AP Chemistry common mistakes
AP Chemistry · Understand the traps and how to avoid them.
When writing the electron configuration of a transition-metal cation, the student removes 3d electrons before 4s electrons - for example writing Fe²⁺ as [Ar]3d⁴4s² or Cu⁺ as [Ar]3d⁹4s¹.
The 4s orbital is taught as filling before 3d, so students infer that it must also empty last; in the actual cation the 4s level lies above 3d and is emptied first.
What to do instead
Write the neutral noble-gas configuration first, then remove electrons from the highest principal quantum number before touching d electrons: Fe²⁺ = [Ar]3d⁶, Cu⁺ = [Ar]3d¹⁰. Verify by counting the remaining electrons against the ion's charge.
The student reads a photoelectron spectrum by treating the tallest peak as the highest binding energy, or counts peaks to report the total number of electrons in the atom.
The graph's conventions are inverted relative to ordinary plots - binding energy increases toward the left, and peak height scales with how many electrons occupy that subshell - so the habitual 'higher means more' reading produces wrong conclusions.
What to do instead
State the two rules before answering: horizontal position gives binding energy and therefore identifies the subshell and n; peak height gives relative electron count and the heights must sum to the atom's total electrons. Self-check against a known spectrum such as Ne, which shows three peaks in a 2:2:6 ratio.
The student names molecular geometry directly from the total number of electron domains, calling H₂O 'tetrahedral' and SO₂ 'trigonal planar' instead of bent.
Geometry charts list electron-domain geometries first, and lone pairs that are drawn on the central atom get dropped when the student counts only bonded atoms, so the second step of the decision procedure is skipped.
What to do instead
Use a fixed two-step routine: count all domains to get the electron-domain geometry, then re-count only bonded atoms to name the molecular shape. Add an angle check - CH₄ 109.5°, NH₃ 107°, H₂O 104.5° - because each lone pair compresses the bond angle.
The student claims hydrogen bonding for any substance containing hydrogen (HCl, H₂S, CH₃Cl) and ranks London dispersion forces as dominant for large nonpolar molecules without comparing them against a hydrogen-bonding compound of similar size.
The rule collapses into 'H present means hydrogen bond' instead of 'H covalently bonded to N, O, or F', and students substitute molar mass comparisons for comparisons of intermolecular force type.
What to do instead
Before ranking, write each species' force inventory: hydrogen bonding only when H is directly bonded to N, O, or F; otherwise dipole–dipole if polar or London dispersion if nonpolar. Rank force type first (H-bond > dipole–dipole > dispersion), and use size or polarizability only as a tiebreaker within the same type.
The student writes net ionic equations by splitting weak acids, weak bases, and insoluble solids into separate ions, and finishes without checking that total charge balances on both sides.
Strong-electrolyte splitting is over-applied to every formula containing ions or an acidic hydrogen, and the mass balance check (atoms) is performed while the charge balance check is forgotten entirely.
What to do instead
Apply three rules in order: split only strong acids, strong bases, and soluble salts; keep weak electrolytes, gases, and precipitates as intact formulas; then verify atoms and total charge on each side. Circle the charge column specifically, since atom counts alone do not catch this error.
The student derives a rate law from a proposed mechanism by using the coefficients of the overall balanced equation as exponents, and leaves an intermediate species in the final rate expression.
The overall equation is the most prominent piece of given information and it superficially resembles the coefficient-to-exponent pattern the student already uses for equilibrium constants, so mechanism steps are bypassed.
What to do instead
Take the rate law from the slow step only, using that step's coefficients as exponents. If an intermediate appears, solve the fast equilibrium step for it and substitute it out. Add the reminder that reaction orders are established by experiment, never by the overall balanced equation.
The student reports the heat absorbed by the solution as ΔH of the reaction with the same sign, or reports q for the experiment without dividing by the moles of limiting reactant consumed.
q_solution and q_reaction are computed in a single line and the sign inversion q_rxn = −q_surroundings is silently dropped, while ΔH is a per-mole quantity that students treat as a per-experiment quantity.
What to do instead
Write the sign step as its own line, then divide: ΔH = −q_solution / n_limiting. Check the final units are kJ/mol and apply a plausibility test - a temperature rise (positive q_solution) must correspond to a negative ΔH.
The student computes the reaction quotient Q from the stock concentrations of two solutions that were just combined, ignoring the dilution that occurs on mixing.
The problem presents two labeled bottles with clean molarities, and students plug those numbers directly into the Q expression without accounting for the new total volume in which the ions now find themselves.
What to do instead
Insert an explicit dilution line before any equilibrium work: c_new = c_stock × V_stock / V_total for each ion. Then build Q from those diluted values with stoichiometric coefficients as exponents and compare to Ksp or K. Never write a Q or K expression straight from bottle labels.
The student applies pH = pKa + log([base]/[acid]) using the starting buffer amounts even after a strong acid or strong base has been added, skipping the neutralization stoichiometry.
Henderson–Hasselbalch is the most rehearsed buffer formula, so it is deployed on every buffer question, including those where a strong reagent has already consumed part of the buffer pair and changed the ratio.
What to do instead
Use a two-stage layout: first a stoichiometry table in which the strong acid converts base to conjugate acid (limiting reactant decides how much), then substitute the post-reaction moles into Henderson–Hasselbalch. If the strong reagent is in excess, declare the buffer destroyed and compute pH from the leftover strong acid or base.
The student calculates E°cell by adding the two tabulated reduction potentials without reversing the anode half-reaction, or uses the sign of E°cell to predict spontaneity at nonstandard concentrations.
Every tabulated potential is written as a reduction, so both numbers look ready to add as printed; and E°cell gets treated as a universal spontaneity test in the same way as ΔG.
What to do instead
Use E°cell = E°cathode − E°anode with both values taken as reduction potentials, or flip the anode value and add. For nonstandard conditions, apply the Nernst equation to get E before judging direction, and link E > 0 to ΔG = −nFE < 0 as the spontaneity criterion.