# Are electron orbital orientations filled in any particular order? [duplicate]

This question is about the magnetic quantum numbers and their corresponding orbital orientations, e.g., px, dyz, etc. Are the electrons in any given subshell distributed across the orbitals in any particular order as the number of electrons increases? Such as m = -1 first, then m = 1, etc.?

• No. Also, there is no one-to-one correspondence between $p_x, p_y, p_z$ and $m=-1,0,1$. Jan 30 at 15:21
• a little more detail would be super helpful. How ARE they distributed then? Randomly? Why is there no correspondence? What difference do the notations represent? Jan 30 at 15:41
• Let me explain with the example of p-subshells. The first electron entering the p-subshell could occupy anything - x or y or z. Suppose it occupied x, then the next will occupy either y or z but not x again yet. Suppose z is occupied now, so then the electron occupies y. Now after all three are occupied by one each, pairing occurs again randomly. From my memory, all -x,y,z are of same energy but only their orientation in space is different. Jan 30 at 16:15
• Consider that x, y and z are completely arbitrary (other than being orthogonal to each other), and m_l was arbitrarily chosen to align with the arbitrarily chosen z (ie it could have been the y component just as easily). Also recall that electrons aren't connected to individual orbitals until we measure them. An atom with a single electron in p orbitals is better thought of as a superposition of the states with px occupied, py occupied and pz occupied, and even that is a greater degree of localization than is "true". Jan 30 at 16:16

No, you cannot say that there is a particular order in which the shells are filled.

### Relation between $$\mathrm{p_x,p_y,p_z}$$ with $$\mathbb{m_l}$$ values

The magentic quantum number $${m_l}$$ represents the projection of the angular momentum $$\vec{l}$$ of an electron in an orbital, on an axis, usually taken to be the z-axis. The orbitals, and their $$\mathrm{m_l}$$ values, come from the solution of the Schroedinger's equation for a Hydrogen atom.

Now, the problem is, the solution for the orbitals with $${l=1}$$ (i.e. p orbitals) produces one real wavefunction (corresponding to $$m_l=0$$), and two other complex wavefunctions (corresponding to $$m_l=1$$ and $$m_l=-1$$). If the main axis is taken to be z, then the $$m_l=0$$ orbital has the usual dumb-bell type shape and points in the z-direction, so it is the $$\mathrm{p_z}$$ orbital. However, the complex orbitals are conjugates of each other, and if you plot the real part, they both look like donuts. (You can find more on this here and here). So the other two orbitals, let's call them $$\mathrm{p_1}$$ and $$\mathrm{p_{-1}}$$, do not really correspond to anything we usually see in the chemistry textbooks.

However, wavefunctions themselves don't hold any physical meaning, because we cannot observe them. We can only observe things like electron energy, electron density etc. So, we can take linear combinations of the two complex orbitals to get two new real p orbitals, as long as it conserves the total energy. However, doing this means that the new $$\mathrm{p_x}$$ and $$\mathrm{p_y}$$ orbitals are no longer eigenfunctions of the $$\hat{L_z}$$ operator (i.e. they have no well defined angular momentum along the z-axis, so no well defined $$m_l$$ value). In most cases angular momentum is not really important, so we can safely use $$\mathrm{p_x}$$ and $$\mathrm{p_y}$$.

I suspect that there is a similar case with the d orbitals as well.

### Which orbital is filled in first?

Inside one shell (e.g. 2p) the electrons are indistinguishable, so it is not possible to say which orbital is occupied by which electron. However, if a shell is not completely filled, i.e. there are unpaired electrons, then there would be an total orbital angular momentum and total spin angular momentum coming from all of those electrons combined. It is possible to identify states which have different angular momenta, as they have different energies. This forms the basis of the coupling schemes (LS coupling or jj coupling). However, you still cannot say exactly which electrons are residing in orbitals with which $$m_l$$ values, because multiple arrangements (microstates) can contribute to the same energy level (term).