Download What We Need to Know About Electrons

What We Need to Know About Electrons
1.
Electrons are moving charged particles therefore they are magnetic.
2.
Electrons move in two ways –
a. Spin on their axis – given quantum number S
b. Orbit the nucleus – given quantum number L
Can define a total angular momentum – quantum number J
J=S+L
In simplest case L can be neglected as electron is an orbitally nondegenerate state. Therefore can consider “Spin Only” behaviour.
Spin only is a good approximation for some transition metal ions:
Ni2+ in an octahedral field; high spin Mn2+ or Fe3+; octahedral Cr3+ or
Mn4+; Cu2+ and Mn3+ as Jahn-Teller distorted.
Terrible approximation for others – e.g. octahedral Co2+
Curie Law
1.
As single electrons are magnets, if you place them in a magnetic field they’ll
align with the field. However the energy difference between aligned with
field and against field is << thermal energy at room temp.
Get random orientation – equal populations of alignment with/against field.
2.
As you lower T, energy difference becomes more important and population
changes – more align anti-parallel to the field.
3.
To explain this behaviour Curie invented a parameter – called “Magnetic
Susceptibility”, χ, – which is a measure of how attracted a sample is to a
magnetic field. Normally measured as an apparent mass increase. As more
electrons align anti-parallel to the field at low temperature, χ increases. In
fact χ is inversely proportional to field: this is the Curie Law
1/χ = CT
C = “The Curie Constant”
Curie Law Plots
Slope = C
The Curie Constant and Magnetic Moment
Curie Law:
1/χ = CT
The molecular information is in the slope of the line,
i.e. C – “the Curie constant”. C = χT
Traditionally this is converted into “magnetic moment”, µ
µ = (8C)1/2 –
8 comes from a bunch of constants
Also:
µ = g[J(J+1)]1/2 which becomes
µSO = g [S(S+1)]1/2 if L = 0
For systems where unpaired electrons don’t communicate to each other
(magnetically dilute) µ is a good parameter as the Curie Law applies.
For systems where unpaired electrons communicate (magnetically non-dilute) µ
has little meaning as the Curie Law is not obeyed:
Reason:
1/χ vs. T is not a straight line, therefore a parameter derived
from the slope, C, is meaningless
χT
C and µ are pretty redundant. If we combine all the previous equations we get:
χT = C = µ2/8 = [g2S(S+1)]/8
This is much better because it relates the measured parameters directly to
molecular properties.
If there is more than one spin centre, you simply include that as a parameter
giving:
χT = g2 n. S (S+1)
8
Where n is the number of centres. If there is more than one type of centre or
spin level, it is easy to adjust:
χT = ga2 na. Sa (Sa+1) + gb2 nb. Sb (Sb+1)
8
8
χT – Simple Examples
χT = g2 n. S (S+1)
8
Take three systems containing five electrons (assume g = 2):
Five Ti3+
One H.S. Fe3+
One L.S. Fe3+
n = 5, S = ½
n = 1, S = 5/2
n = 1, S = 1/2
χT = 15/8
χT = 35/8
χT = 3/8
Non interacting
Spin pairing
disfavoured
Spin pairing
favoured
≡ Ferromagnetic
≡ Antiferromagnetic
Less Simple Example – A Copper Dimer(1)
Put unpaired electrons on different metal centres, e.g. two Cu2+
ions. Let them interact. Get two possibilities:
1. Align parallel i.e.
S=½+½=1
2. Align anti-parallel, i.e.
S=½-½=0
YOU ALWAYS GET BOTH
CASES. QUESTION IS WHICH IS
GROUND STATE
Less Simple Example – A Copper Dimer(2)
Assume S = 1 and S = 0 have same energy, then use:
χT = ga2 na. Sa (Sa+1) + gb2 nb. Sb (Sb+1)
8
8
Complication – must allow for degeneracy of states. Degeneracy given by
2S+1. If S = 1 and S = 0 have same energy, three times as many molecules
will be in the S = 1 state as S = 0.
χT = 1/4 1.ga2 0 (0+1) + 3 gb2 1 (1+1)
{
8
χT = 3g2/16
8
}
If we had two non-interacting S = ½ then
χT = g2 n. S (S+1)
8
χT = 3g2/16
At high T often find χT ≈ value for non-interacting spins. This is not
because there is no interaction – simply says interaction energy < kT
Less Simple Example – A Copper Dimer(3)
If S = 0 has lowest energy then:
χT = ga2 na. Sa (Sa+1) + gb2 nb. Sb (Sb+1)
8
8
becomes
χT = g2 n. S (S+1)
8
With S = 0, i.e. χT = 0
At low T will always find χT tends to 0 if S = 0 is the ground state
because at very low temperature interaction energy >kT
S = 1 (threefold degenerate)
S=0
At low T Boltzmann gives all
population down here
Less Simple Example – A Copper Dimer(4)
If S = 1 has lowest energy then:
χT = g2 n. S (S+1)
8
With S = 1, i.e. χT = g2.1.(2)/8 = g2/4
At low T may find χT tends to g2/4 if S = 1 is the ground state because
at very low temperature interaction energy >kT
S=0
S=1
At low T Boltzmann gives all
population down here
Less Simple Example – A Copper Dimer(5)
Summarise:
1.
For two Cu2 ions if no interaction then χT = 3g2/16
this is likely to be the limiting value at high temperature
even if there is an interaction
2.
At low T the value of χT will tend to 0 if S = 0 is the ground state
this is the case if anti-ferromagnetic exchange is the major
interaction, i.e. aligning neighbouring spins anti-parallel
3.
At low T the value of χT will tend to g2/4 if S = 1 is the ground state
this is the case if ferromagnetic exchange is the major
interaction, i.e. aligning neighbouring spins parallel
Nomenclature
S=1
= 2J or J
S=0
Sane version:
J negative = AF
J positive = F
Treating Raw Data
Not said anything about experiment – and not going to (until we have a SQUID).
Experimental data: from a SQUID will normally get a column of temperature vs
“long moment” at a specific field (also a whole pile of columns you don‘t need).
Field (Oe)
Temperature (K)
Long Moment
(EMU)
1.000000e+002
1.198679e+002
2.952116e-004
1.000000e+002
9.020821e+001
3.642680e-004
1.000000e+002
6.241134e+001
4.590345e-004
1.000000e+002
3.206844e+001
7.022155e-004
1.000000e+002
1.003232e+001
3.274861e-003
χ comes from long moment.
Very simple equation:
χ=
(Long moment/field)
(mass of sample/MW)
[You then need to add a diamagnetic correction – calculated from Pascal’s
constants (Rubber Handbook and O’Connor review).]
Most common problem – getting MW correct. If there’s solvent and you don’t allow
for it, χ will seem low. If you include solvent and it ain’t there, χ will be high.
Diamagnetic correction vital for weakly paramagnetic samples (e.g. Cu2+
complexes); less vital for strongly paramagnetic samples (e.g. Fe3+).
Result – “absolute” values are more likely to be wrong than shape of curves
Plots for AF Copper Dimer - χT
J=0
J = -10
J = -50
0.8
ChiT
0.6
J = -100
0.4
J = -200
0.2
Obvious: decline
in χT faster for
larger AF J value
0.0
0
50
100
150
T
200
250
300
Less obvious: lack of maximum makes fitting data imprecise. Also, as χT
zero at low T, fit will be very sensitive to impurities.
Plots for AF Copper Dimer - χ
J=0
Maximum in χ arises because Curie law says
χ rises as T falls, while energy levels means
S = 1 becoming depopulated.
J = -10
When fitting for AF system should fit χ vs T
first, as the maximum is very sensitive to
value of J for moderate J.
Chi
0.02
(For large J fit will always be approximate).
J = -50 J = -100
J = -200
0.00
0
100
T
200
300
Plots for F Copper Dimer - χ
0.10
Plots shown for J = 0, +10, +20, +50,
+200
0.08
Virtually indistinguishable!
Plot of χ vs T useless for getting size of J
(or even telling F exchange from no
interaction).
Chi
0.06
0.04
0.02
0.00
0
100
T
200
300
Plots for F Copper Dimer - χT
1.1
J = 200
ChiT
1.0
J = 50
0.9
J = 20
J = 10
J=0
0.8
0
50
100
150
T
200
250
300
χT very dependent on J: still no maximum so it will be more
difficult to get a precise J cf. moderate negative J value.
Why AF or F for Copper?
Both F and AF exchange interactions are observed for copper(II). One classic
piece of magnetochemistry (due to Hatfield) demonstrated what controls this
for hydroxide bridged dimers:
OH
Cu
Cu
OH
J value depends on Cu-O-Cu angle:
Angle > 98.4°, J > 0 i.e. F
Angle < 98.4°, J < 0 i.e. AF
Why AF or F for Copper?
Easily explained from MO picture:
Single upe on Cu(II) is in dx2-y2 orbital – which has a lobe on
the Cu...O vector.
Imagine O uses a p-orbital to bond to each Cu(II) centre.
x
y
OH
Cu
Cu
OH
If Cu-O-Cu angle = 90° then:
O-atom will bind to one Cu
with px orbital and the other
with py orbital.
Why AF or F for Copper?
x
y
OH
Cu
Cu
OH
dx2-y2
px, py
Look at bonding: get two
pairs of bonding/anti-bonding
orbitals:
Add electrons: 2 from each
O orbital + 1 from each dx2y2, get 6.
Feed into orbitals gives two
upe in ground state, i.e. F
exchange
Why AF or F for Copper?
x
y
Cu
OH
Cu
Change angle to ca. 180°
Now O-atom only uses one
orbital. Energy levels
become:
Only need include 4
electrons (2 from O and 1
each Cu): gives 0 upe, i.e.
AF exchange
dx2-y2
py
Why AF or F for Copper?
M.O. picture tells us about extremes: doesn’t tell us where F Æ
AF transition should occur.
First part (F exchange) is a specific case of the orthogonality
principle: if the unpaired electrons are in orthogonal orbitals,
then the exchange will be ferromagnetic.
Angle of 98.4° only has importance for planar hydroxide bridged
copper(II) dimers.
Not uncommon for assumption to be made that this cross-over
applies to other systems – it might, but that will be coincidence.
Does appear to be common that the more obtuse the angle, the
stronger the AF exchange found.
Bigger Clusters(1)
The bigger the cluster the more complicated the magnetic properties.
The basic interpretation (F or AF) is normally fairly simple.
First complication: AF exchange does not always lead to S = 0 ground state
For example – consider a linear trinuclear cage with S spin on each ion:
Ferromagnetic
exchange.
Anti-ferromagnetic
exchange.
SGS = 3S
SGS = 1S
Ferro’ exchange will always give a larger SGS than AF exchange,
but AF exchange can give very large SGS if the shape is correct.
Bigger Clusters(2)
For example:
Ferromagnetic
exchange.
Anti-ferromagnetic
exchange.
SGS = 7S
SGS = 5S
Most high spin cages involve AF exchange as the predominant
interaction: e.g. all Mn12 derivatives, Fe19, Fe14, Cr12
Bigger Clusters(3)
Second complexity: more than one exchange path, therefore more
than one J value (in general)
e.g. have centreedge (black) and
edge-edge (red)
interactions
Separating the two components can be difficult. Can normally
work out sign of stronger interaction.
Bigger Clusters(4)
Suppose black
interaction
strongly AF.
Suppose red
interaction weakly
AF.
Result SGS = 5S
Notice: black interaction overcomes red and aligns spins parallel.
Does not matter whether red is AF or F.
Bigger Clusters(5)
Suppose red
interaction
strongly AF.
Suppose black
interaction weakly
AF.
Result SGS = ?
Problem: ring couples to give Sring = 0, but centre wants to be
coupled AF to two different spins. Central spin is “frustrated”.
Result: probably ground state is S.
Examples of Bigger Clusters(1)
Ni12 Æ SGS = 12
1.
χT never falls – sign
that exchange is
ferromagnetic.
2.
χT close to Curie law
(red line) – shows
coupling very small
3.
Low T χT does not
reach value for S = 12:
even at 1.8 K excited
states are occupied.
Confirms point 2.
4.
Shape of χT curve more
important than low T
value in showing F or
AF exchange.
Examples of Bigger Clusters(2)
Fe10 Æ SGS = 11
1.
χT starts below value for
10 S = 5/2 ions. Says
exchange large and AF.
2.
Low T χT plateaus near
value for S = 11.
Confirms exchange large.
3.
Real problem with
sample: working out
molecular weight.
Crystals heavily solvated,
but solvent lost during
SQUID measurement.
Examples of Bigger Clusters(3)
Fe9 Æ SGS = unknowable
?
??
Any attempt to assign spin
rapidly becomes impossible
– and impossible many
times!