General remarks

This seminar deals only with paramagnetic coordination compounds.

Complicated mathematics will be avoided, where possible!

TMn+ ions have pure d-electron configurations (recall: s electrons are lost first,

as the diffuse s-orbitals are destablized in complexes)

Cr2+: d4

Fe3+: d5

Ni2+: d8

metal organic compounds have also dn

Fe2+, Cr(CO)6, Cr(h6-C6H6)2 d6

Fe3+, V(CO)6, V(h6-C6H6)2 d5

General remarks

Constants and units

cm molar magnetic suceptibility [cm3·mol–1] (cgs/emu)

[emu·mol–1] „

[m3·mol–1] (SI)

Conversion factor: cm(cgs) × 4p10–6 = cm (SI)

K3[Fe(CN)6] cm = -122.7×10–6 emu/mol

cm = -1.542×10–9 m3/mol

----------------------------------------------------------------------------------------------------------

N = 6.023×1023 mol–1

mB = 0.92731×10–20 erg/Gauss;

mB = 9.27×10–24 J/T

kB = 1.38×10–23 J/K

NmB2/(3kB) = 0.125 cm3/(K·mol)

)(828.23

32 Kcm

mol

N

k

B

B m

cgs units, N = 6.023×1023 mol–1

mB = 0.92731×10–20 erg/Gauss; 9.27×10–24 J/T

TBeff cmm 828.2/

1. Spin-Only-Paramagnetism

Effective magnetic moment, meff, of 3d metal complexes can be estimated

to a first approximation with the spin-only formula

Beeff SSg mm )1(

)1( SSgn e

B

eff

effm

m

mB = Bohr Magneton = eħ/(2me) =9.27408×10–24 J/T

meff = effective magnetic moment

neff = effective magnetic moment in units of mB

ge = 2.00232

S = Ssi (Total spin quantum number)

si = spin quantum number (+1/2 or -1/2)

i S neff

1 ½ 1.73

2 1 2.83

3 3/2 3.88

4 2 4.90

5 5/2 5.92

Note: in the OCP text book meff is represented as meff

Spin-Only-Formula spin-state of complex and

number of unpaired electrons can be determined

d3: CrIII, MoIII, MnIV, VII: 3.88 mB

d5: MnII, FeIII: 5.92 mB

neff data (~ 300 K) for selected compounds of d3 and d5 ions

d3

CrCl3 3.90 K3[Cr(ox)3].3H20 3.62

[Cr(NH3)6]Br3 3.77 KCr(SO4)2.12H2O 3.84

[Cr(en)3]Br3 3.82 K3[MoCl6] 3.79

[Cr(bpy)3]Cl3 3.81 K2[MnCl6] 3.84

K3[Cr(CN)6] 3.87 [V(en)3]Br2 3.81

K3[Cr(NCS)6].4H2O 3.79 [V(bpy)3]Cl2 3.67

K3[Mo(NCS)6].4H2O 3.70 [Mo(bpy)3]Cl3 3.66

(NnBu4)3[Cr(N3)6] 3.76 K4[V(CN)6] 3.78

d5

MnCl2 5.79 FeCl3 5.73

MnBr2 5.82 (Et4N)[FeCl4] 5.88

(NH4)2Mn(SO4)2.6H2O 5.88 (NH4)Fe(SO4)2.12H2O 5.89

[Mn(NH3)6]Cl2 5.92 K3[Fe(ox)3].3H2O 5.90

(Et4N)2[MnCl4] 5.94

neff(theor.)

Spin-Only-Formula spin-state and

number of unpaired electrons can be determined

d3: CrIII, MoIII, MnIV, VII: 3.88 mB

d5: MnII, FeIII: 5.92 mB

[nBu4]2[Mn(CH3)6] 3.90 mB MnIV, d3

V(Cp)2, Vanadocene 3.78 mB VII, d3

Mn(Cp)2, Manganocene 5.86 mB MnII, d5

this is true also for more exotic compounds

Spin-Only Formula only valid for the following conditions:

room temperature (295 K)

for 3d TM ions (i.e. K2[ReIVCl6] = 3.25 mB (expected = 3.88 mB)

for mononuclear complexes (polynulcear complexes may show

cooperative phenomena (antiferro- or ferromagnetic interactions))

for totally quenched orbital momentum (= TM ions with E or A ground terms)

ferrocene

Fe

vandocene

V

manganocene

Mn Mn CH3H3C

CH3

H3C

CH3

CH3

2

I- < Br- < S2- < SCN- < Cl- < N3- < F- < OH- < O2

- < OH2 <

NCS- < NH3 ~ py < en < bpy < NO2- < CH3

- < CN- < CO

spectrochemical series:

Orbital contribution to the magnetic moment

Beeff SSg mm )1(

Spin-only formula

the orbital angular momentum L has also

a magnetic moment associated with it, for free ions with L and S,

Beeff SSgLL mm )1()1(2

spinorbit

Orbital contributions to the magnetic moment do explain the deviations from the spin-only values

the orbital contribution to the magnetic moment is not totally quenched

Two prominent examples:

CoCl2 5.47 mB

CoCl42─ 4.67 mB

expected 3.88 mB h.s.-CoII has d7 (3 unp. electrons)

general trends:

d6 to d9: larger values than calculated

d1 to d4: smaller values than calculated

only d5 is well behaved

This is readily explained

a) by the fact that

> 0 for d1-d4 and

< 0 for d6-d9

= spin-orbit coupling constant

b) Fe3+ (S=5/2), L = ML = Sml = 0

L

S

L

S

Spin-orbit coupling can cause temperature dependent magnetic moments (Ti3+, d1)

Orbital momentum in transition metal ions and complexes

In coordination compounds orbital momentum means:

electron can move from one d orbital to another degenerate

d orbital. However, dxy, dxz, dyz, and dzz, dx2-y2 are no longer

degenerate in a complex.

In an octahedral complex, e– can only move within an

open t2g shell (first order orbital momentum => of importance in magnetochemistry)

d1, d2, (l.s.)-d4, (l.s.)-d5, etc have first order orbital momentum (T ground terms),

d3, d4 have no first order orbital momentum (A, E ground terms)

Terms with T symmetry

exhibit orbital angular momentum

can show spin-orbit coupling

This rule is only applicable in Oh

Symmetry.

Terms with T symmetry

exhibit L = 1,

HSO = -ALS

EJ = -1/2A[J(J+1)-L(L+1)-S(S+1)

For (t2g)n less than half occupied: positive

more than half occupied: negative

dx2-y2

dxy

(leer)

Orbital contribution to the magnetic moment

Quenching of the orbital contribution, T-term and A, E-term ions

Quenching of the orbital contribution, to the magnetic moment, due to ligand field

n ground t2gneg

m ligand field quenching

term term

1 2D t2g1 2T2g No

2 3F t2g2 3T1g No

3 4F t2g3 4A2g Yes

4 5D t2g3eg

1 5Eg Yes

t2g4 3T1g No

5 6S t2g3eg

2 6A1g Yes

t2g5 2T2g No

6 5D t2g4eg

2 5T2g No

t2g6 1A1g Yes

7 4F t2g5eg

2 4T1g No

t2g6eg

1 2Eg Yes

8 3F t2g6eg

2 3A2g Yes

9 2D t2g6eg

3 2Eg Yes

These ions

actually

have L = 1

and thus a

„residual“

contribution

(not full

contribution)

to the

spin moment

Octahedral symmetry

Typical Ions: Ti3+ (d1), V3+ (d2), l.s-Mn3+ (d4), l.s.-Fe3+ (d5, i.e. K3[Fe(CN)6])

h.s-Fe2+ (d6), h.s.-Co(2+)

Magnetic moment depends also on C.N.

Nickel(II), d8

octahedral (3A2g) 2.9 – 3.4 mB

tetrahedral (3T1) 3.2 – 4.0 mB

trigonal bipyramidal 3.2 – 3.8 mB or 0

square pyramidal 3.2 – 3.4 mB or 0

square planar 0

Ni CNNC

CN

NC

2

Ni

Cl

ClCl

Cl

2

N

NiH2N

NH

NH2

NH2

Ni

N

NH2

NH2

H2N

Cl

2

Orbital momentum

quenched

not quenched

CoII, tetr. 4.4-4.8 4A2

CoII, oct., 4.8-5.3 4T1g

tetr. [NiX4]2- (X = Cl, Br, I)

tetr. [Ni(SPh)4]2─

[Ni(PPh3)2Br2] 3.27 mB

Spin equilibria

NiII(tetr.) NiII(sq.pl) (in solution)

High-spin and low-spin complexes

possible for d4-d7 electronic configurations (in octahedral complexes)

possible for d3-d6 electronic configurations (in tetrahedral complexes)

Examples (all are low-spin):

d4 [Cr(bpy)3]2+ , [Cr(CN)6]

4–, [Mn(CN)6]3– t2g

4 S = 1

3.20 mB

d5 [Fe(CN)6]3–, [Fe(en)3]

3+, [Mn(CN)6]4– t2g

5 S = 1/2

2.25 mB 2.40 mB 2.18 mB

d6 [Fe(CN)6]4–, [Co(NH3)6]

3+, [Cr(CO)6] t2g6 S = 0

d7 [Co(diars)3]2+, [Co(NO2)6]

4–, [NiF6]3– t2g

6eg1 S = ½

1.84 mB

the deviations from the ideal values are again attributable to orbital

contributions to the magnetic moment

AsPh2

AsPh2

diars

High-spin → low-spin transitions, spincrossover

Become feasible for d4 to d7 in octahedral case, if Do(h.s.) ~ Do(l.s.)

h.s.->l.s transitions can be affected by

variation of temperature or pressure

At lower temperature the l.s-form

always dominates

l.s. and h.s. form can be present in

an equilibrium (in solution as well as

in solid state)

Prominet examples:

Fe, d5: [Fe(S2CNR)3]

High-spin → low-spin transitions

Fe, d5: [Fe(S2CNR)3]

High T meff → 4.7 mB (h.s., S = 5/2)

Low T meff → 2.25 mB (l.s., S = ½)

S S-

N Fe SS

S

S

S

S

Spin-equilibria are rare.

Abrupt spincrossover more often

encounteredc(50%L.S./50%H.S.) = c(L.S.) + c(H.S.)

High-spin and low-spin tetrahedral complexes

d3: K3FeVO4 3.71 mB S = 3/2 (e)2(t2)1 high-spin

ReIV(o-tolyl)4 1.31 mB S = ½ (e)3(t2)0 low-spin

MnIV(1-nor)4 3.78 mB S = 3/2 (e)2(t2)1 high-spin

h.s. l.s.

n 3 1

h.s. l.s.

4 0

h.s. l.s.

5 1

d3 d4 d5 d6

h.s. l.s.

4 2

M1-nor

d4: [CoV(1-nor)4]+ S = 0 (e)4(t2)

0 low-spin

[FeIV(1-nor)4]+ S = 0 (e)4(t2)

0 low-spin

[MnIII(1-nor)4]- S = 2 (e)2(t2)

2 high-spin

MR4-complexes with 4d and 5d elements and sterically demanding ligands

are low-spin

High-spin → low-spin transitions

Fe, d6: [FeII(bpy)2(NCS)2]

High T meff → 5.2 mB (h.s., S = 4/2)

Low T meff → 2.25 mB (l.s., S = 0)

Fe NCSN

NCS

N

N

N

N

N

cis-[FeII(NCS)2(phen)2]

Phenanthrolin (phen)

High-spin and low-spin tetrahedral complexes

d5: [NEt4][FeCl4] 5.88 mB S = 5/2 (e)2(t2)3 high-spin

[NEt4][Fe(SPh)4] 5.73 mB S = 5/2 (e)2(t2)3 high-spin

CoIV(1-nor)4 1.89 mB S = 1/2 (e)4(t2)1 low-spin

h.s. l.s.

n 3 1

h.s. l.s.

4 0

h.s. l.s.

5 1

d3 d4 d5 d6

h.s. l.s.

4 2

M1-nor

d6: [CoIII(1-nor)4]– 3.18 mB S = 1 (e)4(t2)

2 low-spin

General observations:

low-spin tetrahedral complexes are rare (Dt = –4/9 Do)

a tetrahedral complex with low-spin configuration requires:

strong ligand field, a high-metal oxidation state, sterically demanding ligands

(particularly for bigger 4d 5d elements) to prevent the formation of M-M bonds

or adoption of coordination number 6

High-spin and low-spin complexes

l.s.-d4, l.s.-d5, and l.s.-d7 display positive and commonly large deviations

from the spin only expectation (for the first transition series)

Explanation: (t2g)n configurations behave magnetically like (p)n configs;

when more than half-filled subshell (as is the case in d4-d7); S and L

are parallel; and any orbital contribution increases meff beyond the spin-only

value

All octahedral complexes of 4d and 5d elements are low-spin

Gleichungen zur Berechnung magnetischer Momente

a) Spin-Only-Formel (Näherung für 3d-Ionen mit A- und E-Ligandenfeldgrundtermen,

vgl. Tab. 1)

)1(.. SSgB

os

g = 2.0023 ~ 2, S = Gesamtspinquantenzahl, S = n/2, n = Anzahl ungepaarte Elektronen

b) Berechnung des magnetischen Momentes aus der experimentell bestimmten molaren

magnetischen Suszeptibilität (m)

TN

km

Ba

B

B

2

.exp 3 = Tm828.2

kB = Boltzmann-Konstante, Na = Avogadro-Zahl, B = Bohrsches Magneton

m = molare magnetische Suszeptibilität (Dimension: cm3∙mol–1)

im cgs-System: 1 J = 10–7 erg, 1 T(esla) = 104 Gauss, 1 erg = Gauss2∙cm3

kB = 1.38∙10–16 erg/K, B = 9.274∙10–21 erg/Gauss

Na∙B2/(3kB) = 0.125 cm3∙K∙mol–1

c) bei nicht-unterdrücktem Bahn-Moment wäre der Erwartungswert für :

)1()1(4 LLSSB

S = Gesamtspinquantenzahl, S = n/2, n = Anzahl ungepaarte Elektronen

L = Gesamtbahndrehimpulsquantenzahl = Summe der ml-Werte, L = ml

d) für Lanthanoid-Ionen ( ist groß) gilt

)1(2

)1()1()1(1

)1(

JJ

JJLLSSg

JJg

J

J

B

, gJ = Lande-Faktor

f) Russel-Saunders-Symbole

Termsymbol für Grundzustand eines Atoms oder Ions: 2S+1LJ

S = Gesamtspinquantenzahl, S = n/2, n = Anzahl ungepaarte Elektronen

L = Gesamtbahndrehimpulsquantenzahl = Summe der ml-Werte, L = ml

J = Quantenzahl des Gesamtdrehimpulses,

bei weniger als halbgefüllten Schalen gilt für den Grundterm: J = (L-S)

bei mehr als halbgefüllten Schalen gilt für den Grundterm: J = (L+S)

Tab.1 Berechnete und gemessene effektive magnetische Momente (in B) oktaedrischer

Komplexe der 3d-Metalle.

Zahl der d-

Elektronen

Grundterm EK Ligandenfeld-

grundterm

eff(ber.) eff(gem.) Bahnmoment

erwartet?

1 (Ti3+) 2D t2g1 2T2g Ja

2 (V3+) 3F t2g2 3T1g Ja

3 (Cr3+) 4F t2g3 4A2g Nein

4 (Mn3+) 5D t2g3eg

1 5Eg Nein

t2g4eg

0 3T1g Ja

5 (Fe3+) 6S t2g3eg

2 6A1g Nein

t2g5eg

0 2T2g Ja

6 (Fe2+) 5D t2g4eg

2 5T2g Ja

t2g6eg

0 1A1g Nein

7 (Co2+) 4F t2g5eg

2 4T1g Ja

t2g6eg

1 2Eg Nein

8 (Ni2+) 3F t2g6eg

2 3A2g Nein

9 (Cu2+) 2D t2g6eg

3 2Eg Nein

Aufgaben:

1) Bei Raumtemperatur ist der gemessene Wert von eff für [Cr(en)3]Br2 4.75 B. Ist der

Komplex low-spin oder high-spin?

[Cr(en)3]Br2 enthält ein Cr2+-Ion (d4). In der low-spin-Konfiguration hätte der Komplex 2

ungepaarte Elektronen. In der high-spin-Konfiguration 4. Mit der Spin-Only-Formel ergeben

sich effektive magnetische Momente von 2.83 bzw. 4.90. Der letztere Wert liegt deutlich

näher am experimentell bestimmten als der erstere. Der Komplex hat daher eine high-spin-

Konfiguration.

2) Das magnetische Moment eines oktaedrischen Co(II)-Komplexes beträgt eff = 4.0. Welche

Elektronenkonfiguration hat dieser Komplex?

Oktaedrische Co(II)-Komplexe (d7-Konfiguration) können low- (1 ungepaartes Elektron) oder

high-spin (3 ungepaarte Elektronen) konfiguriert sein. Nimmt man an, dass die Spin-Only-

Formel in erster Näherung Gültigkeit besitzt, ist:

eff (high-spin) = 3.90

eff (low-spin) = 1.73

Der Komplex ist daher high-spin Komplex.

3) Das magnetische Momente der Komplexe [Mn(H2O)6]2+, [Fe(H2O)6]

3+, [MnCl4]2− und

[FeCl4]− beträgt jeweils ungefähr 5.92. Was sagt Ihnen das über die geometrische und

elektronische Struktur der Komplexe? Warum ist die Spin-Only-Formel in diesen Fällen so

genau?

[Mn(H2O)6]2+: d5, t2g

3eg2, high-spin

[Fe(H2O)6]3+: d5, t2g

3eg2, high-spin

[MnCl4]2–: d5, e2t3, high-spin, tetraedrisch gebaut

[FeCl4]–: d5, e2t3, high-spin, tetraedrisch gebaut

Die Komplexe besitzen einen A-Ligandenfeldgrundterm (L = 0).

4) Bestimmen Sie das Russel-Saunders-Termsymbol 2S+1LJ für den jeweiligen Grundzustand

der folgenden Ionen:

a) Ti3+, b) Ni2+, c) Cr3+, d) Ce3+, e) Ho3+, f) Er3+, g) La3+, h) Lu3+

Lösungsweg:

Schritt 1) n, Anzahl ungepaarter Elektronen bestimmen

=> S = n/2

Schritt 2) L bestimmen:

Summe der ml-Werte für die Anordnung mit niedrigster Energie (Hundsche

Regeln), Zuordnung des bestimmten Wertes für L nach folgendem Schema:

L 0 1 2 3 4 5 6

S P D F G H I

Schritt 3) J bestimmen:

J = (L–S) bei weniger als halbgefüllten Schalen

J = (L+S) bei mehr als halbgefüllten Schalen

Ti3+, d1 , n = 1, S = ½, Ms = 2∙1/2 + 1 = 2

ml 2 1 0 -1 -2

L = 2, => 2D

J = L – S = 2 – 1/2 = 3/2 => 2D3/2

Ni2+, d8, n = 2, S = 1, Ms = 2∙1 + 1 = 3

ml 2 1 0 -1 -2

L = (2 · 2) + (2 · 1) +(2 · 0) – 1 – 2 = 3, => F

J = L + S = 3 + 1 = 4 => 3F4

Cr3+, d3, n = 3, S = 3/2, Ms = 2∙3/2 + 1 = 4

ml 2 1 0 -1 -2

L = (1 · 2) + (1 · 1) +(1 · 0) = 3, => F

J = L – S = 3 – 3/2 = 3/2 => 4F3/2

Ce3+, f1 n = 1, S = 1/2, Ms = 2∙1/2 + 1 = 2

ml 2 1 0 -1 -23 -3

L = (1 · 3) = 3, => 2F

J = L – S = 3 – 1/2 = 5/2 => 2F5/2

Ho3+, f10 n = 4, S = 2, Ms = 2∙2 + 1 = 5

ml 2 1 0 -1 -23 -3

L = (2 · 3) + (2 · 2) + (2 · 1) +(1 · 0) – (1 · 1) – (1 · 2) – (1 · 3) = 6, => I

J = L + S = 6 + 2 = 8 => 5I8

La3+: f0, n=0, S=0, MS=1, L = 0, J = 0, => 1S0

Lu3+: f14, n=0, S=0, MS=1, L = 0, J = 0, => 1S0

5) Für welche der folgenden Ionen erwarten Sie einen Bahnbeitrag zum magnetischen

Moment:

a) V3+, b) Cr3+, c) Ti4+, d) Fe3+ (low-spin), e) Fe3+ (high-spin)

Für Ionen mit T1g- oder T2g-Ligandenfeldgrundtermen. Das sind: V3+ (3T1g) und Fe3+, low-

spin (2T2g), vgl. Tabelle 1.

6) Berechnen Sie den Wert des effektiven magnetischen Moments der folgenden Ionen:

a) Ti3+, Ce3+, Yb3+, Ho3+, Eu3+.

Ti3+: Spin-Only-Formel: eff = 1.73

Für Lanthanoide (Ce3+-Eu3+) gilt:

)1(2

)1()1()1(1

)1(

JJ

JJLLSSg

JJg

J

J

B

,

wobei S, L und J aus den Termsymbolen bestimmt werden können

Ce3+: 2F5/2, S = 1/2, L = 3, J = 5/2, gJ = 0.857, = 2.535

Yb3+: 2F7/2, S = ½, L = 3, J = 7/2, gJ = 1.143, = 4.536

Ho3+: 5I8, S = 2, L = 6, J = 8, gJ = 1.25, = 10.607

Eu3+: = 0

7) Berechnen das magnetische Moment von Dy2(SO4)3∙8H2O.

8) Berechnen Sie den Wert der Curie-Constanten C für die Verbindung

(NH4)2Mn(SO4)2∙6H2O in cgs-Einheiten.