Singapore A-Level H2 Chemistry central themes overview quiz (SEAB 9729) quiz

The kinetic theory of gases, which relates the temperature of a gas to the motion of its particles

Valence shell electron pair repulsion (electron-pair repulsion theory)

Le Chatelier's principle, which predicts how an equilibrium responds to a change in conditions

The Bohr model of the atom, which places electrons in fixed circular orbits around the nucleus

Energetics: how fast; equilibria: how feasible; kinetics: how far

Energetics: how far; equilibria: how fast; kinetics: how feasible

Energetics: how feasible; equilibria: how far; kinetics: how fast

All three answer how fast a reaction goes

$2.50\ \text{mol}$

$0.100\ \text{mol}$

$4.00 \times 10^{-3}\ \text{mol}$

$2.50 \times 10^{-3}\ \text{mol}$

The nuclear charge increases while electrons fill the same shell, so the radius falls and electrons are held more strongly

The atomic radius increases steadily across the period, so the outer electron is held more tightly each time

Electrons are added to successively higher principal shells, each one lying steadily further from the nucleus than the last

Shielding by inner electrons increases faster than the nuclear charge, pulling the outer electron in strongly

$K_a = \dfrac{[\text{HA}]}{[\text{H}^+][\text{A}^-]}$

$K_a = [\text{H}^+][\text{A}^-][\text{HA}]$

$K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$

$K_a = \dfrac{[\text{H}^+]}{[\text{A}^-]}$

They emit all wavelengths of visible light equally, so the eye perceives a strong mixture of colours

Their d orbitals split, and an electron absorbs visible light to move between the levels

Their s orbitals overlap with those of neighbouring atoms, and this overlap absorbs the visible light

They contain no unpaired electrons at all, and fully paired d electrons are what produces the colour

Infrared spectroscopy (functional-group absorptions)

Proton NMR spectroscopy (hydrogen environments)

A flame test (characteristic flame colour)

Mass spectrometry (molecular ion peak)

Infrared spectroscopy

Mass spectrometry

Carbon-13 NMR spectroscopy

A precipitation test for halide ions

The reaction must already be sitting at equilibrium, which is why no observable change is occurring

Energetics and kinetics are independent: feasibility does not guarantee a fast rate

The standard enthalpy change for the reaction must in fact be strongly positive rather than negative

The activation energy for the reaction must be zero, since a feasible reaction faces no energy barrier

Both groups become steadily more reactive down the group as the atoms grow larger and less tightly bound

Atomic radius decreases down both groups, which changes how strongly each atom holds its outer electrons

Down Group 2 it is easier to lose electrons, while down Group 17 it is harder to gain an electron

Nuclear charge decreases down both groups, weakening the attraction the nucleus has for its outer electrons

As a single one-step reaction that converts the starting material straight into the final target

As a random set of unrelated reactions chosen with no particular plan linking one step to the next

As an equilibrium that never quite reaches the product, so the target can only ever form in trace amounts

As a sequence of functional-group interconversions

Combine all three: molecular mass, functional groups and hydrogen environments, into one consistent structure

Use only the mass spectrum and ignore the other two, since the molecular mass alone fixes the structure

Pick whichever single spectrum is easiest to interpret and stop there, treating the others as unnecessary

Assume the compound is ionic and simply work out the formula from the charges on the ions rather than the spectra