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: \(Al(g) \rightarrow Al^+(g) + e^-\) 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

First Ionization Energy

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

Second Ionization Energy

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

Third Ionization Energy

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

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.

Key concepts

Aluminum Atom

Aluminum is a widely known element represented by the symbol Al. It is the 13th element on the periodic table and belongs to the group of elements known as "post-transition metals." Aluminum is a soft and lightweight metal, often used for its corrosion resistance and conductivity. In its ground state, an aluminum atom has 13 protons, 13 electrons, and commonly 14 neutrons.
The electrons are distributed in different energy levels or shells around the nucleus. For aluminum, the electron configuration is

• 1s yen superscript 2
• 2s yen superscript 2
• 2p yen superscript 6
• 3s yen superscript 2
• 3p yen superscript 1

...showing that the outermost shell has three electrons. These valence electrons are involved in the ionization process where they are removed to form ions.

First Ionization Energy

The first ionization energy refers to the energy required to remove the outermost electron from a neutral atom in its gaseous state, forming a cation. For aluminum, this can be represented as
\( Al(g) \rightarrow Al^+(g) + e^- \)
This process is relatively straightforward because it only involves removing one electron from the neutral aluminum atom. It requires less energy compared to subsequent ionizations because there is no positive charge that increases the attraction for the remaining electrons. Factors affecting the first ionization energy include:

• Atomic size: Larger atoms have lower ionization energies.
• Nuclear charge: Higher nuclear charges increase ionization energy.
• Shielding: More inner electron shells reduce the effective nuclear charge and thus, ionization energy.

Second Ionization Energy

After the first electron is removed, you are left with an ion (\( Al^+ \)) which already has a positive charge. The second ionization energy involves removing another electron from this ion:
\( Al^+(g) \rightarrow Al^{2+}(g) + e^- \)
This step requires more energy than the first ionization. The remaining electrons are more strongly attracted to the nucleus due to the positive charge of the ion, which comes from the loss of one electron. Therefore, more energy is required to overcome this attraction and remove an additional electron.
As the nuclear charge effectively increases, each successive electron becomes harder to remove.

Third Ionization Energy

Following the removal of the second electron, we have \( Al^{2+} \) as the ion. The third ionization energy is the energy required to remove yet another electron from this doubly charged ion:\( Al^{2+}(g) \rightarrow Al^{3+}(g) + e^- \)This step sees the highest increase in ionization energy. Due to the even higher positive charge (\( Al^{2+} \) ), the attraction force on the remaining electrons is significantly stronger. More energy must be applied to remove this electron compared to the first and second ionizations.
In summary, as you increase the positive charge of an ion by removing more electrons, the harder it becomes to remove further electrons due to increased electrostatic attraction towards the nucleus.