### 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: Soon!