The Rules for Assigning Them

Problems #1 - 10

Examples and Problems only (no solutions)

**Problem #1:** Which of the following is a possible set of quantum numbers that describes an electron?

(a) n = 3, ℓ = 2, m _{ℓ}= −3, m_{s}= −^{1}⁄_{2}(d) n = 3, ℓ = 1, m _{ℓ}= −1, m_{s}= +^{1}⁄_{2}(b) n = 0, ℓ = 0, m

_{ℓ}= 0, m_{s}= +^{1}⁄_{2}(e) n = 4, ℓ = −3, m _{ℓ}= −1, m_{s}= +1(c) n = 4, ℓ = 2, m

_{ℓ}= −1, m_{s}= 0(f) none

**Solution:**

The solution is to look at each one and see if there is a violation of the rules for each quantum number. My technique is to look at n and m_{s} first.

1) Look at n and see that . . .

. . . (b) violates the rule that n starts at 1, so (b) is not the answer.

2) Look at m_{s} and see that . . .

. . . (c) and (e) both fail. m_{s}can only take on one of two values: +^{1}⁄_{2}or −^{1}⁄_{2}

3) Now you look at combos. The first choice is to examine n, ℓ pairs:

choice (a) ---> When n = 3, the ℓ values can be 0, 1, 2. So, (a) is good.

choice (d) ---> again, n = 3, so the ℓ value of 1 is an acceptable value.Nothing has been eliminated by looking at n, ℓ values as a pair, so we move on.

4) The next pairing to consider is the ℓ, m_{ℓ} pair:

choice (a) ---> when ℓ = 2, the m_{ℓ}values are allowed to be −2, −1, 0, 1, 2. The value in choice (a) is −3, which is not allowed.

5) Since choice (f) is none, we must examine our last remaining choice to see if it is possible or not.

answer choice (d):n = 3 ---> meets the rule for n

m_{s}---> meets the rule for m_{s}

n, ℓ ---> 3, 1 meets the rule

ℓ, m_{ℓ}---> meets the ruleThe answer to the question is choice (d).

**Problem #2:** Each electron orbital is characterized by 3 quantum numbers: n, ℓ, and m_{ℓ}.

n is known as the ____ quantum number.

ℓ is known as the ____ quantum number.

m_{ℓ}is known as the ____ quantum number.

**Solution:**

n is the principal quantum number

ℓ is the azimuthal quantum number

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

**Problem #3:** Each electron orbital is characterized by 3 quantum numbers: n, ℓ, and m_{ℓ}.

n specifies ___.

ℓ specifies ___.

m_{ℓ}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_{ℓ}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

**Solution:**

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

(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.(c) n = 3, ℓ = 2ℓ = 1 tells us that it will be the 'p' subshell.

The orbital designation is 2p.

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_{ℓ}?

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

**Solution:**

The rule for m_{ℓ}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_{ℓ}values for a given ℓ: 2ℓ + 1.(a) For ℓ = 5, the m

_{ℓ}values are −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, a total of eleven values.(b) Seven values of m

_{ℓ}resulting from 2(3) + 1 = 7(c) 2ℓ + 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, +^{1}⁄_{2}) and (4, 3, −1, +^{1}⁄_{2}).

**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

_{ℓ}values.The two electrons have the same spin (m

_{s}= +^{1}⁄_{2}).

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

(a) n = 2, ℓ = 1, m_{ℓ}= 1, m_{s}= +^{1}⁄_{2}

(b) n = 3, ℓ = 2, m_{ℓ}= 2, m_{s}= +^{1}⁄_{2}

(c) n = 4, ℓ = 1, m_{ℓ}= −1, m_{s}= −^{1}⁄_{2}

(d) n = 4, ℓ = 3, m_{ℓ}= 3, m_{s}= −^{1}⁄_{2}

(e) n = 5, ℓ = 0, m_{ℓ}= 0, m_{s}= +^{1}⁄_{2}

**Solution:**

The solution procedure involves looking at the n, ℓ pairings. The m_{ℓ}and m_{s}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_{ℓ}, m_{s}) are possible for an electron in an atom?

Select all that apply:

(a) 3, 2, 2, − ^{1}⁄_{2}(f) 5, 3, −3, + ^{1}⁄_{2}(b) 2, 1, 3, + ^{1}⁄_{2}(g) 3, 1, −2, − ^{1}⁄_{2}(c) −3, 2, 2, − ^{1}⁄_{2}(h) 5, 3, 0, + ^{1}⁄_{2}(d) 3, 3, 1, − ^{1}⁄_{2}(i) 3, 2, −1, ± ^{1}⁄_{2}(e) 3, 2, 1, −1 (j) 3, 2, −1, 0

**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 m_{s} values. Since this value can only be +^{1}⁄_{2} or −^{1}⁄_{2}, we can quickly remove any incorrect ones and not have to analyze their n, ℓ, and m_{ℓ} values.

We see that (e) with m_{s}= −1, (i) with m_{s}= ±^{1}⁄_{2}, and (j) with m_{s}= violate the rule for m_{s}. Remember, m_{s}can ONLY be EITHER positive^{1}⁄_{2}or negative^{1}⁄_{2}in the set of four quantum numbers, not any other value. And certainly 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_{ℓ}

In (b), there is a violation of the rule for generating permissible m_{ℓ}from the given ℓ. When ℓ = 1, the permitted values for m_{ℓ}are −1, 0, and 1. A value of 3 is not allowed. (g) shows the same mistake. When ℓ = 1, m_{ℓ}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_{ℓ}values to determine how many orbitals are present:by enumeration: 3, 2, 1, 0, −1, −2, −3 ---> 7 orbitals

by formula: 2ℓ + 1 leads to 2(3) + 1 = 7

**Problem #10:** The maximum number of electrons that can have principal quantum number n = 3 and spin −^{1}⁄_{2} is?

**Solution:**

We need to determine the total number of m_{ℓ}values possible. With n = 3, ℓ can be 0, 1, 2.

ℓ ---> 0 1 2 2ℓ + 1 ---> 1 3 5 <--- nine orbitals total Each orbital can have one electron with m

_{s}= −^{1}⁄_{2}Nine orbitals, each with one electron of −^{1}⁄_{2}gives the answer to be 9.And, just as a reminder, notice that the total number of m

_{ℓ}states can be given by the formula (2ℓ + 1).

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

(a) n = 4

(b) n = 5, m_{ℓ}= +1

(c) n = 5, m_{s}= +^{1}⁄_{2}

(d) n = 3, ℓ = 2

(e) n = 1, ℓ = 0, m_{ℓ}= 0

**Solution:**

(a) n = 4

The total number of m_{ℓ}values (derived from all possible ℓ states for a given n value) is given by n^{2}. This gives us 16 m_{ℓ}values when n = 4. Here is a listing of the m_{ℓ}values when n = 4:

ℓ m _{ℓ}total m _{ℓ}0 0 1 1 −1, 0, 1 3 2 −2, −1, 0, 1, 2 5 3 −3, −2, −1, 0, 1, 2, 3 7 4 −4, −3, −2, −1, 0, 1, 2, 3, 4 9 Since you also have to include m

_{s}, you multiply n^{2}by 2 to get the maximum amount of electrons in the entire energy level:2n^{2}2(4)

^{2}32

(b) n = 5, m_{ℓ} = 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_{ℓ}being equal to 1.When ℓ = 0, m

_{ℓ}can only equal 0. ℓ = 0 is not part of the correct answerWhen ℓ = 1, m

_{ℓ}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

_{ℓ}values:

ℓ = 2 m _{ℓ}= −2, −1, 0, 1, 2ℓ = 3 m _{ℓ}= −3, −2, −1, 0, 1, 2, 3ℓ = 4 m _{ℓ}= −4, −3, −2, −1, 0, 1, 2, 3, 4Each ℓ value has an m

_{ℓ}of 1 allowed.Four different orbitals (n, ℓ, m

_{ℓ}just below) are possible for n = 5 and m_{ℓ}= 1:

5, 1, 1 5, 2, 1 5, 3, 1 5, 4, 1 Each orbital can hold two electrons (m

_{s}= +^{1}⁄_{2}and m_{s}= −^{1}⁄_{2}), so the total number of electrons is 8.

(c) n = 5, m_{s} = +^{1}⁄_{2}

The problem is very similar to (b), except that m_{s}is given. This value corresponds to a single electron within an m_{ℓ}value. Therefore, we need to find out how many numbers of m_{ℓ}we have in order to know the number of max electrons.Like before, ℓ = 0, 1, 2, 3, 4. Since m

_{ℓ}is not specified we have to take all possible m_{ℓ}values for each ℓ value and add them together:

ℓ 2ℓ + 1 Total m _{ℓ}0 2(0) + 1 1 1 2(1) + 1 3 2 2(2) + 1 5 3 2(3) + 1 7 4 2(4) + 1 9 Notice that the total number of m

_{ℓ}states can be given by the formula (2ℓ + 1).Thus, the total number of m

_{ℓ}states is (1 + 3 + 5 + 7 + 9) = 25. This is the highest number of electrons with m_{s}= +^{1}⁄_{2}.You might be required to enumerate all the m

_{ℓ}values as opposed to just stating how many there are using the (2ℓ + 1) formula. In that case:

ℓ m _{ℓ}valuesTotal m _{ℓ}0 0 1 1 −1, 0, 1 3 2 −2, −1, 0, 1, 2 5 3 −3, −2, −1, 0, 1, 2, 3 7 4 −4, −3, −2, −1, 0, 1, 2, 3, 4 9

(d) n = 3, ℓ = 2

The only numbers not specified are m_{ℓ}and m_{s}. You have to determine all the possibilities for them:# of m_{ℓ}= 2ℓ + 1 = 2(2) + 1 = 5There will be two electrons per m

_{ℓ}state (m_{s}= +^{1}⁄_{2}and −^{1}⁄_{2}) therefore, maximum number of electrons is 5 x 2 = 10.

(e) n = 1, ℓ = 0, m_{ℓ} = 0

The only number not specified is m_{s}. Consequently, there are only two electrons that can have the 3 values given. The quantum numbers for those two electrons are:1, 0, 0, +^{1}⁄_{2}

1, 0, 0, −^{1}⁄_{2}

**Bonus Problem #2:** What is the Principal Quantum Number of the first shell to have:

(a) s orbitals?

(b) p orbitals?

(c) d orbitals?

(d) f orbitals?

**Discussion:**

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 to (a):**

The PQN of the first shell to have an 's' orbital is 1.Since the 's' orbital is associated with an ℓ equal to 0, apply the rule for ℓ to the PQN of 1 and get 0.

**Solution to (b):**

The PQN of the first shell to have a 'p' orbital is 2.Since the 'p' orbital is associated with an ℓ equal to 1, apply the rule for ℓ to the PQN of 2 and get 0 and 1. The 0 is associated with the 's' orbital in the second shell and the value of 1 is the 'p' orbital.

**Solution to (c):**

The PQN of the first shell to have a 'd' orbital is 3.Since the 'd' orbital is associated with an ℓ equal to 2, apply the rule for ℓ to the PQN of 3 and get 0, 1, and 2. The 0 is associated with the 's' orbital in the third shell and the value of 1 is associated with the 'p' orbital in the third shell. The ℓ of 2 denotes the 'd' orbital.

**Solution to (d):**

The PQN of the first shell to have an 'f' orbital is 4.Since the 'f' orbital is associated with an ℓ equal to 3, apply the rule for ℓ to the PQN of 4 and get 0, 1, 2, 3. The 0 is associated with the 's' orbital in the fourth shell, the value of 1 is associated with the 'p' orbital in the fourth shell. The ℓ of 2 denotes the 'd' orbital in the fourth shell and the ℓ of 3 signifies the first occurrence of 'f' orbitals.