Why does the 4s subshell fill before the 3d subshell?

When filling subshells in an electron configuration, why do you get to level 4s, then jump back to level 3d and so forth?

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Quantum theory shows that each atom's electronic structure is a unique compromise between several different effects. Electronic configurations of the fourth period elements can be appreciated by considering these effects: 

Raising n raises orbital energy. Electrons are attracted to the nucleus. To pull an electron farther away from the nucleus, you have to work against that attraction. That means an electron farther from the nucleus has more energy than electron closer in; energy is required to move the electron out, and energy can be released when the electron moves in. So we expect outer shells to have higher energies than inner shells, because increasing n increases the average distance between the nucleus and the electron.For atoms heavier than copper this effect dominates, and 4s electrons have higher energy than 3d electrons.

Raising l raises orbital energy. Higher l values result in orbitals with more nuclear nodes (a node being a place where the probability of finding the electron is zero). We say high-l orbitals are "less penetrating" because their electrons have a lower probability of being found at or near the nucleus. That gives high-l orbitals (like d orbitals) more energy than low-l orbitals (like s orbitals) within the same shell.This effect causes 4s orbitals to have lower energy than 3d orbitals for elements lighter than copper. (Although for hydrogen, the unoccupied 4s and 3d orbitals have nearly identical energies).

Within a subshell, more unpaired spins means a lower overall energy. Quantum mechanics predicts that the motions of electrons with paired spins are "correlated". Paired electrons move together, while electrons with unpaired spins can stay farther away from each other on average. Since electrons repel each other, paired electrons have more energy than unpaired electrons, all other things being equal. This spin correlation effect explains why Cr has a [Ar] 4s13d5 configuration rather than a [Ar]4s23d4 configuration- the former has more unpaired electron interactions than the latter.

Author: Fred Senese [email protected]

Hi Jestin,

Aufbau principle is right that electrons of an atom are arranged in its orbitals starting from the lower to the higher energy level of the orbitals so that Mn atom has [Ar]4s23d5 electron configuration. However, potential energy of electron is tightly correlated to the atom shell rather than subshell (orbital), and the potential energy of fourth shell of atom is higher than the third shell since the distance from its atomic nucleus is farther than the third shell. As the consequence, electron(s) in 4s will be prefer released first than that in 3d so that we get [Ar]3d5 electron configuration for Mn2+ ion, not [Ar]4s23d3 unstable electron configuration due to the magnitic field of electron spin close to the vacant orbitals. In addition, it is actually we can write [Ar]4s23d5 or [Ar] 3d54s2 for Mn electron configuration, the both are right.

3d electron will go 1st.

4s electrons will go first... 

According to the Aufbau principle, within the 3d period a 3d electron is added every time a proton is added to the nucleus. After the 3d electron has been added its energy turns out to be lower than that of the 4s electrons that are already present. Therefore the latter are removed more easily than the 3d electrons that have just been added.

4s will go first

The often tricky thing about is: the Aufbau principle describes correctly the ground state of a metal atom in the d series. Following the principle in the 4th period first 4s, then 3d shells were filled.

However, this is only the case for isolated metal atoms and frequently also for the bulk metals. In compounds of the d block metals, this is different. As a general rule you can say, that any valence electron ion such a compound has d character. Thus, as an example, all Fe(II) complexes have a d6 configuration and NOT a d4s2 (or s2d4) and Pt(II) complexes are d8 configured! Even for Ti(III) its clearly a d1 configuration (you can see this experimentally e.g. from the Jahn-Teller distortion.

There are almost no exceptions from this rule. Only for e.g. Y(II) or La(II) there are ambiguities about the character of the single valence electron. So, the rule is usually:

The valence electrons of a d block element in a chemical compound are d electrons!

Which, from the linguistic viewpoint, is a quite simple rule.

The reason for this is much harder to explain. My simple explanation is, that the d electrons are very prone for interactions with the atomic orbitals of bonding partners (in the case of complexes, interaction with the ligand atoms). The reason for this might be their diffuse character. Note that the label "d" means "diffuse". The interaction with the surrounding atoms stabilises the d shells relatively to the s or p shells. Thus, all remaining valence electrons find themselves in d shells.

Thanks Axel, So do you mean that d orbitals can better handle electronic correlation than s and p orbitals !!

Generally yes, it seems they can handle correlations better. The reason might lie in the symmetry of the d orbitals. I have the impression that d orbitals can cope with much more different geometries compared with s and p orbitals. Complexes show a big variety of coordination numbers and coordination geometries. This is what we can see, at least.

4s are removed first, they are further away from the nucleus than the 3d.

I hate this expression which is used by chemists "we fill electrons" we donot fill electrons, electrons are present in atoms, we can remove e's to make cations . Experiments show that the 4s e's are removed first , Aquated Fe(III) ions have 5 unpaired e's.

The electrons will be removed first from 4s-orbital and not from 3d orbital. Because 3d orbital is less in energy than the 4s orbital. Electron gets removed first from higher energy level. Kindly refer to Peter Atkins Chapter 1, this will give you better insight.

Electron will remove from 3d orbital first then the nuclear charge become more and this will take 3d orbital in lower energy than 4s. At this time electron from 4s would come to 3d level.

Dear Prof. Klein, Can you please comment on Dr. Seyyed Javad's answer!!!

Dear jestin.

Seyyad javed is absolutely right electrons remove from 3d first it will raise nuclear charge which pulls all the subshells down n reduce diameter of each shells n subshells as well,both 4s and 3d orbitals have very small energy differences , increase of nuclear charge caused by removal of electron from 3d ,pull down  d subshell more than s subshell which cause 3d less energetic then 4s now electron fall down to 3d orbital obeying afbua rule leaving 4s orbital vacant as prof Klein  says so both the professors sayyed javad and Klein are right ,actually phenomenon become so complicated at quantum levels. 

Hope it will help you to understand. 

Regards 

Muhammad Imran Khattak.