Chemistry QUESTION------TOP SCORE---electron configuration problem

[Toothguy80]

What is the correct electron configuration of Zn2+

so Zn has 30 electrons - 2 = 28

so it has to be this, right?
[Ar] 4s2 3d8

but TOP SCORE says that the answer is this

[Ar] 4s0 3d10

this cant be right, 4s comes before 3d, doesnt it?
.
.
.
.
.
.
.
.
.
Another question,
An electron going from 3p5 4s2 to 3p6 4s1 will emit energy, why? what is the reasoning behind this. Are electrons being taken away, given?

---

Members don't see this ad.

 

[sam12]

For the first one the answer is right, its a transition metal right, so you can expect it to fill the higher orbital first.

For the second one, when electrons jump to a high energy level they need energy to make that jump, thus they absorb energy. Here the electron is falling back to a lower energy level, thus energy is being released.

 

[AaBbZz]

This is a tricky question. Transition metals like to give up there s electrons first especially when they have a full d orbital like Zn.

Also the 5d elements like Mn, Tc, Re give up their s orbital electrons first forming ex) Mn 4s0 3d5 a +2 cation
Manganese also forms a +7 cation which takes away all the 4s and 3d5 electrons, think KMnO4 (Mn here is +7)
This characteristic of being able to form multi cations gives transition metals the property to form colors.

What you should know is that Mn is not going to make a +3 or +4 cation easily because it would have to get rid of it's stable half filled orbital.

Also, Transition elements love this half stage and full stage state so elements like Iron(Fe) like to form +3 cations which would give them a 4s0 3d5 configuration as well.

 

[TimeforDAT]

you take 2 electrons off the 4s orbital because when taking electrons away, you need to take them away from the orbital with the higher principal value 👎, in this case it is taken from the 4s orbital

and if for example it were to say something like what is the electron config for Zn4+ then it would be 4s0 3d8


and 4s fills before 3d, 4s is lower in energy.

 

[restrential]

For an excellent explanation, see erotetica's 2 responses here. He seems to have an excellent grasp on the transition metal concepts.

 

[mhamilton23]

I agree, 4s electrons will be removed first. When you are filling period 4, you actually fill the 4s orbital before the 3d orbitals due to the fact that 4s orbital "penetrates" very close to the nucleus, whereas the 3d orbitals don't very much. This leads to the 4s orbitals being lower in energy. This seems to be contradicting the logic behind removing 4s electrons before 3d electrons when you start ionizing a transition-metal. If 4s orbitals are more stable, then why would you remove 4s electrons before the "less" stable 3d electrons? The reason is that once you start filling the d oribtals, you are actually placing them in orbitals that are between the nucleus and the 4s electrons. The fact that they lie interior to the 4s electrons allows them shield/repel the 4s electrons. Now the 4s electrons become higher in energy, and that's why it's easier to remove the 4s electrons.