---
title: "The atomic number and ground-state electron configuration for neutral atoms of \\(\\text{Na}\\) and \\(\\text{Mg}\\) are shown in the table below.  | Element | Atomic number (\\(Z\\)) | Ground-state electron configuration | | :— | :— | :— | | \\(\\text{Na}\\) | \\(11\\) | \\(1s^2 2s^2 2p^6 3s^1\\) | | \\(\\text{Mg}\\) | \\(12\\) | \\(1s^2 2s^2 2p^6 3s^2\\) |  Which of the following correctly identifies the element whose \\(1s\\) electrons have the greater binding energy in a photoelectron spectrum, and provides the correct justification?"
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url: "https://nerd-notes.com/ubq/121397/"
date_modified: "2026-08-23T05:02:05+00:00"
---

# The atomic number and ground-state electron configuration for neutral atoms of \(\text{Na}\) and \(\text{Mg}\) are shown in the table below.

| Element | Atomic number (\(Z\)) | Ground-state electron configuration |
| :— | :— | :— |
| \(\text{Na}\) | \(11\) | \(1s^2 2s^2 2p^6 3s^1\) |
| \(\text{Mg}\) | \(12\) | \(1s^2 2s^2 2p^6 3s^2\) |

Which of the following correctly identifies the element whose \(1s\) electrons have the greater binding energy in a photoelectron spectrum, and provides the correct justification?

The atomic number and ground-state electron configuration for neutral atoms of \(\text{Na}\) and \(\text{Mg}\) are shown in the table below.

| Element | Atomic number (\(Z\)) | Ground-state electron configuration |
| :--- | :--- | :--- |
| \(\text{Na}\) | \(11\) | \(1s^2 2s^2 2p^6 3s^1\) |
| \(\text{Mg}\) | \(12\) | \(1s^2 2s^2 2p^6 3s^2\) |

Which of the following correctly identifies the element whose \(1s\) electrons have the greater binding energy in a photoelectron spectrum, and provides the correct justification?

- **A.** \(\text{Na}\), because having only one \(3s\) valence electron minimizes electron-electron repulsion throughout the atom, allowing the \(1s\) electrons to be held more tightly by the nucleus.
- **B.** \(\text{Mg}\), because \(\text{Mg}\) has a greater nuclear charge and outer-shell electrons do not shield the \(1s\) electrons, resulting in a stronger Coulombic attraction to the nucleus.
- **C.** \(\text{Mg}\), because the completely filled \(3s\) subshell provides additional quantum mechanical stability that lowers the potential energy of all electrons in the atom.
- **D.** \(\text{Na}\), because the larger atomic radius of \(\text{Na}\) reduces electron density near the nucleus, decreasing shielding between the core \(1s\) electrons.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/121397/*