Državna Matura Chemistry common mistakes
Državna Matura Chemistry · Understand the traps and how to avoid them.
When asked to write the electron configuration of a transition-metal cation such as Fe²⁺ or Cu⁺, the candidate removes electrons from the 4s orbital only after emptying 3d, or writes [Ar]3d⁶ for Fe²⁺ but [Ar]3d⁹4s² for Cu⁺, mixing the neutral-atom order with the cation order.
Candidates memorize the aufbau filling order (4s fills before 3d) and apply it unchanged to ions, not realizing that once 3d is occupied, 4s lies higher in energy and electrons are removed from 4s first. The 4s/3d crossover is genuinely counterintuitive and textbooks often state the filling rule without restating the removal rule.
What to do instead
Drill a fixed two-step procedure: write the neutral-atom configuration, then remove electrons from the highest-n principal quantum number first (4s before 3d for first-row transition metals). Practise the full set Fe, Fe²⁺, Fe³⁺, Cu, Cu⁺, Cu²⁺, Cr, Cr³⁺ and check each against the rule that the ion has the same electron count as the preceding noble gas plus the stated d electrons. For Cu²⁺ confirm d⁹ paramagnetism and for Zn²⁺ confirm d¹⁰ diamagnetism as self-checks.
The candidate predicts molecular shape by counting all electron pairs around the central atom, including lone pairs, and reports e.g. H₂O as linear or NH₃ as trigonal planar, or omits the lone pair entirely and reports SO₂ as linear rather than bent.
VSEPR is taught as 'count pairs and match a shape name', so candidates memorize a table of total pair counts instead of separating bonding domains from lone pairs. They also forget that a lone pair occupies more space and compresses bond angles below the ideal value.
What to do instead
Use a written template for every molecule: (1) total valence electrons, (2) Lewis structure, (3) count sigma bonds + lone pairs = steric number, (4) name the electron-domain geometry, (5) remove lone pairs to get the molecular geometry, (6) state the expected angle and whether it is compressed. Apply it to H₂O (bent, ~104.5°), NH₃ (trigonal pyramidal, ~107°), SO₂ (bent) and CH₄ (tetrahedral, 109.5°) side by side so the difference between domain geometry and molecular geometry becomes automatic.
When balancing a redox equation in acidic or basic solution, the candidate balances atoms but not charge, or adds H₂O and H⁺ to the wrong side, and in basic medium forgets to convert H⁺ to OH⁻ by adding OH⁻ to both sides. The final equation has unequal total charge on the two sides.
The half-reaction method has many bookkeeping steps and candidates lose track of which species they are currently balancing (atoms, then O with H₂O, then H with H⁺, then charge with electrons). The basic-medium conversion is an extra step that is easily skipped under time pressure.
What to do instead
Follow the fixed order every time: balance all atoms other than O and H, balance O with H₂O, balance H with H⁺, balance charge with electrons, multiply half-reactions to equalize electrons, add and cancel. Then perform an explicit charge audit by summing charges on each side and writing the totals. For basic medium, after the acidic balance add one OH⁻ per H⁺ to both sides, combine H⁺ + OH⁻ into H₂O, and cancel waters; verify the charge audit again after conversion.
In limiting-reagent problems the candidate compares the given masses (or moles) of the two reactants directly and declares the smaller number limiting, ignoring the stoichiometric coefficients, or computes the theoretical yield from the excess reagent.
The phrase 'limiting reagent' suggests the reactant present in the smallest amount, and candidates shortcut the mole-ratio division. Problems are also often set with the numerically smaller amount being the excess reagent precisely to test this.
What to do instead
Always convert both reactants to moles, divide each by its stoichiometric coefficient, and compare the quotients; the smallest quotient is limiting. Then compute product moles from the limiting reagent only, using its coefficient ratio. Practise with a deliberately misleading case (e.g. 4 mol H₂ with 1 mol O₂ for water: 4/2 = 2 versus 1/1 = 1, so O₂ limits) and state the comparison explicitly in the answer.
The candidate calculates the enthalpy of a reaction using Hess's law but adds or subtracts formation enthalpies with the wrong signs, or applies ΔH = ΣΔHf(products) − ΣΔHf(reactants) without multiplying each term by its stoichiometric coefficient, and reports an answer with the wrong sign for an exothermic reaction.
Formation enthalpies are tabulated as signed values and candidates either flip signs when reversing a reaction or forget that the elements in their standard states have ΔHf = 0. The multiplication by coefficients is a separate arithmetic step that is easily dropped.
What to do instead
Write the formation-enthalpy expression with explicit coefficients before substituting any numbers, e.g. ΔH = [1·ΔHf(CO₂) + 2·ΔHf(H₂O)] − [1·ΔHf(CH₄) + 2·ΔHf(O₂)], and mark O₂ as 0. When reversing a step in a Hess cycle, write the reversed equation and change the sign of ΔH on the same line. Sanity-check the sign against whether the reaction is combustion or bond-forming, which should be exothermic.
When asked to write the electron configuration, formula or reactivity trend down a group, the candidate confuses group and period numbering, or predicts that reactivity of halogens increases down the group, or that metallic character decreases down a group.
Periodic trends are memorized as isolated sentences without the underlying cause (effective nuclear charge, shielding, atomic radius, ionization energy). Without the causal chain, the direction of each trend is guessed, and group/period labels are swapped because both are small numbers.
What to do instead
Learn each trend through one causal sentence: down a group, radius increases because shells are added and shielding rises, so ionization energy falls and metallic character rises; across a period, effective nuclear charge rises, so radius falls and ionization energy rises. For halogens specifically, oxidizing power falls down the group because the larger atom holds the incoming electron less tightly. Draw the periodic table with arrows for each trend and recite the causal sentence, then test with the alkali metals and halogens explicitly.
The candidate names or draws an organic structure that violates valence or omits a functional group, for example writing an ester as R–O–CO–R with the carbonyl carbon bonded to two oxygens and one carbon but also to a hydrogen, or naming CH₃CH₂COOH as 'ethyl methanoate' or 'propanal'.
Functional groups are memorized as abbreviations rather than as connectivity, so the candidate cannot check that carbon has four bonds and oxygen two. Prefix/suffix rules (e.g. -oic acid, -al, -one, -oate) are learned as lists without mapping each suffix to a specific atom arrangement.
What to do instead
For every structure, draw the full skeleton with all bonds and count bonds on each carbon and oxygen before naming. Build a one-page functional-group table that pairs the suffix/prefix with a drawn fragment (carboxyl –COOH, aldehyde –CHO, ketone –CO–, ester –COO–, amide –CONH₂) and the rule for which carbon gets the lowest locant. Practise converting a name to a structure and back for the same molecule to expose mismatches.
In equilibrium calculations the candidate substitutes initial concentrations into the equilibrium constant expression, or omits the change row of the ICE table, and consequently obtains a K or concentration that does not satisfy the equilibrium condition. In acid–base problems the same candidate uses the initial acid concentration as [H⁺] for a weak acid.
The ICE table is taught as a layout but candidates treat the 'E' row as optional, and weak-acid problems look like simple concentration problems. The distinction between initial, change and equilibrium is glossed over because strong acids behave that way.
What to do instead
Always build a three-row ICE table and write the equilibrium expression only from the E row. For weak acids, write Ka = x²/(c − x), state the approximation x ≪ c only when Ka/c is small, and check by substituting the answer back into the Ka expression. Include at least one problem where the approximation fails (relatively large Ka or very dilute acid) and solve the quadratic, so the check becomes habitual.
The candidate writes the cell diagram or calculates E°cell with the anode and cathode reversed, or uses the reduction potential of the oxidized species with the wrong sign for the oxidation half-reaction. The resulting E°cell is negative for a cell that should be spontaneous.
Both electrodes are written as reductions in the table, and the candidate forgets that the anode reaction must be reversed and its potential sign changed. Cell notation (anode on the left, salt bridge in the middle) is memorized as a picture without the oxidation-on-the-left rule.
What to do instead
Use the rule E°cell = E°cathode − E°anode with both values taken as reduction potentials from the table, never flipping signs manually. Identify the cathode as the species with the higher reduction potential. Write the cell diagram left to right as anode | anode solution || cathode solution | cathode and check that oxidation appears on the left. Confirm spontaneity by the sign of E°cell and of ΔG° = −nFE°cell.
In titration data analysis the candidate averages all burette readings including the rough trial, or averages the volumes rather than the titres, and reports the result to more significant figures than the burette can support (e.g. 23.4567 cm³ from a 0.1 cm³ burette).
Averaging feels like the safe statistical move, so the rough titration and outliers are included. Significant figures are treated as a formatting afterthought rather than a consequence of instrument precision.
What to do instead
Adopt a fixed data-handling routine: discard the rough trial, identify concordant titres within 0.10 cm³, average only those, and carry the precision of the burette (two decimal places, 0.05 cm³) through the calculation. State the number of significant figures allowed by the least precise measurement in the chain. Practise with a table containing one rough value and one outlier so the selection step is explicit.