Question

Write equations that show the processes that describe the first, second, and third ionization energies of an aluminum atom. Which process would require the least amount of energy?

Short answer

The first, second, and third ionization energies of an aluminum atom are represented by the following equations: 1. First Ionization Energy: \(Al \rightarrow Al^+ + e^-\) 2. Second Ionization Energy: \(Al^+ \rightarrow Al^{2+} + e^-\) 3. Third Ionization Energy: \(Al^{2+} \rightarrow Al^{3+} + e^-\) The process that requires the least amount of energy is the first ionization process: \(Al \rightarrow Al^+ + e^-\).

Step-by-step solution

First Ionization Energy

The first ionization energy corresponds to removing one electron from a neutral aluminum atom: \[Al \rightarrow Al^+ + e^-\] Here, a neutral aluminum atom loses an electron, forming an Al^+ ion and a free electron.

Second Ionization Energy

The second ionization energy corresponds to removing another electron from the Al^+ ion formed in the first ionization process: \[Al^+ \rightarrow Al^{2+} + e^-\] In this step, the Al^+ ion loses another electron, forming an Al^2+ ion, and another free electron.

Third Ionization Energy

The third ionization energy corresponds to removing yet another electron from the Al^(2+) ion formed in the second ionization process: \[Al^{2+} \rightarrow Al^{3+} + e^-\] In this step, the Al^(2+) ion loses another electron and forms an Al^3+ ion, along with a free electron.

Least Energy Requirement

Generally, as more electrons are removed, the ionization energy increases. This means that the first ionization energy requires the least energy, followed by the second ionization energy, and the third ionization energy requires the most energy. Therefore, the process requires the least energy is the first ionization process: \[Al \rightarrow Al^+ + e^-\]