### Quantum Number Problems#11 - 25

Problem #11: What is the maximum number of electrons that can be identified with the following set of quantum numbers? n = 4, ℓ = 0, m = 0, ms = +½

Solution:

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.

Problem #12: How many electrons can have n = 3, m = +2, ms = +½?

Solution:

One electron.

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

3, 2, 2, +½

Problem #13: How many electrons can have n = 3, m = 0, ms = +½?

Solution:

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 = +½. Three electrons is the answer.

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 +½ II. 2 2 -1 +½ III. 2 0 1 -½ IV. 4 2 2 +½

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 +½ or -½. For 3d, ms will take on the same values. So, 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, -½
(b) 3, 1, -1, -½
(c) 1, 0, 0, +½
(d) 2, 0, 0, +½
(e) 1, 1, 0, +½

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.

(e) is the answer to the question. Its problem lies in the n, ℓ pair.

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

(a) 14, 9, -3, -½
(b) 9, 5, -1, 0
(c) 15, 2, -6, ½
(d) 7, 10, 0, ½
(e) 10, 9, 1, ¾

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, -½ (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. -½ 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, m0 equal to zero is not allowed.

(c) 15, 2, -6, ½ (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, ½ (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, ¾ (incorrect)

n = 10 is allowed. ℓ = 9 is allowed. m = 1 is allowed. ms = ¾ 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, ½ (e) 3, 2, 0, -2 (b) 4, 3, -2, ½ (f) 4, 3, 4, -½ (c) -2, 1, 0, -½ (g) 2, 1, 0, ½ (d) 2, 2, 2, ½ (h) 4, 2, -2, ½

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 = ½

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, ½
(c) 3, 1, 0, -½
(d) 4, 2, -1, -32
(e) 2, 1, -2, ½
(f) -1, 0, 0, -½
(g) 4, 2, 1, -½
(h) 2, 1, 3, ½

Solution:

(a) 3, 2, 0, -2 -- not possible, ms must be +½ or -½
(b) 3, 4, 0, ½ -- not possible, ℓ cannot be greater than n
(c) 3, 1, 0, -½ -- possible
(d) 4, 2, -1, -32 -- not possible, ms must be +½ or -½
(e) 2, 1, -2, ½ -- not possible, m cannot be -2 when ℓ = 1
(f) -1, 0, 0, -½ -- not possible, n must be 1, 2, 3...
(g) 4, 2, 1, -½ -- possible
(h) 2, 1, 3, ½ -- not possible, m 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.