ELEMENTS AND PHYSICAL CHEMISTRY
PRACTICE EXAMINATION 2026
QUESTIONS WITH ANSWERS GRADED A+
◍ explain why the first ionisation energy of Mg is greater than the first
ionisation energy of Sr.
Answer: 1. Mg removes electron from shell closer to the nucleus - smaller
atomic radius2. greater attraction between nucleus and outer electron
◍ explain why the first ionisation energy of Mg is greater than the first
ionisation energy of Sr 2.
Answer: 3. Sr+ ion smaller than Sr atom4. greater nuclear attraction between
ion and outer electron
◍ the student reacts 1.00g of calcium carbonate with an excess of dilute nitric
acid explain why the students two reactions produce different volumes of
gas.
Answer: 1. Mr of Srco3 is diff to Mr of caco3 2. Mr of strontium carbonate
is greater than Mr of calcium carbonate3. more moles from caco3
◍ a student reacts an excess of magnesium with 25.0cm3 of 0.500 moldm-3 of
HCl the student also reacts an excess of magnesium with 25.0cm3 of 0.500
moldm-3 ethanoic acid explain why these two reactions of magnesium
produce the same volume of gas but at different rates.
Answer: 1. HCl is a strong acid and CH3COOH is a weak acid2. greater H+
conc in HCl and more frequent collisions3. both acids are monobasic and
have the same number of moles of acid
◍ define monobasic acid.
, Answer: one mole of acid dissociates to form one mole of H+
◍ define dibasic acid.
Answer: an acid that produces two hydrogen ions per molecule of the acid
◍ a buffer solution has a pH of 4.50when a small volume of water is added,
the pH does not change explain why.
Answer: ratio of [HA]/[A-] is the same
◍ explain how the successive ionisation energies provide evidence for the
electron shells in sodium atoms.
Answer: 1. large rise between 1st and 2nd IEs and 9th and 10th IEs
◍ explain how the first ionisation energies of magnesium and aluminium give
further details of electron structure.
Answer: 1. Mg has outer electron in 3s sub shell2. Al has outer electron in
3p sub shell3. 3p sub shell has higher energy than 3s sub shell
◍ why does the reaction proceed with reference to entropy [Co(NH3)6]3+ +
3H2NCH2CH2NH2 → [Co(H2NCH2CH2NH2)3]3+ + 6NH3.
Answer: 1. 4 particles on the reactant side and 7 particles on the product side
2. entropy increases over course of reaction so it proceeds3. entropy more
likely to increase if there are more particles in products than reactants -
chelate effect
◍ why does the coordination number change.
Answer: 1. the chloride ions are bigger than the nitrogen atoms on the
ethanediamine ligand2. 6 nitrogens can bond to the cobalt as compared to 4
chloride ions
◍ why can't a substance with a double bond act as a bidentate ligand.
Answer: 1. the nitrogens on the opposite sides of the double bond - rotation
around a double bond is restricted 2. only one of the nitrogen atoms would
be in a position to donate a lone pair of electrons
◍ Periodic table.
Answer: A tabular representation of all known elements, organised and
, classified in order, based on atomic number.
◍ Early periodic tables.
Answer: •1817: Döbreiner's triads - 3 elements w/ regularly varying
properties: S Se Te•1865: Newlands - "law of octaves", about 55
elements.•Early tables were based on mass number (A) or "combining
weight"
◍ explain how Fe2+ ions catalyse the reaction between s2o82- and i- ions.
Answer: 1. Fe2+ ions and s2o82- ions attract because they have opposite
charges2. similarly Fe3+ and I- ions attract because they have opposite
charges
◍ Modern periodic table.
Answer: •1869: Mendeleev and Meyer – "properties of the elements are a
periodic function of their atomic weights;" 63-element table.•1913: Moseley
– X-ray emission spectra vary with atomic number (Z)
◍ Group ( # of valence electrons).
Answer: A vertical column in the periodic table containing elements with
similar chemical properties.
◍ explain how Fe2+ ions catalyse the reaction between s2o82- and i- ions 2.
Answer: 3. Ea much lower than Ea for reaction between s2o82- and iodide
ions without a catalyst because like charges repel4. irons variable oxidation
states allow it to react with either reactant and thereby can be regenerated
◍ contact process sulfur 2so2 + o2 --> 2so3heterogenous catalysis.
Answer: 1. so2 + v2o5 --> so3 + v2o4 2. v2o4 + 1/2o2 --> v2o5
◍ Period (# of shells).
Answer: A horizontal row going across in the periodic table.
◍ homogenous catalysis s2o82- + 2I- --> 2so42- + I2.
Answer: 1. s2o82- + 2fe2+ --> 2so42- + 2fe3+ 2. 2fe3+ + 2I- --> 2Fe2+ + I2
◍ Metals e.g. Na and Cu.
Answer: The elements found to the left of the periodic table.