[Pourya]

Why electron configuration of Fe2+ is [Ar]3d6 instead of [Ar]3d5 4s1 (like Cr)?
Is there any explanation for this?!

[chenbeier]

It is more stable to have  a empty s-orbital as a half filled one and the d-orbital will be only touched after the s electrons are used.

[Enthalpy]

It is more stable to have  a empty s-orbital as a half filled one and the d-orbital will be only touched after the s electrons are used.

As Cr and Fe²⁺ differ only by their nucleus, this explanation can't suffice.

[...] electron configuration of Fe2+ is [Ar]3d6 [...]

Do you have a credible source for the claim? A quick Web search brought me only garbage, with this configuration resulting from wrong reasoning. Like: "remove the outermost two electrons", which is a mistake at transition elements.

[coolman50544]

effective nuclear charge increases as the atom is ionized. the effective nuclear charge lowers the potential energy of all electrons but some more than others. since 3d electrons are closer to the nucleus than 4s, 3d electrons have a relatively lower potential energy than if they were in the 4s orbital. thus, they remain in the 3d orbital.

[hypervalent_iodine]

It is more stable to have  a empty s-orbital as a half filled one and the d-orbital will be only touched after the s electrons are used.

As Cr and Fe²⁺ differ only by their nucleus, this explanation can't suffice.

They also differ in the fact that one is a neutral metal and the other is an ion. This can have huge implications on the relative stability of the orbitals due to orbital contraction and changes in ionic radius. Unlike most other first row transition metals, chromium will only partially fill the 4s as there is an energy pay off for having all 5 d orbitals partially filled. The changes I mentioned that come about from ionisation of Fe to Fe²⁺ have more of an effect on the d orbitals than the 4s, which increases the difference in energy between the two and means its harder for an electron to jump from the 3d to the 4s. IOW, there simply isn't the same energetic pay-off.

[...] electron configuration of Fe2+ is [Ar]3d6 [...]

Do you have a credible source for the claim? A quick Web search brought me only garbage, with this configuration resulting from wrong reasoning. Like: "remove the outermost two electrons", which is a mistake at transition elements.

Their claim is based on basic principles of electronic configuration / ionisation. When removing electrons, you remove from the highest energy orbital first. For first row transition metals those are the electrons in the 4s orbital, thus neutral Fe goes from [Ar]3d⁶4s² to [Ar]3d⁶ when it becomes Fe²⁺.

[Enthalpy]

[...] When removing electrons, you remove from the highest energy orbital first [...]

And the remaining electrons never rearrange, especially at transition elements?

[hypervalent_iodine]

[...] When removing electrons, you remove from the highest energy orbital first [...]

And the remaining electrons never rearrange, especially at transition elements?

Depends on the context, I suppose. Formally, this is how it is described and taught, but it isn't a hard and fast rule and really it is better to consider the relative energies of electronic configurations rather than atomic orbital energies in order to predict what will happen. Take vanadium, for example, which is [Ar]3d³4s². When you remove one electron, it doesn't become [Ar]3d³4s¹, but rather [Ar]3d⁴. The configurations of Co⁺ and Ni⁺ are similar in this regard. This document explains some of it: http://www.chemistry.uoguelph.ca/educmat/chm2060_preuss/L9-2013.pdf

Obviously the exact mechanism is rooted in much more complicated quantum phenomena, and I am not in a position to try and explain it. I never paid more attention than was absolutely necessary to inorganic and physical chemistry in undergrad (which I regret), so I am a bit out of my depth.

[Enthalpy]

What we could agree on is that interactions among the electrons (repulsion and antisymmetry) (appearing in "effective charge") make the difference.

With one single electron in a hydrogen-like ion, and neglecting corrections from relativity, finite nucleus size and so on, P protons make all orbitals P times smaller. The potential energy scales like P² because the nucleus charge scales like P, and the kinetic energy scales like P² because it depends on k². All 3d 4s... orbitals, just scaled down, remain solutions of Schrö's equation, the sequence of energies is kept, and the world is simple.

The repulsion energy among electrons does not scale as P² because they keep their charge q. If the orbitals kept their shape and scaled down by P, the electron repulsion energy would scale only by P. Changing P can't keep the solution for the same number of electrons, even if scaling the dimensions.

So the electronic configurations of Cr and Fe²⁺ may differ.

[Enthalpy]

Effective nuclear charge increases as the atom is ionized. the effective nuclear charge lowers the potential energy of all electrons but some more than others. since 3d electrons are closer to the nucleus than 4s, 3d electrons have a relatively lower potential energy than if they were in the 4s orbital. thus, they remain in the 3d orbital.

Why? From K to Zn, 4s fills before 3d, with Cr and Cu as the only exceptions. Click for full size on the appended picture cropped from Nist.

Can someone find a reliable source about the electronic configuration of Fe²⁺?

[hypervalent_iodine]

Effective nuclear charge increases as the atom is ionized. the effective nuclear charge lowers the potential energy of all electrons but some more than others. since 3d electrons are closer to the nucleus than 4s, 3d electrons have a relatively lower potential energy than if they were in the 4s orbital. thus, they remain in the 3d orbital.

Why? From K to Zn, 4s fills before 3d, with Cr and Cu as the only exceptions. Click for full size on the appended picture cropped from Nist.

Can someone find a reliable source about the electronic configuration of Fe²⁺?

I think it’s important to note here for the OP that there is nothing wrong with the electronic configuration for Fe²⁺ that they have given. It is correct and it is in complete agreement with what they would have been taught.

To answer your question, it does seem like a contradiction when you consider it in plain terms: you fill the 4s first, so the 4s must be lower in energy than the 3d, yet you remove from the 4s first, thus it must be higher. These are heuristics. They don’t account necessarily for mechanism, just the end result, and even then there are notable exceptions. In any case, it has to do with shielding and penetration, which I think that document I linked to goes over to some extent. If you have access to a copy of Shriver and Atkins (heck, even Blackman’s Chemistry text), I’m sure it explains it as well.

[coolman50544]

Effective nuclear charge increases as the atom is ionized. the effective nuclear charge lowers the potential energy of all electrons but some more than others. since 3d electrons are closer to the nucleus than 4s, 3d electrons have a relatively lower potential energy than if they were in the 4s orbital. thus, they remain in the 3d orbital.

Why? From K to Zn, 4s fills before 3d, with Cr and Cu as the only exceptions. Click for full size on the appended picture cropped from Nist.

Can someone find a reliable source about the electronic configuration of Fe²⁺?

Because in iron cation, the effective nuclear charge increase from the loss of two electrons is more drastic for 3d orbitals than the 4s orbital causing 3d electrons to have lower potential energy relative to 4s. So 4s electrons are ionized.

[Enthalpy]

I think it’s important to note here for the OP that there is nothing wrong with the electronic configuration for Fe²⁺ that they have given. It is correct and it is in complete agreement with what they would have been taught.

I would like to see a reliable document about that. No book here, I'm alone with my PC. An online course maybe. My first thought would be like Pourya's one: same electron configuration as Cr. And if more protons favour the deeper levels, then Fe²⁺ should have more electrons on 4s and fewer on 3d than Cr has, yes.

[Enthalpy]

[...] the effective nuclear charge increase from the loss of two electrons is more drastic for 3d orbitals than the 4s orbital [...]

Why? 4s orbitals are lower than 3d. They should be more sensitive to additional protons.

[Enthalpy]

Different reasoning.

Maybe an Fe²⁺ alone in vacuum does adopt the Cr electron configuration, or even one that puts more electrons on 4s and fewer on 3d. But this does not apply to a solid.

In FeO, Fe₂O₃, MnO and more, each metal atom has 6 oxygen neighbours in directions orthogonal or nearly
https://en.wikipedia.org/wiki/Iron(II)_oxide
https://en.wikipedia.org/wiki/Iron(III)_oxide
https://en.wikipedia.org/wiki/Manganese(II)_oxide
https://en.wikipedia.org/wiki/Cobalt(II)_oxide
https://en.wikipedia.org/wiki/Nickel(II)_oxide
https://en.wikipedia.org/wiki/Corundum
and a spherical orbital can do that easily, a d orbital not so much. The best d candidate shape, upper right on the drawing there
https://winter.group.shef.ac.uk/orbitron/AOs/3d/index.html
extends little in the xy plane. Not all stationary wave functions are represented, but a sum of +z and -z from the upper right z² would have only two lobes in the xy plane, and the others are zero along one axis.

Consistently, the bivalent oxides I've checked are cubic.

[hypervalent_iodine]

I think it’s important to note here for the OP that there is nothing wrong with the electronic configuration for Fe²⁺ that they have given. It is correct and it is in complete agreement with what they would have been taught.

I would like to see a reliable document about that. No book here, I'm alone with my PC. An online course maybe. My first thought would be like Pourya's one: same electron configuration as Cr. And if more protons favour the deeper levels, then Fe²⁺ should have more electrons on 4s and fewer on 3d than Cr has, yes.

Without meaning to sound facetious, it is a trivial thing to Google. It is content taught in high school and university level chemistry - I myself have taught it at both levels. Your issue here is that you are conflating ions and neutral elements. I explained why they are different in this context a few posts back.