67 exam-style questions on Structure 3: Classification of matter with answers and explanations. Six real samples below — the full set is free with an account.
Which species has the smallest radius: Na+, Mg2+, F-, or O2- (all isoelectronic with neon)?
Answer: Mg2+
All four are isoelectronic (10 electrons). The radius decreases as nuclear charge increases because more protons pull the same electron cloud inward. Mg2+ has 12 protons, the most of the four, so it has the smallest radius.
What is the correct IUPAC name for the compound CH3-CH2-CH2-OH?
Answer: Propan-1-ol
The chain has 3 carbons (prop-) and an -OH group on the end carbon (position 1), so it is an alcohol named propan-1-ol. Propan-2-ol would have the OH on the middle carbon.
Methane reacts with chlorine in UV light. What type of mechanism is this, and what is the reactive intermediate?
Answer: Free radical substitution; free radicals
The reaction CH4 + Cl2 -> CH3Cl + HCl in UV light proceeds by free radical substitution. UV light homolytically splits Cl2 into chlorine radicals (initiation), which then react in propagation steps. The intermediates are highly reactive free radicals with unpaired electrons.
A tertiary halogenoalkane undergoes nucleophilic substitution. Which mechanism does it favour and what is the rate-determining step?
Answer: SN1; formation of a tertiary carbocation
Tertiary halogenoalkanes favour SN1 because the tertiary carbocation formed is stabilised by three electron-donating alkyl groups. The rate-determining (slow) step is the heterolytic loss of the halide to form this carbocation, so the rate depends only on [halogenoalkane]. Primary halogenoalkanes favour SN2.
Ethene reacts with bromine water. What type of reaction is this and what is observed?
Answer: Electrophilic addition; orange bromine decolourises
The C=C double bond in ethene is electron-rich and attracts the electrophile Br2, undergoing electrophilic addition to form 1,2-dibromoethane. The orange/brown bromine is decolourised, which is the standard test for unsaturation (C=C double bonds).
When a primary alcohol is oxidised by acidified potassium dichromate under reflux, what is the final organic product?
Answer: A carboxylic acid
A primary alcohol is oxidised first to an aldehyde, then further to a carboxylic acid. Under reflux (which keeps the aldehyde in the flask), oxidation goes all the way to the carboxylic acid. Distilling would allow the aldehyde to be collected instead. The orange dichromate turns green (Cr3+).
67 Structure 3: Classification of matter questions — free with an account. Spaced repetition, streaks and full exam simulations included.
Practise them freeThis platform is independently developed and is not endorsed by, affiliated with, or sponsored by the International Baccalaureate Organization. “International Baccalaureate” and “IB” are registered trademarks of the IBO.