The outer valence electrons of Mn are 3d5 4s2, but why not 3d7?

In short, if manganese is arranged in 3d7, the total energy of these 7 electrons will be greater than the total energy of 7 electrons at "3d5 4s2". Now the key question is why the last 2 electrons of manganese are in the 4s orbital, which is lower than the 3d orbital occupied by 5 electrons.

First of all, it is necessary to know that no matter which layer of electrons, its distribution is throughout the space where the atom is located (therefore, the name "electron cloud" is quite graphic), and the difference is only in the specific distribution. For example, the 4S electron, although its most common spherical shell is larger than that of the 3d electron, which is far from the nucleus, at the same time, the 4S electron penetrates deeper into the inner shell more often than the 3D electron (this is called "orbital penetration", and it can also be considered that the orbit of the 3d electron is closer to the circular orbit, while the orbit of the 4S electron is a flat ellipse, so that its far-nucleus point is farther away, and the near-nuclear point is closer), so the 4S electron is generally attracted more by the nucleus and less repelled outward by other electrons in the inner shell, So its energy tends to be lower. Another factor is the "polarization of the atom solid", which is also due to the fact that the 4S electron is relatively closer to the nucleus, which makes the atom real (the nucleus and the inner electrons make up the atomic solid) more polarized, so the attraction of the two is also more, which also reduces the energy.

However, after all, the fact that "the spherical shell where 4s electrons most often occur is larger than the spherical shell where 3d electrons most often appear" makes the 4S electron energy higher. In summary, the first two points make the energy of 4s electrons lower than that of 3d electrons, but the latter point is the opposite, which is a competitive relationship. It is necessary to be specific to the atoms in order to derive from the very complex calculations which factor prevails.

For example, for 1 valence electron of sodium, the 4s orbital has lower energy, so it "crosses" the 3d orbital and occupies the 4s orbital. For the 7 valence electrons of manganese, the situation is a little more complicated. For the first 5 electrons, the 3d orbital energy is lower.

Note that the 3d orbital has 5 branches, and the 5 electrons occupy these 5 branches, and the distance between them is as much as possible by the 5 "evenly divided" branches, so that the potential energy corresponding to the repulsion between these 5 electrons is smaller. If the 6th and 7th electrons are then filled into the 3d orbital, they must be squeezed into one of the 5 branches of the 3d orbital, so that the two electrons are squeezed into these two branches, and they are closer to each other than the spacing of the aforementioned 5 electrons, and the corresponding electric potential energy is greatly increased. In this way, if the 6th and 7th electrons fill the 4s orbital, they will be far away from the 5 electrons and have lower energy.

The answer upstairs doesn't seem to feel very good, and the 3d5 is a semi-full structure, which is relatively stable.

The electrons are generally in the outermost shell, and the electron pairs are shifted to show electrical properties due to the formation of shared electrons, rather than being charged by the gain and loss of electrons.

Theoretically, the electronegativity of manganese is very low, that is, the ability to get electrons is very weak, and when it is difficult to get one electron, how can you expect it to get five electrons.