Gen Chem Problem

[Simon Chiang]

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What would be the Correct electron configuration for Zn+2?

[Ar]4s2.4d8
or
[Ar}4s0. 4d10

 

[NumbaOneStunna]

This is actually what the answer is: (I think you copied the correct answer choice incorrectly)
[Ar]4s0.3d10

I just took Topscore three and this was the first problem - and I missed due to carelessness. 🙂 My understanding of it is that when strip an atom of its electrons it'll lose its outermost electrons or valence electrons first. The higher principle shell number (4>3) means that the 4s electrons are removed before the 3d electrons.


I related question that I have is:
What's Zn+3

[Ar]4s1.3d10
or [Ar]4s0.4d9

Does this 4s1d10 thing only apply to the certain transition metals in VIB and IB or to anything that is similar? Thanks!

 

[sixkiller]

This is actually what the answer is: (I think you copied the correct answer choice incorrectly)
[Ar]4s0.3d10

I just took Topscore three and this was the first problem - and I missed due to carelessness. 🙂 My understanding of it is that when strip an atom of its electrons it'll lose its outermost electrons or valence electrons first. The higher principle shell number (4>3) means that the 4s electrons are removed before the 3d electrons.


I related question that I have is:
What's Zn+3

[Ar]4s1.3d10
or [Ar]4s0.4d9

Does this 4s1d10 thing only apply to the certain transition metals in VIB and IB or to anything that is similar? Thanks!

Remember that electrons are lost according to potential energy; those with the highest potential energy are lost first. Notice that this is the opposite of Aufbau ordering in which orbitals/subshells are filled...

Thus, 4s0.3d9 is correct for Zn+++

For transition metals remember that an electron in the 4s orbital is lower than an energy a single electron in a 3d orbital. Thus, according to aufbau ordering, 4s fills before the 3d subshell.

However, once the 3d subshell BEGINS TO FILL (first electron is added to the 3d subshell), the 3d electrons exert a shielding effect on the 4s electrons....

Now, the 4s electrons are HIGHER in potential energy than the 3d electrons...this is why 4s electrons are lost before 3d electrons.

**Exceptions to this rule are: Cr, Cu, Mo, Ag, and Au (notice their positions to one another on the periodic table; you can think of them as analogs of one another). At the very least you should memorize the electron configurations of Cr and Cu**

***Cr: 4s1 3d5, Cu: 4s1 3d10....(n-1)d subshell fills before ns***
***Mo: 5s1 4d5, Ag: 5s1 4d10......see the pattern?****
***Au: 6s1 4f14 5d10....d and f fill before s; also notice that f is (n-2).***

***for the above exceptions, ns electrons are still lost first, the only exception is the filling order; inner shell fills before outer shell***

For transition metals, ns electrons (where n = to the period number of the transition metal) is lost before the (n-1)d electrons.

Main group elements are no exception either; electrons with the highest potential energy are lost first.

Recall the following order of potential energy for subshells WITHIN a GIVEN SHELL 👎...

ns < np < nd < nf...

Applying this rule to p-block elements, np is lost before ns....

 

[NumbaOneStunna]

Sixkiller thanks for the informative post, learned a few things there. I wasn't thinking when I wrote my question lol, the wrong answer choice didn't even have the right # of electrons 🙁.

 

[jorasic]

Nice.