Castilla y León PAU Chemistry common mistakes
Castilla y León PAU Chemistry · Understand the traps and how to avoid them.
Writing electron configurations by filling orbitals strictly in the order 1s 2s 2p 3s 3p 4s 3d 4p without checking anomalies, so Cr and Cu come out as [Ar]4s²3d⁴ and [Ar]4s²3d⁹, and ions such as Cu⁺ or Fe²⁺ are built by removing electrons from the wrong subshell.
The aufbau diagram is memorised as a fixed list rather than as an energy-ordering rule; candidates also forget that once the 3d subshell is occupied the 4s orbital is no longer lower in energy, so 4s electrons are not always the first to leave.
What to do instead
After writing any configuration for a neutral atom of group 6 or 11, apply the half-filled/filled stability check and write the exception explicitly. For transition-metal cations, remove electrons from the highest-n subshell first (4s before 3d) and write the configuration with the 3d block before 4s to show the correct occupation. Practise with Cr, Cu, Fe²⁺, Fe³⁺, Cu⁺, Cr³⁺ until each is written correctly from memory.
Answering periodic-property questions by quoting a single trend ('electronegativity increases to the right and upwards') without handling the crossover cases, so ionisation energy is claimed to rise smoothly across period 2 and the first ionisation energy of O is stated to be higher than that of N.
The trend is learned as a slogan, and the subshell half-filled stability argument (and the small size effect in period 2/3 comparisons) is not applied.
What to do instead
For every comparison, justify with two arguments: effective nuclear charge and shielding, and the electron configuration of the species losing the electron. Learn the standard anomalies (Be > B, N > O, Mg > Al, P > S) and the reason for each. For group comparisons, state that the new shell outweighs the increased Z.
Drawing a Lewis structure that satisfies the octet but then assigning molecular geometry from the number of electron pairs around the central atom without subtracting lone pairs, so H₂O is called linear, NH₃ is called trigonal planar and SO₂ is given a 180° bond angle.
The VSEPR procedure is compressed into 'count the bonds', and the distinction between electron-pair geometry and molecular geometry is not made explicit.
What to do instead
Use a fixed four-step routine: total valence electrons, skeleton, distribute as bonding and lone pairs, then count electron domains and lone pairs separately. Write both geometries down (e.g. NH₃: tetrahedral electron-pair geometry, trigonal pyramidal molecular geometry, ~107°). Include one species with a lone pair on the central atom in every practice set.
Treating all molecules with polar bonds as polar molecules, so CO₂, CCl₄, BF₃ and SO₃ are labelled polar, and predicting the wrong relative boiling points for substances of similar molar mass.
Polarity is judged bond by bond rather than from the vector sum, and molecular geometry is not used to check whether the bond dipoles cancel.
What to do instead
For each molecule, first determine the geometry from VSEPR, then draw the bond dipoles with their directions and add them as vectors. State the rule explicitly: identical bonds arranged symmetrically about the central atom cancel. Then rank boiling points using the dominant intermolecular force (dispersion < dipole–dipole < hydrogen bonding) and check whether hydrogen bonding is possible (H bonded to N, O or F).
Adding enthalpies of formation directly without reversing the sign of the formation enthalpy of the reactants, or forgetting to multiply each term by its stoichiometric coefficient, when applying Hess's law or ΔH°r = ΣΔH°f(products) − ΣΔH°f(reactants).
The formula is memorised as a single subtraction but the physical meaning (products minus reactants, formation from elements) is not, and the coefficients in the balanced equation are treated as decoration.
What to do instead
Always balance the equation first and write the coefficients above each species. Then write the sum explicitly with signs shown, e.g. ΔH° = [1(−394) + 2(−286)] − [1(−75) + 0], before calculating. Check plausibility: combustion and neutralisation of strong acid–strong base must come out negative.
Judging spontaneity from the sign of ΔH alone ('exothermic means spontaneous'), or claiming a reaction with positive ΔH and positive ΔS is never spontaneous, instead of evaluating ΔG = ΔH − TΔS and finding the temperature threshold.
Enthalpy is the familiar quantity and entropy is treated as a vague 'disorder' idea, so the temperature dependence of ΔG is never applied.
What to do instead
For every spontaneity question, build a four-case table (ΔH, ΔS signs) and state the temperature condition for each. When both are positive, actually solve T > ΔH/ΔS and give the numerical threshold with units in kelvin. Remember ΔG = 0 defines equilibrium, not 'no reaction'.
Writing equilibrium expressions that include pure solids or liquids, or using initial concentrations instead of equilibrium concentrations in Kc, and then computing pH or composition from those wrong values.
The equilibrium table is set up but the 'change' row is not subtracted/added correctly, and the rule that heterogeneous pure condensed phases have unit activity is not internalised.
What to do instead
Always draw an ICE table with initial, change and equilibrium rows, and cross out any pure solid or liquid in the K expression before substituting. After solving, back-substitute the equilibrium values into Kc to verify. For weak acids, state the approximation x ≪ c₀ and check it is valid (x less than about 5% of c₀).
Calculating the pH of a weak acid with pH = −log c₀ as if it were strong, or forgetting to convert between Ka and pKa (or Kb and pKb) and using pOH and pH interchangeably.
Strong-acid formulas are over-practised, the square-root relation [H₃O⁺] = √(Ka·c₀) is used without checking its validity, and the pKa/pKb conversion pKa + pKb = 14 is not applied at 25 °C.
What to do instead
Classify the acid first (strong/weak, and note polyprotic behaviour). For weak acids write Ka = x²/(c₀ − x), state the approximation, solve, and verify the 5% condition; otherwise solve the quadratic. Always label whether a number is pH, pOH, pKa or pKb, and convert with pH + pOH = 14 only at 298 K.
Computing solubility from Ksp by writing s directly as the molar solubility without accounting for the stoichiometry, e.g. giving s = Ksp for AgCl but also s = Ksp for Ag₂CrO₄ instead of s = (Ksp/4)^(1/3), and ignoring the common-ion effect when the solution already contains one ion.
The Ksp expression is written correctly but the relationship between s and the ion concentrations is not derived from the stoichiometric coefficients, and the common-ion case is treated as an ordinary solubility problem.
What to do instead
For every salt, write the dissolution equation, then express each ion concentration in terms of s with its coefficient before substituting into Ksp. For common-ion problems, put the initial concentration of the common ion in the ICE table and solve the resulting expression (often a simple linear or cubic equation). Compare the result with the pure-water solubility and comment on the direction of the change.
Balancing redox equations in acidic or basic medium without adding H₂O and H⁺/OH⁻ to close the oxygen and hydrogen balance, or building a galvanic cell with the anode and cathode reversed and reporting a cell potential with the wrong sign.
The half-reaction method is applied only to electrons and charge, and the conventional notation (oxidation at the anode, reduction at the cathode, E°cell = E°cathode − E°anode) is confused with the sign convention of the standard reduction potentials.
What to do instead
Use the strict half-reaction sequence: balance atoms other than O and H, balance O with H₂O, balance H with H⁺ (or add OH⁻ in basic medium), balance charge with electrons, equalise electrons, add and cancel. For cells, write the two half-reactions with their E° values, identify which is reversed (oxidation), and compute E°cell = E°cathode − E°anode; a positive value confirms the spontaneous direction. Never multiply E° values by stoichiometric coefficients.
Naming or drawing organic compounds without applying the priority rules for the principal functional group and the lowest-locant rule, so a compound is called 3-methylbutan-1-ol instead of 3-methylbutan-1-ol's correct isomer, or an aldehyde is named as a ketone and a carboxylic acid as an ester.
The suffix is chosen from the most familiar group rather than from the IUPAC priority order (acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine > alkene/alkyne > alkane), and numbering is started from the wrong end.
What to do instead
Before naming, list all functional groups, pick the highest-priority one as suffix, and number the chain to give that group the lowest locant; then apply lowest locants to multiple bonds and substituents alphabetically. For drawing from a name, build the parent chain first, number it, then attach substituents and check the total valence of every carbon. Practise a set containing at least one compound with two functional groups.