Crystal Field Theory (Octahedral Geometry) - Coordination Chemistry Video Lecture | Inorganic Chemistry

in this video I'm going to introduce
crystal field theory which is a model of
bonding that helps describe the
brilliant and wide range of colors that
are seen in these coordination compounds
and also the magnetic properties so
remember the metals that we've been
discussing our transition metals and if
we recall from an earlier section in
chemistry atomic orbital Theory the s
block or an S orbital is spherically in
shape and we can have a 1 s when N
equals 1 and a 2 s orbital so lithium's
electron would go in a 2 s orbital and
since only 2 electrons can go in each
orbital the s block is 2 atoms wide so
to speak
the p subshell or the p block or the p
subshell has 3 orbitals and they're all
similar in shape but they are oriented
along the x y&z axis and if we label
that x y&z and there are 6 electrons
that can fit in the sub shell of P
orbitals so two electrons in each
orbital I have a picture of the D
orbitals because I can't draw those but
the transition metals occur in this
region here and the D block has 5
orbitals
and each of those orbitals have
different shapes which I'm going to show
here and two orbitals of particular
importance are going to be the DX
squared minus y squared orbital and the
DZ squared if we look at the difference
these three orbitals they're all D
orbitals these 3 that each have four
lobes the lobes lie
in the plane and the shapes do not point
along the X or Y axis so that's going to
be of importance when we look at the
octahedral geometry so I could try to
draw an octahedron from this similar to
this the octahedral geometry that we've
been drawing where the metal cation I'm
just going to put a plus in the charge
could be plus 1 2 or 3 or 7 at the
central metal which is acting as a Lewis
acid is coordinated at these sites as
well so the four vertices of a
tetrahedron are actually shown here well
the four along the plane and then that
virtus this is the this would be where
the DX squared minus y squared orbitals
actually point right to those vertices
and then the DZ squared points at the
vertices along this vertical axis so
Yi's two orbitals we're going to see
they experience the most repulsion so
remember all five of these D orbitals
would be around the central metal and so
if we consider more of a chemists view
of an electron as a negative smeared out
charge so if that metal has got the
these two orbitals there that negative
charge is going to repel any ligands
that would be attracted to this positive
metal center so let's say I have a
cyanide ligand that negative charge on
the cyanide is actually going to
experience repulsion and what that's
going to do energetically is
cause these two orbitals to be higher in
energy than these three orbitals so that
is when we look at an octahedral complex
so the repulsion between the metals D
orbitals
specifically the d x squared minus y
squared and the DZ squared increase the
energy because that's not a favorable
that's a repulsion instead of an
attraction so even though those
electrons on the cyanide are attracted
to the positive metal the nucleus of
that metal that metal still has these D
orbitals that are pointing in the
positions where those ligands will be
coordinating the metal so this increases
just insane increase in energy with
respect to the other three orbitals so
we're going to see on the slide in just
a moment that we're going to have the
five D orbitals for example with the
free metal so the metal cation just in
solution and then when the ligands are
coming toward it we can think about that
like they're sort of coming in for a
landing on a spaceship but when the
ligands are going to coordinate that
metal that actually causes an increase
in energy because of that repulsion term
and when the electrostatic charges are
attractive that actually decreases
energy what's going to happen with the
metals D orbitals is the three orbitals
that don't experience the repulsion are
going to be lower in energy than the
free metal and then the DX squared and x
squared minus y squared and the DZ
squared orbitals
higher in energy than these three
orbitals that don't point at the
vertices of a tetrahedron
so these experience the most repulsion
so again that's going to be a
high-energy situation and these three
orbitals experience the least repulsion
from the ligands and so they will be
lower in energy than these two if we
remember atomic orbital Theory all of
the D electrons on a free ion or metal
have the same energy so when we put
electrons in we would keep those
electrons unpaired because electrons
repel each other so that repulsion
repulsion between electron-electron
repulsion energy is the reason that it's
more favorable for electrons to stay
unpaired before they go to the next
higher energy level so that concept
we're going to keep and one of the
things we're going to have to do is
remember that the orbital split in this
fashion in an octahedral geometry so I
should write that here so this is an
octahedral geometry which is very common
and we're going to have to remember that
the three orbitals are lower in energy
the d x squared minus y squared and the
DZ squared the orbitals are higher in
energy because of the physical
positioning of those orbitals with
respect to an octahedral geometry now
let's go back to this page
so we're going to be using the periodic
table because if we have for example an
iron that's in the plus 3 oxidation
state that means iron loses three
electrons and so we're going to just
count over to see how many D electrons
and iron plus three would have so we
just cross out there's 1 2 3 electrons
that are gone in an iron plus 3 ion and
then we just count so it would have 1 2
3 4 5 D orbitals left so we would say
that iron plus 3 is a d5 metal and if we
do a quick atomic orbital diagram for
iron just the element iron you know it's
element number 26 we have the 1s 2s and
2p then the 3s and the 3p and the trick
here was remembering that the 4s orbital
is actually lower in energy that sub
shell is lower in energy than the 3d
orbitals so a sphere is that spherical
shape minimize the surface area which is
a more stable shape and remember as we
go increase in energy we're increasing
in this direction the higher we go
within the closer those energy values
get but if we were to put 26 electrons
in to represent iron the atom here's
argon and we fill the 4s before the 3d
so that was the trick on doing an
electron configuration from an orbital
diagram and this is N equals 1 2 3 4 so
we know this is the
for s-block and the DS begin at energy
level three so this is actually the 3d
block and then iron there's 20 21 22 23
24 25 26 so iron the element has 26
electrons but if iron loses three
electrons to become iron plus three the
4s electrons are going to come off first
so 4s are removed before the 3d so we
sort of pretend that we're in charge of
putting electrons in their position and
we're also responsible for removing them
so if we get rid of the 4s electrons
that's those two electrons that's a 4s1
4s2 and then one of these d electrons
will be gone and so we're in that
position and so instead of having to do
this every time we can just count over
after we've eliminated electrons and
count how many positions over we are and
that will be how many D electrons that
that metal ion has in the center again
that positive charge is what attracts
those ligands whether they have a lone
pair or whether they are neutral and or
whether they have a charge so crystal
field theory we're going to talk about
and about out of lead not out of time
here so let's grab a pin
so for crystal field theory what's going
to be important for my students is to
from the formula from any compounds
formula and we'll do an example
determine the charge on the metal so
that is the oxidation state and that
will determine how many D electrons are
gone and so that will be the oxidation
state then determine how many D
electrons that metal ion has and from
there if we have an octahedral geometry
you will just put the electrons in the
orbitals remembering that those orbitals
split in energy and so that's where we
lose electrons then the splitting of the
D orbitals
and this is important this is in an
octahedral geometry we'll talk about
tetrahedral and square planar later in
an octahedral geometry will draw the
five D orbitals and there may be a very
large gap or just a small gap between
those D orbitals so we'll discuss that
on the next slide so our particular
chemistry book just calls this Delta but
this is Delta e now energy is increasing
going up the paper so the free metal
oops the free metal is here and then in
the presence of the ligands so we can
have this situation or we could have a
very large gap so I didn't draw that
very well I think I'll make this one
smaller so there may be a very small
difference in energy so this is still
our d x squared minus y squared and the
DZ squared those two orbitals that point
exactly at the metals ligands the chord
the coordination sites so we could have
a small energy gap or we could have a
large energy gap and that depends on
multiple things but primarily the
spectrochemical series which we will
look at on the next slide okay and so if
electrons if there were four electrons
here on the if we had a D for metal for
example in this situation the fourth
electron
would be in a higher energy orbitals we
kind of messed that up putting the label
there but the electron-electron
repulsions are actually greater than the
Delta II so nature is going to always do
the lowest energy most stable
configuration so these electrons instead
of pairing up because the
electron-electron repulsion is greater
than this energy gap the electron the
fourth electron will be here we actually
call this a high spin complex to me that
seems backwards because but that's
because the electron is in a high energy
level and so we talk about the span of
an electron if we have another D for
metal and depending on the ligands which
we'll discuss later these four electrons
would look would be in a different
arrangement so we call this a low spin
complex because all the electrons are in
the lowest energy levels and what causes
the difference between a large gap and a
small gap primarily depends on the
ligands and we will look at the
spectrochemical series and discuss that
in the next video