Transcript Chapter 8

Chapter 8
Magnetic and optical properties
Magnetic Properties
Railgun
Magnetization and Magnetic Susceptibility
If a body is placed in a homogeneous magnetic field Ho,
the magnetic field with the body varies from free-space
value.
That is,
B = Ho + 4p M
B : magnetic induction (the field within the body)
M : intensity of magnetization
B/Ho = 1 + 4p (M/Ho) = 1 + 4p cv
M/Ho : dimensionless
cv : magnetic susceptibility per volume
Other definition of magnetic susceptibility
Gram Susceptibility: cg = cv/d
(unit: cm3/g)
d = density
Molar Susceptibility: cm = cg.(MW)
(unit: cm3/mol, or emu)
MW: molecular weight
Macroscopic Point of View
Magnetic moment, M, can be related to the rate
the energy change of a body in the magnetic
field, Ho.
-E
M=
Ho
The sign of the induced magnetic moment, M, defines
two types of magnetic behaviors:
Diamagnetism and Paramagnetism
-E
M=
Ho
M (and cv, cg, cm) is negative: diamagnetism
Materials with only paired electrons (In fact, all materials
exhibit diamagnetism)
M (and cv, cg, cm) is positive: paramagnetism
Materials with unpaired electrons (In fact, these materials
exhibit both paramagnetism and diamagnetism)
Ferromagnetism, antiferromagnetism
and ferrimagnetism
Ferromagnetism, antiferromagnetism
and ferrimagnetism can be considered
as special forms of paramagnetism.
Diamagnetism
a property of all materials
It arises from the interactions of
electron pairs with magnetic field,
generating a small magnetic field
opposing the applied magnetic field, Ho.
Diamagnetism
N
S
N
B = Ho + 4p M
S
Diamagnetism
Diamagnetic materials move to the region of lower field strength
(repelled by Ho)
M=
-E
Ho
For diamagnetic materials, M < 0,
∴
E
>0
Ho
Ho
That is, the energy of the system increases with the applied
magnetic field, Ho. ∴ the body moves in the direction of lower
energy (i.e. lower field). The process is exothermic.
Superconductors are perfect
diamagnetic materials
How to obtain diamagnetism (cdia)?
cT = cpara + cdia
To study the paramagnetic behaviors of
materials, cdia must be subtracted from cT
Pascal’s constants
Calculation of cdia from Pascal’s constants:
cdia = i cA + i cB
i
A: atoms
B: bonds
i
Table of Selected Pascal’s Constants
atom
cA (x10-6 emu)
atom
cA (x10-6 emu)
atom
cA (x10-6 emu)
H
-2.93
F
-63
Na+
-6.8
C
-6.00
Cl
-20.1
K+
-14.9
C(aromatic)
-6.24
Br
-30.6
bond
cB (x10-6 emu)
N
-5.57
I
-44.6
C=C
+5.5
N(aromatic)
-4.61
Mg2+
-5
C≡C
+0.8
N(monamide)
-1.54
Zn2+
-15
C=N
+8.2
N(diamide, imide)
-2.11
Pb2+
-32.0
C≡N
+0.8
O
-4.61
Ca2+
-10.4
N=N
+1.8
O2(carboxylate)
-7.95
Fe2+
-12.8
N=O
+1.7
S
-15.0
Cu2+
-12.8
C=O
+6.3
P
-26.3
Co2+
-12.8
anions
c (x10-6 emu)
Hg2+
-40.0
Ni2+
-12.8
C≡N-
-13.0
Example 1:
5 x C (ring) = 5 x (-6.24) = -31.2
5 x H = 5 x (-2.93) = -14.6
1 x N (ring) = 1 x (=4.61) = -4.61
H
H
H
N
H
H
pyridine
i cA
i
= -31.2 + (-14.6) + (-4.61) = -50.3 x 10-6 cm3/mol (or emu)
In this case,  cBi is zero, because the “ring values” of C and N are used.
i
Example 2:
O
3 x C = 3 x (-6) = -18
6 x H = 6 x (-2.93) = -17.6
1 x O = -4.61
H3C
CH3
acetone
 cA
= -18 + (-17.6) + (-4.61) = -40.21 x 10-6 emu
i cB
= 1 x (C=O) = +6.3 x 10-6 emu
i
i
i
cdia = i cA + i cB
i
i
= -33.9 x 10-6 emu
Example 3:
K4Fe(CN)6 Transition metal complex
4 x K+ = 4 x (-14.9) = -59.6
1 x Fe2+ = -12.8
6 x (C≡N-) = 6 x (-13.0) = -78.0
cdia = i cA + i cB
i
i
= -150.4 x 10-6 emu
How to obtain cdia of Cr(acac)3?
Cr(acac)3 is paramagnetic so it is difficult to measure its
diamagnetism directly.
Method I.
Calculate cdia from Pascal’s constants.
Method II.
Synthesize Co(acac)3, Co3+: d6 low spin.
Use the cdia value of Co(acac)3 as that of Cr(acac)3.
Method III.
Measure the cdia value of Na(acac), to obtain the cdia value of
acac-.
Then include this value in Pascal’s constant calculation.
H
H
Expt’l
Agreement
-50.4 x 10-6
-49 x 10-6
good
H
N
H
Calculated
H
AsMe2
-147 x 10-6
-194 x 10-6
poor
AsMe2
However, since cpara >> cdia , the disagreement
usually does not cause too much problem.
Disagreement of cdia
Paramagnetism
Paramagnetism arises from the
interaction of Ho with the magnetic field
of the unpaired electron due to the spin
and orbital angular momentum.
Derivation of M and cm
from microscopic point of view
For S = 1/2 system (Spin only, no orbital contribution)
 = -gS

: magnetic moment
S
: spin angular momentum
ge
: electron g-factor (ge = 2.0037 for free e-)
 ( or B) : Bohr magneton of the electron
 = 9.3 x 10-21 erg/Gauss
Interaction energy between
magnetic moment and applied field
The Hamiltonian describing the
interaction energy of this moment with
the applied magnetic field, Ho, is:
H = -.H = gS.H
The energies for S = ½ system
The energies (eigen values) for S = 1/2 (ms = +1/2, -1/2) system are:
E = msgHo
E1/2 = (1/2)gHo and E-1/2 = –(1/2)gHo
ms=1/2
Energy
ms=1/2
E = gHo
ms=-1/2
Zeeman effect
ms=-1/2
Ho
Relative populations of ½ and –½ states
When Ho = 25 kG E ~ 2.3 cm-1
At 300 K, kT ~ 200 cm-1
Boltzmann distribution
P1/2 e-E/kT ~ 1
=
P-1/2
The populations of ms = 1/2 and –1/2 states are almost equal
with only a very slight excess in the –1/2 state.
That means that even under very large applied field, the net
magnetic moment is still very small.
To obtain M ( or cm), we need to consider all the
energy states that are populated.
∵
H = -.H = gS.H
∴ The magnetic moment, n with the direction // Ho, of an
electron in a quantum state n can be obtained by:
n = -(∂En)/(∂H) = -msg
So we consider:
(1) The magnetic moment of each energy state.
(2) The population of each energy state.
That is, M = Nn.Pn
N : Avogadro’s number
Pn : probability in state n.
M = Nn.Pn
H = -.H = gS.H
n = -(∂En)/(∂H) = -msg
Pn =
Nn
NT =
-En/kT
e
-E /kT
e
n
Nn: population of state n
NT: population of all the states
Energy
ms=1/2
ms=-1/2
ms=1/2
n
1/2gH
-1/2g
E = gHo
ms=-1/2
E = msgHo
En
-1/2gH
1/2g
Ho
n = -(∂En)/(∂H) = -msg
M = Nn.Pn
n e
N
ms
-En/kT
M=
Since gH<<kT when Ho ~ 5kG
e
-En/kT
ms
±x
e
g/2kT
=
N [ g/2 e
-g/2kT
-g/2e
g/2kT
-g/2kT
[e
+e
N g [ 1 + g/2kT (1
= 2 1 + g/2kT + (1
=
N g2
4kT
]
~ 1±x
when x << 1
]
g/2kT)
]
g/2kT)
= M
M for S=1/2 system
Curie Law of paramagnetic materials
cm =
cm
M

=
slope = C
N g2
4kT
Curie Law: cm = C/T
N g2
∴
1/T
C=
4k
Curie-Weiss Law
 is called Weiss constant.
If  is positive, then the
magnets tend to align
parallelly.
If  is negative, then the
magnets tend to align
antiparallelly.
cm
 1
cm = C/(T-)
T
If the system is not magnetically dilute (pure paramagnetic), the
neighboring magnetic moments may exhibit an overall tendency of
parallel alignment or antiparallel alignment. (still considered as
paramagnetic, not ferromagnetic or antiferromagnetic)
For general S values (not only S = 1/2)
En = msgbH ms = -S, -S+1, …. , S-1, S
N m=-S (-msg)e
S
M=
-msgH/kT
=
s
e
-msgH/kT
N g2H S(S+1)
3kT
ms
cm =
M

=
N g2 S(S+1)
3kT
spin only
cm =
M

=
N g2 S(S+1)
3kT
S=3/2
For S=1/2
cm =
N g2
For S=1
cm =
2N g2
3kT
For S=3/2
cm =
5Ng2
4kT
S=1
cm
S=1/2
slope = Curie const.
4kT
1/T
c for S = 1/2, 1, 3/2
Definition of eff
cm =
eff
N g2 S(S+1)
3kT
=(
3k
N
) ( cmT ) ~ 2.828( cmT )
1/2
1/2
1/2
eff = g[S(S+1)]1/2 (BM, Bohr Magneton)
or eff = [n(n+2)]1/2 where n= number of unpaired e-
Saturation of Magnetization
The Curie-Wiess law does not hold where the system is
approaching saturation.
An approximation has been made:
gH << kT so that
±x
e
~ 1±x
If the applied magnetic field is very large,
Curie-Weiss law is not valid.
(M is not proportional to H)
cm =
M

=
N g2 S(S+1)
3kT
Saturation of Magnetization
follow Curie -Weiss law
S=2
4
S=3/2
M/ 3
mol-1
S=1
2
Energy
ms=1/2
ms=-1/2
ms=1/2
E = gHo
ms=-1/2
S=1/2
1
0
1
H/kT
2
M = -msg  M =  for 1 magnet.
If H is large enough, the
probability of ms= -1/2
populated is close to 100%.
Ho
Plot of eff vs Temperature
3.87
eff
S=3/2
eff = 2[S(S+1)]1/2
(BM)
eff = 2.828(cT)1/2
temperature
All the molecules are in the S=3/2 state at all temperatures.
Plot of eff vs Temperature
3.87
eff
S=3/2
(BM)
1.7
S=1/2
temperature
The eff value of the system gradually decreases from hightemperature value of 3.87 BM (S=3/2) to low-temperature
value 1.7 BM (S=1/2)
Spin equilibrium and Spin Crossover
3.87
eff
S=3/2
(BM)
3.87
eff
S=3/2
(BM)
1.7
S=1/2
temperature
1.7
S=1/2
temperature
Calculation of eff for f-block elements
Now, we consider spin-only cases.
For f-block elements, spin-orbit coupling is very large
eff = g[J(J+1)]1/2
S(S+1)-L(L+1)+J(J+1)
g = 1+
2J(J+1)
g-value for free ions
Example: calculate the eff of Nd3+ (4f3)
ml +3
+2
+1
0
-1
-2
-3
g = 1+
S(S+1)-L(L+1)+J(J+1)
2J(J+1)
Lmax = 3+2+1 = 6
Smax = 3 x 1/2 = 3/2 2S+1= 2 x 3/2 + 1 = 4
Ground state J = L-S = 6-3/2 = 9/2
Ground state term symbol: 4I9/2
3/2(3/2+1)-6(6+1)+(9/2)(9/2+1)
g = 1+
= 0.727
2x(9/2)(9/2+1)
eff = g[J(J+1)]1/2 = 0.727[(9/2)(9/2+1)]=3.62 BM
eff values of d-block elements
For d-block elements, spin-orbit coupling
is less important. In many cases, eff =
g[S(S+1)]1/2 is valid.
The orbital angular momentum is often
“quenched” by special electronic
configuration, especially when the
symmetry is low.
Spin-Orbit Coupling
z
y
y
ex
dx2-y2
x
dxy
For example, if an electron can move back and forth
between dx2-y2 and dxy orbitals, it can be considered
as circulating about the z-axis, giving significant
contribution to orbital angular momentum.
Spin-Orbit Coupling
dx2-y2
E
dxy
If dx2-y2 and dxy orbitals have different energies in a
certain electron configuration, electrons cannot go back
and forth between them.
∴ little contribution from orbital angular momentum.
Spin-Orbit Coupling
dx2-y2 dxy
E
Electrons have to change directions of spins
to circulate  Little contribution from orbital
angular momentum.
Spin-Orbit Coupling
dx2-y2 dxy
E
Orbitals are filled.  Little contribution
from orbital angular momentum.
Spin-Orbit Coupling
dx2-y2
E
dx2-y2 dxy
dxy
E
Spin-orbit couplings are significant in the
above two cases.
Other orbital sets that may give
spin-orbit coupling.
dx2-y2/dxy
dxz/dyz
dxz/dxy
dyz/dxy
dz2/dxz
dz2/dyz
rotate about z-axis
rotate about z-axis
rotate about x-axis
rotate about y-axis
rotate about y-axis
rotate about x-axis
There are no spin-orbit coupling contribution
for dz2/dx2-y2 and dz2/dxy
Magic pentagon
Related to “strength” of spin-orbit coupling
dz2
6
2
dxz
2
dx2-y2
6
2
dyz
2
8
2
dxy
Spin-orbit coupling influences g-value
g = 2.0023 +
n
E1-E2
2.0023: g-value for free ion
+ sign for <1/2 filled subshell
- sign for >1/2 filled subshell
n: number of magic pentagon
: free ion spin-orbit coupling constant
For example: CuIIL4 system (Cu2+: d9)
L
dx2-y2
dxy
L
CuII
dz2
dxz dyz
g = 2.0023 +
L
n
E1-E2
L
D4h point group
(axial symmetry)
 = -829 cm-1
E(dxy)-E(dx2-y2) = 15000 cm-1
(obtained by UV spectroscopy)
gz = g// = 2.0023-8(-829)/15000 = 2.44
gx = gy = g⊥ = 2.0023
spin-orbit coupling has little contribution to x and y directions.
Now, we can predict if the angular momentum will be quenched.
Example: check all the electron configurations in an octahedral field.
Which ones of the above electron configuration in Oh field have little spin-orbit
contribution (with g ~ 2.0)?
d3, d4(HS), d5(HS), d6(LS), d7(LS), d8, d9, d10
In low-symmetry field, spin-orbit coupling
are quenched
Remember that Oh is a high-symmetry field.
If the symmetry is lowered, degeneracy will be
destroyed and the orbital contribution will be quenched.
Oh
D4h
D4h: all are quenched except d1 and d3
For low-symmetry field, all are quenched and therefore
eff = g[S(S+1)]1/2 (spin-only) is valid.
Van Vleck Equation
In some cases, the paramagnetic behaviors
are more complicated due to
(1)
mixing of low-lying excited state
(2)
zero field splitting (ZFS)
(3)
higher order Zeeman effect.
These problems can be treated using Van
Vleck Equation.
Van Vleck Equation
En : the energy of state n can be expressed as a power series of H.
En = En(0) + H･En(1) + H2･En(2) + H3･En(3) + …..
En(0) : En of zero field; can be set to zero by convention.
H : applied magnetic field
En(1) : 1st order Zeeman coefficient
En(2) : 2nd order Zeeman coefficient (related to mixing of
low- lying excited state)
En(3) : 3rd order Zeeman coefficient (very small and can
be neglected)
Van Vleck Equation
En = En(0) + H･En(1) + H2･En(2) + H3･En(3) + …..
-En
n = H =- En-2HEn
o
(1)
(0)
-En/kT
e
(1)
(2)
-En-HEn-H2En
=e
kT
(2)
(0)
-En
=e
kT
(1)
(2)
-HEn-H2En
xe
kT
(0)
(1)
-En
HEn)
(1= kT x e kT
∵ when x is small, e-x ~ 1-x, and H2En(2) << kT
M=
N n 
n
(0)
-En/kT
e
e
-En/kT
n) x
kT
Nn (-En-2HEn) (1- HE
e
kT
(1)
=
n
-En
(1)
(2)
 (1- HE ) x e
-E(0)
n
(1)
n
kT
n
kT
For a paramagnetic system, when H = 0, M = 0
(0)
∴
 En e = 0
(1)
-En
kT
n
∴
cm =
and (En(2))2 and (En(2)˙En(1)) are very small
and can be neglected. cm = M/H
N n [
(1) 2
(En)
-En
-2En]e
-E
e
(2)
kT
(0)
n
n
kT
Van Vleck Equation
(0)
kT
Application of Van Vleck Equation
Example 1: S = 1/2, ms=+1/2, -1/2, no excited state  En(2) = 0
En = En(0) + H･En(1) + H2･En(2) + …….
E = msgHo
n = -(∂En)/(∂H) = -msg
(0)
Energy
ms=1/2
ms=-1/2
ms=1/2
Energy En
E(1)n
1/2g
0
1/2g
-1/2g
Ho
0
-1/2g
E = gHo
ms=-1/2
(0)
cm =
N n [
(1) 2
(En)
-2En]e
-E
e
(2)
kT
[
N
n
ms=1/2
Energy
kT
ms=1/2
n
kT
(1) 2
(En)
kT
e
0
]e
0
=
ms=-1/2
N[
2
(g/2)
kT
E(1)n
1/2g
0
1/2g
-1/2g
Ho
0
-1/2g
E = gHo
ms=-1/2
(0)
n
cm =
-E(0)
n
Energy En
2
+
(-g/2)
kT
]
1+1
n
cm =
Ng2
4kT
cm =
M

=
N g2 S(S+1)
3kT
this result is the same as what we obtained from simple
boltzmann distribution.
Application of Van Vleck
Equation S=1/2
Example 2: Cr3+, d3, S = 3/2 with
zero field splitting (ZFS) = D
(0)
ms=+3/2
ms=
ms=
_
+
1/2
E=ZFS=D
0
Energy
ms= +_ 1/2
0
D+3/2gH
D
3/2g
D-3/2gH
D
-3/2g
ms=+1/2
1/2gH
0
1/2g
ms=-1/2
-1/2gH
0
-1/2g
_ 3/2
+
ms= +_ 3/2
Energy En E(1)n
ms=-3/2
H
In this case, En(2) = 0.
cm =
N n [
(1) 2
(En)
-En
- ]e
-E
e
kT
(2)
2En
(0)
(0)
ms=+3/2
kT
(0)
n
ms=
_
+
1/2
E=ZFS=D
0
Energy
ms= +_ 1/2
0
2
2
2
(-g/2)
[
e
e
cm =
e e e
0
kT
0
cm =
D+3/2gH
D
3/2g
D-3/2gH
D
-3/2g
ms=+1/2
1/2gH
0
1/2g
ms=-1/2
-1/2gH
0
-1/2g
ms= +_ 3/2
ms= +_ 3/2
n
kT
Energy En E(1)n
ms=-3/2
H
2
]
(g/2) 0 (3g/2) -D/kT (-3g/2) -D/kT N
+ kT
+ kT
+ kT
0
-D/kT
-D/kT
+
+
+
e
e
e
-D/kT
Ng2
4kT
[
1 + 9e
1+
-D/kT
e
]
Example 2
cm =
-D/kT
Ng2
4kT
[
1 + 9e
1+
]
-D/kT
e
When ZFS is very small (D0) or at high temperature (kT >> D),
e-D/kT  1
cm =
Ng2
4kT
x 10/2
=
5Ng2
4kT
ms=+3/2
g2
cm = M = N3kT
S(S+1)

with S = 3/2
ms= +_ 3/2
ms=+1/2
ms= +_ 1/2
ms=-1/2
ms=-3/2
Example 2 when D 0 or at high temperature
When D ∞ or at low temperature (kT<<D), e-D/kT  0.
cm =
-D/kT
Ng2
4kT
cm =
[
1 + 9e
1+
Ng2
4kT
-D/kT
e
]
ms= +_ 3/2
ms= +_ 1/2
The system behaves like:
cm =
M
N g2 S(S+1)
= 3kT

with S = 1/2
ms= +_ 3/2
E=ZFS=D
If D is very large
only ms = +1/2, -1/2
are populated
ms= +_ 1/2
Value of ZFS can be obtained by curve-fitting
3.87
eff
S=3/2
S=1/2
cm
(BM)
1.7
S=1/2
temperature
cm =
S=3/2
HT
1/T
LT
-D/kT
Ng2
4kT
[
1 + 9e
1+
-D/kT
e
]
ZFS (D) can be obtained
by curve-fitting.
Example 3. S =1 system
(0)
Energy En En
_1
+
ms=+1
E=ZFS=D
ms=-1
ms=
(1)
D+gH
D
g
D-gH
D
-g
0
0
ms= +_ 1
ms= 0
0
ms= 0
0
ms=0
Energy
0
cm =
N [
n
(1) 2
(En)
H
-E(0)
n
- ]e
-E
e
kT
(2)
2En
(0)
n
n
kT
Van Vleck equation
kT
2
(-g) 2 -D/kT (g) -D/kT
+ kT
kT
]
e
e
cm =
-D/kT
-D/kT
e +e +e
N[
0
N[
2
(-g) 2 -D/kT (g) -D/kT
+ kT
kT
e ]
e
cm =
-D/kT
-D/kT
e +e +e
0
[ 2N
g2 -D/kT
e ]
cm =
-D/kT
1+2e
kT
At high temperature, or ZFS is very small (D<<kT) then e-D/kT  1
cm =
2Ng2
3kT
The system appears to be like S = 1
with no ZFS.
cm =
M

=
N g2 S(S+1)
3kT
At low temperature or very large ZFS (D>>kT), then e-D/kT  0.
cm  0
The system appears to be diamagnetic because only
ms = 0 state is populated.
e.g. 3
Interactions of micromagnets
Paramagnetic materials
No interactions
between magnets
cm
These magnets are
oriented randomly
under zero applied
magnetic field.
paramagnetic
T
If there are interactions between these micromagnets, these
materials are ferromagnetic, antiferromagnetic or ferrimagnetic.
With interactions among micromagnets
Ferromagnetic, Antiferromagnetic and Ferrimagnetic
ferromagnetic
cm
+ Curie temperature
' temperature
Neel
cm
+
paramagnetic
paramagnetic
antiferromagnetic
T
T
Magnetic domain
H
Average domain size: 20 ~200 nm
Hysteresis curve of M vs H
M
Mr
Ms
Ms : saturation magnetization
Mr: remanence magnetization
Hc: coercive magnetic field
Hc
H
Magnetic interaction of polynuclear clusters
Cu2(OAc)4.(H2O)2
cm
O
O
O
H2O
O
Cu
O
Cu
O
paramagnetic
O
T
OH2
' temperature
Neel
O
cm
+
paramagnetic
antiferromagnetic
T
Antiferromagnetic coupling complexes
Some terms to define:
Magnetic orbital:
orbital containing an unpaired electron
Exchange interactions:
magnetic interactions between metal
centers
HO
2
Exchange parameter: J, coupling
constant
O
O
O
Cu
O
O
Cu
O
O
O
OH2
Magnetic behavior of d1-d1 dimer
O
O
O
H2O
O
J: coupling constant
J > 0 : antiferromagnetic coupling
J < 0 : ferromagnetic coupling
O
S=1
S=0
J>0
J<0
S=1
J
S=0
Cu
S=0
-J
S=1
Cu
O
O
O
OH2
The energy diagram of d1-d1 dimer system
En = En(0) + H･En(1) + H2･En(2) + H3･En(3) + …..
(0)
Energy En
ms=+1
S=1
J
0
Energy
S=0
0
E(1)n
J+gH
J
g
ms= 0
J
J
0
ms=-1
J-gH
J
-g
0
0
0
ms=0
H
Antiferromagnetic coupling system
Van Vleck equation
cm =
[
N
n
(1) 2
(En)
kT
cm =
cm =
ms=+1
(0)
-En
- ]e
-E
e
(2)
2En
S=1
kT
(0)
n
N[
(0)
n
kT
J
0
Energy
2
(-g) 2 -J/kT (g) -J/kT
+ kT
kT
e
e
0
+
3e
-J/kT
S=0
[3+ e ]
J/kT
J
g
ms= 0
J
J
0
ms=-1
J-gH
J
-g
0
0
0
ms=0
H
0
2 
kT
' temperature
Neel
+
cm
paramagnetic
Bleany-Bowers Equation
antiferromagnetic
The value of J can be obtained by curve-fitting.
E(1)n
J+gH
2g  -J/kT ]
]
e = N[ e
1 + 3 e -J/kT
2Ng2
kT
Energy En
T
Two extreme conditions of the d1-d1 system
cm =
2Ng2
[3+ e ]
kT
J/kT
2g2 -J/kT
N[ kT e
= 1 + 3 e -J/kT
]
O
O
O
H2O
O
Cu
O
Cu
O
When J >> kT or T  0, cm  0,
this system is diamagnetic or a Cu-Cu bond is formed.
When J << kT or at high temperature, cm = Ng22/2kT
Ng22/2kT can be considered as two Ng22/4kT,
i.e., d1-d1 can be considered as two independent d1 system
O
O
OH2
The relationship between J and TN
' temperature
Neel
cm
cm =
+
paramagnetic
antiferromagnetic
T
2Ng2
[3+ e ]
kT
J/kT
Solve (cm)/  (T) = 0
TN ≒ (5/8) J/K
Direct couping through metal-metal bonding
or through superexchange via ligands?
Magnetic orbital
dx2-y2
dz2
dxy
dxz, dyz
O
O
O
H2O
O
Cu
O
Cu
O
O
OH2
z-axis
O
Weak -bond is formed between the interaction of magnetic orbitals.
Is the d1-d1 interaction through
() -bonding ?
(2) superexchange via ligands ?
Goodgame’s experiment in 1969
[Cu2(O2CH)4(NCS)2]2d(Cu…Cu)
= 2.716 Å
[Cu2(O2CCH3)4(NCS)2]2d(Cu…Cu) = 2.643 Å
S=1
J = 485 cm-1
S=1
J = 305 cm-1
S=0
S=0
This results indicate that the d1-d1 interaction
O
is not through metal-metal bonding only.
HO
Cu
O
Superexchange mechanism through ligands
O
may dominate.
O
O
Cu
2
O
O
O
OH2
Comparison of V4+-V4+ and Mo5+-Mo5+ dimers
O
L
O
L
V
L
O
L
L
V
L
L
Mo
L
d(V…V) = 3.20 Å
L
z
O
L
y
L
x
Mo
L
d(Mo…Mo) = 2.78 Å
S=1
dz2
S=1
J = 200 cm-1
S=0
Paramagnetic at HT
Antiferromagnetically coupled
At LT
J = 3000
cm-1
S=0
Diamagnetic compound
dx2-y2
dxz,dyz
dxy
Magnetic phenomena in 1D crystal
Variation of flux density in diamagnetic
and paramagnetic substances in a
magnetic field
Magnetic susceptibility
Plot of reciprocal susceptibility against
temperature
Some properties of ferromagnetic materials
Some Values of magnetic moments
Antiferromagnetic coupling of spins of d
electrons on Ni2+ ions through p electrons
of oxide ions
Ferromagnetic ordering in bcc Fe, fcc Ni and
hcp Co
Electronic constitution of iron, cobalt and
nickel
4.8 + 2.6 = 7.4
4.8 up + 2.6 down
4.8 - 2.6 = 2.2
Occupied energy levels and density of
states N(E) for 3d and 4s, 4p bands
Pauli paramagnetism
Electrons near Fermi
level are paired.
 temperature independent
H
Excess unpaired
electrons
Induced by
magnetic field
Pauli paramagnetism is temperature independent but field dependent.
Curie and Néel temperatures in
lanthanides
Magnetic structure of antiferromagnetic
and ferromagnetic spinels
Spinel
[M2+]tet[Fe3+]2octO4
Inverse Spinel
[Fe3+]tet[M2+, Fe3+]octO4
Partial inverse Spinel
[Fe3+1-xZn2+x]tet[M2+1-xFe3+1+x]octO4
Variation in saturation magnetization
with composition for ferrite solid solution
M1-xZnxFe2O4
Partial inverse Spinel
[Fe3+1-xZn2+x]tet[M2+1-xFe3+1+x]octO4
M
M
O
When x is large, the
antiferromagmnetic
coupling is destroyed.
Garnets
Atomic coordinates for Y3Fe5O12 (YIG)
x3
x3
x2
Variation of magnetic moment at 0 K of
garnets
Spontaneous Magnetization in Dy3Fe5O12 garnet
Dy3Fe3Fe2O12
The spins on rare earth (Dy3+) sublattice randomize much
rapidly than those on Fe3+ sublattice.
Crystallographic data for ilmenite
Schematic representation of the luminescence
Schematic design a fluorescent lamp
Luminescence spectra of activated ZnS
phosphors after irradiation with UV light
Various types of electronic
transition in activator ions
Ground state potential energy diagram for a
luminescent center in an ionic host crystal
Ground and excited state potential energy
diagrams for a luminescent center
Non-radiative energy transfer involved in
operation of a sensitized phosphor
Some lamp phosphor materials
Schematic representation of anti-Stokes
and normal luminescence phenomena
Energy levels of the Cr3+ ion in ruby
crystal and laser emission
Design of a ruby laser
Ga-As
laser
Energy levels of the Nd3+ ion in
neodymium lasers