Coordination Compunds

When two or more compounds combine in a stoichiometric ratio without any change in their ordinary valencies, they form addition or molecular compounds.  These addition compounds can be divided into two categories. i.e.,
1) Double salts
2) Complex (or) coordination compounds
1) Double salts:  These salts are formed by mixing of saturated solutions of the two (or more) compounds.  They lose their identity in solution.  When dissolved in water or melted, they furnish their constituent ions.
For example:
a) Mohr salt → Ferrous sulphate + Ammonium sulphate
${\mathrm{FeSO}}_4+{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4+6{\mathrm{H}}_2\mathrm{O}\to {\mathrm{FeSO}}_4{\left({\mathrm{NH}}_4\right)}_2{\mathrm{SO}}_4\cdot 6{\mathrm{H}}_2\mathrm{O}$
b) Alum → Potassium sulphate + Aluminium sulphate
${\mathrm{K}}_2{\mathrm{SO}}_4+{\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3+24{\mathrm{H}}_2\mathrm{O}\to {\mathrm{K}}_2{\mathrm{SO}}_4\cdot {\mathrm{Al}}_2{\left({\mathrm{SO}}_4\right)}_3\cdot 24{\mathrm{H}}_2\mathrm{O}$
Karnallite → Potassium chloride + Magnesium chloride
$\mathrm{KCl}+{\mathrm{MgCl}}_2+6{\mathrm{H}}_2\mathrm{O}\to {\mathrm{KCl}.\mathrm{MgCl}}_2\cdot 6{\mathrm{H}}_2\mathrm{O}$
The shapes and sizes of crystals of Mohr salt is different than that of ferrous sulphate and ammonium sulphate.  In solution Mohr salts gives the tests of the NH₄⁺, Fe²⁺ and SO₄^2–.

2) Coordination or complex salts:
These are the compounds which retain their identities even when dissolved in water.  On dissolving in water or melting they donot give all of their constituent ions.
For example:
Adding KCN into the ferrous cyanide solution results into potassium ferrocyanide.
$4\mathrm{KCN}+\mathrm{Fe}(\mathrm{CN}{)}_2⟶{\mathrm{K}}_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]$
When poto. Ferrocyanide is dissolved in water, it does give the test of CN^– or Fe²⁺ ions instead it gives a characteristic test of a new ion [Fe(CN) ₆] ^4–, the ferrocyanide ion.
$\begin{array}{rl}& {\mathrm{K}}_4\left[\mathrm{Fe}\right(\mathrm{CN}){]}_6\to 4{\mathrm{K}}^{+}+{\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]}^{4-}\\ & {\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]}^{-4}\to {\mathrm{Fe}}^{+2}+6{\mathrm{CN}}^{-}\\ & {\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]}^{-4} \text{involves two things:}\end{array}$

(i) An acceptor:  Acceptor is the atom that can accept e^– pairs from the donors.  It is generally called as central metal ion or atom.  Usually one or more neutral molecules or negative ions are attached to it by the dative bond. e.g., Fe²⁺ ion in [Fe(CN)₆]^4–.
(ii)  A donor :  A donor is an atom or molecule which can donate a pair of electrons.  The molecules or ions having donor atoms are called ligands.  They are attached to central metal atom.
e.g., CN^– in [Fe(CN)₆]^4–.

Ligands:
Neutral molecules (CO, NH₃, H₂O etc), cation (NO⁺, N₂H₅⁺ etc) and anions (Cl^–, OH^–, F^– etc) which donate pair (s) of electron to any metal and forms covalent bond are called ligands.  (Latin lizare = bind).  The common donor atoms in most of the ligands are nitrogen, oxygen and sometimes carbon, halogens, phosphorus, sulphur etc.  These atoms, who donate pairs are called donor atoms or donor sites of the ligand.
More examples:

Complex salt	Central atom/ion	Ligand	Donor atom
${\mathrm{K}}_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_8]$	Fe2+	CN–	C
$\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]\mathrm{Cl}$	Ag+	NH3	N
$\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4\right]{\mathrm{Cl}}_3$	Cr3+	H2O	O

Types of ligands:
Depending on the no. of binding sites, the ligands have been classified as follows.

Chelating ligands:  (Greek chele = claw):
When a polydentate ligand binds with a central metal atom by more than one binding site and thus forms a closed ring type structure involving metal atom, the resulting complex is called chelating complex and the ligand, chelating ligand.
e.g., Cu²⁺ complexes with ethylene diammine (bidentate), it binds by both the N atoms and form a chelate: 

Characteristics of chelates:

1. Complexes of chelating ligands are more stable than non chelated complexes (chelating effect).

2. Chelating ligands with large groups form unstable rings in comparison to smaller groups (steric effect).

3. Chelating ligands without double bonds can form five membered ring (including metal as one of the member) while those with double bonds with six membered.
e.g.,  5 membered.

6 membered → m
(ac = acetyl acetone)

Bridging ligand:  A bridging ligand is one ligand which because of presence of more than 1 e^– pair can bind with more than one metal atom at the same time.
e.g.,  ${\mathrm{NH}}_2-{\mathrm{NH}}_2,{\mathrm{NH}}_2^{-},{\mathrm{OH}}^{-},{\mathrm{NO}}_2^{-}$ etc.

Ambidentate ligand:  A ligand which has more than one donor atoms, but at a time only either of them is used for complexation, is ambidentate ligand.  e.g., -CN, – SCN, – NO₂.
Note:  Polydentate ligands have flexidentate character i.e., they are free either to use all of their available donor sites or only some of them.
e.g., EDTA through uses all the six donor sites while complexation but in some cases, it uses less than six also.

Coordination Sphere:
While writing the formula of complexes, the central metal atom and the non-ionizable ligands bound with it are closed in square bracket [  ].  Because this part is formed due to coordination between metal and ligands, it is called coordination sphere and the remaining part, outside the square bracket, which is bound by electrostatic forces and ionizes into cations or anions in solution is called ionization sphere.

e.g.,   ${\mathrm{K}}_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_8]\to 4{\mathrm{K}}^{+}+\underset{\text{Coordination sphere}}{{\left[\mathrm{Fe}\right(\mathrm{CN}{)}_B]}^{4-}}$
$\left[\mathrm{Cu}\left({\mathrm{Cn}}_2\right)\right]{\mathrm{Cl}}_2\to \underset{\text{Coordination sphere}}{{\left[\mathrm{Cu}\right(\mathrm{Cn}{)}_2]}^{2+}}+2{\mathrm{Cl}}^{-}$

Coordination Number:
It is the total number of ligands attached to the central metal atom in the coordination sphere.  E.g., in, the C.N. of Pt is 6 and that of Ni in  is 4.
Complexes in different coordination states and their stereochemistry.

Coordination No.	Hybridization	Example	Stereochemistry
2	sp	${\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}{\left[{\mathrm{CuCl}}_2\right]}^{-}$	Linear
3	sp2	${\left[{\mathrm{Hgl}}_3\right]}^{-}$	Trigonal planar
4	sp3	${\left[{\mathrm{CoCl}}_4\right]}^{2-},\left[\mathrm{Ni}\right(\mathrm{CO}{)}_4]$	Tetrahedral
4	dsp2	${\left[\mathrm{Ni}\right(\mathrm{CN}{)}_4]}^{2-},{\left[\mathrm{Cu}\right(\mathrm{en}{)}_2]}^{2+}$	Square planar
5	sp3d	$\left[\mathrm{Cul}\right(\mathrm{dipy}{)}_2]$	Trigonal bipyramidal
6	sp3d2	${\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]}^{2+},{\left[\mathrm{Ni}\right(\mathrm{en}{)}_3]}^{2+}$	Octahedral
7.	sp3d3	${\left[\mathrm{Z}{\mathrm{rF}}_7\right]}^{3-},{\left[{\mathrm{HgF}}_7\right]}^{3-}$	Pentagonal pyramidal

Werner theory:
Werner studied the structures of complexes like CoCl₃.6NH₃ and CuSO₄.4NH₃ and put forth his theory in 1893.  The main points of his theory are:
1. Each complex compound contains two types of valences:
(i)  Primary valency or principal valency or ionizable valency.
(ii)  Secondary valency or non-ionizable valency.
2. Primary valency: it corresponds to the oxidation no. of central metal atom (ion) and is always satisfied by anions. It is represented by dotted lines.
3. Secondary valency: It is the coordination no. of the central atom (ion) and may be satisfied by either neutral or negative (rarely by positive) ligands. It is represented by thick lines.
4. Each metal has a fixed no. of secondary valencies (co-ordination no.) which depends on the size and charge of that central metal atom.
5. The ions that are attached by primary valencies are ionic in nature and dissociate into the solution while the secondary valencies being coordinate, do not.
6. Primary or ionic valencies are non directional while secondary are directional and they give rise to stereoisomerism in the complexes.
7. Each central metal atom tends to satisfy both of its valencies viz. primary and secondary.
8. Werner has pointed the coordination no. of a metal to be either four (tetrahedral or square planar arrangement) or six (in octahedral shape).
Note:  Now, it has been established that coordination can be any integar between 2 to 9 for a metal atom.
While in attempt to satisfy their valencies, some groups (generally anions) are found to be joined as primary as well as secondary valencies.
On above basis Werner suggested following structures for various complexes of cobalt.

The valencies which are represented by thick as well as dotted lines have both ionic & covalent character and do not ionize in the solution.  So the last compound i.e., CoCl₃.3NH₃ will not give white precipitate with AgNO₃ solution.

Compound	No. of ionizable chlorine atoms
CoCl3.6NH3	3
CoCl3.5NH3	2
CoCl3.4NH3	1
CoCl3.3NH3	0

In modern terms the complex ion which is formed by coordination bonds called coordination sphere is written in square brackets and the ionizable ions outside the bracket.  Hence the above four salts (complexes) can be written as

Werner formula	Modern formula	Cation	Anion	Total No. of Ions
CoCl3.6NH3	[Co(NH3)6]Cl3	[Co(NH3)6]3+	3Cl–	4
CoCl3.5NH3	[Co(NH3)5Cl]Cl2	[Co(NH3)5Cl]2+	2Cl–	3
CoCl3.4NH3	[Co(NH3)4Cl2]Cl	[Co(NH3)4Cl2]+	Cl–	2
CoCl3.3NH3	[Co(NH3)3Cl3]	Non electrolyte		0

Effective Atomic Number (EAN):
It is the resultant no. of e^– of the central metal atom (ion) of an complex after gaining e^– from secondary valancies (ligands).
While coordinate bond formation there is gain of 2e^– per biding of donor sites of the ligands.
Hence,
EAN = Atomic number of metal atom – No. of e^– lost in cation formation + No. of e^– gained from donor atoms.
EAN = Z – Ox. St. + 2L
L = Total no. of ligands or No. of donor sites of ligand(s)
Usually the EAN of an metal is the atomic no. of next inert gas of that very period, thus explaining the stability of the complexes so formed.  Those with EAN not that of an inert gas are relatively less stable.

Illustration 1: EAN of Fe in K₄ [Fe(CN)₆ Atomic no. of Fe = 26
Solution: Oxidation no. of the central metal atom, Fe
4(+1) + (x) + (-1) 6 = 0 ⇒ x = + 2
No. of e^– gained from 6CN^– = 2 × 6 ⇒ 12
EAN = 26 – 2 + 12
= 36 

Illustration 2:  EAN of Cr in [Cr(NH₃)₆]Cl₃
Solution: Z = 24
Ox. st. = + 3
2L = 12
EAN = 24 – 3 + 12
= 33           [three short of Kr]

Characteristics of coordination compounds:
1) Change in solubility:  Solubility of a sparingly soluble salt may increase with the complex formation provided that the ion is involved in the complexation.
e.g., AgCl is precipitated in aq. solutions when AgNO₃ is added to any solution having Cl^– ions.  This precipitate (sparingly soluble salt) becomes soluble if NH₄OH is added to it due to complex, [Ag (NH₃)₂]Cl formation.
Similarly many other soluble ions bring about their precipitation with the formation of a complex.
e.g., Ni²⁺ are soluble in water and when, DMG (dimethyl glyoxime) is added to it, a red coloured precipitate, ${\left[\mathrm{Ni}\right(\mathrm{dmg}{)}_2]}^{-1}$ is obtained:

2) Change in conductivity:
Since, there is a change in the number of ions while complex formation, the conductivity changes abruptly.
e.g., $\underset{5\text{ions}}{\underbrace{{\mathrm{Cu}}^{2+} + 4{\mathrm{Cl}}^{-}}}⟶\underset{1 \mathrm{ion}}{{\left[{\mathrm{CuCl}}_4\right]}^{2-}}$ (Conductivity decreases).
3) Change in colour:  Almost all the complexation reactions are colour sensitive and show a variation in colour.  Anhydrous CuSO₄ is white, when it absorbs water or dissolved in water the colour changes to blue due to complexation of water molecules with CuSO₄.
The colour change is observed due to changed energy states of the d orbitals of the central metal atom.
4) Change in pH:  If complex formation is associated with either consumption or production of H⁺, the reaction is pH sensitive and change of pH informs complex formation.
e.g., when metal complexes with EDTA or Oxalate, the [H⁺] of the solution increases (due to change of -CO-O-H bond) and hence the pH decreases.
5) Change in chemical properties:  When complex has been formed by the central metal atom and the ligand, both of them lose their chemical properties since metal atom achieves stable inert gas configuration by binding the ligand with it making them to behave as a different chemical identity.  e.g., The ammonical solution of AgNO₃, which involves formation complex [Ag(NH₃)₂]⁺ doesn’t give white ppt of AgCl with KCl solution, since Ag⁺ are no longer available.

Change in physical properties:  The properties of the solutions of complexes which depend on the number of particles get changed because no. of particles decreases.  The properties like osmotic pressure, lowering of vapour pressure, depression in freezing point and elevation of boiling point, shows a sharp decrease than the theoretical value.  These properties are more widely used for the study of complexes in comparison to conductivity measurements which are not possible with neutral ligands.

Nomenclature of coordination compounds:
The International Union of Pure & Applied chemists has given a systematic nomenclature for coordination compounds in 1957 with few modifications in 1962 & 1971, they are exclusively used except some of more studied or familiar compounds, which will retain their trivial names e.g., K₄[Fe(CN)₆] as potassium ferrocyanide, K₃[Fe(CN)₆ potassium ferricyanide, Ni(CO)₄ nickel carbonyl etc.

Following are some of the important rules for the nomenclature of inorganic complexes:
1. Order of listing central metal atom (ion) and ligand:
Ligands are named before the central metal atom, but while writing their chemical formula the order is reversed.
e.g., [Fe(CN)₆]^4–
Hexacyano ferrate (II) ion, cyano follows ferrate (II)

2. Order of listing ions:
If complex is salt (ionic), cation(s) is named before the anion(s)
e.g,

NaCl	Sodium (I) chloride	[Not a complex]
[Cr(NH3)6]Cl3	Hexa ammonie chromium (HI) Chloride
[K2[PtCl6]	Potassium hexachloroplastinate (IV)

Oxidation no. of central metal atom:
In every compound, the oxidation no. of central atoms is to be mentioned by Roman numeral written in parenthesis just after the name of the metal atom.  Zero oxidation state is indicated by (O) and negative by putting negative sign before the Roman numeral (e.g., – I).
K₃[Fe(CN)₆]                                            Potassium hexacyanoferrate (III)
Fe(Co)₅ Tetracarbonyl iron (O)
Na{Co(CN)₄]                                          Sodium tetracyanocobaltate (-I)

Termination (ending) of names of complexes:
Anionic complexes, those having negatively charged coordination sphere, end with the name of metal suffixing -ate and in cationic or neutral complexes, the name of metal is retained.
K₂[Ni(CN)₄]                                            Potassium tetracyanonickelate (II)
[Co(H₂O)₆](SO₄)₃                                  Hexa aqua cobalt (III)     sulphate
[Pt(NH₃)₂Cl₂]                                         Diamminedichloroplatinum (II)

Order of ligands:
If different types of ligands are present in a complex, they are arranged in the order of:
i) Negative ligand
ii) Neutral ligand
iii) Positive ligand

e.g., [Cr (H₂O)₅Cl] Cl₂                       Chloro apintaoquochromium (III) chloride
Note;  Whole of the coordination sphere is always written as a single word without any break, hyphen or comma.
In case, these are several negative ligands in a complex, they are named alphabetically.
e.g., Na₂ [Pt (NO₂)₂Cl₄]                           Sodium tetrachlorodinitroplatinate(IV)

Similarly, if several neutral ligands are present, they are enlisted as:
i) H₂O
ii) NH₃
iii)  then all others in alphabetically
e.g., [Pt(NH₃)₄ (NO₂)Cl]SO₄                     Chloronitrotetraammineplatinum (IV) sulphate

Name of ligands:  Neutral ligands are named as molecules with few exception enlisted under column B.

A : General		B : Exceptions
Ligand	Name	Ligand	Name
(CH2NH2)2	Ethylene diammine	H2O	Aquo
		NH3	Ammine
(C6H5)3P	Triphenyl phosphine	CO	Carboxyl
(C2H5)2NH	Diethyl ammine	CS	Thiocarbonyl

Positive ligands end in ‘- ium’.

Ligand	Name
NO+	Nitrosylium
N2H5+	Hydrazinium
Cl+	Chloronium

All negative ligands end with ‘0’.

Anion (–ide)	Symbol	Ligand name (–0)	Anion	Symbol	Ligand name –ate)
Chloride	Cl–	Chloro	Carbonate	CO32–	Carbonato
Oxide	O2–	Oxo	Oxalate	${\mathrm{C}}_2{\mathrm{O}}_4^{2-}$	Oxalato
Oxanide	CN–	Cyano	Nitrate	${\mathrm{NO}}_3^{-}$	Nitrato
Hydride	H–	Hydrido	Acetate	CH3COO–	Acetato
			Nitrite	${\mathrm{NO}}_2^{-}$	Nitro or nitrite

When ${\mathrm{NO}}_2^{-}$ group bond with N atom (N-donor) it is called nitro and when by O atom (O-donor), nitrito is the name.
Note:  The spelling of ammine with two m’s differentiate it with organic amines. Its so since derived from ammonia.

Numerical prefixes: 
(i)  When more than one ligands of same kind, are present, their number is indicated by greek prefixes, di, tri, tetra, penta, hexa et.
(ii)  If chelating ligands are the ligand already having di, tri, tetra etc in their names are present 2, 3, 4….. times, the prefixes bis, tris, tetra etc are used before the names of ligands, which are written in paranthesis.
e.g.,  ${\mathrm{K}}_3\left[\mathrm{Al}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right]$        Potassium trioxalatoalluminate (III)
$\left[\mathrm{Co}\right(\mathrm{en}{)}_2{\mathrm{Cl}}_2]{\mathrm{SO}}_4$           Dichlorobis (ethylenediamine) cobalt (IV) sulphate

Point of attachment of ligands:
For ambidentate ligands, the donor atom which forms coordinate bond is designated by the symbol of its element in italics and hyphened between ligand and metal atom or so.
e.g.,
${\left({\mathrm{NH}}_4\right)}_2\left[\mathrm{Pt}\right(\mathrm{SCN}{)}_6]$           Ammonium hexathiocyanato-S-platinate(IV)
${\left[{\left({\mathrm{CH}}_3\right)}_4\mathrm{N}\right]}_2\left[\mathrm{Co}\right(\mathrm{NCS}{)}_4]$  Tetramethylammonium tetrathiocyanato-N-cobaltate (II)

Molecules (solvent) of crystallization: 
Generally water or any other solvent molecule, when present in the crystal of a complex as lattice component, is written after the name of complex and the no. of molecules is indicated by Arabic numerals between the complex and solvent molecule.
e.g.,  ${\mathrm{AlCl}}_3{\left({\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\right)}_3$         Aluminium (III) chloride – 4- ethanol
$\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}2{\mathrm{H}}_2\mathrm{O}$   Dichlorotetra aquochromium (III) chloride-2- water

(Note) Bridged Polynuclear complexes:
Bridging ligands, which join the two centres of coordination, are prefixed by Greek letter μ.  Each briding ligand is prefixed by μ, however if they are of same kind di, tri etc can be used before the name of the ligands.

Illustration 3:  Write the IUPAC names of the following complexes.
Cationic complexes

Problem	Solution
$\left[\mathrm{Co}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{SO}}_4$	Hexaaquacobalt (II) sulphate
$\left[\mathrm{Zn}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{Cl}}_2$	Tetraamminezine (II) chloride
$\left[{\mathrm{PdCl}}_4\right]{\mathrm{SO}}_4$	Tetrachloropalladium (VI) sulphate
$\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}$	Dichlorotetraaquochromium (III) perchlorate
$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\mathrm{ONO}\right]{\mathrm{Cl}}_2$	Nitritopentaammine cobalt (III) chloride

Anionic complexes

Problem	Solution
$\mathrm{K}\left[{\mathrm{AgCl}}_2\right]$	Potassium dichloroargentate (II)
Ba[(CuCl4]	Barium tetrachlorocuprate (II)
Na [Mn(CO)5]	Sodium pentacarbonylmanganate (-I)
(NH4)2 [TiCl6]	Ammonium hexachlorotitanate (VI)
Na[Fe(EDTA)]	Sodium ethylene diammine tetra acetatoferrate (III)
K [BF4]	Potassium tetrafluroborate(III)
[Pt (NH3)4Cl2] [PtCl4]	Dichlorotetra ammine platinum (IV) tetrachloroplatinate

 Neutral complex

Problem	Solution
Fe(CO)5	Pentacarbonyl iron (O)
[Pt (NH3)2Cl2]	Dichlorodiammine platinum (II)
[Co(en)3]	Tris(ethylene diammine) cobalt (O)

Isomerism in Co-ordination Compounds:
Similar to the isomerism exhibited by organic compounds, specific orientation of ligands in space around the central atom gives rise to isomerism in co-ordination compounds. The isomerism in co-ordination compounds can be classified as:

1. Structural isomerism             2. Stereo isomerism

1. Structural isomerism
a) Ionization isomerism:
Complexes having same empirical formula but producing different ions in solution state are ionization isomers and the phenomenon is ionization isomerism.
Such an isomerism is due to the interchange of groups between the co-ordination sphere of the metal ion and the ions outside the sphere.
Example:

$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\mathrm{Br}\right]{\mathrm{SO}}_4$                         and                        $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{SO}}_4\right]\mathrm{Br}$
Pentaamminebromocobalt(Ill) sulphate            Pentaamininesulphatocobalt(III) bromide
(Violet)                                                                                     (Red)

The violet compound in aqueous solution gives precipitate with BaCI₂ and not with AgNO₃ solution, whereas red compound gives a precipitate with AgNO₃ solution and no precipitate with BaCI₂ solution.
b) Hydration isomerism: This isomerism arises when different number of water molecules are present within and outside the co-ordination sphere. For example, 3 hydration isomers of CrCl₃ 6H₂O are;
$\begin{array}{ll}\underset{\text{Green}}{\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_3},\underset{\text{Green}}{\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2\cdot {\mathrm{H}}_2\mathrm{O}},& \underset{\text{Green}}{\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}.2{\mathrm{H}}_2\mathrm{O}}\end{array}$
c) Salt or Linkage isomerism: This isomerism arises when ligand has two possibilities in its mode of attachment to the metal atom. For example, the ligand —NO₂ binds with metal atom through N or 0 in the following example.

$\begin{array}{rl}& \underset{\text{Penta amminenitritocobalt(III)chloride(Red)}}{\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5\mathrm{ONO}\right]{\mathrm{Cl}}_2} \underset{\text{Penta ammin enitrocobalt (III) chloride (Yellow)}}{\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{NO}}_2\right]{\mathrm{Cl}}_2}\end{array}$

d) Polymerisatlon isomerism: This isomerism arises when compounds have the same stoichiometric composition but different molecular composition (which is an integer multiple of stoichiometric composition), e.g., $\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_2{\mathrm{Cl}}_2\right] \text{and }\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_4\right]\left[{\mathrm{PtCl}}_4\right]$. This is not considered as true isomerism.
e) Co-ordination isomerism: This isomerism exist in polynuclear complexes (where both complex anions and complex cations are present) by partial or complete exchange of ligands. For example,
$$\begin{gathered} \begin{array}{l}\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_{\theta }\right]\left[\mathrm{Cr}\right(\mathrm{SCN}{)}_6] \text{and }\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_2\right(\mathrm{SCN}{)}_4] \\ \\ \left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]\left[\mathrm{Cr}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right] \text{and }\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_8\right]\left[\mathrm{Co}{\left({\mathrm{C}}_2{\mathrm{O}}_4\right)}_3\right] \\ \\ \left[\mathrm{Co}\right(\mathrm{en}{)}_3]\cdot \mathrm{Cr}(\mathrm{CN}{)}_{6 }\text{and }\left[\mathrm{Cr}\right(\mathrm{en}{)}_3]\left[\mathrm{Co}\right(\mathrm{CN}{)}_8]\end{array} \end{gathered}$$

Stereoisomerism:
Two compounds are called stereoisomers when they contain the same ligands in their co-ordination sphere but differ in their l arrangement. Stereoisomerism is further classified as geometrical and optical isomerism.
1) Geometrical isomerism: This type of isomerism arises due to different positions occupied by ligands around the metal When two similar ligands are adjacent to each other, then isomer is referred as cis-isomer and when they are at opposite position then trans-isomer. Geometrical isomerism of square planar (Co-ordination number = 4) and octahedral complex has been discussed below.

a) Square planar complex: Positions A – B, B – C, C – D and A – Dare adjacent to each other and are Cis-with respect to each other. Positions A – C and B – D are trans to each other. (i) Complexes of type ${\mathrm{M}}_{A_4},{\mathrm{M}}_{{\mathrm{A}}_3\mathrm{B}}$ do not show geometrical isomerism. (ii) Square planar complexes having formula ${\mathrm{M}}_{{\mathrm{A}}_2{\mathrm{B}}_2}, {\mathrm{M}}_{{\mathrm{A}}_2\mathrm{BC}}, {\mathrm{M}}_{\mathrm{ABCD}}$ show geometrical isomerism and exist in two forms.
[(NH₃)₂Cl₂] resembling to formula ${\mathrm{M}}_{{\mathrm{A}}_2{\mathrm{B}}_2}$ exists in cis and trans forms.

[Pt(Py)₂NH₃Cl]⁺ resembling to formula ${\mathrm{M}}_{{\mathrm{A}}_2\mathrm{BC}}$ exists as

(iii)  Complexes of M_ABCD type exist in 3-geometrical isomeric forms e.g.,
${\left[\mathrm{Pt}\left({\mathrm{NH}}_3\right)\left({\mathrm{NH}}_2\mathrm{OH}\right)\right(\mathrm{py}\left)\left({\mathrm{NO}}_2\right)\right]}^{+}$

(b) Octahedral Complexes:
Positions A – B, B – C, C – D A – D etc. which are at 90° to each other are cis with respect to each other while A – F, B – D, C – E  (at 180⁰ to each other) are trans positions.

(I) Complexes of type ${\mathrm{M}}_{{\mathrm{A}}_4{\mathrm{B}}_2},{\mathrm{M}}_{{\mathrm{A}}_4\mathrm{BC}}$ will show geometrical isomerism.
Example: ${\left[\mathrm{Fe}\right(\mathrm{CN}{)}_4{\left({\mathrm{NH}}_3\right)}_2]}^{-2}$

(ii) Facial (fac) and meridional (mer) isomers:
Such type of geometrical isomers are in compounds of type ${\mathrm{M}}_{{\mathrm{A}}_2{\mathrm{B}}_3}$ type. If three ligands of same type are at the corners of a triangular face then it will be fac-isomer. If they are at the corners of a square plane then it will be mer-isomer.

e.g., [RbCl₃(py)₃]

2) Optical Isomerism: Optical isomerism is due to the absence of elements of symmetry (plane of symmetry, axis of symmetry or centre of symmetry) in the complex. If the molecule possesses any of these symmetry elements, the compound will not be optically active. An optically active compound cannot be superimposed on its mirror image. It has the property of rotating the plane of polarized light either to its right (called dextro) or to its left (called Iaevo). If polarized light remains unaffected, the compound is inactive or racemic, i.e., it is composed of 50% dextro and 50% Iaevo form.

(a) Optical isomerism in octahedral complexes:
(i)  Complexes of type ${\mathrm{M}}_{{\mathrm{A}}_3}$ type show optical isomerism where A is symmetrical bidentate ligand.
 e.g.,${\left[\mathrm{Cr}\right(\mathrm{Ox}{)}_3]}^{3-}, \mathrm{A}=\underset{CO{O}^{–}}{\overset{CO{O}^{–}}{|}}$Oxalato (symmetrical bidentate ligand)

(ii) Complexes of ${\mathrm{M}}_{{\mathrm{AB}}_2{\mathrm{C}}_2}$ type also shows optical isomerism.  Here A = symmetrical bidentate ligand while B and C are monodentate ligands.
e.g.,  ${\left[\mathrm{Co}\right(\mathrm{en}\left)\right(\mathrm{Cl}{)}_2{\left({\mathrm{NH}}_3\right)}_2]}^{+}$

(iii) Complexes of hexadentate ligands like EDTA, also shows optical isomerism.
eg., [Co(EDTA)]^–

(b) Optical Isomerism In tetrahedral complexes:
Tetrahedral complexes also show optical isomerism but in complexes of type of the M_ABCD resolution is not possible. However complexes of type may be resolved where A is unsymmetrical bidentate ligand.
e.g., [Ni(gly)₂] glycinate ligand.
In analogy with tetrahedral carbon, it is expected that a tetrahedral metal complex with four different groups attached to the metal ion,$\left[{\mathrm{M}}_{\mathrm{abcd}}\right]$ should exhibit optical activity. However, it has not been possible to isolate optically active dextro and laevo forms of such a complex because of its labile nature.

Illustration 4: How many geometrical isomers are possible for the following complex? ${\left[\mathrm{PI}\left({\mathrm{NH}}_3\right){\mathrm{Cl}}_5\right]}^{-}$
Solution: No isomers are possible.
Because this is an octahedral complex of the type MAB₅.

Illustration 5: How many Geometrical isomers are possible for the complex ${\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_4\mathrm{ClBr}\right]}^{+2}$ ?Write the structures?
Solution: This is of the type of MA₄BC and Cis, trans isomers (two isomers) are possible.

Bonding in Co-ordination Compounds:
Werner was the first to explain the bonding in complex compounds. Later on Sidgwick gave his concept. However, these were not able to explain all the facts about complexes. Four distinct approaches have been given to explain properties of complex compounds such as colour, geometry and magnetic properties. These are:

1. Valence bond theory

2. Crystal field theory

3. Molecular orbital theory

4. Ligand field theory

Valence bond theory:
Since, majority of the complexes are formed by the transition metals which have their ‘d’ orbitals incomplete, valence bond theory takes into account the hybridization of orbitals. Since, the d-orbitals of the penultimate quantum shell, in the case of transition metals, are near in energy to s- and p-orbitals of the outermost shell, various kinds of hybridization are possible.

Concepts of Valence bond theory:

• The central metal atom provides a number of empty orbitals equal to its co-ordination number for the formation of covalent bonds (or more accurately co-ordinate bonds) with ligand orbitals.

• The empty orbitals of the metal ion hybridize to give an equal no. of’ hybrid orbitals of equivalent energy.

• The metal atom or ion can use (n-1) d, ns, np, nd orbitals for hybridization to yield square planar, tetrahedral or octahedral geometry.

• These hybridized orbitals then overlap with ligand orbitals which can donate an electron pair for bending.

• The overlapping may result in a bond or possibly a ligand → metal co-ordinate bond.

• The ligand’s electron pair enters the metal ion orbital while still maintaining the electronic configuration originally present in ligand.

• Each ligand donates a pair of electrons to the central metal ion.

• The non-bonding metal electrons present in the inner orbitals do not take part in chemical bonding.

• If the complex contains unpaired electrons, the complex is paramagnetic in nature whereas, if it does not contain unpaired electrons, the complex is diamagnetic in nature.

• Under the influence of a strong ligand (like, NH CN^– , CO, etc.), the electrons can be forced to pair up against the Hund.’s rule of maximum multiplicity.

Geometry (shape) and magnetic nature of the complexes (Application of valence bond theory):

Limitations of valence bond theory: The theory could not explain:
(i) The colour of transition metal complexes.
(ii) The optical absorption spectra.
(iii) The detailed magnetic properties.

Crystal field theory:

• The ligands, are assumed to be point charges.

• The metal cation is surrounded by the ligands with the lone pair of electrons.

• The attractions between metal ion and ligands are purely ionic.

• The valency electrons of metal are repelled by the negatively charged ligands, so that they occupy those d-orbitals which have their lobes away from the direction of ligands. This effect of ligands is particularly marked on d-electrons and it depends on number of electrons, their arrangement and on nature of ligands.

• The energy difference between two sets of d-orbitals is determined by Δ₀ and called crystal field splitting energy.

• The field strength of some ligands is in the order of $\begin{array}{l}\mathrm{CO}>{\mathrm{CN}}^{-}>{\mathrm{NO}}_2^{-}>\mathrm{O}-\text{Phenol}>\text{dipyridil}>\mathrm{en}>{\mathrm{NH}}_3=\text{pyridine}>\mathrm{EDTA}\end{array}$
$>{\mathrm{H}}_2\mathrm{O}>{\mathrm{C}}_2{\mathrm{O}}_4^{-2}>\text{ethylalcohol}>{\mathrm{OH}}^{-}>{\mathrm{F}}^{-}>{\mathrm{NO}}_3^{-}>{\mathrm{Cl}}^{-}>{\mathrm{S}}^{-2}>{\mathrm{Br}}^{-}>{\mathrm{I}}^{-}$

• In octahedral field, d-subshell split into e_g, t_2g sets, here energy of e_g, set is greater than the energy of t_2g set

• In tetrahedral field t_2g set have more energy than the e_g set of orbitals (explained in solved subjective problem 9).

• We have Δ_t = $\frac{4}{9}$Δ₀   , where Δ_t is crystal field splitting energy in tetrahedral complex
Δ₀  is crystal field splitting energy in octahedral complex

Inner orbital complex:
The complex, which is formed by the use of inner d-orbitals for hybridization is called inner orbital complex.
Strong ligands like CO, CN^–, NO₂^– forces the unpaired electrons to pair up and forms inner orbital complexes.

Outer orbital complexes: 
The complex which is formed by the use of outer d-orbitals for hybridization is called an outer orbital complex.
Weak ligands like I^–, Br^–, Cl^–, F^–, H₂O ….. do not disturb the electronic configuration of the metal and forms outer orbital complexes.

Uses of co-ordination compounds:
1) Used in electroplating industry.
2) Micronutrients for plants, like Zn, Fe, Mg, etc, ….. are supplied in the form of soluble chelate complexes.
3) The formation of soluble complexes is used in the extraction of metals like Ag & Au.
4) Estimation of Ca⁺² and Mg⁺² in water is done volumetrically using complexing agent EDTA
5) The platinum complex, [Pt Cl₂(NH₃)₂], known as cis platina is used in chemotheraupetic treatement of cancer.
6) Many bio-catalysts are complex enzymes containing metal ions.
7) Applications In qualitative analysis
(a) The separation of Ag⁺ and ${\mathrm{Hg}}_2^{2+}$ in the first group of qualitative chemical analysis is based upon the formation of a soluble complex of silver chloride with ammonia.
$\mathrm{AgCl}\left(\mathrm{s}\right)+2{\mathrm{NH}}_3\to \underset{\text{Soluble complex}}{{\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}\mathrm{Cl}}$

(b) The detection of copper in the presence of cadmium in the second group of chemical analysis can be made by cyanide complexes,

${\mathrm{K}}_2\left[\mathrm{Cu}\right(\mathrm{CN}{)}_4] \text{and }{\mathrm{K}}_2\left[\mathrm{Cd}\right(\mathrm{CN}{)}_4]$.

The copper complex is more stable than the cadmium complex and, therefore, when H₂S gas is passed through a mixture of these two complexes, only Cd(II) gets precipitated as CdS (yellow) leaving a blue coloured solution of copper complex.
(c) Ferric ions can be detected by means of the red coloured sulphocyanide complex.
${\mathrm{Fe}}^{3+}+{\mathrm{SCN}}^{-}\to \left[\mathrm{Fe}\right(\mathrm{SCN}){]}^{2+}$
(d) Some complexes serve as the confirmatory tests for various cations and anions. For example,
(i) Test for sulphide:  Sodium nitroprusside gives a red colour with sulphides.
$\underset{\text{Sodium nitroprusside}}{{\mathrm{Na}}_2\left[\mathrm{Fe}\right(\mathrm{CN}{)}_5\mathrm{NO}]}+{\mathrm{S}}^{2-}\to {\mathrm{Na}}_2{\left[\mathrm{Fe}\right(\mathrm{CN}{)}_5\mathrm{NOS}]}^{2-}$

(ii) Test for ${\mathrm{NH}}_4^{+}$:  Nessler’s reagent, K₂[HgI₄], form a brown complex (NH₄)₂[HgI₄] with ammonium ions.
(iii) Copper salts give blue colour with ammonia owing to the formation of blue ${\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}$ complex.
(iv)  Potassium ferrocyanide gives a red coloured ppt. of copper ferrrocyanide with copper salts.
${\mathrm{Cu}}^{2+}+\underset{\text{Potassium ferrocyanide}}{{\mathrm{K}}_4\left[\mathrm{Fe}\right(\mathrm{CN}{)}_4]}\to \underset{\text{Red ppt}}{{\mathrm{Cu}}_2\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]}+2{\mathrm{K}}^{+}$
(v) Nickel gives a violet or a brick red ppt. with dimethylglyoxime (DMG).
(vi) Applications in quantitative analysis: Complex compounds have an important use for quantitative estimation of various cations and anions. For example, Ni can be estimated gravimetrically as nickel dimethyl glyoximate. EDTA (ethylenediamminetetraacetic acid) has been used in the estimation of zinc by volumetric methods.

Illustration 6: What is the order of wavelength of absorption in the visible region of the following?
${\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+2},{\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]}^{+2} \text{and }{\left[\mathrm{Ni}{\left({\mathrm{NO}}_2\right)}_6\right]}^{-4}$
 Solution: The increasing field strength of the ligands is in the order of H₂O < NH₃ < NO₂^–
$\therefore$ The energies absorbed for excitation will be in the order
${\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+2}<{\left[\mathrm{NI}{\left({\mathrm{NH}}_3\right)}_6\right]}^{+2}<{\left[\mathrm{Ni}{\left({\mathrm{NO}}_2\right)}_6\right]}^{-4}$
and $E\propto \frac{1}{\lambda } \therefore$ The wavelength order is 
${\left[\mathrm{Ni}{\left({\mathrm{NO}}_2\right)}_6\right]}^{-4}>{\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_6\right]}^{+2}>{\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+2}$

Illustration 7:  ${\left[{\mathrm{NiCl}}_4\right]}^{-2}$  is paramagnetic while Ni(CO)₄ is diamagnetic though both are tetrahedral.  Why?
Solution: In ${\left[{\mathrm{NiCl}}_4\right]}^{-2}, {\mathrm{Cl}}^{-}$ are weak and the electronic configuration of Ni⁺² is

In Ni(CO)₄, the electronic configuration of Ni(0) is

Illustration 8: Explain why Ni(CO)₄ posses tetrahedral geometry, while [Ni(CN)₄]^–2 is square planar.

Solution: 

$\therefore$ Due to rearrangement of electronic configuration, sp³ hybridisation takes place, hence shape is tetrahedral.

$\therefore$ Here, the hybridization is dsp² and hence square planar.

Organometallic Compounds:
Organic compounds in which metal atom is directly linked to carbon atom are known as organometallic compounds. For example, NaC = CNa is an organometallic compound as Na is directly linked to carbon whereas C₂H₅ONa, Ti(OC₂H₅)₄ are not organometallic compounds since the metal atom is linked to carbon through oxygen.
A wide varieties of metals have been used to prepare organometallic compounds, important ones are magnesium, zinc, cadmium, alkali metals (sodium, potassium and lithium) and lead. Some representative organometallic compounds are,
$\begin{array}{rl}& \underset{\mathrm{Ethyimagnesium}bromide}{{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{MgBr},}\underset{\mathrm{Diethyl}zinc}{{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_2\mathrm{Zn},}\underset{ \mathrm{Phenyl}lithium}{{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Li},}\underset{\mathrm{Tetaethyl}lead}{{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_4\mathrm{Pb}}\end{array}$
Alkyl or aryl magnesium halides (RMgX or ArMgX) are also called Grignard reagents.

ClassifIcation of organometallic compounds:
1) Ionic compounds of electropositive metals: These, compounds are mostly formed between the electropositive metals and the carbon compounds which are mostly acidic in nature. Thus, organometallic compounds of alkali metals and alkalme earth metals consist of ions or ion pairs.
$\begin{array}{rl}& \begin{array}{lll}\underset{\mathrm{Alkysodium}}{{\mathrm{R}}^{-}{\mathrm{Na}}^{+}}& \underset{\mathrm{Alky}\mathrm{magnesium}halide}{{\mathrm{R}}^{-}{\mathrm{Mg}}^{2+}{\mathrm{X}}^{-},}& \underset{ \mathrm{Phenyl}lithium}{{\mathrm{C}}_6{\mathrm{H}}_5^{-}{\mathrm{Li}}^{+}}\end{array}\end{array}$
These compounds are normally insoluble in hydrocarbon solvents. They are very reactive towards air and water. The stability of these compounds depends upon the structure of the carbon containing part of the compound.
2) σ – bonded complexes: In a a-bonded complex, a metal and a carbon atom of the ligand are joined together with a sigma bond. This means that the ligand contributes one electron and is therefore, called one electron donor. Tetramethyl tin, (CH₃)₄ Sn and trimethyl aluminium, (CR₃)₃ Al are σ-bonded organometallic compounds.
3) π-bonded complexes: These are organometallic compounds which involve the use of its π-bonds present in organic compounds. For example, Zeise’s salt, ferrocene and dibenzene chromium are organometallic compounds of this type. In all these compounds the π-electrons of the organic compound interact with the metal ion and thus, occupy one of the co-ordination sites. For example, in ferrocene and dibenzene chromium, the iron and chromium atoms are sandwiched between two aromatic rings. The number of

carbon atoms involved in the formation or π-complexes with metals is indicated by the power of h (pronounced as eta). For example, ferrocene is represented as $\mathrm{Fe}{({\eta }^5-{\mathrm{C}}_5{\mathrm{H}}_5)}_2$ indicating that 5-carbon atoms or cyclopentadienyl anion are involved in π-complexation with the metal. Similarly, one can write dibenzene chromium as indicating that all the six carbons of benzene are involved in n-complexation with chromium.

4) Metal carbonyls: These are the complexes where carbon of carbon monoxide donates a pair of electrons to the metal. Nickel carbonyl and iron carbonyl are the common examples.
In metal carbonyls the oxidation state of the metal is zero. These metal carbonyls may be monomeric, bridged or polynuclear.

(i) Bonding in metal carbonyls: The metal-carbon bond in metal carbonyls has σ as well as p character.

(a) σ-Overlap: In a sigma bonded complex, the lone pair of electrons is present on the bonding orbital of carbon monoxide. This bonding orbital containing lone pair interacts with the empty d-orbital of the metal to form a metal-carbon bond as shown below:

(b) π-Overlap: In addition to this, the antibonding orbitals of CO can also overlap with the filled d-orbitals of the metal resulting in back bonding as explained earlier. Thus, metal carbonyls become much more stable compounds due to multiple bonding in them.

Synthesis of organometallic compounds:
Some of the important methods to generate metal-carbon bond are given below:
1) By direct reaction of metals:
$$\begin{gathered} \begin{array}{rl}& \underset{\mathrm{n}\text{-buty bromide}}{\mathrm{n}-{\mathrm{C}}_4{\mathrm{H}}_9\mathrm{Br}}+2\mathrm{i}\stackrel{\text{Ether}}{\to }\underset{\text{n-buty lithium}}{\mathrm{n}-{\mathrm{C}}_4{\mathrm{H}}_9\mathrm{Li}}+2\mathrm{LiCl} \\ 4{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}+\underset{\text{alloy}}{\underset{\text{Sodium-lead}}{4\mathrm{NaPb}}}\to \underset{\mathrm{Tetra} \mathrm{ethyl} \mathrm{lead}}{{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_4\mathrm{Pb}}+4\mathrm{NaCl}+3\mathrm{Pb} \\ \mathrm{Mg}+\mathrm{R}-\mathrm{X}⟶\underset{\mathrm{Grignard} \mathrm{reagent}}{\mathrm{Mg}\begin{array}{c}⋰\mathrm{R}\\ \ddots \mathrm{X}\end{array}} \\ 2\mathrm{Zn}+2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Br}\to {\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_2\mathrm{Zn}+{\mathrm{ZnBr}}_2\end{array} \end{gathered}$$

2) By using an alkylating agent: Grignard reagent and alkyl lithium on reaction with most of the metal and non-metal halides in the presence of ether as solvent yield other organometaflic compounds.

$$\begin{gathered} {\mathrm{PCl}}_3+3{\mathrm{C}}_8{\mathrm{H}}_5\mathrm{MgCl}\stackrel{\mathrm{Ether}}{\to }\underset{\text{Triphenyl phosphine}}{\mathrm{P}{\left({\mathrm{C}}_6{\mathrm{H}}_5\right)}_3}+3{\mathrm{MgCl}}_{2 \\ } \\ \begin{array}{rl}& {\mathrm{SnCl}}_4+4\mathrm{n}-{\mathrm{C}}_4{\mathrm{H}}_9\mathrm{Li}\to \underset{\text{Tetrabutyl tin}}{{(\mathrm{n}-{\mathrm{C}}_4{\mathrm{H}}_9)}_4\mathrm{Sn}}+4\mathrm{LiCl} \\ \begin{array}{rl}{\mathrm{PbCl}}_4+4{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{MgBr}\to & \underset{\text{Tetra ethyl lead}}{{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_4\mathrm{Pb}}+4\mathrm{MgBrCl}\end{array}\end{array} \end{gathered}$$

3) Preparation of π-complexes:
(a) Preparation of ferrocene : Ferrocene is prepared by the reaction of iron oxide with cyclopentadiene at 523 K.

$2{\mathrm{C}}_5{\mathrm{H}}_6+\mathrm{FeO}\stackrel{523\mathrm{K}}{\to }\underset{\text{Ferrocene}}{\left[{\left({\mathrm{C}}_5{\mathrm{H}}_5\right)}_2\mathrm{Fe}\right]}+{\mathrm{H}}_2\mathrm{O}$
It can also be prepared by treatment of cyclopentadienyl magnesium bromide with ferrous chloride.
$\begin{array}{rll}\underset{\text{Magnesium bromide}}{\underset{\text{Cyclopentadieny}}{{\mathrm{C}}_5{\mathrm{H}}_5\mathrm{MgBr}}}& +{\mathrm{FeCl}}_2& \to \underset{\text{Ferrocene}}{\left[\mathrm{Fe}{({\eta }^5-{\mathrm{C}}_5{\mathrm{H}}_5)}_2\right]}+2\mathrm{MgBrCl}\end{array}$
(b) Preparation of Zeise’s salt:
$\underset{\text{Ethene}}{{\mathrm{C}}_2{\mathrm{H}}_4}+\underset{\text{platinate(IV)}}{\underset{\text{Potassium tetrachloro}}{{\mathrm{K}}_2\left[{\mathrm{PtCl}}_4\right]}}\to \underset{\text{Zeise’s salt}}{\mathrm{K}\left[{\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{PtCl}}_3\right]}+\mathrm{KCl}$
(c) Preparation of dibenzene chromium:
$2{\mathrm{C}}_6{\mathrm{H}}_6+\mathrm{Cr}\left(\text{vapour}\right)\stackrel{\text{HighT}}{\to }\underset{\text{Dibenzen e chromium}}{\left[{\left({\mathrm{C}}_6{\mathrm{H}}_6\right)}_2\mathrm{Cr}\right]}$

4) Preparation of metal carbonyls: (a) Nickel carbonyl is formed when finely divided nickel reacts with CO at room temperature.
$\mathrm{Ni}+4\mathrm{CO}\to \mathrm{Ni}(\mathrm{CO}{)}_4$

(b) Iron carbonyl is formed when iron reacts with CO at high pressure and temperature.
$\mathrm{Fe}+5\mathrm{CO}\to \left[\mathrm{Fe}\right(\mathrm{CO}{)}_5]$

Applications of organometallic compounds:
(1) [Ni(CO)₄] is formed during the purification of nickel in metallurgy.
(2) Grignard reagent and several other organometallics are good reagents for the synthesis of compounds.
(3) Ethyl mercury chloride is used to prevent the infection of young plants.
(4) Many organometallic compounds are found to be good catalyst for hydrogenation of unsaturated compounds.
(5) Triethyl aluminium in conjuction with TiCl₄ is a well-known catalyst for Ziegler-Natta polymerisation.
(6) Organoarsenic compounds are used in the treatment of the disease syphilis.
(7) As homogeneous catalyst, e.g., (Ph₃P₃) RhCI (Wilkinson’s catalyst) for hydrogenation of alkane to alkane.
(8) Tetraethyl lead (TEL) is an important anti-knock compound present in leaded gasoline. TEL is a colourless, strongly smelling liquid (b.p, about 200°C), insoluble in water but soluble in ether, gasoline, etc. It is used as an ‘anti-knock’ for motor fuels. When mixed with petrol (1 part in 4000 parts) it raises the octane number of petrol and thus, prevents the knocking in internal combustion engine.
(9) Butyl lithium is an excellent catalyst for anionic polymerisation.

Illustration 9: Out of Zn(CN)₂, Ni(CO)₄, NiCO₃, which is an organo metallic compound?  Why?
Solution: In Ni(CO)₄, Ni is attached to carbon of carbonyl group, NiCO₃ is ionic
$\therefore$Ni(CO)₄ is O.M.C.

Illustration 10: Cr(H₂O)₆]Cl₃ has a magnetic moment of 3.83 B.M.  Write the correct distribution of 3d electrons in the Cr present in the complex.
Solution: Magnetic moment value indicates that there are 3 unpaired e^– s present in Cr.
As the complex is octahedral, the 3 electrons will be present in 3d_xy, 3d_yz, 3d_xz orbitals.
$\therefore$The configuration is  $3{\mathrm{d}}_{xy}^{1}3d_{xy}^{1}3d_{xz}^1$

SOLVED PROBLEMS-1

Prob 1. State the oxidation numbers of the metals in the following complex compounds.

a) ${\left[\mathrm{Co}\left(\mathrm{CN}\right)\left({\mathrm{H}}_2\mathrm{O}\right)(\mathrm{en}{)}_2\right]}^{+2}$                    b) ${\left[{\mathrm{CrCl}}_2(\mathrm{OX}{)}_2\right]}^{-3}$

Sol: a)  x + 1 (-1) + 0 + 2 (0) = + 2 $\Rightarrow$ x = + 3

b) x + 2 (-1) + 2 (-2) = -3        $\left(\because \mathrm{OX}\to –2\right)$

x = + 3

Prob 2. A metal complex having composition Cr (NH₃)₄Cl₂Br has been isolated in two forms A & B.  The A form reacts with AgNO₃ to give a white ppt, readily soluble in dilute aqueous NH₃.  The B form gives a pale yellow ppt., soluble in conc. NH₃.  Write the formulae of A & B forms?

Sol: A form forms white ppt. $\Rightarrow$ Cl is out side the co-ordination sphere so that it forms AgCl (white ppt) and the coordination number of Cr is 6.

$\therefore \mathrm{A}\to \left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_4\mathrm{ClBr}\right] \mathrm{Cl}$

B form gives yellow ppt (AgBr) $\Rightarrow$ Br is out side the coordination, sphere

$\therefore \mathrm{B}\to \left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Br}$

Prob 3. Why hexacyanocomplexes of metals in their +2 oxidation state are usually yellow, but the corresponding hexa aqua complexes are often blue or green?

Sol: CN^– ligand is stronger and hence it generates large energy gap between the two sets of d-orbitals.  Hence they absorb high energy indigo light and appears as yellow.

H₂O ligand generates less energy gap between the two sets of d-orbitals, while the crystal field splitting.  Hence they absorb orange (or) red light and appears as blue (or) grey.

Prob 4. A 0.01m complex of CoCl₃ + NH₃ (1:4 ratio) is found to give freezing point of -0.0372⁰C.  Then suggest the formula of the complex, (K_p of water = 1.86)

Sol: $\Rightarrow$ The depression in freezing point = 0.0372⁰C

$\Rightarrow {\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{iK}}_{\mathrm{f}}\mathrm{m}\Rightarrow 0.0372=\mathrm{i}(1.86)(0.01)$

$\Rightarrow \mathrm{i}=\frac{0.0372}{1.86\times 0.01}=2$

For 100% dissociation, we got I = 2 $\Rightarrow$ two ions are produced from the complex.

$\therefore$ The possible formula is $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}$

Prob 5. Arrange the following in the increasing order of electrical conductivity

$\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}, {\mathrm{K}}_2\left[{\mathrm{PtCl}}_6\right], \left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right], {\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$

Sol: The conductivity of solution depends on the number of ions produced by the complex.

$\therefore$ the order is $\left[\mathrm{Ni}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]<\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl}<{\mathrm{K}}_2\left[{\mathrm{PtCl}}_6\right]<{\mathrm{K}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]$

Prob 6. x, y and z are three complexes of chromium (III) with the molecular formula CrCl₃H₁₂O₆.  All the three complexes have water and chloride ions as ligands.  Complex ‘X’ does not react with conc. H₂SO₄, where as complexes y and z lose 6.75% and 13.5% of their original weight respectively on reacting with conc., H₂SO₄.  Identify x, y & z?

Sol: No reaction with H₂SO₄ $\Rightarrow$ NO H₂O is outside the coordination sphere and coordination number of Cr should be 6.

$\Rightarrow \mathrm{X}\to \left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]{\mathrm{Cl}}_3$

b) For y $\Rightarrow$ Molecular wt. of complex = 266.5

$\Rightarrow$100g complex looses $\to$ 6.75 gr . wt (H₂O)

266.5gr complex $\to$ ?

$=\frac{266.5\times 6.75}{100}=18\mathrm{gm}\left(1{\mathrm{H}}_2\mathrm{O}\text{ molecule }\right)$

$\therefore$ formula of y is $\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{Cl}\right]{\mathrm{Cl}}_2{\mathrm{H}}_2\mathrm{O}$

c)  for z $\Rightarrow$ 100g complex looses $\to$ 13.5gm wt. (H₂O)

$\Rightarrow$ 266.5g complex $\to$ ?

$\Rightarrow \frac{266.5\times 13.5}{100}=36\mathrm{gm}\left(\mathrm{i}.\mathrm{e}.,2{\mathrm{H}}_2\mathrm{O}\text{ molecules }\right)$

$\therefore$ formula of z is $\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_4{\mathrm{Cl}}_2\right]\mathrm{Cl} 2{\mathrm{H}}_2\mathrm{O}$

Prob 7. The EAN of a metal carbonyl M(CO)_x is 36.  The atomic number of metal is 24.  The value of x is?

Sol: EAN = 24 + x (2) = 36  $\Rightarrow$ x = 6

Prob 8. What is Wilkinson’s catalyst?

Sol: A mixture of TiCl₄ and Al(C₂H₅)₃ is known as Wilkinson’s catalyst.

Prob 9. Draw a figure to show the splitting of degenerate of orbitals in an

a) Octahedral crystal field                              b) Tetrahedral crystal field?

Sol: a) In an octahedral crystal field, the ligands will approach the central metal, along the axis.

Hence, e_g orbitals will suffer more electrostatic repulsions and hence their energy will be greater than t_2g orbitals.

b) In tetrahedral field, t_2g orbitals are pointing close to the direction in which ligands are approaching.  As a result energy of t_2g orbitals will be more than that of e_g orbitals.

Prob 10.  How the colour of complex ions can be explained?

Sol: In, complex ion, in the presence of ligand atmosphere, d-orbitals will split into ${\mathrm{t}}_{2\mathrm{g}}\left({\mathrm{d}}_{\mathrm{xy}}, {\mathrm{d}}_{\mathrm{yz}}, {\mathrm{d}}_{\mathrm{xz}}\right)$ set and e_g set  $\left({\mathrm{d}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}, {\mathrm{d}}_{z^2}\right)$ orbitals, each set having different energy.  The electrons which are usually present in the lower set of orbitals absorbs suitable light and are promoted to set having more energy.  These intra d-d transitions are responsible for the colour.

SOLVED PROBLEMS-2

Prob 1: AgCl dissolves in a solution of NH₃ but not in water because

(A)  NH₃ is better solvent than H₂O                 

(B)  Ag⁺ forms a complex ion with NH₃

(C)  NH₃ is a stronger base than H₂O              

(D)  the dipole moment of water is higher than NH₃

Sol: (B) $\mathrm{AgCl}+2{\mathrm{NH}}_4\mathrm{OH}⟶\underset{\mathrm{Soluble}}{\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}\mathrm{Cl}+2{\mathrm{H}}_2\mathrm{O}$

Prob 2: Which will not show geometrical isomerism?

(A) [Co (NH₃)₄]Cl₂           (B) [Co (en)₃]³⁺           (C) [Co (en)₂ Cl₂]Cl           (D) [Cr (NH₃)₄Cl₂]Cl

Sol: (B)  It is of the type MA₃

Prob 3: Which of the following compound is used to detect the presence of  Fe³⁺ ions in solution?

(A) KCN               (B) KNO₃               (C) K₄[Fe (CN)₆]               (D) K₃[Fe (CN)₆]

Sol: (C) It gives ferric ferrocyanide, which is having characteristic color (Prussian blue)

Prob 4: The geometry of Ni (CO)₄ and Ni (PPh₃)₂Cl₂ are

(A) both square planar
(B) tetrahedral and square planar respectively
(C) both tetrahedral
(D) square planar and tetrahedral respectively

Sol: (B) In Ni(CO)₄, sp³ and in Ni (PPh₃)₂Cl₂, dsp² hybridization takes place.

Prob 5: The structure of iron penta carbonyl is

(A) square planar
(B) trigonal bipyramidal
(C) triangular planar
(D) none of the above

Sol: (B) In FeCO₅, Fe undergoes dsp² hybridization.

Prob 6: The two compounds sulphatopentaammine cobalt (III) bromide and sulphato pentaammine cobalt (III) chloride are

(A) linkage isomers
(B) ionization isomers
(C) co-ordination isomers
(D) no isomers

Sol: (D)  Both are not having same ligands.

Prob 7: Formula of ferrocene is

(A) ${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{4–}$          (B) ${\left[\mathrm{Fe}(\mathrm{CN}{)}_6\right]}^{3+}$          (C) $\left[\mathrm{Fe}(\mathrm{CO}{)}_5\right]$          (D) $\left[\mathrm{Fe}({\eta }^2–{\mathrm{C}}_2{\mathrm{H}}_5{)}_2\right]$

Sol: (D) Factual

Prob 8: The total number of possible isomers of the compound [Cu (NH₃)₄] [PtCl₄] are        

(A) 3               (B) 4               (C) 5               (D) 6

Sol: (B)  i.e., $\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]\left[{\mathrm{PtCl}}_4\right], \left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_3\mathrm{Cl}\right] \left[{\mathrm{PtCl}}_3\left({\mathrm{NH}}_3\right)\right]$

$\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_4\right]\left[{\mathrm{CuCl}}_4\right]\text{ and }\left[\mathrm{Pt}{\left({\mathrm{NH}}_3\right)}_3\mathrm{Cl}\right]\left[{\mathrm{CuCl}}_3\left({\mathrm{NH}}_3\right)\right]$

Prob 9:  In case of [Ti (H₂O)₆]³⁺ ion, the net CFSE value will be

(A) –0.4 $∆$_o               (B) –0.6 $∆$ _o               (C) zero               (D) 0.2 $∆$ _o

Sol: (A) One unpaired electron is present in lower energy ${\mathrm{t}}_{2_{\mathrm{g}}}$ orbital