UEE Chemistry common mistakes
UEE Chemistry · Understand the traps and how to avoid them.
When asked to write the electron configuration of a transition-metal ion such as Fe²⁺ or Cu⁺, candidates remove electrons from the 3d subshell before emptying the 4s subshell, writing Fe²⁺ as [Ar]3d⁴4s² instead of [Ar]3d⁶.
The aufbau order fills 4s before 3d, so candidates memorise '4s comes first' and apply that ordering to ionisation as well. They do not realise that once 3d is occupied, 4s orbitals lie higher in energy, so 4s electrons are lost first.
What to do instead
After filling, re-rank the subshells by energy for the actual species before removing electrons: for first-row transition metals remove 4s electrons before 3d. Drill a short set (Fe, Fe²⁺, Fe³⁺, Cu, Cu⁺, Cr, Cr³⁺) and write the full configuration each time, checking the 4s count is zero before any 3d electron is removed.
Candidates draw a Lewis structure, count electron domains correctly, then assign molecular geometry using only bonding pairs, so H₂O is called linear and NH₃ trigonal planar rather than bent and trigonal pyramidal.
Textbook diagrams often show only the shape of the bonded atoms, and the names 'tetrahedral electron geometry' and 'bent molecular geometry' get conflated. Lone pairs are drawn but not counted as domains in the final step.
What to do instead
Use a fixed four-step routine every time: total valence electrons, Lewis structure, count all electron domains (bonding + lone pairs), then name electron geometry and molecular geometry separately. Practise the series CH₄, NH₃, H₂O, and SF₄, SO₂, XeF₄, stating both names aloud so the distinction becomes automatic.
In limiting-reagent problems candidates compare the raw moles of the two reactants and call the smaller number limiting, ignoring the stoichiometric coefficients, or they compute the theoretical yield from the excess reagent.
'Smallest number of moles' is a simple rule that works for 1:1 reactions and is remembered without its condition. Multi-step problems also tempt candidates to stop after finding moles of one reactant.
What to do instead
Always divide moles of each reactant by its coefficient in the balanced equation and compare those quotients; the smallest quotient identifies the limiting reagent. Then convert moles of limiting reagent to product using the mole ratio, and finish by computing percentage yield against that theoretical value. Do three or four problems with coefficients other than 1:1 until the division step is habitual.
When using the ideal gas law with mass or density data, candidates substitute grams directly for n, or mix units so that pressure is in kPa while R is taken as 8.314 J K⁻¹ mol⁻¹ with volume in dm³, producing answers wrong by a factor of 1000.
The symbols in pV = nRT look unit-free on the page, and candidates focus on rearranging rather than on unit bookkeeping. R has several numerical values and the wrong one is easy to pick.
What to do instead
Before substituting, write a unit line for every quantity and convert to a consistent set (p in Pa, V in m³, T in K, n in mol, R = 8.314). Convert mass to moles with n = m/M first, and always convert Celsius to kelvin. Check the answer's order of magnitude against a rough estimate to catch factor-of-1000 slips.
Candidates calculate ΔH for a reaction using average bond enthalpies of all bonds in the reactants and products without distinguishing bonds broken from bonds formed, or they apply Hess's law with a sign error when reversing an equation.
Bond breaking is endothermic and bond forming exothermic, but both are often tabulated as positive bond enthalpies, so the subtraction ΔH = Σ(bonds broken) − Σ(bonds formed) is applied inconsistently. Reversing an equation also requires reversing the sign of ΔH, which is easily forgotten under time pressure.
What to do instead
Write the enthalpy cycle explicitly with arrows: bonds broken on one side (positive), bonds formed on the other (negative), then sum. For Hess's law, rewrite each given equation with its ΔH, mark any that you reverse with a sign change, and add them line by line. Use formation enthalpies as a cross-check, since ΔH = ΣΔHf(products) − ΣΔHf(reactants) gives the same sign convention every time.
When calculating the pH of a weak acid, candidates use [H⁺] = c directly instead of solving with Ka, or they calculate pH of a buffer by ignoring the Henderson–Hasselbalch ratio and simply taking −log Ka.
Strong-acid formulas are practised far more, and the approximation that weak acids are only slightly ionised is stated in class but not operationalised. Buffer problems look like simple acid problems, so the ratio term is dropped.
What to do instead
Classify the acid first: strong (complete ionisation, [H⁺] = c) or weak (set up Ka = x²/(c − x) and solve, or use x ≈ √(Kac) only when ionisation is under about 5%). For buffers, write pH = pKa + log([base]/[acid]) and substitute the actual concentrations. Check that a weak-acid pH is higher than the same concentration of strong acid, and that a buffer pH lies near pKa.
Candidates state that increasing temperature increases reaction rate because particles collide more often, omitting the dominant effect that the fraction of collisions with energy above the activation energy rises sharply.
Collision frequency is the first factor taught and is easy to visualise, while the Boltzmann distribution argument is more abstract. The distinction between the two effects is not tested often enough in class to stick.
What to do instead
Answer rate questions in two parts: collision frequency (concentration, pressure, surface area) and collision energy (temperature, catalyst). For temperature, always mention that the Maxwell–Boltzmann distribution shifts so a much larger fraction of molecules exceeds Ea, and that this exponential effect dominates over the modest increase in collision frequency. Sketch the two curves to anchor the explanation.
In balancing redox half-equations in acidic or alkaline solution, candidates balance atoms but forget to add H₂O and H⁺/OH⁻, or they add electrons to the wrong side so the charges do not balance.
The atom-balancing steps are mechanical and easy to follow, but charge balancing is a separate check that is often skipped. Alkaline conditions require an extra conversion step from the acidic form that is frequently omitted.
What to do instead
Follow a fixed order: balance all atoms except O and H, balance O with H₂O, balance H with H⁺, then balance charge with electrons, and finally equalise electrons before adding. For alkaline solutions, add OH⁻ to both sides to neutralise H⁺ and simplify. Verify by summing charges on each side; they must be equal before combining half-equations.
Candidates describe the reaction of an alkene with HBr or the oxidation of a primary alcohol without specifying the mechanism or product class, giving 'it adds HBr' rather than naming the carbocation intermediate and Markovnikov product, or writing a carboxylic acid where an aldehyde is expected under distillation.
Organic reactions are often memorised as one-line summaries without the mechanistic reasoning that determines regiochemistry and product class. The role of reaction conditions (reflux versus distillation, oxidising agent strength) is under-emphasised.
What to do instead
For each functional-group reaction, write the mechanism arrows and state the regiochemical rule (Markovnikov, anti-Markovnikov with peroxides) and the product class. Attach conditions to outcomes: primary alcohol with acidified dichromate under distillation gives the aldehyde, under reflux gives the carboxylic acid. Build a one-page reaction map per functional group and redraw it from memory weekly.
In titration calculations candidates use the burette reading directly as the volume of analyte, or they read the meniscus from the top of the liquid rather than the bottom, and they report results to more decimal places than the apparatus justifies.
The recorded titre is the volume delivered, not the volume in the flask, and this distinction is lost when candidates copy numbers into a formula. Burette graduations increase downwards, which makes misreading common, and significant figures are treated as a formality.
What to do instead
Label every volume in the calculation as 'titre delivered' or 'aliquot in flask' before substituting. Read the burette at eye level from the bottom of the meniscus, record to ±0.05 cm³, and average concordant titres only. Carry significant figures through and round the final answer to match the least precise measurement, typically three or four significant figures for a burette.