Multiple Choice
When the second ionization energy is supplied to an atom, which ion is formed as a result of removing the second electron?
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10. Periodic Properties of the Elements
Periodic Trend: Successive Ionization Energies
Multiple Choice
Which of the following atoms will be expected to have the smallest second ionization energy?
A
K
B
Al
C
Mg
D
Na
Verified step by step guidance1
Understand what second ionization energy means: it is the energy required to remove the second electron from an atom after the first electron has already been removed.
Recall the electron configurations of the atoms involved:
- K (Potassium): [Ar] 4s¹
- Al (Aluminum): [Ne] 3s² 3p¹
- Mg (Magnesium): [Ne] 3s²
- Na (Sodium): [Ne] 3s¹
Consider the stability of the resulting ion after the first electron is removed. The second ionization energy will be higher if removing the second electron disrupts a stable noble gas configuration.
Analyze each atom after the first electron is removed:
- K⁺: configuration becomes [Ar], a noble gas configuration, so removing the second electron means removing from a stable noble gas core, which requires very high energy.
- Al⁺: configuration becomes [Ne] 3s², not a noble gas core, so second ionization energy is moderate.
- Mg⁺: configuration becomes [Ne] 3s¹, also not a noble gas core, so second ionization energy is moderate.
- Na⁺: configuration becomes [Ne], a noble gas core, so removing the second electron means removing from a stable noble gas core, which requires very high energy.
Compare the second ionization energies: atoms that reach a noble gas configuration after the first ionization will have a much higher second ionization energy. Therefore, the atom whose second ionization energy is smallest is the one where the second electron is removed before reaching a noble gas core, which is Na in this case.
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