Why is element 29 not 9 electrons in the 3D layer and 2 electrons in the 2s layer

The experiment shows that the copper is 3d10 4s1, which is semi-filled. In addition, the 2S layer pit must be full, otherwise it will not be able to continue to row upward.

3d104s1 meets half full and will be more stable than 3d94s2.

Element 29 is copper!

Atomic electron configuration 3d94s2 ion electron configuration 3d9

Is there a problem with this?

Hunter's rule, full full, half full, and full empty is relatively stable.

3d10 satisfies full and 4s1 satisfies half full. Whereas, 3d9 is neither fully or half-full or empty.

In addition, the 2S sublayer must be lined with 2 electrons, believe you are writing it wrong.

Because 3D 5 conforms to the semi-full principle ... It's been 2 years since I graduated from college. But I still love chemistry.

In fact, the energy of the s, p, d, and f of the electron shell increases sequentially, resulting in the energy of the 3d layer sometimes being greater than that of the 4s layer. Resulting in staggered energy levels.

Just as water flows downward, electrons preferentially enter orbits with low energy levels.

Arranged according to the principle of lowest energy.

It is arranged according to the principle of the most stable.

Copper is most stable bivalent in aqueous solutions and monovalent in solids.

The valence of the first subfamily is very strange, see University Inorganic Chemistry.

The outer nuclear electron shell of copper is copper is a common metal, it should be +1 I don't know very well, it's in front of zinc and zinc is +3. It may be affected by this.

The number of electrons in the n-lamina is 1 or 2

The n-layer is the fourth layer, and 5 electrons on 3d may have 2 electrons on 4s or 1 electron (semi-full) on 4s

I think Cu is [Ar] 3D10 4S1 where all 10 electrons in the 3d orbital are full, and 4S is also semi-full with one electron.

If the 4S track is slowed down first, then [AR] 3D9 and 4S2 are the full charge of the 4S track.

Because the energy difference between 3d and 4s energy levels is small, when there is a 3d orbital with 10 electrons fully filled, and 4s an electron is also semi-filled, it is lower energy and more stable than 3d4 4s2 is the full charge of the 4s orbital!

According to the principle of lowest energy! So use the first arrangement!

Adopt it! Hee-hee.

According to Hunt's rule of extranuclear electron filling arrangement, when the sublayer orbit is fully filled, empty or semi-filled, it is relatively more stable, and the actual 3d10 of copper is more conducive to the reduction of system energy than the theoretical 3d9.

Follow Pauli's principle of incompatibility and Hunt's rule.

Because the outer orbital of the atom is relatively stable in the state of full full, all empty or half full, the outermost half-full is a little more stable.

Because that's what is stated on the energy distribution chart.