The Rules for Assigning Them

Fifteen Examples

Just below, I am going to introduce four rules for assigning quantum numbers. Many problems in this area will ask you to identify an incorrect set of quantum numbers. Below the discussion that introduces the rules, I have 10 examples with solutions. Keep the rules close at hand as you go through the examples and my solutions.

There are four quantum numbers; their symbols are n, ℓ, m_{ℓ} and m_{s}. EVERY electron in an atom has a specific, unique set of these four quantum numbers. The story behind how these numbers came to be discovered is a complex one. Articles and books about those events in the early to mid-1900's are still being published today.

A warning before I proceed: this is a 100% non-mathematical discussion. The equations governing electron behavior in atoms are complex. This area of study generated LOTS of Nobel Prizes and the reasoning leading to the above mentioned equations is sophisticated and sometimes quite subtle.

Just keep this in mind: EVERY electron's behavior in an atom is governed by a set of equations and that n, ℓ, m_{ℓ}, and m_{s} are values in those equations. EVERY electron in an atom has a unique set of quantum numbers.

Lastly, I'm going to let the Internet discuss what these numbers mean. I will just describe their existence and the rules for how to determine them in this tutorial. The next tutorial will start with hydrogen and assign quantum numbers to its electron, then proceed to helium and do the same, then lithium, beryllium, and so on.

Lastly, the quantum numbers can be grouped into shells, subshells and orbitals. For example, there are three 3p orbitals and that all have n = 3 and ℓ = 2. There is a 4f subshell and it has seven orbitals. The 4f subshell has n = 4 and ℓ = 3. Examples of this will play out in the examples and problems.

**I. The Principal Quantum Number (signified by the letter 'n'):** This quantum number was the first one discovered and it was done so by Niels Bohr in 1913. Bohr thought that each electron was in its own unique energy level, which he called a "stationary state," and that each electron would have a unique value of 'n.'

In this idea, Bohr was wrong. It very quickly was discovered that more than one electron could have a given 'n' value. For example, it was eventually discovered that when n = 3, eighteen different electrons could have that value of n.

Keep in mind that it is the set of four quantum numbers that is important. As you will see, each of the 18 electrons just mentioned will have its own unique set of n, ℓ, m_{ℓ}, and m_{s}.

Finally, there is a rule for what values 'n' can assume. It is:

n = 1, 2, 3, and so on.

n will always be a whole number and NEVER less than one.

One point: n does not refer to any particular location in space or any particular shape. It is one component (of four) that will uniquely identify each electron in an atom.

**II. The Azimuthal (or Angular Momentum) Quantum Number (signified by the letter 'ℓ'):** about 1914-1915, Arnold Sommerfeld realized that Bohr's 'n' was insufficient. In other words, more equations were needed to properly describe how electrons behaved. In fact, Sommerfeld realized that TWO more quantum numbers were needed.

The first of these is the quantum number signified by 'ℓ.' When Sommerfeld started this work, he used n' (n prime), but he shifted it to 'ℓ' after some years. I'm not sure why, but it seems easier to print ℓ than n prime and what if the printer (of a textbook) accidently dumps a few prime symbols, leaving just the letter 'n?' Ooops!

The rule for selecting the proper values of 'ℓ' is as follows:

ℓ = 0, 1, 2, . . . , n − 1

ℓ will always be a whole number and will NEVER be as large as the 'n' value it is associated with.

Comment: I have chosen to use a script letter ℓ rather than 'l.' Please be aware that, on the Internet, many, many people will use 'l' rather than 'ℓ.' Be aware of this so as not to be confused.

**III. The Magnetic Quantum Number (signified by m _{ℓ}):** this quantum number was also discovered by Sommerfeld in the same 1914-1915 time frame. I don't think he discovered one and then the other, I think that him realizing the need for two runs together somewhat. I could be wrong in this, so don't take my word for it!

The rule for selecting m_{ℓ} is as follows:

m_{ℓ}starts at negative 'ℓ,' runs by whole numbers to zero and then goes by whole numbers to positive 'ℓ.'

For example, when ℓ = 2, the m_{ℓ} values generated are −2, −1, 0, +1, +2, for a total of five values.

**IV. The Spin Quantum Number (signified by m _{s}):** spin is a property of electrons that is not related to a sphere spinning. It was first thought to be this way, hence the name spin, but it was soon realized that electrons cannot spin on their axis like the Earth does on its axis. If the electron did this, its surface would be moving at about ten times the speed of light (if memory serves correctly!). In any event, the electron's surface would have to move faster than the speed of light and this isn't possible.

In 1925, Wolfgang Pauli demonstrated the need for a fourth quantum number. He closed the abstract to his paper this way:

"On the basis of these results one is also led to a general classification of every electron in the atom by the principal quantum numbernandtwoauxiliary quantum numbersk_{1}andk_{2}to which is added a further quantum numberm_{ℓ}in the presence of an external field. In conjunction with a recent paper by E. C. Stoner this classification leads to a general quantum theoretical formulation of the completion of electron groups in atoms."

In late 1925, two young researchers named George Uhlenbeck and Samuel Goudsmit discovered the property of the electron responsible for the fourth quantum number being needed and named this property spin.

The rule for selecting m_{s} is as follows:

after the n, ℓ and m_{ℓ}to be used have been determined, assign the value +^{1}⁄_{2}to one electron, then assign −^{1}⁄_{2}to the next electron, while using the same n, ℓ and m_{ℓ}values.

For example, when n, ℓ, m_{ℓ} = 1, 0, 0; the first m_{s} value used is +^{1}⁄_{2}. However a second electron can also have n, ℓ, m_{ℓ} = 1, 0, 0; so assign −^{1}⁄_{2} to it.

Here's a table summarizing the rules:

Quantum NumberSymbolPossible ValuesPrincipal n 1, 2, 3, 4, . . . Angular Momentum ℓ 0, 1, 2, 3, . . . , (n − 1) Magnetic m _{ℓ}−ℓ, . . . , −1, 0, 1, . . . , ℓ Spin m _{s}+ ^{1}⁄_{2}, −^{1}⁄_{2}

**Example #1:** Assign a correct set of four quantum numbers for the valence electron in a sodium atom.

**Solution:**

1) Sodium has a total of eleven electrons and one of them is the sole valence electron that sodium has. I propose to assign all ten sets of quantum numbers and build up to the eleventh set, which will be the answer to the question.

2) When we set n = 1, we find that there are two electrons accounted for:

1, 0, 0, +^{1}⁄_{2}

1, 0, 0, −^{1}⁄_{2}These two electrons are in the 1s orbital. s orbitals are always characterized by ℓ and m

_{ℓ}equalling zero.

3) When we set n = 2, this will generate eight sets of quantum numbers:

The 2s orbital:2, 0, 0, +^{1}⁄_{2}

2, 0, 0, −^{1}⁄_{2}The 2p subshell (composed of three 2p orbitals):

2, 1, −1, +^{1}⁄_{2}

2, 1, −1, −^{1}⁄_{2}2, 1, 0, +

^{1}⁄_{2}

2, 1, 0, −^{1}⁄_{2}2, 1, 1, +

^{1}⁄_{2}

2, 1, 1, −^{1}⁄_{2}

3) When we set n = 3, there are eighteen possible quantum number sets. We need only the first of them:

3, 0, 0, +^{1}⁄_{2}This is the answer. By convention, the positive m

_{s}is used first.This electron is in the 3s orbital.

**Example #2:** An electron cannot exist in the energy state described by which set of quantum numbers below?

(a) 3, 2, 2, −^{1}⁄_{2}

(b) 4, 3, 3, +^{1}⁄_{2}

(c) 2, 1, −3, +^{1}⁄_{2}

(d) 2, 0, 0, −^{1}⁄_{2}

(e) 1, 0, 1, −^{1}⁄_{2}

**Solution:**

1) The first step is to scan all the n and m_{s} values, looking for anything that is not +^{1}⁄_{2} or −^{1}⁄_{2}:

All n values are correct.All m

_{s}values are correct.

2) The next step is to scan the n, ℓ pairs. You need to know the rule for ℓ and look to see if a pair violates the rule when compared to the n value.

All the n and ℓ pairs are correct. For example, look at (c). When n = 2, the possible ℓ values are 0, 1, and 2. (c) has a correct n, ℓ pair.

3) Now, we scan the ℓ, m_{ℓ} pairs (this is where the incorrect answer will be). You need to know the rule for m_{ℓ} and look to see if a pair violates the rule when compared to the ℓ value.

Choice (c) is incorrect. When ℓ = 1, the possible m_{ℓ}values are −1, 0, and +1. −3 is not an allowed value when ℓ = 1.

4) I copied this question from Yahoo Answers and the person who answered it gave this as an answer:

Answer is c ---> −3 not possible with PQN 2Notice that this method is comparing the n value to the m

_{ℓ}. The absolute value for m_{ℓ}will always be lesser than n, never greater than or equal to n. In this question, the absolute value of the m_{ℓ}value is 3, which is greater than the n value of 2.Look at this set:

n = 2, ℓ = 1, m_{ℓ}= −2, m_{s}= −^{1}⁄_{2}Looking at the n and m

_{ℓ}pairing, we find that the absolute values are equal. It is an incorrect set of quantum numbers.

**Example #3:** Indicate which of the following quantum states are allowed and which are
disallowed under the rules governing the electronic structure of atoms.

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

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

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

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

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

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

**Solution:**

1) A good starting step in a problem like this is to look at the n and m_{s} values first. The rule for m_{s}:

m_{s}can only take on values of +^{1}⁄_{2}and −^{1}⁄_{2}Now, you look for any that violate that rule, to find:

(d) n = 3, ℓ = 2, m

_{ℓ}= 2, m_{s}= +^{1}⁄_{3}We don't care what the other values in (d) are, the presence of the +

^{1}⁄_{3}makes choice (d) disallowed.Now, scan the n values. Sometimes a value of 0 is used when the question is asking for incorrect sets. In this question, all the n values are allowed.

2) Next, compare n and ℓ values in each remaining choice. Here are the two rules:

n = 1, 2, 3, . . . (all integer values)

ℓ values range from zero to n − 1 (all integer values)The key is that ℓ must ALWAYS be smaller than n. See choice (b):

(b) n = 3, ℓ = 3, m

_{ℓ}= −2, m_{s}= −^{1}⁄_{2}When n equals 3, then the allowed ℓ values will be 0, 1, and end with 2. Three (3) would not be included.

3) Now, compare ℓ and m_{ℓ} values. Here is the m_{ℓ} rule:

m_{ℓ}ranges from −ℓ to zero to +ℓ (all integer values)The point is that the absolute value of m

_{ℓ}cannot be greater than that of ℓ. Look at (e):(e) n = 2, ℓ = 1, m

_{ℓ}= −2, m_{s}= −^{1}⁄_{2}When ℓ = 1, the allowed m

_{ℓ}values are −1, 0, +1. So m_{ℓ}= −2 is disallowed.

4) (a), (c), and (f) are the allowed sets. Compare them step-by-step to the rules for n, ℓ, m_{ℓ}, and m_{s} to verify this.

**Example #4:** Explain why each of the following sets of quantum numbers wouldn't be permissible for an electron according to the rules for quantum numbers.

(a) n = 1, ℓ = 0, m_{ℓ}= 0, m_{s}= +1

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

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

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

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

**Solution:**

1) First, a mention of the rules for the four quantum numbers:

n = principal quantum number = major energy level

Values are 1, 2, 3, . . .ℓ = azimuthal quantum number = energy sublevel

Values are 0 to n − 1.m

_{ℓ}= magnetic quantum number = the orbital in the sublevel

Values are −ℓ, . . . , 0, . . . , +ℓm

_{s}= spin quantum number = electron in orbital

Values are +^{1}⁄_{2}or −^{1}⁄_{2}

2) The problem with (a) is in the m_{s} value:

The error is that m_{s}= +1. The m_{s}value can only be +^{1}⁄_{2}or −^{1}⁄_{2}.

3) The problem with (b) lies in the relationship between n and ℓ:

Since n = 1 in (b), the value for ℓ MUST be 0. Remember, ℓ starts at 0 and goes up by integers to n − 1. So, we start at 0 and work our way up to 1 minus 1, which also equals 0. With n = 1, there is only one possible ℓ value and it is 0.This set also fails a comparison of n and m

_{ℓ}. The absolute value for m_{ℓ}is always smaller than that for n.

4) The problem with (c) lies in the relationship between ℓ and m_{ℓ}:

Since ℓ = 2, the values for m_{ℓ}will be −2, −1, 0, +1, +2. The error in (c) is that m_{ℓ}is 3, when the maximum m_{ℓ}(when ℓ = 2) can only be 2.Choice (c) also fails the n and m

_{ℓ}comparison. The absolute value for m_{ℓ}is always smaller than n, never equal or larger.

5) The problem with (d) is in the n value:

n values start with 1, not 0.

6) The problem with (e) is that it fails in its m_{s} value.

m_{s}is allowed to only be values of +^{1}⁄_{2}or −^{1}⁄_{2}. +^{3}⁄_{2}fails this rule.

**Example #5:** A hydrogen atom has n = 5 and m_{ℓ} = −2. What are the possible values for ℓ in this orbital?

**Solution:**

With n = 5, the possible ℓ values are 0, 1, 2, 3, 4.Since m

_{ℓ}ranges from −ℓ to +ℓ, we need all ℓ that generate a −2.When ℓ = 0, m

_{ℓ}can only equal 0. When ℓ = 1, the most negative m_{ℓ}generated is −1. When ℓ = 0 or 1, an m_{ℓ}value of −2 is not generated.ℓ values of 2, 3, and 4 will each include an m

_{ℓ}value equal to −2Here's one example:

When ℓ = 3, m_{ℓ}values are 3, 2, 1, 0, −1, −2, −3.

**Example #6:** Which of the following is a possible set of quantum numbers in an atom?

(a) 3, 2, −1, +1

(b) 3, 3, −1, +^{1}⁄_{2}

(c) 3, 1, −2, −^{1}⁄_{2}

(d) 3, 1, 0, +^{1}⁄_{2}

**Solution:**

You have to analyze each one according the rules for assigning quantum numbers.

(a) 3, 2, −1, +1 <--- incorrect because m_{s}is either +^{1}⁄_{2}or −^{1}⁄_{2}, not +1(b) 3, 3, −1, +

^{1}⁄_{2}<--- incorrect because ℓ takes on values up to n − 1. So, when n = 3, the permissible values of ℓ are 0, 1, and 2.(c) 3, 1, −2, −

^{1}⁄_{2}<--- incorrect because m_{ℓ}takes on values from −ℓ to +ℓ, by integers. When ℓ = 1, m_{ℓ}takes on the values of 1, 0, and −1. A m_{ℓ}value of −2 is not permitted when ℓ = 1(d) 3, 1, 0, +

^{1}⁄_{2}<--- correct. It fits all the rules for n, then ℓ, then m_{ℓ}, then m_{s}

**Example #7:** An orbital has n = 4 and m_{ℓ} = −1. What are the possible values of ℓ for this orbital?

**Solution:**

All possible ℓ values range from 0 to n − 1 by integers, so:0, 1, 2, 3for all possible ℓ values.

We look for cases when m

_{ℓ}= −1When ℓ − 0, the only possible m

_{ℓ}value is 0.ℓ = 1 is the first value that generates m

_{ℓ}equal to −1. This is because m_{ℓ}values range from −ℓ to 0 to +ℓ by integers. So, with ℓ = 1, we'd have −1, 0, +1 for our m_{ℓ}valuesThe second ℓ value that generates m

_{ℓ}= −1 is 2. This is because, when ℓ = 2, we have m_{ℓ}values of −2, −1, 0 +1, +2.The third ℓ is 3. It generates m

_{ℓ}values of −3, −2, −1, 0, +1, +2, +3So, our answer is 1, 2, 3 for the ℓ values. The only one eliminated is 0.

**Example #8:** In potassium how many electrons will have ℓ = 0 as one of its quantum numbers.

**Solution:**

1) ℓ = 0 is associated with the s orbital and each principal quantum number has an ℓ = 0, so 4 occurences in K gives 2 in the first shell, 2 in the second shell, 2 in the third and 1 in the 4th. Total = 7. The quantum number sets are:

1, 0, 0, + ^{1}⁄_{2}2, 0, 0, + ^{1}⁄_{2}3, 0, 0, + ^{1}⁄_{2}4, 0, 0, + ^{1}⁄_{2}1, 0, 0, − ^{1}⁄_{2}2, 0, 0, − ^{1}⁄_{2}3, 0, 0, − ^{1}⁄_{2}2) ℓ = 0 is associated with one of the three p orbitals. In potassium, the second shell (n = 2) has a p orbital occupied. The third shell (n = 3) also has a p occupied. Two electrons in each for a total of 4. The quantum number sets are:

2, 1, 0, + ^{1}⁄_{2}3, 1, 0, + ^{1}⁄_{2}2, 1, 0, − ^{1}⁄_{2}3, 1, 0, − ^{1}⁄_{2}By the way, K does have an electron in the 4th shell, but it's not in a p orbital. It's in that s orbital of the 4th shell, discussed just above.

3) 11 of the 19 electrons have ℓ = 0

The other 8 electrons have ℓ = −1 (four of them, two in 2nd shell and 2 in 3rd shell) and ℓ = +1 (four of them, two in 2nd shell and 2 in 3rd shell). Writing out these eight quantum number sets is left as an exercise for the student.

**Example #9:** In a single atom, what is the maximum number of electrons that can have the quantum numbers n = 4 and m_{ℓ} = 2

**Solution:**

1) Brief restatement of the quantum number rules:

n is an integer and can range from 1 to infinity

ℓ is an integer and can range from 0 to n − 1

m_{ℓ}is an integer and can range from −ℓ to +ℓ

m_{s}is either +^{1}⁄_{2}or −^{1}⁄_{2}

2) Given that n = 4, we know that ℓ has four permissible values (0, 1, 2, 3). We need the ℓ values that will generate an m_{ℓ} of 2. Let us look at each ℓ in turn:

ℓ = 0Only an m_{ℓ}of 0 can be generated. ℓ = 0 is not part of the answer.ℓ = 1

Three m_{ℓ}values are generated when ℓ = 1. They are −1, 0 , and 1. Since a 2 cannot be generated, ℓ = 1 is not part of the answer.ℓ = 2

The m_{ℓ}values generated are −2, −1, 0, 1, and 2. Since a 2 is generated, this will become part of the correct answer.ℓ = 3

Since a 2 for the m_{ℓ}is also generated here, this is the other part of the correct answer. The m_{ℓ}values generated are −3, −2, −1, 0, 1, 2, and 3.

3) There are two sets of n, ℓ, m_{ℓ} values in the answer. They are:

4, 2, 2 and 4, 3, 2However, we are not yet done.

4) Each orbital described above (the 4, 2, 2 and 4, 3, 2 values) can hold two electrons each. So, this is the answer to the question:

the maximum number of electrons that can have the quantum numbers n = 4 and m_{ℓ}= 2 is four.

5) Here are their quantum number sets:

4, 2, 2, +^{1}⁄_{2}

4, 2, 2, −^{1}⁄_{2}

4, 3, 2, +^{1}⁄_{2}

4, 3, 2, −^{1}⁄_{2}

**Example #10:** Determine which set(s) of quantum numbers is NOT allowed:

(a) n = 5, ℓ = 3, m_{ℓ}= −1, m_{s}= +^{1}⁄_{2}

(b) n = 1, ℓ = 0, m_{ℓ}= 0, m_{s}= −^{1}⁄_{2}

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

(d) n = 4, ℓ = 1, m_{ℓ}= 0, m_{s}= −^{1}⁄_{2}

(e) n = 6, ℓ = 4, m_{ℓ}= −3, m_{s}= +^{1}⁄_{2}

**Solution:**

1) (a) is allowed:

For n = 5, ℓ = 0, 1, 2, 3, 4. Thus, ℓ = 3 is allowed.

For ℓ = 3, m_{ℓ}= −3, −2, −1, 0, 1, 2, 3. Thus m_{ℓ}= −1 is allowed.

m_{s}= +^{1}⁄_{2}is allowed.

2) (b) is allowed:

For n = 1, ℓ = 0 only. Thus, ℓ = 0 is allowed.

For ℓ = 0, m_{ℓ}= 0 only. Thus m_{ℓ}= 0 is allowed.

m_{s}= −^{1}⁄_{2}is allowed.

3) (c) is NOT allowed:

For n = 2, ℓ = 0, 1. Thus, ℓ = 2 is NOT allowed.

4) (d) is allowed:

For n = 4, ℓ = 0, 1, 2, 3. Thus, ℓ = 1 is allowed.

For ℓ = 1, m_{ℓ}= −1, 0, 1. Thus m_{ℓ}= 0 is allowed.

m_{s}= −^{1}⁄_{2}is allowed.

5) (e) is allowed:

For n = 6, ℓ = 0, 1, 2, 3, 4, 5. Thus, ℓ = 4 is allowed.

For ℓ = 4, m_{ℓ}= −4, −3, −2, −1, 0, 1, 2, 3, 4. Thus m_{ℓ}= −3 is allowed.

m_{s}= +^{1}⁄_{2}is allowed.

**Example #11:** All the following sets of quantum numbers describe nonexistent orbitals. Find the mistake in each one.

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

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

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

(d) n = 5, ℓ = 3, m_{ℓ}= −2, m_{s}= −1

**Solution:**

(a) n cannot equal zero. It starts at 1 and goes up by integers from there.(b) n is correct, but ℓ is wrong. ℓ starts at zero and goes up by integers to n − 1.

(c) m

_{ℓ}is incorrect. When ℓ = 2, m_{ℓ}can only take on these five values: −2, −1, 0, 1, 2.(d) m

_{s}is incorrect. Values of +^{1}⁄_{2}and −^{1}⁄_{2}are the only two values allowed.

**Example #12:** An electron in an atom is in the n = 3 and ℓ = 1 quantum state. Identify the possible values of m_{ℓ} that it can have.

**Solution:**

1) We recall the rule for identifying possible m_{ℓ} values:

m_{ℓ}starts at negative 'ℓ,' runs by whole numbers to zero and then goes by whole numbers to positive 'ℓ.'

2) With ℓ = 1, we have the following m_{ℓ} values:

−1, 0, +1

**Example #13:** What are all the possible values of ℓ when n = 3?

(a) ℓ = 0, 1, 2, 3(b) ℓ = −2, −1, 0, 1, 2

(c) ℓ = −3, −2, −1, 0, 1, 2, 3

(d) ℓ = 0, 1, 2

**Solution:**

1) We recall the rule for determining ℓ values:

ℓ = 0, 1, 2, . . . , n − 1

2) Answer choice (d) is the correct answer.

**Example #14:** Which of the following combination of quantum numbers is/are allowed?

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

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

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

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

**Solution:**

1) (a) & (c) are allowed. (b) & (d), therefore, are not. Let's look at (b):

The n value is allowed, but ℓ = 3 when n = 1 is not allowed. When n = 1, the only allowed ℓ value is zero.By the way, that is the case in (a). It has n = 1, so the only allowed ℓ value would be zero, which is what (a) has. When ℓ = 0, the only possible m

_{ℓ}value is also zero and that's what (a) has.

2) Let's look at (d):

The disallowed value in (d) is m_{s}equalling +^{3}⁄_{2}. For all possible sets of n, ℓ, and m_{ℓ}the only choices for m_{s}are +^{1}⁄_{2}and −^{1}⁄_{2}The n, ℓ, and m

_{ℓ}values for (d) follow the rules correctly, however, the m_{s}value makes the set be not allowed.

**Example #15:** Which of the following combinations of quantum numbers are allowed for an electron in a one-electron atom?

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

(b) n = 6, ℓ = 2, m_{ℓ}= 1, m_{s}= +^{1}⁄_{2}

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

(d) n = 6, ℓ = 0, m_{ℓ}= 1, m_{s}= +^{1}⁄_{2}

**Solution:**

1) (a) and (b) are allowed. Let's look at why (c) and (d) are not allowed. First (c):

Rationale #1: ℓ values start at zero and go by integers up to n − 1. When n = 1, the only possible ℓ value is zero. A value of −1 is not allowed in this example.Rationale #2: ℓ values cannot be negative. Ever.

2) Let's look at (d):

The n, ℓ combination of 6, 0 is allowed. The problem comes with the m_{ℓ}value. Remember, m_{ℓ}values go from −ℓ to 0 to +ℓ. When ℓ = 0, the only possible m_{ℓ}value is 0.