What is the basis for whether the outermost electron configuration of ferrous ions is Ar 3d54s1 or A

According to Pauling's approximate energy level diagram, there is a staggered energy level phenomenon between 3d and 4s, and e3d is greater than e4s, so the electronic configuration of the iron atom is two electrons in 4s and six electrons in 3d. When an iron atom loses electrons, it loses 3d electrons first, but according to Hunt's rule, the half-filled D5 and fully filled D10 structures are relatively stable. Therefore, the outermost electron configuration of ferrous ions is ar 3d54s1 instead of ar 3d6

According to the University Press, the third edition of Inorganic Chemistry p99-100 pages.

As can be seen from the Cotton scale atomic orbital energy level diagram, the 4s orbital of the fourth period transition metal element (z>=21), including iron (26), is actually higher than the energy of the 3d orbital. This is due to the presence of 3d electrons, which weakens the attraction of the nucleus to 4s electrons.

So for the element z>=21, 4s electrons are most likely to be lost.

The three states of half-full, full-full, and all-empty are the most stable, and the ferrous ions should be AR 3d44s2, but AR 3d54s1 is more stable.

There is a large gap between the energy of 2S and 2D, and the lowest energy must be met first, ranking in 2S

However, the difference between 3d and 4s is smaller, so since 3d5 and 4s2 are stable structures, of course, 3d close to the nucleus should be selected to de-arrange, so that the electrons are more attracted by the nucleus and will be more stable.

When you go to the back of the country, you will find that the energy level staggering will be more chaotic, for example, the outermost layer of palladium reaches 18 electrons.

Because according to Hunt's rule, the d-shell is stable in a half-full, full, or full-empty state, so the d-shell tends to align 0 or 5 or 10 electrons.

They say that the energy levels are different, so that some of them can be in an excited state and some of them can't.

It depends on whether it is a primary or secondary element.

1. Main family elements.

For the main group elements, the configuration of the outermost electrons and the configuration of the outer electrons mean the same thing.

The outer electrons, i.e., the peripheral electrons, are the characteristic electrons.

2. Paragroup elements.

For subgroup elements, the outermost electrons refer to the electrons of the outermost electron shell;

Outer electrons refer to the outer shell plus part of the electrons of the subouter shell. That is, it is commonly said that the characteristic electrons, peripheral electrons, and outer electrons.

For example, Fe atom:

The outer electron configuration is: 3d6 4s2

The outermost electron configuration is: 4s2

According to the principle of lowest energy, under normal conditions, the electrons in the energy level are completely empty, fully full, and the energy is the lowest and most stable in the half-full state. 3d54s1 is more stable than 3d6. Ferrous ions, on the other hand, lose 2 electrons from 3d64s2 to 3d6

When ions are formed, follow the Cotton level diagram (don't worry if you don't know), for the elements of the first 4 periods, the outer electrons are always lost first, and there is no electron rearrangement while the atoms follow the Pauling energy level diagram, Cu, Cr are special cases of Hunt's rule.

In other words, the results of the spectral experiment are like this, and you have to accept it.

The electronic configuration of the iron atom is two electrons in the first row of 4s and six electrons in the last row of 3d; When the iron atom loses electrons, the outermost electron configuration of the ferrous ion is ar 3d54s1 instead of ar 3d6According to the university.