# Why do metal valence electrons have low energies?

I know that metallic bonding happens because metal valence atoms have low energy so it can move free from atom to atom. But is there a reason why the have such low energies or is it just a fact that makes an element a metal.

• Bloch states extend throughout a crystal, whether it is a metal or a semiconductor or an insulator. What makes a crystal one or the other is the band structure and how many electrons are where in the band structure. – Jon Custer Sep 18 '17 at 22:20
• @FintanRoche do you know any quantum mechanics? – CR Drost Sep 19 '17 at 13:51

The reason for this is actually somewhat simple: the energy states are very nearly degenerate; they all have roughly the same energy.

The following review of the Hydrogen atom may be review and you may skip it, but I am including it for the readers who may not know. So we have one model which we use to understand such things and it is the quantum mechanical model of the Hydrogen atom, which we can solve exactly. When we solve it exactly, we find that our solution really has a certain universality: the most important part, the angular component of the wavefunction, holds for any spherically symmetric potential. Furthermore when we add the spin degeneracy what we get explains the entire structure of the periodic table: we should start by adding a row with 2 elements, then the next "widening" must add 6 elements, then the next "widening" must add 10 elements, then the next must add 14 and then 18.

What is happening here is that electrons cannot simultaneously occupy the same state; at most two can simultaneously occupy one orbital state, and only if they have opposite spin. So if we divide those mystery numbers by this factor of 2 we find the odd numbers: 1, 3, 5, 7, 9.The orbital states have higher and higher total orbital angular momentum $L^2 = \hbar^2 \ell (\ell + 1)$, which can be more or less focused along one particular direction $L_z = \hbar m$: the condition that $L_z^2 < L^2$ means that for the $\ell=0$ state we only have $m=0$ but for $\ell=1$ we are allowed a value of $m$ in $\{-1,0,1\}$ and so forth. Combined with the overall energy level $n$ which is allowed to take the values $\{1,2,3,\dots\}$, which basically sets the "distance" of the orbit as well as the maximum angular momentum $\ell < n$, we have a full description of the Hydrogen atom. For historical reasons we call the $\ell=0$ states the "s-orbitals" and the $\ell=1$ states the "p-orbitals" and $\ell=2$ the "d-orbitals" and the $\ell=3$ the "f"-orbitals. The reason that adding the new elements sometimes skips a row comes down to what is the lowest energy state between the higher-$\ell$, lower-$n$ states and the lower-$\ell$, higher-$n$ states; so the $4s$ states happen to get filled before the $3d$ states for example: this is why you do not see the $d$ states appear until the fourth row even though they begin with the third radial energy.

Now if you bring two atoms near each other, these orbitals overlap and generally speaking they "hybridize" to some sort of shared orbital states. But as we have seen, if the electrons do not fill the current shell particularly well, then there will be lots of similar-$\ell$, similar-$n$ states which are all trying to hybridize with each other.

Imagine if the electron explicitly had to hop from atom to atom: when the electron shells are very sparsely filled, when an electron "hops" from one atom to another, there will be a lot of same-energy states that the electron could occupy on the other, so the hopping simply doesn't take much energy. In order to prevent this, the atom that we're building a crystal out of needs to have its current electron shell be relatively "full". This is why the left hand side of the periodic table is all metals and you have to go to the right hand side to find the nonmetals; their shells are more and more full and therefore it costs an electron energy to hop onto a nearest neighbor. When this "hopping" becomes low-energy is when you get these highly spread out electron states; when it is high-energy is when you see a state where the hybridization forms two new orbitals, one with lower energy and one with higher energy, and these do not spread out among more than one atom very easily.

• But this has almost nothing to do with solid state bonding... – Jon Custer Sep 18 '17 at 23:25
• @JonCuster I mean I can do some nearest neighbor Hamiltonians and the mean field theory assuming that my undergrad physics is not too rusty, but the comparative irrelevance of $m$ to the energy creating a lot of hybridized states is as far as I know the reason why there are no semiconductor gaps on the left hand side of the periodic table. I am just trying to knock down the complexity of that answer to the pre-undergraduate level because the asker has not indicated any familiarity with QM. Do you have any suggestions towards this goal? – CR Drost Sep 19 '17 at 12:53
• Sadly, solid state physics is built on QM, so it is just plain hard. Somehow perhaps one could argue about the number of free electrons and the Fermi surface without delving too deeply in to Bloch solutions. My bigger issue had to do with electrons 'hopping' from atom to atom - to me that will make it much harder to understand Bloch states. – Jon Custer Sep 19 '17 at 13:13