# Why does a neutral atom of beryllium not have any electrons in a p orbital?

Here's what I understand about quantum number and orbitals, please correct me if anything is wrong:

Electrons enter into these different types of orbitals because they have a higher/lower amount of energy, and we know that the number of electrons varies from element to element along with the energy levels of those electrons in a pretty linear way, conveniently allowing us to organize them into blocks on the periodic table where their highest energy level electron is in one of the many types of orbitals (spdf). I know the energy shell occupied by an electron (the principle quantum number) corresponds with the angular momentum quantum number which determines the magnetic quantum number, which determines the type of orbital an electron is in. I know that all of that is because the electrons in higher energy shells/levels have more energy are thus more likely to be in an orbital such as d or f as opposed to an s or p orbital, for example.

Note that I don't know why the orbitals exist in the exact shapes they do (although I imagine it has something to do with the electrons speeding up or slowing down at certain points making it more likely for us to discover it in a certain spot), or what the threshold is for the amount of energy required to make an electron move into the next type of orbital (d to f, for example).

But, I have one question: why does a neutral beryllium atom with 4 electrons in 2 energy shells/levels not have any electrons in a p orbital?

My reasoning is as follows:

If $n = 2$ where $n$ is the energy shell occupied by beryllium's fourth electron, then $l = 0,1$ and $m_l = \text{either }0 \text{ or }-1,0,1$. If $m_l = -1,0,1$. Then, why isn't this fourth electron in a p orbital?

• chemistry.stackexchange.com/questions/152/… Mar 2 '18 at 17:21
• You are correct in all your analysis in the first two paragraphs (except that the "energy levels of those electrons in a pretty linear way, " follow the $(n+l)$-rule so $\ce{4s<3d}$ in energy, but it doesn't matter). But I don't understand your final question. Two electrons go into $\ce{1s}$ subshell. The remaining two electrons go into $\ce{2s}$ subshell. Is there any more electron left for the $p$ orbital? Mar 2 '18 at 17:22
• In the ground state of the atom the orbitals fill up with electrons going into the lowest energy orbitals. Up to two in s orbitals and up to 6 in p orbitals. The 4 electrons in Be go into the 1s and 2p orbitals first. The 2p orbitals are at slightly higher energy and so do not have any electrons. If you were to excite the Be atom then it is possible to have an electron in a 2p orbital. Mar 2 '18 at 17:24
• I seem to have forgotten that each orbital has 2 electrons in it, not one. Bit of a facepalm here... sorry guys! Thanks for the answers. Mar 2 '18 at 17:26
• This is funny because from the Aufbau principle, this is all correct, but quantum chemistry predicts some admixture of p orbitals into the ground state. I miss general chemistry... Mar 2 '18 at 22:02

Beryllium has 4 electrons in neutral atom. Lowest energy levels are filled first. So firstly 1s2 (ist shell now filled) That leaves 2 left over. So they go into the 2nd shell and the 2s orbital. Therefore electron configuaration is 1s2 2s2. That's it, no electrons left over to put into 2p.

We need to first consider n, the quantum number for the shell, and then l (~the classical angular momentum of the electron), which tells us the shape of the electron density. The first shell (n=1) contains two s electrons with paired spins. That shell is closed, and you begin over again with n=2: the possibilities for n=2 are l=0 (s electron shape, spherical) and l=1 (p electron shape, lobed). Then, since a lobed shape can point in different directions, when l=1, ml can equal +1, 0 or -1 , defining px, py, pz electron shapes, each pointing in one of three orthogonal directions.

So the filling process puts the first two (lowest energy) electrons in the 1s level (when full, it is 1s2); then the next level is 2s (filling to 2s2), then 2p. Since we have run out of electrons, the 2p levels do not get filled, so we don't have to consider 2px, 2py or 2pz designations.

Now, for boron, there will be a p electron. Whether you call it px, py or pz will depend on how you define the directions in a molecule. In an individual free boron atom, 2px, 2py and 2pz all have the same energy and are degenerate.

This is all theoretical. If you ever heated up beryllium so that there were free atoms, they would be so energetic that quantum levels up to 10 or 100 would be occupied, and there would be beryllium ions floating around too!