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Online one can find many pictures of d-orbitals.

enter image description here

I know that these states correspond to : n = 3, l = 2, m = -2, ...,2

but I don't know which one is which and I couldn't find a clear asignment anywhere. What are the magnetic quantum numbers for each of the above displayed states?

I feel like I should be able to find this everywhere, but I couldn't find any explicit statement.

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  • $\begingroup$ The shapes should be related to the spherical harmonic functions, are you familiar with those? $\endgroup$
    – user108787
    Commented Nov 15, 2016 at 16:12

3 Answers 3

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There is actually no 1:1 relation between $m_l$ and your notation. The wavefunctions like $\psi_{3d_{xy}}$ are constructed as a superpositions of $\psi_{n,l,m_l}$ with different $m_l$.

The (hydrogen-like) wavefunction, depending on $n$, $l$ and $m_l$, looks like

$$\psi_{n,l,m_l}(r,\theta,\varphi) = R(r) Y_l^{m_l}(\theta,\varphi)$$

$Y_l^{m_l}$ is a complex function. Since (or as long as) they are degenerate, you can make linear combinations of these and still get a solution to your original problem. It is possible to get real-valued wavefunction that way:

$$ \psi_{3d_{z^2}} = \psi_{3,2,0} \propto {3z^2-r^2\over r^2}$$ $$ \psi_{3d_{xz}} = {1\over{\sqrt2}}(\psi_{3,2,-1}-\psi_{3,2,+1}) \propto {xz\over r^2}$$ $$ \psi_{3d_{yz}} = {1\over{\sqrt{2i}}}(\psi_{3,2,-1}+\psi_{3,2,+1}) \propto {yz\over r^2}$$ $$ \psi_{3d_{x^2-y^2}} = {1\over{\sqrt{2}}}(\psi_{3,2,-2}+\psi_{3,2,+2}) \propto {x^2-y^2\over r^2}$$ $$ \psi_{3d_{xy}} = {1\over{\sqrt{2i}}}(\psi_{3,2,-2}-\psi_{3,2,+2}) \propto {xy\over r^2}$$

It should be clear why these combinations are named that way ($d_{z^2}$ is sometimes actually called $d_{3z^2-r^2}$ because of this).

Because drawing the imaginary part of $Y_l^{m_l}$ is a hassle, pictures oftentimes silently ignore that and actually draw the linear combinations, and since you always combine $-m_l$ and $+m_l$ (for higher $l$ too), the difference is often not relevant (because a lot of effects depend on the absolute value of $m_l$). But when you attend your solid state (or molecular) physics lectures, you will need that.

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  • $\begingroup$ So the picture posted by Spectra is wrong? $\endgroup$
    – Stein
    Commented Nov 15, 2016 at 19:12
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    $\begingroup$ @Stein The picture is correct, just the interpretation that $m_l=2 \Leftrightarrow d_{xy}$ is not. The basic shape stays the same, just the imaginary part isn't shown, and the axis are arbitrary for a free atom. You might get in trouble when you have fixed axis (e.g. when your atom is in a solid crystal), because then it will get relevant to understand some things. And when you have to calculate stuff. $\endgroup$
    – Solarflare
    Commented Nov 15, 2016 at 19:49
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I got this one. Does this help ? The subscript of d represents the m value. enter image description here

Link :http://study.com/academy/lesson/electron-orbital-definition-shells-shapes.html

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enter image description here

Image Source: Spherical Harmonic Functions

The shapes are related to the spherical harmonic functions, and as you can see from the picture above, given the correct $l $ and $m $ numbers, you should be able to match up the orbitals to the appropiate functions.

Source: Spherical Harmonic Functions And Electron Orbitals

Spherical harmonics are important in many theoretical and practical applications, e.g., the representation of multipole electrostatic and electromagnetic fields, computation of atomic orbital electron configurations,[my emphasis] representation of gravitational fields, geoids, and the magnetic fields of planetary bodies and stars, and characterization of the cosmic microwave background radiation.

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