Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: nozo on April 01, 2006, 04:55:36 PM
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Ok guys, I'm now in the Wave function section, I just wanted to make sure if my reasoning is correct... :-\
Now in the Sr atom in the ground state
How many electrons have ml = -1 (for Sr atom)?
The answer here is 8, but I don't know why that is
Tia!
PS: I edited a lot cuz I've figured it out (next time I should take the time to read the book) :P Basically for n, just use n^2 to figure out the number of orbitals and from there multiply e-(2); then for l, just use 2(l)+1 to figure out the # of orbitals, etc.
My prob is just the one above... say given just the ml, how can I know the # of orbitals?
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Hi,
Ok the electron configuration for Sr is
1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^2, 3d^10, 4p^6, 5s^2
Now, according to the quantum number rules two electrons in the p, d, or f sublevels can have a ml value of -1, this would just be two electrons in the same orbital. This is fine as long as they have opposite spins. So Sr has 2 electrons in the 2p, 3p, 3d, and 4p orbitals that have that magnetic quantum number of -1.
Hope this helps :)
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For Sr :
n = 5
l = 4 , 3 , 2 , 1 , 0
If we consider the the values of the magnetic quantum number ml for each angular momentum qunatum number :
For l = 4
ml = -4, -3, -2, -1, 0 , 1 , 2 , 3 , 4 ( 5g subshell)
For l = 3
ml = -3, -2, -1, 0 , 1 , 2 , 3 ( 5f subshell )
For l = 2
ml = -2, -1, 0 , 1 , 2 ( 5d subshell )
For l = 1
ml = -1, 0 , 1 ( 5p subshell )
For l = 0
ml = 0 ( 5s subshell )
Therefore, p , d , f, and g subshels have each one orbital with ml = -1 , then the max number of electrons having ml = -1 in an Sr atom is 8 electrons. ( each orbital has 2 electrons at most )
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Thanks guys, ok let me see if I get this straight...
In general to find the max # of e-, suppose n = 4, ml = +1
Then l could be 3, 2, 1, 0 (f, d, p, s orbitals)
So, would that mean I have 4 orbitals present * 2 e- = 8 e-? Right?
Here's another example: n = 4, l = 3, ml = -2
So here, l is the f orbital, so it can only hold up to 2 e-?
Tia!
PS: Oh and thanks for solving the other problem I had Vant_H :) It makes more sense now :)
BTW, I did the calculations and coincidentally, it came out to the same answers I had for each molarity... -of course I did not understand the logic behind it, but I do now! Thanks again!!
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In general to find the max # of e-, suppose n = 4, ml = +1
Then l could be 3, 2, 1, 0 (f, d, p, s orbitals)
correct
So, would that mean I have 4 orbitals present * 2 e- = 8 e-? Right?
Nop, in this case, only l = 3, 2, and 1 ( p,d, and f ) subshells can each have one orbital with ml = +1 . Notice that S subshell can never have any orbital with ml = +1, -1, or any other number except 0. Therefore, we have 3 orbitals with ml = +1 ====> max no of e's = 6
In the previous case, n was 5, and so g subshell was taken into account. Hence there were max 8 e's.
Try for n=5, how many electrons can have the designation ml = 0 ? How about for n=4?
PS: Oh and thanks for solving the other problem I had Vant_
You're most welcome! :)
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Try for n=5, how many electrons can have the designation ml = 0 ? How about for n=4?
Ok, umm.. would the answer be 1 orbital * 2 e- = 2 e- max? Since l can only be 0?
For n = 4, ml = 0.. the answer would be the same... 2 e- max?
So if n = 4, l = 3, ml = -2... would that also mean it can only have 2 e- as well?
Thanks.. I hope I'm getting it... :idiot:
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For n =5, it's 10e
For n =4, it's 8e.
Remember : just count the ml's . You got it , I guess
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Thank you again Van_H, yes I FINALLY got it!! THank youu :'(