INORGANIC CHEMISTRY (Paper III)

Coordination Chemistry: IUPAC nomenclature, bonding theories – review of Werner’s theory and Sidgwick’s concept of coordination, Valence bond theory, geometries of coordination numbers 4-tetrahedral and square planar and 6-octahedral and its limitations, crystal filed theory, splitting of d-orbitals in octahedral, tetrahedral and square-planar complexes – low spin and high spin complexes – factors affecting crystal-field splitting energy, merits and demerits of crystal-field theory. Isomerism in coordination compounds – structural isomerism and stereo isomerism, stereochemistry of complexes with 4 and 6 coordination numbers. 10 hr

It was known for many years that two salts can be dissolved together and a new third compound can be crystallised out of this mixture. These are of two kinds… Double salts and complexes. For example, if KCN and Fe(CN)2 are mixed together, in a solution and then crystallized, the resulting compound is a complex K4[Fe(CN)6], whereas, if K2SO4 and Al2(SO4)3 are mixed together, in a solution and crystallized, they form a double salt. When K4[Fe(CN)6] is dissolved, it ionizes to give four K+

ions and [Fe(CN)6]4- ion and there is no Fe2+ ion. When a double salt is dissolved in water, it ionizes to give the same ions as the original two compounds ie K+ ion, Al3+ ion, and SO4

2- ions. The coordination compounds or complexes are different from double salts. Hence Werner realized that compounds like K4[Fe(CN)6] were different and these are known as coordination compound or a complex.

A complex (coordination compound) has a central metal ion and ligands surrounding it. The number of ligands bonded to the metal is the coordination number of the metal. Together, the metal and the ligands make the coordination sphere which is represented in a pair of square brackets.

Coordination compounds can be cationic, anionic or neutral eg [Co(NH3)6]Cl3 the complex is cationic [Co(NH3)6]3+

K4[Fe(CN)6] the complex is anionic [Fe(CN)6]4-

A neutral complex may have metal in the zero oxidation state and neutral ligands like [Ni(CO)4]Or be neutral because the ligands have a equal and opposite charge to that of the metal ion [Co(NH3)3Cl3]If the complex is ionic, it has a counterion of the opposite charge. The oxidation state of the metal can be calculated. Eg:- In [Co(NH3)3Cl3], the valencies are Co (x) NH3 (neutral so 0) Cl- (-1)

x +(3 x 0) + (3 x -1) = 0 since it is overall neutral.x = 0 – 0 – (-3) = +3

Ligands bind to the metal ion by donating a pair of electrons. In the ligand, the atom that actually donates the electron is the donor atom Ammonia binds through the nitrogen atom and hence N is the donor atom in ammonia similarly, water binds

through oxygen atom using the lone pair of electrons on N and O :NH3 and :OH2

H2

O

H3N NH3

H2O OH2

M M H2O OH2

H3N NH3

OH2

1

Some ligands are bidentate—they bind through two donor atoms eg ethylenediamine NH2CH2CH2NH2

CH2- CH2

NH2 NH2

M

Oxalate ion O=C –O M

O=C –O

Acetylacetonate CH3COCH2COCH3+ CH2

CH3 C C CH3

O O

MThese donor atoms may be the same kind as in the examples given above, or of different kinds as in Glycinate ion,

Ligands can have higher denticity— diethylenetriamine is tridentate

NH2CH2CH2NHCH2CH2NH2

CH2CH2 CH2

NH CH2

NH2 NH2

M

Glycinate ion is a bidentate ligand with 2 different donor atoms O and N

H2NCH2COO − CH2

H2N C O

M O

EDTA ion is a hexadentate ligand.Ligands can be monodentate, bidentate, tridentate etc.They are also classified as oxygen donors, nitrogen donors, sulphur donor, etc depending on the donor atom.Some ligand form strong bonds to a particular metal and others may form weaker bonds. They can therefore be classified as strong and weak ligands. However, a ligand that is strong for one metal may, in some cases, be a weak ligand for another metal..

IUPAC nomenclature:-

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The IUPAC has rules for naming complexes. However, there are more than one method of naming allowed by the IUPAC. The most commonly used method uses the following rules—

A complex (coordination compound) has a central metal ion and ligands surrounding it.Coordination compounds can be cationic, anionic or neutral eg [Co(NH3)6]Cl3 the complex is cationic [Co(NH3)6]3+

K4[Fe(CN)6] the complex is anionic [Fe(CN)6]4-

A neutral complex may have metal in the zero oxidation state and neutral ligands like [Ni(CO)4]Or be neutral because the ligands have a equal and opposite charge to that of the metal ion [Co(NH3)3Cl3]If the complex is ionic, it has a counterion of the opposite charge.

Coordination compounds are named according to IUPAC rules. 1.As with any inorganic compound, the cation is named first whether it is the complex or the counterion.2.In the complex, the ligands are named first in alphabetical order, then the metal is named.3.The oxidation state of the metal is mentioned in Roman numbers inside a bracket.4.If the complex is anionic, the metal has a suffix -ate for example - cobaltate(III). If it is cationic, the metal

retains its name eg. Cobalt(III). For elements like iron, silver, mercury etc, the roman name of the metal is used for the anionic form and the common name for the cationic form eg. Iron(III) but ferrate(III) ; silver(I) and argentate(I).

5.The ligands are named before the metal. 6.The ligands are named in alphabetical order. 7.Ligands that are anionic have a suffix with -o eg. nitrato, sulphato, fluoro, iodo, glycinato, etc. 8. Neutral ligands retain their names- ethylene diamine

However, some ligands have special names H2O is aqua, NH3 is ammine. NO is nitrosyl, CO is carbonyl. 9. There are very few cationic ligands. These have a suffix -ium eg. hydrazinium NH2NH3

+ 10. If there are more than one ligand of each kind, it is prefixed by di- tri- tetra- etc.

diammine, trichloro, etc11. If the ligand has numerical affixes in its name, then to denote number of ligands, the affix becomes bis-, tris-

tetrakis- eg bis(ethylenediamine) bis(triethylenediamine)

[Co(NH3)6]Cl3

Complex Cation anion (counter ion)

Here the complex is cationic and the counterion is anionic. The counterion is named as in any salt. The complex is named as per the rules above whether is the cation or the anion.

[Co(NH3)6]Cl3 hexa ammine cobalt(III) chloride

K4[Fe(CN)6]

Cation Complex anion ----- here the cation is the counter ion

Some examples:-K4[Fe(CN)6] potassium hexacyano ferrate(II)[Co(NH3)4BrCl] Cl tetraammnine bromo chloro cobalt(III) chloride[Pt(NH3)6]Cl4 hexa ammine platinum(IV) chloride K3[Fe(CN)6] potassium hexacyano ferrate(III)[Co(en)3](NO3)3 tris ethylenediamine cobalt(III) nitrate[Co(NH3)3ClBrI] triammine bromo chloro cobalt (III) (no counterion since it is neutral)K3[Fe(C2O4)3] potassium trioxalato ferrate(III)[Cr(H2O)6]Cl3 Hexa aqua chromium(III) chlorideK2[PtCl4] Potassium tetrachloro platinate(II)K2[PtCl6] Potassium hexachloro platinate(IV)NaBH4 Sodium tetrahydrido borate(III)

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LiAlH4 Lithium tetrahydrido aluminate(III)K3[Nb(O2)4] potassium tetraperoxo niobate(V)K3[Ti(H2O)6] Potassium hexa aqua titanate(III)[Co(NH3)3ClBrI] triamminebromochloroiodo cobalt(III)

[Pt(NH2CH2COO)2] diglycinato platinum (II) [Co(en)3]3+ tris(ethylenediamine)cobalt(III) cation[Cr(en)2Cl2]Cl dichloro bis(ethylenediamine)chromium(III)chloride.

Naming bridged compounds:- Some complexes have more than one metal ion. These are binuclear, trinuclear etc complexes. The two metal may be linked by metal-metal bonds or through a ligand tat acts as a bridge between the two metals. Bridging ligands are named with a prefix μ- eg μ-oxo or μ-nitrato, μ- amido etc.

H O

(H2O)4 Fe Fe(H2O)4 (SO4)2 The complex is named di-μ- hydroxo- octaaqua diiron(III) sulfate O H

H2 4+ N

(NH3)4 Co Co(NH3)4 μ- amido- μ- nitrito-N-octaammine dicobalt(III) cation.

N O2

Theories of bonding in coordination compounds:-

Werner’s hypothesis (postulates):-Alfred Werner realized that some compounds like those formed between potassium sulphate and aluminum sulphate were different from those formed between potassium cyanide and ferrous cyanide. In the early 19th century (1823), before the discovery of electrons, bonding was considered in terms of valencies- an atom bonded to another to fulfill its valency. AlfredWerner studied some compounds of cobalt and platinum and came to the conclusion

` that atoms of metals showed two kinds of valencies— primary valency and secondary valency.` Every metal had a fixed number of primary and a fixed number of secondary valencies and it satisfies both types of valencies..` The primary valency is ionisable. It is satisfied by negative ions.` The secondary valency is non ionisable and is satisfied by both negative ions or by neutral molecules. Sometimes a negative ion can satisfy both primary and the secondary valency of the metal.` The primary valency is non directional` The secondary valency is directional and determines the shape of the molecule.

Werner’s work:-Werner came to these conclusions after studying compounds of cobalt and platinum with ammonia. He prepared compounds of cobalt chloride with ammonia in different ratiosand determined experimentally,

` the number of free chloride ions by precipitating them with silver nitrate.` the electrical conductivity of their solutions. From this he determined the total number of ions in the compound.

He also tried to prepare the different isomers of these compounds and predict their geometries.CoCl3.6NH3:- In this compound with the molar ratios of 1 CoCl3 : 6 NH3, Werner found there were 3 Cl- ions precipitated by AgNO3. Thus there are 3 ionisable chloride ions in this compound. The molar conductance of 300

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Sm2mol-1 showed that the compound had 4 ions. Hence the formula of [Co(NH3)6]Cl3 was postulated with 3 Cl- ions and 1 cation.

Werner represented this as

Cl NH3

NH3 NH3

CoCl NH3 NH3

NH3

Cl

Where the dashed lines represented the primary valencies and the solid lines the secondary valencies.The primary valencies represent the oxidation state of the metal and the secondary valency represents the coordination number of the metal as per the modern terminology.

Similar studies are on CoCl3.5NH3 CoCl3.4NH3 and CoCl3.3NH3 gave results as shown in the table below.

compound No of free chloride ions

Molar conductance and Total number of ions

Modern formula

CoCl3.6NH3 3 300 S m2 mol-1

4 ions[Co(NH3)6]Cl3

CoCl3.5NH3 2 262 S m2 mol-1

3 ions[Co(NH3)5Cl]Cl2

CoCl3.4NH3 1 102 S m2 mol-1

2 ions[Co(NH3)4Cl2]Cl

CoCl3.3NH3 0 No conductance 0 ions

[Co(NH3)3 Cl3]

The compounds were represented by Werner as

CoCl3.5NH3

Cl NH3

NH3 NH3

CoCl NH3 NH3

Cl

CoCl3.4NH3

Cl

NH3 NH3

CoCl NH3 NH3

Cl

CoCl3.3NH3

Cl

NH3

Cl Co

NH3 NH3

Cl

In these compounds, the chloride ion satisfies both the primary valency and the secondary valency of cobalt.He carried out similar work with compounds between PtCl4 and NH3

Werner had postulated that the secondary valencies determine the geometry of the complexes. The compounds he prepared had a secondary valency (coordination number) of six.

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Their shape could be `planar hexagonal`prismatic or`octahedral

He prepared isomers of the compounds. He compared the number of isomers possible in each of these geometries with the number of isomers he could synthesise and came to the conclusion that six coordinated complexes were octahedral in shape. This was later confirmed by structural studies.

compound No of isomers possiblePlanar hexagon

No of isomers possiblePrism

No of isomers possibleOctahedron

Exptally obs isomers

CoCl3.6 NH3 1 1 1 1CoCl3.5 NH3 1 1 1 1CoCl3.4 NH3 3 3 2 2CoCl3 3 NH3 3 3 2 2

The shape must be octahedral

Drawbacks:- Werner’s theory predated the electronic theory of valency. It was proposed before the discovery of the electron. Hence it is not a complete theory of bonding. It does not explain the bonding characteristics in complexes.

Sidgwick’s concept of bonding in coordination compounds:-Sidgwick proposed that metal ions formed complexes by bonding with as many ligands as resulted in the metal acquiring an effective atomic number of the nearest noble gas. (The effective atomic number or EAN is the total number of electrons that the metal ion has around it after it has bonded to the ligands)Eg [Co(NH3)6]3+ Co is in the +3 oxidation state. Its atomic number is 27Co3+ = 27 -3 = 24 electrons. Each ligand donates 2 electrons 6 NH3 = 2 x 6 = 12 Total = 36The Cobalt ion has an EAN of 36 which is that of the nearest noble gas KrEAN for some other compounds—Calculate the EAN for the following and sate whether they follow Sidgwick’s EAN rule.

(i) K3[Fe(CN)6](ii) K4[Fe(CN)6](iii) [Pt(NH3)4]Cl2

(iv) [Pt(NH3)6]Cl4

K4[Fe(CN)6] Fe atomic no -- 26. Oxidation state- +2

Therefore Fe2+ has 24 electrons around it. After bonding with 6 ligands, it acquires 12 electronsTotal number of electrons around Fe3+ ion is 24 + 12 = 36The complex follows Sidgwick’s EAN rule.

K3[Fe(CN)6] Fe atomic no -- 26. Oxidation state- +3

Therefore Fe3+ has 23 electrons around it. After bonding with 6 ligands, it acquires 12 electronsTotal number of electrons around Fe3+ ion is 23 + 12 = 35 The EAN is 35 which is NOT that of a noble gas. Hence the EAN rule fails in this and in some other cases.

Another drawback of this theory is that each metal atom is surrounded by 12 electrons more than it usually has. How is it able to hold so many extra electrons?

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Hence the EAN rule was discarded except in the case of metal carbonyls. These compounds follow the EAN rule. (refer to the chapter on carbonyls).When the valence bond theory was proposed, it was used to explain bonding in complexes.

Modern theories of bonding:- The modern theories explain bonding in coordination compounds and also explain many of the properties of these compounds. Valence Bond theory:-The valence bond theory was used to explain bonding in coordination compounds.Postulates of the Valence Bond Theory of bonding in Coordination compounds:-`The metal is the central atom and is surrounded by ligands.`The ligands donate electron pairs to the metal to form coordinate bonds.`The ligand orbitals that contain these electron pairs overlap with the metal orbitals. `The metal uses vacant orbitals of suitable energy to accommodate these electrons`The vacant orbitals are usually hybrid orbitals.`In the case of d block metals, the inner (n-1) d or the outer n d orbitals are used in hybridisation.`The geometry of the complex will depend on the hybridisation. This is in accordance with the VSEPR rules.`The electrons of the metal may occupy the hybrid orbitals in a paired formation or singly. This depends on the nature of the ligand that is forming the complex. This leads to the formation of high spin and low spin

VBT approach to Bonding in complexes with coordination number 4:-

According to the VBT, the metal orbitals are Hybridised by interaction with the approaching ligand. These hybrid orbitals are then used for bonding with the ligand orbitals.An example – the case of Ni2+ ion

Atomic number of Ni = 28Ni = 1s2 2s2 2p63s23p6 3d8 4s2

Ni2+ = 1s2 2s2 2p63s23p6 3d8

A 4-coordinate complex can have either `a square planar geomet ry or `a tetrahedral geometry. This depends on the hybridisation of the metal orbitals.

Ni2+ = 1s2 2s2 2p63s23p6 3d8

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General Postulates of the Valence Bond Theory:- i Atoms form bonds with each other by sharing electrons. ii The valence electrons are involved in bonding.iii The corresponding orbitals are considered to overlap.iv Greater the extent of overlap, stronger the bond.v The molecule can be represented by a Lewis structurevi The orbitals used for bonding may be hybridised orbitals or unhybridised orbitals. vii The hybridisation detrmines the geometry of the compound and the structure is determined by considerations

of the VSEPR theory viii There can be more than one Lewis structure possible. The actual structure is taken to be a resonance hybrid

of all the feasible Lewis structures.

It must be noted that once filled, the 4s orbital is higher in energy. Therefore, when a first row d block metal atom is ionised, the 4s electrons are ionised first. Similarly for the 2nd and 3rd row d block elements, the ns electrons are ionised first. .

↑↓ ↑↓ ↑↓ ↑↓ ↑↓

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑

1s 2s 2p 3s 3p 3d 4s 4p 4dAccording to the Valence Bond theory, the inner orbitals are not used for bonding . Hence, only the outer orbitals may be considered.

↑↓ ↑↓

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑

3s 3p 3d 4s 4p 4d

The available empty orbitals are 4s,4p,4d and if the electrons can pair up, one 3d orbital is also available.If the ligand that is forming the bond with the metal is a strong one, then it can force the electrons to pair up against the Hund’s rule. (this happens in many cases, but not in all cases.)If the ligand is weak, it cannotWith a strong ligand, like CN- pairing-up of electrons occurs, and we get dsp2 hybridisation.

Square Planar

↑↓ ↑↓

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↑

3s 3p 3d 4s 4p 4d

↑↓ ↑↓

↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓

3s 3p 3d 4s 4p 4d

Pairing dsp2hybridisation

Hence K2 [Ni(CN)4] is a square planar complex (dsp2 hybridisation gives rise to square planar geometry)

CN CN Ni

CN CN

With a weak ligand like Cl-, there is no pairing of electrons. Hence the hybridisation is sp3.

The complex K2 [NiCl4] is tetrahedral in shape (sp3 hybridisation gives rise to tetrahedral geometry)

Tetrahedral ↑↓ ↑

↓↑↓ ↑ ↑

3d 4s 4p 4d Cl

No Pairing sp3 hybridisation

The dashed line represents a ligand electron Ni Cl Cl

ClSome other 4 coordinate complexes-

Square planar - [Cu(NH3)4]2, [PtCl4]2- , [Pt(NH3)4]2+

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Tetrahedral:- [CuCl4]2- , [ZnCl4]2-, [MnCl4]2-, [CoCl4]2- Some compounds of s and p block elements also show this geometry . eg BH4

- BeF42- etc

Bonding in some other Square planar complexes:- [Pt(NH3)4]2+ and [PtCl4]2-

Pt (78) 5d8 6s2

Pt2+ 5d8

↑↓

↑↓ ↑↓ ↑↓

5d 6s 6p Pairing up of d electrons d sp2 hybridisation

[Cu(NH3)4]2+ Cu (29) 3d94s2 or 3d10 4s1

Cu2+ = 3d9

↑↓

↑↓

↑↓ ↑↓

↑

3d 4s 4p According to the VBT, this complex is expected to have a tetrahedral geometry, but has been found experimentally to have a square planar geometry. To explain the observation, various modifications are proposed.

One such modification is that the 9th electron in the 3d orbital is promoted to the vacant 4p orbital and the vacated 3d orbital is used for hybridisation to give d sp2 hybridisation.

↑↓

↑↓

↑↓ ↑↓

↑

3d 4s 4p This explanation has been criticised since it involves a large increase in the energy of the electron. This is one of the criticism for the VBT.

Valence bond treatment of Tetrahedral complexes:-

Cl

M Cl Cl[ZnCl4]2- Cl

↑↓

↑↓ ↑↓

↑↓ ↑↓

3d 4s 4p sp3hybridisation.

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[CuCl4]2- (this complex has been reported to exist in both tetrahedral as well as in square planar geometry)

↑↓ ↑↓ ↑↓ ↑↓ ↑

3d 4s 4p

sp3hybridisation.

Similarly for Ni2+ ion, complex with the strong ligand CN- , the electrons can pair up in the d orbitals vacating one of the d orbitals. This leads to a dsp2 hybridisation.Ni = 28 -- 3d8 4s2

Ni2+ 3d8

↑↓

↑↓ ↑↓

↑ ↑

3d 4s 4p

↑↓

↑↓ ↑↓

↑↓

3d 4s 4p d sp2 hybridisation.

However, with a weak ligand like Cl-, no pairing occurs, and the d orbitals are not available for hybridisation. Hence it forms sp3 hybridisation.

↑↓

↑↓ ↑↓

↑ ↑

3d 4s 4p sp3hybridisation

Coordination number 6-octahedral complexes :-All complexes with coordination number 6 are octahedral irrespective of the nature of the ligand.[Co(NH3)6]3+ cation is a commonly used example. It exists as its chloride usually [Co(NH3)6]Cl3 Co – 27 --- 3d74s2

Co3+ has the electronic configuration 3d6 since it ionises by losing the electrons in the 4s and on electron from the 3d orbitals.

↑↓

↑ ↑ ↑ ↑

3d 4s 4p 4d.Ammonia is a strong ligand. There are six ligands therefore, six empty orbitals are needed for the electrons. The electrons in the 3d sub shell pair up Therefore, d2sp3 hybrid orbitals are formed.

↑↓

↑↓ ↑↓ .. ..

.. .. ..

..

3d 4s 4p 4dd2sp3 hybridisation

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Since the inner d orbitals (ie (n-1)d orbitals) are used, such complexes are known as inner orbital complexes. Since the electrons are paired up, they are also known as spin paired complexes. There are smaller number of unpaired electrons in these complexes. Hence they are also known as low spin complexes. [CoF6]3- -- This complex has F- a weak ligand. Hence the d electrons do not pair up in the Co3+ ion. The complex has to therefore use sp3d2 hybridisation, using the nd orbitals which are empty. Hence it is an outer orbital complex / spin free complex /high spin complex.

3d 4s 4p 4d

sp 3 d2 hybridisation

Magnetic properties:-Since the two kinds of complexes have different number of unpaired electrons, they differ in their magnetic properties. For example, the inner orbital complex [Co(NH3)6]3+ has no unpaired electrons, and hence it has zero magnetic moment; while the outer orbital complex [CoF6]3- has 4 unpaired electrons and a magnetic moment of μ= √n(n+1) = √4(4+1) = √20 = 4.47 BMwhere μ is the magnetic moment in Bohr magnetons, n is the number of unpaired electrons.Other complexes that are six coordinated also have either sp3d2 or d2sp3 hybridisation and the complex is octahedral in shape.

[Cr(NH3)6]3+ Cr = at no 24 3d54s1. Cr3+ = 3d3

3d 4s 4p 4d

d2sp 3 hybridisationIt is an inner orbital complex, magnetic moment is √3(3+2) = √15= 3.88 BM

In this case, since two of the 3d orbitals are free anyway, the hybridisation is always d2sp3 irrespective of whether it is a

complex with a strong ligand or a weak ligand.

NH3

NH3 NH3

Cr NH3 NH3

NH3

The VBT is able to explain bonding in most complexes. It can explain their structure and magnetic properties. However, it has a few drawbacks.

Limitations of Valence Bond Theory:-The VBT is a qualitative theory.It does not explain the fact that many complexes absorb visible radiation and are hence colored and that sometimes, the same metal can form different colored complexes with different ligands.It cannot explain the bonding in square planar copper complexes satisfactorily.

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↑↓ ↑ ↑ ↑ ↑ .. .. ..

.. .. ..

↑ ↑ ↑ .. .. .. .. ..

..

It does not explain the relative stabilities of different complexes.

Crystal field theory:-Another theory was proposed by Bethe and van Vleck to explain the bonding in coordination compounds.Postulates:-

`The metal atom and the ligands are assumed to be point charges.`The interaction between them is assumed to be purely electrostatic.`The interaction between the metal and the ligand causes the degenerate d-orbitals of the metal ion to split and

lose their degeneracy.`The splitting depends on the geometry of the complex ie the direction of the approach of the ligands.

Shapes of d orbitals:- y z z

x x y

dxy dxz dyz

y z

x x

dx2-y2 dz2

It can be seen that the d orbitals are oriented in specific manner.The dxy, dxz and the dyz are oriented between the coordinate axes, while the dx2 – y2 and the dz2 are oriented along the coordinate axes. Thus the five degenrate d orbitals belong to two sets…one set of three orbitals oriented between the coordinate axes, and one set of two orbitals oriented along the coordinate axes.

`When the ligand approaches the metal ion, these d orbitals interact with the approaching ligand. `Depending on the geometry of the complex to be formed, the ligands approach the metal from different angles.

M M M

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`Those orbitals that are close to the approaching ligands are affected more by the ligands and are hence comparatively, raised in energy, while those orbitals that are farther away from the approaching ligands are lowered in energy.

Splitting of d orbitals in an octahedral field:- In an octahedral complex, the ligands approach the metal ion along the three coordinate axes.

Ligands approaching the metal and the positions of the dx2-y2 and dz2 orbitalsdx2-y2

dz2 Z

ligand

X Y

Ligands approaching the metal and the position of the dxy orbital

Z

X Y

Similar geometry can be seen for the dxz and dyz orbitals.

It can be seen from the above figures that the dx2-y2 and the dz2 orbitals are directly in the path of the approaching ligands whereas, the dxy is at an angle to the approaching ligand. Hence the ligands affect the dx2-y2 and the d z2 orbitals more than the dxy or dxz or dyz orbitals. Thus the two sets of orbitals are affected differently by the approach of the ligands.

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The 5 degenerate d orbitals are therefore no longer degenerate. The set of orbitals along the axes, ie the dx2-y2 and the dz2 orbitals are slightly raised in energy and the dxy, dxz, dyz orbitals are slightly lowered in energy.The three lower energy levels are known as the t2g levels and the two upper levels are eg levels.(the nomenclature is from the group theoretical cosiderations of the molecular symmetry)The energies can be represented in a diagram.

Splitting of d orbitals in an octahedral field

— — dx2-y2, dz2 (belonging to the class eg)

10Dq or ΔO.

— — — — —

Degenrate d orbitals — — — dx dxz, dyz (belonging to the class t2g)in the presence of ionsor ligands around it. Splitting of d orbitals The presence of negative ions in the presence of the ligands

— — — — — or ligands raises the energy bonding with the metal ionOf all the d orbitals.

Degenerate d orbitals in a free metal ion

This is called Crystal field splitting. (this is because such effects exist in a crystal where a metal is surrounded by negative ions in the lattice)The difference in the energy of the two sets of d orbitals is termed 10Dq or ΔO.Since the nett energy must be conserved, the amount by which the three d levels are lowered is the same as the amount by which the two d levels are raised.Hence the energy by which the two d orbitals (dx2-y2 and dz2) are raised must be 0.6 ΔO

The amount by which the three d orbitals are lowered must be 0.4 ΔO ( 2 x 0.6 = 3 x 0.4) . The weighted mean value is the Bari Centre.The value of ΔO depends on various factors—the nature of the metal ion as well as the nature of the ligands.In general, for the same metal in the same oxidation state, some ligands can cause a greater splitting than others. A ligand that causes greater splitting, ie higher value of ΔO is known as a strong ligand. The presence of electrons in the lower d levels leads to stabilisationWhereas the presence of electrons in the higher d levels leads to destabilisation.The nett value is known as the Crystal Field Stabilisation Energy or CFSE . For example, in the case of Cr3+ ion, forming a complex with 6 ligands. Cr3+ ion has 3d3 configuration.

eg

6 Dq or 0.6 Δo Bari centre

↑ ↑ ↑

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4 Dq or 0.4 Δo

t2g

It may also be represented as

t2g eg

Calculation of CFSE:-Since the amount by which the energy is lowered is 0.4 ΔO for each electron in the t2g level, the nett stabilisation (lowering of energy) is 3 x(- 0.4) ΔO= -1.2 ΔO Since there is no electron in the eg levels, there is no destabilisation. Therefore CFSE = 1.2 ΔO

In the case of Co3+ ion, forming a complex with 6 ligands. Co3+ ion has 3d6 configuration. Here, there are two

possibilities

Case 1When the ligand is weak eg F- ion as in [CoF6]3- the value of ΔO is small. The difference between the t2g and the eg levels is not much. Hence the electrons fill up the energy levels as per th Hund’s rule.

eg

↑↓ ↑ ↑ ↑ ↑ ΔO (small)

t2g

Calculation of CFSE:-There are 4 electrons in the t2g levels. Hence the energy is lowered by 4 x 0.4 ΔO ie there is stabilisation of 1.6 ΔO

There are 2 electrons in the eg levels. This raises the energy by 2 x 0.6 ΔO ie it destabilises the system by 1.2 ΔO CFSE = 4 x (-0.4) ΔO + 2 x 0.6 ΔO = (- 1.6 + 1.2) ΔO = - 0.4 ΔO

From the distribution of electrons in the different d orbitals, we can determine the magnetic properties of the complexIn this case, there are 4 unpaired electrons. Hence the magnetic moment of the complex is

Magnetic moment = √ n(n+2) = √ 4 x 6 = √24 = 4.9 BM (Bohr magneton)

Case 2 where the ligand is strong eg NH3 as in [Co(NH3)6]3+

The value of ΔO is large . The difference between the t2g and the eg levels is great . Hence the electrons fill up the lower first before occupying the higher levels. This is contrary to the Hund’s rule.

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↑ ↑ ↑

↑ ↑ ↑

↑ ↑

↑↓ ↑ ↑

eg

ΔO (large)↑↓ ↑ ↑ ↑ ↑

t2g

Calculation of CFSE:-There are 6 electrons in the t2g levels. Hence the energy is lowered by 6 x (-0.4) ΔO ie there is stabilisation of 1.6 ΔO

CFSE = 6 x (-0.4) ΔO = - 2.4 ΔO

In this case, there are no unpaired electrons. Hence the magnetic moment of the complex is zero.

Assignment :- Calculate the CFSE and magnetic moments for complexes with d1, d2 , d3 , d4 , d5

, d6 , d7 , d8 , d9, d10 in a strong field an in a weak field.

Splitting of d orbitals in square-planar complexes:-A square planar complex can be considered to be an octahedral complex whose two axial ligands have been removed.

M M

This means that the z axis does not have any ligand. Hence the orbitals that have a z component are affected less by the approaching ligands whereas, the orbitals having only x and y components are much more affected by the approaching ligands. Therefore, the orbitals having z component are lowered in energy compared to the orbitals that have x and y components.

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↑↓ ↑↓ ↑↓

Splitting of d orbitals in a square planar field

— dx2-y2

dx2-y2 and dz2

— —

— dxy

— — — — — — dz2

— — — dxy dxz dyz

Degenerate d orbitals d orbitals in — — dxz dyz

an octahedral field d orbitals in a square planar field

The dz2 orbital has no x y component. Hence it is much lower in energy then the dx2-y2 orbital. Hence the energy levels in a square planar complex is different from that of an octahedral complex.

Splitting of d orbitals in a tetrahedral field:-In a tetrahedral complex, the ligands approach the metal from the corners of a tetrahedon.

The approaching ligands are closer to the dxy, dxz, dyz orbitals than to the dx2-y2 dz2 orbitals. Hence the three orbitals dxy, dxz , dyz levels are higher in energy than the dx2-y2 dz2 orbitals. The splitting is the reverse of that in an octahedral complex. However, the extent of splitting is much less since

`only four ligands are causing the effect`none of the orbitals is directly in the path of the ligands.

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Hence the value of Δt is much smaller than the value of ΔO.In fact, for similar metal ions and ligands, Δt ≈ 4 ΔO

9

The splitting of the d orbitals in a tetrahedral field can be represented as follows

— — — dx dxz and dyz

Δt

— — — — —

Degenerate d orbitals — — dx2-y2 and dz2

in the presence of ionsor ligands around it. Splitting of d orbitals The presence of negative ions in the presence of the ligands

— — — — — or ligands raises the energy bonding with the metal ionof all the d orbitals.

Degenerate d orbitals

In the case if a tetrahedral complex, the value of Δt is small and hence whether the ligand is strong or weak, the electrons occupy the orbitals in accordance with the Hund’s rule.

Magnetic Properties:- Octahedral field-Low spin and high spin complexes -As in the case of the valence bond theory, the crystal field theory can explain the magnetic properties of complexes. In a strong field, the value of ΔO is large. Hence the electrons fill up the t2g levels before occupying the eg level. The complex is therefore low spin, ie the number of unpaired electrons is minimum, whereas, in a weak field, the electrons occupy the orbitals in accordance with the Hund’s rule and the number of unpaired electrons is higher.

For example, in the case of a Co3+ ion, which has the electronic configuration 3d6

Case 1 complex with a strong ligand like NH3,

↑↓

↑↓ ↑↓

t2g eg

No unpaired electrons therefore the magnetic moment is zero. This is a low spin complex.μ = 0

Case 2 weak ligand like F-

↑↓ ↑ ↑ ↑ ↑t2g eg

4 unpaired electrons therefore it is a high spin complex. The magnetic moment is

μ = √4 x 6 = √ 24 = 4.9 BM

18

This is in agreement with experimental evidence.

Factors affecting crystal-field splitting energy:- The extent to which the d orbitals are split, depends both on the i) the nature of the ligandsii) and the nature of the metalas well as on the geometry of the complex.

Nature of the Ligands:-

The ligands play a major role in determining the extent to which the d orbitals are split due to complex formation. A strong ligand like CN- ion or NH3 molecule can split the d orbitals to a greater extent… ie, the value of Δo will be large for a complex with such ligands. The ligands can be arranged in order of decreasing ability to split the d orbitals of the metal. Such a series is known as the spectrochemical series (reason for the name—see the spectroscopic properties of complexes**)

CO ≈ CN- > NO2- > en > NH3 > H2O > F- > Cl- > Br- > I-

The above series shows that ligands like CO cause a greater split in the d orbitals of the metal atom. The value of Δo will be higher in such complexes. The value of ΔO will be small in complexes formed with iodide or bromide ions as ligands. Hence the ligands like CN- , en, ammonia are called strong field ligands or strong ligands , while the ligands like halide ions are weak field or weak ligands.For example, the complex [Co(H2O)6] 3+ has a Δo value of 18,600 cm-1 while the complex [Co(NH3)6] 3+ has a ΔO value of 23,000 cm-1 . (The values are determined from the UV-visible absorption spectrum of the complex and are hence reported in cm-1)

Nature of the metal:-

The ability to form a complex depends on the size and the charge on a metal ion. Hence the crystal field splitting depnds on the oxidation state of the metal ion. It also depends on whether it is a first row d-block element or a second/ third row d-block element.Oxidation state:- Higher the oxidation state of the metal ion, greater the extent of splitting of the d orbitals. In the case of cobalt, which forms complexes in +2 and in +3 oxidation states, the complexes show remarkable difference in their absorption spectrum.

[Co(H2O)6]3+ ΔO = 18,600 cm-1

[Co(H2O)6]2+ ΔO = 9,400 cm-1

Size:- The elements in the first row of the d block show smaller splitting compared to the corresponding element in the second and the third row. The heavier elements have d orbitals that are extended farther in space. Hence they interact more with the incoming ligands and are therefore split to a greater extent. This can be seen by the value of ΔO for the complexes formed by Co, Rh, Ir all metals in the same group,having the configuration nd6 and in the same oxidation state.

[Co(NH3)6]3+ ΔO = 23,000 cm-1 [Rh(NH3)6]3+ ΔO = 34,000 cm-1 [Ir(NH3)6]3+ ΔO = 41,000 cm-1

Shape of the complex:- As mentioned before, the extent of crystal field splitting for a tetrahedral complex is much smaller than for an octahedral complex. ΔO > ΔT This is because only 4 ligands are interacting with the metal ion and none of these approach the metal orbitals directly. Hence the extent of splitting depends on the shape of the complex.

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Merits and demerits of crystal-field theory.:- `The CFT explains the splitting of d orbitals in the field imposed by the approaching ligands.`The CFT is able to explain the spectral and magnetic properties of many complexes. `It is a simple theory since it assumes that the atoms are point charges. `The CFT applies to complexes of all shapes.`The CFT assumes that the metal and the ligands are point charges. This is of course nt true and is only an approximation.`The CFT assumes that the interaction between the metal and the ligand is purely electrostatic in nature. However, it is seen* that there is a considerable covalency in the metal ligand bonds, specially amongst the more polarisable atoms.`The CFT is able to explain the electronic spectrum of most complexes, but in some cases, the intensity and the frequency of some absorption babnds cannot be predicted from the CFT.(* Evidence of covalency can be had from vibrational spectra- since the formation of a complex affects the vibration frequancies of the ligand, it can be assumed that a metal ligand covalent bond is formed. Other evidence include ESR, NMR etc)

Isomerism in coordination compounds –

Definition of isomerism:- Assignment --Write all definitions. Isomerism is of two kinds—Structural isomerism and stereoisomerism.

Structural isomerism:-As in organic compounds, coordination compounds show structural isomerism. These are of many kinds:(i) Ionisation isomerism (ii) Hydrate isomerism(iii) Linkage isomerism (iv) Coordination isomerism(v) Coordination position isomerism (vi) Polymerisation isomerism Ionisation isomerism:-An anion can act as aligand as well as a counterion. Hence in a complex with the same molecular formula, we can have two isomers –one with an anion acting as a ligand and another with the same anion acting as a counter ion.Eg [Co(NH3)5Br]SO4 and [Co(NH3)5 SO4]BrThe first complex has the Br- ion as a ligand while in the second complex, the Br- ion is a counter ion.Such isomers are ionisation isomers. The two isomers ionise differently.

[Co(NH3)5Br]SO4 [Co(NH3)5 Br]2+ + SO42-

[Co(NH3)5 SO4]Br [Co(NH3)5 SO4] + + Br-

Another example -- [Pt(NH3)4Cl2]Br and [Pt(NH3)4Br2]Cl

Hydrate isomerism:- This is similar to the above kind of isomerism. Water can be present as water of crystallisation or water can act as a ligand. Example:- [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl ]Cl2.H2OThe two complexes will ionise differently..the first will give 3 chloride ions and the second only 2 chloride ions. They also differ in their colour[CrCl2(H2O)4]Cl.2H2O bright-green[CrCl(H2O)5]Cl2.H2O grey-green[Cr(H2O)6]Cl3 violet

Linkage isomerism:-Some ligands have more than one donor atoms and can bind to the metal through either of them. Ligands like CNS- and NO2

- ions can bind to the metal either one of the donor atoms like N or S in CNS- and N or O in NO2-. Such ligands

20

are called ambidentate ligands. Complexes with ambidentate ligands can be formed with either one of the atoms as the donor atom. Hence two isomers are possible.Example:- [Co(NO2)(NH3)5]Cl2 and [Co(ONO)(NH3)5]Cl2

The first complex has the N atom linked to the Co atom while in the second complex, the O atom is linked to the Co atom.

Coordination isomerism:- This type of isomerism is seen in compounds where the cation and the anion are both complex. Example :- [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6] In the first compound the NH3 is in the coordination sphere of the cobalt ion and the cyanide ion in the coordination sphere of the chromium ion. While in the second, the coordination is interchanged.

Coordination Position isomerism:- In bridged complexes, where there are two metal ions, the ligand can be linked to the two different metal atoms.

OH

(NH3)4Co Co(NH3)2Cl2

OH

and

OH

Cl(NH3)3Co Co(NH3)3Cl OH

Are coordination position isomers.

Polymerisation isomerism:- When two complexes have the same empirical formula, but different molecular formula, they are polymerisation isomersExample:-

[Pt(NH3)2Cl2] and [Pt(NH3)4] [PtCl4] are polymerisation isomers.

Stereoisomerism in coordination compounds:- Some complexes have the same formula, but the difference is in the relative arrangement of their atoms. Such isomers are stereo isomers. Stereoisomers can be (i) Geometrical isomers or (ii) Optical isomers

Geometrical isomerism:-

FourCoordinate Complexes:These can be tetrahedral or square planar. There can be no cis –trans isomeism in a tetrahedral complex.A square planar complex can exhibit geometrical isomerism if it has different ligands. M a2b2 or Ma2bcExample Ma2b2 -- [Pt(NH3)2Cl2]

NH3 NH3 NH3 ClPt Pt

Cl Cl Cl NH3

Cis trans

21

In the case of a complex of the type Ma2bc

NH3 NH3 NH3 BrPt Pt

Cl Br Cl NH3

Cis trans

Mabcd also has geometric isomers (draw the different possibilities) eg [PtBrCl (NH3)(pyr)]

In complexes with bidentate ligands like glycinate ion, where the donor atoms differ, we can have cis trans isomerism.

NH2 NH2 NH2 O CCH2 Pt CH2 CH2 Pt CH2

C O O C C O NH2

Cis trans

Geometrical isomerism in 6 coordinate complexesAll six coordinate complexes are octahedral in shape. Complexes of the types Ma6 or Ma5b can have no isomers. Complexes of the types Ma4b2 Ma4bc Ma3b3 can all have geometrical isomers. Complexes with bidentate ligands can also have geometrical isomers.

Ma4b2 and Ma4bc NH3 NH3

NH3 NH3 NH3 ClPt Pt

Cl Cl Cl NH3

NH3 NH3

Cis (i) trans (ii)

NH3 NH3

NH3 NH3 NH3 BrPt Pt

Cl Br Cl NH3

NH3 NH3

Cis trans

NH3 NH3

NH3 NH3 Cl NH3

Pt Pt

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Cl NH3 NH3 Cl

Cl NH3

Cis transThese two structures are tha same as (i) and (ii). Draw different variations and see for yourself by just rootating the paper.

Geometrical Isomers in Ma3b3:-In the compound [Co(NH3)3Cl3], we can have 2 geometrical isomers. The first with all the Cl atoms forming a face of the octahedron (all the NH3 also form another face of the octahedron). In the second case, all the Cl form a meridinial plane of the octahedron (all NH3 also form another meridinial plane) The first isomer is called a Facial or Fac isomer and the second a meridinial or Mer isomer.

Cl Cl

Cl NH3 NH3 ClCo Co

Cl NH3 Cl NH3

NH3 NH3

Facial Meridinial

It can be seen in the Mer isomer, 2 Cl atoms are cis to each other and 2 are trans to each other. Hence the cis trans terminology is not usedin this case.

Geometrical isomerism in 6 coordinated complexes with bidentate ligands:-Complexes of the type M(aa)2 b2 exhibit cis trans isomerism. Example [Co(en)2Cl2]

Note ;- (aa) indicates a bidentate ligand with 2 domor atoms of the same kind, while (ab)denotes a bidentate ligand with 2 different donor atoms.

Cl Cl

NH2 Cl NH2 NH2

Co CoNH2 NH2 NH2 NH2

NH2 Cl Cis trans

Optical Isomerism:-4 coordinate complexes:- 4 coordinate complexes can be tetrahedral or square planar. For a compound to show optical isomerism it should not have a plane of symmetry. Square planar complexes are symmetric about the plane of the molecule. Hence square planar complexes do not show optical isomerism.

a b M

c d

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However, exceptions can be seen if the ligand has substituents that destroy the plane of symmetry.

Bis(isobutylene diamine meso stilbene diamine) Platinum(II) shows optical isomerism since the substituents –methyl and phenyl groups are at an angle and destroy the plane of symmetry.

φ

C NH NH C Pt CH3

C NH NH C

φ CH3

A tetrahedral complex can show m optical isomerism only if all the ligands attached to the metal ion are different ie Mabcd. It is very difficult to synthesise such a complex. Hence there are no real examples of such isomerism

d

M a c

b

An example of optical isomerism in a tetrahedral complex is seen is bis(benzoyl acetonato) Beryllium (II). This is optically active because the substituents Ph and Me groups on the ligand cause the loss of symmetry . Hence on losing the plane of symmetry, it is optically active.

Eg bis(benzoylacetonato) Beryllium(II)

CH3 O CH3

C O CBe CH2

H2C O C

C O C6H5

C6H5

6 coordinate complexes:-Octahedral complexes of the type Ma6, Ma5b, Ma4b2 and Ma3B3 all have a plane of symmetry. (Draw them) Hence they do not exhibit optical isomerism. However, complexes with bidentate ligand ie M(aa)2b2 (only the cis isomer) and M(aa)3 show optical isomerism.

[Co(en)3]Cl3

NH2 NH2

NH2 NH2 NH2 NH2

Co CoNH2 NH2 NH2 NH2

24

NH2 NH2

The cis isomer of [Co(en)2Cl 2] Cl

NH2 NH2

NH2 NH2 NH2 NH2

Co CoNH2 Cl Cl NH2

Cl Cl The trans isomer does not show optical isomerism since it has a plane of symmetry. (draw it and see)

Complexes of the kind Mabcdef will show both geometrical and optical isomerism. The compound [Pt(py)(NO2) NH3ClBrI] has 15 geometrical isomers and each of these has one optical isomer—totally 30 isomers.Same for the compound [Co Cl Br I NH3 NO2 SCN]2- (not yet characterised)Draw the different isomeric structures of the following mentioning the type of isomerism exhibited.(i) K3[Fe(ox)3] (ii) [Pt(NH2CH2COO)2] (iii) [ Cr(en)3]Cl3,

Spectral and magnetic properties of metal complexes: Electronic absorption spectrum of [Ti(H2O)6]3+

ion. Types of magnetic behavior, spin-only formula, calculation of magnetic moments, experimental determination of magnetic susceptibility – Gouy method.

Spectral properties of metal complexes:

Spectral and magnetic properties of metal complexes: Electronic absorption spectrum of [Ti(H2O)6]3+ ion. Types of magnetic behavior, spin-only formula, calculation of magnetic moments, experimental determination of magnetic susceptibility – Gouy method.

Theories of bonding can explain some of the properties of coordination compounds.

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Many coordination compounds show absorption of radiation in the visible region. The valence bond theory is unable to explain this phenomenonHowever, the CFT does explain it qualitatively.In the CFT model, in an octahedral complexfor example, the d orbitals are split into two setsthe t2g and the eg sets of orbitals.The difference in the energies of these two setsis known as ΔO which is 10 Dq.The lower t2g orbitals have electrons and if there isa vacant energy level in the eg set, the electroncan absorb energy and jump from the t2g

to the eg levels. This absorption of energy comesfrom the incident light. The energy required topromote an electron from the t2g to the eg levels is ΔO. Hence the complex absorbs radiation.It therefore shows an absorption spectrum.This means complexes are coloredsince they absorb radiation in the visible region.of the electromagnetic spectrum. The complexes are colored due to a d-d transition of the electrons. . However, according to Laporte selection rule, transitions between orbitals of the same subsidiary quantum numbers are forbidden. Hence d-d transitions are weak…ie the compounds are not strongly colored., they are pale colored….eg MnSO4 , FeSO4 etc.Another selection rule is the spin selection rule- the transition can occur only if the spin multiplicity is retained—ie a transition from a singlet state to a triplet state or vice versa are not allowed. Charge transfer spectra:-In addition to d-d transitions, complexes may show charge transfer absorptions. These maybe metal ligand charge transfer (MLCT) or ligand metal charge transfer(LMCT).In the first case, the electron in the d orbital of the metal gets excited to the antibonding MO of the ligand. Since these are nor d-d transitions, they are allowed transitions and hence these complexes are strongly coloured. In the case of LMCT, the ligand bonding electrons are excited to the metal eg orbital. Again this too shows absorption of high intensity. eg Ce(SO4)2, KMnO4 etc.

Electronic absorption spectrum of [Ti(H2O)6] 3+ ion: The complex [Ti(H2O)6]3+ has Ti in the +3 oxidation state. Ti 3d24s2 Ti3+ is 3d1

It has 1 electron in the 3d orbital. When it forms a complex, its electronic configuration is t2g1 eg

0

When exposed to light, the complex absorbs light of the wavelength λ whose energy is equal to ΔO

— — eg

ΔO ie, ΔO = E = hc/λ = hc ν where ν is 1/ λ — — — — —

— — — t2g

The absorption of energy promotes the electron to the eg level. That is, a d-d transition takes place. Hence the compound shows an absorption spectrum

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White light is a mixture of many wave lengths…traditionally the seven colours. If any of its component wavelengths are missing, it would no longer be white. It would be coloured. A band of a particular wave length range, has a particular colour for example, 450-475 nm is blue light while 620-750 nm range is red. If white light somehow missed the band 620-750, it would lose the red component and appear a color complementary to red…cyanMany coordination compounds are coloured in nature. This is because when white light falls on the compound, it absorbs some wavw lengths and reflects the remaining wave lengths. Thus the reflected light is different from the incident light. The reflected light has a few wave lengths missing. Hence the reflected light has color and the compound looks colored.

(From the higher level, the electron just loses energy by collision with other molecules, or through vibrations and gets back to the t2g level. This process is called relaxation. It can again get excited.)

The absorption spectrum of a complex can be measured. Most complexes show maximum absorption in a small band of wavelength. This is the absorption maxima λmax. The value of λmax depends on the crystal field splitting. Greater the value of Δo, higher the energy absorbed and shorter the λmax . Spectra are measured in terms of wavelength in angstrom units, or more commonly, in terms of wave number ν where ν is 1/ λ. It is reported in cm-1

In the case of the hexaquatitanium (III) complex, the single electron in the t2g orbital is promoted to the e g

Orbital by the absorption of energy ofE = h x c x 20,300

(1 cm-1 = 0.0028591 kcal/mol)

Other complexes of Ti3+ show a different absorption maximum. It depends on the strength of the ligand.[Ti(CN)6]3- has a maxima at about 22000 cm-1 while the weaker ligand Cl- gives a complex [TiCl6]3- which shows a maximum at about 13000 cm-1

An illustrative diagram of the absorption spectra of the complexes of a metal with different ligands.

absorbance

27

Wave number

This is the origin of the spectrochemical series. Ligands are arranged in order of decreasing energy or wave number at which they absorb radiation. (see Coordination Chemistry)

Factors affecting λmax The extent to which the d orbitals are split, depends both on the iii) the nature of the ligandsiv) and the nature of the metalas well as on the geometry of the complex.

Nature of the Ligands :- The ligands play a major role in determining the extent to which the d orbitals are split due to complex formation. A strong ligand like CN- ion or NH3 molecule can split the d orbitals to a greater extent… ie, the value of Δo will be large for a complex with such ligands. The ligands can be arranged in order of decreasing ability to split the d orbitals of the metal. Such a series is known as the spectrochemical series (reason for the name—see the spectroscopic properties of complexes**)

CO ≈ CN- > NO2- > en > NH3 > H2O > F- > Cl- > Br- > I-

The above series shows that ligands like CO cause a greater split in the d orbitals of the metal atom. The value of Δo will be higher in such complexes. The value of ΔO will be small in complexes formed with iodide or bromide ions as ligands. Hence the ligands like CN- , en, ammonia are called strong field ligands or strong ligands , while the ligands like halide ions are weak field or weak ligands.For example, the complex [Co(H2O)6] 3+ has a Δo value of 18,600 cm-1 while the complex [Co(NH3)6] 3+ has a ΔO value of 23,000 cm-1 . (The values are determined from the UV-visible absorption spectrum of the complex and are hence reported in cm-1)

Nature of the metal:- The ability to form a complex depends on the size and the charge on a metal ion. Hence the crystal field splitting depnds on the oxidation state of the metal ion. It also depends on whether it is a first row d-block element or a second/ third row d-block element.Oxidation state:- Higher the oxidation state of the metal ion, greater the extent of splitting of the d orbitals. In the case of cobalt, which forms complexes in +2 and in +3 oxidation states, the complexes show remarkable difference in their absorption spectrum.

[Co(H2O)6]3+ ΔO = 18,600 cm-1

[Co(H2O)6]2+ ΔO = 9,400 cm-1

Size:- The elements in the first row of the d block show smaller splitting compared to the corresponding element in the second and the third row. The heavier elements have d orbitals that are extended farther in space. Hence they interact more with the incoming ligands and are therefore split to a greater extent. This can be seen by the value of ΔO for the complexes formed by Co, Rh, Ir all metals in the same group,having the configuration nd6 and in the same oxidation state.

[Co(NH3)6]3+ ΔO = 23,000 cm-1 [Rh(NH3)6]3+ ΔO = 34,000 cm-1 [Ir(NH3)6]3+ ΔO = 41,000 cm-1

Shape of the complex:-

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As mentioned before, the extent of crystal field splitting for a tetrahedral complex is much smaller than for an octahedral complex. ΔO > ΔT This is because only 4 ligands are interacting with the metal ion and none of these approach the metal orbitals directly. Hence the extent of splitting depends on the shape of the complex.

Magnetic properties of metal complexes

An atom has electrons that are spinning. A charged particle that spins,generates a magnetic field. Similarly, a charged particle that rotates, also generates a magnetic field. Hence an electron has magnetic moment due to its spinning motion as well as due to its rotation. An atom therefore shows magnetic moment if it has electrons. However, since there are many electrons, the rsulting magnetic moment due to all the electrons in the atom may lead to many kinds of magnetic properties.

Types of magnetic behavior :-Magnetic properties: A substance can be paramagnetic diamagnetic, ferromagnetic, antiferromagnetic or ferrimagnetic. When placed in a magnetic field, a paramagnetic substance is attracted to the field, whereas a diamagnetic substance is repelled. Paramagnetic substances have unpaired electrons in them while substances having only paired electrons are diamagnetic. Ferromagnetic, ferrimagnetic and antiferromagnetc substances also have unpaired electrons(paramagnetic) but in addition the unpaired electrons have certain characteristic arrangements.

1) paramagnetic substances have unpaired electrons . On applying an external field, the spins of all these electrons

are aligned in one direction. They will therefore have a net magnetic moment.

no field in amagnetic field

2) Ferromagnetic substances also have all the magnetic moments aligned exactly even in the absence of an external

field. These also have unpaired electrons, but the net moment is very high . The alignment persists even after the external field is removed. They can be magnetised.

3) In some cases, the magnetic moments of half the atoms are aligned in one direction and those of other half in exactly the opposite direction. The net magnetic moment is zero. These are antiferromagnetic substances.

4) In ferrimagnetic substances, the magnetic moments of some of the atoms are in one direction and the others are opposing but not of equal magnitude. Therefore, there is a resulting net magnetic moment. (eg MnO)

Paramagnetism can be observed only in an external field.29

The magnetic moments can give a lot of information regarding the nature of bonding in metal complexes. Paramagnetic high spin complexes are more ionic in character. The magnetic moment can also give some information about the structure of the complex

Coordination compounds exhibit magnetic propeties. The compounds contain d or f block metals. Take the case of compounds of d block metals—the metal has electrons in the d orbitals. Some of these are unpaired. Hence, compounds of d block elements are paramagnetic in nature. The upe are randomly oriented. In an external magnetic field, the field inside the sample will be > or < the applied field. H = B ─ H0 where B is field inside the substancce. H0 is the free field and difference H = 4 I where I is the magnetic moment / unit volume.

Theory of magnetismThe classical theory of magnetism was well developed before quantum mechanics. Lenz's Law (~1834), states that: when a substance is placed within a magnetic field, H, the field within the substance, B, differs from H by the induced field, 4πI, which is proportional to the intensity of magnetization, I. That is; B = H + 4πI where B is the magnetic field within the substance H is the applied magnetic field and I is the intensity of magnetisation This can also be written as B/H = 1 + 4π I/H,   or    B/H = 1 + 4πκ where B/H is called the magnetic permeability of the material and κ is the magnetic susceptibility per unit volume, (I/H) By definition, κ in a vacuum is zero, so under those conditions the equation would reduce to B=H. It is usually more convenient to measure mass than volume and the mass susceptibility, χg, is related to the volume susceptibility, κ, through the density. χg = κ /ρ where ρ is the density. Finally to get our measured quantity on a basis that can be related to atomic properties, we convert to molar susceptibility χm =χg * RMM Since this value includes the underlying diamagnetism of paired electrons, it is necessary to correct for the diamagnetic portion of χm to get a corrected paramagnetic susceptibility. χ'

m = χm + χdia There are numerous methods for measuring magnetic susceptibilites, including, the Gouy, Evans and Faraday methods. These all depend on measuring the force exerted upon a sample when it is placed in a magnetic field. The more paramagnetic the sample, the more strongly it will be drawn toward the more intense part of the field.

Curie LawNormal paramagnetic substances obey the Curie Law χ = C/Twhere C is the Curie constant. Thus a plot of 1/ χ versus T should give a straight line of slope 1/C passing through the origin (0K).Whereas many substances do give a straight line it often intercepts just a little above 0K and these are said to obey the Curie-Weiss Law: χ =C/(T+Φ) where Φ is known as the Weiss constant.

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Spin-only formula, calculation of magnetic moments :-

An electron is a charged particle that is spinning as well as moving along its orbit. Both these movements lead to magnetic moments. Hence an electron has magnetic moment due to its spin an also due to its orbital motion

In the elements of the d-block , the differentiating electron enters the penultimate d orbitals. There are 5 d orbitals Hence these have unpaired electrons and are therefore paramagnetic . The magnetic moments in these elements are usually very close to the spin-only magnetic moment.

μ = 4S(S+1)

or = n(n+2). B is the magnetic moment and is expressed in Bohr Magneton. B where

1B = e h /4 m

To calculate the magnetic moment of complex ions of the 1st row d block elements, we only need to kow the number of upe

The orbital contribution to the magnetic moment is negligible because the d-orbitals are spread out and exposed to the environment in which the metal ion is present. The various ions present in the environment or the crystal field can quench the orbital contribution to the magnetic moment. Hence the d-block elements exhibit spin only magnetic moments. Quenching Phenomenon in which a very strong electric field, such as a crystal field, causes the orbit of an electron in an atom to precess rapidly so that the average magnetic moment associated with its orbital angular momentum is reduced to zero is known as quenching.In complexes of the f block elements, quenching of the orbital contribution does not occur and the nett magnetic moment are a combination of both spin and orbital contributions.

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Complex M(+n) Elec. config No. of upe

μ calc μobs

K3[Fe(CN)6 Fe (+3) 3d5 (low spin) 1 1.73BM

2.25BM

K4[Fe(CN)6 Fe(+2) 3d6 (low spin) 0 0 0

[Co(NH3)6]Cl3 Co(+3) 3d6 (low spin) 0 0 0

Na3[CoF6] Co(+3) 3d6 (high spin) 4 4.90BM

5.30BM

Na3[FeF6] Fe(+3) 3d5(high spin) 5 5.92BM

5.85BM

K3[TiF6] Ti(+3) 3d1 1 1.73BM

1.70BM

μ = 4S(S+1) + L(L+1)

Experimental determination of magnetic susceptibility – Gouy method.

Magnetic moments are measured using a Guoy balance. This is a balance with a magnet below. One of the pans of the balance is substituted by a hook from which a sample tube can be hung. The sample tube is weighed and weighed again in the magnetic field.Then filled with water up to the mark and weighed. This is used to determine the volume of the tube.

The sample is then filled in the sample tube and weighed. The electromagnet is switched on. The sample is then weighed in the field. Similarly the standard Hg[Co(CNS)4] is also weighed in the field and without the field. The molar susceptibility is given by χ x molecular mass

Fom the magnetic susceptibility, we can find the magnetic moment

μeff = 2.828 (χ T)1/2

Sample

electromagnet

HgCo(CNS)4 has a standard = 16.44 × 10-6 cgs units per gram at 20C,If the sample is paramagnetic, the second weight is greater than the first since the sample is attracted to the field. The increase in weight is compared to that of a standard ( Hg[Co(CNS)4] ) W1 Weight of empty tube, field off __________gW2 Weight of empty tube, field on __________gW3 Weight of tube filled, field off __________gW4 Weight of tube filled, field on __________g

32

W5 Weight of tube filled with water (once only) __________gFor each determination, repeat the measurements (as you proceed through the table) untilW2-W1 and W4-W3 give constant values (0.05 mg.) are obtained.The weights recorded are used as follows:V = (W5-W1) where d is the density of water in g mL-1 at Td= W2-W1

= W4-W3

m = W3-W1

In the equation:(m) - (0.029 10-6) V = (- )The magnetic susceptibility χ for the sample can be calculated.

Reactivity of metal complexes Labile and inert complexes, ligand substitution reactions – SN1 and SN2, substitution reactions of square planar complexes – Trans effect and applications of trans effect.

Labile and inert complexes:-Thermodynamic and kinetic factors are both important in the dissociation / formation of a complex.The tendency of a complex to dissociate, is determined by the free energy change for the dissociation reaction. The reaction

ML4 M + 4 L has a certain value of ΔGIf thius value is negative, the complex has a tendency to dissociate and is unstable.If the reaction

M + 4 L ML4 has a negative ΔG, then the complex is stable

In aqueous media, this may be written as a ligand exchange

ML4 + 4 H2O M(H2O)4 + 4 L ΔG -ve

However, even if the complex has a tendency to dissociate, ie even if the ΔG is negative for the dissociation/ligand exchange reaction, it needs to have a mechanistic path to undergo the reaction. In the absence of this path, the complex will remain undissociated…ie it is inert.

When a complex is in solution, it may undergo ligand exchange. ML4 + H2O M L3(H2O) + L

The ligand may be replaced by the water molecules. When this exchange is favourable, the complex is said to be labile. ie a labile complex is one which undergoes ligand exchange easily.

That is, if the half life of the complex is 1 minute or less, at a concentration of 0.1 M, the complex is labile.

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An inert complex is one that does not undergo ligand dissociation or ligand substitution easily…ie If the half life of the complex is greater than 1 min at 0.1 M concentration it is an inert complex.(This definition was given by Taube.)

The complex [Co(NH3)6]3- is thermodynamically unstable to acid hydrolysis..ie in acid medium, the NH3 ligands are substituted by H2O. The complex is inertThe complex [Ni(CN)4]2- is thermodynamically stable, but undergoes ligand exchange very quickly. The complex is labile.

This means that the The kinetics of the ligand substitution reaction in various complexes cannot necessarily be predicted by their thermodynamic stability. Secondly, the nature of the reaction cannot always be decided from the stoichiometric equation.

The rate at which a complex undergoes reaction depends on various factors- the nature of the metal ion, ligand, stereochemistry etc. The study of reaction mechanism in complexes is more difficult than in organic compounds since the reactions are usually much faster. But still, some generalised mechanistic pathways can be proposed for transition metal complexes since they are similar in nature. It has already been pointed out that a complex can be thermodynamically unstable, but kinetically inert or it can be fairly stable but labile. That means that even if the free energy change for the dissociation of the complex is favourable (unstable) , it may still not dissociate if the kinetic pathway is not available (inert). A complex is labile if it undergoes reactions with a half life of about one minute or less. The kinetics of different reactions are studied by different methods depending on whether they are fast or slow. Some methods used areDirect analysis like titrations etc, electrometric methods, spectrophotometric methods are used for slower reactions while for fast reactions, flow methods, stopped flow methods, relaxation techniques, NMR are used.First the reaction stoichiometry is determined. Then the reaction intermediates are characterized by different means… by isolating them if they are fairly stable, making them undergo a different reaction by adding other substrates, studying the isomers obtained if any. The reaction mechanism is then put together from many such evidence. In general, mechanisms of inorganic reactions are more difficult to determine than those of organic reactions since the inorganic reactions are usually fast.

The reactions that complexes undergo are

1) substitution eg [Cu(H2O)4]2+ + 4 NH3 [Cu(NH3)4]2+ + 4 H2O

2) elimination 3) electron transfer4) polymerization

Ligand substitution reactions:-

Substitution reactions can occur in different ways—1a Nucleophilic 1b Electrophilic substitution1c Substitution by oxidative addition followed by reductive elimination.

Nucleophilic substitution:-

From the study of the rate law for some reactions, it is seen that substitution reactions likeML4X + Y ML4Y occur in one of the four mechanisms—

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Substitution of X by Y in MXL4 can occur either by the dissociation of the M- X bond first and then the formation of M-Y bond. It can also occur by the formation of M---Y weak bond and then the breaking of M-X bond.The mechanism can even be somewhere in between these two extreme options .These are known as the Dissociative ligand substitution or SN1 mechanism, Associative ligand substitution or SN2 mechanism, Between these two extremes are the Interchange-associative Ia and Interchange-dissociative Id mechanisms

Of these, the two important ones are- associative or SN2 and dissociative or SN1

Associative or SN 2 mechanism :-In these reactions, a definite reaction intermediate is formed. The rate determining step is the attack of the entering ligand E on the complex ML3D where D is the departing ligand.The rate of the reaction is therefore dependent on both the concentration of the complex as well as on the concentration of the entering ligand

Rate = k [ML3D] [E]

In this mechanism, the M-D bond is not yet fully broken , when the E group enters. The transition state obtained in this mechanism is of a higher coordination number than the original complex.

L L LX M L + Y X------ M------Y L M Y + X

L L L LThe transition state is

trigonal bipyramidal in shape

The energy profile diagram shows the transition state E---ML3----D which is relatively stable.

E E-ML3 -D

ML3D + E

35

ML3E + D

Reaction coordinates

An example Trans-[PtCl2(py)2] + L trans-[PtClL(py)2] + Cl-

[Pt Cl(dien)]+ + I- [Pt I(dien)]+ + Cl- dien=diethylene triamine

Ligand exchange is observed in cyanide complexes using radioactive *CN-

[Ni(CN)4]2- + *CN- [Ni *CN (CN)3]2- + CN-

Asociative mechanism is common in 4 coordinate complexes rather than in complexes with higher coordination numbers. This is because the intermediate has a coordiantion number higher than the original complex and is sterically not favourable in such cases—for example, a 6 coordiante complex would have a transition state that has 7 ligands attached to the metal. The rate of the reaction in these cases depends on the nucleophilicity of the attacking ligand. Greater the nucleophilicity of the attacking ligand E, faster the reaction.

CN- ≈ CO > I- > py > Cl- > OH- etc.In the reactionTrans-[PtCl2(py)2] + L trans-[PtClL(py)2] + Cl-

The rate varies from 4 x 10-4 to 6 for different ligands and can therefore be said to follow an associative or SN2 mechanism. The kinetics of ligand substitution reactions do not always follow what is expected from their thermodynamic stability. The mechanistic pathway determines the kinetics irrespective of the thermodynamic factors.The stoichiometry of the reaction is also not always an indicator of the mechanism.

Dissociative Mechanism SN 1 :-

In complexes that follow this mechanism, the departing ligand D dissociates, leaving the metal with a coordination number less than before. The entering ligand then attacks the metal and forms the new complex.

For example, in a 6 coordinated complex, ML5D ML5

ML5 + E ML5EOr

ML5D + E ML5E + D

The transition state ML5 is 5 coordinated.The 5 coordinated intermediate can be square pyramid (as is usual) or even trigonal bipyramidal (only if π bonding stabilises this)

L L L

L L L L L L M M + D M

L L L L L L

D L E Not favoured

L36

M L L

L

However, since the tbp intermediate requires a lot of rearrangement of M-L bonds, it is not usually formed. The reaction ususally goes through a square pyramidal intermediate. The tbp intermediate is formed only if such a structure is favourable for π bond formation.In this mechanism, the rate is given by the equation

Rate = k[ML5D] That is, the rate depends only on the concentration of the complex and is independent of the concentration or nature of the entering group E.

Complexes of higher coordination numbers usually follow SN1 or dissociative mechanism.

The energy profile is as follows

E E + ML3 + D

ML3D + E

ML3E + D

Reaction coordinates

Some reactions follow a path that is neither SN1 nor SN2. These are said to follow the Interchange mechanism I

Trans effect:- This phenomenon is seen in square palnar complexes.In complexes having different ligands attached to the same metal, and if a ligand undergoes substitution, the new ligand will have a preferred position.This is explained by an example.

To prepare [Pt(NH3)2Cl2], we can either start from [Pt(NH3)4]2+ or from [PtCl4]2-.

Cl Cl Cl NH3 + NH3 Cl NH3

Pt + NH3 Pt Pt ie cis diammine dichloroPlatinum(II)Cl Cl Cl Cl Cl NH3

NH3 NH3 Cl NH3 + Cl- Cl NH3

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Pt + Cl- Pt Pt ie trans diammine dichloroPlatinum(II) NH3 NH3 NH3 NH3 NH3 Cl

That is, the new incoming ligand will always bond trans to a Cl- ion and not trans to ammonia.This phenomenon where a ligand already attached to a metal atom directs further substituents to a position trans to itself is called the trans effect.In the above examples, the Cl- group has a trans effect or is trans directing.

A ligand that binds to a metal and is electronegative, has a tendency to withdraw electron density away from the metal, towards itself. This makes the metal vulnerable to a nucleophilic attack from the opposite side since there, the electron density is decreased. Hence the next substituent will be directed to a position trans to this ligand.

Strongly π acid ligands are also strongly trans directing since they withdraw electron density from the metal due to their π acid nature.

Stability of metal complexes:

Stability of metal complexes: Thermodynamic stability and kinetic stability, factors affecting the stability of metal complexes, chelate effect, determination of composition of complex by Job’s method and mole ratio method.

As discussed earlier, a complex maybe stable or unstable depending on the free energy change for the formation reaction. The tendency of a complex to dissociate, is determined by the free energy change for the dissociation reaction. The reaction

ML4 M + 4 L has a certain value of ΔGIf this value is negative, the complex has a tendency to dissociate and is unstable.If the reaction

M + 4 L ML4 has a negative ΔG, then the complex is stable

In aqueous media, this may be written as a ligand exchange

M(H2O)4 + 4 L ML4 + 4 H2O ΔG -ve

A stable complex may be inert. It may also be labile. Similarly an unstable complex may be labile or inert.The kinetc inertness and lability depend upon the mechanistic pathways present for the complex to undergo dissociation. For example,

[Ni(CN)4]2- is a stable complex, but it undergoes ligand substitution very fast ie. it is a stable, but labile complex.On the other hand, [Co(NH3)6]3+ is an unstable complex, but stays unaltered for a long time. It is unstable, but inert. Consider the formation of a 4 coordinate complex

M(H2O)n + 4 L ML4 + n H2O

Ignoring the water, we can represent it as

M + 4 L ML4

This has an equilibrium constant β

β = [ML4] [M] [L]4 **

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The free energy for the formation of the complex is given by

Δ G = − RT ln β

Hence the formation constant of a complex is a thermodynamic property of the complex.

Formation constants:-Let us consider the formation of a complex ML4

M + 4 L ML4 which has a n equilibrium constant β

β = [ ML 4]

[M] [L]4

Suppose the complex forms in stages or steps

M + L ML this has an equilibrium constant k1

ML + L ML2 equilibrium constant k2

ML2 + L ML3 equilibrium constant k3

ML3 + L ML4 equilibrium constant k4

k1 = [ML] so [ML] = k1 [M] [L] [M] [L]

k2 = [ML2] substituting for [ML] k2 = [ML2] [ML] [L] k1[M] [L]2

k3 = [ML3] substituting for [ML2] k3 = [ML3] [ML2] [L] k1k2 [M] [L]3

k4 = [ML4] substituting for [ML3] k4 = [ML4]

[ML3] [L] k1k2 k3 [M] [L]4

Therefore k1k2k3k4 = [ML4] [M] [L]4

But β = [ML4] [M] [L]4

Therefore

k1k2k3k4 = β

The overall formation constant is the product of the stepwise formation constants.

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Factors affecting stability:-The stability of a complex depends both on the ligand and on the metal atom. I addition, certain complexes are affected by the environment.

i) Environmental effects---(a) A complex with a volatile ligand may decompose easily at high temperatures and low pressures.(b) The pH also affects stability of some complexes.

ii) Nature of the Ligand—The nature of the ligand can affect the stability of a complex. Strong ligands form strong complexes. Hence the value of the Δo and the CFSE determines the stability of a complex.

However, the soft acids ie metals in the third row d block or the heavier metals in their lower oxidation states, show preference for the softer bases

iii) Nature of the Metal—The nature of the metal also affects the stability of the complex. In general, greater the positive charge on the metal ion, and smaller the size of the ion, stronger the complex formed. That means the first row d block elements in their higher oxidation states from stronger complexes.

Oxidation state--K3[Fe(CN)6] has a stability constant of the order 1031 while K4[Fe(CN)6] has a stabilty constant of the order of 106 . Similarly, [Co(CN)6]3 – is much more stable than [Co(CN)6]4– Size—The first row d block elements form more stable complexes than the second row d block elements.The general stability sequence of high spin octahedral metal complexes for the replacement of water by other ligands is: ( Known as the Irving William Series)Mn(II) < Fe(II) < Co(II) < Ni(II) < Cu(II) > Zn(II) (though for soft bases, the trend will be the reverse)

iv) Chelation:-- The formation of a chelate introduces greater stability to the complex. This is partly due to entropy effects.

M(H2O)6 + 3 en M(en)3 + 6 H2O 4 molecules have given rise to 7 molecules. Hence the entropy is considerably raised by chelation.Chelation is also facilitated by the fact that once one of the donor atoms of a bidentate ligand is attached to the metal, the other being just adjacent to the metal, is attached much more easily than if it were a free ligand.However, only chelation leading to 5 or 6 membered ring formation is favourable.

H2N CH2

CH2

40

M NH2

v) Hard-hard and soft-soft interactions---Stability of those complexes formed by a hard metal with a hard base or by a soft metal with a soft base are more stable compared to those involving hard-soft interactions.

Temperature dependenceAll equilibrium constants vary with temperature according to the van 't Hoff equation [18]

d lnK = ΔH dT RT2

R is the gas constant and T is the thermodynamic temperature . Thus, for exothermic reactions, (the standard enthalpy change, ΔH ,

is negative) K decreases with temperature, but for endothermic reactions (ΔH is positive) K increases with temperature.ENTROPY Chelation

Equilibrium log β ΔG ΔH /kJ mol−1 −TΔS /kJ mol−1

Cd2+ + 4 MeNH2 Cd(MeNH2)42+ 6.55 -37.4 -57.3 19.9

The Irving-Williams series refers to high-spin, octahedral, divalent metal ion of the first transition series. It places the stabilities of complexes in the order

Mn < Fe < Co < Ni < Cu > Zn

This order was found to hold for a wide variety of ligands.[29] There are three strands to the explanation of the series.

1. The ionic radius is expected to decrease regularly for Mn2+ to Zn2+. This would be the normal periodic trend and would account for the general increase in stability.

2. The crystal field stabilisation energy (CFSE) increases from zero for manganese(II) to a maximum at nickel(II). This makes the complexes increasingly stable. CFSE returns to zero for zinc(II).

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Hard and soft acids bases (HSAB)

Classification, Pearson’s concept of hardness and softness, application of HSAB principles – Stability of compounds / complexes, predicting the feasibility of a reaction.

The formation of a complex may be regarded as an acid – base interaction. The ligand donates electron pairs and may hence be regarded as a Lewis base while the metal ion that receives electron pairs regarded as a Lewis acid. It was observed that for a given ligand, the stability of the complex with a dipositive metal ion is in the order called the Irving William series. Ba2+ < Sr2+ < Ca2+ < Mg2+ < Mn2+ < Fe2+ < Co2+ < Ni2+ < Cu2+ < Zn2+

This order is partly due to the decrease in size . It was also observed that some ligands form stable complexes with metal ions like Ag+ Hg2+ etc while others form stable complexes with Ti4+ Co3+ etc.On the basis of these observations, Chatt, Ahrland and Davies classified metals and ligands as class a or class b Pearson made the classification Hard acids and Soft acids to describe these observations.He suggested that metal ions that are not easily polarised are hard acids ie ions with small size and high nuclear charge – that is class a metal ions. since these metals have tightly held electrons. Larger ions that are more easily polarised are soft acids or class b metals. The same metal in different oxidation state behaves differently. Lower oxidation states are softer and higher oxidation states are hard acids.Ligands that combine more readily with hard acids are hard bases Ligands that combine more readily with soft acids are soft basesHighly electronegative ligands bases are class a bases eg ligands with oxygen donor atom, nitrogen donor atom and fluoride ion. Ligands that are capable of backbonding are class b ligands or soft bases. These have low lying vacant MO which can accept electrons donated back by the metal ion.

Hard acids (class a metals) Borderline acids Soft acids (class b metals)H+ , alkali metal ions , alkaline earth metal ions, Sc3+ La3+ Ce4+ and other lanthanide ions.Ti4+, Zr4+ Hf4+ VO2+ Cr3+ Cr6+ MoO3+ WO4+ Mn2+ Mn7+ Fe3+ Co3+

Some of the p block metals and halogen cations like Cl3+ or I3+ are also hard acids

Fe2+ Co2+ Ni2+ Cu2+ Zn2+

Rh3+ Ir3+ Ru3+ Os2+ Pd2+ Pt2+ Cu+ Ag+ Au+ Cd2+ Hg2

2+ Metals in their zero oxidation state

Hard bases Borderline bases Soft basesF- ion, O donors and N donors-NH3 , RNH2 N2H4 H2O OH─ O2─ ROH RO─

CH3COO─ CO32─ NO3

─ PO4 3─ SO4

2─

ClO4─

F ─

Br─ NO2─ N3

─ S donors, P donors *1

H─ R─ SCN- R3P R2S RS- S2O3

2-

1 * in the case of As or Sb, the bond length is too much for it to be strong.42

Pearson suggested a simple rule--Hard acids prefer to bind to hard bases and soft acids to soft bases.

This is simply a rule based on observations made on various complexes. It does not mean that complexes of soft acids with hard bases or hard acids with soft bases are not formed – only that they are relatively less stable.There is some theoretical basis for this however.

`In the case of small metal ions (hard acid-hard base interactions), ionic forces predominate and since the ions are small, and hence closer, there is greater attraction between them.

`In the case of the larger metal ions (soft-soft interactions), the covalent forces predominate. Here the polarising power of the base and the polarisability of the acid are the main considerations.

`Soft metal ions are larger in size and have lower charge on them. Hence there is greater tendency for back bonding in their complexes That is M→L bonding takes place in addition to the weaker L→ M σ bond

`Class a (hard acids) do not take part in backbonding since their electrons are tightly held.

`Hard acids have higher electronegativity. Eg ions of electropositive metals are highly electronegative eg Li+ Na+ and these are hard acids.while Cu+ or Pd2+ which has lower tendency to pull electrons are softer acids.

`To determine if an base B is hard or soft, the reaction is observed.BH+ + CH3Hg+ CH3HgB+ + H+ 

If the above equilibrium shifts to the right, the base is soft ( if Keq is large) while a shift to the left indicates that the base is hard (Keq is small). A hard base prefers H+ ion rather than bind to Hg. Similarly, for an acid,

A+ + X─ AX where X is a halide.

If A is a hard acid, the equilibrium is to the right if X = F─ and to the left if X = I─ the reverse indicates that A is a soft acid. For any metal or ligand, existing substituents affect its hardness or softness. Eg BF3 is a hard acid while BH3 is a soft acid. Electronegative substituents harden an acid since the substituents withdraw electron density away from the metal atom.

Applications:-`The best application of the HSAB rule is with an ambidentate ligand like CNS- The nature of the coordinating atom can be predicted. The ligand binds through S when it complexes soft metals like Pt(II), Pd(II) etc [Hg(SCN)4]2- while it binds through N in complexes with hard acids like Co(III) Cr(III) etc [Co(NCS)4]2- .

`From the HSAB theory, we can predict which kinds of interactions lead to more stable compounds.It can be predicted that the reaction

HgF2 + BeI2 BeF2 + HgI2 Soft-hard hard-soft hard-hard soft-soft is very likely to occur ie has a high feasibility since the products are more stable than the reactants. It is observed that this is so. The reaction is exothermic and has H = - 397 kJmol-1

`The stability of complexes can be predicted----hard-hard or soft-soft combinations give stable complexes. The synthesis of complexes can therefore be designed based on this. If a hard-soft complex is to be synthesised, a hghly stabilising counter ion would be needed to stabilise the complex. Complexes like [CoI6]3- are not stable while [CoF6]3- is stable. Sometimes the use of suitable ligands can stabilise a metal in an unusual oxidation state.

43

K4 [Ni(CN)4] has Ni in its zero oxidation state The low Os is stabilised by the soft ligand CN─ . Copper is not stable +1 oxidation state, but [Cu(CH3CN)]4NO3 can be prepared because the low oxidation state is stabilised by the CN- ion.

`The hydrolysis behaviour of complexes can be predicted. Complexes of hard acid –soft base and soft acid –hard base are likely to undergo hydrolysis easily.

` Soft metals like Hg have an affinity for soft bases like S donor ligands. While hard metals do not. This is used to selectively complex metals when they occur in a mixture along with other metals. This process is useful in analysis eg by solvent extraction.

`Metals occur in nature as minerals. The nature of the compound in these minerals depends on the softness/hardness of the metal.

`In reactions catalysed by metals, the choice of metal depends on their softness/hardness .

`The HSAB theory can also explain the greater solubility of certain substances eg AgCl is sparingly soluble but much more soluble in water than AgI . Ag+ is a soft acid and forms a stronger bond with I─ a soft base Therefore , the Ag-I bond is stronger than the Ag-Cl bond and ionises less easily.

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Bioinorganic chemistry: Essential elements, biological significance of Na, K, Mg, Ca, Fe, Co, Ni, Cu, Zn and chloride (Cl -). Metalloporphyrins – hemoglobin, structure and function, Chlorophyll, structure and role in photosynthesis.

Definitions--Simple salts - a compound formed by the reaction of an acid with a base containing two ions of opposite charge is a salt. In solution or I the molten state it ionises to give a cation (+ve ion) and an anion (-ve ion). Double salts—These salts are formed when two simple salts crystallise together to give crystals of one type containing both the salts. The compound on dissolution gives ions of both the salts. Eg Mohr’s salt FeSO4.(NH4)2SO4.6H2O consists of the 2 salts FeSO4 and (NH4)2SO4 which have crystallised together alon with 6 molecules of water of cystallisation. On ionisation, one mole of the salt gives one mole of Fe2+ ions , Three moles of SO4

2- ions and two moles of NH4+ ions.

Coordination compounds— it was observed that when a solution of CuSO4 was treated with excess of ammonia solution, Ligand—In a coordination compound, the central metal atom is linked to electron donating groups or ions. These groups or ions that bond to the metal by donating electron pairs, are known as ligands. Eg NH3, H2O, Cl- , NO2

- and many more. A ligand is a Lewis base.Ligands can be neutral molecules like ammonia, water, ethylene diamine or anions like halides, sulfate, nitrite, nitrate. Very rarely a positive ion can act as a ligand eg hydrazinium ion H2N NH3

+

Donor atom –

Chelate—

Coordination sphere—

so that the octahedral potential (truncated at ℓ = 4) gets the form:

45

where the real constants A and D may be used as adjustable parameters (it is tradition to use the

symbol D here).

Transformation of V to a basis of functions of T2g and Eg symmetry gives the matrix the following

diagonal form

Because this matrix is diagonal it means that the eigenvalues (the diagonal elements) and the

eigenvectors (the group theoretical functions) have been found. The first eigenvalue (energy of the

state of T2g symmetry) is 3-fold degenerate and the energy of the state of Eg is 2-fold degenerate.

The energy difference between the two states is (A + Y) −(A + X) = Y − X ≡ 10 Dq; the value q is

proportional to the radial part of the matrix elements and D is the potential parameter introduced

above. The factor 10 is introduced for convenience.

Dq:-Dq = 1/6 (ze2r4/a5)r = radius of d orbital of the central metal ion a = distance between ligand and central atomThe energy of interaction between dz

2 and the approaching ligand is given byE(φdz2) = ∫ ( φdz2)* Voct ( φdz2) dτ φdz2 is the wave function for the dz

2 orbital .

The solutions to this gives 3 factors--- a set of geometric factors D and q

Where D = 35 ze2 / 4 a 5

And q = 2< r4> r= distance of e from the nucleus.

Dq gives a measure of the splitting of the d orbitals ie 10Dq =Δo

E(φdz2) = + 6Dq

Similar calculations for the other d orbitals gives the termsE(φdx2-y2) = + 6DqE(φdxy) = − 4DqE(φdyz) = − 4DqE(φdxz) = − 4Dq

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