Quantum Mechanics & Sub-shells(I understand it but need help)

[FrankP]

Hello,

 

So I am in general chemistry and I have had a long absence since my last chemistry class which was high school back in 2006 so I have forgotten general principles.

 

My teacher has her PHD in it so to her this seems simple, my question is two fold.

 

When using Aufbau Principle and mapping out the 1s^2-2s²-2p⁶-3s²-3p⁶-4s²-3d¹⁰-4p⁶-5s²-Etc...

 

Is there an easy way to determine the valence election shells? My book mentions in passing that the valence shells are determined by the group/ period that they are located in but it does not explain that.

 

The homework I am fine with but I have a test coming up next week and so I am preparing in advance. I have to do a few problems where I find valence shells I can do this by writing out the whole Aufbau chart and then doing the math to add up to the total number of electrons. Just wanted to know if this is the only way.

 

My second question is when you determine the atom you wish to map out so say for example my teacher does this (which I know she will because she told me she will lol)

 

n=3 L=2 m_L = +2 M_s= -¹/₂

 

We determined that Z=30 and therefore the element is Zn but I don't fully grasp how to reason my wa to getting that answer.

 

Thanks in advance for the help!

 

Frank

[Sensei]

Is there an easy way to determine the valence election shells? My book mentions in passing that the valence shells are determined by the group/ period that they are located in but it does not explain that.

 

For group 1-2 valence electrons are 1-2 (so they are from s),

for group 13-18 valence electrons are 3-8 (simply subtract 10) (so they are from p). Except Helium.

 

f.e. Al is in 13 group, and has 3 valence electrons. 3s2,3p1.

f.e. Na is in 1 group, and has 1 valence electron. 3s1.

 

For groups 3-12, there is no such simple way to tell.

 

Well, you should read wikipedia for a start

https://en.wikipedia.org/wiki/Valence_electron

Edited October 7, 2016 by Sensei

[hypervalent_iodine]

Just to perhaps clarify some of Sensei's post, he is including the D block elements in his group numbering. Some places will teach that there are 8 groups, not 18. In that case, the group number corresponds to the number of valence electrons in the neutral species (e.g. Fluorine is group 7 and has 7 valence electrons).

 

As for your second question, you can't determine the identity of the atom based on the quantum numbers of one of its electrons. That makes far too many assumptions and flies in the face of pretty basic principles. You firstly have to assume that the electron it is describing is the last one to be placed into an orbital, meaning it occupies the highest energy position. You can determine that this electron is in a 3d orbital, since n = 3 (the energy level) and l = 2 (this describes the type of orbital; l = 0 for s, 1 for p and 2 for d). The ml numbers tells you the orientation along the Cartesian axes, but since this is an atom, the d orbitals are degenerate and you have no way of knowing which of those 5 is filled in what order. That means your electron could have been 1st d electron, or it could be the last. I'm not sure how you're expected to solve that question. Are you given more information?

[FrankP]

Just to perhaps clarify some of Sensei's post, he is including the D block elements in his group numbering. Some places will teach that there are 8 groups, not 18. In that case, the group number corresponds to the number of valence electrons in the neutral species (e.g. Fluorine is group 7 and has 7 valence electrons).

 

As for your second question, you can't determine the identity of the atom based on the quantum numbers of one of its electrons. That makes far too many assumptions and flies in the face of pretty basic principles. You firstly have to assume that the electron it is describing is the last one to be placed into an orbital, meaning it occupies the highest energy position. You can determine that this electron is in a 3d orbital, since n = 3 (the energy level) and l = 2 (this describes the type of orbital; l = 0 for s, 1 for p and 2 for d). The ml numbers tells you the orientation along the Cartesian axes, but since this is an atom, the d orbitals are degenerate and you have no way of knowing which of those 5 is filled in what order. That means your electron could have been 1st d electron, or it could be the last. I'm not sure how you're expected to solve that question. Are you given more information?

 

 

Were not given any more information then this unfortunately I don't know how I am supposed to solve it either.

 

I follow your reasoning when solving the problem and it actually makes more sense but like you said I am not sure how we could solve that given the title info we got.

 

So we determined in class that the values that I posted were for Z=30

 

The only way I can reasonably explain what my teacher has done here is to say that since she states m_L = +2 M_s= -¹/_2... That would mean that +2 would place the last electron in the 5th sub shell and being that the spin is -1/2 that would mean that all electrons were paired in this 5th sub shell since this was the 5th sub shell in the d orbital we could assume that this element would have the maximum number of electrons in its outer shell, 5d¹⁰ and then we would have to find the element that has such a electron structure? does that make sense or am I wrong here too lol

 

I think given your explanation it makes more sense now for me to follow along.

 

If I could ask a stupid question when you said "and l = 2 (this describes the type of orbital; l = 0 for s, 1 for p and 2 for d)." how did you know this? I feel like I am struggling because ML and L never seem to make sense to me. My book says L is n-1 but thats never the case... and then ML I obviously have no clue how to get that because I can't even get L

[hypervalent_iodine]

The l number describes the shape of the orbital, so the orbital it refers to can be figured based on how many orientations the l number can have, which is just simply how many ml numbers are possible. So:

 

For l = 0 there is only one possible ml number (0). There is only one type of s orbital, therefor l = 0 corresponds to the s orbitals.

 

For l = 1 you can have 3 ml numbers (-1, 0, +1). These correspond to the three p orbitals - px, py, and pz.

 

For l = 2 you have 5 ml numbers, which correspond to the 5 d orbitals.

[FrankP]

This makes sense you know what is actually helping me make this click. I watched a quick video lecture and in there they explained that the M_L value is actually related to how to graph it mathematically.. Pz Py Px etc. That made it make more sense to me and so L is explaining that shape/type of orbital that the e- will have, while M_L States how many ways that orbital can be "graphed" or tracked idk if graphed is the proper term.

 

And then we use those -1,0,1 values to place electron spin within in order to calculate M_s which is either +1/2 when unpaired and -1/2 when unpaired.

 

See my problem is memorizing things is hard for me so I like to try and make it make sense in a grander picture that helps me remember things when things make sense to me. I find I have the hardest time with memorization without application. I think I am getting it to finally click though however you guys will let me know if is of any truth haha

[hypervalent_iodine]

I suck at rote learning as well. Much like you, this section of general chemistry didn't fully click for me until I had something tangible and visual to relate all the information too. I'm glad it helped.

[FrankP]

Yea it is really difficult for me to do that i'm glad I am not the only one who doesn't find this to be super simple!

 

 

Anyhow I had a part 2 of my question one of my worksheets says this...

 

How many sub-shells in the 7th main energy level? How many electrons in this shell? What is the difference between the P_x and P_y orbitals?

 

 

My attempt to answer this:

 

a)

When N = 7

L = 0, 1, 2, 3, 4, 5, 6 (meaning 7 sub-shell's)

0=s, 1=p, 2=d, 3=f, 4=g, 5=h, 6=i

 

b)

max 26 electrons in i subshell

Max e- allowed in 7th Main energy level (98 max e-)

s=2

p=6

d=10

f=14

g=18

h=22

i=28

 

 

c)

The difference between Px and Py orbitals is the way they occupy a 3D space. A 3D space is graphed using an X, Y & Z axis. When an orbital such as Px is graphed it occupies/aligns itself with the X axis. VS a Py orbital which would align itself with he Y axis of the 3D space/graph

 

 

 

How am I doing? I found this question to be tough but as I reasoned through it just now I believe I am confident in the answer...

 

Problem I have found on this is that my book and internet sources have never really explained what goes on past N=4

 

We have made mention of G orbitals but have never quantified a maximum number of e- for it but I tried to recognize a trend of +4 per sub shell and therefore guessed that the max for i sub shell would 28.

[FrankP]

Bump