6
$\begingroup$

The form of the p orbitals that we are familiar with are the $\mathrm{p}_x$, $\mathrm{p}_y$, and $\mathrm{p}_z$ orbitals:

enter image description here

(source: ChemTube 3D)

I also know that the p subshells have the quantum number $l = 1$, meaning that the magnetic quantum number can take the three values $m_l = -1, 0, +1$. Which orbital has which value of $m_l$?

$\endgroup$
24
$\begingroup$

The answer is... it is not so simple. Some quantum mechanics follow, but the TL;DR version is that while $m_l=0$ corresponds to $p_z$, the orbitals for $m_l=+1$ and $m_l=-1$ lie in the $xy$-plane, but not on the axes. The reason for this outcome is that the wavefunctions are usually formulated in spherical coordinates to make the maths easier, but graphs in the Cartesian coordinates make more intuitive sense for humans. The $p_x$ and $p_y$ orbitals are constructed via a linear combination approach from radial and angular wavefunctions and converted into $xyz$. Thus, it is not possible to directly correlate the values of $m_l=\pm1$ with specific orbitals. The notion that we can do so is sometimes presented in introductory courses to make a complex mathematical model just a little bit simpler and more intuitive.

From Physical Chemistry by Atkins and DePaula, the three wavefunctions for $n=2$ and $l=1$ are as follows.

$$\begin{align} &\Psi_{2,1,0}&&=r\cos{\theta}f(r)\\ &\Psi_{2,1,+1}&&=-\dfrac{r}{\sqrt{2}}\sin{\theta}\mathrm{e}^{\mathrm{i}\phi}f(r)\\ &\Psi_{2,1,-1}&&=\dfrac{r}{\sqrt{2}}\sin{\theta}\mathrm{e}^{-\mathrm{i}\phi}f(r)\\ &\end{align}$$

The notation is $\Psi_{n,l,m_l}$, $r$ is the radius, $\theta$ is the angle with respect to the $z$-axis and $\phi$ is the angle with respect to the $xz$-plane. $$f(r)=\sqrt{\dfrac{Z^5}{32\pi a_0^5}}\mathrm{e}^{-Zr/2a_0}$$

in which $Z$ is the atomic number (or probably better nuclear charge) and $a_0$ is the Bohr radius.

In switching from spherical to Cartesian coordinates, we make the substitution $z=r\cos{\theta}$, so: $$\Psi_{2,1,0}=zf(r)$$

This is $\Psi_{2p_z}$ since the value of $\Psi$ is dependent on $z$: when $z=0;\ \Psi=0$, which is expected since $z=0$ describes the $xy$-plane.

The other two wavefunctions are unhelpfully degenerate in the $xy$-plane. An equivalent statement is that these two orbitals do not lie on the $x$- and $y$-axes, but rather bisect them. Thus it is typical to take linear combinations of them to make the equation look prettier. Linear combinations are allowed by the maths of quantum mechanics. If any set of wavefunctions is a solution to the Schrödinger equation, then any set of linear combinations of these wavefunctions must also be a solution. We can do this because orbitals and the wavefunctions that describe them are not real physical objects. They constitute a mathematical model.

In the equations below, we're going to make use of some trigonometry, notably Euler's formula:

$$\mathrm{e}^{\mathrm{i}\phi}=\cos{\phi}+\mathrm{i}\sin{\phi}$$ $$\sin{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi}}{2\mathrm{i}}$$ $$\cos{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi}}{2}$$

We're also going to use $x=\sin{\theta}\cos{\phi}$ and $y=\sin{\theta}\sin{\phi}$.

$$\begin{align} \Psi_{2p_x}=\frac{1}{\sqrt{2}}\left(\Psi_{2,1,+1}-\Psi_{2,1,-1}\right)=\frac{1}{2}\left(\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\cos{\phi}f(r)=xf(r) \\ \Psi_{2p_y}=\frac{\mathrm{i}}{\sqrt{2}}\left(\Psi_{2,1,+1}+\Psi_{2,1,-1}\right)=\frac{1}{2\mathrm{i}}\left(\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\sin{\phi}f(r)=yf(r)\\ \end{align}$$

So, while $m_l=0$ corresponds to $\Psi_{p_z}$, $m_l=+1$ and $m_l=-1$ cannot be directly assigned to $\Psi_{p_x}$ and $\Psi_{p_y}$. Rather $m_l=\pm1$ corresponds to $\{\Psi_{p_x},\Psi_{p_y} \}$. Put another way, I suppose we could say that $m_l=+1$ might correspond to $\Psi_{p_{x+y}}$ and $m_l=-1$ might correspond to $\Psi_{p_{x-y}}$.

$\endgroup$
  • $\begingroup$ Why did we assign the value of $m=0$ to the $p_z$ orbital, is it conventional or it was solved using equations (i.e. In the equation, was it like this that putting $m=0$, we get $\Psi=z \cdot f(r) $?) $\endgroup$ – Jaideep Khare May 31 '17 at 6:56
  • $\begingroup$ The angular wavefujctions are functions of two angular variables relative to the axes. They are defined relative to the z axis. However, the assignment of the three axes is arbitrary and all possible assignments are degenerate. $\endgroup$ – Ben Norris Jun 4 '17 at 22:57
  • $\begingroup$ Is there any definition of z-axis quantum mechanics (I hope it's called so), since axes are arbitrary, and how we would define that in a sample of, say Hydrogen atoms, where zillions of molecules are randomly oriented. $\endgroup$ – Jaideep Khare Jun 4 '17 at 23:04
  • $\begingroup$ @BenNorris Do you mind elaborating on the case for $d$ orbitals and $f$ orbitals as well? Thanks! $\endgroup$ – Gaurang Tandon Mar 11 '18 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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