How is the Pauli Exclusion Principle important in regards to the “octet rule”?
Answer:
The Pauli Exclusion Principle is in fact the main reason why we have the idea of an “octet rule” and why some transition metals like Chromium have access to 12 bonding electrons, and sometimes up to even 18.
The Pauli Exclusion Principle essentially states:
No two electrons may have entirely identical quantum states; at least one quantum number must be different. I’ve given a formal explanation of the octet rule here. Please read that before proceeding, as I will be furthering that discussion. Following that, we then realize that the octet rule is centered around the Pauli Exclusion Principle.
EXCEPTIONS TO THE OCTET RULE
We can then determine how Chromium, for example, can use 12 valence electrons, sometimes, for the same reason, and not just 8. Here’s what I mean.
Chromium’s electron configuration is:
(the original diagram is different, but it was wrong, because it had electrons in the \(4p\), which would mean \(30\) electrons, not \(24\). It also has the \(3d\) higher in energy than the \(4s\), but it’s actually not, for Chromium, according to Eric Scerri, replying to David Talaga.)
where the blue atomic orbitals are the valence orbitals.
QUINTUPLE BONDS?!
The energy levels are so close together, however, that Chromium actually sometimes has access to 12 valence electrons, rather than just 6. That’s why Chromium can sometimes make SIX bonds. Just take a look at this!
One single bond and one quintuple bond, and one interaction (dashed bond)! Okay, so how in the world?!
QUANTUM NUMBER CONSIDERATIONS
We can realize that Chromium sometimes has access to its \(3p\) orbitals as well, as that would give it \(12\) electrons, and the \(3p\) is closest in energy to the \(3d\) orbital (when moving downwards in energy).
So, we can consider the following quantum numbers:
n = 3:
\(\begin{aligned}&l=1,2 \\
&m_{l}=-2,-1,0,+1,+2 \\
&m_{s}=\pm 1 / 2
\end{aligned}\)
(covering the \(3p\) and \(3d\) orbitals)
n = 4:
\(\begin{aligned}&l=0 \\
&m_{l}=0 \\
&m_{s}=\pm 1 / 2
\end{aligned}\)
(covering the \(4s\) orbital)
In the same type of atomic orbital (examining only the \(3d\) for instance, or examining only the \(4s\) etc), all quantum numbers are the same, except for \(m_{l}\) and \(m_{s}\), which CAN be different.
UNIQUE QUANTUM STATES
As a result, for the \(3d\) orbital, we have five unique quantum states for the five available spin-up electrons \(\left(m_{l}=-2,-1,0,+1,+2 \text { with } m_{s}=+1 / 2\right) . m_{l}\) just ultimately tells us that there are five different \(3d\) orbitals \(\left(3 d_{z^{2}}, 3 d_{x^{2}-y^{2}}, 3 d_{x y}, 3 d_{x z} \text {, and } 3 d_{y z}\right) \text {. }\)
This, however, doesn’t include the spin-down electrons due to Hund’s rule of favoring the maximum spin state, which, for Chromium’s \(3d\) orbitals, is +5/2, and due to how there are exactly 5 electrons here.
Next, for the \(3p\) orbitals, we have three pairs of electrons. We have \(m_{l}=-1,0,+1 \text { with } m_{s}=-1 / 2 \text {, }\) as well as \(m_{l}=-1,0,+1\) with \(m_{s}=+1 / 2\) That makes for a total of six unique quantum states, and thus six possible electrons that can exist in the \(3p\) orbitals.
Finally, the \(4s\) orbital has only one valence electron, which obviously can only exist in one possible way at a time. The quantum numbers corresponding to it are l = 0, \(m_{l}=0\) with \(m_{s}=+1 / 2\) Whether you believe that a single electron can flip its spin or not, either way, that means one unique quantum state.
TAKE-HOME MESSAGE
Hence, Chromium could sometimes have 5 + 6 + 1 = 12 unique quantum states for each of the 12 electrons available for bonding, following the Pauli Exclusion Principle.
Each electron can only occupy one state at a time (like how one twin can only be that twin for all time), so with 12 electrons, 12 states are implicitly possible. That’s how I would rationalize why in the world Chromium can make SIX bonds sometimes. 🙂