Reading a Periodic Table Pattern from Ionisation Energies
A student is analysing a data sheet for an element Q. The successive ionisation energies of Q, in kJ mol⁻¹, are: 1st 578; 2nd 1817; 3rd 2745; 4th 11578; 5th 14831; 6th 18378. She also has Period 3 first ionisation energies (Na 496, Mg 738, Al 578, Si 786, P 1012, S 1000, Cl 1251, Ar 1521) and atomic radii (Na 186, Mg 160, Al 143, Si 117, P 110, S 104, Cl 99, Ar 94 pm). Her draft notes contain the sentence: 'Ionisation energy always increases across a period because the nuclear charge increases.' She cannot explain why the third ionisation energy of Q is so much smaller than the fourth, and she wants to use the argon radius as proof that the radius trend is perfectly regular. She must submit a short written report defending or correcting both claims.
1. Deduce the group of the periodic table in which Q is found. Justify your answer using the numerical data, and state the likely electron configuration of Q.
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The size of the jump between successive ionisation energies shows when an electron is being pulled from a new, inner shell.
- The first three values (578, 1817, 2745 kJ mol⁻¹) rise steadily. That is what happens when electrons are removed from the same outer shell while the effective nuclear charge on the remaining electrons increases.
- Between the 3rd (2745) and 4th (11578) there is a jump of about 8830 kJ mol⁻¹ - roughly 4.8 times the 2nd-to-3rd increment of 928. The 4th electron is therefore coming from a shell much closer to the nucleus and much less shielded from it.
- Three relatively easily removed outer electrons means three valence electrons, so Q is in Group 3.
The first ionisation energy of 578 kJ mol⁻¹ is exactly the Period 3 value for aluminium, so the most likely identification is Al, configuration 1s²2s²2p⁶3s²3p¹. The 4th electron would be removed from the filled 2p level, which is why it costs so much more.
2. Correct or defend the student's generalisation that ionisation energy 'always increases across a period', using the Mg/Al and P/S data.
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The general trend is upward, but it is not monotonic, so the word 'always' is wrong. Two of the supplied values fall rather than rise, and both have a structural explanation:
- Mg (738) → Al (578): the electron removed from Al is a 3p electron, which is higher in energy and more shielded by the two 3s electrons than the 3s electron removed from Mg. The extra proton does not compensate, so Al's first ionisation energy is lower.
- P (1012) → S (1000): phosphorus has a half-filled 3p subshell (3p³), a configuration of extra stability. In sulfur the fourth 3p electron must pair with an electron already in a 3p orbital, and the resulting electron–electron repulsion makes that electron easier to remove.
A defensible statement is: first ionisation energy generally increases across a period because nuclear charge increases while shielding stays roughly constant, but local dips occur where a new subshell is entered (Group 2 → 13) or where electron pairing begins (Group 15 → 16).
3. The value quoted for argon (94 pm) is included in the same list as the metals and non-metals. Explain why it should be treated with caution, and state one limitation of using atomic radius as evidence for the trend.
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The radii for Na–Al are metallic radii and those for Si–Cl are covalent radii: each is obtained from the distance between bonded or neighbouring atoms in a real structure. Argon forms no ordinary bonds, so its value cannot be measured the same way - it is a calculated value (or one derived from very weak van der Waals contacts). Comparing it directly with the others is therefore not strictly like-for-like, and the apparent smooth continuation of the trend is partly an artefact of how the number was produced.
Limitation: atomic radius is not a single fixed property. It depends on the type of bonding in the sample and on the method used to measure it, so values for a metal, a covalent non-metal and a noble gas are only approximately comparable. The trend is still real and useful, but it should not be presented as exact.