I'm sorry I can draw this thing out on the board but I'll try to explain it. First, however, you must learn the rules.
n = principal quantum number and can have any whole number beginning with 1.
l = any whole number beginning at 0 but can be no larger than n-1
ml = any whole number (including zero) and it can be as small as -l and as large as +l
ms can be +1/2 or -1/2.
Get all of this in your mind and we go to the set up.
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Here is what we do. We count by n until we get to n=3, l=2, ml = 0 and ms = 1/2
n = 1
l can be 0 (only)
ml can be 0 (only)
ms can be +1/2 and -1/2. So we put two electrons in for the +1/2 and -1/2 and that spells He.
n=2
l = 0................or 1
ml = 0............or -1...0...+1
ms = +/-1/2
We put electrons into each of these holes or 8 electrons total. Together with the two we had above makes 10 and we are to Ne.
n = 3
l = 0........or 1..........or 2
ml = 0.....-1, 0, +1.....-2,-1,0,+1,+2
ms +/-1/2
Add electrons in pairs (two for the ml = 0, 6 for the ml of -1,0,+1, and 2 for ml of -2, 2more for ml = of -1 and ONE more for ml = 0.
Remembering we had 10 from n = 2, that plus what we have added here makes 13 more or 23 total. Must be V.
It might help if you drew something like this on a sheet of paper; however, it get laborious to do this every time and you don't need to do that. You can take a shortcut. How?
We take a little stock and look at the periodic table.
n = 1 (period 1 takes you through He)=2e
n = 2 (period 2 takes you through Ne)=10e
n = 3(you wanted n = 3 so we know the element is somewhere after Na.
For l=2, that's a d electron (s electrons are l = 0, p electrons are l =1, d electrons are l = 2, f electrons are l = 3). So a d electrons means it must be somewhere in the 3d transition series (n = 3, l =2). Remembering to fill the 4s next, we add those to get us through K and Ca, then we count what we have so far. Through Ca is 20, adding 1 more gives us Sc, another gives Ti, and the last one gives V. What makes Ti and Sc different.
Sc is n = 3, l = 2, ml = -2
Ti is n = 3, l = 2, ml = -1
V is n = 3, l = 3, ml = 0
etc.
I hope this helps. If you draw something like I did above two or three times I can assure you it gets easier.
The four quantum numbers for the last electron placed in the orbitals of a certain element are: n=3 l=2 m of l=0 and m of s=-1/2. What is the element?
I need help with this part of chemistry I have NEVER understood this ever. I know how to do the configurations but not this. I need some explanation on how to do it.
2 answers
Well I'm studying for my final and he gave us an answer key and with a spin down it says that the answer is nickel.. SOO I'm trying to make sense of it but I understand what n is what l is but how do you figure out what the elements are from ml? And all of that put together?
2 answers