IB Chemistry SL common mistakes
IB Chemistry SL · Understand the traps and how to avoid them.
For an ion such as Fe³⁺, they calculate neutrons as 56 − 3 = 53 and remove three electrons from the 3d subshell before removing from 4s.
They treat the mass number as a particle count and memorise '4s fills before 3d' without applying the rule that electron removal follows the highest principal quantum number.
What to do instead
Use A = Z + N, so neutrons = 56 − 26 = 30. Write the neutral atom configuration [Ar]3d⁶4s², then remove 4s electrons first: [Ar]3d⁵ for Fe³⁺.
In ideal-gas calculations they substitute V in cm³ into PV = nRT while using R = 8.31 J K⁻¹ mol⁻¹, or they use the mass of solvent rather than the mass of solution when calculating concentration.
They recall the formula but do not track units or distinguish concentration (amount of solute per volume of solution) from molality (amount of solute per mass of solvent).
What to do instead
Convert volume to dm³ and pressure to kPa, or use SI units consistently, and write the units beside every substitution. For concentration use n(solute)/V(solution) in dm³; for molality use kg of solvent.
They predict SF4 as tetrahedral by counting the lone pair as a bonding domain, or they describe H2O as linear because they count only the two O–H bonds.
They memorise shapes for AX4 and AX2 without counting lone pairs on the central atom and without applying lone-pair repulsion.
What to do instead
Count σ bonds plus lone pairs to get electron domains, assign AXnEm, and apply repulsion order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. SF4 is seesaw; H2O is bent.
They state that HCl forms hydrogen bonds because H is bonded to a highly electronegative atom, or they explain the boiling point of water by breaking O–H covalent bonds.
They apply a simplified 'H plus electronegative atom' rule and do not separate intramolecular covalent bonds from intermolecular attractions.
What to do instead
Hydrogen bonding requires H directly bonded to N, O or F. Boiling breaks intermolecular forces, not covalent bonds; HCl has dipole–dipole forces, not hydrogen bonds.
They classify brass as a compound and a clear salt solution as a pure substance because it looks uniform.
They use visual appearance and everyday language instead of composition and separation behaviour.
What to do instead
A pure substance has fixed composition and a sharp melting point; an alloy is a mixture of elements, and a homogeneous solution is a mixture that can be separated by physical means such as evaporation.
They explain the decrease in atomic radius across Period 3 only as 'more protons' without mentioning that electrons enter the same shell, and they cannot explain why Al has a lower first ionisation energy than Mg.
They memorise trend directions without the effective nuclear charge and shielding model, and they ignore subshell energies.
What to do instead
Write Zeff = Z − shielding; across a period shielding is roughly constant while Z increases, so radius falls and ionisation energy rises. The Mg–Al dip occurs because the 3p electron is higher in energy and better shielded than the 3s electron.
They calculate ΔH by adding the bond enthalpies of bonds formed to those broken, or they give bonds broken a negative sign.
They confuse bond breaking (endothermic, ΔH positive) with bond formation (exothermic, ΔH negative).
What to do instead
Use ΔH = Σ(bonds broken) − Σ(bonds formed), with both sums as positive magnitudes; sketch an energy cycle and check that bond formation lowers the enthalpy.
They say adding a catalyst shifts the equilibrium position to the products, or that adding an inert gas at constant volume increases product yield by increasing total pressure.
They conflate the rate of attainment with the position of equilibrium and do not apply partial pressures.
What to do instead
A catalyst speeds the forward and reverse rates equally and does not change K or the equilibrium position. At constant volume, an inert gas does not change the partial pressures of the reacting gases, so there is no shift; only changes in their partial pressures matter.
In an SN2 mechanism they draw the curly arrow from the nucleophile to a hydrogen atom on the haloalkane rather than to the carbon bearing the halogen, or they show a tertiary haloalkane undergoing SN2.
They memorise the overall substitution product without following electron pairs and steric access.
What to do instead
Draw the arrow from the nucleophile lone pair to the δ+ carbon and the C–X bond arrow to X. SN2 is favoured at primary carbons; tertiary carbons react by SN1 via a carbocation.
They add absolute uncertainties directly when multiplying or dividing quantities, and they report a burette reading such as 24.50 cm³ without ±0.05 cm³.
They do not distinguish absolute from percentage uncertainty and do not match the reported precision to the instrument.
What to do instead
For multiplication or division, convert to percentage uncertainties and add them; for addition or subtraction, add absolute uncertainties. Quote the final uncertainty to one significant figure and match the decimal places of the value.