Two atoms have the electron configurations \(1 s^{2} 2 s^{2} 2 p^{6}\) and \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}\). The first ionization en ergy of one is \(2080 \mathrm{~kJ} / \mathrm{mol},\) and that of the other is \(496 \mathrm{~kJ} / \mathrm{mol}\). Pair each ionization energy with one of the given electron configurations. Justify your choice.

Understanding electron configurations is critical for interpreting the behavior of atoms in chemical reactions and physical processes. An electron configuration is a notation that shows the distribution of electrons in the atomic orbitals. The configuration illustrates how the electrons are organized around the nucleus, starting from the lowest energy level to the highest.

For example, the electron configuration notation of a neon atom is represented as \(1s^{2} 2s^{2} 2p^{6}\), which tells us that there are two electrons in the 1s orbital, two in the 2s orbital, and six in the 2p orbitals. Each orbital can hold a specific maximum number of electrons; two for an s orbital and six for a p orbital. When an orbital is filled to its maximum capacity, it's considered closed and thus contributes to the atom's stability.

Referring to the exercise, the atom with configuration \(1s^{2} 2s^{2} 2p^{6}\) has a stable, closed shell of electron orbitals, which typically corresponds to inert or noble gases. On the opposite side, an atom with a \(1s^{2} 2s^{2} 2p^{6} 3s^{1}\) configuration has an additional electron in the 3s sublevel, representing the beginning of a new energy level and suggesting that the atom is more reactive and less stable. This additional electron is more easily ionized, which aligns with the lower ionization energy observed.

Atomic stability is a fundamental concept explaining why certain elements are more reactive than others. Atoms gain stability when they have a full valence shell - either by filling it with their own electrons or by sharing, gaining, or losing electrons through chemical bonding.

An atom with a full outer shell of electrons, like the noble gases, exhibits great stability and low reactivity, because these full-shell configurations are energetically favorable. The energy required to add or remove an electron significantly increases when dealing with a full valence shell, hence the high ionization energies for such atoms.

In contrast, atoms that lack a full valence shell tend to be more reactive. This is because they can achieve stability by reacting with other atoms to complete their outer shell, which often involves less energy compared to the ionization of an atom with a complete shell. For instance, as explained in the solution, the atom with the electron configuration ending in \(3s^{1}\) will have lower ionization energy, indicating that it is less stable than its counterpart with a full \(2p^{6}\) shell.

The periodic table of elements is a tabular display of known chemical elements, organized by their atomic number, electron configurations, and recurring chemical properties. Elements within the same column, or group, often have similar properties and share the same number of electrons in their valence shell.

The organization of the periodic table allows us to predict the chemical behavior of elements. For instance, elements on the far right column are the noble gases; they have full valence electron shells, making them highly unreactive. Meanwhile, moving from the top to the bottom of a group, the ionization energy generally decreases because the valence electrons are further away from the nucleus and therefore less tightly bound.

Moreover, across a period (row) from left to right, the ionization energy usually increases, as electrons are added to the same energy level while the nucleus gains more protons, thereby increasing the positive charge which more strongly attracts electrons. This trend can be interrupted by the start of a new sublevel, as seen in the exercise where the electron in the \(3s^{1}\) orbital is more easily removed than those in the \(2p^{6}\) orbitals due to the lower effective nuclear charge, and this is why this element has a lower ionization energy compared to an element with a completed \(2p^{6}\) electron configuration.