Which of the following atoms will be expected to have the smallest second ionization energy?

• A. K
• B. Al
• C. Mg
• D. Na

Answer: (D)

Verified step by step guidance

1. 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.
2. 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
3. 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.
4. 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.
5. 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.