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A-Level The make-up of the atom and how its electrons are arranged

What the A-Level syllabus expects for The make-up of the atom and how its electrons are arranged, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to explain. About 4% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (2 marks)

Physical properties of the elements vary in regular patterns across the Periodic Table. Account for the fact that oxygen has a smaller first ionisation energy than nitrogen.

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Nitrogen has the electron arrangement ...2p3, with each of the three 2p orbitals singly occupied (a stable half-filled sub-shell, no paired 2p electrons). Oxygen has ...2p4, so one 2p orbital now holds a pair of electrons. In that paired orbital the two electrons experience extra electron-electron (inter-electron) repulsion, so the electron removed from oxygen is easier to remove than an unpaired electron from nitrogen. Therefore oxygen's first ionisation energy is lower than nitrogen's.

Example 2 (6 marks)

Give a short explanation for the fact that removing the fourth electron from cobalt requires less energy than removing the fourth electron from iron (that is, why cobalt's fourth ionisation energy is smaller than iron's).

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The fourth ionisation energy corresponds to M3+(g) -> M4+(g) + e-, i.e. removing an electron from the M3+ ion. Iron: Fe is [Ar]3d6 4s2, so Fe3+ is [Ar]3d5 - a stable, exactly half-filled 3d subshell. Cobalt: Co is [Ar]3d7 4s2, so Co3+ is [Ar]3d6 - one d orbital is doubly occupied (a spin-paired pair) while the rest are singly filled. For iron, the 4th electron must be pulled out of the specially stable half-filled 3d5 arrangement, which is energetically unfavourable, so a lot of energy is needed. For cobalt, the 4th electron removed is the second (paired) electron in a doubly-occupied d orbital; losing it relieves the extra electron-electron (spin-pairing) repulsion and produces Co4+ = 3d5, the stable half-filled configuration. Both factors make removal easier, so Co's 4th IE is lower than Fe's.

Example 3 (2 marks)

Iron ranks fourth in abundance among crustal elements and was largely delivered as metal by meteorites. It has several allotropes, one being alpha-iron. Two magnetic behaviours are relevant: in ferromagnets the magnetic moments of unpaired electrons all line up, so the material acts as a magnet, whereas in paramagnets those moments point randomly and align only under an applied field. Alpha-iron is ferromagnetic at room temperature yet turns paramagnetic above 770 C. Using the electron configuration of 26Fe, explain this change.

Show the worked answer

Fe (Z=26) has the configuration [Ar]3d⁶ 4s², giving 4 unpaired electrons in the 3d subshell; the magnetic moments arise from these unpaired electrons. Below 770 C the moments of neighbouring Fe atoms are locked parallel to one another, so the moments reinforce and the metal is ferromagnetic. Above 770 C the extra thermal (vibrational) energy is enough to overcome this coupling, randomising the orientations of the moments so there is no net long-range alignment; the metal is then only paramagnetic (the moments line up only when an external field is applied). The number of unpaired electrons is unchanged - it is the loss of alignment that causes the change.

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More A-Level H2 Chemistry topics

How atoms bond and how that governs a substance's behaviour · Ideal gases and working with gas mixtures · Competing definitions of acids and bases · Trends in the elements across a period and down a group · The mole and reacting-quantity calculations · Enthalpy, entropy and the feasibility of reactions · all of A-Level H2 Chemistry