The Rules for Assigning Them

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_{ℓ} = 0, m_{s} = +^{1}⁄_{2}

**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

_{ℓ}, m_{s}) uniquely describes one electron.

**Problem #12:** How many electrons can have n = 3, m_{ℓ} = +2, m_{s} = +^{1}⁄_{2}?

**Solution:**

One electron.Based on the m

_{ℓ}of +2, the ℓ must be 2, so the full set of quantum numbers is:3, 2, 2, +^{1}⁄_{2}

**Problem #13:** How many electrons can have n = 3, m_{ℓ} = 0, m_{s} = +^{1}⁄_{2}?

**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

_{ℓ}= 0.In each of those orbitals, (the 3s, 3p, and 3d) there can exist one electron with m

_{s}= +^{1}⁄_{2}. Three electrons is the answer.

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

(a) n = 1, ℓ = 1, m_{ℓ}= 0

(b) n = 3, ℓ = 0, m_{ℓ}= 0

(c) n = 1, ℓ = 0, m_{ℓ}= −1

(d) n = 2, ℓ = 1, m_{ℓ}= 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

_{ℓ}of −1 not possible when ℓ = 0 (only m_{ℓ}= 0 is allowed when ℓ = 0)(d): m

_{ℓ}= 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

_{ℓ}.

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

n ℓ m _{ℓ}m _{s}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 m_{s} for any disallowed numbers such as n = 0 or m_{s} = +^{1}⁄_{3}.

All n and m_{s}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_{ℓ} violations:

I. When ℓ = 1, m_{ℓ}can be −1, 0, 1. I is allowed.

III. When ℓ = 0, m_{ℓ}can only be 0. III is not allowed.

IV. When ℓ = 2, m_{ℓ}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_{ℓ}

IV. m_{s}

(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, m

_{s}can be +^{1}⁄_{2}or −^{1}⁄_{2}. For 3d, m_{s}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

_{ℓ}= +1, 0, −1 and for 3d, m_{ℓ}= +2, +1, 0, −1, −2. Some of those are the same, so we have to eliminate (c).The correct answer is (b).

**Problem #17:** Which of the following is not a valid set of four quantum numbers? (Order ---> n, ℓ, m_{ℓ}, m_{s}) Why?

(a) 2, 1, 0, −^{1}⁄_{2}

(b) 3, 1, −1, −^{1}⁄_{2}

(c) 1, 0, 0, +^{1}⁄_{2}

(d) 2, 0, 0, +^{1}⁄_{2}

(e) 1, 1, 0, +^{1}⁄_{2}

**Solution:**

1) Scan all n and m_{s} for disallowed values:

All n and m_{s}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_{ℓ} 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, −^{1}⁄_{2}

(b) 9, 5, −1, 0

(c) 15, 2, -6, +^{1}⁄_{2}

(d) 7, 10, 0, +^{1}⁄_{2}

(e) 10, 9, 1, +^{3}⁄_{4}

**Solution:**

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

(a) 14, 9, −3, −^{1}⁄_{2} (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_{ℓ}values would extend from -9 to 9, so −3 is allowed.−^{1}⁄_{2}is an allowed m_{s}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_{ℓ}values would span −5 to +5. Therefore, −1 is an allowed value for m_{ℓ}. However, m_{s}equal to zero is not allowed.

(c) 15, 2, -6, +^{1}⁄_{2} (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_{ℓ}of −6. The only allowed values are −2, −1, 0, 1, 2.

(d) 7, 10, 0, +^{1}⁄_{2} (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, +^{3}⁄_{4} (incorrect)

n = 10 is allowed. ℓ = 9 is allowed. m_{ℓ}= 1 is allowed. m_{s}= +^{3}⁄_{4}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 m_{s} values for ones not allowed:

(c) has n = −2, which is not allowed.

(e) has m_{s}= −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_{ℓ} pairs for ones not allowed:

(a) has an m_{ℓ}that is not allowed. When ℓ = 2, the m_{ℓ}runs from −2 to 2. A value of −3 is not allowed.

(f) has an m_{ℓ}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_{ℓ}= 0, m_{s}= +^{1}⁄_{2}

**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 = 5That 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, 4Only 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_{ℓ} values when ℓ = 3:

m_{ℓ}takes on the values of −3, −2, −1, 0, 1, 2, 3; a total of seven values.The m

_{ℓ}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_{ℓ} 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 m_{s} 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 m_{s}) was required.

**Problem #22:** Which of the following set of quantum numbers (ordered n, ℓ, m_{ℓ}, m_{s}) are possible for an electron in an atom?

(a) 3, 2, 0, −2

(b) 3, 4, 0, +^{1}⁄_{2}

(c) 3, 1, 0, −^{1}⁄_{2}

(d) 4, 2, −1, −^{3}⁄_{2}

(e) 2, 1, −2, +^{1}⁄_{2}

(f) −1, 0, 0, −^{1}⁄_{2}

(g) 4, 2, 1, −^{1}⁄_{2}

(h) 2, 1, 3, +^{1}⁄_{2}

**Solution:**

(a) 3, 2, 0, −2 -- not possible, m_{s}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, m_{s}must be +^{1}⁄_{2}or −^{1}⁄_{2}

(e) 2, 1, −2, +^{1}⁄_{2}-- not possible, m_{ℓ}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, m_{ℓ}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_{ℓ}= 3

(b) n = 2, ℓ =1, m_{ℓ}= −1

(c) n = 3, ℓ = 1, m_{ℓ}= −1

(d) n = 4, ℓ = 3, m_{ℓ}= 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_{ℓ} values:

Remember the rule for m_{ℓ}: 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_{ℓ}values:

ℓ m _{ℓ}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

_{ℓ}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_{ℓ}, m_{s})

(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 mOther 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!_{s}

(ℓ) ---> the error is in n

(d) and (k) ---> the error is in ℓ (which can never be negative)

(a) this set is allowed

When n= 3, ℓ can assume the values of 0, 1, 2. When ℓ = 1, the values of m_{ℓ}can be −1, 0, 1. +^{1}⁄_{2}is an allowed value for m_{s}.

(b) this set is not allowed

m_{s}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_{ℓ}will be −2, −1, 0, 1, 2, so 1 is allowed. +^{1}⁄_{2}is allowed for m_{s}

(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_{ℓ}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_{ℓ}values are −1, 0, 1. +^{1}⁄_{2}is allowed for m_{s}

(i) this set is allowed

When n = 4, the allowed ℓ values are 0, 1, 2, 3. When ℓ = 3, the allowed m_{ℓ}values are −3, −2, −1, 0, 1, 2, 3. −^{1}⁄_{2}is allowed for m_{s}

(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_{ℓ}, m_{s}) 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 m_{s}= −1. This is not a possible value for m_{s}.

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

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

_{ℓ}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

_{ℓ}can be −3, −2, −1, 0, 1, 2, 3. The values of 5, 3, −3, +^{1}⁄_{2}correctly meet all the requirements.(e) has an error in the ℓ & m

_{ℓ}combo. When ℓ equals 2, the only allowed m_{ℓ}values are −2, −1, 0, 1, 2. A m_{ℓ}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.

**Bonus Problem:** 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_{ℓ}= −1

**Solution to (a):**

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

−4, −3, −2, −1, 0, +1, +2, +3, +4A 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_{ℓ} value, there are two allowed m_{s} values:

+^{1}⁄_{2}and −^{1}⁄_{2}

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

**Solution to (b):**

m_{s}can take on two different values: +^{1}⁄_{2}and −^{1}⁄_{2}There are 2 electrons which can have n = 6, ℓ = 4, m

_{ℓ}= −1.Comment: the answer would be two if m

_{ℓ}were any of the other eight allowed values seen in the solution to (a).