Ionization energy and the shell model of the atom

Consider the reaction describing the loss of an electron by a neutral atom:

$${\bf X} \longrightarrow {\rm X}^+ + e^-$$

Note that the new element of charge balance has come into chemical equations: the total charge on both sides is the same, zero in this case. The energy required for this reaction to occur is the change in energy $\Delta E$ between the products and reactants:

$$\Delta E = E({\rm products}) - E({\rm reactants}) = IE_1$$

where $IE_1$ stands for the first ionization energy. The term ``first'' or the ``1'' subscript is to distinguish this energy from the second ionization energy, which is the energy to remove an electron from the ionized species X$^+$:

$${\bf X}^+ \longrightarrow {\rm X}^{2+} + e^-\;\;\;\;\;\;\;\;\;\;\Delta E = IE_2$$

The energies $IE_1$ and $IE_2$ are always positive and measure the stability of so called ``outer shell'' electrons of an atom. The higher the ionization energy, the greater the energy ``cost'' to remove an electron, hence the lower the tendency for such an atom to act as an electron donor in a chemical bond.

In general, $IE_n$ is the $n$th ionization energy. The first ionization energy $IE_1$ tends to increase across each period, with the highest value occurring for the inert gases. This suggests that in this group, the elements have an especially high stability. On the other hand, if we look at the trend in $IE_n$ as a function of $n$ for a given element, an interesting phenomenon occurs. Consider the element sodium (Na). In units of MJ/mol, the first 10 ionization energies are $IE_1=0.5$, $IE_2=4.56$, $IE_3=6.91$, $IE_4=9.54$, $IE_5=13.35$, $IE_6=16.61$, $IE_7=20.11$, $IE_8=25.49$, $IE_9=28.93$, $IE_{10}=141.37$. If we plot the log base 10 of these values vs. $n$, the plot appears as shown below:

Figure: Plot of ionization energies for Na.

From the plot, one sees three regions of stability. The first electron is removed easily, electrons 2-9 fall into an intermediate range, while electrons 10 and 11 are especially difficult to remove. In an attempt to explain this ``shell''-like structure, we next consider the so-called classical shell model of the atom.

In order to appreciate how the shell structure arises, we must start by writing down the total energy of a molecule, viewed classically simply as a set of positive and negative charges. Suppose a molecule is composed of $N$ nuclei at positions ${\bf R}_1,...,{\bf R}_N$ with charges $Z_1e,...,Z_Ne$ and masses $M_1,...,M_N$ and $M$ electrons at positions ${\bf r}_1,...,{\bf r}_M$. If the nuclei at these positions have instantaneous momenta ${\bf P}_1,...,{\bf P}_N$ and the electrons have momenta ${\bf p}_1,...,{\bf p}_M$, then the total energy is

$$E = K + V$$

Considering the kinetic energy of the nuclei and the electrons, the Coulomb repulsion between the electrons, the Coulomb repulsion between the nuclei, and the Coulomb attraction between the electrons and nuclei, the total energy can be written as

$$E = \sum_{I=1}^N {P_I^2 \over 2M_I} + \sum_{i=1}^M {p_i^2 \o... ...over R_{IJ}} - \sum_{i=1}^M\sum_{I=1}^N {kZ_Ie^2 \over R_{iI}}$$

where $r_{ij} = \vert{\bf r}_i-{\bf r}_j\vert$, $R_{IJ} = \vert{\bf R}_I-{\bf R}_J\vert$ and $R_{iI} = \vert{\bf r}_i-{\bf R}_I\vert$. This energy is a universal that describes all of ordinary matter. Since any chemical system, from proteins, cells, organisms, semiconductors, computer chipts, polymers,... can all be considered as collections of nuclei and electrons, this energy is always the starting point for a microscopic description. The only parameters that distinguish one system from another are the numbers and types of nuclei and the number of electrons.

Let us apply the potential energy formula above to a single atom with nuclear with charge $Ze$ and $M$ electrons. Since there is only one nucleus (only one positive charge), there is no nuclear-nuclear repulsion term so that $V_{nn}=0$. If the nucleus is fixed at the origin, then the potential energy $V$ can be written simply as

$$V = -{Ze^2 \over 4\pi\epsilon_0}\sum_{i=1}^M {1 \over r_i} +... ...4\pi\epsilon_0}\sum_{i=1}^{M-1}\sum_{j=i+1}^M {1 \over r_{ij}}$$

where we have used $k=1/(4\pi\epsilon_0)$. Here $r_i = \vert{\bf r}_i\vert$ is simply the distance of each electron from the origin. If we consider the case of the Li atom, for example, for which $Z=3$ and $M=3$, the formula becomes

$$V = -{3e^2 \over 4\pi\epsilon_0}\left({1 \over r_1} + {1 \ov... ...{1 \over r_{12}} + {1 \over r_{13}} + {1 \over r_{23}}\right)$$

The formula requires knowing the distances of each of the 3 electrons from the origin as well as the three distances $r_{12}=\vert{\bf r}_1-{\bf r}_2\vert$, $r_{13}=\vert{\bf r}_1-{\bf r}_3\vert$ and $r_{23}=\vert{\bf r}_2-{\bf r}_3\vert$ between the three electrons.

The potential energy, even for lithium, is a complicated function of the electron position variables. As a consequence, we cannot solve for the stable orbits of the electrons using classical mechanics. However, a simple approximation serves to explain why some electrons are more stable than others. In the shell model, we imagine fixing one of the electrons, say electron 1, and then letting all of the other electrons move around classically. The effect of the moving electrons on the fixed electron is then taken into account in an average sense, averaging over a long time period. The fixed electron will then see an effective potential that is a simple Coulomb form given by

$$V_{\rm eff}(r) = -{Z_{\rm eff}e^2 \over 4\pi\epsilon_0 r}$$

where $Z_{\rm eff}$ is an effective charge that results from the nucleus and the averaged motion of the electrons we did not fix. We can now generate an effective potential for electron 2 by first removing electron 1 entirely, fixing electron 2 and letting the remaining electrons move around to generate an effective potential. It is clear from this idea of successively removing electrons and generating effective potentials for one of the remaining electrons that each electron will have a different effective potential!

So now let's consider what happens for the case of sodium (Na), for which $Z=11$. If we consider the effective potential of electron 1 by averaging over the motion of the remaining 10 electrons (this can be done relatively easily using a computer), we obtain a value of $Z_{\rm eff}\approx 1$. It's not exactly 1, but close enough to 1 that we will call it 1. For electron 2, we now remove electron 1, and compute an effective potential by averaging over the motion of the remaining 9 electrons, and we obtain a value of $Z_{\rm eff} \approx 5$. It's not exactly 5 but pretty close. Repeating this procedure for electrons 3, 4, 5,... up to 9, we find that $Z_{\rm eff}$ does not change much from 5. It's a little different for each electron, but the value for electrons 2-9 is fairly close to 5. However, if we now consider electrons 10, the value of $Z_{\rm eff}$ changes suddenly to a value quite close to 11. Obviously, for electron 11, the value of $Z_{\rm eff}=11$ exactly because the only charge electron 11 sees is the nuclear charge (which is exactly 11). Thus, we see very roughly, that three effective charge values $Z_{\rm eff}=1,5,11$ emerge from the shell model, and these values correspond to the shells observed in the energy plot above.

Electrons in the innermost shells with the highest values of $Z_{\rm eff}$ are called core electrons, and they tend to be unimportant for chemical reactivity. Electrons in the outermost shells, with low values of $Z_{\rm eff}$ are called valence electrons, and these are the most important in chemical reactions. The importance of electrons in shells with intermediate values of $Z_{\rm eff}$ in chemical reactions depends on the chemical process under consideration. Their role is likely to be important but indirect in that they can affect the distributions of valence electrons and influence their role in chemical bonding without, themselves, participating in chemical bonds.