Part (a) · 1 point
Identify the element and write its complete ground-state electron configuration. Justify your identification using the spectrum.
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AP Chemistry · Short Response0 / 4 pts self-scored
Scenario
The complete photoelectron spectrum (PES) of a pure sample of an unknown element shows five peaks. The binding energies and relative peak heights are: a peak at $239 \text{ MJ/mol}$ (relative height 2), a peak at $22.1 \text{ MJ/mol}$ (relative height 2), a peak at $16.0 \text{ MJ/mol}$ (relative height 6), a peak at $2.0 \text{ MJ/mol}$ (relative height 2), and a peak at $1.0 \text{ MJ/mol}$ (relative height 4). In a complete PES spectrum, peak height is proportional to the number of electrons in each subshell.
Part (b) · 1 point
Explain why the peak at $239 \text{ MJ/mol}$ corresponds to a much greater binding energy than the peak at $22.1 \text{ MJ/mol}$, even though both peaks arise from electrons in s subshells.
Part (c) · 2 points
The first ionization energy of phosphorus ($1012 \text{ kJ/mol}$) is greater than the first ionization energy of this element ($1000 \text{ kJ/mol}$), even though this element has the greater nuclear charge. Explain this observation in terms of the electron configurations of the two atoms.
Teaching notes
This question connects photoelectron spectroscopy to electron configuration and Coulomb's law. The key skills are (1) reading a PES spectrum quantitatively — peak height counts electrons, binding energy orders subshells by distance from the nucleus and shielding — and (2) explaining the famous Group 15/16 ionization-energy inversion. Both explanations must be grounded in Coulombic attraction and electron-electron repulsion, not memorized slogans. On the AP exam, answers that invoke 'octet stability' or 'half-filled shells are stable' without a repulsion-based mechanism do not earn points.