The ionization energies for element X are listed in the table above. On the basis of the data, element X is most likely to be

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The ionization energies for element X are listed in the table above. On the basis of the data, element X is most likely to be

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The question asks us to identify element X based on its ionization energy data. Since the answer options are not provided in the prompt, I’ll focus on explaining why Aluminum (Al) is consistent with a characteristic ionization-energy pattern and how that pattern differs from nearby elements.

- Understanding ionization energy patterns: In atoms, the first ionization energy (IE1) is the energy required to remove the first electron from a neutral atom. For elements in a given group, the number of valence electrons often shows up as a notable change (a large jump) after removing all valence electrons. For Group 13 elements (boron, aluminum, gallium, etc.), there are three valence electrons outside a noble-gas core. This leads to relatively lower IE1 and IE2 values compared to inner-shell electrons, and a substantial increase (a large jump) at IE3 when attempting to remove the third electron and approach the noble-gas core configuration.

- Why Al fits a typical Group 13 pattern: Aluminum has three valence electrons. The first and second ionization energies are moderate because you are removing valence electrons. The third ionization energy is considerably larger because you are removing an electron from a stable, closed-shell (Ne-like) configuration. A large increase between IE2 and IE3 is a hallmark of Group 13 elements like Al, indicating the removal of all three valence electrons to reach a noble-gas core. If the data in the table show this kind of pattern—two smaller successive ionizations followed by a notably larger third ionization—Al is a plausible match.

- How this would differ for nearby elements: If X were in Group 2 (e.g., Mg), you’d expect IE1 and IE2 to be relatively low-to-moderate with a notable but smaller jump after the second electron removal, since there are only two valence electrons. If X were a non-metal in Group 14 (e.g., Si), the pattern would reflect four valence electrons, with the significant increase occurring after the removal of the fourth electron. In contrast, a pronounced large jump after removing the third electron strongly points to a Group 13 element with three valence electrons, like Al.

- Considerations about the missing options: Without the explicit list of answer choices, it’s not possible to critique each alternative directly. However, the distinctive IE pattern described above is a reliable diagnostic for identifying Group 13 elements, and Aluminum is the common representative matching that pattern in many problems of this type.

- Summary of reasoning flow: The moderate IE1 and IE2 values reflect the relatively easy removal of valence electrons, the sizable increase at IE3 signals penetration into a noble-gas core after removing all three valence electrons, and this combination is characteristic of Group 13 elements such as Aluminum. This reasoning distinguishes Al from elements with only one or two valence electrons (which would show different jump points) or from elements with four or more valence electrons (where the large jump occurs after removing more electrons).

AP Chemistry A Unit 1 quiz

The ionization energies for element X are listed in the table above. On the basis of the data, element X is most likely to be

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