Quantum Number Problems
#1 - 10

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QN Probs 11-25

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Problem #1: Give the maximum number of electrons in an atom that can have these quantum numbers:

(a) n = 4
(b) n = 5, m = +1
(c) n = 5, ms = +½
(d) n = 3, ℓ = 2
(e) n = 1, ℓ = 0, m = 0

Solution:

(a) n = 4

The total number of m values (derived from all possible ℓ states for a given n value) is given by n2. This gives us 16 m values when n = 4. Here is a listing of the m values when n = 4:
m total m
0   01
1   -1, 0, 13
2   -2, -1, 0, 1, 25
3   -3, -2, -1, 0, 1, 2, 37
4   -4, -3, -2, -1, 0, 1, 2, 3, 49

Since you also have to include ms, you multiple n2 by 2 to get the maximum amount of electrons in the entire energy level:

2n2

2(4)2

32

(b) n = 5, m = 1

The value of ℓ has not been specified; therefore, we need to take into account all the possibilities for ℓ. When n = 5, the permitted values for ℓ are 0, 1, 2, 3, 4. Let's look at each ℓ in the context of m being equal to 1.

When ℓ = 0, m can only equal 0. ℓ = 0 is not part of the correct answer

When ℓ = 1, m can take on the values of -1, 0, 1. ℓ = 1 is part of the correct answer.

In like manner, ℓ = 2, 3, 4 are all part of the correct answer. Here are the m values:

ℓ = 2    m = -2, -1, 0, 1, 2
ℓ = 3    m = -3, -2, -1, 0, 1, 2, 3
ℓ = 4    m = -4, -3, -2, -1, 0, 1, 2, 3, 4

Each ℓ value has an m of 1 allowed.

Four different orbitals (n, ℓ, m just below) are possible for n = 5 and m = 1:

5, 1, 1    5, 2, 1    5, 3, 1    5, 4, 1

Each orbital can hold two electrons (ms = ½ and ms = -½), so the total number of electrons is 8.

(c) n = 5, ms = ½

The problem is very similar to (b), except that ms is given. This value corresponds to a single electron within an m value. Therefore, we need to find out how many numbers of m we have in order to know the number of max electrons.

Like before, ℓ = 0, 1, 2, 3, 4. Since m is not specified we have to take all possible m values for each ℓ value and add them together:

   2ℓ + 1Total m
0   2(0) + 11
1   2(1) + 13
2   2(2) + 15
3   2(3) + 17
4   2(4) + 19

Notice that the total number of m states can be given by the formula (2ℓ + 1).

Thus, the total number of m states is (1 + 3 + 5 + 7 + 9) = 25. This is the highest number of electrons with ms = ½.

You might be required to enumerate all the m values as opposed to just stating how many there are using the (2ℓ + 1) formula. In that case:

   m valuesTotal m
0   01
1   -1, 0, 13
2   -2, -1, 0, 1, 25
3   -3, -2, -1, 0, 1, 2, 37
4   -4, -3, -2, -1, 0, 1, 2, 3, 49

(d) n = 3, ℓ = 2

The only numbers not specified are m and ms. You have to determine all the possibilities for them:
# of m = 2ℓ + 1 = 2(2) + 1 = 5

There will be two electrons per m state (ms = ± ½) therefore, maximum number of electrons is 5 x 2 = 10.

(e) n = 1, ℓ = 0, m = 0

The only number not specified is ms. Consequently, there are only two electrons that can have the 3 values given. The quantum numbers for those two electrons are:
1, 0, 0, ½
1, 0, 0, -½

Problem #2: Each electron orbital is characterized by 3 quantum numbers: n, ℓ, and m.

n is known as the ____ quantum number.
ℓ is known as the ____ quantum number.
m is known as the ____ quantum number.

Solution:

n is the principal quantum number
ℓ is the azamuthal quantum number
m is the magnetic quantum number

By the way, you sometimes see n labeled as the Principle Quantum Number. This is an incorrect usage of the word principle. ℓ can also be (correctly) called the angular momentum quantum number.


Problem #3: Each electron orbital is characterized by 3 quantum numbers: n, ℓ, and m.

n specifies ___.
ℓ specifies ___.
m specifies ___.

(a) The subshell or orbital shape.
(b) The energy and average distance from the nucleus.
(c) The orbital orientation.

Solution:

n specifies the energy level and average distance from nucleus
ℓ specifies the subshell or orbital shape
m specifies the orbital orientation

Problem #4: Give the orbital designation (1s, 2p, 3d, etc.) of electrons with the following combination of principal and azimuthal quantum numbers.

(a) n = 1, ℓ = 0
(b) n = 2, ℓ = 1
(c) n = 3, ℓ = 2
(d) n = 5, ℓ = 3
(e) n = 6, ℓ = 0
(f) n = 4, ℓ = 2

A handy guide to the ℓ values and subshell/orbital names (s, p, d, f, and so on) is this:

ℓ --->    0   1   2   3   4
subshell --->   s   p   d   f   g

Solution:

(a) n = 1, ℓ = 0

n = 1 tells us that the shell number will be 1.

ℓ = 0 tells us that it will be the 's' subshell.

The orbital designation is 1s.

(b) n = 2, ℓ = 1

n = 2 tells us that the shell number will be 2.

ℓ = 1 tells us that it will be the 'p' subshell.

The orbital designation is 2p.

(c) n = 3, ℓ = 2
n = 3 tells us that the shell number will be 3.

ℓ = 2 tells us that it will be the 'd' subshell.

The orbital designation is 3d.

(d) n = 5, ℓ = 3

n = 5 tells us that the shell number will be 5.

ℓ = 3 tells us that it will be the 'f' subshell.

The orbital designation is 5f.

(e) n = 6, ℓ = 0

n = 6 tells us that the shell number will be 6.

ℓ = 0 tells us that it will be the 's' subshell.

The orbital designation is 6s.

(f) n = 4, ℓ = 2

n = 4 tells us that the shell number will be 4.

ℓ = 2 tells us that it will be the 'd' subshell.

The orbital designation is 4d.


Problem #5: For the quantum number ℓ values below, how many possible values are there for the quantum number m?

(a) 5; (b) 3; (c) 2; (d) 1

Solution:

The rule for m is that, given the ℓ value, we start at -ℓ and go by integers to zero and then to +ℓ. We can use this formula to determine how many m values for a given ℓ: 2ℓ + 1.

(a) For ℓ = 5, the m values are -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, a total of eleven values.

(b) Seven values of m resulting from 2(3) + 1 = 7

(c) 2(2) + 1 gives five values. The enumeration is -2, -1, 0, 1, 2.

(d) Three values (-1, 0, 1) or 2(1) + 1 = 3


Problem #6: What does a set of four quantum numbers tell you about an electron? Compare and contrast the locations and properties of two electrons with quantum number sets (4, 3, 1, +½) and (4, 3, -1, +½).

Solution:

The two electrons exist in the same shell (n = 4), same subshell (ℓ = 3). The particular subshell involved is the 4f.

The two electrons are in different orbitals with the 4f subshell. We know this by the differing m values.

The two electrons have the same spin (ms = +½).


Problem #7: Identify the shell/subshell that each of the following sets of quantum numbers refers to.

(a) n = 2, ℓ = 1, m = 1, ms = ½
(b) n = 3, ℓ = 2, m = 2, ms = ½
(c) n = 4, ℓ = 1, m = -1, ms = -½
(d) n = 4, ℓ = 3, m = 3, ms = -½
(e) n = 5, ℓ = 0, m = 0, ms = ½

Solution:

The solution procedure involves looking at the n, ℓ pairings. The m and ms do not play a role in answering this question.

The n values tells us the first half of the answer. The ℓ values tells us the second half of the answer. Follow this guide:

ℓ --->    0   1   2   3   4
subshell --->   s   p   d   f   g

(a) 2, 1 ---> 2p
(b) 3, 2 ---> 3d
(c) 4, 1 ---> 4p
(d) 4, 3 ---> 4f
(e) 5, 0 ---> 5s


Problem #8: Which of the following set of quantum numbers (ordered n, ℓ, m, ms) are possible for an electron in an atom?

Select all that apply:

(a) 3, 2, 2, -½   (f) 5, 3, -3, ½
(b) 2, 1, 3, ½   (g) 3, 1, -2, -½
(c) -3, 2, 2, -½   (h) 5, 3, 0, ½
(d) 3, 3, 1, -½   (i) 3, 2, -1, ±½
(e) 3, 2, 1, -1

Solution:

Let us find the correct ones by removing all the sets that are incorrect.

1) When scanning for incorrect sets, the first step is to scan the ms values. Since this value can only be +½ or -½, we can quickly remove incorrect ones and not have to analyze their n, ℓ, and m values.

We see that (e) with ms = -1 and (i) with ms = ±½ violate the rule for ms. Remember, ms can be EITHER positive ½ or negative ½ in the set of four quantum numbers, not both at the same time.

2) Next, scan for incorrect n values:

We see that (c) with n = -3 violates the rule for n.

3) Scan for problems with the relationship between n and ℓ:

In (d), there is a violation of the rule for generating permissible ℓ from the given n. When n = 3, ℓ may only take on the values of 0, 1, and 2.

4) Scan for problems with the relationship between ℓ: and m

In (b), there is a violation of the rule for generating permissible m from the given ℓ. When ℓ = 1, the permitted values for m are -1, 0, and 1. A value of 3 is not allowed. (g) shows the same mistake. When ℓ = 1, m cannot equal -2.

5) Examination of (a), (f), and (h) will show that these sets of quantum numbers adhere to all the rules. Verification of this is left to the student.


Problem #9: For principal quantum number n = 4, the total number of orbitals having ℓ = 3 is?

Solution:

With ℓ = 3 we examine the m values to determine how many orbitals are present:
by enumeration: 3, 2, 1, 0, -1, -2, -3 ---> 7 orbitals
by formula: 2(3) + 1 = 7

Problem #10: The maximum number of electrons that can have principal quantum number n = 3 and spin -½ is?

Solution:

We need to determine the total number of m values possible. With n = 3, ℓ can be 0, 1, 2.
 0  1  2 
2(ℓ) + 1 1  3  5 

Each orbital can have one electron with ms = ½. Nine orbitals, each with one electron of ½ gives the answer to be 9.


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