For Exercise 2.29 (Chapter 2 The Quantum Mechanics in Action: Atoms)
How many electrons can have the following quantum numbers in an atom:
a) n = 2, l = 1
b) n = 4, l = 2, ml = 2
c) n = 2
d) n = 3, l = 2, ml = +1
Does anyone know how to solve this problem?
Thank you in advance
Exercise 2.29
Moderators: Chem_Mod, Chem_Admin
Re: Exercise 2.29
I will cover this in class in detail. This question is about 20 pages ahead of what we have covered.
Re: Exercise 2.29
For Problem 2.49
"Predict the number of valence electrons present in each of the following atoms (include the outermost delectrons: (a) N; (b) Ag; (c) Nb; (d) W."
What trend are we expected to see in order to identify the number of valence electrons? The atomic number? Is there also something else?
"Predict the number of valence electrons present in each of the following atoms (include the outermost delectrons: (a) N; (b) Ag; (c) Nb; (d) W."
What trend are we expected to see in order to identify the number of valence electrons? The atomic number? Is there also something else?
Re: Exercise 2.29
In 2.29 part c), I don't understand why the maximum number of electrons is 8 if n=2. Can someone please explain?
Re: Exercise 2.29
Chem_Mod wrote:I will cover this in class in detail. This question is about 20 pages ahead of what we have covered.
I am confused with this question which asks "How many electrons can have the the following quantum numbers in an atom,
a) n=2, l1
b)n=4, l=2, m1=2
c) n=2
d) n=3, l=2, m1=+1 ?"
and with 2.25 which is similar and asks, "How many electrons can occupy, a) 4porbitals; b) 3dorbitals; c) 1sorbital; d)4forbital ?"

 Posts: 36
 Joined: Fri Sep 28, 2018 12:26 am
Re: Exercise 2.29
104785906 wrote:Chem_Mod wrote:I will cover this in class in detail. This question is about 20 pages ahead of what we have covered.
I am confused with this question which asks "How many electrons can have the the following quantum numbers in an atom,
a) n=2, l1
b)n=4, l=2, m1=2
c) n=2
d) n=3, l=2, m1=+1 ?"
and with 2.25 which is similar and asks, "How many electrons can occupy, a) 4porbitals; b) 3dorbitals; c) 1sorbital; d)4forbital ?"
A bit confused about this one as well...

 Posts: 30
 Joined: Fri Sep 28, 2018 12:24 am
Re: Exercise 2.29
For Exercise 2.29 (Chapter 2 The Quantum Mechanics in Action: Atoms)
How many electrons can have the following quantum numbers in an atom:
a) n = 2, l = 1
b) n = 4, l = 2, ml = 2
c) n = 2
d) n = 3, l = 2, ml = +1
Does anyone know how to solve this problem?
Thank you in advance
The question varies in how specific the wave equation is solved to.
In part a, we know the subshell in question is a 2p subshell, because n=2 and for p orbitals (l=1). We do not know which porbital (px,py,pz) we're speaking of,so there are 6 possible electrons that could have n=2, l=1 quantum numbers (ml and ms would make the question more specific).
In Part B, we're told this time we're in the 4d subshell (n=4, l=2) but specifically the dxy orbital (ml=2). Each orbital can hold two electrons; by Pauli Exclusion Principle they must spin in opposite directions (the last little bit is irrelevant to the question at hand).
In Part C, all we know is that it is energy level 2. For all we know, it could be 2s or 2p, but between the two subshells, there are a total of 8 electrons present.
Part D can be solved the same way.
How many electrons can have the following quantum numbers in an atom:
a) n = 2, l = 1
b) n = 4, l = 2, ml = 2
c) n = 2
d) n = 3, l = 2, ml = +1
Does anyone know how to solve this problem?
Thank you in advance
The question varies in how specific the wave equation is solved to.
In part a, we know the subshell in question is a 2p subshell, because n=2 and for p orbitals (l=1). We do not know which porbital (px,py,pz) we're speaking of,so there are 6 possible electrons that could have n=2, l=1 quantum numbers (ml and ms would make the question more specific).
In Part B, we're told this time we're in the 4d subshell (n=4, l=2) but specifically the dxy orbital (ml=2). Each orbital can hold two electrons; by Pauli Exclusion Principle they must spin in opposite directions (the last little bit is irrelevant to the question at hand).
In Part C, all we know is that it is energy level 2. For all we know, it could be 2s or 2p, but between the two subshells, there are a total of 8 electrons present.
Part D can be solved the same way.

 Posts: 30
 Joined: Fri Sep 28, 2018 12:24 am
Re: Exercise 2.29
404768057 wrote:In 2.29 part c), I don't understand why the maximum number of electrons is 8 if n=2. Can someone please explain?
At energy level 2, there are two possible subshells: s and p. (Remember that 2d and 2f do not exist).
The s subshell can hold 2 electrons, while the 3 p orbitals combined hold 6 electrons.
2+6=8.

 Posts: 30
 Joined: Fri Sep 28, 2018 12:24 am
Re: Exercise 2.29
derek1d wrote:For Problem 2.49
"Predict the number of valence electrons present in each of the following atoms (include the outermost delectrons: (a) N; (b) Ag; (c) Nb; (d) W."
What trend are we expected to see in order to identify the number of valence electrons? The atomic number? Is there also something else?
Valence electrons (for nontransition metals) are relatively simple: It increases per row from left to right.
So, column 1 elements have 1 valence electron. Column 2 elements have 2 valence electrons. Skipping over the transitions, Column 13 has 3 valence electrons, Column 14 4 and so on up to column 18, which has 8 (noble gases).
Since it says include outermost delectrons, for transitions take however many (n+1)s electrons are present and add them to the number of (n)d orbital electrons present.
Ex: W has an electron configuration of [Kr] 6s2 4f14 5d3. The outermost valence electrons would be 6s (2) and 5d(3) for a total of 5.
Re: Exercise 2.29
To solve this problem, we need to know that each orbital holds 2 electrons and that ml= l > l
So, in this problem we are focusing on the magnetic quantum number. For example a) n=2, l=1 so ml = 1,0,1. Given that there are three orbitals in example a and that each orbital holds 2 electrons, there would be a total of 6 electrons for example a.
For example b, n = 4, l=2,ml=2. ml=2 is the specified orbital for the quantum numbers, and since each orbital holds 2 electrons, there would be 2 electrons in the quantum numbers.
For example c, we are only given n=2. So l = 0,1. l=0 is the s orbital and l=1 is the p orbital. So for l=0 in the s orbital, ml=0. (there is one orbital so 2 electrons are present in s orbital). And for l=1, ml= 1, 0,1 (there are 3 orbitals here so there are 6 electrons in the p orbital). So when n=2, there is a total of 8 electrons present.
So, in this problem we are focusing on the magnetic quantum number. For example a) n=2, l=1 so ml = 1,0,1. Given that there are three orbitals in example a and that each orbital holds 2 electrons, there would be a total of 6 electrons for example a.
For example b, n = 4, l=2,ml=2. ml=2 is the specified orbital for the quantum numbers, and since each orbital holds 2 electrons, there would be 2 electrons in the quantum numbers.
For example c, we are only given n=2. So l = 0,1. l=0 is the s orbital and l=1 is the p orbital. So for l=0 in the s orbital, ml=0. (there is one orbital so 2 electrons are present in s orbital). And for l=1, ml= 1, 0,1 (there are 3 orbitals here so there are 6 electrons in the p orbital). So when n=2, there is a total of 8 electrons present.

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 Joined: Wed Sep 21, 2016 2:57 pm
Re: Exercise 2.29
Hello!
I'm having a bit of trouble with part D of this problem. The question states that n = 3, l = 2, and ml = +1. So from that, you are able to determine that it is the 3d orbital and ml is +1. I drew out a diagram with the different electrons in 2, 1, 0, +1, +2 sections, with up and down arrows to illustrate the spin. However, the answer key says the number of electrons in this specific atom are 2, however I calculated it out to be 4. Can anyone assist? Thank you!
I'm having a bit of trouble with part D of this problem. The question states that n = 3, l = 2, and ml = +1. So from that, you are able to determine that it is the 3d orbital and ml is +1. I drew out a diagram with the different electrons in 2, 1, 0, +1, +2 sections, with up and down arrows to illustrate the spin. However, the answer key says the number of electrons in this specific atom are 2, however I calculated it out to be 4. Can anyone assist? Thank you!
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