Érettségi Chemistry common mistakes
Érettségi Chemistry · Understand the traps and how to avoid them.
When writing electron configurations, the candidate stops at the noble-gas core (e.g. writes Fe as [Ar] 3d6 4s2 but then fills 4s before 3d for later elements, or writes Cu as [Ar] 3d9 4s2) and does not apply the anomalous half-filled/filled stability for Cr and Cu.
The aufbau order (4s before 3d) is memorized as a rigid rule, so the candidate does not realize that once 3d is occupied the 4s orbital lies higher in energy, and that Cr ([Ar] 3d5 4s1) and Cu ([Ar] 3d10 4s1) are exceptions driven by exchange-energy stabilization.
What to do instead
Draw the orbital energy ladder for each transition element and re-derive the configuration from the previous noble gas; explicitly memorize the two exceptions Cr and Cu and their ions (e.g. Fe2+ = [Ar] 3d6, not 3d4 4s2) and practise removing 4s electrons first when forming cations.
When drawing a Lewis structure the candidate counts only the outer-shell electrons of each free atom and forgets to add or subtract electrons for the charge of a polyatomic ion (e.g. draws SO4^2- with 24 electrons instead of 32, or NH4+ with 8 instead of 8 but with a lone pair on N).
The candidate treats the ion as a neutral molecule and does not systematically apply the total-electron formula (sum of valence electrons minus cation charge plus anion charge) before placing bonds and lone pairs.
What to do instead
Adopt a fixed 4-step routine: count total valence electrons including charge, place the central atom, form single bonds, then distribute remaining electrons as lone pairs and add multiple bonds only to satisfy octets; check formal charges sum to the ion charge before finishing.
The candidate predicts molecular polarity from the presence of polar bonds alone, concluding that CO2, CCl4 or BF3 are polar because they contain polar bonds, and cannot explain why the individual bond dipoles cancel.
Bond polarity and molecular polarity are conflated, and the candidate does not consistently apply VSEPR to obtain the geometry (linear, tetrahedral, trigonal planar) before vector-summing the dipoles.
What to do instead
Always determine the electron-domain geometry first, then the molecular shape, then sum bond dipoles as vectors; practise the pairs CO2 vs SO2, CCl4 vs CHCl3 and BF3 vs NF3 so the role of symmetry becomes automatic.
In stoichiometry problems the candidate uses the mass of the impure or wet sample directly as if it were pure substance, or forgets to convert between the mass of a solution and the mass of dissolved solute using the mass fraction, giving results that are wrong by the purity or concentration factor.
The candidate reads numbers off the problem and immediately divides by molar mass without first writing a clean 'given → unknown' map that separates sample mass, pure-substance mass, amount of substance and solution mass.
What to do instead
Start every calculation with a labelled data table (m(sample), w, m(pure), n, M) and a written unit path; only after the pure amount in moles is established should the candidate use the reaction equation, and finish with a mass-fraction or yield correction where required.
When calculating the enthalpy change of a reaction, the candidate adds or subtracts formation enthalpies without multiplying by the stoichiometric coefficients, or reverses the sign of ΔH when reversing an equation in a Hess's-law cycle.
The candidate treats ΔHf values as per-reaction quantities rather than per-mole-of-substance values, and Hess's-law manipulation is done by eye rather than by explicitly writing the reversed and scaled equations.
What to do instead
Write the full formation equation for each compound with its coefficient, apply ΔHr = ΣνΔHf(products) − ΣνΔHf(reactants), and for Hess cycles physically rewrite each step equation with its direction arrow and sign before summing.
The candidate calculates pH from the analytical concentration of a weak acid (e.g. pH = −log(0.1) for 0.1 M acetic acid) without using Ka, or confuses the equivalence point with the neutral point in a titration and assumes pH 7 at equivalence for a weak acid–strong base titration.
Strong-acid formulas are over-learned and applied indiscriminately; the candidate also does not connect the equivalence point to the composition of the salt formed and its hydrolysis.
What to do instead
Classify the acid/base as strong or weak before choosing a formula; for weak acids use Ka = x²/(c−x) with the approximation check, and determine the equivalence-point pH from the hydrolysis of the resulting salt (e.g. acetate gives pH > 7).
When balancing a redox equation in acidic or basic medium, the candidate balances atoms but not charge, or adds H2O and H+ on the wrong side, so the final equation has unequal total charge on the two sides.
The candidate balances by inspection rather than by the half-reaction method, and forgets that in basic solution the H+ must be neutralized by adding OH− to both sides to form water.
What to do instead
Use the strict half-reaction procedure: balance all atoms except O and H, balance O with H2O, balance H with H+, balance charge with electrons, equalize electron counts, add the halves, and for basic medium add OH− to both sides to remove H+; always verify that total charge is equal on both sides.
The candidate writes the cell diagram or calculates the cell potential with the anode and cathode reversed relative to the spontaneous direction, or forgets that E°cell = E°cathode − E°anode and that a positive E°cell corresponds to a spontaneous reaction.
The candidate memorizes 'oxidation at the anode' as a phrase but does not link it to the sign convention and to which electrode has the higher standard reduction potential in the given pair.
What to do instead
For each cell, list both standard reduction potentials, identify the more positive one as the cathode, write the two half-equations in the spontaneous direction, and compute E°cell = E°cathode − E°anode; check that the result is positive before writing the final cell notation.
When a change is applied to a system at equilibrium, the candidate states only 'the equilibrium shifts right' without saying what happens to the concentrations or the value of the equilibrium constant, and incorrectly claims that adding a catalyst or changing pressure shifts K.
Le Chatelier's principle is applied as a slogan, and the distinction between the position of equilibrium (Q vs K) and the equilibrium constant itself (temperature-dependent only) is not consolidated.
What to do instead
For every disturbance answer three questions in order: which side is favoured, how the concentrations of each species change, and whether K changes (only with temperature); explicitly note that a catalyst changes only the rate of approach, not the position, and that pressure changes matter only when the number of gas moles differs.
In organic nomenclature and reaction writing, the candidate names a compound by listing the substituents without the lowest-locant rule, or writes a substitution reaction where an addition is required (e.g. adding HBr to ethene but writing a substitution product, or chlorinating ethane and writing an addition product).
The candidate learns functional groups in isolation without linking the hydrocarbon skeleton's saturation to the reaction type, and does not systematically number the parent chain from the end giving the lowest locants.
What to do instead
Before writing any organic reaction, classify the substrate as saturated, unsaturated or aromatic and match it to the reaction type (addition for alkenes/alkynes, substitution for alkanes/aromatics, elimination for haloalkanes); for naming, always number from both ends and choose the lowest locant set, then assemble the name alphabetically.