electron configuration questions?

Question

1.) 1st 2 quantom numbers (n,l) of the highest energy e- in the ground state of Sc^3+ are 3,1..true or false

my work is: Sc = [AR] 4s2 3d1 so ..
Sc^3+ = [AR] 3p6

how is it equal to 3,1?

2.) I don't understand the difference between ground and excited states. All i know is that when an e- gets excited, it moves to a higher (excited state).

if the e- config for Cr = 1s2 2s2 2p6 3s2 3p6 4s2 3d4..what would the excited state be?

Answer 1

The electron configuration of Sc3+ is [Ne] 3s2 3p6
The value of "n" is the energy level, which is 3 here.
The value of "l" is the orbital type. For a "p" orbital, l = 1.
So, the statement is true.

Chromium has an unusual electron configuration.
It is actually 1s2 2s2 2p6 3s2 3p6 4s1 3d5.
It seems that the atom is more stable when it has all of the orbitals in a group filled the same way, ie. half-full.

Ground state means that the electrons are in the lowest energy levels available. In an excited state, energy has been added to the atom, and the electrons are (temporarily) at higher levels. Which ones depends on how much energy you put into the atom. Too much, and you remove the electron entirely.
So, it's not really predictable without the energy values.

Answer 2

I cant clearly understand ur question. U need to tell what 3,1 are........

Scandium(21) =1s2 2s2 2p6 3s2 3p6 4s2 3d1
Refer to the Moeller diagram for your doubt and the (n+l value)

Cr(24)=1s2 2s2 2p6 3s2 3p6 4s2 3d5
Cu(29)=1s2 2s2 2p6 3s2 3p6 4s1 3d10
Chromium and copper are two extraordinary cases where the skip a sub orbital. They do this as the find the resultant state more stable

The other person who has answered has given the wrong electronic configuration for Cr