9.

COORDINATION COMPOUNDS
9.1 Introduction : A coordination compound ligands depending on the number of electron
consists of central metal atom or ion surrounded donor atoms they have.
by atoms or molecules. For example, a
chemotherapy drug, cisplatin, Pt(NH3)2Cl2, is 9.2.1 Monodentate ligands : A monodentate
a coordination compound in which the central ligand is the one where a single donor atom
platinum metal ion is surrounded by two shares an electron pair to form a coordinate
ammonia molecules and two chloride ions. The bond with the central metal ion. For example:
species surrounding the central metal atom or the ligands Cl , OH or CN attached to metal
ion are called ligands. The ligands are linked have electron pair on Cl, O and N, respectively
directly to central metal ion through coordinate which are donor atoms :
bonds. A formation of coordinate bond occurs Cl O-H C≡N
when the shared electron pair is contributed
by ligands. A coordinate bond is conveniently Use your brain power
represented by an arrow →, where the arrow Draw Lewis structure of the
head points to electron acceptor. The central following ligands and identify the
metal atom or ion usually an electron deficient donor atom in them : NH3, H2O
species, accepts an electron pair while the
9.2.2 Polydentate ligands : A polydentate
ligands serve as electron donors. Coordination
ligand has two or more donor atoms linked
compounds having a metal ion in the centre are
to the central metal ion. For example,
common. In cisplatin two ammonia molecules
ethylenediamine and oxalate ion. Each of these
and two chloride ligands utilize their lone pairs
ligands possesses two donor atoms. These are
of electrons to form bonds with the Pt(II).
bidentate ligands.
NH3 i. Ethylenediamine binds to metal using
Cl Pt NH3 electron pair on each of its two nitrogens.
Cl
The donor nitrogen and chlorine atoms of H 2N NH2
CH2 - CH2
the ligands are directly attached to and form
bonds with platinum. Similarly oxalate ion (C2O4)2 utilizes
electron pair on each of its negatively charged
Can you recall ? oxygen atoms upon linking with the metal.
What are Lewis acids and bases ?
O O
Formation of a coordination compound C-C
O O
can be looked upon as the Lewis acid-base
ii. Ethylenediaminetetraacetate ion
interaction. The ligands being electron pair
(EDTA)4 binds to metal ion by electron pairs
donors are Lewis bases. The central metal ion
of four oxygen and two nitrogen atoms. It is a
being electron pair acceptor serves as Lewis
hexadentate ligand.
acid.
O O
9.2 Types of ligands : The ligands can O - C - H2C CH2 - C - O
be classified as monodentate and polydentate N - CH2 - CH2 - N
O - C - H2C CH2 - C - O
O O
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9.2.3 Ambidentate ligand : The ligands which [Fe(CN)6]4 has charge number of -4. It can be
have two donor atoms and use the electron utilised to calculate O.S. of Fe. Thus,
pair of either donor atoms to form a coordinate charge number of complex = -4
bond with the metal ion, are ambidentate
ligands. For example, the ligand NO2 links to = (O.S. of Fe + charge of ligands)
metal ion through nitrogen or oxygen. = (O.S. of Fe + 6 × charge of CN ion)
O = (O.S. of Fe + 6 × (-1))
M N or M O-N=O Therefore, O.S. of Fe = -4 + 6 = +2.
O
Can you tell ?
Similarly, SCN has two donor atoms A complex is made of Co(III)
nitrogen and sulfur either of which links to and consists of four NH3 molecules
metal depicted as M ← SCN or M← NCS. and two Cl ions as ligands. What is the
9.3 Terms used in coordination chemistry: charge number and formula of complex ion ?
The following terms are used for describing
coordination compounds. 9.3.3 Coordination number (C.N.) of central
9.3.1 Coordination sphere : The central metal metal ion : Look at the complex [Co(NH3)4Cl2]⊕.
ion and ligands linked to it are enclosed in a Here four ammonia molecules and two chloride
square bracket. This is called a coordination ions, that is, total six ligands are attached to
sphere, which is a discrete structural unit. When the cobalt ion. All these are monodentate since
the coordination sphere comprising central each has only one donor atom. There are
metal ion and the surrounding ligands together six donor atoms in the complex. Therefore,
carry a net charge, it is called the complex the coordination number of Co3⊕ ion in the
ion. The ionisable groups shown outside the complex is six. Thus, the coordination number
bracket are the counter ions. For example, of metal ion attached to monodentate ligands
the compound K4[Fe(CN)6] has [Fe(CN)6]4- is equal to number of ligands bound to it.
coordination sphere with the ionisable K⊕ Consider the bidentate ligand C2O42
ions representing counter ions. The compound or ethylenediamine (en). The complexes,
ionizes as : [Fe(C2O4)3]3 and [Co(en)3]3⊕, have three
K4[Fe(CN)6] 4K⊕ + [Fe(CN)6]4 bidentate ligands each. The total donor atoms
in three of ligands is six and the C.N. of Fe3⊕
Try this... and Co3⊕ in these complexes is six.

Can you write ionisation of C.N. of metal ion in a complex is the

[Ni(NH3)6]Cl2? Identify coordination number of ligand donor atoms directly
sphere and counter ions. attached to it or the number of electron
pairs involved in the coordinate bond.
9.3.2 Charge number of complex ion and
oxidation state of metal ion : Use your brain power
The net charge residing on the complex Coordination number used
ion is its charge number. It is algebraic sum in coordination of compounds in
of the charges carried by the metal ion and somewhat different than that used in
the ligands. The charge carried by the metal solid state. Explain
ion is its oxidation state (O.S.). The complex

193
 The secondary valencies for a metal ion are
Can you tell ?
fixed and satisfied by either anions or neutral
What is the coordination
ligands. Number of secondary valencies is
number of Co in [CoCl2(en)2]⊕,
equal to the coordination number.
of Ir in [Ir(C2O4)2Cl2]3⊕ and of Pt in
[Pt(NO2)2(NH3)2]? Postulate (iv) The secondary valencies have
a fixed spatial arrangement around the metal
9.3.4 Double salt and coordination complex ion. Two spheres of attraction in the complex
Combination of two or more stable [Co(NH3)6]Cl3 are shown.
compounds in stochiometric ratio can give two
types of substances, namely, double salt and NH3
H 3N NH3
coordination complexes.
Co Cl3
Double salt : A double salt dissociates in water
H 3N NH3
completely into simple ions. For example (i)
NH3
Mohr's salt, FeSO4(NH4)2SO4.6H2O dissociates
as : coordination ionization
water
FeSO4(NH4)2SO4.6H2O Fe2⊕(aq) (secondary) sphere (primary) sphere
+ 2NH4⊕(aq) +2SO42 (aq)
ii. Carnalite KCl.MgCl2.6H2O dissociates as Remember...
water
When a complex is brought
KCl.MgCl2.6H2O K⊕(aq) +
into solution it does not dissociate
Mg (aq) + 3Cl (aq)
2⊕
into simple metal ions. When [Co(NH3)6]
Coordination complex : A coordination Cl3 is dissolved in water it does not give the
complex dissociates in water with at least test for Co3⊕ or NH3. However, on reacting
one complex ion. For example, K4[Fe(CN)6] with AgNO3 a curdy white precipitate of
dissociates as the complex ion and counter ion. AgCl corresponding to 3 moles is observed.
K4[Fe(CN)6] 4K⊕(aq) + [Fe(CN)6]4
Problem 9.1 : One mole of a purple coloured
(counter ion) (complex ion) complex CoCl3 and NH3 on treatment with
9.3.5 Werner theory of coordination excess AgNO3 produces two moles AgCl.
complexes : The first attempt to explain nature Write the formula of the complex if the
of bonding in coordination compounds was coordination number of Co is 6.
put forth by Werner. The postulates of Werner Solution : One mole of the complex gives
theory are as follows. 2 moles of AgCl. It indicates that two Cl
Postulate (i) Unlike metal salts, the metal in ions react with Ag⊕ ions. The complex has
a complex possesses two types of valencies two ionisable Cl ions. The formula of the
: primary (ionizable) valency and secondary complex is then [Co(NH3)5Cl]Cl2.
(nonionizable) valency.
Postulate (ii) The ionizable sphere consists Can you tell ?
of entities which satisfy the primary valency One mole of a green coloured
of the metal. Primary valencies are generally complex of CoCl3 and NH3 on
satisfied by anions. treatment with excess of AgNO3
produces 1 mole of AgCl. What is the
Postulate (iii) The secondary coordination
formula of the complex ? (Given : C.N. of
sphere consists of entities which satisfy the
Co is 6)
secondary valencies and are non ionizable.
194
A complex with coordination number six has anionic sphere complex. For example,
octahedral structure. When four coordinating [Ni(CN)4] and K3[Fe(CN)6] have anionic
2

groups are attached to the metal ion the coordination sphere; [Fe(CN)6]3 and
⊕
complex would either be with square planar three K ions make the latter electrically
or tetrahedral structure. neutral.
9.4 Classification of complexes: The iii. Neutral sphere complexes : A neutral
coordination complexes are classified coordination complex does not possess cationic
according to types of ligands and sign of or anionic sphere. [Pt(NH3)2Cl2] or [Ni(CO)4]
charge on the complex ion. have neither cation nor anion but are neutral
sphere complexes.
9.4.1 Classification on the basis of types of
ligands
i. Homoleptic complexes : Consider Use your brain power
[Co(NH3)6]3⊕. Here only one type of ligands Classify the complexes
surrounds the Co3⊕ ion. The complexes in as cationic, anionic or neutral.
which metal ion is bound to only one type of Na4[Fe(CN)6], Co(NH3)6Cl2,
ligands are homoleptic. Cr(H2O)2(C2O4)23 , PtCl2(en)2 and
Cr(CO)6.
ii. Heteroleptic complexes : Look at the
complex [Co(NH3)4Cl2]⊕. There are two types 9.5 IUPAC nomenclature of coordination
of ligands, NH3 and Cl attached to Co3⊕ compounds : Tables 9.1, 9.2 and 9.3 summarize
ion. Such complexes in which metal ion is the IUPAC nomenclature of coordination
surrounded by more than one type of ligands compounds.
are heteroleptic.
Rules for naming coordination compounds
recommended by IUPAC are as follows:
Use your brain power
1. In naming the complex ion or neutral
Classify the complexes molecule, name the ligand first and then the
as homoleptic and heteroleptic metal.
[Co(NH3)5Cl]SO4, [Co(ONO)(NH3)5]Cl2 2. The names of anionic ligands are obtained
[CoCl(NH3)(en)2]2⊕ and [Cu(C2O4)3]3 . by changing the ending -ide to -o and -ate to
-ato.
9.4.2 Classification on the basis of charge on 3. The name of a complex is one single word.
the complex There must not be any space between
i. Cationic complexes : A positively charged different ligand names as well as between
coordination sphere or a coordination ligand name and the name of the metal.
compound having a positively charged 4. After the name of the metal, write its
coordination sphere is called cationic sphere oxidation state in Roman number which
complex. appears in parentheses without any space
For example: the cation [Zn(NH3)4]2⊕ and between metal name and parentheses.
[Co(NH3)5Cl]SO4 are cationic complexes. The 5. If complex has more than one ligand of the
latter has coordination sphere [Co(NH3)5Cl]2⊕; same type, the number is indicated with
the anion SO42 makes it electrically neutral. prefixes, di-, tri-, tetra-, penta-, hexa- and
ii. Anionic sphere complexes : A so on.
negatively charged coordination sphere or a 6. For the complex having more than one type
coordination compound having negatively of ligands, they are written in an alphabetical
charged coordination sphere is called order. Suppose two ligands with prefixes

195
 Table 9.1 : IUPAC names of anionic and neutral ligands
Anionic ligand IUPAC name Anionic ligand IUPAC name
Br , Bromide Bromo CO3 , Carbonate
2
Carbonato
Cl , Chloride Chloro OH , Hydroxide Hydroxo
F , Fluoride Fluoro C2O42 , Oxalate Oxalato
I Iodide Iodo NO2 , Nitrite Nitro (For N - bonded ligand)
CN , Cyanide Cyano ONO , Nitrite Nitrito(For O-bonded ligand)
Thiocyanato (For ligand do-
SO42 , Sulphate Sulphato SCN , Thiocyanate
nor atom S)
Isothiocyanato (For ligand
NO3 , Nitro Nitrato NCS , Thiocyanate
donor atom N)
Neutral ligand IUPAC name Neutral ligand IUPAC names
Ammine (Note the
NH3, Ammonia H2O, water Aqua
spelling)
CO, Carbon monoxide Cabonyl en, Ethylene diamine Ethylenediamine
Table 9.2 : IUPAC names of metals in anionic complexes
Metal IUPAC name Metal IUPAC name
Aluminium, Al Aluminate Chromium, Cr Chromate
Cobalt, Co Cobaltate Copper, Cu Cuprate
Gold, Au Aurate Iron, Fe Ferrate
Manganese, Mn Maganate Nickel, Ni Nickelate
Platinum, Pt Platinate Zinc, Zn Zincate
Table 9.3 : IUPAC names of some complexes
Complex IUPAC name
i. Anionic complexes :
a.[Ni(CN)4]2 Tetracyanonickelate(II) ion
b. [Co(C2O4)3]3 Trioxalatocobaltate(III) ion
c. [Fe(CN)6]4 Hexacyanoferrate(II) ion
ii. Compounds containing complex anions and metal cations :
a. Na3[Co(NO2)6] Sodium hexanitrocobaltate(III)
b. K3[Al(C2O4)3] Potasium trioxalatoaluminate(III)
c. Na3[AlF6] Sodium hexafluoroaluminate(III)
iii. Cationic complexes :
a. Cu(NH3)42⊕ Tetraamminecopper(II) ion
b.[Fe(H2O)5(NCS)]2⊕ Pentaaquaisothiocyanatoiron(III) ion,
c.[Pt(en)2(SCN)2]2⊕ Bis(ethylenediamine)dithiocyanatoplatinum(IV).

iv. Compounds containing complex cations and anion :

a. [PtBr2(NH3)4]Br2 Tetraamminedibromoplatinum(IV) bromide,
b. [Co(NH3)5CO3]Cl Pentaamminecarbonatocobalt(III) chloride,
c. [Co(H2O)(NH3)5]I3 Pentaammineaquacobalt(III) iodide.

v. Neutral complexes :
a. [Co(NO2)3(NH3)3] Triamminetrinitrocobalt(III)
b. Fe(CO)5 Pentacarbonyliron(0)
c. [Rh(NH3)3(SCN)3] Triamminetrithiocyanatorhodium(III)

196
 are tetraaqua and dichloro. While naming rule states that a metal ion continues to accept
in alphabetical order, tetraaqua is first and electrons pairs till it attains the electronic
then dichloro. configuration of the next noble gas. Thus if the
7. If the name of ligand itself contains EAN is equal to 18 (Ar), 36 (Kr), 54 (Xe), or
numerical prefix then display number 86 (Rn) then the EAN rule is obeyed.
by prefixes with bis for 2, tris for 3, EAN can be calculated with the following.
tetrakis for 4 and so forth. Put the ligand Formula
name in parentheses. For example, EAN = number of electrons of metal ion + total
(ethylenediamine)3 or (en3) would appear number of electrons donated by ligands
as tris(ethylenediamine) or tris(ethane-1, = atomic number of metal (Z) - number
2-diamine). of electrons lost by metal to form the
8. The metal in cationic or neutral complex ion (X) + number of electrons donated
is specified by its usual name while in the by ligands (Y).
anionic complex the name of metal ends =Z-X+Y
with 'ate'. Cosider Co[NH3]63⊕
Oxidation state of Cobalt is +3, six ligands
Try this... donate 12 electrons.
Write the representation of Z = 27; X = 3; Y = 12
EAN of Co3⊕ = 27 - 3 + 12 = 36.
• Tricarbonatocobaltate(III) ion.
• Sodium hexacyanoferrate(III). Try this...

• Potassium hexa cyanoferrate (II) Find out the EAN of

Zn(NH3)4 ,[Fe(CN)6]
2⊕ 4

• Aquachlorobis(ethylenediamine)
cobalt(III). Cr(CO)6 and [Fe(CN)6]4 are some examples
• Tetraaquadichlorochromium(III) of coordination compounds which obey
chloride. the EAN rule. Certain other coordination
compounds however do not obey the EAN
• Diamminedichloroplatinum(II).
rule. For example, [Fe(CN)6]3 and Cu[NH3]42⊕
9.6 Effective Atomic Number (EAN) Rule : have EAN 35.
An early attempt to explain the stability
of coordination compounds was made by Use your brain power
Sidgwick who proposed an empirical rule Do the following complexes follow
known as effective atomic number (EAN) rule. the EAN rule ? Cr(CO)4, Ni(CO)4,
EAN equals total number of electrons around Mn(CO)5, Fe(CO)5.
the central metal ion in the complex. EAN
Isomers

Stereo isomers Constitutional or structural

isomers

Geometric or Optical isomers or

Distereoisomers enantiomers

Linkage isomers Coordination isomers Ionization isomers Solvate isomers

Fig. 9.1 : Classification of isomers in coordination compounds

197
9.7 Isomerism in coordination compounds : Here the cis isomer is more soluble in
One of the interesting aspects of water than the trans isomer. The cis isomer
coordination chemistry is existence of isomers. named cisplatin is an anticancer drug while
Isomers are different compounds that have the trans isomer is physiologically inactive.
the same molecular formula. Their chemical The cis isomer is polar with non-zero dipole
reactivities and physical properties such moment. The trans isomer has zero dipole
as colour, solubility and melting point are moment as a result of the two opposite Pt - Cl
different. and two Pt-NH3 bond moments, which cancel
Broadly speaking, isomers are classified each other.
into two types namely stereoisomers and [Pt(NH3)(H2O)Cl2] (MA2BC type)
constitutional (or structural) isomers as Cl H2O Cl H2O
displayed in Fig. 9.1. Pt Pt
9.7.1 Stereoisomers : Stereoisomers have the Cl NH3 H3N Cl
same links among constituent atoms however cis isomer trans isomer
the arrangements of atoms in space are Four coordinate tetrahedral complexes do
different. not show cis and trans isomers.
There are two kinds of stereoisomers in iii. Cis and trans isomers in octahedral
coordination compounds: (a) geometric complexes : The octahedral complexes of
isomers or distereoisomers and (b) enantiomers the type MA4B2, MA4BC and M(AA)2B2 exist
or optical isomers. as cis and trans isomers. (AA) is a bidentate
ligand.
a. Geometric isomers or distereoisomers :
[Co(NH3)4Cl2]⊕, (MA4B2 type)
These are nonsuperimposable mirror image ⊕ ⊕
isomers. These are possible in heteroleptic Cl Cl
H3N Cl H3N NH3
complexes. In these isomers, there are cis and Co Co
trans types of arrangements of ligands. H3N NH3 H3N NH3
NH3 Cl
Cis-isomers : Identical ligands occupy
cis isomer trans isomer
adjacent positions.
Trans-isomer : Identical ligands occupy the [Pt(NH3)4ClBr]2⊕, (MA4BC type)
opposite positions.
2⊕ 2⊕
Cis and trans isomers have different NH3 Cl
Cl NH3 H3N NH3
properties. Cis trans isomerism is observed in
square planar and octahedral complexes. Pt Pt
Br NH3 H 3N NH3
i. Cis and trans isomers in square planar NH3 Br
complexes : The square planar complexes of cis isomer trans isomer
MA2B2 and MA2BC type exist as cis and trans
isomers, where A, B and C are monodentate [Co(en)2Cl2]⊕, (M(AA)2B2 type)
ligands, M is metal. For example : Pt(NH3)2Cl2,
(MA2B2 type) Cl ⊕
Cl ⊕
Cl
Cl en Co en Co en
NH3 Cl NH3 en Cl
Pt Pt
Cl NH3 H 3N Cl cis isomer trans isomer
cis isomer trans isomer
198
 2⊕ 2⊕
Try this... Cl Cl Cl
Cl
Draw structures of the cis and en Pt Pt en
trans isomers of [Fe(NH3)2(CN)4]
en en
d cis isomer l
b. Optical isomers (Enantiomers) :
2⊕
The complex molecules or ions that are Cl
nonsuperimposable mirror images of each
other are enantiomers. The nonsuperimposable en Pt en
mirror images are chiral. (A more elaborate Cl
discussion on chirality and optical isomerism
trans isomer
is included in Chapter 10.)
Enantiomers have identical properties Square planar complexes do not show
however differ in their response to the plane- enantiomers since they have mirror plane and
polarized light. The enantiomer that rotates axis of symmetry.
the plane of plane-polarized light to right Try this...
(clockwise) is called the dextro (d) isomer, 1. Draw enantiomers of
while the other that rotates the plane to left [Cr(ox)3]3
(anticlockwise) is called laevo (l) isomer.
2. Draw enantiomers and cis and trans
i. Optical iomers in octahedral complexes isomers of [Cr(H2O)2(ox)2] (where
[Co(en)3]3⊕ ox = C2O4 )
2

3⊕ 3⊕
en en 9.7.2 Structural isomers (Constitutional
en Co Co en isomers) : Structural isomers possess different
linkages among their constituent atoms and
en en have, their chemical formulae to be the same.
d mirror l They can be classified as linkage isomers,
ionization isomers, coordination isomers and
solvate isomers.
Remember...
a. Linkage isomers : These isomers are formed
Our hands are non
when the ligand has two different donor atoms.
superimposable mirror images.
It coordinates to the metal via different donor
When you hold your left hand upto a
atoms. Thus the nitrite ion NO2 having two
mirror the image looks like right hand.
donor atoms show isomers as :
[Co(NH3)5(NO2)]2⊕ and [Co(NH3)5(ONO)]2⊕
The nitro complex has Co-N bond and the
ii. Octahedral complexes existing as both nitrito complex is linked through Co-O bond.
geometric and optical isomers These are linkage isomers.
[PtCl2(en)2]2⊕
Can you tell ?
In this type of complex, only the cis isomer
exists as pair of enantiomers Write linkage isomers of
[Fe(H2O)5SCN]⊕. Write their IUPAC
names.

199
b. Ionization isomers : Ionization isomers the free solvent molecule. I and II represent
involve exchange of ligands between solvate (hydrate) isomers.
coordination and ionization spheres. For 9.8 Stability of the coordination compounds:
example:
The stability of coordination compounds
[Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4 can be explained by knowing their stability
(I) (II) constants. The stability is governed by metal-
In compound I, anion SO42 , bonded to ligand interactions. In this the metal serves
Co is in the coordination sphere while Br is as electron-pair acceptor while the ligand as
in the ionization sphere. In compound II, anion Lewis base (since it is electron donor). The
Br is in the coordination sphere linked to Co metal-ligand interaction can be realized as the
while SO42 is in the ionisation sphere. These Lewis acid-Lewis base interaction. Stronger
complexes in solution ionize to give different the interaction greater is stability of the
ions. complex.

[Co(NH3)5SO4]Br [Co(NH3)5SO4]⊕ + Br Consider the equilibrium for the metal

ligand interaction :
[Co(NH3)5Br]SO4 [Co(NH3)5Br]2⊕+ SO42
Ma⊕ + nLx [MLn]a⊕ + nx
I and II are examples of ionization isomers.
where a, x, [a⊕ + nx ] denote the charge on the
Can you tell ? metal, ligand and the complex, respectively.
Now, the equilibrium constant K is given by
Can you write IUPAC names of
[MLn]a⊕ + nx
isomers I and II? K=
[Ma⊕][Lx ]n
c. Coordination isomers : Coordination Stability of the complex can be explained in
isomers show interchange of ligands between terms of K. Higher the value of K larger is the
cationic and anionic spheres of different metal thermodynamic stability of the complex.
ions. For example : The equilibria for the complex formation
[Co(NH3)6] [Cr(CN)6] [Cr(NH3)6] [Co(CN)6] with the corresponding K values are given
(cationic) (anionic) (cationic) (anionic) below.
(I) (II) Ag⊕ + 2CN [Ag(CN)2] K = 5.5 ×1018

In isomer I, cobalt is linked to ammine Cu2⊕ + 4CN [Cu(CN)4]2 K = 2.0 ×1027

ligand and chromium to cyanide ligand. In Co3⊕ + 6NH3 [Co(NH3)6]3⊕ K = 5.0 ×1033
isomer II the ligands coordinating to metals
From the above data, [Co(NH3)6]3⊕ is more
are interchanged. Cobalt coordinates with
stable than [Ag(CN)2] and [Cu(CN)4]2 .
cyanide ligand and chromium to NH3 ligand.
I and II are examples of coordination isomers. 9.8.1 Factors which govern stability of the
complex : Stability of a complex is governed
d. Solvate isomers (Hydrate isomers when
by (a) charge to size ratio of the metal ion and
water is solvent) : These are similar to
(b) nature of the ligand.
ionization isomers. Look at the complexes.
a. charge to size ratio of the metal ion
[Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2 .H2O
(I) (II) Higher the ratio greater is the stability.
In compound I the solvent water is directly For the divalent metal ion complexes their
bonded to Cr. In compound II, H2O appears as stability shows the trend : Cu2⊕ > Ni2⊕ > Co2⊕ >
Fe2⊕ > Mn2⊕ > Cd2⊕. The above stability order

200
is called Irving-William order. In the above list iii. The number of vacant hybrid orbitals
both Cu and Cd have the charge +2, however, formed is equal to the number of ligand
the ionic radius of Cu2⊕ is 69 pm and that of donor atoms surrounding the metal ion
Cd2⊕ is 97 pm. The charge to size ratio of Cu2⊕ which equals the coordination number of
is greater than that of Cd2⊕ . Therefore the Cu2⊕ metal.
forms stable complexes than Cd2⊕. iv. Overlap between the vacant hybrid
b. Nature of the ligand. orbitals of the metal and filled orbitals of
the ligand leads to formation of the metal-
A second factor that governs stability
ligand coordinate bonds.
of the complexes is related to how easily the
ligand can donate its lone pair of electrons to v. The hybrid orbitals used by the metal ion
the central metal ion that is, the basicity of the point in the direction of the ligand.
ligand. The ligands those are stronger bases vi. The (n-1)d or nd orbitals used in
tend to form more stable complexes. hybridisation allow the complexes to be
classified as (a) outer orbital and (b) inner
Use your brain power orbital complexes.
The stability constant K of the vii. For hybridisation in the outer orbital
[Ag(CN)2] is 5.5 × 1018 while that complex nd orbitals are used, whereas
for the corresponding [Ag(NH3)2]⊕ is 1.6 × in the inner orbital complexes (n-1)d
107. Explain why [Ag(CN)2]2 is more stable. orbitals are used.
Type of hybridisation decides the structure
9.9 Theories of bonding in complexes : of the complex. For example when the
The metal-ligand bonding in coordination hybridisation is d2sp3 the structure is octahedral.
compounds has been described by Valence Steps to understand the metal-ligand bonding
bond theory (VBT) and Crystal field theory include :
(CFT). i. Find oxidation state of central metal ion
9.9.1 Valence bond theory (VBT) ii. Write valence shell electronic configuration
of metal ion.
Can you recall ?
iii. See whether the complex is low spin or
What is valence bond theory and
high spin. (applicable only for octahedral
the concept of Hybridisation?
complexes with d4 to d8 electronic
The hybridized state is a theoretical step that configurations).
describes how complexes are formed. VBT iv. From the number of ligands find the
is based on the concept of hybridization. The number of metal ion orbitals required for
hybrid orbitals neither exist nor can be detected bonding.
spectroscopically. These orbitals, however, v. Identify the orbitals of metal ion
help us to describe structure of coordination available for hybridisation and the type of
compounds. The steps involved in describing hybridisation involved.
bonding in coordination compounds using the vi. Write the electronic configuration after
VBT are given below. hybridisation.
i. Metal ion provides vacant d orbitals
vii Show filling of orbitals after complex
for formation of coordinate bonds with
formation.
ligands.
viii.Determine the number of unpaired
ii. The vacant d orbitals along with s and
electrons and predict magnetic behaviour
p orbitals of the metal ion take part in
of the complex.
hybridisation.
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 vi. Six orbitals available for hybridisation
Remember...
are two 3d, one 4s, three 4p orbitals
Complete the missing entries.
3d 4s 4p
Coordination Geometry Hybridization
number of complex
2 sp
d2sp3
4 Tetrahedral
The orbitals for hybridization are decided
Square from the number of ammine ligands which
4
planar is six. Here (n-1)d orbitals participate in
6 d2sp3/ sp3d2 hybridization since it is the low spin complex.
vii. Electronic configuration after complex
formation.
Try this...
3d 4s 4p
Give VBT description of
bonding in each of following
complexes. Predict their magnetic
behavior. d2sp3

a. [ZnCl4]2 3⊕
b. [Co(H2O)6]2⊕ (high spin) NH3
c. [Pt(CN)4]2 (square planar) H3N NH3
Co
d. [CoCl4]2 (tetrahedral) H3N NH3
e. [Cr(NH3)6]3⊕ NH3

9.9.2 Octahedral, complexes viii. As all electrons are paired the complex is
a. [Co(NH3)6]3 ⊕ low spin diamagnetic.
i. Oxidation state of Cobalt:3⊕ b. [CoF6]3 high spin
ii. Valence shell electronic configuration i.Oxidation state of central metal Co is 3+
of Co3⊕ is represented in box diagram as ii.Valence shell electronic configuration of
shown below : Co3⊕ is
3d 4s 4p 3d 4s 4p

iii. Number of ammine ligands is 6, number iii. Six fluoride F ligands, thus the number of
of vacant metal ion orbitals required for vacant metal ion orbitals required for bonding
bonding with ligands must be six. with ligands would be six.
iv. Complex is low spin, so pairing of electrons iv. Complex is high spin, that means pairing
will take place prior to hybridisation. of electrons will not take place prior to
v. Electronic configuration after pairing would hybridisation. Electronic configuration would
be remain the same as in the free state shown
3d 4s 4p above.

202
v. Six orbitals available for the hybridisation. one 4s, three 4p. The complex is tetrahedral.
Those are one 4s, three 4p, two of 4d orbitals 3d 4s 4p
3d 4s 4p 4d

sp3
sp3d2
The four metal ion orbitals for bonding with
Six metal orbitals after bonding with six Cl ligands are derived from the sp3
F ligands led to the sp3d2 hybridization. The d hybridization.
orbitals participating in hybridisation for this vi. Four vacant sp3 hybrid orbital of Ni2⊕
complex are nd. overlap with four orbitals of Cl ions.
vi. Six vacant sp3d2 hybrid orbital of Co3+ vii. Configuration after complex formation
overlap with six orbitals of fluoride forming would be.
Co - F coordinate bonds. 3d 4s 4p
vii. Configuration after complex formation.
3d 4s 4p 4d
sp3
viii.The complex has four unpaired electrons
and hence, paramagnetic.
sp3d2
viii. The complex is octahedral and has four 2
Cl
unpaired electrons and hence, is paramagnetic.
Ni Cl
3 Cl
F
F F Cl
Co 9.9.4 Square planar complex
F F [Ni(CN)4]2
F i. Oxidation state of nickel is +2
ii. Valence shell electronic configuration of
Ni2⊕
9.9.3 Tetrahedral complex
3d 4s 4p
[Ni(Cl)4]2
i. Oxidation state of nickel is +2
ii. Valence shell electronic configuration of
Ni2+ iii. Number of CN ligands is 4, so number
3d 4s 4p of vacant metal ion orbitals required for
bonding with ligands would be four.
iv. Complex is square planar so Ni2⊕ ion uses
iii. number of Cl ligands is 4. Therefore dsp2 hybrid orbitals.
number of vacant metal ion orbitals v. 3d electrons are paired prior to the
required for bonding with ligands must be hybridisation and electronic configuration
four. of Ni2⊕ becomes :
iv. Four orbitals on metal available for
hybridisation are 3d 4s 4p

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vi. Orbitals available for hybridisation are In an isolated gaseous metal ion the five d
one 3d, one 4s and two 4p which give dsp2 orbitals, d x2-y2 ,dz2, dxy,dyzd,zx have the same
hybridization. energy i.e. they are degenerate.
vii. Four vacant dsp2 hybrid orbitals of Ni2⊕ ii. When ligands approach the metal ion they
overlap with four orbitals of CN ions to form create crystal-field around the metal ion. If
Ni - CN coordinate bonds. it were symmetrical the degeneracy of the d
vii. Configuration after the complex formation orbitals remains intact.
becomes. Usually the field created is not symmetrical
3d 4s 4p
and the degeneracy is destroyed. The d orbitals
thus split into two sets namely, (dxy, dyz, dxz)
usually refered by t2g and ( d x2-y2 ,dz2) called
dsp2
as eg. These two sets of orbitals now have
viii.The complex has no unpaired electrons
different energies. A separation of energies of
and hence, dimagnetic.
these two sets of d orbitals is the crystal field
2 splitting parameter. This is denoted by Δo (O
for octahedral).
NC CN iii. The Δo depends on strength of the ligands.
Ni The ligands are then classified as (a) strong
NC CN field and (b) weak field ligands. Strong field
ligands are those in which donor atoms are
C,N or P. Thus CN , NC , CO, NH3, EDTA,
Try this...
en (ethylenediammine) are considered to be
Based on the VBT predict strong ligands. They cause larger splitting of
structure and magnetic behavior of d orbitals and pairing of electrons is favoured.
the [Ni(NH3)6]3⊕ complex. These ligands tend to form low spin complexes.
Weak field ligands are those in which donor
9.9.5 Limitations of VBT atoms are halogens, oxygen or sulphur. For
i. It does not explain the high spin or low spin example, F , Cl , Br , I , SCN , C2O42 . In
nature of the complexes. In other words, strong case of these ligands the Δo parameter is
and weak field nature of ligands can not be smaller compared to the energy required for
distinguished. the pairing of electrons, which is called as
ii. It does not provide any explanation for the electron pairing energy. The ligands then can
colour of coordination compounds. be arranged in order of their increasing field
iii. The structure of the complexes predicted strength as
from the VBT would not always match
I < Br < Cl < S2 < F < OH < C2O42
necessarily with those determined from the
<H2O<NCS <EDTA< NH3,< en< CN < CO.
experiments.
To overcome these difficulties in VBT, the Let us understand splitting of d orbitals and
Crystal field theory has been proposed which formation of octahedral complexes
has widely been accepted. In octahedral environment the central metal
9.9.6 Crystal Field theory (CFT) ion is surrounded by six ligands.
C.F.T. is based on following assumptions Ligands approach the metal ion along the x, y,
i. The ligands are treated as point charges. z axes. As the ligands approach the metal ion
The interaction between metal ion and ligand the degeneracy of d orbitals is resolved.
is purely electrostatic or there are no orbital
interactions between metal and ligand.
204
 eg

eg - The higher energy set of orbitals (dz2 and dx2 - y2 )

t2g - The lower energy set of orbitals (dxy,dyz and dzx)
3/5 Δ0
Energy

Δ0 or 10 Dq - The energy separation between the two

Δ0 levels
The eg orbitals are repelled by an amount of 0.6 Δ0.
t2g The t2g orbitals to be stabilized to the extend of 0.4 Δ0.
2/5 Δ0

Fig. 9.2 : Crystal field Splitting in an octahedral complex

With closer approach of ligands along the 9.9.7 Factors affecting Crystal Field Splitting
axes, the doubly degenerate dx2-y2 and dz2 (eg) parameter (∆0)
orbitals experience larger repulsion than the a. The magnitude of crystal field splitting
triply degenerate, t2g orbitals. As shown in Fig. depends on strength of the ligands.
9.2, the eg set has higher energy than the t2g set The strong ligands those appear in
by the amount Δo. The Δo parameter is equal spectrochemical series approach closer to
to 10 Dq units of splitting of t2g and eg levels. the central metal which results in a large
For electronic configurations d1, d2, d3 the crystal field splitting.
electrons occupy t2g orbitals and obey the
Hund’s rules. For electronic configurations d1, b. Oxidation state of the metal : A metal with
d2, d3 and d8, d9, d10 the high spin and low spin the higher positive charge is able to draw
configurations cannot be distinguished. Only ligands closer to it than that with the lower
the electronic configurations d4 to d7 render one. Thus the metal in higher oxidation
the high and low spin complexes.These are state results in larger separation of t2g and
depiciated in Fig 9.5. eg set of orbitals. The trivalent metal ions
Table 9.5 : d orbital diagrams for high spin cause larger crystal field splitting than
and low spin complexes. corresponidng divalent ones.
d orbital 9.9.8 Colour of the octahedral complexes
electronic High spin Low spin
configuration As discussed above, the formation of
an octahedral complex is accompanied by
d4 eg
splitting of d orbitals into t2g and eg sets. A
t2g separation of these two sets of orbitals is Δo,
which can be measured from experiments.
d5 eg The Δo corresponds to a certain frequency
of electromagnetic radiation usually in the
t2g visible region. A colour complementary to the
absorbed frequency is thus observed. Consider
d6 eg
the [Ti(H2O)6]3⊕ complex. The central metal
ion titanium has electronic configuration 3d
t2g
and the electron occupies one of the t2g orbitals
d7 eg (Figure 9.3).

t2g

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eg eg Try this...
Energy Sketch qualitatively crystal
t2g t2g field d orbital energy level diagrams
Ground state Excited state for each of the following complexes:
Fig. 9.3 : d - d transition in d' system a. [Ni(en)3]2⊕ b. [Mn(CN)6]3
The absorption of the wavelength of light c. [Fe(H2O)6]2⊕
corresponding to Δo parameter promotes an Predict whether each of the complexes is
electron from the t2g level. Such energy gap in diamagnetic or paramagnetic.
case of the [Ti(H2O)6]3⊕ complex is 20,300 cm-1
(520 nm, 243 kJ/mol) and a complimentary The dxy, dyz, dzx orbitals with their lobes
colour to this is imparted to the complex. A lying in between the axes point toward the
violet color of the [Ti(H2O)6]3⊕ complex arises ligands. On the other hand, dx2-y2 and dz2
from such d-d transition. orbitals lie in between metal-ligand bond
9.9.9 Tetrahedral complexes axes. The dxy, dyz and dzx orbitals experience
more repulsion from the ligands compared to
A pattern of splitting of d orbitals, that by dx2-y2 and dz2 orbitals.
which is a key in the crystal field theory, is
dependent on the ligand field environment. Due to larger such repulsions the dxy, dyz
This is illustrated for the tetrahedral ligand and dzx orbitals are of higher energy while the
field environment. dx2-y2 and dz2 orbitals are of relatively lower
z energy.
Each electron entering in one of the
dxy,dyz and dzx orbitals raises the energy by
4 Dq whereas that accupying d x2-y2 and dz2
orbitals lowers it by 6 Dq compared to the
M x
energy of hypothetical degenerate d orbitals
in the ligand field.
A splitting of d orbitals in tetrahedral
crystal fields (assumed to be 10 Dq) thus is
y much less (typically 4/9) compared to that
Fig. 9.4 : Tetrahedral structure for the octahedral environment. The crystal
The tetrahedral structure having the field splitting of d orbitals in a tetrahedral
metal atom M at the centre and four ligands ligand field is compared with the octahedral
occupying the corners is displayed along with one in Fig. 9.5. Thus the pairing of electrons
in Fig. 9.4. is not favoured in tetrahedral structure. For
example, in d4 configuration an electron
Tetrahedral Octahedral
would occupy one of the t2g orbitals. The low
eg
spin tetrahedral complexes thus are not found.
t2g Typically metal complexes possessing
Δtet Δo the cetral metal ion with d8 electronic
eg
configuration, for example, Ni(CO)4, favours
Free-ion t2g the tetrahedral structure.

Fig. 9.5 : Splitting of d orbitals in tetrahedral

and octahedral complexes
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9.10 Applications of coordination c. To estimate hardness of water : Hardness
compounds of water is due to the presence of Ca2+ and
a. In biology : Several biologically important Mg2⊕ ions. The ligand EDTA forms stable
natural compounds are metal complexes. They complexes with Ca2⊕ and Mg2⊕. It can,
play important role in a number of processes therefore, be used to estimate hardness.
occuring in plants and animals. d. In electroplating : Usually stable
For example, chlorophyll present in coordination complexes on dissolution
plants is a complex of Mg. Haemoglobin dissociate to small extent and furnish a
present in blood is a complex of iron. controlled supply of metal ions. The metal
ions when reduced clump together to form
b. In medicines the clusters or nanoparticles. When the
i. Pt complex cisplatin is used in the treatment coordination complexes are used the ligands
of cancer. in the complex keep the metal atoms well
separated from each other. These metal atoms
ii. EDTA is used for treatment of lead
tend to form a protective layer on the surface.
poisoning.
Certain cyanide complexes K[Ag(CN)2]
and K[Au(CN)2] find applications in the
electroplating of these noble metals.

Exercises
1. Choose the most correct option.
i. The oxidation state of cobalt ion in the 3. [PtCl2Br2]2 (square planar)
complex [Co(NH3)5Br]SO4 is
4. [FeCl2(NCS)2]2 (tetrahedral)
a. +2 b. +3
c. +1 d. +4 a. 1 and 3 b. 2 and 3
ii. IUPAC name of the complex c. 1 and 3 d. 4 only
[Pt(en)2(SCN)2] is
2+
v. Which of the following complexes are
a. bis (ethylenediamine chiral ?
dithiocyanatoplatinum (IV) ion
1. [Co(en)2Cl2]⊕ 2. [Pt(en)Cl2]
b. bis (ethylenediamine)
dithiocyantoplatinate (IV) ion 3. [Cr(C2O4)3]3 4. [Co(NH3)4Cl2]⊕
c. dicyanatobis (ethylenediamine) a. 1 and 3 b. 2 and 3
platinate IV ion
d. bis (ethylenediammine)dithiocynato c. 1 and 4 d. 2 and 4
platinate (IV) ion vi. On the basis of CFT predict the number of
iii. Formula for the compound sodium unpaired electrons in [CrF6]3 .
hexacynoferrate (III) is a. 1 b. 2 c. 3 d. 4
a. [NaFe(CN)6] b. Na2[Fe(CN)6] vii. When an excess of AgNO3 is added to the
c. Na[Fe(CN)6] d. Na3[Fe(CN)6] complex one mole of AgCl is precipitated.
The formula of the complex is
iv. Which of the following complexes exist
as cis and trans isomers ? a. [CoCl2(NH3)4]Cl

1. [Cr(NH3)2Cl4] 2. [Co(NH3)5Br]2⊕ b. [CoCl(NH3)4] Cl2

207
 c. [CoCl3(NH3)3] v. Write formulae of the following complexes
d. [Co(NH3)4]Cl3 a. Potassium amminetrichloroplatinate
viii. The sum of coordination number and (II)
oxidation number of M in [M(en)2C2O4]Cl b. Dicyanoaurate (I) ion
is vi. What are ionization isomers ? Give an
a. 6 b. 7 c. 9 d. 8 example.
2. Answer the following in one or two vii. What are the high-spin and low-spin
sentences. complexes ?
i. Write the formula for viii. [CoCl4]2 is tetrahedral complex. Draw its
tetraammineplatinum (II) chloride. box orbital diagram. State which orbitals
ii. Predict whether the [Cr(en)2(H2O)2]3+ participate in hybridization.
complex is chiral. Write structure of its ix. What are strong field and weak field
enantiomer. ligands ? Give one example of each.
iv. Name the Lewis acids and bases in the x. With the help of crystal field energy-
complex [PtCl2(NH3)2]. level diagram explain why the complex
v. What is the shape of a complex in which [Cr(en)3]3⊕ is coloured ?
the coordination number of central metal 4. Answer the following questions.
ion is 4 ?
i. Give valence bond description for the
vi. Is the complex [CoF6] cationic or anionic bonding in the complex [VCl4] . Draw
if the oxidation state of cobalt ion is +3 ? box diagrams for free metal ion. Which
vii. Consider the complexes [Cu(NH3)4][PtCl4] hybrid orbitals are used by the metal  ?
and [Pt(NH3)4] [CuCl4]. What type of State the number of unpaired electrons.
isomerism these two complexes exhibit? ii. Draw qualitatively energy-level
viii. Mention two applications of coordination diagram showing d-orbital splitting in
compounds. the octahedral environment. Predict
the number of unpaired electrons in the
3. Answer in brief. complex [Fe(CN)6]4 . Is the complex
i. What are bidentate ligands ? Give one diamagnetic or paramagnetic? Is it
example. coloured? Explain.
ii. What is the coordination number and iii. Draw isomers in each of the following
oxidation state of metal ion in the complex a. Pt(NH3)2ClNO2
[Pt(NH3)Cl5] ?
b. Ru(NH3)4Cl2
iii. What is difference between a double salt
and a complex ? Give an example. c. [Cr(en2)Br2]⊕

iv. Classify following complexes as iv. Draw geometric isomers and enantiomers
homoleptic and heteroleptic of the following complexes.

[Cu(NH3)4]SO4, [Cu(en)2(H2O)Cl]2⊕, a. [Pt(en)3]4⊕ b. [Pt(en2)ClBr]2⊕

[Fe(H2O)5(NCS)]2⊕, tetraammine zinc (II)
nitrate.

208
v. What are ligands ? What are their types ? vii. How stability of the coordination
Give one example of each type. compounds can be explained in terms of
vi. What are cationic, anionic and neutral equilibrium constants ?
complexes? Give one example of each. viii. Name the factors governing the
equilibrium constants of the coordination
compounds.

Activity :
1. The reaction of chromium metal with H2SO4 in the absence of air gives blue
solution of chromium ion.
Cr(s) + 2H⊕(aq) Cr2⊕(aq) + H2(s)
Cr2⊕ forms octahedral complex with H2O ligands.
a. Write formula of the complex
b. Describe bonding in the complex using CFT and VBT.
Draw crystal field splitting and valence bond orbital diagrams.
2. Reaction of complex [Co(NH3)3(NO2)3 with HCl gives a complex
[Co(NH3)3H2OCl2]⊕ in which two chloride ligands are trans to one another.
a. Draw possible stereoisomers of starting material
b. Assuming that NH3 groups remain in place, which of two starting isomers would give
the observed product ?

209