The periodic table organizes information about the ground states of the elements. These states are chemically most important because most atoms spend most of the time in their ground states. All the chemical ideas of valence, bonding, reactivity, and so on are consequences of these ground-state atomic structures. But the periodic table does not tell us anything about the excited states of atoms. It is the excited states that hold the key to understanding atomic spectra, and that is the topic to which we turn next.

Sodium $(Z=11)$ is a multielectron atom that we will use as a prototypical atom. The ground-state electron configuration of sodium is $1s^{2}2s^{2}2p^{6}3s$. The first 10 electrons completely fill the $n=1$ and $n=2$ shells, creating a neon core, while the $3s$ electron is a valence electron. It is customary to represent this configuration as $[\text{Ne]}3s$ or, more simply, as just $3s$.

The excited states of sodium are produced by raising the valence electron to a higher energy level. The electrons in the neon core are unchanged. Thus the excited states can be labeled $[\text{Ne]}nl$ or, more simply, just nl. FIGURE 41.23 is an energy-level diagram showing the ground state and some of the excited states of sodium. Notice that the $1s$, $2s$, and $2p$ states of the neon core are not shown on the diagram. These states are filled and unchanging, so only the states available to the valence electron are shown.

FIGURE 41.23 The $[\text{Ne]}3s$ ground state of the sodium atom and some of the excited states.

Figure 41.23 has a new feature: The zero of energy has been shifted to the ground state. As we have discovered many times, the zero of energy can be located where it is most convenient. For analyzing spectra it is convenient to let the ground state have $E=0$. With this choice, the excited-state energies tell us how far each state is above the ground state. The ionization limit now occurs at the value of the atom’s ionization energy, which is 5.14 eV for sodium.

The first energy level above $3s$ is $3p$, so the first excited state of sodium is $1s^{2}2s^{2}2p^{6}3p$, written as $[\text{Ne]}3p$ or, more simply, $3p$. The valence electron is excited, while the core electrons are unchanged. This state is followed, in order of increasing energy, by $[\text{Ne]}4s\text{,}[\text{Ne]}3d$, and $[\text{Ne]}4p$. Notice that the order of excited states is exactly the same order ($3p$–$4s$–$3d$–$4p$) that explained the fourth row of the periodic table.

Other atoms with a single valence electron have energy-level diagrams similar to that of sodium. Things get more complicated when there is more than one valence electron, so we’ll defer those details to more advanced courses. Nevertheless, you already can utilize the information shown on an energy-level diagram without having to understand precisely why each level is where it is.