Why do iron(II) ions and chromium have different electronic configurations?

Question

What I know:

1. Empty 3d orbitals are higher in energy than empty 4s orbitals
2. Aufbau Principle (electrons always go into an empty orbital with the lowest energy)
3. Partially/half/fully filled 3d orbital are lower in energy than 4s orbitals
4. $\ce{Cr}$ is an exception: $\ce{[Ar] 3d^5 4s^1}$. It breaks the "rule" because the extra stability from half-filled 3d orbitals is favoured.
5. From (1) and (2) => After the $\ce{[Ar]}$ core, electrons will fill 4s before occupying 3d
6. When forming ions, electrons of the highest energy are removed first.
7. Iron(II) ions exist as $\ce{[Ar] 3d^6}$ in reality.

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What I understand:

I understand why Fe is $\ce{[Ar] 3d^6 4s^2}$. [from (5)]
I understand why Cr is $\ce{[Ar] 3d^5 4s^1}$.

From (4), (5) and (3), I believe for $\ce{Cr}$ the 4s is filled before 3d too, i.e. it is $\ce{[Ar] 3d^3 4s^2}$ before the last electron is added. After the last electron is added to form $\ce{[Ar] 3d^4 4s^2}$, a 4s electron immediately moves to the 3d orbital which is of lower energy.

From (6) and (3), for transition metals, 4s electrons are removed first. i.e., I understand why iron(II) ions may be $\ce{[Ar] 3d^6}$

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What I don't understand:

I know $\ce{[Ar] 3d^6}$ is correct for iron(II), but I don't understand why?

Why doesn't iron ions adjust itself as Cr does to achieve the extra stability? Is it because, in this case, it needs to excite an electron to an orbital of higher energy, whereas in Cr, the 4s electron just moves down to an orbital of lower energy?

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P.S.

• I've read What is the electronic configuration of Fe(II) ion? and I am not satisfied with the answers in that thread.

• I've read Do ions also form special electron configurations? (just as Copper and Chromium do), and it only tells me "it is possible for ions to have special electronic configurations too!"

Answer 1

For starters, it isn’t nearly as simple as it is usually put in chemistry classes to explain chromium’s ground state electron configuration.

$\ce{Cr(0)}$ and $\ce{Fe(II)}$ have the same number of valence electrons. You might have heard of the concept of isolobality, which, briefly explained, suggests that compounds with similar electronic configurations (isolobal configurations) will react similarly; so you could replace $\ce{F}$ with $\ce{OMe}$ or even $\ce{[Co(CO)4]}$. The concept has its values and its strong simplifications which I will not go into here.

However, even assuming the isolobality concept’s validity, it would not count $\ce{Cr(0)}$ and $\ce{Fe(II)}$ as isolobal to each other. The reasoning is rather simple: One is a neutral metal, the other is a cation with $+2$ charge. For every electron an atom looses, the atom’s orbitals will contract and the atomic (or ionic) radius will decrease. This also changes the energy levels between orbitals and will affect lower d-orbitals differently than higher s-orbitals.

The result is that the $\ce{4s}$ and $\ce{3d}$ orbitals are now so dissimilar in energy, that electrons hopping back or forth is now no longer an option.

Answer 2

Another way to look at the transition in electronic configuration is to consider species with the same number of electrons but different numbers of protons.

Start with $\ce{Cr^0}$ with its electron configuration $\ce{[Ar] 3d^5 4s^1}$. We add a proton to make $\ce{Mn^+}$, throwing in any necessary neutrons to keep the nucleus stable. We repeat this process to make $\ce{Fe^{2+}},$ $\ce{Co^{3+}}$, etc., and eventually we get to let us say, $\ce{Xe^{30+}}$. In all these species the electromagnetic interactions between the 24 electrons are intrinsically the same, but the excess positive nuclear charge is building up and is bound to overwhelm the electron-electron interactions at some point. By the time we get to the huge excess positive charge in the xenon ion above*, the orbitals become roughly what they would be with isolated electrons, like the familiar hydrogen spectrum multiplied by the effect of a greater positive charge. The 3d orbital energy level sinks down to meet its shell-mates at 3s and 3p, leaving the higher-shell 4s orbital at a higher level. (* -- actually XPS binding energy data reveal that $\ce{Br^{11+}}$ already shows this effect.)

So at some point the outer electrons all fall into the 3d orbitals and the electronic structure shifts from whatever it was in the neutral chromium atom to $\ce{[Ar] 3d^6}$. In my example I predicted that we might have to go all the way to $\ce{Xe^{30+}}$ (or maybe $\ce{Br^{11+}}$ given the above-cited XPS binding energy information) for this transition, but in reality the 3d and 4s orbitals are so closely balanced in the first row of transition metals that we need not go that far. Adding two protons, and a few neutrons for nuclear stability, to chromium is enough.

Answer 3

I've had the same question and this is how my chemistry teacher answered: 3d is a sub energy level of the 3rd energy level and 4s is a sub energy level of the 4th energy level. When electrons fill into the Fe atom they fill according to the Auffbau principle, in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d. But when 2 electrons are to be removed to form the Fe2+ ion, the electrons are not removed from the 3d orbitals. 3d belongs to the 3rd energy level which becomes an inner filled energy level when electrons are present in the 4th energy level. So electrons are removed from the 4s sub energy level, without disturbing the fully filled 3rd energy level.