Bagrut Chemistry common mistakes
Bagrut Chemistry · Understand the traps and how to avoid them.
When asked to write the electron configuration of a transition-metal ion, the candidate fills orbitals in the order of the aufbau diagram for the neutral atom and then removes electrons from the highest-energy subshell (e.g., writes Fe²⁺ as [Ar]3d⁴4s² instead of [Ar]3d⁶).
The aufbau order applies to building neutral atoms; once 4s and 3d are both occupied, 4s is higher in energy, so ionization removes 4s electrons first. Candidates memorize a single filling sequence and never distinguish filling order from removal order.
What to do instead
Drill the rule explicitly: build the neutral atom, then for cations remove from the outermost shell (highest n), which means 4s before 3d. Practice with Fe²⁺, Fe³⁺, Cu⁺, Zn²⁺, Cr³⁺ and check each against the known anomalous configurations of Cr and Cu.
The candidate draws a Lewis structure for a molecule with an expanded octet on a period-2 atom (e.g., puts 10 electrons around N in NO₂ or around O in a sulfate resonance form with S=O double bonds written as O–S–O with extra lone pairs on O), or omits formal-charge bookkeeping entirely.
Textbooks show sulfate and phosphate with double bonds, and candidates generalize this to all central atoms without checking whether the atom has accessible d orbitals. Formal charge is treated as an optional decoration rather than a decision tool.
What to do instead
Apply a checklist: count valence electrons, place the octet on period-2 atoms absolutely, and only expand the octet on period 3 and below. Then compute formal charge on every atom and prefer the structure minimizing formal charge with negative charge on the more electronegative atom.
In stoichiometry problems the candidate treats the limiting reactant as the species with the smallest number of moles, ignoring the coefficients in the balanced equation, and then computes percent yield from the theoretical yield of that wrong reactant.
Comparing raw mole counts is a shortcut that works only for 1:1 stoichiometry; exam questions deliberately use coefficients like 2:3 so the shortcut fails. Candidates also skip balancing or balance only the metals and oxygens.
What to do instead
Always divide moles by the stoichiometric coefficient and compare those quotients; the smallest quotient identifies the limiting reactant. Then compute theoretical yield from that reactant and report yield as actual/theoretical × 100. Do three mixed-coefficient drills before the exam.
When a gas problem changes temperature, the candidate plugs Celsius values directly into the ideal gas law or into P₁V₁/T₁ = P₂V₂/T₂, and reports pressures or volumes off by a large factor.
Everyday intuition uses Celsius, and the formulas are often memorized without the note that T must be absolute. The error is silent because the arithmetic looks correct.
What to do instead
Make 'convert to kelvin' the first written line of every gas calculation, before substituting anything. Check plausibility: doubling the kelvin temperature at constant volume should roughly double the pressure, which a Celsius substitution will not produce.
The candidate predicts the direction of a shift after adding an inert gas at constant volume, or after adding a catalyst, and claims the equilibrium shifts toward the products; similarly, they confuse a shift in position with a change in the value of K.
Le Chatelier's principle is learned as a list of slogans ('more pressure shifts to fewer moles') without the condition that the partial pressures or concentrations must actually change. Catalysts and inert gases are memorized as 'affecting equilibrium'.
What to do instead
State the mechanism first: a shift requires a change in the reaction quotient Q relative to K. Inert gas at constant volume changes no partial pressure, so Q is unchanged; a catalyst changes only the rate of approach to the same K. Only temperature changes the value of K, and its direction follows from ΔH.
When calculating the pH of a weak acid, the candidate uses pH = −log C directly, or sets up the Ka expression but then approximates the equilibrium concentration of the acid as unchanged even when the degree of dissociation is large, and cannot handle a buffer without the Henderson–Hasselbalch relation.
Strong-acid formulas are taught first and stick; the ICE table is seen as optional paperwork. Candidates also forget that in a buffer the conjugate base concentration comes from the salt, not from the weak acid's own dissociation.
What to do instead
Write the Ka expression and an ICE table every time, then test the approximation: if C/Ka is less than about 100, solve the quadratic. For buffers, identify the acid and its conjugate base as separate added amounts, then use pH = pKa + log([base]/[acid]) and check the ratio is between 0.1 and 10.
The candidate computes ΔH for a reaction using bond enthalpies of the bonds in the products minus those in the reactants (reversed sign), or mixes Hess's law with bond energies in the same calculation, and reports an endothermic reaction as exothermic.
Bond breaking releases energy is a persistent misconception, and the two methods look interchangeable. Hess's law sums are also done with the wrong sign when an equation is reversed.
What to do instead
Anchor on the physical picture: breaking bonds absorbs energy, forming bonds releases it, so ΔH = Σ(bonds broken) − Σ(bonds formed). When reversing an equation under Hess's law, flip the sign of its ΔH. Sanity-check the sign against whether the products are more or less stable than the reactants.
In an electrochemical cell the candidate assigns the anode and cathode by left/right position in the diagram rather than by oxidation and reduction, and writes the cell potential as E°cell = E°cathode + E°anode with the anode value kept positive.
Diagrams are drawn with the anode on the left by convention, so position is used as a proxy for chemistry. Candidates also forget that oxidation potentials and reduction potentials have opposite signs.
What to do instead
Label every half-cell with its reaction and its role: oxidation at the anode, reduction at the cathode, always. Use E°cell = E°cathode − E°anode with both values taken as reduction potentials from the table, and check the sign: a spontaneous galvanic cell must give a positive E°cell.
The candidate claims that raising the temperature always increases the reaction rate by increasing the collision frequency, and cannot explain why a catalyst increases rate without changing ΔH or the equilibrium constant; they also treat the rate-determining step as the fastest step.
The Arrhenius factor of the fraction of molecules with energy above the activation energy is under-emphasized relative to collision frequency, and the energy profile diagram is read as a picture rather than a quantitative model.
What to do instead
For each rate question, name the three levers: collision frequency, orientation, and fraction of collisions exceeding Ea. Explain temperature mainly through the exponential Ea/RT term, and a catalyst through a lower Ea on a new pathway. Identify the rate-determining step as the slowest step, the highest barrier on the profile.
When naming or drawing an organic compound, the candidate picks the wrong principal functional group for the suffix (naming an ester as an ether or an aldehyde as a ketone), numbers the chain from the end nearest a substituent instead of the end nearest the principal group, and misses cis/trans or optical isomerism because it does not check for a chiral center or a restricted double bond.
Priority rules are memorized as a list without practice in mixed molecules, and isomerism is studied as a separate topic rather than as a property to check on every structure.
What to do instead
Apply the priority order (carboxylic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine > alkene > alkyne > halide) before numbering, then number to give the principal group the lowest locator. For every structure drawn, run three checks: restricted rotation (cis/trans), four different groups on a carbon (chirality), and same molecular formula with different connectivity (structural isomers).