General remarks. This seminar deals only with paramagnetic coordination compounds. Complicated mathematics will be avoided, where possible!

Transkript

1 General remarks This seminar deals only with paramagnetic coordination compounds. Complicated mathematics will be avoided, where possible! TM n+ ions have pure d-electron configurations (recall: s electrons are lost first, as the diffuse s-orbitals are destablized in complexes) Cr 2+ : d 4 Fe 3+ : d 5 Ni 2+ : d 8 metal organic compounds have also d n Fe 2+, Cr(CO) 6, Cr(h 6 -C 6 H 6 ) 2 d 6 Fe 3+, V(CO) 6, V(h 6 -C 6 H 6 ) 2 d 5

2 General remarks Constants and units c m molar magnetic suceptibility [cm 3 mol 1 ] (cgs/emu) [emu mol 1 ] [m 3 mol 1 ] (SI) Conversion factor: c m (cgs) 4p10 6 = c m (SI) m eff / m B ct K 3 [Fe(CN) 6 ] c m = emu/mol B 2.828( ) c m = m 3 /mol 3 Nm cm K B N = mol 1 m B = erg/gauss; m B = J/T k B = J/K Nm B2 /(3k B ) = cm 3 /(K mol) 3k mol cgs units, N = mol 1 m B = erg/gauss; J/T

3 1. Spin-Only-Paramagnetism Effective magnetic moment, m eff, of 3d metal complexes can be estimated to a first approximation with the spin-only formula m g S( S 1) m n eff eff m m e eff B g e B S( S 1) m B = Bohr Magneton = eħ/(2m e ) = J/T m eff = effective magnetic moment n eff = effective magnetic moment in units of m B g e = S = Ss i (Total spin quantum number) s i = spin quantum number (+1/2 or -1/2) i S n eff 1 ½ / / Note: in the OCP text book m eff is represented as m eff

4 Spin-Only-Formula spin-state of complex and number of unpaired electrons can be determined n eff (theor.) d 3 : Cr III, Mo III, Mn IV, V II : 3.88 m B d 5 : Mn II, Fe III : 5.92 m B n eff data (~ 300 K) for selected compounds of d 3 and d 5 ions d 3 CrCl K 3 [Cr(ox) 3 ].3H [Cr(NH 3 ) 6 ]Br KCr(SO 4 ) 2.12H 2 O 3.84 [Cr(en) 3 ]Br K 3 [MoCl 6 ] 3.79 [Cr(bpy) 3 ]Cl K 2 [MnCl 6 ] 3.84 K 3 [Cr(CN) 6 ] 3.87 [V(en) 3 ]Br K 3 [Cr(NCS) 6 ].4H 2 O 3.79 [V(bpy) 3 ]Cl K 3 [Mo(NCS) 6 ].4H 2 O 3.70 [Mo(bpy) 3 ]Cl (N n Bu 4 ) 3 [Cr(N 3 ) 6 ] 3.76 K 4 [V(CN) 6 ] 3.78 d 5 MnCl FeCl MnBr (Et 4 N)[FeCl 4 ] 5.88 (NH 4 ) 2 Mn(SO 4 ) 2.6H 2 O 5.88 (NH 4 )Fe(SO 4 ) 2.12H 2 O 5.89 [Mn(NH 3 ) 6 ]Cl K 3 [Fe(ox) 3 ].3H 2 O 5.90 (Et 4 N) 2 [MnCl 4 ] 5.94

5 Spin-Only-Formula spin-state and number of unpaired electrons can be determined d 3 : Cr III, Mo III, Mn IV, V II : 3.88 m B d 5 : Mn II, Fe III : 5.92 m B this is true also for more exotic compounds [nbu 4 ] 2 [Mn(CH 3 ) 6 ] 3.90 m B Mn IV, d 3 V(Cp) 2, Vanadocene 3.78 m B V II, d 3 Mn(Cp) 2, Manganocene 5.86 m B Mn II, d 5 ferrocene Fe vandocene V spectrochemical series: I - < Br - < S 2- < SCN- < Cl - < N 3 - < F - < OH - < O 2 - < OH 2 < NCS - < NH 3 ~ py < en < bpy < NO 2 - < CH 3 - < CN - < CO CH 3 CH 3 Mn H 3 C Mn CH 3 H 3 C manganocene CH 3 2 Spin-Only Formula only valid for the following conditions: room temperature (295 K) for 3d TM ions (i.e. K 2 [Re IV Cl 6 ] = 3.25 m B (expected = 3.88 m B ) 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)

6 Orbital contribution to the magnetic moment Spin-only formula m g S( S 1) m eff e B the orbital angular momentum L has also a magnetic moment associated with it, for free ions with L and S, eff 2 e m L( L 1) g S( S 1) m B orbit spin

7 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: CoCl m B CoCl m B expected 3.88 m B h.s.-co II has d 7 (3 unp. electrons) general trends: d 6 to d 9 : larger values than calculated d 1 to d 4 : smaller values than calculated only d 5 is well behaved This is readily explained a) by the fact that > 0 for d 1 -d 4 and < 0 for d 6 -d 9 = spin-orbit coupling constant b) Fe 3+ (S= 5 / 2 ), L = M L = Sm l = 0 L S L S Spin-orbit coupling can cause temperature dependent magnetic moments (Ti 3+, d 1 )

8 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, d xy, d xz, d yz, and d zz, d x2-y2 are no longer degenerate in a complex. In an octahedral complex, e can only move within an open t 2g shell (first order orbital momentum => of importance in magnetochemistry) d 1, d 2, (l.s.)-d 4, (l.s.)-d 5, etc have first order orbital momentum (T ground terms), d 3, d 4 have no first order orbital momentum (A, E ground terms) d xy d x2-y2 (leer) Terms with T symmetry exhibit orbital angular momentum can show spin-orbit coupling This rule is only applicable in O h Symmetry. E J = -1/2A [J(J+1)-L(L+1)-S(S+1) For (t 2g ) n less than half occupied: positive more than half occupied: negative Terms with T symmetry exhibit L = 1, H SO = -A LS

9 Orbital contribution to the magnetic moment

10 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 Octahedral symmetry n ground t 2gn e m g ligand field quenching term term 1 2 D t 1 2g 2 T 2g No 2 3 F t 2 2g 3 T 1g No 3 4 F t 3 2g 4 A 2g Yes 4 5 D t 2g3 e 1 g 5 E g Yes t 4 2g 3 T 1g No 5 6 S t 2g3 e 2 g 6 A 1g Yes t 5 2g 2 T 2g No 6 5 D t 2g4 e 2 g 5 T 2g No t 6 2g 1 A 1g Yes 7 4 F t 2g5 e 2 g 4 T 1g No t 2g6 e 1 g 2 E g Yes 8 3 F t 2g6 e 2 g 3 A 2g Yes 9 2 D t 2g6 e 3 g 2 E g Yes These ions actually have L = 1 and thus a residual contribution (not full contribution) to the spin moment Typical Ions: Ti 3+ (d 1 ), V 3+ (d 2 ), l.s-mn 3+ (d 4 ), l.s.-fe 3+ (d 5, i.e. K 3 [Fe(CN) 6 ]) h.s-fe 2+ (d 6 ), h.s.-co( 2+ )

11 Magnetic moment depends also on C.N. Nickel(II), d 8 Orbital momentum octahedral ( 3 A 2g ) m B tetrahedral ( 3 T 1 ) m B trigonal bipyramidal m B or 0 square pyramidal m B or 0 square planar 0 quenched not quenched 2 Cl 2 NC NC Ni CN CN Cl Ni Cl Cl Co II, tetr A 2 Co II, oct., T 1g H 2 N N H N Ni NH 2 N H 2 2 H 2 N N H 2 N Ni Cl NH 2 tetr. [NiX 4 ] 2- (X = Cl, Br, I) tetr. [Ni(SPh) 4 ] 2 [Ni(PPh 3 ) 2 Br 2 ] 3.27 m B Spin equilibria Ni II (tetr.) Ni II (sq.pl) (in solution)

12 High-spin and low-spin complexes possible for d 4 -d 7 electronic configurations (in octahedral complexes) possible for d 3 -d 6 electronic configurations (in tetrahedral complexes) Examples (all are low-spin): diars AsPh 2 AsPh 2 d 4 [Cr(bpy) 3 ] 2+, [Cr(CN) 6 ] 4, [Mn(CN) 6 ] 3 t 4 2g S = m B d 5 [Fe(CN) 6 ] 3, [Fe(en) 3 ] 3+, [Mn(CN) 6 ] 4 t 2g 5 S = 1/ m B 2.40 m B 2.18 m B d 6 [Fe(CN) 6 ] 4, [Co(NH 3 ) 6 ] 3+, [Cr(CO) 6 ] t 2g 6 S = 0 d 7 [Co(diars) 3 ] 2+, [Co(NO 2 ) 6 ] 4, [NiF 6 ] 3 t 2g6 e g1 S = ½ 1.84 m B the deviations from the ideal values are again attributable to orbital contributions to the magnetic moment

13 High-spin low-spin transitions, spincrossover Become feasible for d 4 to d 7 in octahedral case, if D o (h.s.) ~ D o (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, d 5 : [Fe(S 2 CNR) 3 ]

14 High-spin low-spin transitions S S N S Fe S S S - S S Fe, d 5 : [Fe(S 2 CNR) 3 ] High T m eff 4.7 m B (h.s., S = 5 / 2 ) Low T m eff 2.25 m B (l.s., S = ½) c (50%L.S./50%H.S.) = c (L.S.) + c (H.S.) Spin-equilibria are rare. Abrupt spincrossover more often encountered

15 High-spin and low-spin tetrahedral complexes d 3 d 4 d 5 d 6 M 1-nor h.s. l.s. h.s. l.s. h.s. l.s. h.s. l.s. n d 3 : K 3 Fe V O m B S = 3/2 (e) 2 (t 2 ) 1 high-spin Re IV (o-tolyl) m B S = ½ (e) 3 (t 2 ) 0 low-spin Mn IV (1-nor) m B S = 3/2 (e) 2 (t 2 ) 1 high-spin d 4 : [Co V (1-nor) 4 ] + S = 0 (e) 4 (t 2 ) 0 low-spin [Fe IV (1-nor) 4 ] + S = 0 (e) 4 (t 2 ) 0 low-spin [Mn III (1-nor) 4 ] - S = 2 (e) 2 (t 2 ) 2 high-spin MR 4 -complexes with 4d and 5d elements and sterically demanding ligands are low-spin

16 High-spin low-spin transitions N NCS N N Phenanthrolin (phen) N N Fe N NCS cis-[fe II (NCS) 2 (phen) 2 ] Fe, d 6 : [Fe II (bpy) 2 (NCS) 2 ] High T m eff 5.2 m B (h.s., S = 4 / 2 ) Low T m eff 2.25 m B (l.s., S = 0)

17 High-spin and low-spin tetrahedral complexes d 3 d 4 d 5 d 6 M 1-nor h.s. l.s. h.s. l.s. h.s. l.s. h.s. l.s. n d 5 : [NEt 4 ][FeCl4] 5.88 m B S = 5/2 (e) 2 (t 2 ) 3 high-spin [NEt 4 ][Fe(SPh) 4 ] 5.73 m B S = 5/2 (e) 2 (t 2 ) 3 high-spin Co IV (1-nor) m B S = 1/2 (e) 4 (t 2 ) 1 low-spin d 6 : [Co III (1-nor) 4 ] 3.18 m B S = 1 (e) 4 (t 2 ) 2 low-spin General observations: low-spin tetrahedral complexes are rare (D t = 4/9 D o ) 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

18 High-spin and low-spin complexes l.s.-d 4, l.s.-d 5, and l.s.-d 7 display positive and commonly large deviations from the spin only expectation (for the first transition series) Explanation: (t 2g ) n configurations behave magnetically like (p) n configs; when more than half-filled subshell (as is the case in d 4 -d 7 ); S and L are parallel; and any orbital contribution increases m eff beyond the spin-only value All octahedral complexes of 4d and 5d elements are low-spin

19 Gleichungen zur Berechnung magnetischer Momente a) Spin-Only-Formel (Näherung für 3d-Ionen mit A- und E-Ligandenfeldgrundtermen, vgl. Tab. 1) s. o. g S( S 1) B g = ~ 2, S = Gesamtspinquantenzahl, S = n /2, n = Anzahl ungepaarte Elektronen b) Berechnung des magnetischen Momentes aus der experimentell bestimmten molaren magnetischen Suszeptibilität ( m) exp. 3kB mt 2 = m T N B a B k B = Boltzmann-Konstante, N a = Avogadro-Zahl, B = Bohrsches Magneton m = molare magnetische Suszeptibilität (Dimension: cm 3 mol 1 ) im cgs-system: 1 J = 10 7 erg, 1 T(esla) = 10 4 Gauss, 1 erg = Gauss 2 cm 3 k B = erg/k, B = erg/gauss N a B 2 /(3k B) = cm 3 K mol 1 c) bei nicht-unterdrücktem Bahn-Moment wäre der Erwartungswert für : B 4S( S 1) L( L 1) S = Gesamtspinquantenzahl, S = n / 2, n = Anzahl ungepaarte Elektronen L = Gesamtbahndrehimpulsquantenzahl = Summe der m l-werte, L = m l d) für Lanthanoid-Ionen ( ist groß) gilt gj J ( J 1) B, gj = Lande-Faktor S( S 1) L( L 1) J ( J 1) gj 1 2J ( J 1) f) Russel-Saunders-Symbole Termsymbol für Grundzustand eines Atoms oder Ions: 2S+1 LJ S = Gesamtspinquantenzahl, S = n / 2, n = Anzahl ungepaarte Elektronen L = Gesamtbahndrehimpulsquantenzahl = Summe der m l-werte, L = m l 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)

20 Tab.1 Berechnete und gemessene effektive magnetische Momente (in B) oktaedrischer Komplexe der 3d-Metalle. Zahl der d- Elektronen Grundterm EK Ligandenfeldgrundterm eff(ber.) eff(gem.) Bahnmoment erwartet? 1 (Ti 3+ ) 2 D t2g 1 2 T2g Ja 2 (V 3+ ) 3 F t2g 2 3 T1g Ja 3 (Cr 3+ ) 4 F t2g 3 4 A2g Nein 4 (Mn 3+ ) 5 D t2g 3 eg 1 5 Eg Nein t2g 4 eg 0 3 T1g Ja 5 (Fe 3+ ) 6 S t2g 3 eg 2 6 A1g Nein t2g 5 eg 0 2 T2g Ja 6 (Fe 2+ ) 5 D t2g 4 eg 2 5 T2g Ja t2g 6 eg 0 1 A1g Nein 7 (Co 2+ ) 4 F t2g 5 eg 2 4 T1g Ja t2g 6 eg 1 2 Eg Nein 8 (Ni 2+ ) 3 F t2g 6 eg 2 3 A2g Nein 9 (Cu 2+ ) 2 D t2g 6 eg 3 2 Eg Nein Aufgaben: 1) Bei Raumtemperatur ist der gemessene Wert von eff für [Cr(en)3]Br B. Ist der Komplex low-spin oder high-spin? [Cr(en)3]Br2 enthält ein Cr 2+ -Ion (d 4 ). 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 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 (d 7 -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:

21 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 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+ : d 5, t2g 3 eg 2, high-spin [Fe(H2O)6] 3+ : d 5, t2g 3 eg 2, high-spin [MnCl4] 2 : d 5, e 2 t 3, high-spin, tetraedrisch gebaut [FeCl4] : d 5, e 2 t 3, high-spin, tetraedrisch gebaut Die Komplexe besitzen einen A-Ligandenfeldgrundterm (L = 0). 4) Bestimmen Sie das Russel-Saunders-Termsymbol 2S+1 LJ für den jeweiligen Grundzustand der folgenden Ionen: a) Ti 3+, b) Ni 2+, c) Cr 3+, d) Ce 3+, e) Ho 3+, f) Er 3+, g) La 3+, h) Lu 3+ 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 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 Ti 3+, d 1, n = 1, S = ½, M s = 2 1 /2 + 1 = 2 m l L = 2, => 2 D J = L S = 2 1/2 = 3/2 => 2 D3/2

22 Ni 2+, d 8, n = 2, S = 1, M s = = 3 m l L = (2 2) + (2 1) +(2 0) 1 2 = 3, => F J = L + S = = 4 => 3 F4 Cr 3+, d 3, n = 3, S = 3/2, M s = 2 3/2 + 1 = 4 m l L = (1 2) + (1 1) +(1 0) = 3, => F J = L S = 3 3/2 = 3/2 => 4 F3/2 Ce 3+, f 1 n = 1, S = 1/2, M s = 2 1/2 + 1 = 2 m l L = (1 3) = 3, => 2 F J = L S = 3 1/2 = 5/2 => 2 F5/2 Ho 3+, f 10 n = 4, S = 2, M s = = 5 m l L = (2 3) + (2 2) + (2 1) +(1 0) (1 1) (1 2) (1 3) = 6, => I J = L + S = = 8 => 5 I8 La 3+ : f 0, n=0, S=0, M S =1, L = 0, J = 0, => 1 S0 Lu 3+ : f 14, n=0, S=0, M S =1, L = 0, J = 0, => 1 S0 5) Für welche der folgenden Ionen erwarten Sie einen Bahnbeitrag zum magnetischen Moment: a) V 3+, b) Cr 3+, c) Ti 4+, d) Fe 3+ (low-spin), e) Fe 3+ (high-spin) Für Ionen mit T1g- oder T2g-Ligandenfeldgrundtermen. Das sind: V 3+ ( 3 T1g) und Fe 3+, lowspin ( 2 T2g), vgl. Tabelle 1. 6) Berechnen Sie den Wert des effektiven magnetischen Moments der folgenden Ionen:

23 a) Ti 3+, Ce 3+, Yb 3+, Ho 3+, Eu 3+. Ti 3+ : Spin-Only-Formel: eff = 1.73 Für Lanthanoide (Ce 3+ -Eu 3+ ) gilt: gj J ( J 1) B, S( S 1) L( L 1) J ( J 1) gj 1 2J ( J 1) wobei S, L und J aus den Termsymbolen bestimmt werden können Ce 3+ : 2 F5/2, S = 1/2, L = 3, J = 5/2, gj = 0.857, = Yb 3+ : 2 F7/2, S = ½, L = 3, J = 7/2, gj = 1.143, = Ho 3+ : 5 I8, S = 2, L = 6, J = 8, gj = 1.25, = Eu 3+ : = 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.