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.
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.
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!
