Atomic Sub-shell question

Question

Which orbital of sub-shells $s$, $p$, $d$ and $f$ have Magnetic Orbital Quantum Number $m_l=0$. Like in p sub-shell, which orbital from $p_x, p_y, p_z$ will have $m_l$ as Zero? Also, how to determine the $m_l$ for any orbital?

Answer 1

It's not as well-known as it should be that $p_x$ and $p_y$ are superpositions of the $m_l=\pm1$ orbitals. That means the $p_x$ and $p_y$ orbitals no longer have $m_l$ as a quantum number. This is true for most pictures you will see of other shells' $m\ne0$ orbitals. They do this so that the wave functions are purely real and easy to draw.

So when you see those three figure-eights in different orientations, those are $p_x$, $p_y$, and $p_z$. Sometimes, they don't even want to show a lobe is the negative of the other, so they show you the absolute value of the superposition.

This means that $m=0$ is always the one they label with a single $z$.

You'll know you are actually looking at sets of determinate orbitals and not superpositions if all orbitals have a continuous axial symmetry about the $z$-axis. But since they are complex, if you see these images, you might be looking at $|\Psi|^2$. Check out Griffiths or Alastair I. M. Rae. Griffiths plots $|\Psi|^2$ and tells you how to recognize them from their nodes. Rae instead plots $|\Psi|$.

Answer 2

The shapes of atomic orbitals are related to the spherical harmonics, and their probability densities have the form

$$ Y_\ell^m \propto e^{im\phi} P_\ell^m(\cos\theta) $$

Here we're describing position vectors with the two angles $\theta,\phi$. The angle $\theta$ is the angle between the position vector and the positive $z$-axis, while the angle $\phi$ is the "cylindrical" angle between the projection of a vector onto the $x$-$y$ plane and the positive $x$-axis. That is, the relationship between the spherical coordinates and the Cartesian coordinates is

\begin{align} z &= r\cos\theta \\ x &= r\sin\theta\cos\phi \\ y &= r\sin\theta\sin\phi \\ \end{align}

The associated Legendre polynomials $P_\ell^m(x)$ (where $x$ is just a basic algebraic stand-in, not a position coordinate) are polynomials in $x$ multiplied by powers of $(1-x^2)^{-1/2}$. Our expressions $P_\ell^m(\cos\theta)$, therefore, are polynomials in $\cos\theta$ multiplied by powers of $\sqrt{1-\cos^2\theta}=\sin\theta$.

Let's look specifically at a table for the $p$ orbitals, with $\ell=1$:

\begin{alignat}{2} Y_1^0 &\propto &\cos\theta &\propto \frac zr \\ Y_1^{+1} &\propto{}& e^{+i\phi} \sin\theta &\propto \frac{x+iy}{r} \\ Y_1^{-1} &\propto{}&e^{-i\phi} \sin\theta &\propto \frac{x-iy}{r} \\ \end{alignat}

You can read the second column of these straight from the spherical coordinate definitions, if you know the Euler identity $e^{i\phi}=\cos\phi+i\sin\phi$.

These are the forms of the spherical harmonics that physicists like, because they have well-defined $m$. But chemists like to use linear combinations of these where the complex numbers are removed:

\begin{alignat}{2} Y_p^z &= \quad Y_1^0 &&\propto \frac zr\\ Y_p^x &\propto Y_1^{+1} + Y_1^{-1} &&\propto \frac xr \\ Y_p^y &\propto Y_1^{+1} - Y_1^{-1} &&\propto \frac yr \\ \end{alignat}

These are the "$p$ orbitals" you usually see in chemistry textbooks, where you have blobs of electron probability density corresponding to each of the three axes. If you like arithmetic, you can work through the normalizations and see that they are completely symmetric with each other. There isn't any physical difference between a $p_z$ orbital and a $p_x$ orbital; you turn one into the other by turning your head and looking at it differently.

So, let's think about what you're seeing when you look at an image like this one:

source: Зефр, CC0, via Wikimedia Commons

The $m=0$ orbitals, arranged in the vertical column in the middle of the table, are the ones without any variation in the $x$-$y$ plane, with all of the variation along the $z$-axis. The real $|m|=1$ orbitals are associated with the $x$-$z$ and $y$-$z$ planes. Each increase in $m$ adds another subdivision to the $x$-$y$ plane, while removing a subdivision from the $z$-direction. Once you get to $|m|=\ell$, along the diagonals of the pyramid, then all of the variation is in the $x$-$y$ plane.

But the $m$ in this figure isn't the $m$ corresponding to the projection of the orbital angular momentum, because in this figure the densities are already separated into real cosine and real sine terms. Those wavefunctions are complex. The projections in this figure are their real and imaginary components, which combine with opposite phases. Sorting out which kind of visualization you are seeing is tricky, because different authors and illustrators find different conventions "natural" and "obvious," and use the same quantum numbers to mean slightly different things.

For $|m|=\ell$, you also remove all but a single maximum in the radial probability density, and you recover Bohr-like orbits.

Answer 3

All the subshells have a orbital with $m_l=0$ . $m_l$ signifies orientation of orbitals. Its value lies always between $\pm l$, i.e. $-l,...,0,...l$.

s only can have one orientation. (Only one $m_l$ value, which is 0)

p has three orientations: $(-1,0,1)$

d has 5 $(-2,-1,0,1,2)$

f has 7 $(-3,-2,-1,0,1,2,3)$