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## Electron configurations

In this blog, I discussed the “electron configuration” nomenclature which is so loved by chemists (strange people that they are….). Just to remind you, the noble gas neon, which is at number 10 in the periodic table, may be written as $1s^{2} \; 2s^{2} \; 2p^{6}$. If you add together the superscripts you get $2+2+6=10$, the number of electrons in neutral Helium. Titanium, which is at number 22 in the periodic table may be written as $1s^{2} \; 2s^{2} \; 2p^{6} \; 3s^{2} \; 3p^{6} \; 3d^{2} \; 4s^{2}$. Again, if you add together the superscripts you get $2+2+6+2+6+2+2=22$, the number of electrons in neutral Titanium. I explained in the blog that the letters s,p,d and f refer to “sharp, principal, diffuse” and “fine“, as this was how the spectral lines appeared in the 1870s when spectroscopists first started identifying them.

But, what I didn’t address in that blog on the electron configuration nomenclature is why do electrons occupy different shells in atoms? In hydrogen, the simplest atom, the 1 electron orbits the nucleus in the ground state, the n=1 energy level. If it is excited it will go into a higher energy level, n=2 or 3 etc. But, with a more complicated atom like neon, which has 10 electrons, the 10 do not all sit in the n=1 level. The n=1 level can only contain up to 2 electrons, and the n=2 level can only contain up to 8 electrons, the n=3 level can only contain up to 18 electrons, and so on. This leads to neon having a “filled” n=1 level (2 electrons), and a filled n=2 level (8 electrons), which means it does not seek additional electrons. This is why it is a noble gas.

Titanium on the other hand, with 22 electrons, has a filled n=1 level (2 electrons), a filled n=2 level (8 electrons), a partially filled n=3 level (8 electrons out of a possible 18), and a partially filled n=4 level (2 electrons out of a possible 32). Because it has partially filled n=3 and n=4 levels, and it wants them to be full, it will seek additional electrons by chemically combining with other elements.

What is the reason each energy level has a maximum number of allowed electrons?

It is all due to something called the Pauli exclusion principle.

Wolfgang Pauli, after whom the Pauli exclusion principle is named. He came up with the idea in 1925. In addition to this principle, he also came up with the idea of the neutrino.

## The energy level n

Niels Bohr suggested in 1913 that electrons could only occupy certain orbits. I go into the details of his argument in this blog, but to summarise it briefly here, he suggested that something called the orbital angular momentum of the electron had to be divisible by $\hbar \text{ where } \hbar = h/2\pi, \text{ } h$ being Planck’s constant. We now call these the energy levels of an atom, and we use the letter n to denote the energy level. So, an electron in the second energy level will have $n=2$, in the third energy level it will have $n=3$ etc.

As quantum mechanics developed over the next 15-20 years it was realised that an electron is fully described by a total of four (4) quantum numbers, not just its energy level. The energy level $n$ came to be known as the princpical quantum number. The other three quantum numbers needed to fully describe the state of an electron are

• its orbital angular momentum, $l$
• its magnetic moment, $m_{l}$ and
• its spin, $m_{s}$

## The orbital angular momentum $l$ quantum number

As I mentioned above, spectroscopists noticed that atomic lines could be visually categorised into “sharp”, “principal”, “diffuse” and “fine“, or $s,p,d \text{ and } f$. It was found that the following correspondence existed between these visual classifications and the orbital angular momentum $l$. This is the second quantum number. $l$ can only take on certain values from $0 \text{ to } (n-1)$. So, for example, if $n=3, \; l \text{ can be } 0,1 \text{ or } 2$.

spectroscopic name and orbital angular momentum
Spectroscopic Name letter orbital angular momentum $l$
sharp s $l=0$
principal p $l=1$
diffuse d $l=2$
fine f $l=3$

As this table shows, the reason a line appears as a “sharp” (s) line is because its orbital angular momentum $l=0$. If it appears as a “principal” (p) line then its orbital angular momentum must be $l=1$, etc.

## The magnetic moment quantum number $m_{l}$

The third quantum number is the magnetic moment $m_{l}$, which can only take on certain values. The magnetic moment only shows up if the electron is in a magnetic field, and is what causes the Zeeman effect, which is the splitting of an atom’s spectral lines when an atom is in a magnetic field. The rule is that the magnetic moment quantum number can take on any value from $-l \text{ to } +l$, so e.g. when $l=2, \text{ } m_{l}$ can take the values $-2, -1, 0, 1 \text{ and } 2$ (5 possible values in all). If $l=3 \text{ then } m_{l} \text{ can be } -3, -2, -1, 0, 1, 2, 3$ (7 possible values).

## The spin quantum number $m_{s}$

The final quantum number is something called the spin. Although it is only an analogy (and not to be taken literally), one can think of this as the electron spinning on its axis as it orbits the nucleus, in the same way that the Earth spins on its axis as it orbits the Sun. The spin can, for an electron, take on two possible values, either $+1/2 \text{ or } -1/2$.

## Putting all of this together

Let us first of all consider the $n=1$ energy level. The only allowed orbital angular momentum allowed in this level is $l=0$, which means the only allowed values of $m_{l}$ is also 0 and the allowed values of the spin are $+1/2 \text{ and } -1/2$. So, in the $n=1$ level, the only allowed state is $1s$, and this can have two configurations, with the electron spin up or down (+1/2 or -1/2), meaning the $n=1$ level is full when there are 2 electrons in it. That is why we see $1s^{2}$ for Helium and any element beyond it in the Periodic Table. But, what about the $n=2, n=3$ etc. levels?

The number of electrons in each electron shell
State Principal quantum number $n$ Orbital quantum number $l$ Magnetic quantum number $m_{l}$ Spin quantum number $m_{s}$ Maximum number of electrons
1s 1 0 0 +1/2, -1/2 2
n=1 level Total = 2
2s 2 0 0 +1/2, -1/2 2
2p 2 1 -1,0,1 +1/2, -1/2 6
n=2 level Total = 8
3s 3 0 0 +1/2, -1/2 2
3p 3 1 -1,0,1 +1/2, -1/2 6
3d 3 2 -2,-1,0,1,2 +1/2, -1/2 10
n=3 level Total = 18
4s 4 0 0 +1/2, -1/2 2
4p 4 1 -1,0,1 +1/2, -1/2 6
4d 4 2 -2,-1,0,1,2 +1/2, -1/2 10
4f 4 3 -3,-2,-1,0,1,2,3 +1/2, -1/2 14
n=4 level Total = 32
5s 5 0 0 +1/2, -1/2 2
etc.

The astute readers amongst you may have noticed that the electron configuration for Titanium, which was $1s^{2} \; 2s^{2} \; 2p^{6} \; 3s^{2} \; 3p^{6} \; 3d^{2} \; 4s^{2}$, suggests that the $n=4$ level starts being occupied before the $n=3$ level is full. After all, the $n=3$ level can have up to 18 electrons in it, with up to 10 electrons in the $n=3, l=2$ (d) state. In the $n=3$ level the (s) and (p) states are full, but not the (d) state. With only 2 electrons in the $n=3, l=2$ (d) state, the $4s$ state starts being populated, and has 2 electrons in it. Why is this?

I will explain the reason in a future blog, but it has to do with the “shape” of the orbits of the different states. They are different for different values of orbital angular momentum $l$.