Reading a Photoelectron Spectrum: Group Number, Shielding, and an Ionization-Energy Anomaly
A sample of an unidentified main-group element is analyzed by photoelectron spectroscopy. The spectrum shows five peaks whose relative areas are 2 : 2 : 6 : 2 : 3, listed in order of decreasing binding energy, with measured binding energies of approximately 2150 eV, 190 eV, 135 eV, 19 eV, and 11 eV. A student argues that the element must be in Group 13 because its highest-energy (outermost) peak contains three electrons, and further predicts that the element directly below it in the periodic table will show its outermost p peak at essentially the same binding energy, 11 eV, since both atoms would have three valence p electrons.
1. Assign each peak to a subshell, give the complete electron configuration, and state the element's period and group. Then evaluate the student's "outermost peak has three electrons, so Group 13" reasoning.
Read the worked answer
The peak areas sum to 2 + 2 + 6 + 2 + 3 = 15 electrons, so the atom has Z = 15 and the peaks fill subshells in order of increasing n and l:
- 1s² at 2150 eV, 2s² at 190 eV, 2p⁶ at 135 eV, 3s² at 19 eV, 3p³ at 11 eV
Configuration: 1s²2s²2p⁶3s²3p³, or [Ne]3s²3p³. The element is in Period 3, Group 15 (phosphorus).
The student's shortcut confuses the outermost subshell with the valence shell. Here the valence shell is n = 3, which contains 3s² + 3p³ = 5 electrons, consistent with Group 15. A Group 13 element would show a 3p peak of area 1 (ns²np¹), and a Group 17 element a 3p area of 5. The robust procedure is: sum all areas to get Z, order subshells by increasing n then l, then count electrons in the highest n level to get the group.
2. Explain why the 3s peak (19 eV) lies at higher binding energy than the 3p peak (11 eV) even though both electrons have n = 3. Then predict whether the corresponding outermost p electrons in the element directly below (Z = 33) will appear at 11 eV, above 11 eV, or below 11 eV, and justify.
Read the worked answer
Both electrons are in n = 3, but they differ in penetration. A 3s orbital has appreciable electron density close to the nucleus, inside the [Ne] core, so a 3s electron feels a larger effective nuclear charge and is held more tightly (higher binding energy). A 3p orbital is more shielded by the 1s, 2s, and 2p electrons and has a node structure that keeps it farther from the nucleus, so it is easier to remove (lower binding energy). The 19 eV vs 11 eV gap is a penetration/shielding effect, not a difference in n.
For Z = 33 (arsenic), the outermost p electrons are 4p, not 3p. They sit one shell farther out and are additionally shielded by the full 3d¹⁰ subshell, so Z_eff at the 4p level is lower than at 3p. The peak should therefore appear below 11 eV (roughly 9–10 eV), not at the same position. Binding energies of the outermost p electrons decrease down a group.
3. Predict whether this element's first ionization energy is larger or smaller than (i) the element immediately to its right and (ii) the element directly below it. Give a specific electronic reason for each comparison.
Read the worked answer
(i) Larger than the element to its right. The element to the right (sulfur, Z = 16) has the configuration 3p⁴, in which one 3p orbital holds a pair of electrons. That pairing introduces additional electron–electron repulsion, and the half-filled 3p³ arrangement of the original element gains extra exchange stabilization. The result is the well-known trend break: the Group 15 element has the higher first ionization energy (about 1012 kJ/mol) than the Group 16 neighbor (about 1000 kJ/mol), despite the neighbor's larger Z. Trend lines drawn purely on "IE increases across a period" miss this.
(ii) Larger than the element below. The element below (arsenic) removes a 4p electron, which is in a higher-n shell, farther from the nucleus and screened by the intervening 3d¹⁰ electrons. Its first ionization energy is therefore lower (about 947 kJ/mol). Note that both predictions contradict a naive reading of the trend, which is why the electronic configuration must be used.