Alexander Fufaev
My name is Alexander FufaeV and here I will explain the following topic:

Quantum Numbers (n, l, m) and the Electron Configuration

Table of contents
  1. Principal Quantum Number
  2. Azimuthal Quantum Number
  3. Magnetic Quantum Number
  4. Spin Quantum Number
  5. Electron Configuration of an Atom
  6. Exercises with Solutions

Quantum numbers \(\class{red}{n}\), \(\class{green}{l}\), \(\class{violet}{m}\), and \(\class{brown}{s}\) describe the distribution of electrons in an atom. This distribution of electrons is called the electron configuration.

The quantum numbers characterize different solutions \(\Psi_{\class{red}{n},\class{green}{l},\class{violet}{m},\class{brown}{s}}\) of the Schrödinger equation. The various solutions \(\Psi_{\class{red}{n},\class{green}{l},\class{violet}{m},\class{brown}{s}}\) of the Schrödinger equation describe different orbitals of an atom.

An orbital is a region of space around an atom where the electron described by the wave function \(\Psi_{\class{red}{n},\class{green}{l},\class{violet}{m},\class{brown}{s}}\) is located with a certain probability. The electron with the electron configuration \(\class{red}{n}, \class{green}{l}, \class{violet}{m}, \class{brown}{s})\) can only exist in the corresponding orbital. The probability of finding it anywhere else is zero.

Principal Quantum Number

Principal Quantum Number \(\class{red}{n}\) determines the energy level, or the shell, in which an electron resides.

  • \(\class{red}{n} = 1\) is the K shell.
  • \(\class{red}{n} = 2\) is the L shell.
  • \(\class{red}{n} = 3\) is the M shell.
  • \(\class{red}{n} = 4\) is the N shell.
  • and so on.
Aluminum Electron Shell Model

The principal quantum number can therefore take integer values starting from 1 (for the first shell) and increases in increments of 1 for subsequent shells. The \(n\)-th shell can be occupied by a maximum of \(2\class{red}{n}^2\) electrons.

Azimuthal Quantum Number

Azimuthal Quantum Number (Angular Momentum Quantum Number) \(\class{green}{l}\) describes the orbital angular momentum of the electron and the shape of the orbital in which the electron resides.

The angular momentum quantum number takes non-negative integer values from 0 to \(\class{red}{n} - 1 \). The values of \(\class{green}{l}\) determine the subshell within a shell:

  • \(\class{green}{l} = 0\) corresponds to the s orbital.
  • \(\class{green}{l} = 1\) corresponds to the p orbital.
  • \(\class{green}{l} = 2\) corresponds to the d orbital.
  • \(\class{green}{l} = 3\) corresponds to the f orbital.
  • and so on up to \(\class{green}{l} = \class{red}{n} - 1\).

If the principal quantum number of an electron is \( \class{red}{n} = 2 \), then its azimuthal quantum number \(\class{green}{l}\) can be either 0 or 1. In this case, the azimuthal quantum number takes two possible values: \(\class{green}{l} = 0,1\).

Magnetic Quantum Number

Magnetic Quantum Number \(\class{violet}{m}\) describes the number of orbitals in a subshell and their spatial orientation.

The magnetic quantum number takes integer values from \(\class{violet}{m}=-\class{green}{l}\) to \(\class{violet}{m}=\class{green}{l}\).

  • For an s-electron with \(\class{green}{l} = 0\), \(\class{violet}{m} = 0\).
  • For a p-electron with \(\class{green}{l} = 1\), \(\class{violet}{m} = -1,0,1\).
  • For a d-electron with \(\class{green}{l} = 2\), \(\class{violet}{m} = -2,-1,0,1,2\).
  • For an f-electron with \(\class{green}{l} = 3\), \(\class{violet}{m} = -3,-2,-1,0,1,2,3\).
  • and so on.

Spin Quantum Number

Spin Quantum Number \(\class{brown}{s}\) describes, in simple terms, the angular momentum of an electron. The spin quantum number of an electron can take two possible values: \(\class{brown}{s} = \frac{1}{2}\) (we say: spin up) or \(\class{brown}{s} = -\frac{1}{2}\) (we say: spin down).

Electron Configuration of an Atom

The electron configuration of an atom describes how its electrons are distributed among the various atomic orbitals. It follows the Aufbau principle, the Pauli exclusion principle, and Hund's rule, which we will learn about in the next lessons. The electron configuration is typically represented with notation indicating the number of electrons in each subshell: $$ (\class{red}{\text{Principal Quantum Number}}) \, (\class{green}{\text{Azimuthal Quantum Number}})^{\text{Z}} $$ In the \(\class{red}{n}\)-th shell with the angular momentum quantum number \(\class{green}{l}\), there are \(Z\) electrons. \(Z\) is the number of electrons in the respective shell and subshell. Let's look at some concrete examples:

  • Hydrogen (H) has \( Z = 1 \) electron. Its electron configuration is \( \class{red}{1}\class{green}{s}^1 \). This means: One electron is in the \( \class{red}{1}\class{green}{s}\)-orbital.
  • Helium (He) has \( Z = 2 \) electrons. Its electron configuration is \( \class{red}{1}\class{green}{s}^2 \). This means: Two electrons are in the \( \class{red}{1}\class{green}{s}\)-orbital.
  • Lithium (Li) has \( Z = 3 \) electrons. Its electron configuration is \( \class{red}{1}\class{green}{s}^2\,\class{red}{2}\class{green}{s}^1 \). This means: Two electrons are in the \( \class{red}{1}\class{green}{s}\)-orbital, and one electron is in the \( \class{red}{2}\class{green}{s}\)-orbital. We can also write the electron configuration of lithium as \( \text{[He]}\,\class{red}{2}\class{green}{s}^1 \), because two electrons of the lithium atom are distributed exactly as in the helium atom.
  • Beryllium (Be) has \( Z = 4 \) electrons. Its electron configuration is \( \class{red}{1}\class{green}{s}^2\,\class{red}{2}\class{green}{s}^2 \). Or more compactly: \( \text{[He]}\,\class{red}{2}\class{green}{s}^2 \).
  • Boron (B) has \( Z = 5 \) electrons. Its electron configuration is \( \class{red}{1}\class{green}{s}^2\,\class{red}{2}\class{green}{s}^2\,\class{red}{2}\class{green}{p}^1 \). Or more compactly: \( \text{[Be]}\,\class{red}{2}\class{green}{p}^1 \).

Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

Exercise #1: Quantum Numbers and Orbitals of an Excited Hydrogen Atom

The electron of the hydrogenium atom (hydrogen atom) cannot have an arbitrary binding energy, but can only take on discrete energies \( W_n \), which are characterized by the principal quantum number \( n \). The individual energy states of the electron are designated by letters K, L, M, N etc., especially in chemistry.

  1. What is the energy of the electron in the K, L, and M shells?
  2. How many orbitals \( N \) does the Hydrogen atom have if the energy of the electron is \( W_n \lt -0.5 \,\mathrm{eV} \)?
  3. Enumerate all possible orbitals by specifying their \( (n, l, \class{violet}{m}) \) state. How many orbitals of these have the minor quantum number \( l=2 \)?
  4. What energy must the electron have so that the Hydrogen atom has at least 200 orbitals but at most 300 orbitals?

Tips: Use the Rydberg formula to determine the individual energy states for the respective shells: \[ W_n = -W_{\text R} \, \frac{1}{n^2} \] where \( W_{\text R} = 13.6 \, \text{eV} \) is the amount of energy of the ground state \( n = 1 \), i.e. the amount of the lowest energy that the electron can take on in the H atom.

The shell model is a simplification of the orbital model. The shell K stands for \( n = 1 \). The shell L stands for \( n = 2 \). The shell M stands for \( n = 3 \).

For exercise #1.2: Number of orbitals \( N \) per principal quantum number \( n \) is given by \( n^2 \).

For exercise #1.3: The secondary quantum number \( l \) has the restriction \( l \lt n \) and starts with \(l=0\). The magnetic quantum number \( \class{violet}{m} \) can take the values \( -l \leq \class{violet}{m} \leq l \).

Solution to the exercise #1.1

The energy of the electron is given by the Rydberg formula: 1 $$W_n = -W_{\text R} \, \frac{1}{n^2}$$

Here, \( W_{\text R} = 13.6 \, \text{eV} \) is the Rydberg energy, which gives the energy of an electron in the ground state (\( n = 1 \)) of the Hydrogen atom. The minus sign in Eq. 1 says that the electron is bound in the H atom. This means that the energy must be absorbed to ionize the H atom (to remove the electron from the H atom).

Using 1, you can determine the energy of the electron in the K shell (\( n = 1 \)): $$W_n = -W_{\text R} \cdot \frac{1}{1^2} = -13.6 \, \text{eV}$$

Energy of the electron in the L-shell (\( n = 2 \)): $$W_n = -W_{\text R} \cdot \frac{1}{2^2} = -3.4 \, \text{eV}$$

Energy of the electron in the M shell (\( n = 3 \)): $$W_n = -W_{\text R} \cdot \frac{1}{3^2} = -1.51 \, \text{eV}$$

Solution to the exercise #1.2

The number \( N_{n} \) of orbitals (without spin) for just one energy level \( n \) is given by the principal quantum number squared and is indirectly in the Rydberg formula #1.1 Eq.1: 1 $$N_{n} ~=~ n^2$$

The principal quantum number \( n \), if you look at the Rydberg formula #1.1 Eq.1, is larger the larger the energy of the electron is. Thus according to #1.2 Eq. 1 also the number of orbitals is larger. According to the problem the energy is given by \(W_n\) and should be smaller than \( W_n \lt -0.5 \, \text{eV} \). "Less than" means that you have to use the less sign instead of the equal sign. Insert in this inequality for \(W_n\) the Rydberg formula: 2 $$- W_{\text R} \, \frac{1}{n^2} ~\lt~ -0.5 \, \text{eV}$$

Rearrange the equation for \(n\): 3 $$n \lt \sqrt{ \frac{13.6 \, \mathrm{eV} }{ 0.5 \, \mathrm{eV} } }$$

Notice that the less sign has flipped because both sides have been multiplied by "-". Calculate the square root: 4 $$n ~\lt~ 5.2$$

Thus, the largest principal quantum number according to Eq. 4 is: \( n = 5 \). Consider that such principal quantum numbers as \( n=5.2\) do NOT exist. From this you determine the orbital number with the help of eq. #1.2 Eq. 1. You sum up the number from 1 to 5: 5 $$N ~=~ 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55$$

You can calculate the sum of squares \( n^2 \) as in Eq. 5, which tells you the number of orbitals, generally as follows: 6 $$N ~=~ \frac{n(n+1)(2n+1)}{6}$$

Solution to the exercise #1.3

After determining the number of orbitals in Exercise #1.2, you can also ask yourself, what are the exact states (i.e., quantum numbers) of these 55 orbitals? An orbital is determined by the three quantum numbers:
  • Principal quantum number \( n \), where it can only take integers \(n = 1,2,3... \).
  • Angular moment quantum number \( l \) is always smaller than the principal quantum number. And the maximum principal quantum number is: \( l = n-1 \). Like any other quantum number, it can also take only integers. It can have the value \( l = 0 \) in contrast to \( n \).
  • Magnetic quantum number \( \class{violet}{m} \) is between minimum angular momentum quantum number \(-l \) and maximum angular momentum quantum number \(l\) \( -l \leq \class{violet}{m} \leq l \).

To determine the concrete orbitals, you have to specify the number triplet \( (n,l,\class{violet}{m})\) and in this case 55 number triplets, because of 55 possible orbitals. In (b) you have determined the maximum major quantum number \( n = 5 \) for the energy \( E \lt -0.5 \, \text{eV} \). Thus, the maximum angular momentum quantum number is \( l = n-1 = 5-1 = 4 \).

The magnetic quantum number can only take the following values for \( l = 4 \): $$\class{violet}{m} = \{-4,-3,-2,-1,0,1,2,3,4\}$$

The magnetic quantum number for \( l = 3 \), on the other hand: $$\class{violet}{m} = \{-3,-2,-1,0,1,2,3 \}$$

The magnetic quantum number for \( l = 2 \): $$\class{violet}{m} = \{-2,-1,0,1,2 \}$$

The magnetic quantum number for \( l = 1 \): $$\class{violet}{m} = \{-1,0,1 \}$$

The magnetic quantum number for \( l = 0 \) is \( \class{violet}{m} = \{ 0 \} \).

If you go through all possibilities, considering the above conditions for the quantum numbers, then you get exactly 55 orbitals. So start at \( n = 1 \) and go through the conditions. Then go to \( n = 2 \) and go through the conditions and so on. The following table summarizes all possible triplets up to \(n=5\):

All orbitals up to the 5th energy level
Principal quantum number \( n\) Angular momentum quantum number \(l\) Orbital \( (n,l,\class{violet}{m})\)
10(1,0,0)
20(2,0,0)
21(2,1,-1)
21(2,1,0)
21(2,1,1)
30(3,0,0)
31(3,1,-1)
31(3,1,0)
31(3,2,1)
32(3,2,-2)
32(3,2,-1)
32(3,2,0)
32(3,2,1)
32(3,2,2)
40(4,0,0)
41(4,1,-1)
41(4,1,0)
41(4,1,1)
42(4,2,-2)
42(4,2,-1)
42(4,2,0)
42(4,2,1)
42(4,2,2)
43(4,3,-3)
43(4,3,-2)
43(4,3,-1)
43(4,3,0)
43(4,3,1)
43(4,3,2)
43(4,3,3)
50(5,0,0)
51(5,1,-1)
51(5,1,0)
51(5,1,1)
52(5,2,-2)
52(5,2,-1)
52(5,2,0)
52(5,2,1)
52(5,2,2)
53(5,3,-3)
53(5,3,-2)
53(5,3,-1)
53(5,3,0)
53(5,3,1)
53(5,3,2)
53(5,3,3)
54(5,4,-4)
54(5,4,-3)
54(5,4,-2)
54(5,4,-1)
54(5,4,0)
54(5,4,1)
54(5,4,2)
54(5,4,3)
54(5,4,4)

The second question was how many orbitals have the angular momentum quantum number \( l =2 \). You can easily count this using the table. Of course the energies with \( n=1\) and \(n=2\) cannot have a state with a angular momentum quantum number \( l=2\), because otherwise they would not fulfill the condition \( l \lt n \). So, only principal quantum numbers greater than 2 can be considered. In our case \( n = 3,4,5\) (in total 3 principal quantum numbers).

Each energy state with \( n \gt 2 \) contains exactly five \(l=2\) states, for example for \( n = 3 \) they are (3,2,-2), (3,2,-1), (3,2,0), (3,2,1) and (3,2,2). Three principal quantum numbers thus have \(3 \cdot 5=15\) orbitals with \(l=2\). Alternatively, you can calculate the number \(N(l)\) of orbitals with any \(l\) as follows: $$N(l) ~=~ (n_{\text{max}}-l) \cdot (2l + 1)$$

Of course, the condition \( n \gt l \) must always be fulfilled.

Solution to the exercise #1.4

We are looking for the energy \( W_n \) where the minimum orbital number is \( N_{\text{min}} = 200 \) and the maximum orbital number is \( N_{\text{max}} = 300 \). Thus, the energy \(W_n\) must be in a certain energy range for these two conditions to be satisfied. For this you need the formula from exercise #1.2, which generally gives you the number of orbitals in the H atom. Here is the formula: $$N ~=~ \frac{n(n+1)(2n+1)}{6}$$

We are looking for the principal quantum number \( n \), so that you can determine the minimum energy with the Rydberg formula. Rearrange the equation in such a way that the equation has the following form: $$0 ~=~ 2n^3 + 3n^2 + n - 6N$$

As you can see, you cannot simply solve for \( n \) because it is a 3rd degree polynomial. Use your calculator or the internet, with which you can find the zeros (i.e. the solutions). You get the following real solution, if you use \(N=N_{\text{min}}=200\): n=7.944. And, if you use \(N=N_{\text{max}}=300\), you get the following solution: \(n=9.164\). This means: The Hydrogen atom has between 200 and 300 orbitals exactly when the principal quantum number is between 7.944 and 9.164. But since it must be an integer, it is then either 8 or 9.

To calculate the minimum and maximum energy, simply substitute these two quantum numbers into the Rydberg formula: $$\begin{align}W_{\text{min}} &= -13.6 \, \text{eV} \cdot \frac{1}{8^2} = -0.2125 \, \text{eV} \\\\ W_{\text{max}} &= -13.6 \, \text{eV} \cdot \frac{1}{9^2} = -0.1679 \, \text{eV} \end{align}$$

That means: For at least 200 orbitals to be occupied, the energy \( W_n \) of the electron must be greater than \( - 0.2125 \, \text{eV} \). However, there must also be no more than 300 orbitals, so the energy must be less than \( - 0.1679 \, \text{eV} \). Overall, the energy lies between these two values: \( - 0.2125 \, \text{eV} ~\leq~ W_n ~\leq~ - 0.1679 \, \text{eV} \).