angular momentum quantum number

[globy321]

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how can you find the angular momentum quantum number (l) for the orbital from which a Mg atom loses two electrons to form a Mg2+ ion?

 

[V5RED]

http://chemistry.about.com/od/electronicstructure/a/quantumnumber.htm

http://www.angelo.edu/faculty/kboudrea/general/quantum_numbers/Quantum_Numbers.htm

If you read both of these(or a chemistry textbook) you will get your answer.

 

[Anthrax]

Okay, Mg in ground state has an electron configuration of [Ne]3s2.

It loses two electrons from the outermost energy level, the 3rd energy level. So it loses two electrons from the 3s subshell, when Mg 2+ forms.

For 3s
n= 3
l = 0 (for s subshell)
ml = -l to +l for any electron, so it would be 0 in this case

In general, remember that for an electron in
- the s subshell, l = 0
- the p subshell, l = 1
- the d subshell, l = 2
- the f subshell, l = 3

 

[globy321]

Okay, Mg in ground state has an electron configuration of [Ne]3s2.

It loses two electrons from the outermost energy level, 3s subshell, when Mg 2+ forms.

For 3s,

n= 3
l = 0 (for s subshell)
ml = -l to +l for any electron, so it would be 0 in this case

In general remember that, for an electron in
- the s subshell, l = 0
- the p subshell, l = 1
- the d subshell, l = 2
- the f subshell, l = 3

Don't you find "l" by using n-1? since the electrons are being lost from shell 3 (n=3), then should "l" be 2 (3-1) .

 

[Anthrax]

Don't you find "l" by using n-1? since the electrons are being lost from shell 3 (n=3), then should "l" be 2 (3-1) .

No.

n-1 tells you how many levels of subshells(aka orbitals) are possible for a particular energy level. For a particular energy level, the possible l values are 0 to n-1.

Look at the first energy level, n=1. So the possible subshells for n=1, are 1-1 =0. 0 denotes the s subshell. And that makes sense because the first energy level only has the s subshell.

For the second energy level, n=2. So the possible subshells here are 0 to 1 (n-1).
So the second subshell can have the s subshell (denoted by l=0) and the p subshell (denoted by l=1). Again that makes sense because we only have 2s and 2p subshells and NOT 2d.

For n=3, the possibilities are 0 to 2 (n-1). So we can have l=0 for s subshell, l=1 for the p subshell, l=2 for the 3 subshell. And we know we only have 3s, 3p, and 3d subshells.

Does that help?

 

[V5RED]

Don't you find "l" by using n-1? since the electrons are being lost from shell 3 (n=3), then should "l" be 2 (3-1) .

If you read the links I posted, you would not ask a question like this. They describe how to determine quantum numbers, what they mean, and from which level the electrons are taken when forming ions.

 

[globy321]

No.

n-1 tells you how many levels of subshells(aka orbitals) are possible for a particular energy level. For a particular energy level, the possible l values are 0 to n-1.

Look at the first energy level, n=1. So the possible subshells for n=1, are 1-1 =0. 0 denotes the s subshell. And that makes sense because the first energy level only has the s subshell.

For the second energy level, n=2. So the possible subshells here are 0 to 1 (n-1).
So the second subshell can have the s subshell (denoted by l=0) and the p subshell (denoted by l=1). Again that makes sense because we only have 2s and 2p subshells and NOT 2d.

For n=3, the possibilities are 0 to 2 (n-1). So we can have l=0 for s subshell, l=1 for the p subshell, l=2 for the 3 subshell. And we know we only have 3s, 3p, and 3d subshells.

Does that help?

Let me see if I go this right..
There are shells (principal quantum number n)
There are subshells or orbitals (l angular quantum number which is 0 to n-1).
Since Mg is losing electrons from the third shell (n=3) then l can be 0,1 or 2. And since the 2 electrons are being lost from the s orbital l has to be 0. if the electons were being lost from the p orbital then l would be 1 and if they were being lost from the d then l would be 2. is this right?

To find m(l) you would go l-1to l+1...so m(l) would be-1to1? Can m(l) be negative?
To find m(s) it would either be up +1/2 or down-1/2. Does the orbital fill with the +1/2 first before the -1/2?
Thanks for your help.

 

[Courtjester8]

I'd just like to mention something that you may not have realized.

The quantum stuff you're asking about here will at most be 1 question on the mcat if it shows up at all. Unless you're nearing the end of your studying and are solid on everything else, I'd devote my time to higher priority material to maximize your score. This is one more reason why taking a lot of practice tests is such an important part of mcat studying. Its important to figure out what material is priority material and what isn't.

-cj8

 

[globy321]

I'd just like to mention something that you may not have realized.

The quantum stuff you're asking about here will at most be 1 question on the mcat if it shows up at all. Unless you're nearing the end of your studying and are solid on everything else, I'd devote my time to higher priority material to maximize your score. This is one more reason why taking a lot of practice tests is such an important part of mcat studying. Its important to figure out what material is priority material and what isn't.

-cj8

What do you consider priority material to focus on?

 

[Courtjester8]

For physics: translational motion, momentum, force, work-energy, electro statics, circuits, magnetism, optics, harmonic motion...this is off the top of my head. Its best to decide for yourself based on what you already know and how often it shows up on practice tests.

cj8

 

[Anthrax]

Let me see if I go this right..
There are shells (principal quantum number n)
There are subshells or orbitals (l angular quantum number which is 0 to n-1).
Since Mg is losing electrons from the third shell (n=3) then l can be 0,1 or 2. And since the 2 electrons are being lost from the s orbital l has to be 0. if the electons were being lost from the p orbital then l would be 1 and if they were being lost from the d then l would be 2. is this right?

To find m(l) you would go l-1to l+1...so m(l) would be-1to1? Can m(l) be negative?
To find m(s) it would either be up +1/2 or down-1/2. Does the orbital fill with the +1/2 first before the -1/2?
Thanks for your help.

Yes, that is correct. You seem to understand l and ml and yes ml can be negative.

As for ms, I don't think you should really worry about it. It can be either +1/2 or -1/2 for two electrons in an orbital. Doesn't matter which one you put first. But again, don't worry about this too much. There is generally very little from this area on MCAT.

But if you are interested in learning, let me explain ms.

Sorry one thing I mixed up was that subshells and orbitals are the same. Well an energy level has subshells and each subshell has orbitals. Each orbital holds 2 electrons.

s subshell has just one orbital, so it only hold 2 electrons. One gets ms=+1/2 and one get ms=-1/2.

p subshell can hold 6 electrons, so it has 3 orbitals, px, py, pz. Within each of these, there are 2 electrons, so one electron gets gets ms=+1/2 and one get ms=-1/2. It doesn't matter if +1/2 goes first or -1/2. But they both can't have +1/2 and both can't have -1/2. They must be different.

d subshell can hold 10 electrons, so it has 5 orbitals, dx, dy, dz, d x^2-y^2, and d^z2. So each has two electrons and for each electron in one of these orbitals you can either have ms=+1/2 OR ms=-1/2.

Just remember that ms of 2 electrons in an orbital is never the same.

Again, I do think that you seem to get it now and I suggest that you don't waste too too much time on this. However, if you are interested and have time, then by all means go for it.

 

[Anthrax]

http://www.angelo.edu/faculty/kboudrea/general/quantum_numbers/Quantum_Numbers.htm

This is a really helpful website. Check it out if you have time.

 

[zmth]

In so far as relativistic effects such as magnetic couplings like spin - orbit coupling which is a good approximation esp for light atoms, no spin dependant terms in the energy, then the total orbital angular momentum of the atom is ALWAYS zero !! and that applies to an atom or ion with 5 electrons or however many - odd or even as long as the spin dependence is negligible which is generally true for not too high atomic number nucleons - for the ground state. It is also true for a molecule in its ground state.
many people are confused about this because they use the ' shell model' for a multiple electron system. Remember the 'shell ' comes from the excited states of a single
electron in a one nucleon system. Once there is more than one electron this is no longer the case. There are some inferences one can make using the shell model
approximation but concluding that the ground state of an atom which according to the ' shell model' is in an unfilled higher than s state so has non-zero total orbital
angular momentum is NOT one of them. For one thing there is no longer a single reduced mass. The shell
model is a rough approximation for multi electron systems and must be used with understanding its gross approximations and one can certainly NOT assume that
in the absence of any magnetic couplings that the total orbital angular momentum of the system in the ground state will be anything other than zero ! It has no reason
to be anything otherwise - all forces are coulomb interparticle which are independent of any total rotation of the whole system -
any non-zero total orbital angular momentum would just be adding energy.

 

[Altius Premier Tutor]

Two very important points with respect to this thread: 1) The principle quantum number, specifically, will be tested on MCAT-2015 but none of the other quantum numbers, including the one discussed here. We are still giving our students a general overview of all of them for the sake of coherence, but then emphasizing to them that only the principle quantum number will be tested. 2) Momentum and the closely related principle of impulse are definitely removed from the MCAT-2015 topics list (which I mention because momentum was included in a list of physics topics above).