Chapter 8Chapter 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 + 4 M B : magnetic induction (the field within the body)M : intensity of magnetizationB/Ho = 1 + 4 (M/Ho) = 1 + 4v

M/Ho : dimensionlessv : magnetic susceptibility per volume

Other definition of magnetic

susceptibility Gram Susceptibility: g = v/d (unit: cm3/g) d = density Molar Susceptibility: m = g

.(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.

M = -EHo

The sign of the induced magnetic moment, M, defines two types of magnetic behaviors:

Diamagnetism and Paramagnetism

M (and v, g, m) is negative: diamagnetism

Materials with only paired electrons (In fact, all materials exhibit diamagnetism) M (and v, g, m) is positive: paramagnetism

Materials with unpaired electrons (In fact, these materials exhibit both paramagnetism and diamagnetism)

M = -EHo

Ferromagnetism, antiferromagnetism and

ferrimagnetism

Ferromagnetism, antiferromagnetism and ferrimagnetism can be considered as special forms of paramagnetism.

Diamagnetism

It arises from the interactions of electron pairs with magnetic field, generating a small magnetic field opposing the applied magnetic field, Ho.

a property of all materials

Diamagnetism

SN N S

B = Ho + 4 M

Diamagnetism Diamagnetic materials move to the region of lower field strength

(repelled by Ho)

Ho

M = -EHo

Ho

E> 0

For diamagnetic materials, M < 0, ∴

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 (dia)?

T = para + dia

To study the paramagnetic behaviors of materials, dia must be subtracted from T

Pascal’s constants

Calculation of dia from Pascal’s constants:

dia = i iAi + Bi

A: atoms B: bonds

Table of Selected Pascal’s Constants atom A (x10

-6 emu) atom A (x10

-6 emu) atom A (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 B (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 (x10-6

emu)

Hg2+ -40.0 Ni2+ -12.8 C≡N- -13.0

Example 1:

N

HH

H

H H

pyridine

5 x C (ring) = 5 x (-6.24) = -31.25 x H = 5 x (-2.93) = -14.61 x N (ring) = 1 x (=4.61) = -4.61

iAi = -31.2 + (-14.6) + (-4.61) = -50.3 x 10-6 cm3/mol (or emu)

iBi In this case, is zero, because the “ring values” of C and N are used.

Example 2: O

CH3H3C

acetone

3 x C = 3 x (-6) = -186 x H = 6 x (-2.93) = -17.61 x O = -4.61

iAi = -18 + (-17.6) + (-4.61) = -40.21 x 10-6 emu

iBi = 1 x (C=O) = +6.3 x 10-6 emu

dia = i iAi + Bi = -33.9 x 10-6 emu

Example 3: K4Fe(CN)6 Transition metal

complex

4 x K+ = 4 x (-14.9) = -59.61 x Fe2+ = -12.86 x (C≡N-) = 6 x (-13.0) = -78.0

dia = i iAi + Bi = -150.4 x 10-6 emu

How to obtain dia of Cr(acac)3? Cr(acac)3 is paramagnetic so it is difficult to measure its

diamagnetism directly. Method I.Calculate dia from Pascal’s constants.

Method II. Synthesize Co(acac)3, Co3+: d6 low spin.

Use the dia value of Co(acac)3 as that of Cr(acac)3.

Method III.Measure the dia value of Na(acac), to obtain the dia value of aca

c-.Then include this value in Pascal’s constant calculation.

Calculated Expt’l Agreement

N

HH

H

H H

-50.4 x 10-6 -49 x 10-6 good

AsMe2

AsMe2

-147 x 10-6 -194 x 10-6 poor

However, since para >> dia , the disagreement usually

does not cause too much problem.

Disagreement of dia

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 m

from microscopic point of viewFor S = 1/2 system (Spin only, no orbital contribution)

= -gS : 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 = gS.H

The energies for S = ½ systemThe energies (eigen values) for S = 1/2 (ms = +1/2, -1/2) system are: E = msgHo E1/2 = (1/2)gHo and E-1/2 = –(1/2)gHo

Energy

ms=1/2

ms=1/2

ms=-1/2

ms=-1/2

E = gHo

Ho

Zeeman effect

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 e-E/kT ~ 1P1/2

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 m), we need to consider all the energy states that are populated.

∵ H = -.H = gS.H

∴ The magnetic moment, n with the direction // Ho, of an e

lectron in a quantum state n can be obtained by:n = -(∂En)/(∂H) = -msgSo we consider:

(1)   The magnetic moment of each energy state. (2)   The population of each energy state. 

That is, M = Nn.Pn

 N : Avogadro’s number Pn : probability in state n.

n = -(∂En)/(∂H) = -msg

e-En/kT

e-En/kTPn = =Nn

NT

Nn: population of state n

NT: population of all the states

M = Nn.Pn

H = -.H = gS.H

M = Nn.Pn

Energy

ms=1/2

ms=1/2

ms=-1/2

ms=-1/2

E = gHo

Ho

En1/2gH

-1/2gH

n

-1/2g

1/2g

E = msgHo n = -(∂En)/(∂H) = -msg

M for S=1/2 system

=N[g/2 -g/2

g/2kT -g/2kTe

+

e ][eg/2kT e-g/2kT]

M=N

msn e-En/kT

ms

e-En/kT

e± x~ 1 x±

when x << 1

Since gH<<kT when Ho ~ 5kG

= [ 1 + g/2kT 1( g/2kT) ]g/2kT+1 + )g/2kT(1

N g2

= Ng24kT M=

Curie Law of paramagnetic materials

N g2

4kTm = M

=

Curie Law: m = C/T

∴ N g2

4k=C

m

1/T

slope = C

Curie-Weiss Law

m

T1

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)

m = C/(T-)

is called Weiss constant.If is positive, then the magnets tend to align parallelly.If is negative, then the magnets tend to align antiparallelly.

For general S values (not only S = 1/2)

En = msgbH ms = -S, -S+1, …. , S-1, S

M=N

ms=-S

ms

S

(-msg)e-msgH/kT

-msgH/kTe= Ng2H

3kTS(S+1)

N g2

3kTm = M

= S(S+1) spin only

for S = 1/2, 1, 3/2

m

1/T

slope = Curie const.

S=1/2

S=1

S=3/2For S=1/2 N g2

4kTm =

For S=1 m = 2N g2

3kT

For S=3/2 m = 5Ng2

4kT

N g2

3kTm = M

= S(S+1)

Definition of eff

m S(S+1)N g2

3kT=

eff =( )3kN

1/2

(mT)1/2

~1/2

)mT(2.828

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:

gH << kT so that e± x~ 1 x±

If the applied magnetic field is very large, Curie-Weiss law is not valid.

(M is not proportional to H) N g2

3kTm = M

= S(S+1)

Saturation of Magnetization

M/mol-1

H/kT

S=1/2

S=1

S=3/2

S=2

1

2

3

4

1 20

follow Curie -Weiss law

Energy

ms=1/2

ms=1/2

ms=-1/2

ms=-1/2

E = gHo

Ho

If H is large enough, the probability of ms= -1/2

populated is close to 100%.

M = -msg M = for 1 magnet.

Plot of eff vs Temperature

All the molecules are in the S=3/2 state at all temperatures.

temperature

eff

(BM)

3.87

S=3/2 eff = 2[S(S+1)]1/2

eff = 2.828(T)1/2

Plot of eff vs Temperature

temperature

eff

(BM)

3.87

1.7

S=3/2

S=1/2

The eff value of the system gradually decreases from high-te

mperature value of 3.87 BM (S=3/2) to low-temperature value 1.7 BM (S=1/2)

Spin equilibrium and Spin Crossover

temperature

eff

(BM)

3.87

1.7

S=3/2

S=1/2temperature

eff

(BM)

3.87

1.7

S=3/2

S=1/2

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

g = 1+2J(J+1)

S(S+1)-L(L+1)+J(J+1)

g-value for free ions

Example: calculate the eff of Nd3+ (4f3)

+3 +2 +1 0 -1 -2 -3ml

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 = 9/2Ground state term symbol: 4I9/2

g = 1+2x(9/2)(9/2+1)

3/2(3/2+1)-6(6+1)+(9/2)(9/2+1)= 0.727

eff = g[J(J+1)]1/2 = 0.727[(9/2)(9/2+1)]=3.62 BM

g = 1+2J(J+1)

S(S+1)-L(L+1)+J(J+1)

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

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.

x

y

dx2-y2

x

y

dxy

z

e-

Spin-Orbit Coupling

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.

E dxy

dx2-y2

Spin-Orbit Coupling

Electrons have to change directions of spins to circulate Little contribution from orbital angular momentum.

E

dxydx2-y2

Spin-Orbit Coupling

Orbitals are filled. Little contribution from orbital angular momentum.

E

dxydx2-y2

Spin-Orbit Coupling

Spin-orbit couplings are significant in the above two cases.

E

dxydx2-y2

E

dxydx2-y2

Other orbital sets that may give spin-orbit coupling.

dx2-y2/dxy rotate about z-axisdxz/dyz rotate about z-axisdxz/dxy rotate about x-axisdyz/dxy rotate about y-axisdz2/dxz rotate about y-axisdz2/dyz rotate about x-axis

There are no spin-orbit coupling contribution for dz2/dx2-y2 and dz2/dxy

Magic pentagonRelated to “strength” of spin-orbit coupling

dxydx2-y2

dz2

dxz dyz

66

8

2

22 22

Spin-orbit coupling influences g-value

g = 2.0023 +E1-E2

n

2.0023: g-value for free ion+ sign for <1/2 filled subshell- sign for >1/2 filled subshelln: number of magic pentagon: free ion spin-orbit coupling constant

For example: CuIIL4 system (Cu2+: d9)

CuII

L

LL

L

D4h point group

(axial symmetry)

dx2-y2

dxy

dz2

dxz dyz = -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.

g = 2.0023 +E1-E2

n

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 fieldEn

(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 EquationEn = En

(0) + H ･ En(1) + H2 ･ En

(2) + H3 ･ En(3) + …..

n =-En

Ho= En-2HEn

(1) (2)

-En/kTe = e

-En-HEn-H2En(0) (1) (2)

kT = e-HEn-H2En

(0) (1) (2)

kT-En

kT x e

= (1- kTHEn

(1)

) x e kT-En

(0)

∵ when x is small, e-x ~ 1-x, and H2En(2) << kT

-

Van Vleck Equation

M=N n

n e-En/kT

n e-En/kT= Nn

(-En-2HEn)(1)

(1- kTHEn

(1)

) x e kT-En

n (1- kT

HEn(1)

) x e kT-En

(2)

(0)

(0)

For a paramagnetic system, when H = 0, M = 0

and (En(2))2 and (En

(2)˙En(1)) are very small

and can be neglected. m = M/H

m = Nn[(En)

2(1)

kT -2En(2)]e-En

(0)

kT

ne-En

(0)

kT

∴

∴ nEn

(1)e-En(0)

kT = 0

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 = msgHo n = -(∂En)/(∂H) = -msg

Energy

ms=1/2

ms=1/2

ms=-1/2

ms=-1/2

E = gHo

Ho

0 1/2g

-1/2g0

En(1)En

(0)Energy

1/2g

-1/2g

Application of Van Vleck Equati

on S=1/2

m = Nn[(En)

2(1)

kT -2En(2)]e-En

(0)

kT

ne-En

(0)

kT

=N

n[(En)

2(1)

kT]e0

n

0em =

[N (g/2)2

kT +(-g/2)

2

kT]

1 + 1

m = Ng2

4kT

this result is the same as what we obtained from simple boltzmann distribution.

N g2

3kTm = M

= S(S+1)

Energy

ms=1/2

ms=1/2

ms=-1/2

ms=-1/2

E = gHo

Ho

0 1/2g

-1/2g0

En(1)En

(0)Energy

1/2g

-1/2g

Example 2: Cr3+, d3, S = 3/2 with zero field splitting (ZFS) = D

ms= 1/2_+

ms= 3/2_+

ms= 3/2_+

ms= 1/2_+

E=ZFS=D

ms=+3/2

ms=-3/2

ms=+1/2

ms=-1/2

0

Energy0 H

D+3/2gH

D-3/2gH

1/2gH

-1/2gH

En(0)

En(1)

D

D

0

0

3/2g

-3/2g

1/2g

-1/2g

Energy

In this case, En(2) = 0.

Example 2

m = Nn[(En)

2(1)

kT -2En(2)]e-En

(0)

kT

ne-En

(0)

kT

m =[(-g/2) +

2(g/2)

2

kTe0 0e (3g/2)+kT kT

2-D/kTe +

(-3g/2)2

kT e-D/kT]e0 0e e-D/kT e-D/kT+ + +

m = Ng2

4kT [1 + 9e-D/kT

e-D/kT1 +

]

N

ms= 1/2_+

ms= 3/2_+

ms= 3/2_+

ms= 1/2_+

E=ZFS=D

ms=+3/2

ms=-3/2

ms=+1/2

ms=-1/2

0

Energy0 H

D+3/2gH

D-3/2gH

1/2gH

-1/2gH

En(0)

En(1)

D

D

0

0

3/2g

-3/2g

1/2g

-1/2g

Energy

Example 2 when D 0 or at high temperature

m = Ng2

4kT [1 + 9e-D/kT

e-D/kT1 +

]

When ZFS is very small (D0) or at high temperature (kT >> D),

e-D/kT 1

m = Ng2

4kTx 10/2 = 5Ng2

4kT

ms= 3/2_+

ms= 1/2_+

ms=+3/2

ms=-3/2

ms=+1/2

ms=-1/2

N g2

3kTm = M

= S(S+1)

with S = 3/2

When D ∞ or at low temperature (kT<<D), e-D/kT 0.

m = Ng2

4kT [1 + 9e-D/kT

e-D/kT1 +

]

m = Ng2

4kT

ms= 3/2_+

ms= 1/2_+

ms= 3/2_+

ms= 1/2_+

E=ZFS=D

If D is very largeonly ms = +1/2, -1/2are populated

N g2

3kTm = M

= S(S+1)

with S = 1/2

The system behaves like:

Value of ZFS can be obtained by curve-fitting

temperature

eff

(BM)

3.87

1.7

S=3/2

S=1/2

m

1/T

S=1/2

S=3/2

HT LT

m = Ng2

4kT [1 + 9e-D/kT

e-D/kT1 +

] ZFS (D) can be obtained by curve-fitting.

Example 3. S =1 system

E=ZFS=D

0

Energy0 H

D+gH

D-gH

En(0)

En(1)

D

D

0

g

-g

0

Energy

ms= 0

ms= 1_+

ms= 0

ms= 1_+

ms=-1

0

ms=+1

ms=0

m = Nn[(En)

2(1)

kT -2En(2)]e-En

(0)

kT

ne-En

(0)

kT

m =[ (-g)

+2

e (g)kT kT

2

e ]e 0

+ +e-D/kT

-D/kT -D/kT

e-D/kT

N

Van Vleck equation

e.g. 3

m =[ (-g)

+2

e (g)kT kT

2

e ]e 0

+ +e-D/kT

-D/kT -D/kT

e-D/kT

N m =1 + 2

[ e-D/kT

e-D/kT

]2N g2

kT

At high temperature, or ZFS is very small (D<<kT) then e-D/kT 1

m = 2Ng2

3kT

At low temperature or very large ZFS (D>>kT), then e-D/kT 0.

m 0The system appears to be diamagnetic because only ms = 0 state is populated.

N g2

3kTm = M

= S(S+1)

The system appears to be like S = 1with no ZFS.

Interactions of micromagnets

No interactions between magnets

Paramagnetic materials

These magnets are oriented randomlyunder zero applied magnetic field.

If there are interactions between these micromagnets, these materials are ferromagnetic, antiferromagnetic or ferrimagnetic.

m

T

paramagnetic

Ferromagnetic, Antiferromagnetic and Ferrimagnetic

m

T

+Curie temperature

paramagnetic

ferromagnetic

m

T

+Neel temperature'

paramagnetic

antiferromagnetic

With interactions among micromagnets

Magnetic domain

H

Average domain size: 20 ~200 nm

Hysteresis curve of M vs H

Hc

MsMr

H

M

Ms : saturation magnetizationMr: remanence magnetizationHc: coercive magnetic field

Magnetic interaction of polynuclear clusters

OCu

O

O

O

H2O Cu

O

O

OH2

O

O

Cu2(OAc)4.(H2O)2

m

T

paramagnetic

m

T

+Neel temperature'

paramagnetic

antiferromagnetic

Antiferromagnetic coupling complexesAntiferromagnetic coupling complexes

OCu

O

O

O

H2O Cu

O

O

OH2

O

O

Some terms to define:

Magnetic orbital: orbital containing an unpaired electron

Exchange interactions: magnetic interactions between metal centers

Exchange parameter: J, coupling constant

Magnetic behavior of d1-d1 dimerO

CuO

O

O

H2O Cu

O

O

OH2

O

O

J: coupling constantJ > 0 : antiferromagnetic couplingJ < 0 : ferromagnetic coupling

S=1

S=0

S=0

S=1J -J

J > 0 J < 0

S=0

S=1

The energy diagram of d1-d1 dimer system

0

Energy 0 H

J+gH

J-gH

En(0) En

(1)

J

g

-g

0

Energy

ms=-1

0

ms=+1

ms=0

J

S = 0

S = 1 ms= 0

J

0

J J

0

En = En(0) + H ･ En

(1) + H2 ･ En(2) + H3 ･ En

(3) +

…..

Antiferromagnetic coupling system

The value of J can be obtained by curve-fitting.

0

Energy 0 H

J+gH

J-gH

En(0) En

(1)

J

g

-g

0

Energy

ms=-1

0

ms=+1

ms=0

J

S = 0

S = 1 ms= 0

J

0

J J

0

m = Nn[(En)

2(1)

kT -2En(2)]e-En

(0)

kT

ne-En

(0)

kT

Van Vleck equation

em =

[ (-g)+

2

e (g)kT kT

2

e ]e0

+

-J/kT -J/kTN-J/kT3 1 e=

[2g2 ekT]

+

-J/kTN

-J/kT3

m 3 e= [

2Ng2

]+J/kT

kT

Bleany-Bowers Equation

m

T

+Neel temperature'

paramagnetic

antiferromagnetic

Two extreme conditions of the d1-d1 system

m 3 e= [

2Ng2

]+J/kT

kT 1 e=[2g2 ekT

]+

-J/kTN

-J/kT3

When J >> kT or T 0, m 0, this system is diamagnetic or a Cu-Cu bond is formed.

When J << kT or at high temperature, m = Ng22/2kT

Ng22/2kT can be considered as two Ng22/4kT, i.e., d1-d1 can be considered as two independent d1 system

OCu

O

O

O

H2O Cu

O

O

OH2

O

O

The relationship between J and TN

m

T

+Neel temperature'

paramagnetic

antiferromagnetic

m 3 e= [

2Ng2

]+J/kT

kT

Solve m)/ (T) = 0

TN ≒ (5/8) J/K

Direct couping through metal-metal bonding or through superexchange via ligands?

OCu

O

O

O

H2O Cu

O

O

OH2

O

O

dxz, dyzdxydz2

dx2-y2

z-axis

Magnetic orbital

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

OCu

O

O

O

H2O Cu

O

O

OH2

O

O

[Cu2(O2CH)4(NCS)2]2- [Cu2(O2CCH3)4(NCS)2]2-

d(Cu…Cu) = 2.716 Å d(Cu…Cu) = 2.643 Å

S=1

S=0

S=1

S=0

J = 485 cm-1J = 305 cm-1

This results indicate that the d1-d1 interactionis not through metal-metal bonding only.Superexchange mechanism through ligandsmay dominate.

V

L

V

LL

L

L

L

O O

Mo

L

Mo

LL

L

L

L

O O

x

y

Comparison of V4+-V4+ and Mo5+-Mo5+ dimers

z

dxy

dxz,dyzdx2-y2

dz2

S=1

S=0

S=1

S=0

J = 3000 cm-1J = 200 cm-1

d(V…V) = 3.20 Å d(Mo…Mo) = 2.78 Å

Diamagnetic compound

Paramagnetic at HTAntiferromagnetically coupledAt LT

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 = 7.44.8 up + 2.6 down4.8 - 2.6 = 2.2

Occupied energy levels and density of states N(E) for 3d and 4s, 4p

bands

Pauli paramagnetismPauli paramagnetism

H

Pauli paramagnetism is temperature independent but field dependent.

Electrons near Fermi level are paired. temperature independent

Excess unpaired electronsInduced by magnetic field

Curie and Néel temperatures in lanthanides

Magnetic structure of antiferromagnetic and ferromagnetic

spinels

Spinel[M2+]tet[Fe3+]2

octO4

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

Partial inverse Spinel[Fe3+

1-xZn2+x]tet[M2+

1-xFe3+1+x]octO4

M1-xZnxFe2O4

When x is large, the antiferromagmnetic coupling is destroyed.

OM M

Garnets

Atomic coordinates for Y3Fe5O12 (YIG)

x 3

x 3

x 2

Variation of magnetic moment at 0 K of garnets

Spontaneous Magnetization in Dy3Fe5O12 garnet

The spins on rare earth (Dy3+) sublattice randomize much rapidly than those on Fe3+ sublattice.

Dy3Fe3Fe2O12

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