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What We Need to Know About ElectronsWhat 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 non-degenerate 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 LawCurie 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 = CTC = “The Curie Constant”

Curie Law PlotsCurie Law Plots

Slope = C

The Curie Constant and Magnetic MomentThe Curie Constant and Magnetic MomentCurie 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”, µµ = (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) µhas 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

χχTTC 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)

8+ gb

2 nb. Sb (Sb+1)8

χχT T –– Simple ExamplesSimple Examples

χT = g2 n. S (S+1)8

Take three systems containing five electrons (assume g = 2):

≡ Anti-ferromagnetic

≡ Ferromagnetic

Spin pairing favoured

Spin pairing disfavoured

Non interacting

χT = 3/8χT = 35/8χT = 15/8

n = 1, S = 1/2n = 1, S = 5/2n = 5, S = ½

One L.S. Fe3+One H.S. Fe3+Five Ti3+

Less Simple Example Less Simple Example –– A Copper Dimer(1)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 Less Simple Example –– A Copper Dimer(2)A Copper Dimer(2)

Assume S = 1 and S = 0 have same energy, then use:

χT = ga2 na. Sa (Sa+1)

8+ gb

2 nb. Sb (Sb+1)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)8

+ 3 gb2 1 (1+1)

8{ }χT = 3g2/16

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 Less Simple Example –– A Copper Dimer(3)A Copper Dimer(3)

If S = 0 has lowest energy then:

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

χT = g2 n. S (S+1)8

S = 0

S = 1 (threefold degenerate)

At low T Boltzmann gives all population down here

χT = ga2 na. Sa (Sa+1)

8+ gb

2 nb. Sb (Sb+1)8

becomes

Less Simple Example Less Simple Example –– A Copper Dimer(4)A Copper Dimer(4)

If S = 1 has lowest energy then:

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

χT = g2 n. S (S+1)8

S = 1 At low T Boltzmann gives all population down here

S = 0

Less Simple Example Less Simple Example –– A Copper Dimer(5)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

S = 1

S = 0

Nomenclature

= 2J or J

Sane version:

J negative = AF

J positive = F

Treating Raw DataTreating Raw DataNot 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).

3.274861e-0031.003232e+0011.000000e+0027.022155e-0043.206844e+0011.000000e+0024.590345e-0046.241134e+0011.000000e+0023.642680e-0049.020821e+0011.000000e+0022.952116e-0041.198679e+0021.000000e+002

Long Moment (EMU)

Temperature (K)Field (Oe) 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 Plots for AF Copper DimerDimer -- χχTT

0 50 100 150 200 250 300

0.0

0.2

0.4

0.6

0.8

ChiT

T

J = 0J = -10J = -50

J = -100

J = -200

Obvious: decline in χT faster for larger AF J value

Less obvious: lack of maximum makes fitting data imprecise. Also, as χT zero at low T, fit will be very sensitive to impurities.

0 100 200 300

0.00

0.02

Chi

T

Plots for AF Copper Plots for AF Copper DimerDimer -- χχ

J = 0

J = -10

J = -50 J = -100J = -200

Maximum in χ arises because Curie law says χ rises as T falls, while energy levels means S = 1 becoming depopulated.

When fitting for AF system should fit χ vs T first, as the maximum is very sensitive to value of J for moderate J.

(For large J fit will always be approximate).

Plots for F Copper Plots for F Copper DimerDimer -- χχ

Plots shown for J = 0, +10, +20, +50, +200

Virtually indistinguishable!

Plot of χ vs T useless for getting size of J (or even telling F exchange from no interaction).

0 100 200 3000.00

0.02

0.04

0.06

0.08

0.10

Chi

T

Plots for F Copper Plots for F Copper DimerDimer -- χχTT

0 50 100 150 200 250 3000.8

0.9

1.0

1.1

ChiT

T

J = 0

J = 20J = 10

J = 200

J = 50

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?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:

Cu

OH

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?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.

Cu

OH

Cu

OH

x

y 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?Why AF or F for Copper?

Cu

OH

Cu

OH

x

y Look at bonding: get two pairs of bonding/anti-bonding orbitals:

dx2-y2

px, py

Add electrons: 2 from each O orbital + 1 from each dx2-

y2, get 6.

Feed into orbitals gives two upe in ground state, i.e. F exchange

CuOH

Cu

Why AF or F for Copper?Why AF or F for Copper?x

y

Change angle to ca. 180°

Now O-atom only uses one orbital. Energy levels become:

dx2-y2py

Only need include 4 electrons (2 from O and 1 each Cu): gives 0 upe, i.e. AF exchange

Why AF or F for Copper?Why AF or F for Copper?M.O. picture tells us about extremes: doesn’t tell us where F AF transition should occur.

First part (F exchange) is a specific case of the orthogonalityprinciple: 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)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.

SGS = 3S

Anti-ferromagnetic exchange.

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)Bigger Clusters(2)For example:

Ferromagnetic exchange.

SGS = 7S

Anti-ferromagnetic exchange.

SGS = 5S

Most high spin cages involve AF exchange as the predominant interaction: e.g. all Mn12 derivatives, Fe19, Fe14, Cr12

Bigger Clusters(3)Bigger Clusters(3)Second complexity: more than one exchange path, therefore more than one J value (in general)

Separating the two components can be difficult. Can normally work out sign of stronger interaction.

e.g. have centre-edge (black) and edge-edge (red) interactions

Bigger Clusters(4)Bigger Clusters(4)

Notice: black interaction overcomes red and aligns spins parallel. Does not matter whether red is AF or F.

Suppose black interaction strongly AF.

Suppose red interaction weakly AF.

Result SGS = 5S

Bigger Clusters(5)Bigger Clusters(5)

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.

Suppose red interaction strongly AF.

Suppose black interaction weakly AF.

Result SGS = ?

Examples of Bigger Clusters(1)Examples of Bigger Clusters(1)

Ni12 SGS = 121. 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)Examples of Bigger Clusters(2)Fe10 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)Examples of Bigger Clusters(3)

Fe9 SGS = unknowableAny attempt to assign spin rapidly becomes impossible – and impossible many times!? ?

?