Chemistry

Module – 24.3

Isomerism

Two or more substances having the same molecular formula but different

structural or spatial arrangement are called isomers.

They are of two types.

a. Structural isomerism

b. Stereo isomerism

a. Structural isomerism:

Two or more complexes have same molecular composition, but different

properties are known as isomerism. Those are called as isomers. The isomers

differ in the arrangement of ligands within the complest are called structural

isomers and the phenomenon is called structural isomerism.

1. Ionization isomerism:

This is due to difference in ionisable groups.

Example: i) [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl

ii) [Pt(NH3)4Cl2]Br2 and [Pt(NH3)4Br2]Cl2

2. Polymerization isomerism:

Compounds having the same emperial formula, but different molecular weights.

Example: [Pt(NH3)2Cl] and [Pt(NH3)4][PtCl4]

3. Hydrate isomerism:

Isomerism differing in the number of water molecules attached the metal ion as

ligands in coordination sphere

Example: [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2H2O

Violet Green

4. Linkage isomerism:

Isomers of this type have different kinds of linkages of a ligand to the central

mental ion.

Example: [Co(NH3)5(NO2)]Cl

5. Coordination isomerism:

This isomerism caudes by the interchange of ligands between the two complex

ions.

Example: [Co(NH3)6][Cr(CN)

6. Ligand isomerism:

Some ligands capable of existing isomerism can give different isomeric complexes.

Stereo isomerism in complex compounds:

Two or more complexes having same molecular composition but different in

arrangement of atoms (or) groups in space is called as stereo isomerism.

It is of two types.

a. Geometrical isomerism

b. Optical isomerism

Stereo isomerism in coordina


)]Cl2 and [Co(NH3)5CoNO]Cl2

Coordination isomerism:


][Cr(CN)6] and [Cr(NH3)6][Co(CN)6]


in complex compounds:



Stereo isomerism in coordination number 6 compounds:






The arrangement of six ligands in a complex around central metal ion in two

different ways. Those are regular hexagon and regular octahedron.

If the complex Ma4b2 has hexagonal structure it can give three isomers

corresponding to (1, 2), (1, 3) and (1, 4) positions of b, while octahedral

arrangement can give only two isomers. X

coordination number 6 complexes exhibit octahedral arrangement by giving two

isomers of Ma4b2 type. Hence coordination number 6 complexes have octahedral

structure.

1. No isomerism is possible in [Ma

Example: [Co(NH3)6]3+

; [Co(NH

2. In [Ma4b2] and [Ma4bc] type complexes can give two isomers. Those are Cis

and trans isomers.

Example: [Co(NH3)4Cl2]+

3. In [Ma3b3] type of complexes can exhibit geometrical isomerism.


different ways. Those are regular hexagon and regular octahedron.

has hexagonal structure it can give three isomers

(1, 2), (1, 3) and (1, 4) positions of b, while octahedral

arrangement can give only two isomers. X – ray analysis, confirmed that


type. Hence coordination number 6 complexes have octahedral

No isomerism is possible in [Ma6] or Ma5b of complexes

; [Co(NH3)5Cl]2+

bc] type complexes can give two isomers. Those are Cis

+ gives Cis and trans isomers.

] type of complexes can exhibit geometrical isomerism.


has hexagonal structure it can give three isomers

(1, 2), (1, 3) and (1, 4) positions of b, while octahedral

ray analysis, confirmed that


type. Hence coordination number 6 complexes have octahedral

bc] type complexes can give two isomers. Those are Cis

] type of complexes can exhibit geometrical isomerism.

Example: [Co(NH3)3Cl3] gives Cis

4. In [M(aa)2b2] type of complexes having two bidentate ligands,

geometrical isomerism.

Example: [Co(en)2Cl2]+ gives Cis and trans isomers.

Optical isomerism:

1. The Cis isomer of [M(aa)

can give two isomers (d and l)

] gives Cis – trans isomers.

] type of complexes having two bidentate ligands, can exhibit

gives Cis and trans isomers.

The Cis isomer of [M(aa)2b2] type complexes has no plane of symmetry. So it

can give two isomers (d and l)

can exhibit

] type complexes has no plane of symmetry. So it

2. Complexes of [M(aa)3] type having three bidetate ligands are also

unsymmetric and gives optical isomerism

Example: [Pt(en)3]4+

gives optical isomers.

Stereo isomerism in coordination numbers “4” compounds:

Complex compounds of coordination number “4” type gives

square planar structure.

1. No isomers are possible [Ma

Example: Ni(Co)4, [Cu(NH

2. [Ma2b2] and [Ma2bc] complexes of some metals like Pt (II), Pd (II), Ni (II), Cu (II)

can give square plannnar

Example: [Pt(NH3)2(CN)2

The geometrical isomers of [Pt(NH

] type having three bidetate ligands are also

unsymmetric and gives optical isomerism

gives optical isomers.

Stereo isomerism in coordination numbers “4” compounds:

Complex compounds of coordination number “4” type gives either tetrahedral or

No isomers are possible [Ma4], [Ma3b] types complexes

, [Cu(NH3)4]2+

bc] complexes of some metals like Pt (II), Pd (II), Ni (II), Cu (II)

can give square plannnar structure. These gives dsp2 hybridisation

2] gives Cis trans isomers

The geometrical isomers of [Pt(NH3)2Cl2] are

] type having three bidetate ligands are also

either tetrahedral or

bc] complexes of some metals like Pt (II), Pd (II), Ni (II), Cu (II)

hybridisation

Some complex compounds of Cu, Zn, Ni can give tetrahedral structure. These

gives sp3 hybridisation

[Mabcd] type of complex gives optical isomerism, if thay have tetrahedral.

Three isomers are possible, if they have square planner

Example: [Pt(NH3)(Py)ClBr] gives three isomers.

Applications of complex compounds:

I. Applications in equilibrium

a. Separation of AgCl from Hg

In the Ist

group silver ions precipitated as white ppt of AgCl. It is soluble in

ammonia due to the formation of [Ag(NH

On the other hand, white p


] type of complex gives optical isomerism, if thay have tetrahedral.

Three isomers are possible, if they have square planner 5 structure.

)ClBr] gives three isomers.

Applications of complex compounds:

Applications in equilibrium analysis:

Separation of AgCl from Hg2Cl2:


ammonia due to the formation of [Ag(NH3)2]Cl

On the other hand, white ppt of Hg2Cl2 turns black on treatment with ammonia.


] type of complex gives optical isomerism, if thay have tetrahedral.


turns black on treatment with ammonia.

b. Separation of Cu and Cd:

Candmium is separated from a mixture of copper and cadmium ion by adding

KCN. Copper ion forms a stable complex with KCN in the solution.

Where as cadmium complex being unstable, decomposes to give Cd

ions gives yellow precipitate with H

c. Test for Fe+2

(ferrous ions) and Fe

Ferrous salt solution gives deep blue colour precipitate with potassium ferry

cyanide. These tests indicate the presence of ferrous ion.

Ferric salt precipitate with potassium ferro

ferric ion.

d. Test for K+ and ion:

Potassium salt solution gives yellow precipitate with sodium cobalt nitrate. Thie

test indicates the presence of potassium ion.

Nessler’s reagent is used for identificati

brown precipitate.

e. Test for nickel ions:

Nickel salts react with dimethyl glyoxime in presence of ammonium hydroxide to

give red precipitate of nickel dimethyl glyoxime

Separation of Cu and Cd:


KCN. Copper ion forms a stable complex with KCN in the solution.

Where as cadmium complex being unstable, decomposes to give Cd

w precipitate with H2S

(ferrous ions) and Fe+3

(ferric ion):


cyanide. These tests indicate the presence of ferrous ion.

Ferric salt precipitate with potassium ferro cyanide. This indicates the presence of

ion:


test indicates the presence of potassium ion.

Nessler’s reagent is used for identification of ammonium ion. It gives reddish


give red precipitate of nickel dimethyl glyoxime


Where as cadmium complex being unstable, decomposes to give Cd+2

ions. These


cyanide. This indicates the presence of


on of ammonium ion. It gives reddish


f. The complex “Ferroin” is used as indicator in some redox titrations.

[Fe(phenanthroline)3]2+

II. Applications in quantitative analysis:

a. Estimation of several cations like Mg

presence of suitable indicator

Example: Erio chrome black

b. Dimethyl glyoxime is used for estimation of nickel.

Example: 8 – hydroxyl quinolene is used for estimation of zinc

c. Hardness of water can be calculated by EDTA

Assignment questions:

1. Give any three applications of complex compounds in

2. List various types of structural isomerism possible for

compounds, giving one example each

3. Draw the structures of optical isomers of

a. [Pt(en)3]4+

b. [Co(en)2Cl2]+

Example set:

1. [Co(NH3)5Br]SO4 and [Co(NH

a. Hydrate isomerism

” is used as indicator in some redox titrations.

Applications in quantitative analysis:

Estimation of several cations like Mg+2

, Zn+2

, Cu+2

, Ni2+

using EDTA as titrant in

presence of suitable indicator

chrome black – T

Dimethyl glyoxime is used for estimation of nickel.

hydroxyl quinolene is used for estimation of zinc

Hardness of water can be calculated by EDTA

Give any three applications of complex compounds in quatitative analysis

List various types of structural isomerism possible for coordination

compounds, giving one example each

Draw the structures of optical isomers of

and [Co(NH3)5SO4]Br exhibit

” is used as indicator in some redox titrations.

using EDTA as titrant in

quatitative analysis

coordination

b. Ionization isomerism

c. Ligand isomerism

d. Co – ordination isomerism

Solution: b)

2. Blue colour/precipitate will be obtained when K4[Fe(CN)6] reacts with

a. Fe (II) ions

b. Cu (II) ions

c. Fe (III) ions

d. Cu (I) ions

Solution: c)

3. Which one of the following square planner complexes will show Cis – trans

isomerism

a. Ma4

b. Ma3b

c. Ma2b2

d. Mabcd

Solution: c)

4. The number of isomers possible for square plannar complex. K2[PdClBr2(SN)]

a. 2

b. 4

c. 5

d. 6

Solution: a)

Problem set:

1. Silver chloride is soluble in ammonia due to the formation of

a. [Ag(NH3)2]2+

b. [Ag(NH3)2]+

c. [Ag(NH3)]+

d. [Ag(NH3)]+2

Solution: d)

2. Geometrical isomerism would be expected form?

a. [Zn(NH3)4]2+

b. [Cu(CN)4]3-

c. [Pt(NH3)2Cl2]

d. [Au(NH3)2]+

Solution: c)

3. [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl]Cl2H2O exhibit

a. Hydrate isomerism

b. Ionization isomerism

c. Ligand isomerism

d. Coordination isomerism

Solution: a)

4. Which of the following octahedral complexes does not show geometrical

isomerism (A and B are monodentate ligands)?

a. Ma3b3

b. Ma4b2

c. Ma5b

d. Ma2b4

Solution: c)

5. Name the type of isomerism exhibited by the following isomers

a. [Cr(NH3)6] [Co(CN)6] and [Co(CN)6][Cr(NH3)6]

b. [Co(Py)2(H2O)2Cl2]Cl and [Co(Py)2(H2O)Cl3]H2O

c. [Pt(NH3)4Br2]Cl2 and [Pt(NH3)4Br2]Cl2

d. [Co(NH3)5NO2]Cl2 and [Co(NH3)5Ono)Cl2

Solution:

a. Coordination isomerism

b. Hydrate isomerism

c. Ionization isomerism

d. Linkage isomerism

6. Draw structures of geometrical isomers [Fe(NH3)2(CN)4]

Solution:

Exercise questions:

1. Why is geometrical isomerism not possible in tetrahedral complexes having

two different types of unidentate ligands coordinated with the central metal

ion?

2. Give evidence the [Co(NH

3. Indicate the types of isomerism exhibited by the following complexes and

draw the structure for these isomers

a. [Co(NH3)5NO2](NO3)2

b. Pt[(NH3)(H2O)Cl2]

4. Platinum (II) forms square planar complexes and platinum (IV) gives octahedral

complexes. How many geometrical isomers are possibl

following complexes? Describe their structures?

a. [Pt(NH3)3Cl]+

b. [Pt(NH3)Cl5]-

c. [Pt(NH3)2ClNO2]

d. [Pt(NH3)4ClBr]+2

Coordination isomerism

Draw structures of geometrical isomers [Fe(NH3)2(CN)4]

Why is geometrical isomerism not possible in tetrahedral complexes having


Give evidence the [Co(NH3)5Cl)SO4 and [Co(NH3)5SO4]Cl are ionization isomers?

isomerism exhibited by the following complexes and

draw the structure for these isomers

Platinum (II) forms square planar complexes and platinum (IV) gives octahedral

complexes. How many geometrical isomers are possible for each of the

following complexes? Describe their structures?

Why is geometrical isomerism not possible in tetrahedral complexes having


]Cl are ionization isomers?

isomerism exhibited by the following complexes and

Platinum (II) forms square planar complexes and platinum (IV) gives octahedral

e for each of the

Solutions to exercise questions:

1. Tetrahedral complexes do not show geometrical isomerism because the

relative positions of the unidentate ligands attached to the central metal atom

are the same w.r.t each other.

2. When ionization isomers are dissolved in water, they ionize to give different

ions. These ions then react differently with different reagents to give different

products.

3.

a. A pair of optical isomers

It can also show linkage isomerism

[Co(NH3)5(NO2)](NO3)

It can also show ionization isomerism

[Co(NH3)5(NO2)](NO3)

b. Geometrical (Cis-, trans

Solutions to exercise questions:

Tetrahedral complexes do not show geometrical isomerism because the

unidentate ligands attached to the central metal atom

are the same w.r.t each other.

When ionization isomers are dissolved in water, they ionize to give different


A pair of optical isomers

It can also show linkage isomerism

)2 and [Co(NH3)5(ONO)](NO3)2

It can also show ionization isomerism

)2 and [Co(NH3)5](NO3)(NO2)

, trans-) isomers of [Pt(NH3)(H2O)Cl2] can exist

Tetrahedral complexes do not show geometrical isomerism because the

unidentate ligands attached to the central metal atom

When ionization isomers are dissolved in water, they ionize to give different


] can exist

4.

a. No isomers are possible for a square planar complex of th

b. No isomers are possible for an octahedral complex of the type Mab

c. Cis and trans isomers are possible for a square planar complex of the type

Ma2bc

d. Cis and trans isomers are possible for an octahedral complex of the type

Ma4bc

No isomers are possible for a square planar complex of the type Ma

No isomers are possible for an octahedral complex of the type Mab

Cis and trans isomers are possible for a square planar complex of the type

and trans isomers are possible for an octahedral complex of the type

type Ma3

No isomers are possible for an octahedral complex of the type Mab5

Cis and trans isomers are possible for a square planar complex of the type

and trans isomers are possible for an octahedral complex of the type

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