Inorganic | HSC - Wyatt's Notes

HSC chemistry study notes - Inorganic

flowchart TD
    A[Inorganic] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Atomic radius: Decreases across a period (left to right), increases down a group

Ionisation energy: Increases across a period, decreases down a group

Electronegativity: Increases across a period, decreases down a group

Acid-base reactions: $HA + B \rightarrow A^- + BH^+$

Oxidation-reduction: Loss of electrons = oxidation; gain of electrons = reduction

Solubility rules:

• Most nitrates are soluble
• Most Group 1 salts are soluble
• Most chlorides are soluble (except AgCl, PbCl₂)
• Most sulfates are soluble (except BaSO₄, PbSO₄)

Enthalpy change: $\Delta H = H_{\text{products}} - H_{\text{reactants}}$

Hess’s Law: The total enthalpy change is independent of the pathway.

Calorimetry: $q = mc\Delta T$

Cell potential: $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$

Faraday’s laws: $m = \frac{MIt}{nF}$ where $F = 96485\,\text{C/mol}$

Problem: Balance the following equation in acidic solution: $\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$

Solution:

Step 1: Identify oxidation states:

• Mn: +7 $\rightarrow$ +2 (reduction, gain of 5 electrons)
• Fe: +2 $\rightarrow$ +3 (oxidation, loss of 1 electron)

Step 2: Balance electron transfer: multiply $\text{Fe}^{2+}$ by 5

Step 3: Balance oxygen with water and hydrogen with $\text{H}^+$: $\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$

Answer: $\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$

Problem: Calculate the enthalpy change for the combustion of methane: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

Given: $\Delta H_f°(\text{CH}_4) = -74.8\,\text{kJ/mol}$, $\Delta H_f°(\text{CO}_2) = -393.5\,\text{kJ/mol}$, $\Delta H_f°(\text{H}_2\text{O}) = -285.8\,\text{kJ/mol}$

Solution:

Step 1: Apply Hess’s Law: $\Delta H = \sum \Delta H_f°(\text{products}) - \sum \Delta H_f°(\text{reactants})$

Step 2: Calculate: $\Delta H = [(-393.5) + 2(-285.8)] - [(-74.8) + 2(0)]$ $\Delta H = -965.1 + 74.8 = -890.3\,\text{kJ/mol}$

Answer: $\Delta H = -890.3\,\text{kJ/mol}$

Problem: Calculate the cell potential for a Daniell cell: $\text{Zn} | \text{Zn}^{2+} || \text{Cu}^{2+} | \text{Cu}$

Given: $E°(\text{Zn}^{2+}/\text{Zn}) = -0.76\,\text{V}$, $E°(\text{Cu}^{2+}/\text{Cu}) = +0.34\,\text{V}$

Solution:

Step 1: Identify cathode (reduction) and anode (oxidation):

• Cathode: Cu²⁺ + 2e⁻ → Cu (reduction)
• Anode: Zn → Zn²⁺ + 2e⁻ (oxidation)

Step 2: Calculate cell potential: $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 0.34 - (-0.76) = 1.10\,\text{V}$

Answer: $E_{\text{cell}} = 1.10\,\text{V}$

1. In redox balancing, always balance atoms first, then charges
2. Use standard enthalpies of formation for Hess’s Law calculations
3. Positive cell potential means the reaction is spontaneous
4. Remember that oxidation occurs at the anode and reduction at the cathode

1. Balance the following in basic solution: $\text{ClO}^- + \text{Cr(OH)}_4^- \rightarrow \text{Cl}^- + \text{CrO}_4^{2-}$
2. Calculate the enthalpy of neutralisation for HCl + NaOH
3. A cell has $E_{\text{cell}} = 0.80\,\text{V}$. If the anode has $E° = -0.25\,\text{V}$, what is the cathode potential?

Problem: Will a precipitate form when $0.1\,\text{M}$ $\text{AgNO}_3$ is mixed with $0.1\,\text{M}$ $\text{NaCl}$?

Solution:

Step 1: Identify the possible products by ion exchange: $\text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3$

Step 2: Check solubility rules:

• $\text{NaNO}_3$: All nitrates and Group 1 salts are soluble
• $\text{AgCl}$: Most chlorides are soluble, BUT $\text{AgCl}$ is an exception (insoluble)

Step 3: Since $\text{AgCl}$ is insoluble, a precipitate will form.

Answer: Yes, a white precipitate of $\text{AgCl}$ forms.

Common mistake: Assuming all chloride salts are soluble. AgCl, PbCl₂, and Hg₂Cl₂ are notable exceptions.

Problem: $50\,\text{mL}$ of $1.0\,\text{M}$ HCl is mixed with $50\,\text{mL}$ of $1.0\,\text{M}$ NaOH in a calorimeter. The temperature rises from $25.0°\text{C}$ to $31.4°\text{C}$. Calculate the enthalpy of neutralisation. (Assume the density of the solution is $1.0\,\text{g/mL}$ and $c = 4.18\,\text{J/(g K)}$.)

Solution:

Step 1: Calculate total mass of solution: $m = (50 + 50) \times 1.0 = 100\,\text{g}$

Step 2: Calculate heat absorbed: $q = mc\Delta T = 100 \times 4.18 \times (31.4 - 25.0) = 100 \times 4.18 \times 6.4 = 2675.2\,\text{J} = 2.675\,\text{kJ}$

Step 3: Moles of water formed: $n = 0.050 \times 1.0 = 0.050\,\text{mol}$

Step 4: Enthalpy of neutralisation: $\Delta H = -\frac{q}{n} = -\frac{2.675}{0.050} = -53.5\,\text{kJ/mol}$

Answer: $\Delta H = -53.5\,\text{kJ/mol}$ (close to the theoretical value of $-57.1\,\text{kJ/mol}$)

Common mistake: Forgetting the negative sign. The reaction is exothermic, so $\Delta H$ is negative.

Problem: Calculate the cell potential for a Daniell cell at $25°\text{C}$ when $[\text{Zn}^{2+}] = 0.5\,\text{M}$ and $[\text{Cu}^{2+}] = 0.01\,\text{M}$. Given $E°_{\text{cell}} = 1.10\,\text{V}$.

Solution:

Step 1: Write the cell reaction: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$

Step 2: Write the reaction quotient: $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.5}{0.01} = 50$

Step 3: Apply the Nernst equation ($n = 2$ electrons transferred): $E = E° - \frac{RT}{nF}\ln Q = 1.10 - \frac{0.0257}{2}\ln 50$

Step 4: Calculate: $E = 1.10 - 0.01285 \times 3.912 = 1.10 - 0.0503 = 1.050\,\text{V}$

Answer: $E = 1.05\,\text{V}$

Common mistake: Using $\log_{10}$ instead of $\ln$ in the Nernst equation. At $25°\text{C}$, $\frac{RT}{F} = 0.0257\,\text{V}$ with $\ln$, or $\frac{0.0592}{n}$ with $\log_{10}$.

Problem: Arrange the following elements in order of increasing ionisation energy: Na, Mg, Al, Si, P.

Solution:

Step 1: These elements are all in Period 3 of the periodic table.

Step 2: Ionisation energy generally increases across a period due to increasing nuclear charge and decreasing atomic radius.

Step 3: However, there are exceptions:

• Between Mg and Al: Al has a lower IE than Mg because Al’s outer electron is in a higher energy subshell (3p vs 3s)
• Between P and S: S has a lower IE than P due to electron-electron repulsion in the paired 3p orbital

Step 4: Order: Na < Al < Mg < Si < P

Answer: Na < Al < Mg < Si < P

Common mistake: Assuming ionisation energy always increases uniformly across a period. Exceptions occur at Groups 2-13 and 15-16 due to subshell and pairing effects.

Problem: Given the following data, calculate the enthalpy of formation of ethane ($\text{C}_2\text{H}_6$):

1. $\text{C}_2\text{H}_6(g) + \frac{7}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)$, $\Delta H_1 = -1560\,\text{kJ/mol}$
2. $\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$, $\Delta H_2 = -393.5\,\text{kJ/mol}$
3. $\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)$, $\Delta H_3 = -285.8\,\text{kJ/mol}$

Solution:

Step 1: The formation reaction is: $2\text{C}(s) + 3\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)$

Step 2: Using Hess’s Law: $\Delta H_f = 2\Delta H_2 + 3\Delta H_3 - \Delta H_1$

Step 3: Calculate: $\Delta H_f = 2(-393.5) + 3(-285.8) - (-1560)$ $= -787 - 857.4 + 1560 = -84.4\,\text{kJ/mol}$

Answer: $\Delta H_f(\text{C}_2\text{H}_6) = -84.4\,\text{kJ/mol}$

Common mistake: Forgetting to reverse the combustion equation (multiply by -1) when using Hess’s Law. The combustion equation is the reverse of what we need to subtract.

Problem: A galvanic cell is constructed with Fe/Fe²⁺ ($E° = -0.44\,\text{V}$) and Ag/Ag⁺ ($E° = +0.80\,\text{V}$). Write the cell notation, calculate the cell potential, and determine the spontaneous reaction.

Solution:

Step 1: Identify cathode and anode:

• Fe has the lower reduction potential, so it is oxidised (anode)
• Ag has the higher reduction potential, so it is reduced (cathode)

Step 2: Cell notation: $\text{Fe}(s) | \text{Fe}^{2+}(aq) || \text{Ag}^{+}(aq) | \text{Ag}(s)$

Step 3: Cell potential: $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} = 0.80 - (-0.44) = 1.24\,\text{V}$

Step 4: Spontaneous reaction: $\text{Fe}(s) + 2\text{Ag}^{+}(aq) \rightarrow \text{Fe}^{2+}(aq) + 2\text{Ag}(s)$

Answer: $E_{\text{cell}} = 1.24\,\text{V}$, reaction: $\text{Fe} + 2\text{Ag}^{+} \rightarrow \text{Fe}^{2+} + 2\text{Ag}$

Common mistake: Forgetting to balance the number of electrons transferred when writing the overall cell reaction.

The periodic table is a map of elemental behaviour — position predicts chemical personality. Periodic trends arise from a tug-of-war between nuclear charge (pulling electrons inward) and electron shielding (pushing them outward). Redox chemistry is fundamentally about electron transfer, and electrochemistry measures the driving force behind that transfer as cell potential. Hess’s Law is an energy bookkeeping principle: energy is conserved regardless of the path taken, just as distance walked does not depend on the route.

For exothermic reactions, $\Delta H$ is negative (heat is released). For endothermic reactions, $\Delta H$ is positive (heat is absorbed). Students often write the correct numerical value but assign the wrong sign. In calorimetry, $q = mc\Delta T$ gives the heat absorbed by the solution, so the reaction enthalpy is $\Delta H = -q/n$ for an exothermic reaction. The negative sign is essential.

The cell potential is $E_{cell} = E_{cathode} - E_{anode}$, where the cathode is where reduction occurs and the anode is where oxidation occurs. Students sometimes add the two standard potentials instead of subtracting, or misidentify which half-cell is the cathode. The cathode always has the higher (more positive) reduction potential in a galvanic cell.

Faraday’s first law states $m = MIt/(nF)$, where $n$ is the number of electrons transferred per ion. For example, depositing Cu²⁺ requires 2 electrons ($n = 2$), while depositing Ag⁺ requires only 1 ($n = 1$). Students often use $n = 1$ by default, leading to mass calculations that are off by a factor of 2 or more. Always check the ion’s charge to determine $n$.

• Algebra — Logarithmic functions and matrix operations are used in electrochemistry and equilibrium calculations.
• Calculus — Rate equations and integrated rate laws in chemical kinetics require differentiation and integration.
• Organic — Organic reactions involve redox processes and functional group transformations that build on inorganic principles.
• Mechanics — Energy conservation and work-energy concepts connect thermochemistry to mechanical systems.