14. Which of the following species would be expected to have the lowest ionization energy? 1 pointF-NeO2-Mg2+Na+
Question
- Which of the following species would be expected to have the lowest ionization energy? 1 pointF-NeO2-Mg2+Na+
Solution
Ionization energy is the energy required to remove an electron from a gaseous atom or ion. The trend of ionization energy in the periodic table is that it generally increases as you move from left to right across a period and decreases as you move down a group.
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F- (Fluoride) - This is a negative ion (anion) which means it has gained an extra electron. Anions are usually larger and require less energy to remove an electron, so they have lower ionization energies.
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Ne (Neon) - This is a noble gas with a full electron shell. Noble gases have the highest ionization energies because they have stable electron configurations and it requires a lot of energy to remove an electron.
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O2- (Oxide) - This is another anion, so it will also have a lower ionization energy. However, oxygen is higher up on the periodic table than fluorine, so it will have a slightly higher ionization energy than F-.
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Mg2+ (Magnesium) - This is a positive ion (cation) which means it has lost electrons. Cations are usually smaller and require more energy to remove an electron, so they have higher ionization energies.
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Na+ (Sodium) - This is another cation, so it will also have a higher ionization energy. However, sodium is lower on the periodic table than magnesium, so it will have a slightly lower ionization energy than Mg2+.
So, based on these trends, F- would be expected to have the lowest ionization energy.
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