Zn}^{2+}(aq)$).
  2. Gas / Non-Metal Ion Half-Cell: Uses an inert conductor such as platinum. Gas at $1\,atm$ is bubbled over the electrode submerged in an aqueous solution of its corresponding ions (e.g., $\text{H}_2(g)/\text{H}^+(aq)$).
  3. Solution Half-Cell (Metal in Two Oxidation States): Common for transition elements with multiple stable oxidation states (e.g., $\text{Fe}^{2+}(aq)/\text{Fe}^{3+}(aq)$). Utilises an inert platinum electrode. Associated with distinct color changes (e.g., $\text{Fe}^{2+}$ is pale green, $\text{Fe}^{3+}$ is rusty orange).

• Standard Cell Diagrams and Conventions:

  • Cell shorthand representation:     $\text{Zn}(s) \mid \text{Zn}^{2+}(aq) \parallel \text{Cu}^{2+}(aq) \mid \text{Cu}(s)$
  • Notation Rules:
  • Leftmost species is the solid conducting metal of the left-hand cell.
  • Single vertical line ($\mid$) represents a phase boundary (e.g., solid to aqueous solution).
  • Double vertical lines ($\parallel$) represent the salt bridge.
  • Commas (,) separate species in the same phase inside a single half-cell.
  • Rightmost species is the standard conducting electrode of the more positive half-cell.

• Standard Electrode Potentials ($E^\theta$):

  • Definition: The potential difference of a half-cell connected to a Standard Hydrogen Electrode under standard conditions.
  • Standard Conditions:
  • Temperature: $298\,K$ ($25^\circ\text{C}$).
  • Solution Concentration: $1.0\,mol\,dm^{-3}$ for all aqueous ions.
  • Gas Pressure: $1.0\,atm$ ($101\,kPa$) for any gases.

• The Standard Hydrogen Electrode (SHE):

  • Assigned a standard potential of exactly $0.00\,V$
  • Setup: Hydrogen gas at $1\,atm$ bubbled over a black platinum electrode immersed in a $1.0\,mol\,dm^{-3}\,\text{H}^+$ solution (typically hydrochloric acid) at $298\,K$
  • Platinum black is used to maximize surface area for adsorption.

• The Electrochemical Series and EMF Calculations:

  • More negative $E^\theta$ values indicate stronger reducing power (greater ease of losing electrons / oxidation).
  • More positive $E^\theta$ values indicate stronger oxidising power (greater ease of gaining electrons / reduction).
  • Calculating Cell EMF:$\text{EMF}_{\text{cell}} = E^\theta(\text{reduction / most positive}) - E^\theta(\text{oxidation / most negative})$
  • A cell reaction is thermodynamically feasible if its calculated EMF is positive ($\text{EMF}_{\text{cell}} > 0$).

• Worked Examples for Feasibility and Cell Diagrams:

  • Example 1: Feasibility of Chloride Reducing $\text{Zn}^{2+}$

  • Given half-reactions:       $\text{Cl}_2(aq) + 2e^- \rightleftharpoons 2\text{Cl}^-(aq) \quad E^\theta = +1.36\,V$$\text{Zn}^{2+}(aq) + 2e^- \rightleftharpoons \text{Zn}(s) \quad E^\theta = -0.76\,V$

  • Targeted reduction: $\text{Zn}^{2+} \rightarrow \text{Zn}$ ($-0.76\,V$).

  • Targeted oxidation: $2\text{Cl}^- \rightarrow \text{Cl}_2$ ($+1.36\,V$).

  • Calculation:       $\text{EMF} = -0.76 - (+1.36) = -2.12\,V$

  • Conclusion: EMF is negative, so the reaction is not feasible.

  • Example 2: Halide Reduction of $\text{Fe}^{3+}$ and Cell Diagram

  • Given half-reactions:       $\frac{1}{2}\text{I}_2(s) + e^- \rightleftharpoons \text{I}^-(aq) \quad E^\theta = +0.54\,V$$\text{Fe}^{3+}(aq) + e^- \rightleftharpoons \text{Fe}^{2+}(aq) \quad E^\theta = +0.77\,V$$\frac{1}{2}\text{Br}_2(l) + e^- \rightleftharpoons \text{Br}^-(aq) \quad E^\theta = +1.09\,V$$\frac{1}{2}\text{Cl}_2(g) + e^- \rightleftharpoons \text{Cl}^-(aq) \quad E^\theta = +1.36\,V$$\text{Ce}^{4+}(aq) + e^- \rightleftharpoons \text{Ce}^{3+}(aq) \quad E^\theta = +1.45\,V$

  • (i) Identifying the reducing halide for $\text{Fe}^{3+}$:

    • Reduction half-reaction: $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$ ($+0.77\,V$).
    • To yield a positive EMF, the halide oxidation half-reaction must have $E^\theta < +0.77\,V$.
    • Iodide ($\text{I}^-$) is the only valid halide ($E^\theta = +0.54\,V$).
    • $\text{EMF} = +0.77 - (+0.54) = +0.23\,V$ (Feasible).

  • (ii) Cell diagram combining $\text{Fe}^{3+}/\text{Fe}^{2+}$ and $\text{Ce}^{4+}/\text{Ce}^{3+}$:

    • Cerium cell is more positive ($+1.45\,V$) so it undergoes reduction and sits on the right.
    • Iron cell undergoes oxidation: $\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-$.
    • Cell diagram:         $\text{Pt}(s) \mid \text{Fe}^{2+}(aq), \text{Fe}^{3+}(aq) \parallel \text{Ce}^{4+}(aq), \text{Ce}^{3+}(aq) \mid \text{Pt}(s)$
    • Calculated EMF:         $\text{EMF} = +1.45 - (+0.77) = 0.68\,V$

• Fuel Cells:

  • Fuel cells directly convert chemical energy into electrical energy efficiently.

• Hydrogen-Oxygen Fuel Cell Reactions:
  • Anode (Oxidation):       $\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-$
  • Cathode (Reduction):       $\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}$
  • Overall Reaction:       $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$
• Advantages vs Disadvantages:
  • Advantages:
    1. Produces only water as a byproduct; no $\text{CO}_2$ greenhouse gas emissions.
    2. Higher efficiency with far less energy lost as heat compared to internal combustion engines.
    3. Hydrogen can be generated cleanly via the electrolysis of water.
  • Disadvantages:
    1. Hydrogen gas is highly flammable, posing storage and transportation hazards.
    2. Standard industrial hydrogen production currently relies on fossil fuels, causing net energy loss.

Redox Reactions

• Steps to Construct and Combine Half-Equations:

  1. Balance all atoms other than hydrogen and oxygen.
  2. Balance oxygen atoms by adding $\text{H}_2\text{O}$.
  3. Balance hydrogen atoms by adding $\text{H}^+$ ions.
  4. Balance electric charges by adding electrons ($e^-$).
  5. Multiply each half-equation by appropriate integers so electron numbers equalise, then add half-equations together and cancel species appearing on both sides.

• Worked Examples of Balancing Half-Equations:

  • Example 1: Oxidation of Ethanedioate by Acidified Manganate(VII)
  • Manganate(VII) reduction half-equation:       $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$
  • Ethanedioate oxidation half-equation:       $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2e^-$
  • Equalising electrons ($2 \times$ reduction equation, $5 \times$ oxidation equation):       $2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$$5\text{C}_2\text{O}_4^{2-} \rightarrow 10\text{CO}_2 + 10e^-$
  • Overall Ionic Equation:       $2\text{MnO}_4^- + 16\text{H}^+ + 5\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O} + 10\text{CO}_2$

• Common Redox Titrations:

  • Acidified Manganate(VII) Titrations:
  • Equation: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 5\text{Fe}^{3+}$
  • Stoichiometric Ratio: $1\,\text{MnO}_4^- \equiv 5\,\text{Fe}^{2+}$
  • End-point color change: Purple to persistent light pink.
  • Acidified Dichromate(VI) Titrations:
  • Equation: $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}$
  • Stoichiometric Ratio: $1\,\text{Cr}_2\text{O}_7^{2-} \equiv 6\,\text{Fe}^{2+}$
  • Color change: Orange to green.

• Worked Example: Manganate(VII) Titration Calculation:

  • Four $25.0\,cm^3$ samples of $\text{C}_2\text{O}_4^{2-}$ solution were titrated against $0.0200\,mol\,dm^{-3}\,\text{KMnO}_4$.
  • Titres obtained: $28.80\,cm^3$, $27.95\,cm^3$, $28.00\,cm^3$, $27.80\,cm^3$
  • Mean concordant titre calculation (excluding initial rough titre $28.80\,cm^3$):     $\text{Mean titre} = \frac{27.95 + 28.00 + 27.80}{3} = 27.92\,cm^3 = 0.02792\,dm^3$
  • Moles of $\text{MnO}_4^-$ used:     $n(\text{MnO}_4^-) = 0.0200\,mol\,dm^{-3} \times 0.02792\,dm^3 = 5.584 \times 10^{-4}\,mol$
  • Moles of $\text{C}_2\text{O}_4^{2-}$ ($2 : 5$ mole ratio):     $n(\text{C}_2\text{O}_4^{2-}) = \frac{5.584 \times 10^{-4}}{2} \times 5 = 1.396 \times 10^{-3}\,mol$
  • Concentration of $\text{C}_2\text{O}_4^{2-}$:     $\text{Concentration} = \frac{1.396 \times 10^{-3}\,mol}{0.0250\,dm^3} = 0.0558\,mol\,dm^{-3}$

• Determination of Copper(II) Ions via Thiosulfate Titration:

  • Indirect Method: Excess potassium iodide ($\text{KI}$) added to a blue solution of $\text{Cu}^{2+}$, producing solid copper(I) iodide and aqueous iodine:     $2\text{Cu}^{2+}(aq) + 4\text{I}^-(aq) \rightarrow 2\text{CuI}(s) + \text{I}_2(aq)$
  • Liberated $\text{I}_2$ is titrated against sodium thiosulfate ($\text{Na}_2\text{S}_2\text{O}_3$):     $2\text{S}_2\text{O}_3^{2-}(aq) + \text{I}_2(aq) \rightarrow \text{S}_4\text{O}_6^{2-}(aq) + 2\text{I}^-(aq)$
  • Combined Stoichiometric Relationship:     $2\text{Cu}^{2+} \equiv \text{I}_2 \equiv 2\text{S}_2\text{O}_3^{2-} \implies 1\text{Cu}^{2+} \equiv 1\text{S}_2\text{O}_3^{2-}$
  • Worked Example Calculation:
  • Solution volume of $\text{Cu}^{2+} = 25.0\,cm^3$.
  • Thiosulfate volume = $30.25\,cm^3$, Concentration = $0.250\,mol\,dm^{-3}$.
  • Moles of $\text{S}_2\text{O}_3^{2-} = 0.250 \times 0.03025 = 0.0075625\,mol$.
  • Moles of $\text{Cu}^{2+} = 0.0075625\,mol$.
  • Concentration of $\text{Cu}^{2+} = \frac{0.0075625}{0.0250} = 0.303\,mol\,dm^{-3}$.

• Worked Example: Determination of Copper in Ore Sample (EDTA Titration):

  • $19.20\,cm^3$ of $0.010\,mol\,dm^{-3}$ EDTA reacted with $50.00\,cm^3$ of $\text{Cu}^{2+}$ solution (1:1 mole ratio). Ore sample mass = $11.56\,g$ dissolved in $1.00\,dm^3$.
  • (i) Moles of EDTA = $0.010 \times 0.01920 = 0.000192\,mol$
  • (ii) Moles of $\text{Cu}^{2+}$ in $50.00\,cm^3$ = $0.000192\,mol$
  • (iii) Concentration of $\text{Cu}^{2+}$ in $g\,dm^{-3}$:     $\text{Molar conc} = \frac{0.000192}{0.0500} = 0.00384\,mol\,dm^{-3}$$\text{Mass conc} = 0.00384\,mol\,dm^{-3} \times 63.5\,g\,mol^{-1} = 0.244\,g\,dm^{-3}$
  • Percentage of copper in sample:     $\text{Percentage} = \left( \frac{0.244}{11.56} \right) \times 100 = 2.11\%$

Chemistry of the p-block

• p-Block Electronic Configurations:

  • Group 3: $ns^2 np^1$
  • Group 4: $ns^2 np^2$
  • Group 5: $ns^2 np^3$

• Oxidation States Trends:

  • Display two prominent oxidation states: the group number oxidation state and a lower state two units lower.
  • Group 3: $+3$ and $+1$
  • Group 4: $+4$ and $+2$
  • Group 5: $+5$ and $+3$
  • Stability pattern: Higher oxidation states are most stable at the top of a group; lower oxidation states become progressively more stable descending the group.

• Octet Expansion:

  • Period 2 elements are strictly limited to an octet (8 electrons) in their valence shell due to the absence of accessible d-orbitals.
  • Period 3 and heavier elements can expand their octet by utilizing accessible d-orbitals to form more than 4 covalent bonds.
  • Examples: $\text{PCl}_5$ (10 outer electrons), $\text{SF}_6$ (12 outer electrons), $\text{H}_2\text{SO}_4$, $\text{ClO}_4^-$.

• Inert Pair Effect:

  • The tendency of the outer $s^2$ electron pair to remain unbonded as you descend the p-block groups.
  • Caused by decreasing covalent bond energies down a group; the energy recovered from forming two additional bonds is insufficient to promote an s-electron to a p-orbital.
  • Example: $\text{Pb}^{2+}$ is more stable than $\text{Pb}^{4+}$; $\text{SnCl}_4$ is stable, but $\text{PbCl}_2$ is more stable than $\text{PbCl}_4$.

• Metallic and Amphoteric Character:

  • Ionisation energy decreases down p-block groups, causing a transition from non-metallic to metallic character.
  • Oxides along the metal/non-metal boundary display amphoteric behavior (reacting with both acids and bases).
  • Reactions of Amphoteric Oxides:
  • Aluminium Oxide ($\text{Al}_2\text{O}_3$):$\text{Al}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{AlCl}_3 + 3\text{H}_2\text{O}$$\text{Al}_2\text{O}_3 + 2\text{NaOH} + 3\text{H}_2\text{O} \rightarrow 2\text{Na}[\text{Al}(\text{OH})_4]$
  • Zinc Oxide ($\text{ZnO}$):$\text{ZnO} + 2\text{HNO}_3 \rightarrow \text{Zn}(\text{NO}_3)_2 + \text{H}_2\text{O}$$\text{ZnO} + 2\text{NaOH} + \text{H}_2\text{O} \rightarrow \text{Na}_2[\text{Zn}(\text{OH})_4]$
  • Lead(II) Oxide ($\text{PbO}$):$\text{PbO} + 2\text{HNO}_3 \rightarrow \text{Pb}(\text{NO}_3)_2 + \text{H}_2\text{O}$$\text{PbO} + 2\text{NaOH} + \text{H}_2\text{O} \rightarrow \text{Na}_2[\text{Pb}(\text{OH})_4]$

• Group 3 Chemistry (Boron and Aluminium):

  • Compounds such as $\text{BF}_3$, $\text{BCl}_3$, and $\text{AlCl}_3$ form 3 covalent bonds, leaving them with only 6 valence electrons (electron deficient).

  • They readily accept a lone pair to form a co-ordinate (dative) bond.

  • Aluminium Chloride Dimer ($\text{Al}_2\text{Cl}_6$):

  • At lower temperatures, two $\text{AlCl}_3$ monomers combine via two dative covalent bonds from chlorine lone pairs to aluminium, forming $\text{Al}_2\text{Cl}_6$. It also forms $\text{AlCl}_4^-$.

  • Donor-Acceptor Complexes: e.g., $\text{BF}_3\text{NH}_3$, where nitrogen donates its lone pair to boron's vacant orbital.

• Boron Nitride (BN):

  • Isoelectronic with carbon species ($\text{B} + \text{N}$ has 12 valence electrons, matching $\text{C} + \text{C}$).

• Cubic Boron Nitride (c-BN):

  • Diamond-like 3D tetrahedral network.
  • Extremely hard with a high melting point; used as industrial abrasives and wear-resistant coatings.

• Hexagonal Boron Nitride (h-BN):

  • Graphite-like planar hexagonal sheets.
  • Layers lie directly above one another (unlike staggered carbon in graphite).
  • Contains no delocalised electrons (localized lone pairs on nitrogen), rendering it an electrical insulator.
  • Weak intermolecular forces between layers allow sliding, acting as a high-temperature lubricant.

• BN Nanotopes: Can be wrapped around carbon nanotubes to provide nanoscale electrical insulation.

  • Group 4 Chemistry (Carbon, Silicon, Lead):

• Oxides:

  • $\text{CO}$ ($+2$ state): Strong reducing agent. Reduces $\text{Fe}_2\text{O}_3$ in blast furnace: $\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$
  • $\text{CO}_2$ ($+4$ state): Acidic oxide. Reacts with bases: $\text{CO}_2 + 2\text{NaOH} \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O}$
  • $\text{PbO}$ ($+2$ state): Stable yellow/orange amphoteric solid.
  • $\text{PbO}_2$ ($+4$ state): Unstable, strong oxidising agent; readily reduced to $\text{Pb}^{2+}$.

• Tetrachlorides:

  • $\text{CCl}_4$: Liquid, non-polar covalent. Does not react with water; forms a separate immiscible layer due to lack of available d-orbitals.
  • $\text{SiCl}_4$: Liquid, covalent. Reacts violently with water via hydrolysis:       $\text{SiCl}_4(l) + 2\text{H}_2\text{O}(l) \rightarrow \text{SiO}_2(s) + 4\text{HCl}(g)$       Hydrolysis occurs because silicon has vacant $3d$ orbitals accessible for nucleophilic attack by oxygen lone pairs.
  • $\text{PbCl}_2$: White ionic solid, insoluble in cold water.

• Reactions of Aqueous Lead(II) Ions ($\text{Pb}^{2+}$):

  • $\text{OH}^-(aq)$: Forms white precipitate of $\text{Pb}(\text{OH})_2$. Dissolves in excess $\text{OH}^-$ to form colourless tetrahydroxyplumbate(II) ions $[\text{Pb}(\text{OH})_4]^{2-}$.

  • $\text{Cl}^-(aq)$: Forms a heavy white precipitate of $\text{PbCl}_2$.

  • $\text{I}^-(aq)$: Forms a characteristic heavy, bright yellow precipitate of lead(II) iodide ($\text{PbI}_2$).

  • Group 7 Chemistry (Halogens):

• Oxidising Power Trend: Decreases down Group 7 ($\text{Cl}_2 > \text{Br}_2 > \text{I}_2$).

• Reducing Power Trend: Halide reducing power increases down Group 7 ($\text{I}^- > \text{Br}^- > \text{Cl}^-$).

• Reactions of Solid Sodium Halides with Concentrated $\text{H}_2\text{SO}_4$:

  1. Sodium Chloride ($\text{NaCl}$):$\text{NaCl}(s) + \text{H}_2\text{SO}_4(conc) \rightarrow \text{NaHSO}_4(aq) + \text{HCl}(g)$        Acid-base reaction only; steamy fumes of $\text{HCl}$. No redox occurs because $\text{Cl}^-$ cannot reduce conc. $\text{H}_2\text{SO}_4$
  2. Sodium Bromide ($\text{NaBr}$):$\text{NaBr}(s) + \text{H}_2\text{SO}_4(conc) \rightarrow \text{NaHSO}_4(aq) + \text{HBr}(g)$$2\text{HBr}(g) + \text{H}_2\text{SO}_4(conc) \rightarrow \text{SO}_2(g) + \text{Br}_2(g) + 2\text{H}_2\text{O}(l)$$\text{Br}^-$ reduces sulfur from $+6$ to $+4$ ($\text{SO}_2$); brown fumes of $\text{Br}_2$ observed.
  3. Sodium Iodide ($\text{NaI}$):$\text{NaI}(s) + \text{H}_2\text{SO}_4(conc) \rightarrow \text{NaHSO}_4(aq) + \text{HI}(g)$$\text{I}^-$ is a strong reducing agent and reduces $\text{H}_2\text{SO}_4$ down to $\text{SO}_2$ ($+4$), free Sulfur ($0$), and $\text{H}_2\text{S}$ ($-2$).        Observations: Purple fumes of $\text{I}_2$, yellow solid sulfur, rotten-egg gas smell ($\text{H}_2\text{S}$), steamy fumes of $\text{HI}$.

• Disproportionation Reactions of Chlorine with $\text{NaOH}$:

  • Cold, dilute $\text{NaOH}$ ($15^\circ\text{C}$):$\text{Cl}_2(g) + 2\text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{NaClO}(aq) + \text{H}_2\text{O}(l)$       Oxidation states: $\text{Cl}$ goes from $0$ to $-1$ ($\text{NaCl}$) and $+1$ ($\text{NaClO}$).
  • Hot, concentrated $\text{NaOH}$ ($70^\circ\text{C}$):$3\text{Cl}_2(g) + 6\text{NaOH}(aq) \rightarrow 5\text{NaCl}(aq) + \text{NaClO}_3(aq) + 3\text{H}_2\text{O}(l)$       Oxidation states: $\text{Cl}$ goes from $0$ to $-1$ ($\text{NaCl}$) and $+5$ ($\text{NaClO}_3$).

• Water Treatment and Bleaching: Chlorine and $\text{ClO}^-$ act as powerful oxidising agents that destroy bacterial DNA and oxidise organic dye chromophores into colourless forms.

Chemistry of the d-block Transition Metals

• Transition Metal Definition:

  • A d-block element that forms at least one stable ion with a partially filled d-orbital.
  • Scandium and Zinc: Not transition metals. $\text{Sc}^{3+}$ has an empty d-shell ($3d^0$); $\text{Zn}^{2+}$ has a completely full d-shell ($3d^{10}$).
  • Copper: Is a transition metal because $\text{Cu}^{2+}$ has a partially filled d-shell ($3d^9$).

• Electron Configurations:

  • $\text{4s}$ orbitals fill before $\text{3d}$ orbitals, but $\text{4s}$ electrons are lost first during ionisation.
  • Transition metal cations always possess an empty $\text{4s}$ shell.
  • Anomalies: Chromium ($1s^2 2s^2 2p^6 3s^2 3p^6 3d^5 4s^1$) and Copper ($1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$).

• Common Oxidation States:

  • Chromium: $+3$, $+6$
  • Manganese: $+2$, $+4$, $+7$
  • Iron: $+2$, $+3$
  • Cobalt: $+2$, $+3$
  • Copper: $+1$, $+2$

• Complexes and Ligands:

  • Ligand: A species possessing a lone pair of electrons that forms a co-ordinate bond to a central metal cation.
  • Monodentate: Donates one lone pair per ligand (e.g., $\text{H}_2\text{O}$, $\text{NH}_3$, $\text{Cl}^-$, $\text{CN}^-$).
  • Bidentate: Donates two lone pairs per ligand (e.g., 1,2-diaminoethane, ethanedioate).
  • Geometries:
  • Octahedral (6-coordinate): Most common. Bond angles $90^\circ$
    • Examples: $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$ (pale green), $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$ (yellow), $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ (pale blue), $[\text{Cr}(\text{H}_2\text{O})_6]^{3+}$ (dark green), $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$ (pink), $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ (deep royal blue).
  • Tetrahedral (4-coordinate): Formed with larger ligands like $\text{Cl}^-$. Bond angles $109.5^\circ$
    • Examples: $[\text{CuCl}_4]^{2-}$ (yellow-green), $[\text{CoCl}_4]^{2-}$ (blue).

• Origin of Colour in Transition Metal Complexes:

  • Free gaseous metal ions have 5 degenerate $\text{3d}$ orbitals.
  • In a complex, surrounding ligands create an electrostatic field that splits the $\text{3d}$ orbitals into two non-degenerate energy levels (3 lower, 2 higher).
  • Electrons in lower $\text{3d}$ orbitals absorb a specific frequency ($f$) of visible light and jump to higher $\text{3d}$ orbitals:     $\Delta E = h f$
  • The colour transmitted/observed is complementary to the absorbed frequency.
  • Ions with empty d-shells ($\text{Sc}^{3+}, \text{Ti}^{4+}$) or full d-shells ($\text{Zn}^{2+}, \text{Cu}^+$) cannot undergo d-d transitions and are colourless.

• Transition Metal Catalysis:

  • Homogeneous Catalysts: Same phase as reactants. Transition metals transition between variable oxidation states to lower activation energy.
  • Heterogeneous Catalysts: Different phase than reactants. Unpaired d-electrons adsorb reactant molecules onto the solid surface, weakening bonds.
  • Key Examples:
  • Iron ($\text{Fe}$) in the Haber Process.
  • Nickel ($\text{Ni}$) in vegetable oil hydrogenation.
  • Vanadium(V) Oxide ($\text{V}_2\text{O}_5$) in the Contact Process.
  • Manganese(IV) Oxide ($\text{MnO}_2$) in $\text{H}_2\text{O}_2$ decomposition.

• Precipitation Reactions with Sodium Hydroxide ($\text{NaOH}$):

  • $\text{Cr}^{3+}(aq)$ (green) + $\text{NaOH}$ $\rightarrow$ Green-grey precipitate $\text{Cr}(\text{OH})_3(s)$. Redissolves in excess $\text{NaOH}$ to yield a deep green solution of $[\text{Cr}(\text{OH})_6]^{3-}(aq)$.
  • $\text{Fe}^{2+}(aq)$ (pale green) + $\text{NaOH}$ $\rightarrow$ Dark green precipitate $\text{Fe}(\text{OH})_2(s)$. Insoluble in excess $\text{NaOH}$.
  • $\text{Fe}^{3+}(aq)$ (yellow) + $\text{NaOH}$ $\rightarrow$ Rusty red-brown precipitate $\text{Fe}(\text{OH})_3(s)$. Insoluble in excess $\text{NaOH}$.
  • $\text{Cu}^{2+}(aq)$ (pale blue) + $\text{NaOH}$ $\rightarrow$ Pale blue precipitate $\text{Cu}(\text{OH})_2(s)$. Insoluble in excess $\text{NaOH}$.

Chemical Kinetics

• Methods for Measuring Reaction Rates:

  • Volume of gas produced (gas syringe).
  • Change in gas pressure at constant volume.
  • Loss of mass using a digital balance.
  • Colorimetry (measuring absorbance changes over time).
  • Sampling and quenching followed by titration.

• Calculating Rates:$\text{Rate} = \frac{\Delta \text{concentration}}{\Delta \text{time}}$

  • Initial rate is determined from the gradient of a tangent drawn at $t = 0$

• Rate Equations and Orders of Reaction:

  • For reaction $xA + yB \rightarrow \text{products}$:     $\text{Rate} = k [A]^a [B]^b$
  • $k$ = Rate constant (constant at a given temperature).
  • $a, b$ = Orders of reaction with respect to species $A$ and $B$.
  • Zero Order ($a = 0$): Changing concentration has no effect on rate.
  • First Order ($a = 1$): Doubling concentration doubles the rate.
  • Second Order ($a = 2$): Doubling concentration quadruples the rate ($2^2 = 4$).
  • Overall Order: Sum of individual orders ($a + b$

• Determining Orders from Initial Rate Data:

  • Compare runs where only one reactant concentration is varied while keeping others constant.
  • Rearrange rate equation to calculate rate constant $k$ and determine units by dimensional analysis.

• Rate-Determining Step (RDS):

  • The slowest step in a multi-step reaction mechanism.
  • Species appearing in the experimental rate equation must be involved in or prior to the rate-determining step.

• The Arrhenius Equation:$k = A e^{-\frac{E_a}{R T}}$

  • $k$ = Rate constant
  • $A$ = Pre-exponential / frequency factor
  • $E_a$ = Activation energy ($J\,mol^{-1}$)
  • $R$ = Gas constant ($8.314\,J\,K^{-1}\,mol^{-1}$)
  • $T$ = Absolute temperature ($K$)

• Worked Example: Calculating Rate Constant at a New Temperature:

  • Given: $k = 0.0742\,mol^{-1}\,dm^3\,s^{-1}$ at $273\,K$, $E_a = 58.333\,kJ\,mol^{-1} = 58333\,J\,mol^{-1}$.
  • Step 1: Calculate $A$ at $273\,K$$A = \frac{k}{e^{-\frac{E_a}{R T}}} = \frac{0.0742}{e^{-\frac{58333}{8.314 \times 273}}} = \frac{0.0742}{e^{-25.701}} = 1.077 \times 10^{10}\,mol^{-1}\,dm^3\,s^{-1}$
  • Step 2: Calculate $k$ at $279\,K$$k = 1.077 \times 10^{10} \times e^{-\frac{58333}{8.314 \times 279}} = 1.077 \times 10^{10} \times e^{-25.15} = 0.129\,mol^{-1}\,dm^3\,s^{-1}$

Enthalpy Changes for Solids and Solutions

• Hess's Law: The total enthalpy change for a chemical reaction is independent of the route taken, provided initial and final states are identical.

• Standard Enthalpy Definitions:

  • Standard conditions: $298\,K$, $1\,atm$ ($101\,kPa$), $1.0\,mol\,dm^{-3}$.
  • Enthalpy of Formation ($\Delta_f H^\theta$): Enthalpy change when 1 mole of a compound is formed from its elements in standard states.
  • Enthalpy of Combustion ($\Delta_c H^\theta$): Enthalpy change when 1 mole of a substance is completely burned in oxygen.
  • Enthalpy of Atomisation ($\Delta_{at} H^\theta$): Enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state.
  • Lattice Formation Enthalpy ($\Delta_{latt} H^\theta$): Enthalpy change when 1 mole of a solid ionic compound is formed from its constituent gaseous ions.
  • Enthalpy of Hydration ($\Delta_{hyd} H^\theta$): Enthalpy change when 1 mole of gaseous ions dissolves in water to form infinitely dilute aqueous ions.
  • Enthalpy of Solution ($\Delta_{sol} H^\theta$): Enthalpy change when 1 mole of an ionic solid dissolves in water to form a solution at infinite dilution.

• Born-Haber Cycles:

  • Closed thermochemical energy cycles used to derive lattice energies.

• Worked Example: Magnesium Oxide Lattice Formation Enthalpy:
  • Given data:
    • $\Delta_f H^\theta(\text{MgO}) = -602\,kJ\,mol^{-1}$
    • $\Delta_{at} H^\theta(\text{Mg}) = +149\,kJ\,mol^{-1}$
    • $IE_1(\text{Mg}) + IE_2(\text{Mg}) = +736 + 1450 = +2186\,kJ\,mol^{-1}$
    • $\Delta_{at} H^\theta(\text{O}) = \frac{1}{2}(+498) = +249\,kJ\,mol^{-1}$
    • $EA_1(\text{O}) + EA_2(\text{O}) = -141 + 841 = +700\,kJ\,mol^{-1}$
  • Sum of indirect steps = $149 + 736 + 1450 + 249 - 141 + 791 = +3234\,kJ\,mol^{-1}$
  • Calculation:       $3234 + \Delta_{latt} H^\theta = -602$$\Delta_{latt} H^\theta = -602 - 3234 = -3836\,kJ\,mol^{-1}$

Entropy and Feasibility of Reactions

• Entropy ($S$): A quantitative measure of degree of disorder/randomness in a system. Units: $J\,K^{-1}\,mol^{-1}$.

  • Phase progression: $S(\text{solid}) < S(\text{liquid}) < S(\text{gas})$.

• Calculating Entropy Changes of Reaction ($\Delta S$):$\Delta S = \sum S^\theta(\text{products}) - \sum S^\theta(\text{reactants})$

  • Worked Example: Thermal Decomposition of $\text{NaHCO}_3$:$2\text{NaHCO}_3(s) \rightarrow \text{Na}_2\text{CO}_3(s) + \text{CO}_2(g) + \text{H}_2\text{O}(l)$
  • Entropies: $\text{NaHCO}_3(s) = 101.7$, $\text{Na}_2\text{CO}_3(s) = 135.0$, $\text{CO}_2(g) = 213.6$, $\text{H}_2\text{O}(l) = 69.9\,J\,K^{-1}\,mol^{-1}$
  • Products entropy = $135.0 + 213.6 + 69.9 = 418.5\,J\,K^{-1}\,mol^{-1}$
  • Reactants entropy = $2 \times 101.7 = 203.4\,J\,K^{-1}\,mol^{-1}$
  • $\Delta S = 418.5 - 203.4 = +215.1\,J\,K^{-1}\,mol^{-1}$

• Gibbs Free Energy ($\Delta G$):$\Delta G = \Delta H - T \Delta S$

  • Feasibility Criterion: A reaction is thermodynamically feasible if $\Delta G \le 0$.
  • Note: Divide $\Delta S$ by 1000 to convert $J\,K^{-1}\,mol^{-1}$ into $kJ\,K^{-1}\,mol^{-1}$ prior to calculation.

• Worked Example: Transition Temperature for Tin Allotropes:

  • Equation: $\text{Sn(grey)} \rightarrow \text{Sn(white)}$
  • $\Delta H^\theta = 1.92\,kJ\,mol^{-1}$
  • Entropies: Grey tin = $44.8\,J\,K^{-1}\,mol^{-1}$, White tin = $51.5\,J\,K^{-1}\,mol^{-1}$
  • $\Delta S = 51.5 - 44.8 = +6.7\,J\,K^{-1}\,mol^{-1} = 0.0067\,kJ\,K^{-1}\,mol^{-1}$
  • At minimum feasibility temperature, $\Delta G = 0$$T = \frac{\Delta H}{\Delta S} = \frac{1.92\,kJ\,mol^{-1}}{0.0067\,kJ\,K^{-1}\,mol^{-1}} = 286.6\,K$

Equilibrium Constants

• Equilibrium Expressions:

  • Concentration constant ($K_c$):     $aA + bB \rightleftharpoons cC + dD$$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$
  • Partial pressure constant ($K_p$):     $K_p = \frac{P_C^c P_D^d}{P_A^a P_B^b}$

• Significance of $K$ Values:

  • $K \gg 1$: Products heavily favored at equilibrium ($\Delta G$ is negative).
  • $K \ll 1$: Reactants heavily favored ($\Delta G$ is positive).

• Worked Example: Equilibrium Concentration Calculation ($K_c$):

  • Reaction: $A + B \rightleftharpoons C + 2D$
  • Initial concentrations: $[A] = 0.5$, $[B] = 0.5$, $[C] = 0$, $[D] = 0\,mol\,dm^{-3}$
  • Equilibrium concentration: $[C] = 0.2\,mol\,dm^{-3}$
  • From stoichiometry: $[D] = 2 \times 0.2 = 0.4\,mol\,dm^{-3}$
  • Remaining reactants: $[A] = 0.5 - 0.2 = 0.3$, $[B] = 0.5 - 0.2 = 0.3\,mol\,dm^{-3}$
  • Calculation:     $K_c = \frac{[C][D]^2}{[A][B]} = \frac{0.2 \times (0.4)^2}{0.3 \times 0.3} = \frac{0.032}{0.09} = 0.356\,mol\,dm^{-3}$

• Temperature Effects on $K_c$ and $K_p$:

  • Temperature is the only variable that alters the value of an equilibrium constant.
  • Exothermic reactions: Increasing temperature decreases $K_c$
  • Endothermic reactions: Increasing temperature increases $K_c$

Acid-Base Equilibria

• Brønsted-Lowry Theory:

  • Acid = Proton ($\text{H}^+$) donor.
  • Base = Proton ($\text{H}^+$) acceptor.

• pH Calculations:$\text{pH} = -\log_{10}[\text{H}^+]$$[\text{H}^+] = 10^{-\text{pH}}$

• Weak Acid Dissociation Constant ($K_a$):$HA(aq) \rightleftharpoons H^+(aq) + A^-(aq)$$K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$$p K_a = -\log_{10}(K_a)$

  • Weak acid approximation: $[\text{H}^+] = \sqrt{K_a \times [\text{acid}]}$
  • Worked Example: Calculate pH of $0.01\,mol\,dm^{-3}$ ethanoic acid ($K_a = 1.7 \times 10^{-5}\,mol\,dm^{-3}$):     $[\text{H}^+] = \sqrt{1.7 \times 10^{-5} \times 0.01} = \sqrt{1.7 \times 10^{-7}} = 0.000412\,mol\,dm^{-3}$$\text{pH} = -\log_{10}(0.000412) = 3.4$

• Ionic Product of Water ($K_w$):$K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14}\,mol^2\,dm^{-6} \quad (\text{at } 298\,K)$

  • Strong bases: $[\text{OH}^-] = [\text{Base}]$, so $[\text{H}^+] = \frac{K_w}{[\text{OH}^-]}$
  • Worked Example: pH of $0.8\,mol\,dm^{-3}\,\text{NaOH}$:     $[\text{H}^+] = \frac{1.0 \times 10^{-14}}{0.8} = 1.25 \times 10^{-14}\,mol\,dm^{-3}$$\text{pH} = -\log_{10}(1.25 \times 10^{-14}) = 13.9$

• Buffer Solutions:

  • Solutions that resist changes in pH upon addition of small quantities of acid or base.
  • Acidic buffer equation: $[\text{H}^+] = K_a \times \frac{[\text{acid}]}{[\text{salt}]}$
  • Worked Example: Buffer with $0.20\,mol\,dm^{-3}\,\text{CH}_3\text{COOH}$ and $0.40\,mol\,dm^{-3}\,\text{CH}_3\text{COONa}$ ($K_a = 1.75 \times 10^{-5}$):     $[\text{H}^+] = 1.75 \times 10^{-5} \times \left(\frac{0.20}{0.40}\right) = 8.75 \times 10^{-6}\,mol\,dm^{-3}$$\text{pH} = -\log_{10}(8.75 \times 10^{-6}) = 5.1$

• Acid-Base Titration Curves:

  • Strong Acid / Strong Base: Equivalence point at pH 7; vertical section spans pH 3–11.
  • Weak Acid / Strong Base: Equivalence point at pH > 7 (pH 8–9); buffer region present.
  • Strong Acid / Weak Base: Equivalence point at pH < 7 (pH 4–5); vertical section spans pH 3–7.
  • Weak Acid / Weak Base: No distinct vertical region; pH meter required.
  • Indicator Selection: An indicator is suitable if its transition pH range falls within the vertical section of the titration curve.