coordination geometry class 12 very brilliant presentation

Theories Related to Coordination Compounds
There are five theories
that describe the
bonding features in
coordination compounds.

Crystal field theory (CFT)
coordination
Werner’s theory
Valence bond theory (VBT)
Bonding in
compounds
Ligand field theory (LFT)
Molecular orbital theory (MOT)

Werner’s Theory
Werner’s coordination
theory was the 1st
attempt to explain the
bonding in coordination
compounds.
1893

Werfier’s Theory
While adding excess of
AgNO3 solution in a series of
compounds of Co(III) chloride
with NH3, the following
results were found:

Werner’s Theory
Because it has an
ionisable valence of 3
CoCl3.6NH3 +
1 mol
AgNO3
Excess
3AgCl
3 mol
Why this gives
3 mol AgCl

Werner’s Theory
CoCl3.5NH3 +
1 mol
AgNO3
Excess
2AgCl
2 mol
Why this gives
2 mol AgCl
Because it has an

Werner’s Theory
CoCl3.4NH3 +
1 mol, (green)
AgNO3
Excess
AgCl
1 mol
Why this gives
1 mol AgCl
Because it has an

Werner’s Theory
CoCl3.4NH3 +
1 mol, (violet)
AgNO3
Excess
AgCl
1 mol
Why this gives
1 mol AgCl
Because it has an
ionisable valency of 1

The atoms within the
square brackets form
a single entity and do
not dissociate under
the reaction conditions.
Conclusion

Conclusion of Werner’s Experiment
Atoms in square brackets do not dissociate.
Colour Formula
Solution
conductivity
corresponds to
Yellow
3+ _
[Co(NH3)6] 3Cl 1:3 electrolyte
Purple
2+ _
[CoCl(NH3)5] 2Cl 1:2 electrolyte
Green
+ _
[CoCl2(NH3)4] Cl 1:1 electrolyte
Violet
+ _
[CoCl2(NH3)4] Cl 1:1 electrolyte

CoCl3.6NH3 CoCl3.5NH3
CoCl3.4NH3 CoCl3.4NH3
Conductance

Werner’s Theory
Primary Secondary
Metals in coordination
compounds show two types of
valence/linkages

Primary Valency
The primary valence are
normally ionisable and are
satisfied by negative ions.
[ Mn (H2O)6 ] Cl2
Primary valency

H2O
H2O
H2O
H2O
H2O
H2O
Cl
Cl
M
Primary valency

The secondary valence are
non-ionisable. These are
satisfied by neutral molecules
or negative ions.
Secondary Valence
A secondary valency is equal to
the coordination number and is
fixed for a metal.

Ligands linked by
secondary valence
M L
Metal Ligand

[ Mn(H2O)6
] Cl2
Secondary valency
Secondary Valency

Cl
Cl
Mn
Secondary valency
H2O
H2O
H2O
H2O
H2O
H2O

The ions/groups bound by the
secondary linkages to the metal
have characteristic spatial
arrangements corresponding to
different coordination numbers.
Secondary Valency
Coordination
polyhedra

Structure of [Co(NH3)6]Cl3
[Co(NH3)6]Cl3

Structure of [Co(NH3)5Cl]Cl2
[Co(NH3)5Cl]Cl2

Structure of [Co(NH3)5Cl2]Cl
[Co(NH3)4Cl2]Cl

Primary valency Secondary valency
It is ionisable It is non ionisable
It is non-directional
It is directional in
nature
It is equal to oxidation
number of central
atom
It is equal to
coordination number
It is denoted by
dotted/dashed line
It is denoted
by solid line

Most common
geometrical shapes in
coordination compounds
Octahedral Tetrahedral Square planar
Werfier’s Theory
3+
[Co(NH3)6] [Ni(CO)4]
2_
[Pt(Cl)4]

Octahedral
Tetrahedral Square Planar

What you will learn
• Hydrate Isomerism
• Polymerisation Isomerism
• Stereoisomerism in Tetrahedral Complexes
• Geometrical and Optical Isomerism in Tetrahedral
Complexes and in Square Planar Complexes
• Geometrical Isomerism in Octahedral Complexes
Stereoisomerism in Complexes

It arises when
counter ion in a
complex salt itself
is a potential
ligand.
So, it can displace
an actual ligand,
which can then
become a counter
ion.
Ionisation Isomerism

[Pt(NH3)3Br] NO2 [Pt(NH3)3NO2] Br
Counter ion

[Co(NH3)5(SO4)] Br [Co(NH3)5Br] SO4
Counter ion

Ionisation isomers
produce different ions in
aqueous solutions.
[Co(NH3)5(SO4)]Br [Co(NH3)5(SO4)]+ + Br
Red
[Co(NH3)5Br]SO4 [Co(NH3)5Br]2+ + SO4
Violet
Note!
2─
—
H2O
H2O

Isomers that differ
with respect to the
number of solvent
ligand molecules as
well as counter ions
Solvate Isomerism

Ammonia (NH3)
1
Water (H2O)
2
Solvent Ligand

When water is present
as a solvent, the
solvate isomerism is
known as hydrate
isomerism.
.
Hydrate Isomerism

Hydrate Isomerism
CrCl3.6H2O [Cr(H2O)6]Cl3
Bright blue green
[Cr(H2O)4Cl2]Cl.2H2O
Dark green
Produces
3Cl─
Produces
2Cl─
[Cr(H2O)5Cl]Cl2.H2O
Produces
1Cl─
Violet

Note!
Solvate and hydrate
isomerism are kind of
ionisation isomerism.

How are the below
compounds related?
Pt(NH3)2Cl2
and
[Pt(NH3)4] [PtCl4]

Polymerisation Isomerism
A special case of coordination
isomerism but not a true isomerism
In which compounds differ from each
other in molecular formula
& molecular weight.

Polymerisation Isomerism
[Pt(NH3)2Cl2] and [Pt(NH3)4] [PtCl4]
1
[Pt(NH3)Cl3] and [Pt(NH3)3Cl2] [PtCl4]
2

The ionisation isomer of [Cr(H2O)4Cl(NO2)]Cl is:
a
b
c
d
[Cr(H2O)4Cl2](NO2)
[Cr(H2O)4(O2N)]Cl2
[Cr(H2O)4Cl(ONO)]Cl
[Cr(H2O)4Cl2(NO2)].H2O
Ionisation isomers are the
complexes that produce different
ions in solution, i.e., they have ions
interchanged inside and outside
the coordination sphere.
[Cr(H2O)4(O2N)]Cl2
and [Cr(H2O)4Cl2](NO2)
have different ions outside the
coordination sphere and they are
isomers. Therefore, they are
ionisation isomers.
Hence, option b is
the correct answer.

Which of the following pairs represent linkage isomers?
a
b
c
d
[Cu(NH3)4][PtCl4] and
[Pt(NH3)4][CuCl4]
[Pd(PPh3)2(NCS)2] and
[Pd(PPh3)2(SCN)2]
[Co(NH3)5(NO3)] SO4 and
[Co(NH3)5(SO4)] NO3
[PtCl2(NH3)4]Br2 and
[PtBr2(NH3)4]Cl2
Linkage isomerism is shown by
ambidentate ligands like NCS and
SCN. It can be linked through N or S.
[Pd(PPh3)2(NCS)2] and
[Pd(PPh3)2(SCN)2]
Hence, option b is
the correct answer.

The type of isomerism present in nitropentammine
chromium(III) chloride is:
a
b
c
d
Linkage
Hydrate
Ionisation
Polymerisation
Hence, option b is
the correct answer.

Structural
isomerism
Isomerism
Stereo
isomerism
Isomerism in Coordination Compounds

Stereoisomerism
Isomerism due
to different spatial
arrangements of
atoms or groups
(ligands) about
the central metal
atom or ion

Geometrical and Optical Isomerism
Coordination number
4 6
Tetrahedral
Square
planar
Octahedral

Coordination number
4 6
Tetrahedral
Square
planar
Octahedral

Generally, GI is not possible in
tetrahedral complexes.
[Ma4] , [Ma2b2], or [Mabcd]
Geometrical Isomerism in Tetrahedral Complexes

Reason
As the arrangement of every
ligand around the central metal
atom/ion in space is equivalent
in every respect

Optical Isomerism in Tetrahedral Complexes
A complex can show optical
isomerism only when the
attached 4 ligands are different.
But POS & COS
must be absent

Optically
inactive
[Be(acac)2]
acetylacetonato
─ ─

─ ─
[Be(acac)2]
benzoylacetonato
Optically
active

[M(AA)2]n±
[M(AA)a2]n±
[Ma4]n±
[Ma3b]n±
GI in Square Planar Complexes
[Ma2bc]n±
[Mabcd]n±
n±
[M(AB)a2]
[Ma2b2]n±

Where
,
a,b,c,d
Simple monodentate
ligands
AA
Symmetrical
bidentate ligand
AB
Unsymmetrical
bidentate ligand
Glycinato
CN─
, Br─
, NH3 etc.
en,
oxalato,
etc.

[M(AA)2]n±
[M(AA)a2]n±
[Ma4]n±
[Ma3b]n±
[Ma2bc]n±
[Mabcd]n±
n±
[M(AB)a2]
[Ma2b2]n±

[Ma2b2]n±
[Ma2bc]n±
[Mabcd]n±
[Ma3b]n±
[M(AA)2]n±
[M(AA)a2]n±
4
[Ma ]n±
2
[M(AB)a ]n±
Do not show
geometrical
isomerism.

Reason
Since the possible
geometry is only one
Simple
monodentate
ligand
Symmetrical
bidentate
ligand
Unsymmetrical
bidentate
ligand

[Ma3b]n±
[M(AB)a2]n±
[M(AA)2]n±
[M(AA)a2]n±
[Ma4]n±
[Ma2b2]n±
[Ma2bc]n±
[Mabcd]n±
Shows
geometrical
isomerism

[Ma2b2]n±
Same group
opposite side
(Trans)
Same group
same side
(cis)

[Ma2bc]n±
Same group
opposite side
(Trans)
Same group
same side
(cis)

Total possible
isomers: 3
[Mabcd]n±

Are there any other
possibilities for
geometrical isomerism
in square planar
complexes?

n±
[M(AB)2]

Cis Trans
[Pt(NH3)2(Cl)2]

[Pt(NH3)BrCl(py)]

[Pt(gly)2]
Cis Trans

Stereo isomerism in
square planar complexes
Geometrical Optical

Optical Isomerism in Square Planar Complexes
Square planar complexes generally
do not show optical isomerism
This is because all the 4 ligands
and metal cations exist
in the same plane.
Have POS

(Isobutylenediamine)
(meso-diphenylethylenediamine)
Palladium(II) or Platinum(II) complex
2+ 2+
Mirror
Where M = Pd(II) or Pt(II)

[Pd(EDTA)]2─
Acts as a
tetradentate ligand

2- 2-
Mirror
Optically active
compounds

Stereo isomerism in
octahedral complexes
Geometrical Optical

Octahedral complexes
where ligands are
Geometrical Isomerism in Octahedral Complexes
Monodentate
Unsymmetrical
bidentate and
monodentate
type
Symmetrical
bidentate and
monodentate
type

where ligands are
Geometrical Isomerism in Octahedral Complexes
Monodentate
Unsymmetrical
bidentate and
monodentate
type
Symmetrical
bidentate and
monodentate
type
Having no
chiral centre

Case 1(a)
6
[M(a) ]n±
Does not show GI
a
a
Octahedral Complexes Having Monodentate Ligands

b
5
[Ma b]n±
Does not show GI
a
Case 1(b)

[Ma4b2]n±
b
a
Cis
b
b
Trans
Possible number of
geometrical isomers
2
=
Case 1(c)

b
a
n±
[Ma2b2c2]
1
Case 1(d)

a
a
a
a
n±
[Ma2b2c2]
2 3
Case 1(d)

b
b
c
c
n±
[Ma2b2c2]
Possible number of
geometrical isomers
= 5
Oct6hedr6l Complexes H6vifig Mofiodefit6te Lig6fids
4 5
Case 1(d)

n±
[Ma3b3]
a a
b
Facial (Fac)
b
Meridional (Mer)
Possible number of
geometrical isomers
2
=
Have three
identical ligands
on one
triangular face
Have three
identical
ligands in a
plane
bisecting the
molecule
Case 1(e)

n±
[Ma3b3]
a a
b
Facial (Fac)
b
Meridional (Mer)
Possible number of
geometrical isomers
= 2
Have three
identical ligands
on one
triangular face
Have three
identical
ligands in a
plane
bisecting the
molecule
Case 1(e)

What you will learn
• Geometrical Isomerism is Octahedral Complexes
Continued
• Optical Isomerism in Octahedral Complexes
• Problems Based on Stereoismerism
• Applications and Importance of Coordination
Chemistry
Stereoisomerism in Octahedral
Complexes

Case 1(a)
n±
[M(a)6]
Does not show GI
a
a

b
n±
[Ma5b]
Does not show GI
a
Case 1(b)

[Ma4b2]n±
b
a
Cis
b
b
Trans
Possible number of
geometrical isomers
2
=
Case 1(c)

Cis Trans

b
a
Case 1(d)
n±
[Ma2b2c2]
1

a
a
a
a
Case 1(d)
n±
[Ma2b2c2]
2 3

b
b
c
c
n±
[Ma2b2c2]
Possible number of
geometrical isomers
= 5
Case 1(d)
4 5

Case 1(e)
3 3
[Ma b ]n±
a a
b
Facial (Fac)
b
Meridional (Mer)
Possible number of
geometrical isomers
= 2
Have three
identical
ligands on one
triangular
face
Have three
identical
ligands in a
plane
bisecting the
molecule

NH3
NH3
NO2
Facial (Fac)
NH3
Meridional (Mer)

Facial (Fac) Meridional (Mer)

Case 1(f)
[M(abcdef)]n±
Possible number of
geometrical isomers
15
=

Case 2(a)
n±
[M(AA)3]
Does not
show GI
A
A
Symmetrical Bidentate and Monodentate Ligands

Case 2(b) [M(AA)2a2]n±
Possible number of
geometrical isomers
= 2
a
a
A
cis
a
trans

cis trans

2+
2+
[Pt(en)2Cl2]2+
cis trans

cis
trans

a
a
a
b
Case 2(c)
2 2
[M(AA)a b ]n±
Possible number of
geometrical isomers
3
=
b
b

c
b
a
b
Case 2(d)
n±
[M(AA)a2bc]
Possible number of
geometrical isomers
= 4
a
a a
c

Case 3(a)
3
[M(AB) ]n±
Possible number of
geometrical isomers
2
=
Unsymmetrical Bidentate and Monodentate Ligands
mer-isomer Fac-isomer

[Co(gly)3]
Possible number of
geometrical isomers
= 2

fac isomer
mer isomer
Ufisymmetric6l Bidefit6te 6fid Mofiodefit6te Lig6fids

Case 3(b)
n±
[M(AB)2a2b2]
Possible number of
geometrical isomers
= 4

where ligands are
Optical Isomerism in Octahedral Complexes
Monodentate
Unsymmetrical
bidentate and
monodentate
type
Symmetrical
bidentate and
monodentate
type

Case 1(a)
Possible number of
enantiomers
0
=
n±
[M(a4b2)]
Does not
show OI

Case 1(b)
a
b
a
a
a
a b
c
Optically
active
Optically
inactive
Optically
inactive
Optically
inactive
Optically
inactive
n±
[M(a2b2c2)]
All possible
GIs
b c

Case 1(b)
a
b
a
b
n±
[Ma2b2c2]
Possible number of
enantiomeric pairs
1
=
Total possible
stereoisomers
6
=
Optical
isomers

where ligands are
Optical Isomerism in Octahedral Complexes
Monodentate
Unsymmetrical
bidentate &
monodentate
type
Symmetrical
bidentate &
monodentate
type
Having no
chiral centre

Case 2(a)
A A
Possible number of
Enantiomeric pairs
= 1
Optical
[M(AA) ]n±
isomers
3
A A
Total possible
stereoisomers
2
=

Case 2(b)
Possible
GI’s
n±
[M(AA)2ab]
a
A
A
Optically
active
b
Optically
inactive

Case 2(b) [M(AA)2ab]n±
Possible number of
Enantiomeric pairs
1
=
A
A
A
A
Optical
isomers
Total possible
stereoisomers
3
=

Case 3(a)
n±
[M(AB)3]
Optically
active
Optically
active
Possible
GI’s

Possible number of
Enantiomeric pairs
2
=
[Co(gly)3]
Optical
isomers
Total possible
stereoisomers
4
=

Ufisymmetric6l Bidefit6te 6fid Mofiodefit6te Lig6fids

Case 3(b)
2 2
[M(AB) a ]n±
A A
A
A
B
A
A
B
Optical
isomers

Case 3(b)
Possible number of
enantiomeric pairs
3
=
A
a
A
a
Optical
isomers
Total possible
stereoisomers
8
=

Ionisation isomers
a
Coordination isomers
b
Geometrical isomers
c
Linkage isomers
d
A reaction of cobalt(III) chloride and ethylene diamine in a
1 : 2 mole ratio generates two isomeric products A (violet
coloured) and B (green coloured). A can show optical activity,
but B is optically inactive. What type of isomers do A and B
represent?

We know that ethylenediamine is a bidentate ligand and Co3+ forms an
octahedral complex having a co-ordination number 6.
Here, 2 moles of ethylenediamine can satisfy four co-ordination numbers.
The remaining two would be satisfied by existing chloride ions.
The reaction is,
CoCl3 +2C2H8N2 → [CoCl2(en)2]Cl
According to a given ratio, the above product can only be formed. As it
says there are two products, another product should be an isomer. The
possibility is that two Cl ions can be either in cis form or in trans-form. On
observing the cis form, there is no plane of symmetry and hence, it is chiral
and optically active, and the trans will be optically inactive.
Hence, they are Geometrical isomers of each other.
Hence, option c is the correct answer.

Both (A) and (B) can be
optically active.
a
Both (A) and (B) can not be
optically active.
b
(A) can not be optically active, but
(B) can be optically active.
c
(A) can be optically active, but
(B) can not be optically active.
d
Consider the complex ions, trans-[Co(en)2Cl2]+ (A) and
cis-[Co(en)2Cl2]+ (B) The correct statement regarding them is:
(A) is trans-form and shows plane of
symmetry, which is optically
inactive.
(B) is cis-form and does not show
plane of symmetry. Hence, it is
optically active.
Hence, option c is the
correct answer.

—
[Pt(NH3)Cl3]
a
b [PtCl2(NH3)2]
[Ni(CO)4]
c
[Co(NO2)3(NH3)3]
d
Which of the following complexes exhibit facial-meridional
geometrical isomerism?

NH3
NH3
Mer-isomer
NH3
NO2
Fac-isomer
Hence, option d is the correct answer.

List I List II
(a) [Co(NH3)6][Cr(CN)6] (i)
Linkage
isomerism
[Co(NH3)3(NO2)3]
[Cr(H2O)6]Cl3
[CrCl2(ox)2]3−
Match List-I with List-II:

a (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
b (a)-(iv), (b)-(ii), (c)-(iii), (d)-(i)
c (a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
d (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
Match List-I with List-II:
(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
[Co(NH3)6][Cr(CN)6]
– Coordination isomerism
[Co(NH3)3(NO2)3] – Linkage isomerism
[Cr(H2O)6]Cl3– Solvate isomerism
[CrCl2(ox)2]3−– Optical isomerism
Hence, option a is the
correct answer.

Biological
systems
Applications and
importance of
coordination
chemistry in Catalyst
Medicinal
chemistry
Other industrial
applications
Analytical
chemistry
Metallurgy

Why the bonds in coordination
compounds have directional
properties?
2
Why coordination compounds have
characteristic magnetic and optical
properties?
3
Why only certain elements possess
the remarkable property of forming
coordination compounds?
1
Limitations of Werner’s Theory

Valence Bond Theory
The valence bond theory,
VBT, was extended to
coordination compounds by
Linus Pauling.
In 1931

Valence Bond Theory
Postulate 1
The formation of complex involves a
reaction between a Lewis base (ligand)
and a lewis acid (metal
or metal ion)
With the formation of a coordinate
covalent (or dative) bonds.

Postulate 2 VBT utilises the concept of
hybridization, in which (n-1)d, ns, np or
ns, np, nd orbitals of metal atom or ion
are hybridised to yield a set of
equivalent orbitals of definite geometry.
Postulate 3
These hybrid orbitals are allowed to
overlap with ligand orbitals that can
donate electron pairs for bonding.
Valence Bond Theory

Valence Bond Theory
Postulate 4
The number of unpaired electrons
measured by the magnetic
moment of the compounds
determines which
d-orbitals are used.

Valence Bond Theory
The hybridisation and
shape of the complexes
can be predicted.
With the help of some
known properties.
Magnetic
moment

Hybridisation of Coordination Compounds
C.N.
of metal
2 3
Type of
hybridisation
sp sp2
Shape of
complex
Linear
Trigonal
planar
Type of
d-orbital
- -

C.N.
of metal
4
Type of
hybridisation
sp3
Shape of
complex
Tetrahedral
Type of
d-orbital
-

Do you know any other
types of hybridisation
with CN=4?

C.N.
of metal
4 4 4
Type of
hybridisation
sp3 dsp2 d3s
Shape of
complex
Tetrahedral
Square
planar
Tetrahedral
Type of
d-orbital
- d x2 – y2 dxy, dyz, dxz

C.N.
of metal
5 5
Type of
hybridisation
sp3d dsp3
Shape of
complex
Trigonal
bipyramidal
Square
pyramidal
Type of
d-orbital
dz2 dx2 – y2

C.N.
of metal
6 6
Type of
hybridisation
sp3d2 d2sp3
Shape of
complex
Octahedral Octahedral
Type of
d-orbital
d x2 – y2 , dz2 d x2 – y2 , dz2

TheExampleso
f
Different
Coordination
Numbers!
Let’sSee

μex = 0
No unpaired
electrons
Diamagnetic
[Ag(NH3)2]
+
Coordination Number = 2

Ag+
4d10 5s0 5p0
=
NH3
:
NH3
:
4d10
sp-hybridised
orbitals
5s0
5p0

Hybridisation sp
=
Shape Linear
=
+
[Ag(NH3)2]

μex = 0
No unpaired
electron
Diamagnetic
_
[Hg(CN)3]

Hybridisation sp2
=
Shape Trigonal planar
=
_
[Hg(CN)3]

μex = 2.5 BM
2 unpaired
electrons
Paramagnetic
_
[Ni(Cl)4]2

Ni2+
= 3d8 4s0 4p0
_
Cl
_
Cl
_
Cl
_
Cl
3d8
sp3-hybridised
orbitals
4s0
4p0

Hybridisation sp3
=
Shape = Tetrahedral
2
_
[Ni(Cl)4]

μex = 0
Diamagnetic
No unpaired
electron
2_
[Ni(CN)4]

Ni2+
3d8 4s0 4p0
=
Inner
orbital or
low-spin
complex
_
CN
_
CN
_
CN
_
CN
dsp2-hybridised
orbitals
3d8
4s0 4p0

dsp2
Hybridisation =
Shape
Square
planar
=
2
_
[Ni(CN)4]

Inner Orbital Complex
In the complex formation, the
Inner d-orbitals are used in
the hybridisation.

Outer Orbital Complex
In the complex formation, the
outer d-orbitals are used in
the hybridisation.

μex = 0
Diamagnetic
No unpaired
electron
VO
3
4
_
Coordifi6tiofi Number = 4

3d0 4s0 4p0
=
V5+
2_
O
2_
O
2_
O
2
_
O
d3s-hybridised
orbitals
3d0 4s0 4p0
Inner
orbital or
low-spin
complex

Hybridisation d3s
=
Shape Tetrahedral
=
Another
example
KMnO4 & K2Cr2O7
=
VO4
3
_

μex = 0
Fe(CO)5
Diamagnetic
No unpaired
electron

Electron pairs from CO
Inner
orbital or
low-spin
complex
Fe 3d 4s 4p
6 2 0
=
0
dsp3-hybridised
orbitals
3d6 4s0 4p0

Fe(CO)5
dsp3
Hybridisation =
Shape =
Trigonal
bipyramidal

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