Consider the following ionization energies for aluminum: Al(g) ⟶Al^+(g)+e^- I1=580 kJ / m

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Consider the following ionization energies for aluminum: ...

Consider the following ionization energies for aluminum: $$\begin{array}{c}{\operatorname{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} \quad I_{1}=580 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} \quad I_{2}=1815 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} \quad I_{3}=2740 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} \quad I_{4}=11,600 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ a. Account for the trend in the values of the ionization energies. b. Explain the large increase between $I_{3}$ and $I_{4}$

Consider the following ionization energies for aluminum:
$$\begin{array}{c}{\operatorname{Al}(g) \longrightarrow \mathrm{Al}^{+}(g)+\mathrm{e}^{-} \quad I_{1}=580 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{+}(g) \longrightarrow \mathrm{Al}^{2+}(g)+\mathrm{e}^{-} \quad I_{2}=1815 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{2+}(g) \longrightarrow \mathrm{Al}^{3+}(g)+\mathrm{e}^{-} \quad I_{3}=2740 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{Al}^{3+}(g) \longrightarrow \mathrm{Al}^{4+}(g)+\mathrm{e}^{-} \quad I_{4}=11,600 \mathrm{kJ} / \mathrm{mol}}\end{array}$$
a. Account for the trend in the values of the ionization energies.
b. Explain the large increase between $I_{3}$ and $I_{4}$

Key Concepts

Core vs. Valence Electrons

The distinction between core and valence electrons is key in explaining the dramatic increase in ionization energy observed between certain steps. Valence electrons are those in the outermost energy level and are less tightly bound, whereas core electrons, located closer to the nucleus, are held much more strongly due to lower energy levels and less shielding, resulting in a large jump in the energy required to remove them.

Electron Configuration

Electron configuration describes the distribution of electrons among the different atomic orbitals in an atom. The arrangement is crucial for understanding ionization energy trends because electrons in outermost (valence) shells are generally easier to remove compared to those in inner (core) shells, which are more tightly bound to the nucleus.

Effective Nuclear Charge

Effective nuclear charge is the net positive charge experienced by an electron in an atom, taking into account both the actual nuclear charge and the shielding effect of other electrons. As electrons are removed, the effective nuclear charge on the remaining electrons increases, leading to higher ionization energies for subsequent electron removals.

Ionization Energy

Ionization energy is the energy required to remove an electron from a gaseous atom or ion. It is a fundamental property that reflects the attraction between the nucleus and the electrons, and it increases as electrons are removed sequentially, as each successive electron is more strongly attracted to the increasingly positive ion.

Periodic Trends

Periodic trends in ionization energies show that as electrons are removed from an atom, the energy required increases due to a higher effective nuclear charge acting on fewer electrons. The trend is also influenced by the arrangement of electrons in different shells and subshells, which affects the electron-electron repulsions and screening effects within the atom.

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