My chemistry books state without giving any reason that due to its' Bell shaped(2d:planar*) structure p_orbital possess more energy than s -orbital?(speherical/ 2d:circle)

So how does geometry affects energy content of an Orbital?

*See edits for why this notation is used

  • $\begingroup$ I don't know the field but starting at en.wikipedia.org/wiki/Computational_chemistry and going on from there can be quite interesting. $\endgroup$
    – Emil
    Commented Jan 28, 2017 at 6:14
  • $\begingroup$ The p orbitals are not planar. en.wikipedia.org/wiki/Atomic_orbital#Shapes_of_orbitals $\endgroup$
    – MaxW
    Commented Jan 28, 2017 at 7:11
  • $\begingroup$ Well to be specific(err on my part) ..planar is 2d approximation of it's bell structure(my book also use d the same term)..but anyway for the sake of definitness i will edit..thank you @MaxW $\endgroup$
    – Xasel
    Commented Jan 28, 2017 at 7:14
  • $\begingroup$ All orbitals are 3D in shape, none are "planar." $\endgroup$
    – MaxW
    Commented Jan 28, 2017 at 7:16
  • $\begingroup$ l0l...well s ir I meant to say their 2d approximation on the peace of paper..you can't expect a chemistry book to have rigourous details ...I hope you now got what I am trying to say :how the geometry affects energy content and can answer int hat context? $\endgroup$
    – Xasel
    Commented Jan 28, 2017 at 7:19

1 Answer 1


You can calculate a energy $E(\left\vert \alpha \right\rangle)$ of a state $\left\vert \alpha \right\rangle$:

$$ E(\left\vert \alpha \right\rangle) = \left\langle \alpha \vert \mathbf{E} \vert \alpha \right\rangle$$

But I think you don't want to do this ;). Actually, the states using in chemistry are the energy eigenstates, which means you know their energy by their name.

There are three quantum numbers, ($n$, $l$, $m$) which correspond to the electron's principal quantum number, angular momentum and magnetic quantum number respectively. For example, in the term like $3p_x$ state, 3 means n = 3 and $p_x$ means that l = 1 while m = +1 and -1 states are mixed.

In the single electron problem, the energy is determined exclusively by the quantum number $n$, so $4d$ and $4d_{xy}$ have a same energy.

You said a energy of the $p$-orbital is higher than that of $s$-orbital. This is not true when there is a single electron in the atom. However, the atoms with multiple electrons, we should consider the interactions between the electrons. For example, if we compare a energy of $2s$ of $1s^22s$ and a energy of $2p_z$ of $1s^22p_z$, the latter is bigger than the former. This is because $p$-orbital resides in outer region compared to $s$-orbital, so the probability for $1s^2$ electrons to be inside of the state is larger, resulting in smaller pulling force.


The geometry information is included in the quantum numbers. And the energy of electron in the atoms with a single electron is trivial, let me focus on the atoms with multiple electrons. Then, we should guess the interactions between the electrons, and the geometry really matters here.

The volume or position of a state should be considered in the context. For example, $p_x$, $p_y$, $p_z$ have identical energy values. However, provided the orbital $p_x$ is filled with a electron, another electron would come in $p_y$ or $p_z$ not in $p_x$, because of the interaction.

You should draw the orbital at least once, so you know the interactions between the orbitals.

  • $\begingroup$ :-( $n$ is the principal quantum number not the "energy" quantum number. The 3rd orbital has no $f$ suborbital, just $s$, $p$ and $d$. $\endgroup$
    – MaxW
    Commented Jan 28, 2017 at 7:03
  • $\begingroup$ my question is about how it's geometry affects the energy content (energy as a function of geometry) generally .Not why E as a function of n or d(distance).Can you answer in that context? $\endgroup$
    – Xasel
    Commented Jan 28, 2017 at 7:04
  • $\begingroup$ ;-( the energy is determined exclusively by the quantum number n, .. not true // ... so 4s and 4fxy have a same energy. ... not true $\endgroup$
    – MaxW
    Commented Jan 28, 2017 at 7:08
  • $\begingroup$ @MaxW Really? provided the atom includes a single electron? $\endgroup$
    – 0Tech
    Commented Jan 28, 2017 at 7:19

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