Cameroon GCE A-Level Chemistry common mistakes
Cameroon GCE A-Level Chemistry · Understand the traps and how to avoid them.
When asked to write the electronic configuration of a transition metal ion such as Fe²⁺ or Cu²⁺, the candidate removes electrons from the 4s orbital first and writes 3d⁴ for Fe²⁺ instead of 3d⁶, or writes the configuration of the neutral atom and stops there.
The candidate has memorised the 'fill 4s before 3d' rule from building-up neutral atoms and applies it in reverse when ionising. The rule is about orbital energies during filling, not about which electrons leave first; once the 3d subshell is occupied, 4s electrons are lost preferentially because the 3d electrons are lower in energy in the ion.
What to do instead
Practise a fixed two-step routine: write the full configuration of the neutral atom, then remove electrons from the highest principal quantum number shell first (4s before 3d), never from 3d first. Drill the pairs Fe/Fe²⁺/Fe³⁺, Cu/Cu⁺/Cu²⁺, Mn/Mn²⁺ and check each against the 3d count. Also state explicitly that Cu and Cr are exceptions in the neutral atom (3d¹⁰4s¹ and 3d⁵4s¹) and explain the half-filled/full d-subshell stability in one sentence.
When drawing a dot-and-cross diagram or predicting the shape of a molecule such as NH₃, H₂O or SF₆, the candidate draws bonds without showing lone pairs, or counts lone pairs as bonding regions and predicts a trigonal planar shape for NH₃.
The candidate treats the diagram as a bookkeeping exercise for valence electrons rather than a 3D model. Lone pairs are often omitted because they are 'not bonding', and the VSEPR rule that lone pairs repel more strongly than bonding pairs is either unknown or not applied to the bond-angle and shape conclusion.
What to do instead
Use a rigid procedure every time: (1) count valence electrons and total electron pairs; (2) assign bonding and lone pairs on the central atom; (3) count electron domains to get the electron-pair geometry; (4) remove lone-pair positions from the shape name; (5) state the expected compression of bond angles. Memorise the standard outcomes: 4 domains with 1 lone pair = trigonal pyramidal (~107°), with 2 lone pairs = bent (~104.5°), 5 domains with 1 lone pair = see-saw, 6 domains with 2 lone pairs = square planar. Always draw lone pairs as distinct dots or crosses so they are visible in the answer.
In a calculation involving a gas, the candidate substitutes pressure in kPa or volume in cm³ directly into pV = nRT with R = 8.31 J K⁻¹ mol⁻¹, or uses temperature in °C, and reports an answer that is wrong by a factor of 1000 or more.
The candidate focuses on rearranging the equation and treats units as decoration. The mismatch arises because R = 8.31 requires pressure in Pa and volume in m³, while exam data is often given in kPa and cm³; similarly, all gas equations require thermodynamic temperature.
What to do instead
Adopt a mandatory unit-conversion line before substitution: write p in Pa (kPa × 1000), V in m³ (cm³ × 10⁻⁶, dm³ × 10⁻³), T in K (°C + 273). Write the converted values on the line above the substitution so the marker can see them. For reacting-gas problems, use the molar ratio directly from the balanced equation or Avogadro's law at the same T and p rather than converting to moles unnecessarily. Do two or three timed conversion drills per revision session until the conversions are automatic.
When calculating an enthalpy change from bond enthalpies or from standard enthalpy of formation, the candidate adds all the bond enthalpies of the reactants and products instead of subtracting, or uses formation enthalpies with the wrong sign convention, giving a value with the wrong sign or the wrong magnitude.
The candidate has memorised 'products minus reactants' as a phrase without attaching it to a physical meaning. Bond breaking is endothermic and bond making is exothermic, so the sign of a bond-enthalpy calculation depends on which sum is subtracted from which; formation enthalpies are defined for the formation of one mole of compound from elements, which reverses the intuitive direction for reactants.
What to do instead
Before any calculation, write the defining equation with its sign convention in words: ΔH = Σ(bonds broken) − Σ(bonds formed) for bond enthalpies, and ΔH = ΣΔHf(products) − ΣΔHf(reactants) for formation enthalpies. Then draw the full displayed formula of every species and list each bond with its enthalpy before summing, so no bond is double-counted or missed. Sanity-check the sign against the reaction type (combustion must be negative; decomposition usually positive) and against the rough magnitude expected from the number of bonds broken.
When asked to predict the effect of a change on an equilibrium, the candidate says the position 'shifts to the side with fewer moles' for a pressure change without checking whether the number of gaseous moles actually differs on the two sides, or claims a catalyst shifts the equilibrium position.
Le Chatelier's principle is recalled as a set of slogans rather than as a rule about opposing a change. The candidate does not count gaseous moles on each side, and confuses the effect of a catalyst on rate with an effect on position. Similar errors occur with temperature, where the candidate shifts the equilibrium in the endothermic direction regardless of whether the forward reaction is exothermic or endothermic.
What to do instead
For every equilibrium question, write the balanced equation with state symbols and annotate the number of gaseous moles on each side. Then apply a fixed checklist: concentration change (shift away from the added species), pressure change (only if Δn(gas) ≠ 0; shift to the side with fewer moles on increase), temperature change (shift in the direction that absorbs or releases heat, identified from the sign of ΔH), catalyst (no effect on position, only on time to reach equilibrium). State explicitly 'no shift because Δn(gas) = 0' when that is the case, since markers award that reasoning.
In an acid–base titration calculation, the candidate calculates the moles of acid and base and then equates them directly without using the stoichiometric ratio from the balanced equation, or calculates pH from concentration without taking −log₁₀, or forgets to convert concentration to moles using the titre volume in dm³.
The candidate treats all acid–base reactions as 1:1, which is true for HCl + NaOH but false for H₂SO₄ + NaOH or for a diprotic acid. In pH questions, the candidate confuses concentration with pH and skips the logarithm, or uses the initial volume instead of the volume at the equivalence point.
What to do instead
Write the balanced equation first and record the mole ratio explicitly (e.g. H₂SO₄ : NaOH = 1 : 2). Then follow a fixed chain: n(acid) = c × V(dm³); n(base) = n(acid) × ratio; c(base) = n(base) / V(base). For pH, always write pH = −log₁₀[H⁺] and, for a weak acid, use Ka = [H⁺]²/[HA] before taking the log. For buffer questions, use the Henderson–Hasselbalch form pH = pKa + log([salt]/[acid]) and check that the ratio is dimensionless. Keep volumes in dm³ throughout and state the conversion line.
When describing the reaction of a halide with concentrated sulfuric acid or the thermal stability of Group 2 carbonates and nitrates, the candidate gives a single observation for all halides or states that 'Group 2 compounds decompose on heating' without naming the trend and its cause.
The candidate memorises the reaction of one example (usually NaCl or CaCO₃) and generalises it. The underlying trend depends on bond strength and polarisability, which the candidate has not linked to the observations, so the answer lacks the specific products (e.g. H₂S, SO₂, Br₂, I₂) and the explanation.
What to do instead
Build a table for the halides with concentrated H₂SO₄: HCl (steamy fumes, no redox), HBr (brown Br₂, SO₂), HI (purple I₂, H₂S, SO₂), and attach the reason (reducing power increases down the group as the H–X bond weakens and the halide is more easily oxidised). For Group 2 carbonates and nitrates, state that thermal stability increases down the group because the larger cation distorts the anion less (lattice enthalpy and polarisation argument), and quote the decomposition products (oxide + CO₂; oxide + NO₂ + O₂). Practise writing the trend, the reason and one equation in each answer.
When asked to explain why a complex such as [Cu(NH₃)₄(H₂O)₂]²⁺ is coloured or why [Fe(H₂O)₆]³⁺ is more paramagnetic than [Fe(CN)₆]³⁻, the candidate says 'because it is a transition metal' or 'because of d electrons' without referring to d-orbital splitting, ligand field strength or the number of unpaired electrons.
The candidate has learned that transition metals give coloured complexes but has not connected colour to d–d transitions and the ligand-field splitting Δ, nor connected magnetic moment to unpaired electron count. The high-spin/low-spin distinction between weak-field (H₂O) and strong-field (CN⁻) ligands is often missing.
What to do instead
For colour questions, write the full explanation chain: degenerate d orbitals split into two sets by the ligand field; Δ depends on the ligand and the metal; an electron is promoted between the sets absorbing a specific wavelength in the visible region; the complementary colour is observed. For magnetism, draw the d-orbital occupancy for the specific ion and count unpaired electrons, noting that CN⁻ is strong-field (low spin) and H₂O is weak-field (high spin). Practise the [Fe(H₂O)₆]³⁺ vs [Fe(CN)₆]³⁻ pair and the [Cu(NH₃)₄(H₂O)₂]²⁺ case until the explanation is fluent.
When drawing a mechanism for the reaction of an alkene with HBr or the hydrolysis of a halogenoalkane, the candidate draws the product directly with no curly arrows, or draws arrows starting from the wrong atom, or omits the intermediate carbocation and the reason for Markovnikov regioselectivity.
The candidate treats mechanisms as pictures of the products rather than as electron-flow descriptions. Curly arrows are often placed from the electrophile to the nucleophile (the wrong direction) because the candidate has not internalised that arrows show electron movement from electron-rich to electron-poor. The carbocation intermediate is omitted because the candidate does not see it as a required step.
What to do instead
Adopt a fixed arrow discipline: the tail of every curly arrow is on a lone pair or a bond, and the head points to the atom receiving the electrons. For electrophilic addition, draw the alkene π bond attacking the electrophile, then the carbocation intermediate, then the nucleophile attacking the carbocation. Justify Markovnikov by comparing the stability of the two possible carbocations (tertiary > secondary > primary) and state that the more stable one is formed. For nucleophilic substitution, draw the C–X bond breaking and the nucleophile attacking in the same step (SN2) or after carbocation formation (SN1), and label the mechanism.
In practical data analysis, the candidate calculates a mean titre by including a rough titre or an outlier, reports a result to more significant figures than the burette can support, or calculates percentage error against the wrong value.
The candidate treats data analysis as arithmetic rather than as judgement about measurement quality. Rough titres are recorded to find the endpoint and are not accurate, and outliers arise from random error; the candidate averages them because they are numbers on the page. Significant figures are copied uncritically from the calculator display.
What to do instead
Apply a fixed data-handling routine: (1) identify and exclude the rough titre and any reading differing from the others by more than about 0.10 cm³, stating the reason; (2) average the concordant titres only; (3) report the mean to 2 decimal places (burette precision ±0.05 cm³) and carry that precision into the final answer; (4) calculate percentage error as (uncertainty / measured value) × 100 using the apparatus uncertainty, not the mean. Practise on past practical papers by writing the exclusion reason and the precision statement explicitly, since these carry marks.