Reading an Element from Its Successive Ionization Energies
An unknown solid element X is under investigation in a laboratory. Its successive ionization energies, in kJ mol⁻¹, are: IE₁ = 578, IE₂ = 1817, IE₃ = 2745, IE₄ = 11 575, IE₅ = 14 830, IE₆ = 18 376. X burns in oxygen to give a white oxide that dissolves in dilute nitric acid and also in concentrated sodium hydroxide, giving clear solutions in each case. With chlorine, X forms a chloride of formula XCl₃ that sublimes at about 180 °C; the molten chloride conducts electricity only weakly. X is a good conductor of electricity in the solid state. Its first ionization energy is lower than that of the element immediately to its left in the same period (738 kJ mol⁻¹), even though that neighbour has the smaller nuclear charge. The sample is known to come from period 3 or period 2 of the periodic table.
1. Deduce the group, period and identity of X, showing how each piece of data constrains your answer.
Read the worked answer
Group 13. The decisive feature is the size of the jumps: IE₂/IE₁ ≈ 3.1 and IE₃/IE₂ ≈ 1.5, but IE₄/IE₃ = 11 575/2745 ≈ 4.2. Three electrons come off comparatively cheaply, then the fourth is torn out of a much more tightly bound inner shell. Three valence electrons → group 13.
Period 3, X = Al. IE₁ = 578 kJ mol⁻¹ is essentially aluminium's (577.5); boron, the other group-13 candidate, is far higher (801). It also matches the period-3 sequence Na 496, Mg 738, Al 578, Si 786. Metallic conduction in the solid state and a low-sublimation-point, weakly conducting molten chloride point to a covalent, molecular chloride (Al₂Cl₆ vapour) rather than an ionic salt - again Al, not a group-1 or group-2 metal. The amphoteric oxide confirms Al₂O₃.
2. Explain why IE₁ of X is lower than that of the period-3 element to its left, and why IE₄ is roughly four times IE₃.
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The neighbour with 738 kJ mol⁻¹ is Mg (Z = 12); X is Al (Z = 13). Al has the larger nuclear charge, so a naive trend predicts a higher IE₁ - but the electron removed from Mg is a 3s electron, whereas in Al it is a 3p electron. The 3p orbital is higher in energy and is shielded from the nucleus both by the neon core and by the two 3s electrons, so it feels a smaller effective nuclear charge. 3s penetrates closer to the nucleus than 3p; that penetration difference outweighs Al's extra proton. This is the dip at group 13.
After three electrons are removed, the ion is Al³⁺ with a 1s² 2s² 2p⁶ (Ne-like) core. IE₁–IE₃ removed the n = 3 shell; IE₄ must remove a 2p electron from a shell that is much closer to the nucleus, is shielded by only the two 1s electrons, and experiences a large effective nuclear charge. Hence the ~4-fold jump.
3. Predict, with an equation for each case, what happens when the oxide of X is treated with (i) dilute nitric acid and (ii) concentrated sodium hydroxide. Explain the behaviour in terms of charge density.
Read the worked answer
Both give clear solutions - the oxide is amphoteric.
(i) Al₂O₃(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂O(l)
(ii) Al₂O₃(s) + 2OH⁻(aq) + 3H₂O(l) → 2[Al(OH)₄]⁻(aq)
The ion Al³⁺ is small (r ≈ 54 pm) and triply charged, so its charge density is high. A cation of high charge density strongly polarizes a coordinated water molecule, weakening an O–H bond and releasing H⁺: this is the Brønsted acidity of [Al(H₂O)₆]³⁺ that lets the oxide/hydroxide dissolve in base as aluminate. The same high polarizing power lets Al³⁺ bind hydroxide tightly, so the oxide lattice can be opened by OH⁻. Contrast MgO: Mg²⁺ is larger and less charged, so MgO is only basic. Practical consequence: never store an alkaline aluminate solution expecting the oxide to reprecipitate unchanged, and use the amphoteric reaction to distinguish Al₂O₃ from MgO.