### Quantum Numbers:The Rules for Assigning ThemProblems #11 - 25

Problem #11: What is the maximum number of electrons that can be identified with the following set of quantum numbers?

(a) n = 4, ℓ = 0, m = 0, ms = +12
(b) n = 3, m = +2, ms = +12
(c) n = 3, m = 0, ms = +12

Solution to (a):

One electron.

The quantum number set in the question describes an electron in the 4s orbital.

Remember, each correct set of 4 quantum nubers (n, ℓ, m, ms) uniquely describes one electron.

Solution to (b):

One electron.

Based on the m of +2, the ℓ must be 2, so the full set of quantum numbers is:

3, 2, 2, +12

Solution to (c):

Based on n = 3, the ℓ values can be 0, 1, 2.

For each of the three ℓ values (0, 1, 2), there exists an m = 0.

In each of those orbitals, (the 3s, 3p, and 3d) there can exist one electron with ms = +12. Three electrons is the answer.

Problem #12: What is the maximum number of electrons that can have the following sets of quantum numbers?

(a) n = 4, ℓ = 3, m = 3, ms = −12
(b) n = 4, ℓ = 3, m = 4, ms = −12

Solution:

1) Since all four quantum numbers are specified, the answer will be either one or zero. It will be one if all four quantum numbers are correct and zero if any one (or more) of them are incorrect.

A value of n = 4 is allowed. Therefore, ℓ can take on the values of 0, 1, 2, or 3, so ℓ = 3 is allowed. When ℓ = 3, there can be seven values of m (−3, −2, −1, 0, 1, 2, 3), so m = 3 is allowed. ms of −12 is allowed. The correct answer to (a) is one.

The value for m lies outside of the allowed boundaries. Given an ℓ value of 3, the allowed values of m are as listed in the comments on (a). A value for 4 for m is not allowed. The correct answer to (b) is zero.

Problem #13: In an atom, what is the maximum number of electrons that can have the following quantum numbers?

(a) n = 6, ℓ = 4
(b) n = 6, ℓ = 4, m = −1

Solution to (a):

1) We need to determine how many m values are possible when ℓ = 4. Following the appropriate rule (which you should know by now!), m takes on:

−4, −3, −2, −1, 0, +1, +2, +3, +4

A total of nine different values. By the way, the positive signs don't have to be there. I just wanted to put them in.

2) For each of the nine possible m value, there are two allowed ms values:

+12 and −12

3) There can be 18 electrons which have n = 6 and ℓ = 4

Solution to (b):

ms can take on two different values: +12 and −12

There are 2 electrons which can have n = 6, ℓ = 4, m = −1.

Comment: the answer would be two if m were any of the other eight allowed values seen in the solution to (a).

Problem #14: Which of the following combinations of quantum numbers are allowed?

(a) n = 1, ℓ = 1, m = 0
(b) n = 3, ℓ = 0, m = 0
(c) n = 1, ℓ = 0, m = −1
(d) n = 2, ℓ = 1, m = 2

Solution:

The answer is (b). Here's how to figure it out:

You know this because the other combinations do not, at some point, follow the rules. Let's look at (a) as an example.

The rule for the ℓ quantum number is that it starts at zero and runs by integers up to n − 1. When n = 1, we start at zero but n − 1 equals zero. So, when n = 1, the only possible ℓ value is zero. Choice (a) has ℓ = 1 when n = 1. Not possible.

(c): m of −1 not possible when ℓ = 0 (only m = 0 is allowed when ℓ = 0)

(d): m = 2 not possible when ℓ = 0 (only values of −1, 0, 1 are permitted).

Only (b) follows all the rules for n, ℓ, and m.

Problem #15: The following sets of quantum numbers, listed in the order n, ℓ, m, and ms were written for the last electron added to an atom. Identify which sets are valid:

 n ℓ mℓ ms I. 2 1 0 +1⁄2 II. 2 2 -1 +1⁄2 III. 2 0 1 −1⁄2 IV. 4 2 2 +1⁄2

Which of the following sets of quantum numbers is/are allowed?

(a) I and III
(b) I and IV
(c) I, II, and III
(d) II, III, and IV
(e) They are all allowed.

Solution:

1) Scan all the n and ms for any disallowed numbers such as n = 0 or ms = +13.

All n and ms are correct.

2) Scan each n, ℓ pair for rule violations:

I. n = 2, ℓ = 1 is allowed. When n = 2, ℓ can equal 0 and 1.
II. n = 2, ℓ = 2 is not allowed. When n = 2, ℓ can equal 0 and 1, but not 2.
III. n = 2, ℓ = 0 is allowed.
IV. n = 4, ℓ = 2 is allowed. ℓ can take on the values of 0, 1, 2, 3 when n = 4.

3) We now examine I., III, and IV for ℓ, m violations:

I. When ℓ = 1, m can be −1, 0, 1. I is allowed.
III. When ℓ = 0, m can only be 0. III is not allowed.
IV. When ℓ = 2, m can be −2, −1, 0, 1, 2. IV is allowed.

4) I and IV are allowed. Answer choice (b).

Problem #16: Which of the following quantum number cannot be the same for an electron in the 2p orbital and one in the 3d orbital?

I. n
II. ℓ
III. m
IV. ms
(a) I only
(b) I and II only
(c) I, II, III
(d) I, II, IV
(e) I, II, III, IV

Solution:

For 2p, n = 2 and for 3d, n = 3 ---> n cannot be the same but all five answers have n in them, so can't eliminate anything.

For 2p, ms can be +12 or −12. For 3d, ms will take on the same values. That eliminates answers (d) and (e).

For 2p, ℓ = 1 and for 3d, ℓ = 2. That means the correct answer must have II in it. That eliminates (a), leaving (b) and (c) as the remaining possibilities.

For 2p, m = +1, 0, −1 and for 3d, m = +2, +1, 0, −1, −2. Some of those are the same, so we have to eliminate (c).

Problem #17: Which of the following is not a valid set of four quantum numbers? (Order ---> n, ℓ, m, ms) Why?

(a) 2, 1, 0, −12
(b) 3, 1, −1, −12
(c) 1, 0, 0, +12
(d) 2, 0, 0, +12
(e) 1, 1, 0, +12

Solution:

1) Scan all n and ms for disallowed values:

All n and ms values are correct.

2) Look at all n, ℓ pairings for disallowed values:

The pairings in (a) through (d) are all correct. However, (e) is not correct. When n = 1, the only possible ℓ value is 0.

3) Look at the ℓ, m pairings for (a) to (d):

All are correct.

4) This leaves (e) as the answer to the question. Its problem lies in the n, ℓ pair. When n = 1, the only possible ℓ is 0, since we must start at zero and go to 1 − 1, which is also zero.

Problem #18: Determine which sets of quantum numbers are correct and which are incorrect.

(a) 14, 9, −3, −12
(b) 9, 5, −1, 0
(c) 15, 2, -6, +12
(d) 7, 10, 0, +12
(e) 10, 9, 1, +34

Solution:

Instead of scanning n and ms values first, let's start with (a) and look at each one in turn.

(a) 14, 9, −3, −12 (correct)

n = 14 is an allowed value. An ℓ of 9 is allowed. With n = 14, the ℓ values would run from 0 to 13. With ℓ = 9, the m values would extend from -9 to 9, so −3 is allowed.−12 is an allowed ms value.

(b) 9, 5, −1, 0 (incorrect)

n = 9 is an allowed value. An ℓ of 5 is allowed because the allowed ℓ values would run from 0 to 8. With ℓ = 5, the m values would span −5 to +5. Therefore, −1 is an allowed value for m. However, ms equal to zero is not allowed.

(c) 15, 2, -6, +12 (incorrect)

n = 15 is allowed. ℓ would run from 0 to 14, so ℓ = 2 is allowed. However, ℓ = 2 does not allow for an m of −6. The only allowed values are −2, −1, 0, 1, 2.

(d) 7, 10, 0, +12 (incorrect)

n = 7 is allowed. However, ℓ = 10 is not allowed. The highest ℓ value allowed (when n = 7) is 6.

(e) 10, 9, 1, +34 (incorrect)

n = 10 is allowed. ℓ = 9 is allowed. m = 1 is allowed. ms = +34 is not allowed.

Problem #19: Classify each set of quantum numbers as possible or not possible for an electron in an atom.

 (a) 3, 2, −3, +1⁄2 (e) 3, 2, 0, −2 (b) 4, 3, −2, +1⁄2 (f) 4, 3, 4, −1⁄2 (c) −2, 1, 0, −1⁄2 (g) 2, 1, 0, +1⁄2 (d) 2, 2, 2, +1⁄2 (h) 4, 2, −2, +1⁄2

Solution:

1) Examine all the n and ms values for ones not allowed:

(c) has n = −2, which is not allowed.
(e) has ms = −2, which is not allowed.

2) Examine the n, ℓ pairs for ones not allowed:

(d) has a n, ℓ pair of 2, 2. The ℓ value of 2 is not allowed when n = 2.

3) Examine all the ℓ, m pairs for ones not allowed:

(a) has an m that is not allowed. When ℓ = 2, the m runs from −2 to 2. A value of −3 is not allowed.
(f) has an m of 4, when 3 is the maximum allowed value (based on ℓ = 3).

4) (b), (g), and (h) are all correct.

Problem #20: What is wrong with the following set of quantum numbers?

n = 2, ℓ = 2, m = 0, ms = +12

Solution:

ℓ cannot be 2 when n is 2. The d orbital (ℓ = 2) first appears in the third shell (n = 3).

Problem #21: Give the quantum numbers for all orbitals in the 5f subshell.

Solution:

1) The first step is to determine the value of n:

n = 5

That was easy. it comes from the 5 in 5f.

2) The values for ℓ are derived from n:

When n = 5, ℓ can take on the values of 0, 1, 2, 3, 4

Only one of the ℓ values is associated with the f subshell. It is 3.

Remember, s is associated with an ℓ of 0. The p subshell always has an ℓ equal to 1 while the d subshell is always associated with 2. When ℓ equals 4, this is associated with the g subshell. No known element has an electron in the 5g subshell.

3) Now we determine the m values when ℓ = 3:

m takes on the values of −3, −2, −1, 0, 1, 2, 3; a total of seven values.

The m values are associated with how many orbitals are present. In the 5f subshell, there are seven orbitals.

4) The quantum numbers of those seven 5f orbitals follows (the numbers are in n, ℓ, m order):

5, 3, −3
5, 3, −2
5, 3, −1
5, 3, 0
5, 3, 1
5, 3, 2
5, 3, 3

5) The question asked for the quantum numbers of the orbitals and this does not include the ms quantum number. The spin quantum number allows for two electrons per orbital; it plays no role in determining how many orbitals are present.

By the way, the first three quantum numbers were discovered quickly, in the period of about 1913 to 1915. It was not until about 1925 that Wolfgang Pauli realized that a fourth quantum number (now called ms) was required.

Problem #22: Which of the following set of quantum numbers (ordered n, ℓ, m, ms) are possible for an electron in an atom?

(a) 3, 2, 0, −2
(b) 3, 4, 0, +12
(c) 3, 1, 0, −12
(d) 4, 2, −1, −32
(e) 2, 1, −2, +12
(f) −1, 0, 0, −12
(g) 4, 2, 1, −12
(h) 2, 1, 3, +12

Solution:

 (a) 3, 2, 0, −2 not possible, ms must be +1⁄2 or −1⁄2 (b) 3, 4, 0, +1⁄2 not possible, ℓ cannot be greater than n (c) 3, 1, 0, −1⁄2 possible (d) 4, 2, −1, −3⁄2 not possible, ms must be +1⁄2 or −1⁄2 (e) 2, 1, −2, +1⁄2 not possible, mℓ cannot be −2 when ℓ = 1 (f) −1, 0, 0, −1⁄2 not possible, n must be 1, 2, 3... (g) 4, 2, 1, −1⁄2 possible (h) 2, 1, 3, +1⁄2 not possible, mℓ cannot be 3 when ℓ = 1

Problem #23: Which set of quantum numbers cannot occur together to specify an orbital?

(a) n = 3, ℓ = 2, m = 3
(b) n = 2, ℓ =1, m = −1
(c) n = 3, ℓ = 1, m = −1
(d) n = 4, ℓ = 3, m = 3

Solution:

I want to examine all four sets together, as opposed to individually.

1) First, the 'n' values:

Each is a positive integer, greater than zero. All are allowed values.

2) Second, the 'ℓ' values:

Each ℓ value falls within the allowed range. Here is a list:
 n ℓ 2 0, 1 3 0, 1, 2 4 0, 1, 2, 3

3) Third, the m values:

Remember the rule for m: the values are integers and run from -ℓ to zero to +ℓ. Here's a list for the three ℓ values that are dictating m values:
 ℓ mℓ 1 −1, 0, +1 2 −2, −1, 0, +1, +2 3 −3, −2, −1, 0, +1, +2, +3

The set that fails is (a). When ℓ = 2, m cannot take on a value of 3.

Problem #24: Identify which sets of quantum numbers are valid for an electron. Each set is ordered (n, ℓ, m, ms)

 (a) 3, 1, −1, +1⁄2 (g) 4, 3, 4, −1⁄2 (b) 3, 2, −1, 0 (h) 2, 1, 1, +1⁄2 (c) 3, 2, 1, +1⁄2 (i) 4, 3, 1, −1⁄2 (d) 3, −2, −2, −1⁄2 (j) 1, 0, 0, −1⁄2 (e) 2, 3, 1, +1⁄2 (k) 2, −1, 1, −1⁄2 (f) 1, 3, 0, +1⁄2 (ℓ) 0, 1, 1, −1⁄2

Solution:

Comment: you can scan for obvious flaws, such as values that clearly are wrong. I think (b), (d), (k), and (ℓ) would fall into this category.

(b) ---> the error is in ms
(ℓ) ---> the error is in n
(d) and (k) ---> the error is in ℓ (which can never be negative)
Other invalid answers will come about because of a comparison between two values, as opposed to a single number violating the rules (as in n cannot assume a value of zero). Let us compare!

(a) this set is allowed

When n= 3, ℓ can assume the values of 0, 1, 2. When ℓ = 1, the values of m can be −1, 0, 1. +12 is an allowed value for ms.

(b) this set is not allowed

ms cannot assume a value of zero.

(c) this set is allowed

When n = 3, the values for ℓ can range 0, 1, 2, so 2 is an allowed value. When ℓ = 2, the values for m will be −2, −1, 0, 1, 2, so 1 is allowed. +12 is allowed for ms

(d) this set is not allowed

When n = 3, the allowed values for ℓ are 0, 1, 2. A value of −2 is not allowed for ℓ.

In fact, negative values for ℓ are NEVER allowed.

(e) this set is not allowed

When n = 2, the allowed values for ℓ are 0, 1. A value of 3 is not allowed for ℓ when n = 2.

In order for an ℓ equal to 3, the value for n must be 4 or greater.

(f) this set is not allowed

When n = 1, the allowed value for ℓ is only zero. A value of 3 is not allowed for ℓ when n = 1.

In order for an ℓ equal to 3, the value for n must be 4 or greater.

(g) this set is not allowed

When n= 4, ℓ values can be 0, 1, 2, 3. However, when ℓ = 3, the allowed m values are −3, −2, −1, 0, 1, 2, 3. A value of 4 is not allowed when ℓ = 3.

(h) this set is allowed

When n = 2, the allowed ℓ values are zero and 1. When ℓ = 1, the allowed m values are −1, 0, 1. +12 is allowed for ms

(i) this set is allowed

When n = 4, the allowed ℓ values are 0, 1, 2, 3. When ℓ = 3, the allowed m values are −3, −2, −1, 0, 1, 2, 3. −12 is allowed for ms

(j) is allowed

No comment given.

(k) is not allowed

See comment at beginning of answers.

(ℓ) is not allowed

See comment at beginning of answers.

Problem #25: Which of the following set of quantum numbers (ordered n, ℓ, m, ms) are possible for an electron in an atom?

 (a) 4, 3, 4, −1⁄2 (e) 3, 2, −3, +1⁄2 (b) 2, 1, 0, +1⁄2 (f) 2, 2, 2, +1⁄2 (c) −2, 1, 0, −1⁄2 (g) 3, 2, 1, −1 (d) 5, 3, −3, +1⁄2 (h) 4, −2, 1, −1⁄2

Solution:

1) Scan the principal quantum number:

We see that (c) has n = −2. This is not a possible value for n. The values for n are always positive.

2) Scan the azimuthal quantum number:

We see that (h) has ℓ = −2. This is not a possible value for ℓ. The values for ℓ are always positive.

3) Scan the magnetic quantum number:

No values are found to be in error.

5) Scan the spin quantum number:

We see that (g) has ms = −1. This is not a possible value for ms.

6) Now, you have to scan combinations of (1) n & ℓ and (2) ℓ & m So, here we go:

(a) the n & ℓ combo is OK, but the ℓ & m combo is not possible. When ℓ = 3, m takes on the values of −3, −2, −1, 0, 1, 2, 3. An m of 4 is not allowed when m = 3

(b) 2, 1 for the n & ℓ combo is OK. 1, 0 for the ℓ & m is also OK. (b) is a possible set of quantum numbers for an electron in an atom.

(d) is a possible set of quantum numbers. When n = 5, ℓ can be 0, 1, 2, 3, 4. When ℓ = 3, m can be −3, −2, −1, 0, 1, 2, 3. The values of 5, 3, −3, +12 correctly meet all the requirements.

(e) has an error in the ℓ & m combo. When ℓ equals 2, the only allowed m values are −2, −1, 0, 1, 2. A m value of −3 in not possible with the specified value of ℓ being 2.

(f) has an error in the n & ℓ combo. When n = 2, the values for ℓ start at 0, then 1 and that's it. The ℓ values can only go as high as n − 1. A n & ℓ combo of 2, 2 is forbidden.

7) (b) and (d) are the only sets of quantum number possible. All the others have at least one error in them.

Problem #26: Determine the number of electrons that can be described by each of the given quantum numbers:

(a) n = 3, ℓ = 2
(b) n = 3, ℓ = 3, m = +1
(c) n = 4
(d) n = 4, ℓ = 3, m = +1, ms = +12
(e) n = 2, ℓ = 2, m = 2, ms = −12

Solution:

(a) n = 3, ℓ = 2

When ℓ = 2, m takes on 5 values (−2, −1, 0, +1, +2). Each one of those m values is associated with an orbital. Each orbital can hold two electrons (Remember, ms can take on two different values.). The answer is 10. (By the way, these is the set of 3d orbitals.)

(b) n = 3, ℓ = 3, m = +1

One specific m value means one orbital. The answer is 2 electrons.

By the way, when ℓ = 3, m can take on these values: −3, −2, −1, 0, +1, +2, +3

(c) n = 4

ℓ can equal 0, 1, 2, 3. Figure out how many m values with each ℓ value. Remember, each m represents an orbital and each orbital can hold 2 electrons.

Here's the start to answering (c):

When ℓ = 0, m can only equal 0 ---> one orbital
When ℓ = 1, m can equal −1, 0, +1 ---> three orbitals
When ℓ = 2, . . . .
When ℓ = 3, . . . .

(d) n = 4, ℓ = 3, m = +1, ms = +12

This set of 4 quantum numbers is correct. It represents 1 electron.

(e) n = 2, ℓ = 2, m = +2, ms = −12

This set of quantum numbers is incorrect. When n = 2, ℓ can only equal 0 or +1, not +2. This set represents zero electrons.

Bonus Problem #1: All of the following sets of quantum numbers (ordered n, ℓ, m, ms) are not possible for an electron in an atom. Identify the error in each one.

 (a) 0, 0, 0, 1⁄2 (e) −4, 3, 1, 1⁄2 (b) 2, 1, 0, −1 (f) 3, 2, 2, 1⁄3 (c) 5, 3, 4, 1⁄2 (g) 3, 1, 2, −1⁄2 (d) 3.5, 2, 1, −1⁄2 (h) 3, 2, −3, 1⁄2

Solution:

(a) 0, 0, 0, 12

The principal quantum number, n, must start at 1, not zero.

(b) 2, 1, 0, −1

Remember to look at n and ms first. When yo do that, you see that ms is given as −1. Thi is not an allowed value for ms.

(c) 5, 3, 4, 12

The fault here lies in the ℓ, m pairing. When ℓ = 3, the maximum m allowed is 3. The given value of 4 is not allowed.
(d) 3.5, 2, 1, −12
The principal quantum number, n, must be a positive integer, starting at 1. 3.5 is not an integer.

(e) −4, 3, 1, 12

The principal quantum number, n, must be a positive integer. −4 is not positive.

(f) 3, 2, 2, 13

The ms value is not one that is allowed.

(g) 3, 1, 2, −12

Look at the ℓ, m pair. When ℓ = 1, the m values can only be −1, 0, 1. A value of 2 is not allowed.

(h) 3, 2, −3, 12

The fault is with the ℓ, m pair. A −3 value for m is not allowed when ℓ = 2.

Bonus Problem #2: One QN in each set is not allowed. Identify the mistake and replace it with one that is allowed.

(a) n = 3, ℓ = 3, m = +2
(b) n = 2, ℓ = 1, m = −2
(c) n = 1, ℓ = 1, m = 0

Solution:

(a): The ℓ value is wrong, it must be less than n. Replace with ℓ = 2. The ℓ cannot be 0 or 1 because the m is +2. That value can only be generated with ℓ = 2.

(b): The m value is wrong, the absolute value of m must be less than or equal to ℓ. Replace with m = −1. 0 and +1 are also correct replacements.

(c): The ℓ value is wrong, replace with ℓ = 0. This is the only possible correct answer