[danomite15]

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?

3.) what is the max number of e- in an iodine atom that have the principal quantom number n=5, ml=+1

the answer is 2..not really sure how to get there

my work is..if n=5, l=4
so ml = 4,3,2,1,0,-1,-2,-3,-4

how is the answer 2?

[Albireo]

1)3p⁶ is Ar, and essentially, the highest e- for this would equal 3,1 because n dictates the energy level while l has a specific number for each orbital, s = 0 p = 1 d = 2 f = 3 etc etc
2) The difference between ground state and excited states is that the excited state is in a different orbital, such as 1s² 2s² 3s²
3) The answer is two because you’re working in the p orbit which means the ml is limited to 1, 0 , -1

[Aufbau89]

-For your first situation, Sc³⁺'s electron configuration (in ground state) should be the same as Ar's e^- configuration, which is 1s² 2s² 2p² 3s² 3p⁶ ,  so, taken the 3p⁶ in consideration, n should be 3 and l should be 1. 

-For your second situation, excited e^- configurations usually skip orbitals; they often jump to the next orbital without finishing the one they've already started.  So your configuration for Cr is stable because it did not skip any orbitals and is written completely.

-For your third situation, why does l=4?  Doesn't it equal 1 since I is in the p-orbital?  The configuration for Iodine would be [Kr]5s²4d¹⁰5p⁵.  In n=5, there are 2 e- in the 5s orbital, 10 e- in the 4d orbital, and 5 e- in the 5p orbital.  M_l can't be + 1 in the s-orbital, can only be +1 in the d-orbital when it is the 5^th e^-, and only in the p-orbital when it is the 3^rd; and that is why the answer is two.


-Hope this helps!