The Spreadsheet With One Wrong Number: Assigning an Element from Successive Ionisation Energies
A metrology trainee is checking a spreadsheet of successive ionisation energies (kJ mol⁻¹) for an unidentified Period 3 element X used to calibrate an atomic absorption spectrometer. The listed values are:
738, 1451, 7733, 10543, 1363, 17995, 21703, 25666, 31670, 35460
The trainee's supervisor typed the fifth entry "from memory" after the printout jammed, and the raw instrument file cannot be reopened until Monday. A second column in the same sheet holds a comparison element Y:
578, 1817, 2745, 11578, 14831, 18378, 23293, 27460, 31820, 38455
Neither column is labelled. The certificate supplied with the calibration standard says only that the metal is "light, silvery, and burns with a brilliant white flame". The trainee has one hour to submit the identity, group, and block assignment, and is worried that the odd entry will invalidate the whole calibration run.
1. Identify element X and its group and block, correct the suspect value, and justify both the correction and the assignment from the numbers rather than from the flame description.
Read the worked answer
Step 1 – the correction is forced by physics, not by taste. Successive ionisation energies of the same element must rise strictly: each electron is removed from an increasingly positive ion with one fewer electron to screen it. The entry 1363 kJ mol⁻¹ is lower than IE4 (10543), which is impossible. Reading it as 13630 restores a smooth progression (10543 → 13630 → 17995) and is one keystroke away from the typed value - a decimal-place slip, not a real anomaly.
Step 2 – find the jump. Ratios of consecutive values for X:
| Ratio | Value |
|---|---|
| IE2/IE1 | 1451/738 = 1.97 |
| IE3/IE2 | 7733/1451 = 5.33 |
| IE4/IE3 | 10543/7733 = 1.36 |
| IE5/IE4 | 13630/10543 = 1.29 |
The factor-of-5.3 step between IE2 and IE3 dwarfs everything else. That is the signature of breaking into a new, inner shell: only two electrons are held in the outermost shell, so the third electron comes from n = 2.
Step 3 – assignment. Two valence electrons → Group 2, s-block. Twelve electrons → 1s² 2s² 2p⁶ 3s², i.e. magnesium. The flame description (brilliant white) is consistent with Mg but is not evidence - it is a confirmatory detail only. The ionisation data are the argument.
2. Explain, in terms of effective nuclear charge, shielding and shell structure, why IE2 is roughly twice IE1 but IE3 is roughly five times IE2.
Read the worked answer
IE1 and IE2 remove electrons from the same subshell.
- IE1: Mg(g) → Mg⁺(g) + e⁻, electron lost from 3s.
- IE2: Mg⁺(g) → Mg²⁺(g) + e⁻, the second 3s electron.
Both electrons sit in n = 3 at similar mean radius, so the energies are of the same order. The rise is real but modest: once the first electron has gone, the remaining 3s electron has no partner to screen it (the two 3s electrons screen each other by roughly 0.35 each), the ion is smaller, and the effective nuclear charge felt by the survivor is therefore higher. Result: a factor of about 2, not a step change.
IE3 breaks into the core. The third electron must come from 2p:
- n falls from 3 to 2, so the electron is on average much closer to the nucleus and far better held.
- The 3s electrons that previously shielded the 2p core have been removed, so the 2p electrons face a much larger effective nuclear charge.
- 2p orbitals penetrate the core and are poorly shielded by other n = 2 electrons.
Together these make IE3 about five times IE2. The size of the jump is the diagnostic: a small factor (≈2) between consecutive values means same subshell; a large factor (≥4) means a new shell.
3. A classmate claims "first ionisation energy always increases across Period 3". Use the data for X and Y to evaluate this claim, and explain what determines the trend.
Read the worked answer
The claim is false, and the sheet itself disproves it.
X = Mg (IE1 = 738) and Y = Al (IE1 = 578). Aluminium has the larger nuclear charge (Z = 13 vs 12), yet its first ionisation energy is lower. The explanation is subshell energy and shielding, not nuclear charge alone:
- Mg loses a 3s electron; Al loses a 3p electron.
- 3p orbitals are less penetrating than 3s, so a 3p electron is on average further from the nucleus and is shielded by the two 3s electrons as well as by the other 3p electron.
- The extra proton in Al does not compensate for this extra shielding, so the 3p electron is easier to remove.
A second, smaller dip occurs at sulfur (IE1 ≈ 1000 kJ mol⁻¹) compared with phosphorus (≈1012 kJ mol⁻¹): removing one of the paired electrons in a 3p orbital relieves electron–electron repulsion, which offsets the higher nuclear charge.
Conclusion for the trainee's report: the general trend across a period is upward, but it is not monotonic. Ionisation energy follows the energy and occupancy of the orbital from which the electron is taken - subshell type, pairing, and shielding - so any trend statement must be qualified.