Question

Which ionization energy ( \(I E_{1}, I E_{2}\), or \(I E_{3}\) ) corresponds to the following process? $$ \mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} $$

Short answer

The ionization energy corresponds to the process is the third ionization energy, IE_3.

Step-by-step solution

Understanding Ionization Energy

Ionization energy is defined as the energy required to remove an electron from an atom or an ion. The first ionization energy (IE_1) corresponds to the energy needed to remove the first electron, the second ionization energy (IE_2) is the energy to remove the second electron, and so forth.

Identifying the Ionization Energy

Here, Aluminum starts as Al^2+ and loses an electron to become Al^3+. This means that two electrons have already been removed from the Aluminum atom to get to Al^2+. Therefore, the ionization energy required to remove another electron to form Al^3+ corresponds to the removal of the third electron, so it is IE_3.

Key concepts

First Ionization Energy (IE_1)

Understanding the concept of first ionization energy (IE_1) is foundational in studying the electronic structure of elements. Ionization energy generally refers to the energy needed to remove an electron from a neutral atom in its gaseous state. Specifically, IE_1 represents the amount of energy required to remove the first electron from a neutral atom.

When an electron is removed, it leaves behind a positively charged ion. This first electron is typically the easiest to remove because it experiences the full attractive force of the nucleus with minimal repulsion from other electrons. An important thing to note is that IE_1 varies across the periodic table: it tends to increase as you move from left to right across a period due to the increasing nuclear charge. It also generally decreases as you move down a group because the outer electrons are further from the nucleus and are more shielded by inner electrons.

In learning about IE_1, it's valuable to look at trends and exceptions in the periodic table, such as why the first ionization energy of aluminum is lower than that of magnesium despite being to its right. Such anomalies often relate to electron configurations and subshell structures.

Second Ionization Energy (IE_2)

Following the notion of first ionization energy, the second ionization energy (IE_2) is the energy required to remove a second electron from an atom. After the first electron is removed, the atom becomes a positively charged ion, and removing a second electron means overcoming the increased electrostatic pull from the positively charged ion.

Key Factors Influencing IE_2

• Nuclear charge: With one less electron, the effective nuclear charge on the remaining electrons increases, making them more tightly bound to the nucleus.
• Electron-electron repulsion: With one electron gone, there's less electron-electron repulsion, which also contributes to the increased attraction between the nucleus and the remaining electrons.
• Shielding effect: As there is now one less electron, the shielding effect slightly diminishes, leading to a stronger attraction between the remaining electrons and the nucleus.

It's important to grasp that IE_2 is always higher than IE_1 for a given element because the ion is more positively charged, thus requiring more energy to remove another electron. Examining the second ionization energy can give insight into an element's electron configuration, such as why the IE_2 of magnesium is much higher than its IE_1.

Third Ionization Energy (IE_3)

Pushing forward into the realm of third ionization energy (IE_3), this is the energy necessary to remove a third electron from an atom. After the removal of two electrons, the atom in question has become a doubly positive ion. IE_3 involves an even greater amount of energy than both IE_1 and IE_2 due to the factors of increased nuclear charge and reduced electron-electron repulsion acting upon fewer electrons.

One interesting aspect of IE_3 is that it can occasionally present a huge jump in energy compared to IE_1 and IE_2, especially if the removal of the third electron means breaking into an inner electron shell or removing an electron from a very stable electron configuration. For instance, elements with a full s or p subshell will exhibit very high third ionization energies.

Understanding IE_3 is pivotal in explaining properties of elements and their compounds. For example, the dramatic increase in IE_3 for aluminum explains its prevalence as Al^3+ in compounds. This correlates with the exercise showing that the ionization energy required to move from Al^{2+} to Al^{3+} is indeed IE_3, as it pertains to the removal of the third electron from the neutral aluminum atom.