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Chapter 7: Problem 37

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

Expert verified
The equations for the first, second, and third ionization energies of an aluminum atom are: 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, which involves removing the first electron from the 3p orbital: \[Al(g) \rightarrow Al^+(g) + e^-\]

Step by step solution

01

Understanding Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or ion. The first ionization energy is the energy required to remove the first electron, the second ionization energy is the energy required to remove the second electron, and the third ionization energy is the energy required to remove the third electron.
02

Write equations for ionization energies of aluminum

Aluminum has the chemical symbol "Al" and an atomic number of 13. Its electron configuration is \(1s^2 \: 2s^2 \: 2p^6 \: 3s^2 \: 3p^1\). Below are the equations for the first, second, and third ionization energies: 1. First ionization energy (removing the first electron from the 3p orbital): \[Al(g) \rightarrow Al^+(g) + e^-\] 2. Second ionization energy (removing the second electron from the 3s orbital): \[Al^+(g) \rightarrow Al^{2+}(g) + e^-\] 3. Third ionization energy (removing the third electron from the 3s orbital): \[Al^{2+}(g) \rightarrow Al^{3+}(g) + e^-\]
03

Determine which process requires the least energy

As electrons are removed from the atom, the remaining electrons experience a stronger attractive force from the nucleus, making them harder to remove. Consequently, the ionization energy increases with each subsequent electron removed. Thus, the process that requires the least amount of energy is the first ionization energy, which involves removing the first electron from the 3p orbital: \[Al(g) \rightarrow Al^+(g) + e^-\]

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