Hund’s Rule and Magnetism
This idea of spectroscopic notation can also help us explain a variety of different properties and
phenomena seen on the Periodic Table. One of these properties is the idea of magnetism. We
know that some metals respond to a magnetic field and others don’t as well. Why is that? To
answer that question, we first need to understand what magnetism is.
Magnetism is an atoms’ response to a field applied to it. You should remember from Physics
class that a charged particle that spins produces a magnetic field. This is very much like the
earth’s magnetic field. Whichever way the particle spins, the field is generated in a particular
direction.
Here is a particle
that is spinning.
Because it is spinning this direction,
it generates a magnetic field like this:
Here is a particle
that is spinning in
the opposite direction. It’s field is
reversed from the
first:
Let’s set up an experiment. We will place a sample of an element on a scale and zero it with a
magnetic field off. The we will apply a magnetic field and see how much the sample is moved
by the magnetic field. The more it is moved, the greater the number on the scale.
Without getting to specific into the values, let’s collect the data based upon the following scale:
Some response to a field
No response to a field
Lots of response to a field
Let’s then apply the experiment to all the elements in the fourth row of the periodic table. Here
are the results:
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
Doesn’t seem like much of a pattern but let’s look at with the splits for each subshell shown:
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
A couple of things to notice here. First of all, it seems within each subshell, response to a magnetic field starts off small and then gradually increases until the halfway at which point it goes
back down. Second, at the end of every subshell the response is zero:
K
Ca
Sc
Ti
V
Increases to
halfway
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
After halfway
decreases until
zero at the end
Ge
As
Increases to
halfway
Se
Br
Kr
After halfway
decreases until
zero at the end
When we combine all the knowledge from our previous sections and apply it to this, we can
begin to make some sense of the above results. Let’s do it section by section:
K
Ca
For potassium, K, the spectroscopic notation is [Ar] 4s 1 which means there is one electron in
the s orbital. Because this is the “s” subshell, l=0
Thus, ml can only = 0 which means there is only 1 orbital.
Let’s represent this orbital with a box:
s
n=4
l=0
ml = 0
Now what we have to do is fill that orbital with the proper number of electrons. Since potassium is 4s 1, we have
only 1 electron that must have ms = +1/2 or –1/2. It doesn’t matter which you choose. Let’s represent +1/2 with
an up arrow and –1/2 with a down arrow. Let’s start with +1/2 or the up arrow in the orbital box:
s
n=4
l=0
ml = 0
ms = +1/2
What we see is that there is one electron in the box spinning “up” and making a magnetic field. Thus, we get a
small response to the applied field which explains the results seen at the top of the page.
K
Ca
Now let’s move on to Calcium. Ca is [Ar] 4s2. This means it has 2 electrons in the 4s subshell.
The Pauli Exclusion Principle says that no 2 electrons can have the exact same four quantum
numbers so the 2 electrons in calcium must be:
n=4
l=0
ml = 0
ms = +1/2
n=4
l=0
ml = 0
ms = -1/2
Which means that both these electrons are in the same orbital but they must be spinning in opposite directions. To
fill the orbital box we get:
n=4
l=0
ml = 0
ms = +1/2
s
n=4
l=0
ml = 0
ms = -1/2
What we see now is there are 2 electrons with opposite fields spinning in the same location.
The opposite nature of these resultant magnetic fields cancels each other out. Thus, Calcium exhibits no response
to a magnetic field which is exactly what we see on the results. The two results here are typical of what we see in a
variety of different elements so we need terms to identify them. The two terms are as follows:
Paramagnetic = the element responds to a magnetic field due to
unpaired electrons in orbitals
And
Diamagnetic = the element shows no response to a magnetic
field due to completely filled orbitals with no unpaired electrons
s
K
Ca
One unpaired electron so we see
a small response to magnetism
K
No unpaired electrons so we see
zero response to magnetism
Ca
Let’s continue this across the row. Scandium is our next element
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Sc has the notation: [Ar] 4s23d1 which means its valence electrons are in the 3d subshell. Thus:
3d
n=3
l=2
ml = -2 -1 0 +1 +2
which means there are 5 orbitals which means we need to draw 5 orbital boxes past the s:
s
d
We know the 2 electrons in the 4s will cancel out their magnetic fields from having a full orbital box and the third electron will go into a “d” orbital unpaired which gives a small response
to a magnetic field:
s
2
Sc: [Ar] 4s 3d
d
Sc
1
All is following the pattern seen but we’re going to run into a question on the next element, Titanium. Doing the same thing as above, we have 2 choices. We either pair up this next electron
s
d
Paired electrons
in the d2 orbitals
Ti: [Ar] 4s23d2
OR we can put the next electron in the next orbital over:
s
Ti: [Ar] 4s23d2
d
2 electrons now unpaired in 2 different
orbitals
What do we choose? What would be the consequences of each and how would we know?
Think about the magnetism:
If we go with choice #1 which is that the next electron pairs up, what would be the expected
result for the magnetic field?
s
d
Paired electrons
in the d2 orbitals
Ti: [Ar] 4s23d2
Since all the electrons are paired, the response should be zero. Is that what we see?
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
NO! In fact, the response Titanium has is MORE than that of Scandium which means it must
have more than 1 electron that is unpaired AND that unpaired electron must be spinning the
same way as to increase the magnetic field seen:
s
d
Ti: [Ar] 4s23d2
Both of these electrons are unpaired and spinning the same direction, thus giving off the same
magnetic field. Instead of canceling out, they
constructively interfere with each other and
boost the response to the magnetic field!
This trend continues across the row with the next three elements as we see the new electrons
being put in empty orbitals spinning the same direction and increasing the magnetism:
s
d
Sc: [Ar] 4s23d1
Ti: [Ar] 4s23d2
V: [Ar] 4s23d3
Cr: [Ar] 4s23d4
Mn: [Ar] 4s23d5
We see an increase in the number of unpaired
electrons all spinning the same direction and this
coincides with an appropriate increase in the
response to the magnetic field!
What we glean from this is what is known as Hund’s Rule which says:
The most stable arrangement of electrons in an atom is when degenerate (same subshell) orbitals are half-filled spinning the same direction or totally filled.
Thus, we half-fill similar orbitals with electrons before we totally fill any of them. If we continue the pattern from before, however, we have to move past the point where all the d orbitals
are half-filled:
s
d
Sc: [Ar] 4s23d1
Ti: [Ar] 4s23d2
V: [Ar] 4s23d3
Cr: [Ar] 4s23d4
Working up to
half-filled
Increasing
magnetism
Must start filling the orbitals
with electrons
spinning the
other direction
Decreasing
magnetism
Mn: [Ar] 4s23d5
Fe: [Ar] 4s23d6
Co: [Ar] 4s23d7
2
8
Ni: [Ar] 4s 3d
Cu: [Ar] 4s23d9
Zn: [Ar] 4s23d10
And we must start filling the orbitals with electrons spinning the other direction. Every one of
these we add, the opposite magnetic field cancels out some the magnetic response. By the time
we get to Zinc, at the end of the subshell, all the orbitals are completely filled and thus Zn atoms are completely diamagnetic and give no response to magnetism.
This however, leads to two different exceptions that take place. I said before that Chromium
and Copper have the following arrangements:
s
d
Cr: [Ar] 4s23d4
Cu: [Ar] 4s23d9
This is, in fact, NOT true. Look at each notation and remember Hund’s Rule (the most stable
arrangement is when similar orbitals are 1/2 filled or totally filled). In the case of Chromium,
four of the five degenerate orbitals are half-filled. In the case of Copper, all the degenerate d
orbitals are almost totally filled. The atom could gain extra stability if it just had one more electron to put in each. However, these are both metals; they don’t like to gain electrons, they like
to lose them.
Because these orbitals overlap in energy and to gain extra stability, Chromium and Copper each
shift one electron from the s into the d:
s
d
Cr: [Ar] 4s23d4
Cu: [Ar] 4s23d9
Check out what just happened. Now each of the “d” orbitals are all satisfying Hund’s Rule.
They are either half-filled spinning the same direction or totally filled. Thus the true spectroscopic notation of Chromium and Copper are:
s
d
Cr: [Ar] 4s13d5
Cu: [Ar] 4s13d10
There are many different elements that do this. However, these are the only two examples the
AP test people want you to know so they are the only ones we are going to focus on.