Loss from 4s over 3d?

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Loss from 4s over 3d?

Postby MKearney_4G » Sat Nov 02, 2019 10:48 pm

When an electron is added to an n=4 d block atom, it is added to the 4s subshell because it is most stable. Then the 3d block begins to be filled. When removing an e- from an atom with 3d electrons, though, you remove them from the 4s subshell first. Why does it remove these first if they are supposedly the most stable of the 3 subshells?

Alice Chang 2H
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Re: Loss from 4s over 3d?

Postby Alice Chang 2H » Sat Nov 02, 2019 11:58 pm

The way that I understand it is that in electron configuration, you arrange by increasing energy. So, in this case, if only the s-shell is filled, then the s-shell is named first. But otherwise, if there is an electron in the d-shell, then the d-shell is named first before s-shell, because the d-shell will have less energy than s.

To answer your question, the electrons are removed from the s-shell/orbital first due to it being furthest away from the nucleus. The d-orbital is now closer to the nucleus than the s-orbital, so it is easier to remove an electron from s- first rather than d-shell. Therefore, electrons are removed from 4s before 3d.

Hope this makes sense?

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Re: Loss from 4s over 3d?

Postby 805422680 » Sun Nov 03, 2019 12:04 am

After the 3D orbital is filled, it falls to a lower energy than the 4s orbital. Therefore, since electrons are first lost from the highest energy shell or valence shell first, the 4s orbital loses electrons before the 3D orbital.

Tyler Angtuaco 1G
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Re: Loss from 4s over 3d?

Postby Tyler Angtuaco 1G » Sun Nov 03, 2019 12:05 am

It is necessary to think of energy levels when considering the addition and subtraction of electrons. When there are no electrons occupying the 3d orbital, the 4s orbital is lower in energy. However, once the 4s orbital is filled and electrons begin to occupy the 3d orbital in accordance with Aufbau's principle, the 3d orbital becomes lower in energy than the 4s orbital. When removing electrons, remove from the 4s orbital first since it is now higher in energy. Another reason why you must remove electrons in this order is because subshells are most stable when they are half filled or completely filled. Removing from the d-orbital first would grant you a configuration of 3d9 and 4s2, which is less stable than if you were to remove from the 4s orbital first, which would give you 3d10 and 4s1.

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Re: Loss from 4s over 3d?

Postby 405268063 » Sun Nov 03, 2019 10:45 am

I think you remove electrons with the highest electron affinity first, which in this case would be from the 4s orbital.

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