---
title: "A materials chemist compares the energy required for selected successive ionizations of three neighboring period- elements. The gaseous ions are formed by successive removal of valence electrons from isolated atoms.  | Ionization energy | Process | |—|—| | \\(\\operatorname{IE}_1(\\text{P})\\) | \\(\\text{P(g)} \\rightarrow \\text{P}^{+}\\text{(g)}+e^-\\) | | \\(\\operatorname{IE}_2(\\text{S})\\) | \\(\\text{S}^{+}\\text{(g)} \\rightarrow \\text{S}^{2+}\\text{(g)}+e^-\\) | | \\(\\operatorname{IE}_3(\\text{Cl})\\) | \\(\\text{Cl}^{2+}\\text{(g)} \\rightarrow \\text{Cl}^{3+}\\text{(g)}+e^-\\) |  Which of the following correctly ranks the ionization energies from greatest to least and provides the best justification?"
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url: "https://nerd-notes.com/ubq/120032/"
date_modified: "2026-08-21T08:40:42+00:00"
---

# A materials chemist compares the energy required for selected successive ionizations of three neighboring period- elements. The gaseous ions are formed by successive removal of valence electrons from isolated atoms.

| Ionization energy | Process |
|—|—|
| \(\operatorname{IE}_1(\text{P})\) | \(\text{P(g)} \rightarrow \text{P}^{+}\text{(g)}+e^-\) |
| \(\operatorname{IE}_2(\text{S})\) | \(\text{S}^{+}\text{(g)} \rightarrow \text{S}^{2+}\text{(g)}+e^-\) |
| \(\operatorname{IE}_3(\text{Cl})\) | \(\text{Cl}^{2+}\text{(g)} \rightarrow \text{Cl}^{3+}\text{(g)}+e^-\) |

Which of the following correctly ranks the ionization energies from greatest to least and provides the best justification?

A materials chemist compares the energy required for selected successive ionizations of three neighboring period- elements. The gaseous ions are formed by successive removal of valence electrons from isolated atoms.

| Ionization energy | Process |
|---|---|
| \(\operatorname{IE}_1(\text{P})\) | \(\text{P(g)} \rightarrow \text{P}^{+}\text{(g)}+e^-\) |
| \(\operatorname{IE}_2(\text{S})\) | \(\text{S}^{+}\text{(g)} \rightarrow \text{S}^{2+}\text{(g)}+e^-\) |
| \(\operatorname{IE}_3(\text{Cl})\) | \(\text{Cl}^{2+}\text{(g)} \rightarrow \text{Cl}^{3+}\text{(g)}+e^-\) |

Which of the following correctly ranks the ionization energies from greatest to least and provides the best justification?

- **A.** \(\operatorname{IE}_1(\text{P})>\operatorname{IE}_2(\text{S})>\operatorname{IE}_3(\text{Cl})\), because increasing positive charge from \(\text{P}\) to \(\text{Cl}^{2+}\) increases electron-electron repulsion and makes electron removal easier.
- **B.** \(\operatorname{IE}_2(\text{S})>\operatorname{IE}_3(\text{Cl})>\operatorname{IE}_1(\text{P})\), because \(\text{S}^{+}\) uniquely has a half-filled \(3p\) subshell, whereas \(\text{Cl}^{2+}\) retains paired \(3p\) electrons.
- **C.** \(\operatorname{IE}_3(\text{Cl})>\operatorname{IE}_2(\text{S})>\operatorname{IE}_1(\text{P})\), because each species initially has the configuration \([\text{Ne}]3s^2 3p^3\), so electron-electron repulsion is comparable and the species with greater nuclear charge holds its electrons more strongly.
- **D.** \(\operatorname{IE}_3(\text{Cl})>\operatorname{IE}_1(\text{P})>\operatorname{IE}_2(\text{S})\), because \(\text{S}^{+}\) still has a \(3p^4\) configuration with a paired set of electrons, making its electron easiest to remove.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/120032/*