The first and second electrons go in 1s orbital, the 3rd and 4th go in 2s orbital, and 5th to 10th go in 2p orbitals.

My question is how does these additional electrons affect the orbital shapes? Are the orbital shapes somewhat same as the commonly shown orbitals?

I mean do the electrons in 2s orbital behave in same way(i.e the 2s orbital shape doesn't change) even when more electrons come in the 2p orbitals?

Im unable to be very elaborate about my question, so here(attached below) is a picture that comes to my head when I think of a Neon atom. I think of all electrons belonging to the respective orbitals.

How correct or incorrect is this picture? Are the orbitals same as this? Or something else happens when more electrons come into play, i.e this picture is misleading?

enter image description here

  • 1
    $\begingroup$ These are the 5th to 10th electrons which are on the 2p orbitals. $\endgroup$
    – dan
    Jun 1 at 5:38
  • $\begingroup$ The key missing point in this classical representation is the t axis: the x,y & z axis rotate, and most probably not around a fix axis in 3D. $\endgroup$
    – dan
    Jun 1 at 5:53

2 Answers 2


The $s$ orbitals look the same, but the $p$ orbitals can change a little. All these things, however, very much depends upon the choice of basis and the approximation method we are working by.

For example, if you use Density Functional Theory to calculate the orbitals, it will be different from a Hartree-Fock calculation. They are different approximation methods.

Because you are talking about Neon, a noble gas, you can cover the $s$ orbitals normally, and take the $p$ orbitals as separate. By that I mean that you do not have to worry about hybirdising the orbitals. $sp,\ sp^2,\ sp^3$ orbitals all look really different from simple $p$ orbitals.

And within p orbitals, are you going to consider $p_x,\ p_y,\ p_z,$ or $\left|2,1,\pm1\right>$ and $\left|2,1,0\right>$? That is a choice of basis / representation that looks different even though it really does not matter.

Do note that orbitals themselves are an approximate idea, so you do not have to go into it too deeply. Even if you fix yourself to constant number of electrons (and no positrons), the true wavefunction is a correlated mess of all the electrons, and you cannot separate them out into specific orbitals. We continue to talk about orbitals because of their tremendous success in explaining chemical bonds despite being only approximate.


Some of you description is good (e.g., there are shells), but the details are incorrect.

One problem is your picture: it clearly shows 3 preferred directions in the $P$ orbitals. There are clearly 3 preferred axes defining special alignment.

The problem is: how does a neon atom know how you have defined your axis direction? It doesn't. Physics is independent of coordinate choices.

Note that:

$$ \sum_{l=-m}^m Y_{lm}^*(\theta, \phi)Y_{lm}(\theta, \phi) \propto 1 $$

which means any full shell is spherically symmetric.

If you look at chemists' version of orbitals (which are linear combinations of the physicists' version), you have $P_x, P_y, P_z$. Note that (in the angular coordinate):

$$ P_z = Y_{1, 0}(\theta, \phi) \propto r\cos\theta = z$$

Likewise for $P_x$ and $P_y$, so:

$$P_x^2 + P_y^2 + P_z^2 \propto x^2 +y^2+z^2 = r^2$$

which is spherically symmetric.

The neon atom does not care about your coordinates.

(A usefully exercise is get your favorite plotting tools and do a polar plot of $r(\theta) = \cos^2\theta$ and $r(\theta) = \sin^2(\theta)$...they are individually somewhat dumbbell shaped, but the sum is clearly a circle).

Another problem is the labeled electron being in a specific orbit. If you label an orbital by:


and a two particle orbit by:

$$|\psi\rangle_{12} =|n, l, m\rangle_1| n', l', m'\rangle_2$$

and so on for $n$ particles. (The subscript labels the electron).

For brevity, I will ignore spin, and show a 5-particle wavefunction using the Slater determinant:

$$|\psi\rangle=\left | \begin{array}{ccccc} |1,0,0\rangle_1 & |2,0,0\rangle_1& |2,1,-1\rangle_1& |2,1,0\rangle_1& |2,1,1\rangle_1 \\ |1,0,0\rangle_2 & |2,0,0\rangle_2& |2,1,-1\rangle_2& |2,1,0\rangle_2& |2,1,1\rangle_2 \\|1,0,0\rangle_3 & |2,0,0\rangle_3& |2,1,-1\rangle_3& |2,1,0\rangle_3& |2,1,1\rangle_3 \\|1,0,0\rangle_4 & |2,0,0\rangle_4& |2,1,-1\rangle_4& |2,1,0\rangle_4& |2,1,1\rangle_4 \\|1,0,0\rangle_5 & |2,0,0\rangle_5& |2,1,-1\rangle_5& |2,1,0\rangle_5& |2,1,1\rangle_5 \\ \end{array}\right | $$

where the determinant is a convenient notation for creating the fully anti-symmetric wavefunction under particle interchange.

(A 10-particle wave function would have $10!=3,628,800$ terms--so no, not going to write that out).

It is really the formal form of the Pauli Exclusion Principle (PEP). Saying "no electron can be in the same quantum state" is really not sufficient. The correct formulation is that the multi-electron state must be antisymmetric under interchange of any two particle labels.

That is the crux of the PEP: the whole reason you need to fill shells is because you need 5 different states to have a non-zero wavefunction that is antisymmetric under interchange of any pair of electrons.

What that means is that there is no "1st electron" that fills the $S$ state and so on; rather, each electron is entangled in all states.

It is not a result that can be interpreted classically, in which electrons have an individual identity. Electrons are not just identical: they are indistinguishable.


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