Chapter Two, Section Three


  1. Learn about the Bohr atom, Pauli Exclusion Principle, Hund's rule, and the electronic buildup of atoms.
  2. Learn why the periodic table has series of elements with same properties.
  3. Learn about ionization energy.


Niels Bohr proposed a model for the hydrogen atom that explained the spectrum of the hydrogen atom. The Bohr model was based on the following assumptions.

  1. The electron in a hydrogen atom travels around the nucleus in a circular orbit.
  2. The energy of the electron in an orbit is proportional to its distance from the nucleus. The further the electron is from the nucleus, the more energy it has.
  3. As long as the electron stays in a stationary orbit (same distance from the nucleus), it does not emit radiation as classical physics required.
  4. Only a limited number of orbits with certain energies are allowed. This is the quantization of orbits using Planck's hypothesis. Allowed orbits have their electron's angular momentum be an integral multiple of Planck's constant divided by 2pi.
  5. An electron can only pass from one stationary state to another. Radiation is absorbed when an electron jumps to a higher energy orbit and emitted when an electron falls into a lower energy orbit.
  6. The energy of the radiation emitted or absorbed is exactly equal to the difference between the energies of the orbits.

Using the above assumptions, the energy of a hydrogen atom in level n was found to be

E = -RH/n2 where RH = 2.179 x 10-18

Note that the energy has its lowest value when n = 1 and this is called the ground state. Also notice that E cannot have just any value, it must have values corresponding to n = 1, 2, 3, 4, 5, etc. We say that the energy levels are quantized.

The Bohr atom model was proven to be wrong and is called the old quantum theory. The new quantum theory is able to describe atomic and molecular situations acceptably, but we will not go over the details here. Suffice it to say that Quantum Mechanics predicts the three quantum numbers n, l, and ml. Another quantum number is ms, the spin quantum number with values 1/2 or - 1/2. The n, l, and ml values are interrelated as shown below:

The n quantum number is called the principal quantum number and is related to the average distance of the electron from the nucleus.

The l quantum number is called the angular momentum quantum number and it is related to the "shape of the atomic orbitals". We represent values of the l quantum numbers by the following letters:

value of l01234
letter usedspdfg

Thus we will talk about the 1s level (n = 1 and l = 0), the 2p level (n = 2 adn l = 1, etc.

Most books talk about orbitals as the clouds of electrons about the nucleus of the atom. Although such a definition gives me a severe case of the mental hives, we will also use that terminology. The spherical electron clouds are then called the "s orbitals", the dumbell shaped electron clouds are called "p orbitals", and there also the d and f orbitals that will not concern us in this course. The 1s orbital is represented by a smaller electron cloud than the 2s orbital because 2 is greater than 1 and the n quantum number is related to the distance of the electron from the nucleus.

Most books then talk about the principle energy levels corresponding to the n values: the 1st principle energy level (n = 1), the second principle energy level (n = 2), etc. The 1st principle energy level will only contain 1 orbital (1s) for there is no such thing as the 1p, 1d, or 1f orbital or level. The second principle energy level will contain the 2s and 2p orbitals (no such thing as the 2d or 2f level). The third principle energy level will contain the 3s, 3p, and 3d orbitals (no such thing as the 3f level). We call these collections of orbitals in a principle energy level shells..

Noting the interrelations of quantum numbers above, we see that for the 2p (n = 2 and l = 1) level, ml can be -1, 0, +1. Therefore there are three 2p levels. This collection of three sublevels is called a subshell. To complete the 2nd shell, we would have the 2s level that would be another subshell. So the 2nd shell has 2 subshells.

The hydrogen atom has only one electron so its ground state energy state would be the 1s state. If energy is input into the system the electron can be excited into an excited state. However at room temperature the hydrogen atom will have its electron almost exclusively in the ground state.

So we have for the hydrogen atom, energy levels which we can depict as the following:

E3 n=3;l=0,1,23s,3p,3d

In the last column you see that all of the "orbitals" with the same n value have the same energy. In quantum mechanics this phenomena is called degeneracy. So we say that the 4s, 4p, 4d, and 4f are energy degenerate meaning that they all have the same energy. We saw in our last discussion that the energy only depended upon the n value. This is true for hydrogen, but as soon as a second electron is added, like for the helium atom, the degeneracy is removed and the "orbitals" with the same n values no longer have the same energy.

When we try to solve quantum mechanics for helium or for any system with more than one electron we encounter a mathematical problem that nobody has been able to solve exactly (analytically). However the approximation techniques have become very sophisticated and we can solve quantum mechanics for very large molecules with great accuracy now.

When we have a system with two or more electrons the energy is now a function of n and l. So we then use the "n + l" rule to determine which combination of n and l have the lowest energy: "The combination with the lowest value of n + l has the lower energy. If two combinations have the same value of n + l then the combination with the lower n value has the lower energy."

Then using the above n + l rule we can develop the following ordering of energy levels, starting with the lowest energy and proceeding to higher energies:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d <5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p

To check this out, below I have written the n + l value for each n and l value above. Make sure that you understand this.

1< 2< 3 = 3 <4 = 4 <5 = 5 = 5 <6 = 6 = 6 < 7 = 7 = 7 = 7 < 8 = 8 = 8

You see that whenever the n + l value was equal, I choose the pair with the lower n value first. For example, 3p goes before 4s. If you grasp this rule you can use this as a basis to easily write the quantum numbers for any electron in the periodic table. We do have to know a couple of more concepts that we need for such a task:

The Pauli Exclusion Principle which says that no two electrons in the same atom can have all four quantum numbers the same.

The Hund Rule which says that when you are adding electrons to build up an atom the electrons prefer to maintain the same spin (either spin up or spin down) as long as possible.

We will now demonstrate the application of the above three rules to build up the periodic table.

We always start with the lowest possible energy which is the 1s level. So hydrogen will have the electron configuration:

1s ---- n = 1, l = 0, ml = 0, ms = -1/2;

We will always assume that the spin down (ms = -1/2;) quantum number is the one with the lowest energy.

We can form the electron configuration for helium by adding an electron with spin up :

1s2 ---- n = 1, l = 0, ml = 0, ms = +1/2

We used the n + l rule and the Pauli exclusion principle which allowed us to add another electron with the n, l, and ml values because we could make the ms value different.

Now we go to lithium and we can't use the 1s level anymore without violating the Pauli exclusion principle. So the third electron we add will be a 2s electron.

So we have the electron configurations for the first three elements:

H 1s

He 1s2

Li 1s22s

As we added another 1s electron to the hydrogen electron configuration to get the helium electron configuration, we can add another electron to the helium electron configuration to get the beryllium electron configuration;

Be 1s22s2

What are the four quantum numbers for the last electron added to form the beryllium electron configuration? Answer is at the bottom of this page.

Now we are ready for boron. We are through with the 1s and 2s levels without violating the Pauli exclusion principle so we now go to the 2p level:

B 1s22s22p

Now with the 2p level we have three ml values to work with (-1, 0, 1). Again we will start with the negative value for the lowest energy so that the four quantum numbers for the last electron added to boron are

n = 2, l = 1, ml = -1, ms = -1/2

When we go to carbon we can either go to ml = 0 and keep the ms = -1/2 or we can keep ml = -1 and change ms to +1/2. Which should we do? Remember Hund's rule? Yup. So we keep the ms = -1/2 and change ml to 0. Carbon will then have the electron configuration

C 1s22s22p2

and the last electron added has the four quantum numbers

n = 2, l = 1, ml = 0, ms = -1/2

Now we go to nitrogen and we can still keep ms = -1/2 by changing ml to +1. So nitrogen will have the electron configuration

N 1s22s22p3

and the last electron added has the four quantum numbers

n = 2, l = 1, ml = 1, ms = -1/2

Now to determine the electron configuration for oxygen we must change ms to +1/2 and start the variation of ml over again starting with -1. Then the electron configuration for oxygen will be

O 1s22s22p4

and the last electron added has the four quantum numbers

n = 2, l = 1, ml = -1, ms = +1/2

Then for fluorine we can continue the variation of ml so that we have

F 1s22s22p5

and the last electron added has the four quantum numbers

n = 2, l = 1, ml = 0, ms = +1/2

Similarly for neon we will have

Ne 1s22s22p6

and the last electron added has the four quantum numbers

n = 2, l = 1, ml = +1, ms = +1/2

Now we are through with 1s, 2s, and 2p. Make sure that you understand this.

When we fill a shell, we can use the symbol [rare gas] to indicate that portion of the electron configuration so that we don't have to write out all details:

Mg [Ne]3s2



Now we are ready to understand why the periodic table shows periodic properties as it does (the elements in the first column all have similar chemical properties, the elements in the second column all have similar chemical properties, etc.). The chemical properties of an atom are determined by the outer electrons, or the electrons in the outer shell. Looking at the electron configuration of the elements in the first column, we see that they all have only 1 electron in the outer s orbital. All the elements in the second column have 2 electrons in the s orbital. So now we understand what Mendeleev could not understand when he first published his first periodic table (1869). Quantum mechanics was not on the scene until the 1920's.


The ionization energy is the energy required to remove the outer electron from an atom in the gaseous state. Some values of these are presented in your book. Here again we see trends as we would expect due to the similar electron configurations down a column. Since the out electron is getting further away from the nucleus as we go down a column (n increasing), we expect the ionization energy to decrease. Going across a row we see the ionization energy roughly increasing due to the pairing of electrons in the levels. We won't be concerned with the explanations of the irregularity at this point.


  1. How many unpaired electrons does P have? Post your answer, with an explanation, on WebCT.
  2. Here is a good site for an overview of quantum numbers and electron configuration.
  3. Here is a site which explains the building of the periodic table using the quantum numbers.>
  4. wct


After you have studied this material and practiced some problems, take quiz Three. If you score at least 80 on the test then you are ready to continue to the next section.

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Web Author: Dr. Leon L. Combs
Copyright 2001 by Dr. Leon L. Combs - ALL RIGHTS RESERVED