Textbook Exercise 1E.15

[Adam_Ventura_1H]

I stumbled into a question when I was given the task to find what element is being represented by the ground-state of (b) [Ar]3d^3 4s^2. I think it is Vanadium or Manganese. I am not really sure. Hope someone can explain, since the 4s gets filled up before the 3d.

[Kaylee Messick 3J]

Hello! I believe it would be vanadium, though I could be wrong because I am still a little confused on what differentiates "ground-state" electron configurations. I know that we write the 3d^3 before 4s^2 because the d-orbital ends up being lower energy than the s orbital, so we have to write it in that order. Hope this helps!

[Michael Iter 2F]

Hi, I believe ground state just means that is how the electrons are normally configured without being excited. I also think its vanadium. No matter which order the electron configuration is written, it should still add up to however many electrons are in the atom at hand.

[Lauren Mungo 1K]

Whenever you are in doubt, just count the amount of electrons represented and match it to the element on the periodic table with the same number of electrons.

[t_rasul2I]

Ground state refers to having it at a state that is expected of the electrons, meaning they are not exited or anything. Moreover, just count the electrons given despite the configuration to determine the specific electron.

[Yun Su Choi 3G]

But why is the ground state electron configuration [Ar]3d^3 4s^2, not [Ar]3d^5?
For ground state electron configuration, shouldn't all the orbitals in the lowest energy level be filled first, which in this case is 3d before 4s?

For example, for question 1E.13, the ground state electron configuration of silver was [Kr]4d^10 5s^1 instead of [Kr]4d^9 5s^2 because the d-orbital was filled first.

Can someone clarify this difference?

[Anna Yang 1A]

I think it is due to the overlap as the energy levels increase and become closer and closer to each other. When this happens, if you look at a visual you can see that the 4s shell is slightly lower in energy than the 3d shell, and thus, the 4s shell is filled before the 3d. I believe in lecture, it was mentioned that after atomic number 20 (Ca) 3d shifts down in energy and 4s is higher in energy as it is expected to be; this occurs after the 4s shell has been filled (after 4s^2), allowing the 3d shell to fill (with the exceptions of chromium and copper). In regards to silver, the electron configuration is [Kr]4d^105s^1 because it is easier for the s-orbital to give up one electron out of the two electrons needed for the orbital to be filled (less reactive) than in the d-orbital where it has 9 electrons and only needs one more electron to fill its shell (more reactive).