Question

In the transition metals, why do they seem to alternate between 4s1 and 4s2? For example, why is Vanadium ([Ar]4s2, 3d3) proceeded by ([Ar]4s1, 3d5)?

Answer 1

It's not. The element before vanadium is titanium, which has electron configuration [Ar]4s2 3d2.

What you probably mean to ask is why the element succeeding vanadium, chromium, has configuration [Ar]4s1 3d5. Not only has the next electron been added to the 3d subshell, which is expected, but one of the 4s electrons has also moved down to the 3d subshell.

There are two ways to think about this. The most common is just the say that half-filled and filled subshells are unusually stable, so that if it's possible to shift one electron and get a half-filled or filled subshell, it will often happen. In the case of chromium, the a priori configuration of [Ar] 4s2 3d4 can be made more stable by shifting one electron from the 4s to the 3d subshell, making the 3d subshell exactly half-filled. You see a similar effect near the end of the first transition series, where the a priori configuration of copper, [Ar]4s2 3d9 can be made more stable by shifting a 4s electron down, making a completely filled d subshell ([Ar]4s1 3d10).

If you want an answer that's more physical, then the reason is that electrons in d subshells are poorer at shielding outer electrons from the influence of the nucleus, because in a d orbital an electron spends unusually low amounts of time close to the nucleus (where it can do the most shielding).

Hence as electrons are added to the d subshell, and protons are added to the nucleus, as you go across a Period, the effective nuclear charge felt by outer electrons increases steeply. At some point it gets high enough that the energy of the outermost (here 4s) electrons increases above the energy of the 3d subshell, and an electron is transferred down to the d subshell.

The real question then becomes: why doesn't this process continue with the next element? That is, why isn't the electron configuration of manganese [Ar]4s1 3d6 or even [Ar] 3d7 ?

Now we can invoke the issue of the half-filled subshell: up to 5 electrons, you can put each new electron into a separate d orbital, so that they are (through the magic of quantum exchange) automatically kept further apart from each other, reducing electron-electron repulsion. However, as soon as you want to add the 6th or higher electrons, you need to start doubling up electrons, putting more than one in each orbital (with the opposite spin of course). That brings electrons physically closer to each other, on average, and increases electron-electron repulsion. It's this increased energy due to added electron-electron repulsion that brings the energy of the 3d subshell up again after chromium, so that once again the added electrons go first into the 4s subshell.

Answer 2

That was one of the most concise articles about chemistry I've ever read. Thank you.

While we're at it, why is a full "outer shell" (in noble gases) stable (physically)?

Answer 3

Same reason. The effective nuclear force felt by the outer electrons is a maximum, because they're all part of the same shell and do not shield each other well. An additional electron would go outside this shell and be well-shield from the nucleus. Consequently it would be easy to remove. This explains why something like Na+ is easy to form.

Conversely, if we have an outer shell which is missing only one electron, like neutral F, then there is a very strong Zeff attracting any nearby electrons, because the remaining 7 electrons in the outer shell shield the nucleus poorly. (You can see this by the fact that F is a very small atom, meaning the outermost electrons are pulled in tightly towards the nucleus.) Consequently plenty of energy is released if the electron falls into the outer shell and forms an F- anion. So this anion is easy to form.

In short, it's not so much that the complete shell (Ne) is stable, so much as that the incomplete shells nearby (Na and F) are unusually unstable.