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# 7.3: Periodic Table of Elements

We move on now to atoms beyond hydrogen. As a first step, let's consider what happens if we just add more protons to the nucleus. [We can also start adding neutrons, but they do not contribute to the electromagnetic potential that appears in the hamiltonian for the electron. Here we are interested in the chemical properties of atoms, which are determined by their electron structure. Atoms with the set number of protons and electrons can have different numbers of neutrons (these varieties are called isotopes), but while they have different masses and exhibit different nuclear activity, behave in the same manner chemically.] If the number of protons is the integer $$Z$$, then we find that the coulomb potential changes to:

$V\left(r\right)=\dfrac{-Ze^2}{4\pi \epsilon r}$

An atom with this many protons and only one electron will behave like a hydrogen atom with $$e^2$$ replaced everywhere with $$Ze^2$$. Looking back at results from the hydrogen atom, we find that the Bohr radius (Equation 5.1.4) is reduced by a factor of $$Z$$, and the depth of the ground state energy (Equation 5.1.7) is increased by a factor of $$Z^2$$:

$\begin{array}{l} a_Z= \dfrac{4 \pi \epsilon_o \hbar^2}{m_e \left(Ze^2\right)} =\dfrac{1}{Z}a_o\\ E_{Z1} = - \dfrac{m_e \left(Ze^2\right)^2}{2\left(4\pi\epsilon_o\right)^2\hbar^2}=Z^2 E_1\end{array}$

The unsurprising result is that the presence of more protons simply binds the electron more tightly to the atom. Put another way, the ground state energy of a single electron is lower with more protons in the nucleus. This result will play an important role when more electrons are then added, since additional electrons result in higher energy levels, resulting in an interesting conflict as atomic numbers grow.

## Energy Dependence on the Orbital Angular Momentum

Let’s consider a case of helium, where one of the electrons is in the $$1s$$ (ground) state and the other is in the $$2p$$ ($$n = 2$$, $$l = 1$$) state. Recall that for these maximum-$$l$$ states the orbit is “nearly circular,” and the distance from the nucleus of maximum probability is the Bohr orbit. That is, the lower energy electron spends its time mostly around a distance of $$a_o$$ from the nucleus. The higher energy electron, on the other hand, spends its time mostly around a distance of $$4a_o$$ from the nucleus. Thus there is, effectively, a shell of negative charge between the outer electron and the nucleus, and the repulsive force between this shell and the outer electron causes it to be bound more loosely (we say that the inner electron screens the outer one from the pull of the nucleus).

However, now suppose that the outer electron is instead in the $$2s$$ ($$n = 2$$, $$l = 0$$) state. This electron has no angular momentum, and the result is that it has two antinodes, and while it still spends much of its time farther from the nucleus than the inner electron, it does spend more of its time close to the nucleus than it does in the $$2p$$ state. Therefore the shielding for the $$2p$$ state is greater than it is for the $$2s$$ state.

The phrase, "...spends more of its time close to the nucleus..." is quite an inaccurate description of the electron's condition, but (as with so many classical descriptions) gives a somewhat easier description of the phenomenon to visualize than, "...has a probability density that is larger on average close to the nucleus..." I beg your indulgence for this and other such language used here.

More generally, we conclude that the effective potential energy for multiple-electron atoms varies as a function of the $$l$$ quantum number. This removes the “accidental degeneracy” we saw for the hydrogen atom, where the energy levels didn’t depend upon $$l$$. Specifically, we see that the higher the $$l$$ quantum number of an electron is, the less tightly-bound it is to the nucleus (because the shielding of the attractive force is greater). When something is less-tightly-bound, it requires less energy to break free, which means it is in a higher energy state. So higher $$l$$ values correspond to higher energy states (all else being equal).

## Shells and Subshells

If increases in $$n$$ and l both result in increases in the energy state of an electron, which one dominates? That is, if we are given the quantum numbers of two electrons for an atom where electrons fill in all the lowest energy states, how do I determine which electron is in the higher energy state? We can’t solve Schrödinger’s equation exactly for multi-electron atoms, but approximation methods do provide useful rules of thumb:

• First check the value of $$n + l$$. The state with the higher sum is at higher energy.
• If $$n + l$$ is a tie, then $$n$$ is the tiebreaker – the state with the higher $$n$$ value has the higher energy.

So for example, the states $$4p$$, $$4d$$, and $$5s$$, in order of highest-to-lowest energy are:

1. $$4d$$ ($$n+l=4+2=6$$)
2. $$5s$$ ($$n+l=5+0=5$$, wins tiebreak with higher n)
3. $$4p$$ ($$n+l=4+1=5$$, loses tiebreak with lower n)

In one-dimensional quantum mechanics, we got used to thinking of $$n$$ as the "energy quantum number," and this was reinforced by the accidental degeneracy of the hydrogen atom, even though we started calling it the "principle quantum number." Now that the energy level is determined by a combination of the $$n$$ and $$l$$ quantum numbers, we have to make a mental note to discard this bias.

Electrons residing in a state with quantum number n are said to be in the $$n^{th}$$ shell. There are in general also electrons within a given shell with different $$l$$ values, so electrons within the $$n^{th}$$ shell that have the same $$l$$ value are said to be in the $$nl$$ subshell.

So using our rules of thumb, we see that as we construct ground states of atoms by filling-in electrons for the lowest energy levels first, there are subshells of higher shells filling before lower shells are filling (though of course lower full shells always fill in before the higher full shells do). Note that the rule of thumb only applies to adding new electrons (with the balancing protons). The energy levels of electrons in interior shells go back to the $$n$$-decides-the-energy rule when the outer shells become filled. in the example above, the $$4d$$ sub-shell is higher energy than $$5s$$, which means that as we are adding protons and electrons to build the atom, the $$5s$$ subshell fills first. But as the $$n=4$$ shell fills, the energy of the $$4d$$ states drops below those of the $$5s$$ states. Even the $$4f$$ sub-shell, whose initial energy is even greater than $$6s$$, drops below the energy of $$5s$$ as outer subshells populate.

It should also be mentioned that we have certainly not lost all signs of degeneracy. We still have the ml and ms quantum numbers that provide that. Taking these into account provides for the populations of the shells and sub-shells. We now add to our spectroscopic notation the number of electrons in the sub-shell that are allowed thanks to these degeneracies. We use a notation called the electronic configuration of an element. Specifically, if we are talking about boron (Z = 5) in the ground state, we write:

$1 s^2 2s^2 2p^1 \nonumber$

The superscript indicates the number of electrons in the subshell. Naturally this number is limited to the number of available $$m_l$$ values ($$2l+1$$) multiplied by two (since electrons are spin-$$\frac{1}{2}$$ fermions). Also, if we add up all the superscripts, we get the atomic number of the atom (we assume here that we are not talking about ions – atoms for which the number of protons does not equal the number of electrons).

While we may build our description of atoms (and then place them in the periodic table) using their ground states, they (just like the hydrogen atom) can be in excited states. For the hydrogen atom it was easy to tell immediately from the spectroscopic notation if we were talking about an excited state. For multi-electron atoms it is not quite this easy – we need to keep in mind both the maximum subshell populations and the rules of thumb.

## Valence Electrons and the Periodic Table

As we build atoms of ever-larger $$Z$$ values, the increased number of protons of course acts to lower the lowest energy state, because the force of attraction grows (i.e. the potential energy becomes more negative). If it were not for the exclusion principle, added electrons would keep going into the ground state, and atoms would become tighter and tighter bound. But the exclusion principle forces new electrons to occupy higher energy states (remember “higher energy” means “closer to zero energy,” which means “less energy needed to remove the electron”), which prevents this perpetual cascade into lower and lower energies.

This tug-o-war between the reduction of energy due to increasing $$Z$$ values and the increase of energy due to the exclusion principle is what leads to the periodicity of the elements. Let’s start at hydrogen and keep increasing $$Z$$ (for ground states) one step at a time and see what happens:

$$H\rightarrow He\;:\;1s^1\rightarrow 1s^2$$

As we saw above, adding a proton to the hydrogen atom nucleus decreases the energy of the electron by a factor of 4. Assuming no interaction between the two electrons, their energies would therefore be about $$-54.4eV$$. It turns out that the repulsive force between the electrons adds some energy to each electron to make it about $$-40eV$$, but they are still clearly more tightly bound than the electron in the hydrogen atom. If we add another proton and electron (to form lithium), will we see this tighter-bonding trend continue?

$$He\rightarrow Li\;:\;1s^2\rightarrow 1s^2 2s^1$$

While the added proton does reduce the energy for the three electrons, the third electron is precluded from entering the $$n=1$$ shell by the exclusion principle, which means it must be (on average) much farther from the nucleus than the other two electrons. As such, it is repelled by the two electrons in the $$n=1$$ shell, even as it is attracted by the three protons. There is a net attraction, binding this electron, but it is held far less tightly than the $$n=1$$ shell electrons (it is at a must higher energy), which means it can more easily be torn away from the atom than the others. This outer electron is known as a valence electron, and the number of electrons an atom has in this category is referred to as that element's valence (though we will see a slight change in the language of this accounting shortly). Let's move on to Beryllium:

$$Li\rightarrow Be\;:\;1s^2 2s^1\rightarrow 1s^2 2s^2$$

It looks at first like this change might result in another case like helium, since we have added another proton (tighter bonding), and were able to add the electron at the same average distance as the previous electron. Well, these features certainly result in an element that is less reactive (a term that loosely refers to how easily an element can give up or accept spare electrons) than lithium, but this change does not elevate this element to the level of a "noble gas." The best way to see this is to consider the change when the next electron is added. While the next electron needs to go into the $$2p$$ subshell, and this "more circular" orbit will be shielded slightly by the "more elliptical" orbits of the $$2s$$ subshell electrons, this doesn't account for anywhere near as much difference as the huge jump that occurs when the shell number changes. The next inert gas therefore comes about when the $$n=2$$ shell is filled – neon ($$1s^2 2s^2 2p^6$$).

Note that when it comes to valence, it is much more energetically preferred for an atom to add an electron when its $$p$$ subshell is nearly full than it is to give up all of its outer electrons. Valence is therefore defined in terms of the nearest filled shell. That is, if it requires fewer electrons be taken away to get to a stable outer shell than added, the valence number is positive by that number, and if it requires fewer electrons be added, then the valence is negative by that number.

Let's jump ahead a bit, to get to some cases where the ground state energy changes are governed less-trivially by the rules of thumb. When we get to the $$n=3$$ shell, it seems like the next inert gas would occur when this shell is filled. The $$m_l$$ and $$m_s$$ quantum numbers available for the $$3d$$ subshell allow for 10 electrons to occupy it, so filling this shell occurs when $$Z=28$$:

$1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} \nonumber$

But this is not correct. We are restricting ourselves to ground states, and building from lower energy states to higher ones, but according to our rule of thumb, these are not the lowest energy states: $$4s$$ ($$n+l=4$$) is lower energy than $$3d$$ ($$n+l=5$$). Consequently, the next inert gas (argon, $$Z=18$$) occurs after the $$3p$$ subshell is filled (before any electrons occupy the $$3d$$ subshell), because the next jump is to the $$n=4$$ shell:

$1s^2 2s^2 2p^6 3s^2 3p^6 \nonumber$

Excluding the lanthanides and actinides (these feature $$f$$ subshells, which, with the rules of thumb, complicate the organization, so I have left them out, but the reader is encouraged to construct the electronic configurations of these elements from the rules of thumb), the periodic table is organized this way (I have also omitted the myriad of man-made elements in the $$n=7$$ row):

Figure 7.3.1 – Periodic Table and Electronic Structure