Does each value of $m$ have a fixed orbital (given the value of $l$ as well as $n$)?
For example, if I know that $n = 2$, $l = 1$, and $m = -1$, which orbital ($\ce{p_x}$, $\ce{p_y}$ or $\ce{p_z}$) does it represent?
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$\begingroup$ They are assigned arbitarily, no significance whatsoever . For diagrams the the orbital along the x axis will be $p_x$ though similarly for y, z. $\endgroup$– Avyansh KatiyarMar 11, 2018 at 13:43
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2$\begingroup$ It is not arbitrary: $m = -1$ is a linear combination (a quantum superposition) of $p_x$ and $p_y$ orbitals. I agree this is a duplicate. $\endgroup$– orthocresolMar 11, 2018 at 14:13
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1 Answer
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Actually, $n=2 ,l=1$, and $m=0$ represents $\ce{p_z}$ orbital. But if you look carefully into the wave functions, you will find that $\ce{p_x}$ and $\ce{p_y}$ orbitals are additive and subtractive linear combinations of $\ce{$\psi$_{2,1,1}}$ and $\ce{$\psi$_{2,1,-1}}$ respectively.
The wavefunction for $n=2,l=1,m=0$ is, $$\ce{$\psi$_{2,1,0} = Are^{-r/2a_0}cos$\theta$}$$ You can see that this wave function is proportional to $\ce{rcos$\theta$}$ which is actually $z$ in polar coordinates. That's why due to this proportionality to $\ce{rcos$\theta$ or z}$, this $\ce{$\psi$_{2,1,0}}$ corresponds to $\ce{p_z}$.
But if you look into $\ce{n=2,l=1,m=+1}$ or $\ce{n=2,l=1, m=-1}$ wavefunctions, $$\ce{$\psi$_{2,1,1} = A're^{-r/2a_0}sin$\theta$e^{i$\phi$}}$$$$\ce{$\psi$_{2,1,-1} = A're^{-r/2a_0}sin$\theta$e^{-i$\phi$}}$$But if you take additive linear combination of the above two, you will get $\ce{e^{i$\phi$} + e^{-i$\phi$} = 2cos$\phi$}$, and you will get the resultant wave function proportional to $\ce{rsin$\theta$cos$\phi$ }$ which is actually wavefunction for $\ce{p_x}$ orbital( as $\ce{x=rsin$\theta$cos$\phi$ }$ in polar coordinates)
If you take subtractive combination , $\ce{e^{i$\phi$} - e^{-i$\phi$} =2sin$\phi$}$, and you will get resulting wave-function proportional to $\ce{rsin$\theta$sin$\phi$}$ which is the desired wave function of $\ce{p_y}$ orbital ( as $\ce{y=rsin$\theta$sin$\phi$ }$ in polar coordinates.
So, you can see $n=2,l=1,m=-1$ doesn't correspond to any of the three orbital . $m=0$ only corresponds $\ce{p_z}$ orbital. $\ce{p_x or p_y}$ orbital have wavefunctions which are linear combinations of the actual solution of Schrodinger's solutions $\ce{$\psi$_{2,1,1}}$ and $\ce{$\psi$_{2,1,-1}}$(i.e $m=1$ and $m=-1$ orbitals)
answered Mar 11, 2018 at 14:20