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

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The first, second, and third ionization energies of an aluminum atom are represented by the following equations: 1. First Ionization Energy: \(Al(g) \rightarrow Al^+(g) + e^-\) 2. Second Ionization Energy: \(Al^+(g) \rightarrow Al^{2+}(g) + e^-\) 3. Third Ionization Energy: \(Al^{2+}(g) \rightarrow Al^{3+}(g) + e^-\) The process that requires the least amount of energy is the first ionization energy, as it is easier to remove an electron from a neutral atom than from an ionized atom with a higher positive charge.

Step by step solution

01

First Ionization Energy

The process of removing the first electron from the aluminum atom. \(Al(g) \rightarrow Al^+(g) + e^-\)
02

Second Ionization Energy

After the first electron has been removed, the process of removing the second electron from the resultant ion: \(Al^+(g) \rightarrow Al^{2+}(g) + e^-\)
03

Third Ionization Energy

After the removal of the first and second electrons, the process of removing the third electron from the resultant ion: \(Al^{2+}(g) \rightarrow Al^{3+}(g) + e^-\)
04

Least Energy Required

The process that would require the least amount of energy is the first ionization energy. This is because it is easier to remove an electron from a neutral atom than from an ionized atom with a higher positive charge. The higher positive charge attracts the remaining electrons more strongly, requiring more energy to remove them.

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