Balearic Islands PAU Chemistry / Química 2º Bachillerato UIB common mistakes
Balearic Islands PAU Chemistry / Química 2º Bachillerato UIB · Understand the traps and how to avoid them.
When asked to write the electron configuration of a transition-metal cation or an anion, the candidate removes electrons from the highest n shell (e.g., Fe²⁺ written as [Ar]4s²3d⁴ or Cu⁺ as [Ar]4s¹3d⁹) instead of from the 4s orbital first, or adds electrons to the wrong subshell for anions such as O²⁻ or Cl⁻.
Candidates learn the aufbau filling order (4s before 3d) and mechanically apply the reverse order to ions, without realising that once the 3d subshell is occupied the 4s orbital lies higher in energy, so 4s electrons are lost first. Anion errors come from adding to the last-filled subshell instead of continuing the filling sequence.
What to do instead
Adopt a fixed two-step routine: write the neutral-atom configuration, then for cations remove electrons from the outermost n first (4s before 3d), and for anions continue the aufbau sequence into the next subshell. Drill the specific pairs Fe/Fe²⁺/Fe³⁺, Cu/Cu⁺/Cu²⁺, Cr/Cr³⁺, O/O²⁻ and check the result against the noble-gas core each time.
The candidate explains periodic trends (atomic radius, ionisation energy, electronegativity) by invoking only the group number or only the number of shells, or states that ionisation energy always increases across a period without noting the exceptions at group 13→14 (e.g., B > Be? no: Be > B) and group 15→16 (N > O).
Textbook summaries compress the trend into a one-line rule, and candidates memorise the direction rather than the two competing factors (effective nuclear charge and shielding/distance). The anomalous drops are taught as trivia and then forgotten under exam pressure.
What to do instead
For every trend question, force yourself to write both factors explicitly: (1) number of occupied shells (dominant down a group) and (2) effective nuclear charge / shielding (dominant across a period). Then, for period-2 and period-3 elements, memorise the two anomalies with their cause: half-filled 2p³ (N) and the extra 2p–2p repulsion in O, and the 2s² full subshell in Be versus the 2p¹ in B.
When predicting molecular geometry and polarity, the candidate counts only bonding pairs (giving linear for H₂O or trigonal planar for NH₃) or stops at the electron-pair geometry and never converts to the molecular geometry, and then declares a molecule non-polar without checking whether bond dipoles cancel by symmetry.
VSEPR is often reduced to a table of 'number of pairs → shape' that ignores lone pairs, and polarity is treated as 'has polar bonds → polar molecule'. Candidates also forget that lone pairs occupy more space than bonding pairs, which compresses bond angles.
What to do instead
Use a rigid four-column worksheet for every molecule: total valence electrons → Lewis structure → count bonding pairs and lone pairs on the central atom → name the electron-pair geometry, then the molecular geometry, then state the ideal angle and the compression caused by lone pairs. Finish by drawing the dipole vectors and testing for cancellation (CO₂ vs SO₂, CCl₄ vs CHCl₃, NH₃ vs BF₃).
The candidate ranks boiling points by molecular mass alone, or claims hydrogen bonding for any molecule containing H (e.g., H₂S, HCl, CH₃F), and ignores that hydrogen bonds require H bonded directly to N, O or F. They also mix up the strength order of intermolecular forces versus covalent bonds.
'Heavier = higher boiling point' works for the alkane and halogen series, so candidates over-generalise it, and the phrase 'hydrogen bonding' is recalled as a synonym for 'strong dipole' rather than as a specific structural requirement.
What to do instead
Before ranking, classify each substance by its dominant interaction: ionic lattice, metallic, hydrogen bond (only N–H, O–H, F–H), dipole–dipole, or London dispersion. Rank within a family by mass/size, then compare across families by interaction strength. Practise the classic contrasts H₂O vs H₂S, HF vs HCl, ethanol vs dimethyl ether, and butane vs propanone.
When asked how a change of temperature, pressure, volume or catalyst affects a system at equilibrium, the candidate answers with Le Chatelier qualitatively but gets the direction wrong for temperature (treating an exothermic forward reaction as favoured by heating) and wrongly claims a catalyst or a pressure change shifts the position of equilibrium.
Candidates memorise 'increase T shifts right' without tying it to the sign of ΔH, and they conflate the rate effect of a catalyst with a change in the equilibrium constant. Pressure effects are applied even when the number of gaseous moles is equal on both sides.
What to do instead
Use a fixed decision tree: (1) write ΔH with its sign; (2) for temperature, compare with the endothermic direction and state that K changes; (3) for pressure/volume, count gaseous moles on each side and only shift if they differ; (4) for catalysts, state explicitly 'reaches equilibrium faster, K and position unchanged'. Always name which side becomes favoured and why, in one sentence.
The candidate computes the equilibrium constant from initial concentrations without building an ICE (initial–change–equilibrium) table, or omits units and the correct powers, and treats Kc as if it depended on the initial amounts. They also confuse Kc with the reaction quotient Q when deciding the direction of shift.
The algebra is done in one step under time pressure, and the distinction between initial and equilibrium concentrations is glossed over because both appear in the same problem statement. The dimensionless nature of K (activities) is rarely emphasised in class.
What to do instead
Always write the balanced equation, then a three-row ICE table with x, then substitute only the equilibrium row into the Kc expression with the stoichiometric exponents. Compute Q from the current (non-equilibrium) values and compare Q with K to decide the shift direction. State that Kc depends only on temperature.
The candidate calculates pH of a weak acid using [H⁺] = √(Ka·c) even when the degree of dissociation is large, or uses the strong-acid formula [H⁺] = c for a weak acid; they also forget to convert between Ka and pKa, and they compute pH of a buffer with the Henderson–Hasselbalch equation while plugging in initial rather than equilibrium amounts.
The approximation √(Ka·c) is presented as 'the' formula for weak acids, and the condition for its validity (α < 5 %, i.e., c/Ka > ~10⁴) is not stressed. Buffer problems look like simple substitution, so the acid–base reaction that consumes strong acid or base is skipped.
What to do instead
For every weak acid, first compute α = √(Ka/c) and check it is below about 5 % before using the approximation; otherwise solve the quadratic. For buffers, do the limiting-reagent step first (strong acid/base reacts completely with the buffer pair), then substitute the resulting equilibrium amounts into pH = pKa + log([base]/[acid]). Convert Ka ↔ pKa explicitly in each problem.
In titration and salt-hydrolysis questions, the candidate assumes pH = 7 at the equivalence point regardless of the acid/base strengths, and classifies a salt such as NH₄Cl or CH₃COONa as neutral because 'it comes from an acid and a base'.
Strong acid–strong base titrations dominate practice, so 'equivalence = neutral' becomes a reflex. Hydrolysis is taught as a list of rules rather than as a competition between the conjugate acid and conjugate base for water.
What to do instead
At every equivalence point, identify the species present and decide which of them can donate or accept a proton: strong acid + strong base → neutral; weak acid + strong base → basic (conjugate base hydrolyses); weak base + strong acid → acidic (conjugate acid hydrolyses). Write the hydrolysis equation and its Kh = Kw/Ka (or Kw/Kb) before predicting the pH range, and sketch the titration curve shape.
When balancing redox equations in acidic or basic medium, the candidate balances atoms but not charge, or adds H₂O and H⁺ to the wrong side; in electrochemistry they calculate E°cell as the sum of both reduction potentials instead of E°cathode − E°anode, and they cannot decide which electrode is the anode from the sign of the potential.
Half-reaction balancing is a procedural skill that is rushed, and the sign convention for standard reduction potentials is genuinely confusing: tables list reduction potentials, but the oxidation half-reaction must be reversed without changing the tabulated value's sign before subtracting.
What to do instead
Balance in a fixed order: atoms other than O and H → O with H₂O → H with H⁺ (or OH⁻ in basic) → charge with electrons; then multiply to equalise electrons and add. For cells, write E°cell = E°cathode − E°anode using the tabulated reduction values unchanged, state that the more positive electrode is the cathode, and check that E°cell > 0 for a spontaneous reaction.
In organic chemistry, the candidate identifies a functional group but misnames the compound (e.g., calls an ester an ether, an amide an amine, or an aldehyde a ketone), fails to recognise cis–trans or optical isomerism when a chiral carbon is present, and predicts the product of an addition or substitution reaction without applying Markovnikov's rule or the correct reagent.
Functional groups are learned as isolated names without the priority rules for naming, and stereochemistry is treated as a separate chapter rather than a property to check on every structure. Reaction types are memorised as lists of reagents rather than as electron-flow patterns.
What to do instead
For every structure, run a checklist: locate and name the principal group by IUPAC priority (acid > ester > amide > aldehyde > ketone > alcohol > amine), then check for a carbon with four different substituents (chiral → optical) and for a C=C with two different groups on each carbon (→ cis/trans). For reactions, classify as addition (alkene/alkyne, Markovnikov with HX/H₂O), substitution (alkane with halogen, alcohol with HX), elimination or condensation, and write the reagent and the product explicitly.