---
title: "A student compares the photoelectron spectroscopy (PES) data for gaseous phosphorus (ex	ext{P}ex) and sulfur (ex	ext{S}ex) atoms. The table below lists the valence electron configuration and the binding energy corresponding to the lowest-energy peak in the spectrum for each element.  | Element | Valence electron configuration | Lowest binding energy peak (\\(\\text{MJ/mol}\\)) | | :— | :— | :— | | \\(\\text{P}\\) | \\(3s^2 3p^3\\) | \\(1.01\\) | | \\(\\text{S}\\) | \\(3s^2 3p^4\\) | \\(1.00\\) |  Which of the following statements best explains why the lowest binding energy peak for \\(\\text{S}\\) occurs at a lower energy than that for \\(\\text{P}\\)?"
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url: "https://nerd-notes.com/ubq/121463/"
date_modified: "2026-08-23T05:04:42+00:00"
---

# A student compares the photoelectron spectroscopy (PES) data for gaseous phosphorus (ex	ext{P}ex) and sulfur (ex	ext{S}ex) atoms. The table below lists the valence electron configuration and the binding energy corresponding to the lowest-energy peak in the spectrum for each element.

| Element | Valence electron configuration | Lowest binding energy peak (\(\text{MJ/mol}\)) |
| :— | :— | :— |
| \(\text{P}\) | \(3s^2 3p^3\) | \(1.01\) |
| \(\text{S}\) | \(3s^2 3p^4\) | \(1.00\) |

Which of the following statements best explains why the lowest binding energy peak for \(\text{S}\) occurs at a lower energy than that for \(\text{P}\)?

A student compares the photoelectron spectroscopy (PES) data for gaseous phosphorus (ex	ext{P}ex) and sulfur (ex	ext{S}ex) atoms. The table below lists the valence electron configuration and the binding energy corresponding to the lowest-energy peak in the spectrum for each element.

| Element | Valence electron configuration | Lowest binding energy peak (\(\text{MJ/mol}\)) |
| :--- | :--- | :--- |
| \(\text{P}\) | \(3s^2 3p^3\) | \(1.01\) |
| \(\text{S}\) | \(3s^2 3p^4\) | \(1.00\) |

Which of the following statements best explains why the lowest binding energy peak for \(\text{S}\) occurs at a lower energy than that for \(\text{P}\)?

- **A.** Less energy is required to remove an electron from \(\text{S}\) because the valence electrons in \(\text{S}\) experience more core-electron shielding than those in \(\text{P}\).
- **B.** Less energy is required to remove an electron from \(\text{S}\) because the valence electrons in \(\text{S}\) occupy a subshell with a higher principal quantum number than those in \(\text{P}\).
- **C.** Less energy is required to remove an electron from \(\text{S}\) because the electron-electron repulsion between paired electrons in a \(3p\) orbital makes an electron easier to remove despite the greater nuclear charge.
- **D.** Less energy is required to remove an electron from \(\text{S}\) because the effective nuclear charge of \(\text{S}\) is less than that of \(\text{P}\), decreasing the attraction between the nucleus and the valence electrons.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121463/*