Problem 122 - Chemistry The Science in Context

Question

Why is the first ionization energy \(\left(\mathrm{IE}_{1}\right)\) of \(\mathrm{Al}(Z=13)\) less than the IE \(_{1}\) of \(\mathrm{Mg}(Z=12)\) and less than the \(\mathrm{IE}_{1}\) of \(\mathrm{Si}\) \((Z=14) ?\)

Step-by-Step Solution

Answer

Answer: The first ionization energy of aluminum is lower than both magnesium and silicon due to its electron configuration and position on the periodic table. Aluminum has an unpaired electron in the 3p orbital which requires less energy to remove compared to the fully filled 3s orbital in magnesium and the paired electron in the 3p orbital in silicon.

Step 1: Understanding Ionization Energy

Ionization energy is defined as the amount of energy required to remove an electron from an atom in its gas phase. First ionization energy (IE1) is specifically the energy needed to remove the first electron from an atom. This energy is influenced by electron configurations and atomic size. Step 2: Electron Configurations

Step 2: Electron Configurations

We will first identify the electron configurations of the three elements. Al (Z=13): [Ne] 3s^2 3p^1 Mg (Z=12): [Ne] 3s^2 Si (Z=14): [Ne] 3s^2 3p^2 The electron configurations indicate that Al is in the 3p orbitals, Mg has fully filled 3s orbitals, and Si has a paired electron in the 3p orbitals. Step 3: Position on the Periodic Table

Step 3: Position on the Periodic Table

Now, consider the position of these elements on the periodic table: Mg (Z=12) is in Group 2 (alkaline earth metals) Al (Z=13) is in Group 13 (boron group) Si (Z=14) is in Group 14 (carbon group) Going from left to right on the periodic table, the ionization energy generally increases. However, there could be exceptions. Step 4: Influence of Electron Configuration

Step 4: Influence of Electron Configuration

When analyzing the electron configurations, note that Mg has a fully filled 3s orbital, giving it a higher first ionization energy due to increased stability. In addition, Si has a paired electron in its 3p orbital, adding electron-electron repulsion energy, which makes it easier to remove an electron compared to removing an electron from a filled 3s orbital in Mg. On the other hand, Al has an unpaired electron in its 3p orbital. The energy required to remove that electron is less than that required for Mg or Si due to its higher energy level and increased distance from the nucleus, as well as less electron-electron repulsion. Step 5: Conclusion

Step 5: Conclusion

The first ionization energy of Al is less than both Mg and Si due to its electron configuration and position on the periodic table. Al has an unpaired electron in the 3p orbital which requires less energy to remove compared to the fully filled 3s orbital in Mg and the paired electron in the 3p orbital in Si.

Key Concepts

Periodic Table Trends in Ionization Energy

One of the fundamental concepts in chemistry is understanding the periodic table trends, especially when it comes to ionization energy. Ionization energy tends to increase as we move from left to right across a period. This is because of the increasing nuclear charge that more strongly attracts and holds onto electrons. Ions and atoms with a higher nuclear charge have higher ionization energies because the electrons experience a greater pull towards the nucleus.

However, exceptions to this general trend occur due to the unique electron configurations of certain elements. When comparing elements like magnesium (Mg) and aluminum (Al), even though Al is to the right of Mg on the periodic table, the former has a lower first ionization energy. This counterintuitive situation is where understanding electron configurations play a major role. Moving down groups, ionization energy decreases because the outer electrons are farther from the nucleus, experiencing less attraction due to the increased atomic size.

Electron Configurations and Their Impact

The electron configurations of elements profoundly impact their chemical properties, including ionization energy. The arrangement of electrons across different energy levels and orbitals determines how tightly these electrons are bound to the nucleus. For Al (13 electrons), Mg (12 electrons), and Si (14 electrons), the slight variances in their electron configurations lead to distinct differences in their first ionization energies.

Peculiarities of p Orbitals

An unpaired electron in a p orbital, like the one in aluminum, is easier to remove than a paired electron. This is because an unpaired electron in a higher energy state has less nuclear charge attraction and no repulsion from a paired electron, which significantly lowers its ionization energy. In magnesium, the full s orbital signals stability and a lower energy state, making the removal of an electron more energy-intensive. Meanwhile, the paired p electron in silicon introduces electron-electron repulsion, but the combined effect is not enough to lower the ionization energy below that of magnesium.

Understanding Atomic Size

The concept of atomic size is vital when discussing ionization energy. Atomic size generally increases from top to bottom within a group and decreases from left to right across a period on the periodic table. This is due to increased shielding from additional electron layers, making it easier for electrons in higher energy levels to be removed.

Relevance to Ionization Energy

Larger atoms have a lower ionization energy as the outermost electrons are further from the nucleus and thus less tightly bound. This is the underlying reason why Al has a lower fist ionization energy than both Mg and Si, as Si is further to the right (smaller atomic size relative to Al) and Mg is just one position to the left of Al but with a full s orbital (providing greater electron stability due to the fully-filled orbital preference). Thus, Al's larger atomic size compared to Si and less stable electron configuration compared to Mg result in a lower first ionization energy, perfectly illustrating how electron configuration and atomic size interplay within periodic table trends.