ChemTeam: Quantum Numbers (2023)

Quantum Numbers:
The Rules for Assigning Them
Fifteen Examples

Probs 1-10

Probs 11-25

Examples and Problems only (no solutions)

Return to Electrons Menu

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 15 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 and ms. 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, and ms 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 on a physical basis. 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 what follows.

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, and ms.

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.

First 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.

Second Comment: Some sources call this quantum number the "orbital quantum number." While "azimuthal (angular momentum) quantum number" is used more widely than "orbital quantum number," you're going to need to know all of them. This is because sources will use one term, but not usually acknowledge the existence of the other term.

III. The Magnetic Quantum Number (signified by m): 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 somewhat together. I could be wrong in this, so don't take my word for it!

The rule for selecting m is as follows:

m 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 values generated are −2, −1, 0, +1, +2, for a total of five values.

Comment: I have chosen to show an explicit positive sign on m values (and ms values). Across the Internet and the many classrooms of the world, there will be a variety of people (and texts) that do not include an explicit positive sign. Be aware.

IV. The Spin Quantum Number (signified by ms): 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 number n and two auxiliary quantum numbers k1 and k2 to which is added a further quantum number m 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 ms is as follows:

after the n, ℓ and m to be used have been determined, assign the value +12 to one electron, then assign −12 to the next electron, while using the same n, ℓ and m values.

For example, when n, ℓ, m = 1, 0, 0; the first ms value used is +12. However a second electron can also have n, ℓ, m = 1, 0, 0; so assign −12 to it.

Here's a table summarizing the rules:

Quantum NumberSymbolPossible Values
Principaln1, 2, 3, 4, . . .
Angular Momentum0, 1, 2, 3, . . . , (n − 1)
Magneticm−ℓ, . . . , −1, 0, 1, . . . , ℓ
Spinms+12, −12

Fifteen Examples

Example #1: An electron cannot exist in the energy state described by which set of quantum numbers below?

(a) 3, 2, 2, −12
(b) 4, 3, 3, +12
(c) 2, 1, −3, +12
(d) 2, 0, 0, −12
(e) 1, 0, 1, −12

Solution:

1) The first step is to (a) scan all the n values, looking for those numbers to be positive integers only and (b) scan all ms values, looking for anything that is not +12 or −12:

(a) All n values are correct.

(b) All ms 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.

(Video) Quantum Numbers - PHSCS 106 - Ruben Sosa

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 pairs (this is where the incorrect answer will be). You need to know the rule for m 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 values are −1, 0, and +1. −3 is not an allowed value when ℓ = 1.

4) I copied this question from an "answers" website and the person who answered it gave this as an answer:

Answer is c ---> −3 not possible with PQN 2

Notice that this method is comparing the n value to the m. The absolute value for m will always be lesser than n, never greater than or equal to n. In this question, the absolute value of the m value is 3, which is greater than the n value of 2.

Look at this set:

n = 2, ℓ = 1, m = −2, ms = −12

Looking at the n and m pairing, we find that the absolute values are equal. It is an incorrect set of quantum numbers.

Example #2: Identify which sets of quantum numbers are valid for an electron. Each set is ordered (n, ℓ, m, ms)

(a) 2, 2, −1, +12(g) 2, 1, −1, +12
(b) 0, 2, 1, +12(h) 1, 2, 0, +12
(c) 2, 0, 0, −12(i) 1, 0, 0, ±12
(d) 3, −2, −1, −13(j) 4, 3, 1, −12
(e) 3, 2, 1, +12(k) 3.5, 3, 1, +12
(f) 4, 3, −5, −12(o) 3, 2, 1, −1

Solution:

1) Scan the n values first:

We find that (b) and (k) have invalid n values. n starts at 1 [which eliminates (b)] and goes by positive integer values [which eliminates (k)]

2) Scan the ms values second:

We find that (d), (i), and (o) have invalid ms values.

(d) is invalid because of the use of the 13 value. The ms numerical value is only 12, either positive or negative.

(i) is invalid because the ms is EITHER positive or negative, not both. You must use a single positive sign or a single negative sign, not both at the same time.

(o) is invalid becase it has a value of 1. Only the numeral 12 is used for ms values.

3) Next, look at the n, ℓ pairings:

(a) is not valid because ℓ always ends at n − 1. ℓ can never be equal to n.

In addition to failing the definition of ms, (d) also fails with its use of a negative value for ℓ. ℓ is never negative.

(h) has an ℓ which is greater than n. ℓ is always less than n. (h) is not a valid set of quantum numbers.

4) Lastly, we look at the relationship between ℓ and m:

Here's the relationship between ℓ and m:
m starts at negative ℓ, runs by whole numbers to zero and then goes by whole numbers to positive ℓ
(f) fails the relationship between ℓ and m. When ℓ = 3, the allowed m values are −3, −2, −1, 0, +1, +2, +3. −5 is not part of this five allowed values. (h) is not a valid set of quantum numbers.

5) By a process of elimination, we have arrived at the valid sets of quantum numbers for this problem:

(c) 2, 0, 0, −12
(e) 3, 2, 1, +12
(g) 2, 1, −1, +12
(j) 4, 3, 1, −12

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 = 0, ms = +12
(b) n = 3, ℓ = 3, m = −2, ms = −12
(c) n = 4, ℓ = 3, m = −2, ms = +12
(d) n = 3, ℓ = 2, m = 2, ms = +13
(e) n = 2, ℓ = 1, m = −2, ms = −12
(f) n = 3, ℓ = 2, m = −1, ms = −12

Solution:

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

ms can only take on values of +12 and −12

Now, you look for any that violate that rule, to find:

(d) n = 3, ℓ = 2, m = 2, ms = +13

We don't care what the other values in (d) are, the presence of the +13 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 = −2, ms = −12

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 values. Here is the m rule:

m ranges from −ℓ to zero to +ℓ (all integer values)

The point is that the absolute value of m cannot be greater than that of ℓ. Look at (e):

(e) n = 2, ℓ = 1, m = −2, ms = −12

When ℓ = 1, the allowed m values are −1, 0, +1. So m = −2 is disallowed.

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

Example #4: Explain why each of the following sets of quantum numbers would not be permissible for an electron according to the rules for quantum numbers.

(a) n = 1, ℓ = 0, m = 0, ms = +1
(b) n = 1, ℓ = 3, m = 3, ms = +12
(c) n = 3, ℓ = 2, m = 3, ms = −12
(d) n = 0, ℓ = 1, m = 0, ms = +12
(e) n = 2, ℓ = 1, m = −1, ms = +32
(f) n = 4, ℓ = 3, m = 5, ms = +12

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.

(Video) Quantum Chemistry|CSIR NET June 2022 crash course|CSIR NET September 2022 exam|Crash Course

m = magnetic quantum number = the orbital in the sublevel
Values are −ℓ, . . . , 0, . . . , +ℓ

ms = spin quantum number = electron in orbital
Values are +12 or −12

2) The problem with (a) is in the ms value:

The error is that ms = +1. The ms value can only be +12 or −12.

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. The absolute value for m is always smaller than that for n.

4) The problem with (c) lies in the relationship between ℓ and m:

Since ℓ = 2, the values for m will be −2, −1, 0, +1, +2. The error in (c) is that m is 3, when the maximum m (when ℓ = 2) can only be 2.

Choice (c) also fails the n and m comparison. The absolute value for m 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 ms value.

ms is allowed to only be values of +12 or −12. +32 fails this rule.

7) In (f) compare the m to the ℓ value. When ℓ = 3, there are seven permissible m values. They are:

−3, −2, −1, 0, +1, +2, +3

The m value of 5 is not permitted in this set of quantum numbers.

Example #5: A hydrogen atom has n = 5 and m = −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 ranges from −ℓ to +ℓ, we need all ℓ that generate a −2.

When ℓ = 0, m can only equal 0. When ℓ = 1, the most negative m generated is −1. When ℓ = 0 or 1, an m value of −2 is not generated.

ℓ values of 2, 3, and 4 will each include an m value equal to −2

Here's one example:

When ℓ = 3, m 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, +12
(c) 3, 1, −2, −12
(d) 3, 1, 0, +12

Solution:

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

(a) 3, 2, −1, +1 <--- incorrect because ms is either +12 or −12, not +1

(b) 3, 3, −1, +12 <--- 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, −12 <--- incorrect because m takes on values from −ℓ to +ℓ, by integers. When ℓ = 1, m takes on the values of 1, 0, and −1. A m value of −2 is not permitted when ℓ = 1

(d) 3, 1, 0, +12 <--- correct. It fits all the rules for n, then ℓ, then m, then ms

Example #7: An orbital has n = 4 and m = −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, 3

for all possible ℓ values.

We look for cases when m = −1

When ℓ = 0, the only possible m value is 0.

ℓ = 1 is the first value that generates m equal to −1. This is because m values range from −ℓ to 0 to +ℓ by integers. So, with ℓ = 1, we'd have −1, 0, +1 for our m values

The second ℓ value that generates m = −1 is 2. This is because, when ℓ = 2, we have m values of −2, −1, 0 +1, +2.

The third ℓ is 3. It generates m values of −3, −2, −1, 0, +1, +2, +3

So, 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, +122, 0, 0, +123, 0, 0, +124, 0, 0, +12
1, 0, 0, −122, 0, 0, −123, 0, 0, −12

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, +123, 1, 0, +12
2, 1, 0, −123, 1, 0, −12

By the way, K does have an electron in the 4th shell, but it's not in a p orbital. It's in the s orbital of the 4th shell, discussed just above.

3) 11 of the 19 electrons have ℓ = 0

(Video) Quantum numbers, principal , Azimuthal, magnetic,spin Quantum number

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 = 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 is an integer and can range from −ℓ to +ℓ
ms is either +12 or −12

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 of 2. Let us look at each ℓ in turn:

ℓ = 0
Only an m of 0 can be generated. ℓ = 0 is not part of the answer.

ℓ = 1

Three m 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 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 is also generated here, this is the other part of the correct answer. The m values generated are −3, −2, −1, 0, 1, 2, and 3.

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

4, 2, 2 and 4, 3, 2

However, 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 = 2 is four.

5) Here are their quantum number sets:

4, 2, 2, +12
4, 2, 2, −12
4, 3, 2, +12
4, 3, 2, −12

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

(a) n = 5, ℓ = 3, m = −1, ms = +12
(b) n = 1, ℓ = 0, m = 0, ms = −12
(c) n = 2, ℓ = 2, m = 2, ms = +12
(d) n = 4, ℓ = 1, m = 0, ms = −12
(e) n = 6, ℓ = 4, m = −3, ms = +12

Solution:

1) (a) is allowed:

For n = 5, ℓ = 0, 1, 2, 3, 4. Thus, ℓ = 3 is allowed.
For ℓ = 3, m = −3, −2, −1, 0, 1, 2, 3. Thus m = −1 is allowed.
ms = +12 is allowed.

2) (b) is allowed:

For n = 1, ℓ = 0 only. Thus, ℓ = 0 is allowed.
For ℓ = 0, m = 0 only. Thus m = 0 is allowed.
ms = −12 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 = −1, 0, 1. Thus m = 0 is allowed.
ms = −12 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 = −4, −3, −2, −1, 0, 1, 2, 3, 4. Thus m = −3 is allowed.
ms = +12 is allowed.

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

(a) n = 0, ℓ = 3, m = −3, ms = +12(e) n = 1, ℓ = −1, m = 1, ms = −12
(b) n = 3, ℓ = −1, m = 0, ms = +12(f) n = 3, ℓ = 3, m = −2, ms = −12
(c) n = 3, ℓ = 2, m = −3, ms = −12(g) n = 0, ℓ = −2, m = 1, ms = +12
(d) n = 5, ℓ = 3, m = −2, ms = −1(h) n = 3, ℓ = −2, m = 1, ms = +43

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 is incorrect. When ℓ = 2, m can only take on these five values: −2, −1, 0, 1, 2.

(d) ms is incorrect. Values of +12 and −12 are the only two values allowed.

(e) ℓ is incorrect. When n = 1, ℓ can only take on the value of 0.

(f) ℓ is incorrect. When n = 3, ℓ can take on values only up to n minus 1, so ℓ = 3 is not allowed.

(g) n is incorrect. The value of n starts at 1, zero is not allowed.

(h) ℓ is incorrect. It cannot be a negative number. ms also is incorrect.

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

Solution:

1) We recall the rule for identifying possible m values:

m 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 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:

(Video) المجلس الكيميائي | المحاضرة الثانية Organic Chemistry and Life 🔬/ Dr. Hamad Al Mamari

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 = 0, ms = +12
(b) n = 1, ℓ = 3, m = 3, ms = +12
(c) n = 3, ℓ = 2, m = −2, ms = −12
(d) n = 2, ℓ = 1, m = −1, ms = +32

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 value is also zero and that's what (a) has.

2) Let's look at (d):

The disallowed value in (d) is ms equalling +32. For all possible sets of n, ℓ, and m the only choices for ms are +12 and −12

The n, ℓ, and m values for (d) follow the rules correctly, however, the ms 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 = −1, ms = −12
(b) n = 6, ℓ = 2, m = 1, ms = +12
(c) n = 1, ℓ = −1, m = −2, ms = +12
(d) n = 6, ℓ = 0, m = 1, ms = +12

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 value. Remember, m values go from −ℓ to 0 to +ℓ. When ℓ = 0, the only possible m value is 0. Since m is incorrect, (d) is the set that is not allowed.

Bonus 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, +12
1, 0, 0, −12

These two electrons are in the 1s orbital. s orbitals are always characterized by ℓ and m equalling zero.

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

The 2s orbital:
2, 0, 0, +12
2, 0, 0, −12

The 2p subshell (composed of three 2p orbitals):

2, 1, −1, +12
2, 1, −1, −12

2, 1, 0, +12
2, 1, 0, −12

2, 1, 1, +12
2, 1, 1, −12

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

3, 0, 0, +12

This is the answer. By convention, the positive ms is used first.

This electron is in the 3s orbital.

Bonus Example #2: What are the possible values of n and m for an electron in a 5d orbital? Write the n, ℓ, m for each of the orbitals in the 5d subshell.

Solution:

1) Determine the value for n:

The use of 5d provides the answer.

The 5 in 5d is the value of n the question wants.

Other examples: the value for n in the 4p orbital is 4. The value for n in the 2s orbital is 2.

2) Determine the values for m:

We need to determine what ℓ value is associated with d orbitals.

The answer to that is 2. (s orbitals have ℓ = 0, p orbitals have ℓ = 1, d has ℓ = 2 and f has an ℓ of 3.)

We now apply the rule for m and determine the m values for the 5d orbital:

−2, −1, 0, 1, 2

3) The five n, ℓ, m sets are as follows:

5, 2, −2
5, 2, −1
5, 2, 0
5, 2, 1
5, 2, 2

Sometimes, a teacher will insist that positive signs for the last two m be included. It's just a stylistic thing, so go along with it if it happens to you.

Probs 1-10

Probs 11-25

Examples and Problems only (no solutions)

(Video) SCH4U 1.1: Spectroscopy and the quantum model of the atom

Return to Electrons Menu

FAQs

What are the 4 quantum numbers? ›

In atoms, there are a total of four quantum numbers: the principal quantum number (n), the orbital angular momentum quantum number (l), the magnetic quantum number (ml), and the electron spin quantum number (ms).

What are the 4 quantum numbers of 4s? ›

Therefore the quantum no's of the differentiating electron in 4s-orbital are; n = 4, l = 0, m = 0 and s = +1/2.

What are the 4 quantum numbers of carbon 6? ›

Thus, Carbon with electron 6 has four quantum numbers: $2$,$1$,$0$,$ + \dfrac{1}{2}$ .

How do you write 4 sets of quantum numbers? ›

There are four quantum numbers, namely, principal, azimuthal, magnetic and spin quantum numbers.

What is the 4th quantum number called? ›

To distinguish between the two electrons in an orbital, we need a fourth quantum number. This is called the spin quantum number (s) because electrons behave as if they were spinning in either a clockwise or counterclockwise fashion.

Who discovered 4th quantum number? ›

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.

How many sets of 4 quantum numbers are possible? ›

For a multi-electron system, each electron in the system has a unique set of four quantum numbers. Therefore, the system He2− has 4 electrons and has thus four unique sets of four quantum numbers.

Who proposed 4 quantum numbers? ›

Pauli Exclusion Principle: In 1926, Wolfgang Pauli discovered that a set of quantum numbers is specific to a certain electron. That is, no two electrons can have the same values for n, l, ml, and ms.

What quantum number is 5s? ›

- The principal quantum number (n) for the 5s orbital is '5'.

What is the L value for 4s? ›

Answer and Explanation: For the 4s orbital, the principal quantum number n is 4 since it is at the 4th shell of main orbitals. For n = 4, we can have l as 0, 1, 2, and 3, which represent s, p,d and f orbitals. In this question, the electron is in s state, so l = 0.

Is 4f a quantum number? ›

For a 4f orbital, the principal quantum number is n = 4, the azimuthal quantum number is 3. The values of magnetic quantum numbers will be -3, -2, -1, 0, +1, +2, +3. Total 7 orbitals are present in 4f subshell.

What are quantum numbers for 6s? ›

For an electron in the 6s orbital, the quantum numbers n and are 6 and 0, respectively.

Is carbon plus 4 or 4? ›

The valency of carbon is 4 and valency of oxygen is 2 .

What is 4 degree carbon called? ›

4-degree or quaternary carbon atom are those carbon atom in which all 4 valencies of carbon atom is satisfied by other 4 carbon atoms. e.g. (CH3)4-C in 2,2-dimethylpropane in this that C carbon atom is quaternary carbon atom. Q.

Is 4p a pair of quantum numbers? ›

Solution : The designation `4p` indicates that the orbital has a principal quantum number `n = 4` and an angular-momentum quantum number `l = 1`. The magnetic quantum number can have any of the values `-1, 0`, or `+1`.

How many orbitals are in 4f? ›

For any atom, there are seven 4f orbitals. The f-orbitals are unusual in that there are two sets of orbitals in common use.

What are the 4 quantum numbers of an electron in 3d? ›

Hence, the set of quantum numbers for electrons in 3d orbital is n=3 , l=2 , ml={−2,−1,0,1,2} and ms={12,−12} . Note : For an electron in an atom, it is uniquely described by these four quantum numbers.

Is 5f a quantum number? ›

π = 3.14159 approximately. e = 2.71828 approximately. Z = effective nuclear charge for that orbital in that atom. ρ = 2Zr/n where n is the principal quantum number (5 for the 5f orbitals)
...
Table of equations for the 5f orbitals.
FunctionEquation
Y5fy(3x 2-y 2)= √(70/16) × y(3x 2-y 2)/r3 × (1/4π)1/2
34 more rows

Is 4th electron will have four quantum numbers? ›

n l m s. 1.

What is the 1st quantum number? ›

The first quantum number is called the principal quantum number, and it's denoted by the letter n . It represents the energy level/shell. n can equal any integer above and including 1 , since n=1 represents the first principal energy level.

Who is the father of quantum number? ›

Niels Bohr and Max Planck, two of the founding fathers of Quantum Theory, each received a Nobel Prize in Physics for their work on quanta. Einstein is considered the third founder of Quantum Theory because he described light as quanta in his theory of the Photoelectric Effect, for which he won the 1921 Nobel Prize.

Who is the father of principal quantum number? ›

Principal quantum number was proposed by Bohr to explain the hydrogen atomic spectrum. 2. It denotes the main energy level (or shell or orbit).

Which set of quantum numbers is not possible? ›

The value of spin quantum number can never be a zero, because electrons always have spin either positive or negative. Hence, n = 1, l = 0, ml = 0, ms = 0, this set of quantum number is not possible.

Which set of quantum numbers is impossible? ›

When the Principle quantum number (n) is 3, the possible values of Azimuthal quantum number will be only 0, 1 and 2 which represent s, p and d orbitals respectively. Therefore, the set of quantum numbers given in option (b) is impossible.

What is the highest principal quantum number? ›

As 𝑛 increases, the electron will be at a higher energy and less tightly bound to the nucleus. If we look at the principal quantum number of the three electrons, we can see that electron number (2) has the highest principal quantum number and will therefore have the highest energy.

What is origin of quantum number? ›

Quantum numbers arise in the process of solving the Schrodinger equation by constraints or boundary conditions which must be applied to get the solution to fit the physical situation.

Which is higher 5s or 4d? ›

Even though 5s orbitals have a higher principal quantum number than 4d orbitals, (n = 5 compared to n = 4), they're actually lower in energy. As a result, 5s orbitals are always filled before 4d orbitals. Similarly, 6s orbitals are lower in energy than 5d orbitals, so 6s orbitals are always filled first.

Is 7s orbital possible? ›

For any atom there is just one 7s orbital.

Is 5p or 5d higher in energy? ›

The order of the electron orbital energy levels, starting from least to greatest, is as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p.

What is the L value of 3s? ›

The subshell with n=2 and l=1 is the 2p subshell; if n=3 and l=0, it is the 3s subshell, and so on. The value of l also has a slight effect on the energy of the subshell; the energy of the subshell increases with l (s < p < d < f).

What is the L value of 5p? ›

The principal quantum number n = 5 and the azimuthal quantum number l = 1 specify a 5p orbital.

What is the L value for 5d? ›

The value of the azimuthal quantum number, l for 5d orbital is 2. The possible values of magnetic quantum number, ml are -2, -1, 0, +1, and +2.

Does 4f or 5d come first? ›

Orbitals fill in order of energy. So 5D fills before 4F in some cases simply because the 5D energy levels are lower than the 4F levels for some.

What is n and L at 4f? ›

For nth orbital possible values of azimuthal quantum number, l are from 0 to (n−1). Total of n values. In 4f orbital, n=4 and f-orbital has l=3.

Is 5d or 4f higher in energy? ›

That's defined by Aufbau's principle. You will notice that 4f gets filled before 5d. That signifies that 4f is in a lesser energy than 5d.

What is L in 6s orbital? ›

(a) 6s: n =6, l=0 number of orbitals is 1.

Does a 6s orbital exist? ›

For any atom there is just one 6s orbital. The image on the top is deceptively simple as the interesting feature is buried within the orbital. That on the left is sliced in half to show the five spherical nodes of the 6s orbital. The shape on the right shows the nodal structure of the 6s-orbital.

Is 6f a quantum number? ›

The principal quantum number of the 6f subshell is six. The angular momentum quantum number of the 6f subshell is found as three. The subshell's first number or the numerical value depicts the principal quantum number.

Why does carbon not show +4 or 4 valence? ›

It has to either lose or gain 4 electrons in order to gain a stable electronic configuration. It cannot gain four electrons as a carbon atom has a total of 6 protons and is very small to handle ten electrons.

Why is Group 4 called carbon family? ›

Group 14 elements or IV A elements are known as carbon family. These elements shows similarity in the properties and reactivity so they are known as carbon family.

Does carbon lose 4 electrons or gain 4 electrons? ›

The electronic configuration of carbon is 2,4. So, it forms mainly covalent compound as it can neither lose nor gain four-electron to complete its octet. The example of such compound is Methane (CH4).

What is a 5 carbon chain called? ›

Cyclopentane is a ring of five* carbon atoms.

What is 1 degree 2 degree and 3 degree Carbocation? ›

Methyl carbocation: If no carbon is attached to the carbon with the positive charge it is simply called as methyl carbocation. If one, two or three carbon is attached to the carbon with the positive charge it is called the primary carbocation, secondary carbocation, and tertiary carbocation respectively.

Is carbon just C or c2? ›

Carbon (from Latin carbo 'coal') is a chemical element with the symbol C and atomic number 6.

What are the 4 quantum numbers and what do they tell us about electrons? ›

Each electron in an atom is described by four different quantum numbers. The first three (n, l, ml) specify the particular orbital of interest, and the fourth (ms) specifies how many electrons can occupy that orbital.

Who discovered 4 quantum numbers? ›

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.

What are the first 4 quantum shells? ›

Each energy level is given a number called the principal quantum number, n. The closest shell has a value of n=1.
...
Search form.
ShellSubshellTotal Number of Electrons in Shell
1st Shell1s2
2nd Shell2s, 2p2 + 6 = 8
3rd Shell3s, 3p, 3d2 + 6 + 10 = 18
4th Shell4s, 4p, 4d, 4f2 + 6 + 10 + 14 = 32

What quantum numbers are not allowed? ›

Of the set of quantum numbers {n, ℓ, m , m s}, which are possible and which are not allowed? Spin must be either +1/2 or −1/2, so this set of quantum number is not allowed.

Who named quantum dot? ›

Who invented quantum dots? Quantum dots were discovered in solids (glass crystals) in 1980 by Russian physicist Alexei Ekimov while working at the Vavilov State Optical Institute.

What is the first quantum number? ›

The first quantum number is called the principal quantum number(n). The principal quantum number largely determines the energy of an electron. Electrons in the same atom that have the same principal quantum number are said to occupy an electron shell of the atom.

How many 4f orbitals exist? ›

For any atom, there are seven 4f orbitals. The f-orbitals are unusual in that there are two sets of orbitals in common use.

What are the four quantum numbers for 5s? ›

- The spin quantum number (ms ) value for one electron in 5s orbital of the rubidium is either +12or−12 . - Therefore the four quantum number values for the outermost electron in the Rubidium element are 5, 0 , 0, +12or−12 .

Videos

1. Webinar 64 - Q-Chem 6: Dawn of the Next Generation
(QChemSoftware)
2. Physical Spectroscopy|CSIR NET June 2022 crash course|CSIR NET September 2022 exam|Crash Course
(J Chemistry Team)
3. Physical Spectroscopy|CSIR NET June 2022 crash course|CSIR NET September 2022 exam|Crash Course
(J Chemistry Team)
4. Thermodynamics chemistry|CSIR NET June 2022 crash course|CSIR NET September 2022 exam|Crash Course
(J Chemistry Team)
5. UVM Ruggles Mod19 Lecture1
(Erik Ruggles)
6. Webinar 40 What's New in Q Chem 5 3
(QChemSoftware)
Top Articles
Latest Posts
Article information

Author: Dr. Pierre Goyette

Last Updated: 01/11/2023

Views: 6120

Rating: 5 / 5 (70 voted)

Reviews: 85% of readers found this page helpful

Author information

Name: Dr. Pierre Goyette

Birthday: 1998-01-29

Address: Apt. 611 3357 Yong Plain, West Audra, IL 70053

Phone: +5819954278378

Job: Construction Director

Hobby: Embroidery, Creative writing, Shopping, Driving, Stand-up comedy, Coffee roasting, Scrapbooking

Introduction: My name is Dr. Pierre Goyette, I am a enchanting, powerful, jolly, rich, graceful, colorful, zany person who loves writing and wants to share my knowledge and understanding with you.