corrdination chemistry _gsr

Notes for NET & SET - Chemical Sciences

Chemistry of Transition Elements 1

CHEMISTRY OF TRANSITION ELEMENTS

CO-ORDINATION CHEMISTRY

INTRODUCTION :

The branch of inorganic chemistry that deals with the study of coordination compounds

is called coordination chemistry.

A coordination compound is a compound of a metal with a certain number of species

called ligands bound to the metal. An example is [Ni(CO)4].

A coordination compound is the product of a Lewis acid-base reaction in which

neutral molecules or anions (called ligands) bond to a central metal atom (or ion) by

coordinate covalent bonds.

Ligands are Lewis bases- they contain at least one pair of electrons to donate to a

metal atom/ion. Ligands are also called complexing agents.

Metal atoms/ions are Lewis acids - they can accept pairs of electrons from Lewis bases.

Within a ligand, the atom that is directly bonded to the metal atom/ion is called the

donor atom.

A coordinate covalent bond is a covalent bond in which one atom (i.e., the donor atom)

supplies both electrons. This type of bonding is different from a normal covalent bond

in which each atom supplies one electron.

If the coordination complex carries a net charge, the complex is called a complex ion.

Compounds that contain a coordination complex are called coordination compounds.

The coordination sphere of a coordination compound or complex consists of the central

metal atom/ion plus its attached ligands. The coordination sphere is usually enclosed in

brackets when written in a formula.

The coordination number is the number of donor atoms of ligands bonded to the central

metal atom/ion.

COORDINATION NUMBER (C.N.)

The maximum number of atoms, ions or molecules that are directly linked to central

metal atom in complex is called as the coordination number of the metal. It is different for

different metals. The geometry of the complex depends upon the co-ordination number of

metal.



Coordination Number Geometry

2 Linear

3 Triangular or Trigonal planar

4 Tetrahedral or square planar

5 Trigonal bipyramidal

6 Octahedral

Co-ordination number is the characteristic property of the metal. It takes the values

from 2 to 8, where 4 and 6 are the most common coordination numbers of coordination

compounds.

LIGANDS

Ligand is an ion or neutral molecule attached to the central metal ion in a coordination

compound. Within a ligand, the atom that is directly bonded to the metal atom/ion is called

donor atom. Each ligand has filled p orbital that bonds with the metal.

CLASSIFICATION OF LIGANDS

According to the number of bonds a ligand makes with a metal on distinguishes

monodentate ligands (e.g. ammonia NH3) and polydentate ligands. Polydentate ligands are

called

bidenatate, if they interact with a metal through two donor atoms,

tridentate, if they interact with a metal through three donor atoms,

quadridentate, if they interact with a metal through four donor atoms,

pentadentate, if they interact with a metal through five donor atoms,

hexadentate, if they interact with a metal through six donor atoms.

Ligands interacting with one metal through more than six donor atoms are rare.

The examples of all above types are given below :



Ligands can also be classified as shown below :

Classification of Ligands

Monodentate a Lewis base which can form only one name means one tooth

bond to a central metal atom.

Has more than one donor atom than can

form more than one coordinate covalent

bond to the same metal ion

Classified according to the number of

Donor atoms correctly positioned for

Potential binding to a central metal atom.

Chelating non-linear, often with 2 or 3

atoms separating the donor

atoms

bridging can donate more than one pair of

electrons to more than one metal atom

simultaneously

ambidentate has more than one element that can possesses bridging

serve as a donor atom capability but tends to

be monodentate; often

linear in geometry

macrocyclic large ring compound with several donor an example is 18crown6

atoms that can bind a central metal atom

inside the ring

pi-donor donates electrons from a pi bond to a

metal ion

Chelating, bridging and ambidentate ligands are described in detail.

A chelating ligand has several donor atoms arranged in such a way that they can

interact with one metal center. In the following example, the two nirogen atoms of 1,2-diaminoethane (= ethylenediamine, abbreviation en) are bound to the metal. Together with

the metal the ligand forms a five membered chelate ring.

NH2

NH2bidentate

NH2

NH

NH2

tridentate



NH2

NH2

LL

L

L

H2C

CH2

M

A bridging ligand acts as a bridge between two or more metal centers. In di

mhydroxo-bis (tetraaquairon (III), (it may also be called octoaqua-di mhyroxo-

diiron(III)), two hydroxyls bridge the two irons. Bridging ligands are preceded by m.

Important bridging ligands are: OH-, S2-, CO32-, PO4

3-, NH2-.

An ambidentate ligand has two donor atoms but their geometrical arrangement does

not allow them to bind to the same metal, i.e. they cannot form a chelate ring. These ligands

are responsible for linkage isomerism. Examples of ambidentate ligands are CN-, CO, SCN,

(CH3)2SO (dimetylsulfoxide = DMSO), HCON(CH3)2 (dimetylformamide= DMF)

HISTORICAL DEVELOPMENT IN COORDINATION CHEMISTRY

The exact date of preparation of the first co-ordination compound is not exactly

known. The discovery of hexaamminecobalt (III)chloride, CoCl3.6NH3, by Tassaert, in 1798,

is generally regarded as the beginning of co-ordination chemistry. The formation of this orange

coloured comound by Tassaert was quite intriguing in face of valency considerations. It was

difficult for chemists to understand as to why two stable molecules such as COCl3 and NH3could combine to form another stable molecule. Later on, many such compound were prepared

and their properties were studied.

COORDINATION COMPOUNDS OF TRANSITION METALS

Most of the metals form the co-ordination compounds. The two important conditions

that a metal should exhibit to form coordination compounds is

1) small size and

2) ability to exhibit variable oxidation states.

These two conditions are generally met with d block or transition metals. Hence, most

co-ordination compounds are formed by the transition metals.

The d block elements consist of three rows called first, second and third transition

series. The electronic configurations of these elements is shown below :



First Second Third

Sc 21 3d14s2 Y 39 4d15s2 La 57 5d16s2

Ti 22 3d24s2 Zr 40 4d25s2 Hf 72 5d26s2

V 23 3d34s2 Nb 41 4d45s1 Ta 73 5d36s2

Cr 24 3d54s1 Mo 42 4d55s1 W 74 5d46s2

Mn 25 3d54s2 Tc 43 4d55s2 Re 75 5d56s2

Fe 26 3d64s2 Ru 44 4d75s1 Os 76 5d66s2

Co 27 3d74s2 Rh 45 4d85s1 Ir 77 5d76s2

Ni 28 3d84s2 Pd 46 4d105s0 Pt 78 5d106s0

Cu 29 3d104s1 Ag 47 4d105s1 Au 79 5d106s1

Zn 30 3d104s2 Cd 48 4d105s2 Hg 80 5d106s2

NOMENCLATURE OF CO-ORDINATION COMPOUNDS

Introduction :

Thousands of coordination compounds are known. Nomenclature is important in

coordination chemistry because it gives us basic information about the structure of acoordination compound. IUPAC has recommanded certain rules for the nomenclature of

coordination compounds which are discussed below.Coordination Compound

A complex is formed by the interaction of metal atom and ligands. The ligand in acomplex is said to be coordinated to the metal atom or ion that is the center of the coordination

compound. Any neutral compound that contain a metal atom and its associated ligands iscalled a coordination compound. Such a compound may be formed between a complex ion

and other ions, for example [Ag(NH3)2]+Cl or K+2[Pt(NO2)4]

2 or the complex itself maybe neutral, for example, [Pt(NH3)2 (NO2)2]. The formula of the complex is usually enclosed

in square brackets.Rules for Nomenclature of Coordination Compounds

Rule 1In naming a coordination compound, the name of the cation is given first followed by

the name of the anion. This is illustrated by considering the names of the following compouds

K+ [Pt(NH3)Cl5] Potassium amminepentachloroplatinate (IV)

cation Anion



[Co(NH3)4 SO4] + NO3 Tetramminesulfatocobalt (III) nitrate.

Cation Anion

Rule II : Naming coordination SphereWhile naming a coordinatio sphere, ligands are named first and then the metal atom

along with its oxidation number in parentheses.

A) Naming LigandsNames of some common Negative Ligands

Symbol Name Charge Symbol Name ChargeBr bromo 1 SO4 sulphato 2

Cl chloro 1 H hydrido 1

I Iodo 1 NO2 nitrito 1CO3 carbonato 2 ONO nitrito-O 1

CN cyano 1 SCN thiocyanato 1OH hydroxo 1 NCS thiocyanato-N 1

C2O4 oxalato 2 SO3 Sulfito 2O oxo 2 S2O3 thiosulfato 2

N3 azido 1 N nitrido 1NO3 nitrato 1 C6H5 Phenyl 1

O2 Peroxo 2 NH2 amido 1NH imido 2

Names of Neutral LigandsSymbol Name Charge Symbol Name Charge

NH3 ammine 0 H2O aqua 0CO carbonyl 0 NO nitrosyl 0

(Ph3)P triphenylphosphine 0 C2H4 ethylene 0

CH3NH2 methyl amine 0 en ethylenediamine 0N2 dinitrogen 0 C6H6 benzene 0

Rule No. 1 Naming Ligands

Various ligands that are coordinated to the metal ion are listed in alphabetical order. Certain ligands such as SCN, NO2 contain two atoms that can coordinate with the metal

ion. In such cases the symbol of atom that is coordinated to the metal ion is mentionedafter the name of ligand separated by hyphen.



Rule No. 2 : Indicating Number of Ligands

If the same ligand is present more than once then the number of ligands is indicatedby prefixing words such as di, tri, tetra etc to the name of ligands.

[Co(NH3)6]+3 hexaammine cobalt (III)

K2 [Pt Cl6] Potassium hexachloroplatinate (IV)

[Pt(NH3)4Cl2]Cl2 tetraamminedichloroplatinum (IV) Chloride.K [Pt(NH3)Cl5] Potassium amminepentachloroplatinate (IV)

(Always remember that when ligands are listed alphabetically, spellings (alphabet) of

their proper names are taken into consideration and not the spellings of the prefixes of di, trietc. Thus diammine should be listed under a and not under d)

Use of bis, tris, tetrakis etc.

Words such as bis, tris, tetrakis etc are used to denote the number of those ligands. Whosename already includes a number

Eg : ethylene diammine, triphenyl phosphineMany a times ambiguity is created due to use of words such as di, tri etc to indicate

the number of ligands. For example, when two methyl amine molecules are coordinated tometal atom then using di gives dimethylamine. This confuses us whether dimethylamine means

two molecules of methyl amine or one molecule of dimethyl amine. In such cases words suchas bis, tris, tetrakis etc are used to denote numbers of such ligands. Some of such ligands

whose number is denoted by using bis, tris etc are listed below.Benzene

Pyridinemethyl amine, thiosulphato

Whenever the words bis, tris, tetrakis etc are used to specify the number of ligands,the name of the ligand is written in parentheses.

[Co(en)3] Cl3tris (ethylenediamine) cobalt (III) chloride.Na3 [Ag(S2O3)2]

Sodium bis (thiosulfato) argentate (I)[CuCl2(CH3NH2)2]

dichlorobis(methyl amine) copper (II)B) Naming Metal Atom/Ion

Name of the central metal atom is written after the names of ligands. The oxidationnumber of the metal atom is indicated by Roman numerals in parentheses after the name of

the metal atom.



Determining Oxidation State of Metal Atom

Oxidation number of central atom is determined by usual method.Eg. In K4[Fe(CN)6], the oxidation number of Fe is :

4XK +XFe + 6XCN = 0 where X = oxidation number4(+1) + XFe + 6(1) = 0

XFe = +2Eg. In [Cr(en)3]Cl3, the oxidation number of Cr is :

Xcr + 3Xen + 3XCl = 0Xcr + 3(0) + 3(1) = 0

XCr = +3Eg. In [Fe(CN)6]

4, the oxidation number of Fe is :

XFe + 6XCN = 4

XFe + 6(1) = 4XFe = +2

Naming Metal AtomThe name of metal atom depends upon the charge on coordination sphere.

Neutral or Cationic Coordination SphereWhen the coordination sphere is either a cation or a neutral molecule, the name of

the central atom remains as such.Eg : [Co(NH3)6]Cl3

In the above complex Co is in +3 oxidation state and coordination sphere in theabove complex is cationic.

[Co(NH3)6]Charge = XCo + 6XNH3

= +3+0= +3

Thus the coordination sphere bears +3 charge and hence is cationic. Thus the name

of the metal should be written as Cobalt.[Co(NH3)6]Cl3Hexaamminecobalt (III) Chloride.

[Pt(NH3)2 Cl4] Diamminetetrachloroplatinum (IV)[Cr(en)3]Cl3 Tris(ethylenediamine)chromium (III) chloride.

[Pt Cl2(NH3)2] Diamminedichloroplatinum (II)

Anionic Coordination SphereIf the coordination sphere is anion or bears a negative charge then the name of the

central metal atom ends in ATE. (ium of the name of metal atom is replaced by ate).



Eg K3[Co(NO2)6] Potassium hexanitritocobaltate (III).

K2[Pt Cl6] Potassium hexachloroplatinate (IV)Na[AlCl4] Sodium tetrachloroaluminate (III)

Na4[Fe(CN)6] Sodium hexacyanoferrate (II)Li [AlH4] Lithium tetrahydridoaluminate (III)

Ba [BrF4] Barium tetrafluorobromate (III)When there is a Latin name for the metal, it is used to name the metal atom in

negatively charged coordination sphere.

English name Latin name anion nameCopper Cuprum Cuprate

Gold Aurum Aurate

Iron Ferrum FerrateLead Plumbum Plumbate

Tin Stannum Stannate

Names of some Metal Atoms In Negatively Charged Coordination SphereMn Manganate

Fe FerrateCu Cuperate

Co CobaltateZn Zincate

Mo MolybdateSb antimonate

K[Ag (CN)2] Potassium dicyanoargentate (I)

K2[OsCl5N] Potassium pentachloronitridoosmate (VI)

Nomenclature of Complexes Containing Bridging Ligands

For ligands which act as bridge between two metal atoms, the greek letter u is prefixed

to their names. If a coordination compound contains more than one bridging ligand then theprefix m is repeated before the name of each kind of bridging ligand.

Bridging ligands are mentioned alphabetically before the other ligands. This may beillustrated by considering the following examples.



N

N

(NH3)4Co Co (NH3)4 (NO3)4

H2

O2

m -amido- m -nitrito-N octamminedicobalt (III) nitrate

O

O

(H2O)4Fe Fe 4(H2O) (SO4)2

H

H

di- m -hydroxo-octaaquodiiron (III) sulphate

NH

OH

( en)2Co Co ( en)2

3+

m -hydroxo- m -imido-tetrakis (ethylenediamine) dicobalt (III) ion.

Nomenclature of Some Coordination Compounds

Formula Name[Co(CO3)(NH3)4]Cl Pentaamminecarbonatocobalt(III) chloride

K4[Fe(CN)6] Potassium hexacyanoferrate(II)[Co(NH3)6]Cl3 Hexaamminecobalt(III)chloride

Na3[Co(NO2)6] Sodium hexanitrito-Ncobaltate(III)[PtCl4(NH3)2] Diamminetetrachloroplatnium(IV)

[Co(NO2)3(NH3)3] Triamminetrinitrito-Ncobalt(III)

[CoCl(ONO)(en)2]+ Chlorobis(ethylenediammine)nitrito-Ocobalt(III)

[Ag(CN)2]1 Dicyanoargentate(I)

[CoCl2(en)2]SO4 Dichlorobis(ethylenediamine)cobalt(III)sulphate

N

(NH3)4Co Co (NH3)4

2+H

O2m -amido- m -superoxo-octamminedicobalt(III)

Na2[CrOF4] Sodium tetrafluoroxochromate(IV)



[Cu(NH3)4]SO4 Tetraamminecopper(II)sulphate

[Cr(H2O)6]Cl3 Hexaaquachromium(III)chlorideNa2[SiF6] Sodium hexafluorosilicate(IV)

K3[Fe(CN)6] Potassium hexacyanoferrate(III)K4[Mo(CN)8] Potassium octacyanomolybdate(IV)

K3[Fe(CN)5NO] Potassium pentacyanonitrosylferrate(II)[PdI2(NOO)2(H2O)2] Diaquadiiododinitrito-Npalladium(IV)

[Co(en)3]2(SO4)3 Tris(ethylenediamine)cobalt(III)sulphate

NATURE OF METAL-LIGAND BONDING IN COORDINATION COMPOUNDS

Various theories have been proposed to explain various features such as metal-ligand

bonding, colour, geometry and magnetic properties of transition metal complexes. These are

1. The valence bond theory (VBT)

2. The crystal field theory (CFT)

3. The molecular orbital theory (MOT)

All the above theories have been discussed below with appropriate details.

VALENCE BOND THEORY

The valence bond theory was developed by Prof. Linus Pauling. It deals with the

electronic structure of the central metal atom in its ground state and is concerned mainly with

the study of:

1. the kind of bonding,

2. geometry,

3. the gross magnetic properties of the metal complexes.

ASSUMPTIONS OF VALENCE BOND THEORY

This theory involves the following assumptions:

1) The Central metal atom makes available a number of vacant orbitals equal to its

coordination number for the formation of covalent bonds with the ligand orbitals.*1

2) These vacant orbitals hybridize together to form hybrid orbitals *2 which are the same

in number as the atomic orbitals hybridizing together. These hybrid orbitals are vacant,

equivalent in energy and have definite geometry.

3) The ligands have at least one s -orbital containing a lone pair of electrons.



4) Vacant hybrid orbitals of the metal atom or ion overlap with filled s -orbitals of the

ligands to form ligand metal s bond. This type of bond is known as co-ordinate

covalent bond.

5) In addition to the s bond, there is the possibility of a p bond formation due to the

side-ways overlapping of a filled metal orbital with a suitable vacant ligand.

[*1 : - The rearrangement of non bonding electrons of the metal atom or ion takes

place in the following way while making available the empty orbitals for the ligands;

A) The rearrangement of non bonding electrons takes place according to

Hunds rule when the ligands are WEAK.

B) Under the influence of a strong ligand, the electrons can be forced to

pair up against the Hunds rule of maximum multiplicity.]

[*2 : - Numerous combinations of s, p and d orbitals are possible for hybridisation. The

type of hybridisation that the empty orbitals of metal atom undergo decides the geometry

of the resulting complex. Though there are numerous hybridisations possible, in practice

only a few are encountered in metal complexes. The following table gives the co-

ordination number, orbital hybridisation, spatial geometry and examples associated with

each.]

Hybridisation and Geometry

CN Hybridisation Molecular geometry Examples

2 sp Linear [Ag(NH3)2]+, [Ag(CN)2]

3 sp2 Trigonal [HgI3]

4 dsp2 square planar [Ni(CN)4]2 , [Pt(NH3)4]

2+

[PdCl4]2 , [Cu(NH3)4]

2+

sp3 Tetrahedral [Ni(CO)4], [Zn(NH3)4]2+

[NiCl4]2, [Cu(CN)4]

3

5 sp3d Trigonal bipyramidal [Fe(CO)5], [CuCl5]3

dsp3 square pyramidal [SbF5]2, [Ni(CN)5]

3

6 sp3d2 Octahedral [CoF6]3 [Cr(H2O)6]

2+

[Cr(NH3)6]2+ [FeF6]

3

[Fe(H2O)6]3+ [Fe(NH3)6]

2+

7 d2sp3 Octahedral [Cr(CN)6]3 , [CrF6]

3

[Cr(CO)6], [Mn(CN)6]5

[Fe(CN)6]4, [PtCl6]

2



REPRESENTATION OF COMPLEXES BY VBT

We will now discuss a few examples of complex formation to illustrate the norms of VBT.

1) Complexes with co-ordination number = 2

Any complex involving coordination number 2 involves sp hybridisation and has linear

geometry. Consider an example of [Ag(CN)2]-. Its formation by using VBT is shown below:

e configuration of Ag+

4d10 5s0 6p


during approach of two 4d10 5sO 6p

strong CN ligands


in [Ag(CN)2] 4d sp hybridisation

[Ag(CN)2] involves sp hybridisation and thus has linear geometry. Since all electrons

in the electronic configuration of [Ag(CN)2] are paired, it is diamangetic.

Examples to solve :

Q.1 : Describe the hybridisation and geometry of a) [CuCl2]-2 , b) [Cu(NH3)2]

+

Hint : Use following steps.

i) Determine the oxidation state of the central metal ion.

ii) Write the electronic configuration of metal atom in that particular oxidation state.

iii) Rearrange the electrons of central metal ion using the hints given in *1. (Cl is a weak

ligand while NH3 is a strong ligand)

iv) Follow the method illustrated for [Ag(CN)2] to determine the hybridisation and geometry.

2) Complexes with the co-ordination number 3 :

The complexes with co-ordination number 3 involve sp2 hybridisation and have trigonal

geometry. An example of such a complex is [HgI3]. Its formation on the basis of VBT is

illustrated below;



e configuration of Hg+2

5d10 6sO 6pO


during approach of 3I 5d10 6s 6p

weak lignads.


in [HgI3] 5d10 sp2 hybridization

Since [HgI3] involves sp2 hybridisation, it has trigonal geometry.

Magnetic nature- Since all the electrons in [HgI3] are paired, it is diamagnetic.

3) Complexes with the co-ordination number 4 :

There are two possible configurations for metal complexes with co-ordination number

four. These are tetrahedral and square planar. Tetrahedral structure arises from sp3 hybridisation

while the square planar structure is the result of dsp2 hybridisation.

A) Tetrahedral Complexes :

Here we shall discuss the structures of some complexes which have tetrahedral

geometry. Consider [Ni(CO)4] in which Ni is in zero oxidation state*3. Its valence shell

configuration is 3d84s2. The formation of [Ni(CO)4] as per the norms of VBT is explained

below :

e configuration of Ni3d8 4s2

e configuration of Ni

during approach of 3d10 4s 4p

strong CO ligands

e configuration of Ni(CO)4

3d10 sp3 hybridisation

Since [Ni(CO)4] involves sp3 hybridisation, and it has tetrahedral geometry.

Since all the electrons is [Ni(CO)4] are paired, it is diamagnetic.

[*3 : - See the nomenclature of the complexes to find out oxidation state of the metal

atom.]



Consider the other example of [FeCl4] in which Fe is in +3 oxidation state. Its

formation is illustrated below :

e- configuration of Fe+3

e- configuration of Fe+3 during approach ofweak Cl- ligands

e- configuration of Fe+3 in[FeCl4]

-2

3d5 4s 4p

sp3 hybridisation

Cl- Cl- Cl- Cl-

sp3 hybridisation

Thus, [FeCl4]-2 involves sp3 hybridisation and has tetrahedral geometry.

B) Square planar complexes :

Another possible geometry for the 4-coordinated complex is the square planar

involving dsp2 hybridisation. Some examples involving square planar geometries are discussed

below :

Consider [Ni(CN)4]-2 in which Ni is in +2 oxidation state. The formation of

[Ni(CN)4]-2 as per the norms VBT is explained below;

e configuration of Ni3d8 4s2

e configuration of Ni

during approach of 3d10 4s 4p

strong CO ligands

e configuration of Ni(CO)4

3d10 sp3 hybridisation

Since [Ni(CO)4] involves sp3 hybridisation, and it has tetrahedral geometry.

Since all the electrons is [Ni(CO)4] are paired, it is diamagnetic.



Consider the other example of [Pt(NH3)4]2+ in which Pt is in +2 oxidation state and

has valence shell configuration of 5d8. The square planar geometry of [Pt(NH3)4]+2 is explained

by using VBT as

e- configuration of Pt+2

e- configuration of Pt+2 during approach of fourNH3 strong ligands

e- configuration of Pt2- in[Pt(NH3)4]

+2 after gainingfour pairs of electrons from4Cl- ions.

5d 6s 6p

dsp2 hybridisation

NH3 NH3 NH3 NH3

dsp2 hybridisation

Thus, since [Pt(NH3)4]+2 involves dsp2 hybridisation, it has square planar geometry.

Examples to solve :

Q.2 : Predict the hybridisation and geometries of the following complexes by using

VBT.

1) [MnCl4]-2

2) [FeCl4]-2

3) [CoCl4]-2

Complexes with co-ordination number 5 :

There are two possible configurations for metal complexes with co-ordination number

five. These are trigonal bipyramidal and square pyramidal.

Trigonal bipyramidal structure arises form sp3d hybridisation while the square pyramidal

structure is the result of dsp3 hybridisation.

Consider the example of [Fe(CO)5] in which Fe is in zero oxidation state W and

has configuration 3d64s2. The TBP geometry of [Fe(CO)5] is explained by using VBT.



e configuration of Fe

3d6 4s2 5p0


during approach of 3d8 4s0 5p0

during approach of strong

CO lignads


in Fe(CO)5 dsp3 hybridisation

Since [Fe(CO)5] involves dsp3 hybridisation, it is square pyramidal. Since complex

contains paired electrons, it is diamagnetic.

Consider the other example of [Ni(CN)5]3- in which Ni is in +2 oxidation state

and has electronic configuration 3d8. [Ni(CN)5]-3 involves dsp3 hybridisation and has

square pyramidal geometry which is explained by using VBT.

e- configuration of Ni+2

e- configuration of Ni+2 during approach of 5strong CO ligands

e- configuration of Ni+2 in[Ni(CO)5] after gaining5 electron pairs fromCO ligands

3d 4s 4p

dsp3 hybridisation

CO

dsp3 hybridisation

CO CO CO CO

Complexes with the co-ordination number 6 :

Complexes with co-ordination number six are most exclusive and have been studied

on large scale. The complexes with the co-ordination number six involve either d2sp3 or sp3d2

hybridisation. Both the hybridisation give octahedral geometry. The hybridisation and geometry



of some complexes with coordination number 6 is discussed below :

Consider an example of [Fe(CN)6]-3 in which Fe is in +3 oxidation state & has

valence shell configuration of 3d5. The hybridisation and geomery of [Fe(CN)6]-3 is discussed

below by using VBT :

e configuration of Fe+3

3d5 4s0


during approach of six strong 3d5 4s0 5p0

CN ligands


in [Fe(CN)6]3 d2sp3

Since [Fe(CN)6]3 involves d2 sp3 hybridisation, it has octahedral geometry.

Since the complex contains unpaired electron, it is paramagnetic.

Consider the other example of [FeCl6]-3 in which Fe is in +3 oxidation state and has

valence shell configuration of 3d5. The hybridisation and geometry of [FeCl6]-3 is discussed

by using VBT :

e- configuration of Fe+3

e- configuration of Fe+3 during approach of 6weak Cl- ligands

e- configuration of Fe+3 in[FeCl6]

-3 after gaining6 electron pairs from6 Cl- ligands

3d 4s 4p

sp3d2 hybridisation

4d

sp3d2 hybridisation

Cl- Cl- Cl- Cl- Cl- Cl-

Thus, [Fe(Cl)6]-3 involves sp3d2 hybridisation and has octahedral geometry.



INNER ORBITAL AND OUTER ORBITAL OCTAHEDRAL COMPLEXES

Since two d-orbitals used in d2sp3 hybridisation belong to inner shell [i.e. (n-l)th shell],

the octahedral complex compounds resulted from d2sp3 hybridisation are called inner orbital

octahedral complexes. Since these complexes have comparatively lesser number of unpaired

electrons than the outer orbital octahedral complexes (see later on), these complexes are also

called low spin or spin paired octahedral complexes. It is due to the presence of strong

ligands in inner-orbital octahedral complexes of 3d transition series that the electrons present

in 3dz2. and 3dx

2-y

2 orbitals (eg set) are forced to occupy 3dxy, 3dyz and 3dxz orbitals

(t2g set) and thus 3d orbitals of eg set become vacant and hence can be used in 3dx2-y

2,

3dz2, 4s, 4px 4py 4pz (d

2sp3) hybridisation.

Since two d-orbitals are from the outer shell (i.e. nth shell), the octahedral complexes

resulted from sp3d2 hybridisation are called outer orbital octahedral complexes. Since these

complexes have comparatively greater number of unpaired electrons than the inner orbital

octahedral complexes, these are also called high spin or spin free octahedral complexes.

MAGNETISM

Movement of an electrical charge generates a magnetic field in a material. Magnetism

is therefore a characteristic property of all materials that contain electrically charged particles

and for most purposes can be considered to be entirely of electronic origin. In an atom, the

magnetic field is due to the coupled spin and orbital magnetic moments associated with the

motion of electrons. The spin magnetic moment is due to the precession of the electrons about

their own axes whereas the orbital magnetic moment is due to the motion of electrons around

the nucleus. The resultant combination of the spin and orbital magnetic moments of the

constituent atoms of a material gives rise to the observed magnetic properties.

Transition metal complexes are broadly classified as paramagnetic and diamagnetic

on the basis of magnetic properties.

Paramagnetism derives from the spin and orbital angular momenta of electrons. This

type of magnetism occurs only in compounds containing unpaired electrons.

electron spinning on its axis or

gives the spin magnetic moment

electron moving in its orbital creates

an additional magnetic field, leading

to the orbital magnetic moment



Diamagnetism arises when the compound contains no unpaired electrons as the spin

and orbital angular momenta is cancelled out when the electrons exists in pairs.

The value of the magnetic moment associated with paramagnetic substances can be measured

experimentally as well as can be calculated theoretically.

The value of the magnetic moment is calculated experimentally by using Gouy balance.

Gouy balance is used to measure the mass of a sample first in the absence of a magnetic

field, and then when the magnetic field is switched on. The difference in mass can be used

to calculate the magnetic susceptibility of the sample, and from the magnetic susceptibility the

magnetic moment can be obtained.

M2.84 Tm = c

m = magnetic moment in Bohr magnetons (B.M.)

Mc = magnetic susceptibility

T = absolute temperature.

The value of the magnetic moment is theoretically calculated as follows :

The spin and the orbital motion of the electrons are the sources of magnetic moment.

Thus, m is given by the expression :

S L 4S(S 1) L(L 1)+m = + + +

For the 3d transition metal complexes, the orbital moment is not important because

the ligand field quenches the orbital contribution. This can be more easily understood from

the following explanation that comes from CFT.

In order for an electron to contribute to the orbital in which it resides must be able

to transform into an exactly identical and degenerate orbital by a simple rotation (it is the

rotation of the electrons which induces the orbital contribution). For example, in an octahedral

complex, the degenerate t2g set of orbitals (dxz, dyx, dyz) can be inter converted by a 900

rotation. However the orbitals in the eg subset (dz2,dx2-y2) cannot be interconverted by

rotation about any axis as the orbital shapes are different; therefore an electron in the eg set

does not contribute to the orbital angular momentum and is said to be quenched. In the free

icon case the electrons can be transformed between any of the orbitals as they are all

degenerate, but there will still be partial orbital quenching as the orbitals are not identical.

Electrons in the t2g set do not always caontribute to the orbital angular moment. For

example in the d3, t2g3 case, an electron in the dxz orbital cannot be rotation be placed in

the dyz orbital as the orbital already has electron of the same spin. This process is also called

quenching.



Tetrahedral complexes can be treated in a similar way with the exception that we fill

the e orbitals first, and the electrons in these do not contribute to the orbital angular momentum.

Thus, for 3d complexes, the magnetic moment (m s) can be calculated by the following

spin formula:

S 4S(S 1) 2 S(S 1)m = + = +

Where S is the total spin of the complex. In the ground state, S is one-half the number

of unpaired electrons, n.

Therefore spin-only magnetic moment S n(n 2)m = +

Units of the m s is Bohr Magneton (B.M).

Thus, spin only formula can be used to calculate the magnetic moment from the value

of number of spin-free (unpaired ) electrons in the complex.

Number of unpaired electrons Spin-only magnetic moment, B.M.

1 1.7

2 2.8

3 3.9

4 4.9

5 5.9

Solved example :

Q.3 Calculate the magnetic moment of [Fe(CN)6]3-

.

Ans. : The electronic configuration of Fe+3 in [Fe(CN)6]-3 is

d2sp3 hybrid orbitals

3d 4s 4p 4d

CN CN CN CN CN CN

[Fe(CN)6]-3 has one unpaired electron hence the magnetic moment of [Fe(CN)6]

-3 is

S n(n 2) 1(1 1) 1.7 BMm = + = + =



Example to Solve :

Q.4) [CoF6]-4 is

a) outer orbital and diamagnetic

b) inner orbital and paramagnetic

c) inner orbital and diamagnetic

d) outer orbital and paramagnetic

Q.5) Ni(CO)4 is

a) square planar and paramagnetic

b) tetrahedral and diamagnetic

c) square planar and diamagnetic

d) tetrahedral and paramagnetic

DRAWBACKS OF VALENCE BOND THEORY

1. The valence bond theory does not take into account the splitting of the metals d energy

levels.

2. It is unable to account for or predict the relative energies of the different alternative

structures for a complex.

3. It is not helpful in the interpretation of the spectra of complexes.

4. If fails to explain the reaction rates and mechanisms of reactions with complexes.

5. It does not indicate why certain ligands form outer-orbital complexes whereas some

others form inner-orbital complexes.

6. It does not explain why certain 4- coordinated complexes are tetrahedral whereas others

are square-planar.

7. This theory does not account for the detailed magnetic properties of certain complexes.

For these complexes, experimentally determined magnetic moments are slightly higher

than the values theoretically calculated from the spin-only formula. This deviation is due

to the orbital contribution to the magnetic moment, which is not explained by this theory.



SOLVED EXAMPLES ON VBT

Q.6) On the basis of VBT, answer the following questions for 4-coordinated

complexes: [NiCl4]-2, [Ni(CN)4]

-2.

(i) What is the O.S. of the central metal atom/ion.

(ii) What type of hybridisation is involed?

(iii) What is the geometry and magnetic behaviour of the complex?

(iv) Calculate the value of magnetic moment?

Ans. : [NiCl4 ]-2

Since Cl- ion is mononegatively charged, the oxidation state of Ni is +2.

The hybridisation and geometry of [NiCl4]-2 is accounted on the basis of VBT as

follows:

e- configuration of Ni++

e- configuration of Ni++ during approach of 4Cl-weak ligands

e- configuration of Ni++ in[NiCl4]

-2 after gaining4 electron pairs fromCl- ligands

3d 4s 4p

sp3 hybridisation

sp3 hybridisation

Cl- Cl- Cl- Cl-

Thus, [NiCl4]-2 involves sp3 hybridisation and has tetrahedral geometry.

Since [NiCl4]-2 has two unpaired electrons, it is paramagnetic and its magnetic moment is

( )n n 2m = +

( )2 2 2 8 2.828 BM= + = =[Ni(CN)4 ]

-2

Since CN- is mono negatively charged ligand, O.S. of Ni atom is +2.

Thy hybridisation and geometry of [Ni(CN)4]-2 is accounted on the basis of VBT as

follows:



e- configuration of Ni++

e- configuration of Ni++ during approach of fourstrong CN- ligands

e- configuration of Ni++ in[Ni(CN)4]

-2 after gaining4 electron pairs fromCN ligands

3d 4s 4p

dsp2 hybridisation

dsp2 hybridisation

CN- CN- CN- CN-

Thus, [Ni(CN)4]-2 involves dsp2 hybridisation and has square planar geometry. Since

all the electrons in [Ni(CN)4]-2 are paired. it is diamagnetic.

Example to Solve :

Q.7] On the basis of VBT answer the following questions for the co-ordination

complexes.

A) [Ag(CN)2]- B) [HgI3]

- C) [Zn(NH)4]+2

D) [CoCl4]-2 E) [CoBr4]

-2 F) [MnCl4]-2

G) [Ni(CN)4]-2 H) [Fe(CN)6]

-3 I) [FeCl6]-3

J) [Fe(NH3)6]+2 K) [Mn(CN)6]

4- L) [Cr(CO)6]

M) [Co(CN)6]3- N) [Mn(H2O)6]

+2 O) [FeF6]-3

P) [CoF6]-3 Q) [MnF6]

3- R) [Cu(NH3)6]+2

S) [Zn(NH3)6]+2

i) What is the o.s. of the central metal atom.

ii) What type of hybridisation is involved.

iii) What is the geometry and magnetic behaviour of the complexes?

Q.8] Determine the hybridisation and geometry of [Ti(bpy)3]-.

Ans. : bpy is a strong, neutral bidentate ligand. Ti is present in the form Ti in the complex

and has configuration 3d3 4s2. The hybridisation and geometry of [Ti(bpy)3] is deduced by

using VBT as follows:



e- configuration of Ti-1

e- configuration of Ti-1 during approach of threebidentate strong (bpy) ligands

e- configuration of Ti- in[Ti(bpy)3]

- after gainingsix electron pairs from3 bpy ligands

3d 4s 4p

d2sp2 hybridisation

bpy bpy bpy

Since [Ti(bpy)3]- involves d2sp3 byubridisation, it has octahedral geometry.

Ti

bpy

bpy

bpy

-1

Example to solve :

Q.9] Determine the hybridisation and geometry of following complexes that are

formed from polydentate ligands.

A) [Fe(en)3]+2 B) [Ni(DMG)2] C) [Pt(gly)2]

[Hint - en, DMG and gly are strong bidentate ligands.]

Q.10] Determine the hybridisation and geometry of [Fe(H2O)5(NO)]+2

Ans. : In this complex, since NO which acts as ligand is present as NO+ ion, the central

metal atom is present as Fe+. This electronic configuration of Fe+ is 3d6 4s1=3d7. The

hybridisation and geometry of [Fe(H2O)5(NO)]+2 is deduced by using VBT as follows;

e- configuration of Fe+

e- configuration of Fe+ during approach of 5H2Oand one NO ligands

3d 4s 4p

sp3d2 hybridisation

4d



e- configuration of Fe+ in[Fe(H2O)5(NO)]

+2 after gaining six electron pairs fromfive H2O ligands and one NO ligand

sp3d2 hybridisation

Thus, [Fe(H2O)5(NO)]+ involes sp3d2 hybridisation and has octahedral geometry.

Fe

+

H2O

H2OH2O

H2O

H2O

NO

(Since H2O and NO+ are weak ligands, the distributionof 3d7 electrons in five 3d orbitals in

[Fe(H2O)5(NO)]+2 remains the same as it is for Fe+ ion in free state.)

Example to Solve :

Q.11] Determine the hybridisation and geometry of the following complexes.

[Fe(CN)5(NO)]-2

[Co(ONO)(NH3)5]+2

[Co(NH3)4Cl2]+1

[Cr(NH3)2(SCN)4]+1

[Hint : If the complex contains at least one strong ligand then the e- distribution will be against

the Hunds rule of maximum multiplicity. See*1 for more details.]

Q.12) The magnetic moment value of [Mn(CN)6]3- ion is 2.8 BM. Predict the type

of hybridisation and geometry of the ion.

Ans. : We know that m is given by:

( )n n 2m = +

or ( )2.8 n n 2= +or (2.8)2 = n(n+2)

or n2+2n - 7.84 = 0

or n2+2n - 8 = 0

or (n+4)(n-2) = 0

\ n = 4, +2



Thus, [Mn(CN)6]3- ion has two unpaired electrons (n=2). In the given complex ion,

Mn is present as Mn+3 which is a 3d4 ion. Since C.N. of Mn3+ = 6, [Mn(CN)6]3- ion has

octahedral geometry which may arise either from d2sp3 hybridisation (inner orbital) or from

sp3d2 hybridisation (Outer orbital) as shown in Fig.1. Now since d2sp3 hybridisation gives

n = 2 and sp3d2 hybridisation gives n=4, [Mn(CN)6]3- ion has inner orbital octahedral

geometry which results from d2sp3 hyubridisation.

Mn3+ ion (3d44s04p04d0)

[Mn(CN)6]3- ion (d2sp3)

[Mn(CN)6]3- ion (sp3d2)

3d 4s 4p

d2sp3 hybridisation : Innerorbital octahedral geometry

4d

sp3d2 hybridisation: Outerorbital octahedral geometry

(n=4)

(n=2)CN- CN- CN- CN- CN- CN-

CN- CN- CN- CN- CN- CN-(n=4)

Fig. 1. d2sp3 and sp3d2 hybridisation of Mn3+ ion in [Mn(CN)6 ]3- ion.

Q.13) Magnetic moment value of [MnBr4]2- ion is 5.9 B.M. On the basis of VBT,

predict the type of hybridisation and geometry of the ion.

Ans. : We know that,

( )n n 2 B.Mm = +If we put n=5 in the above equation, we get m = 35 B.M. = 5.91 B.M. Thus

[Mn Br4]2- ion has five unpaired electrons (n=5). In [MnBr4]

2- ion, the central atom is Mn2+

ion which is 3d5 ion. Now since C.N. of Mn2+ = 4, [MnBr4]2- ion may have either square

planar (dsp2 hybridisation) or tetrahedral (sp3 hybridisation) geometry as shown below;

Mn2+ ion (3d54s04p0)

3d 4s 4p

(n = 5)



[MnBr4]2-

(Square planar)

[MnBr4]2-

(Tetrahedral)

dsp2 (square planar)

Br- Br- Br- Br-

Br- Br- Br- Br-

sp3 (tetrahedral)

(n = 3)

(n = 5)

Square planar and tetrahedral geometries of [MnBr4 ]2- ion.

Since dsp2 hybridisation (square planar) gives n=3 and sp3 hybridisation (tetrahedral)

gives n=5, [MnBr4]2- ion has tetrahedral geometry and not square planar. Alternatively, since

Br- given is a weak field ligand, [MnBr4]-2 ion is tetrahedral in geometry. Tetrahedral

complexes are given by weak field ligands, since these are HS complexes.

Q.14) Explain : [Co(NH3)6]+3 is diamagnetic while [CoF6]

-3 is strongly paramagnetic.

Ans. : [Co(NH3)6]+3 has d2sp3 hybridisation giving octahedral configuration where all the

electrons are paired so that the molecule is diamagnetic. In case of [CoF6]-3 , F is a weak

ligand so that all the electrons cannot be paired and we find four unpaired electrons so that

the complex is strongly paramagnetic. It is a case of outer orbital complex since the outer

4d orbital is involved in bybridisation.

[Co(NH3)6]+3

[CoF6]-3

3d 4s 4p

sp3d2 hybridisation

d2sp3 hybridisation

3d 4s 4p 4d

corrdination chemistry _gsr

Documents