I was reading about atomic orbitals. From Bohr's model, electron orbitals are considered circular. But as I saw the sub-orbitals like $2p_x$, $3d_{x^2-y^2}$ I noticed something strange. These sub-orbitals aren't circular. How can that be possible if $2p$ and $3d$ are circular? Also why is $3d$ defined like this $(x^2-y^2)$? What does this actually mean?

I don't know how to write mathjax. Also I couldn't decide where to post it.

  • 1
    $\begingroup$ Orbitals such as $2s$ or $2p$ are beyond the Bohr model. They are actual solutions to the single-electron Schrodinger equation... $\endgroup$
    – Jon Custer
    Jul 10, 2018 at 13:22
  • 2
    $\begingroup$ Bohr's model, despite it's historic significance, is not valid and honestly I wish they'd stop teaching it as people have to unlearn the darn thing to learn quantum models, even at the level of an overview. It's the source of lots of confusion. $\endgroup$ Jul 10, 2018 at 13:24

2 Answers 2


Your statement that 2p is circular is wrong. All three 2p orbitals are dumb-bell shaped, the only difference is their orientation is along three different axis.

In the Bohr's Model of Hydrogen atom, there is an electron orbiting around a positive charged nucleus almost similar to the solar system model. Using this model few observations could be explained such as the Rydberg formula for emission lines of hydrogen. But the major problem for this model was that there was no justification about why only orbits of some particular energy level and radii were possible. It also could not explain certain facts such as the splitting of the emission lines of Hydrogen etc.

So this model had to be left behind, and the Hydrogen atom was solved using quantum mechanics which could explain the physically obtained observations much better. The concept of atomic orbitals and their shape is an outcome of the quantum mechanical model which obey's the uncertainty principle. This is obtained from solving the time independent Schrodinger's equation. This model clearly explains why energy levels and angular momentum are quantised. The different atomic orbitals namely 1s, 2s, 2p , 3d etc. are the various stationary states (i.e. solutions) of the quantum mechanical model. Their shape essentially highlights the probability density distribution of an electron occupying a particular stationary state in 3d space.

Now the orbitals are named so, from their orientation in 3d space, and their wavefunction. The angular part of the 3d wave functions are listen below. I hope you will now understand the origin of their names.

$$Y_{3d_{xy}} = \sqrt{(60/4)}xy/r^2 × (1/4π)^{1/2} \\ Y_{3d_{xz}}= \sqrt{(60/4)}xz/r^2 × (1/4π)^{1/2}\\ Y_{3d_{yz}}= \sqrt{(60/4)}yz/r^2 × (1/4π)^{1/2}\\ Y_{3d_{x^2-y^2}} = \sqrt{(15/4)}(x^2 - y^2)/r^2 × (1/4π)^{1/2}\\ Y_{3d_{z^2}}= \sqrt{(5/4)}(2z^2-(x^2 + y^2))/r^2 × (1/4π)^{1/2}$$

This concept will be more clear, after you take a quantum mechanics or a quantum chemistry course.

  • 1
    $\begingroup$ I wouldn't say that there was "no justification about why only orbits of some particular energy level and radii were possible"; the justification was that the angular momentum of the electron only existed in integer units of $\hbar$. But that assumption was pretty much ad hoc, at least until the full solutions of Schrödinger's equation were developed. $\endgroup$ Jul 10, 2018 at 14:46
  • $\begingroup$ @Satwata Hans Just out of curiosity, I was wondering if I could draw the 3D graph of $xy$, then would I get a shape like $3d-{xy}$? I understood what you said about their origin but I was just wondering. $\endgroup$ Jul 10, 2018 at 16:45
  • $\begingroup$ @AsifIqubal Yes absolutely, the probability density distributions are 3d. The 2d representations which we see are the contour maps of the 3d objects. $\endgroup$ Jul 10, 2018 at 17:09

The number in front of the orbital (1, 2, 3, ...) is the prime quantum number $n$ determining the energy of the electron at the orbital (in the ideal case where the electrons do not have spin and only interact with the nucleus), according to, \begin{equation} E_{n}=-\frac{E_1}{n^2}, \;\;\; E_1 = \frac{mZ^2e^4}{2\hbar^2} \;\; (in\;Gauss\;units) \end{equation}

The next letter (s, p, d, f) is the angular momentum number determining the angular momentum eigenvalue of the electron at that orbital. "$s$" means $l=0$, "$p$" means $l=1$, "$d$" means $l=2$ and "$f$" means $l=3$, from which the total angular momentum $L$ is obtained, \begin{equation} L=\hbar\sqrt{l(l+1)} \end{equation} The next axes letter ($x$, $y$, $z$) determines the direction of the angular momentum (that's not precisely true actually). For example "$p_x$" means $\vec{L}=\hbar\sqrt{2}\hat{x}$. For higher angular momenta, the degeneracy makes this analogy not useful. For the $n=3$ orbitals, for example, there are in total 9 possible orbitals (times 2 if we take into consideration the spin of the electron). From those 9 orbitals, 1 of them is the $2s$, 3 of them are the $3p_x$, $3p_y$ and $3p_z$ and 5 of them are the $3d$ orbitals. From those 5 $3d$ orbitals, the convention of symbolizing them as $z^2$, $xz$, $yz$, $xy$ and $x^2-y^2$ is kinda ambiguous but you can figure out the rules if you dig in some more. They have to do with the third quantum number, the magnetic quantum number $m$ determining the z-component of the total angular momentum accordning to, \begin{equation} L_{z}=\hbar m \end{equation} For example, $3d_{z^2}$ means $n=3$, $l=2$ and $m=0$.

Hope this helps!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.