[nozo]

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) 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?

[tamim83]

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

[Hunt]

For Sr :

n = 5
l = 4 , 3 , 2 , 1 , 0

If we consider the the values of the magnetic quantum number m_l for each angular momentum qunatum number :

For l = 4
m_l = -4, -3, -2, -1, 0 , 1 , 2 , 3 , 4 ( 5g subshell)

For l = 3
m_l = -3, -2, -1, 0 , 1 , 2 , 3 ( 5f subshell )

For l = 2
m_l = -2, -1, 0 , 1 , 2 ( 5d subshell )

For l = 1
m_l = -1, 0 , 1 ( 5p subshell )

For l = 0
m_l = 0 ( 5s subshell )

Therefore, p , d , f, and g subshels have each one orbital with m_l = -1 , then the max number of electrons having m_l = -1 in an Sr atom is 8 electrons. ( each orbital has 2 electrons at most )

[nozo]

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!!

[Hunt]

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 m_l = +1 . Notice that S subshell can never have any orbital with m_l = +1, -1, or any other number except 0. Therefore, we have 3 orbitals with m_l = +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 m_l = 0 ? How about for n=4?

PS: Oh and thanks for solving the other problem I had Vant_

You're most welcome!

[nozo]

Try for n=5, how many electrons can have the designation m_l = 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:

[Hunt]

For n =5, it's 10e
For n =4, it's 8e.

Remember : just count the m_l's . You got it , I guess

[nozo]

Thank you again Van_H, yes I FINALLY got it!! THank youu