미해결 질문
For each of the following elements, predict where the "jump" occurs for successive ionization energies (ie).
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10. Periodic Properties of the Elements
Periodic Trend: Successive Ionization Energies
객관식
The successive ionization energies for an unknown element are:
IE1 = 896 kJ/mol IE2 = 1752 kJ/mol IE3 = 14,807 kJ/mol IE4 = 17,948 kJ/mol
To which family in the periodic table does the unknown element most likely belong?
A
1A
B
2A
C
3A
D
4A
E
5A
검증된 단계별 안내1
Understand the concept of ionization energy: Ionization energy is the energy required to remove an electron from an atom in the gaseous state. Successive ionization energies refer to the energy needed to remove each subsequent electron after the first.
Analyze the given ionization energies: Notice the large jump between IE2 and IE3. This indicates that the third electron is being removed from a more stable electron configuration, suggesting the element has two electrons in its outermost shell.
Relate the ionization energy pattern to electron configuration: Elements in Group 2A (alkaline earth metals) have two valence electrons. The large increase in ionization energy after removing the second electron suggests the element has reached a stable noble gas configuration.
Compare with other groups: Elements in Group 1A have one valence electron, so the jump would occur after IE1. Elements in Groups 3A, 4A, and 5A have more than two valence electrons, so the pattern of ionization energies would differ.
Conclude the family: Based on the pattern of ionization energies, the unknown element most likely belongs to Group 2A, as the large increase in ionization energy after the second electron is removed is characteristic of this group.
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