Successive ionisation energies: identifying two Period 3 elements and testing two claims
Two Period 3 elements, X and Y, are supplied as samples with no labels. The class is given only their successive ionisation energies in kJ mol⁻¹:
X: 578, 1817, 2745, 11578, 14831, 18378 Y: 738, 1451, 7733, 10540, 13630, 17995
A student writes: 'Y must be the more metallic of the two, because every one of Y's ionisation energies is higher than the corresponding one for X, so Y holds its electrons more tightly and will form positive ions more readily.' A second student writes: 'X shows a big jump between the third and fourth ionisation energies, which proves X has three electron shells.' The teacher asks the class to use the data to test both claims, to identify the two elements, and to predict how their oxides behave towards hydrochloric acid and towards sodium hydroxide solution.
1. Identify X and Y from the data, and correct both student statements, explaining precisely what a large jump in successive ionisation energies does and does not show.
Read the worked answer
X has a jump of about 4.2× between IE3 (2745) and IE4 (11578); Y has a jump of about 5.3× between IE2 (1451) and IE3 (7733). A large jump means the next electron must be removed from a shell closer to the nucleus, so the number of electrons before the jump equals the number of valence electrons. X therefore has three valence electrons (Group 3, so aluminium, Z = 13) and Y has two (Group 2, so magnesium, Z = 12). The period is already given as 3.
The second student's claim is wrong: the jump shows the number of electrons in the outer shell (the group), not the number of shells. X has three electron shells whether or not a jump occurs; that information comes from its position in Period 3, not from the ionisation energies.
The first student's claim is also wrong on its own terms, because it misreads the data: only Y's first three ionisation energies are higher than X's. From IE4 onwards X's are the higher ones, so the trend reverses. Higher ionisation energies do not make an element more metallic; lower first ionisation energy means easier cation formation, and X has the lower IE1 (578 vs 738), so X is the less metallic of the two.
2. Explain why Y's first three ionisation energies are higher than X's, but X's fourth and later ionisation energies are higher than Y's.
Read the worked answer
Y is magnesium, [Ne] 3s², and X is aluminium, [Ne] 3s² 3p¹.
For the first three electrons, the electron removed from aluminium comes from the 3p subshell, which is higher in energy than the 3s orbital and is additionally shielded by the filled 3s subshell plus the neon core. This shielding outweighs aluminium's extra proton, so X's first three ionisation energies are lower than Y's - the familiar dip at group 3.
After three electrons have been removed, aluminium has lost its 3p and both 3s electrons, so the next electron must come from the n = 2 core (2p). That requires far more energy (11578 kJ mol⁻¹). Magnesium loses its two 3s electrons first, and its third electron also comes from the 2p core (7733 kJ mol⁻¹). Now both elements are removing core electrons from the same shell, so the difference in shielding largely disappears and nuclear charge dominates: aluminium has one more proton than magnesium, so its 2p electrons are held more tightly and its core ionisation energies are higher. The same reasoning explains the consistently higher values of X from IE4 onwards, and is a good illustration that shielding and nuclear charge must be weighed against each other, not applied as slogans.
3. Predict the behaviour of the oxides of X and Y with hydrochloric acid and with sodium hydroxide, and justify the trend in terms of bonding and charge density.
Read the worked answer
Y (magnesium) forms MgO, a basic oxide. It dissolves in hydrochloric acid to give magnesium chloride and water, and does not react with sodium hydroxide.
X (aluminium) forms Al₂O₃, an amphoteric oxide. It dissolves in hydrochloric acid to give aluminium chloride and water, and also dissolves in sodium hydroxide to give a solution containing the tetrahydroxoaluminate ion, [Al(OH)₄]⁻.
Justification: across Period 3 the cation charge increases from 2+ to 3+ and the radius decreases, so the charge density of the cation rises sharply. The higher charge density polarises the oxide ion more strongly, giving the Al–O interaction substantial covalent character; the oxide therefore has no simple, strongly basic oxide ion to donate and can act as an acid or a base depending on the medium. The same effect is seen in the chlorides: MgCl₂ is an ionic solid that conducts when molten, whereas AlCl₃ has a much lower melting/sublimation temperature, is largely covalent in the solid, and hydrolyses in water with evolution of hydrogen chloride. Basicity therefore decreases and acidity increases from left to right across the period.