Electron Configuration of Zn2+

[aDentite]

I thought it was [Ar]4s2 3d8, but the correct answer is [Ar]3d10.

Can someone tell me why the S electrons are removed before the D electrons?

Doesn't 4s fill before 3d ( n + l rule: 4 + 0 = 4, 3 +2 = 5)?

Thanks.

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[UndergradGuy7]

When adding electrons you add to the 4s then to the 3d (there are some exceptions like silver, gold, chromium, Mo).

When ionizing (taking electons off) you take them from the 4s first before the 3d. I believe this is because the ionization energy to remove an electron from a 4s is less than a 3d. Also, an element with a full 3d subshell (like zinc) is more stable than a full 4s subshell. So you want to lose the 4s before the 3d.

I guess an easy way to remember this is to remove electrons the electrons with the largest principal quantum number, n. Which in this case is the "4"s electrons instead of the "3"d.

I usually reason is like this, but am not sure how right this is:
I always thought the 4s are further from the nucleus so they are ionized first (higher principal quantum number, further away), but then why is it lower in energy than 3d if the farther away, the more energy it should have? Which is not the case since 4s are lower in energy than 3d.

 

[aDentite]

According to the "n+L" rule, 4s (4+0=4) is lower than 3d (3+2=5) so 3d is definitely the outer shell and further from the nucleus, even though 3d has a lower principal quantum number. After thinking about it some more, I think the reason that the 4s electrons are removed is because you either have the option of having one full shell and one partially full shell ( [AR] 4s2 3d8) or one empty shell and one full shell ( [Ar] 4s0 3d10 ).

 

[TexasOMFS]

Yes, you add to 4s before 3d. What happens chemically is that as the 3d orbital fills up, it actually stabilizes and falls below the 4s orbital in energy: hence the reason we take from the 4s before the 3d. Pretty sure Chad explains this in his videos somewhere.