x

5
x y

Coordination
y z z

Compounds
dxy dxz z dyz
x

dx 2−y2 dz 2

Shapes of d orbitals.
5.1 | DOUBLE SALTS AND Contents
COORDINATION COMPOUNDS
5.1 Double Salts and
Addition compounds are formed when stoichiometric Coordination Compounds
amounts of two or more stable compounds join together. 5.2 Werner’s Work
For example: 5.3 More Recent Methods of
Studying Complexes
KCl + MgCl 2 + 6H 2 O → KCl MgCl 2 6H 2 O
r r

5.4 Classification of Ligands

carnallite 5.5 Effective Atomic Numbers
K 2 SO4 + Al 2 (SO4 )3 + 24H 2 O → K 2 SO4 Al 2 (SO4 )3 24H 2 O
r r
5.6 Shapes of d Orbitals
5.7 Bonding in Transition
potassium alum
Metal Complexes
CuSO4 + 4NH 3 + H 2 O → CuSO4 4NH 3 H 2 O
r r 5.8 Valence Bond Theory
tetrammine copper(II) 5.9 Crystal Field Theory
sulphate monohydrate 5.10 Effects of Crystal Field
Splitting
Fe(CN)2 + 4KCN → Fe(CN)2 4KCN r
5.11 Tetragonal Distortion of
potassium ferrocyanide Octahedral Complexes
(Jahn-Teller Distortion)
Addition compounds are of two types:
5.12 Square Planar
1. Those which lose their identity in solution (double salts) Arrangements
2. Those which retain their identity in solution (complexes) 5.13 Tetrahedral Complexes
5.14 Magnetism
When crystals of carnallite are dissolved in water, the solution
5.15 Extension of the Crystal
shows the properties of K+, Mg2+ and Cl− ions. In a similar way,
a solution of potassium alum shows the properties of K+, Al3+ Field Theory to Allow for
and SO2− Some Covalency
4 ions. These are both examples of double salts which
exist only in the crystalline state. 5.16 Nomenclature
When the other two examples of coordination com- of Coordination
pounds dissolve they do not form simple ions − Cu2+, or Fe2+ Compounds
and CN− − but instead their complex ions remain intact. 5.17 Isomerism
Thus the cuproammonium ion [Cu(H2O)2(NH3)4]2+ and the

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172 Chapter 5 Coordination Compounds

ferrocyanide ion [Fe(CN)6]4− exist as distinct entities both in the solid and in solution. Complex ions are
shown by the use of square brackets. Compounds containing these ions are called coordination com-
pounds. The chemistry of metal ions in solution is essentially the chemistry of their complexes. Transition
metal ions, in particular, form many stable complexes. In solution ‘free’ metal ions are coordinated either to
water or to other ligands. Thus Cu2+ exists as the pale blue complex ion [Cu(H2O)6]2+ in aqueous solution
(and also in hydrated crystalline salts). If aqueous ammonia is added to this solution, the familiar deep
blue cuproammonium ion is formed:

[Cu(H 2 O)6 ]2 + + 4NH 3 [CoCl 3 r 4NH 3 → 1AgCl  [Cu(H 2 O)2 (NH 3 )4 ] + 4H 2 O

2+

Note that this reaction is a substitution reaction, and the NH3 replaces water in the complex ion.

5.2 | WERNER’S WORK

Werner’s coordination theory in 1893 was the first attempt to explain the bonding in coordination com-
plexes. It must be remembered that this imaginative theory was put forward before the electron had been
discovered by J.J. Thompson in 1896, and before the electronic theory of valency. This theory and his pains-
taking work over the next 20 years won Alfred Werner the Nobel Prize for Chemistry in 1913.
Complexes must have been a complete mystery without any knowledge of bonding or structure. For
example, why does a stable salt like CoCl3 react with a varying number of stable molecules of a compound
such as NH3 to give several new compounds: CoCl3 ∙ 6NH3, CoCl3 ∙ 5NH3 and CoCl3 ∙ 4NH3? What are their
structures? At that time X-ray diffraction, which is the most powerful method of determining the structures
of crystals, had yet to be discovered. Werner did not have at his disposal any of the modern instrumental tech-
niques, and all his studies were made using simple reaction chemistry. Werner was able to explain the nature of
bonding in complexes, and he concluded that in complexes the metal shows two different sorts of valency:
1. Primary valencies: These give rise to first order compounds (e.g. CoCl3 from cobalt and chlorine) and
correspond to the valency (now known as oxidation state) of the central atom. These are non-direc-
tional. In modern terms, the number of primary valen-
cies corresponds to the number of charges on the central
metal atom. The complex [Co(NH3)6]Cl3 actually exists
as [Co(NH3)6]3+ and 3Cl −. Thus the primary valency is
three. Similarly, in case of [Co(H2O)4Cl2]Cl, the primary
valency is also three.
Planar hexagon
2. Secondary valencies: These give rise to the higher order
compounds (e.g. CoCl3⋅nNH3 from CoCl3 and NH3).
These are directional. In modern terms the number of
secondary valencies equals the number of ligand atoms
coordinated to the metal. This is now called the coordina-
tion number. Ligands are commonly negative ions such
as Cl−, or neutral molecules such as NH3. Less commonly,
ligands may be positive ions such as NO+. Each metal has Trigonal prism
a characteristic number of secondary valencies. Thus in
[Co(NH3)6]Cl3 the three Cl− are held by primary valen-
cies. The six NH3 groups are held by secondary valencies.
Werner also attempted to find the shapes of the complexes.
The possible arrangements of six groups round one atom
are a planar hexagon, a trigonal prism, and an octahedron
(Figure 5.1). Werner then compared the number of isomeric Octahedron
forms he had obtained with the theoretical number for each Figure 5.1 Possible geometric shapes
of the possible shapes (Table 5.1). for six-coordination.

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 5.3 More Recent Methods of Studying Complexes 173

Table 5.1 Number of isomers predicted and actually found

Complex Observed Predicted
Octahedral Planar hexagon Trigonal prism
[MX6] 1 1 1 1
[MX5Y] 1 1 1 1
[MX4Y2] 2 2 3 3
[MX3Y3] 2 2 3 3

These results strongly suggested that X

these complexes have an octahedral shape. X X
Only one form
This proof was not absolute proof, as it was M
just possible that the correct experimental MX6 X X
X
conditions had not been found for preparing
Y X
all the isomers. More recently the X-ray struc- Only one form
X X X Y as all six corners
tures have been determined, and these estab- M M
MX5Y are equivalent
lish that the shape is octahedral (Figure 5.2). X X X X
More recently, with a bidentate ligand X X
such as ethylenediamine (1,2-diaminoethane), Y Y
two optically active isomers of octahedral X Y X X Two isomers cis
M M
complexes have been found (Figure 5.3). MX4Y2
and trans
X X X X
In a similar way, Werner studied a range X Y
of complexes which included [PtII(NH3)2Cl2] cis trans
Y Y
and [PdII(NH3)2Cl2]. The coordination number Two isomers and
X X X
is 4, and the shape could be either tetrahedral M
Y
M facial (fac) and
MX3Y3
or square planar. Werner was able to prepare X Y X Y meridional (mar)
two different isomers for these complexes. X Y
A tetrahedral complex can only exist in one Fac- Mer-

form, but a square planar complex can exist Figure 5.2 Isomers in octahedral complexes.
in two isomeric forms. This proved these com-
plexes are square planar rather than tetrahe- Cl NH3
dral (Figure 5.4).
Pt

Cl NH3
M(en)3
cis
Cl NH3

Pt

d form l form NH3 Cl

Mirror trans
Figure 5.3 Optical isomerism in octahedral Figure 5.4 Isomerism in square
complexes. planar complexes.

5.3 | MORE RECENT METHODS OF STUDYING COMPLEXES

The electrical conductivity of a solution of an ionic material depends on:
1. The concentration of solute.
2. The number of charges on the species which are formed on dissolution.
Molar conductivities relate to a 1 M solution and thus the concentration factor is removed. The total number
of charges on the species formed when the complex dissolves can be deduced by comparison of its molar

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174 Chapter 5 Coordination Compounds

conductivity with that of known simple ionic materials (Table 5.2). These conductivities suggest the same
structures for the cobalt/ammonia/chlorine complexes mentioned earlier, as do the results from Werner’s
AgCl experiments, shown in Table 5.3.

Table 5.2 Conductivities of salts and complexes (molar conductivities

measured at 0.001 M concentration)
ohm−1 cm2 mol−1
LiCl → Li+ + Cl− (total of 2 charges) 112.0
−
CaCl2 → Ca + 2Cl 2+
(total of 4 charges) 260.8
CoCl3 · 5NH3 261.3
CoBr3 ∙ 5NH3 257.6
−
LaCl3 → La + 3Cl 3+
(total of 6 charges) 393.5
CoCl3 · 6NH3 431.6
CoBr3 · 6NH3 426.9

Table 5.3 Number of charges related to modern and Werner structures

Charges Primary valency Ionizable chlorines Secondary valency
−
3+
[Co(NH3)6] 3Cl 6 3 3 6NH3 =6
[Co(NH3)5Cl] 2Cl2+ −
4 3 2 5NH3 + 1 Cl− = 6
[Co(NH3)4Cl2]+ Cl− 2 3 1 4NH3 + 2Cl− = 6

The freezing point of a liquid is lowered when a chemical substance is dissolved in it. Cryoscopic mea-
surements involve measuring how much the freezing point is lowered. The depression of freezing point
obtained depends on the number of particles present. Cryoscopic measurements can be used to find if a
molecule dissociates, and how many ions are formed. If a molecule dissociates into two ions it will give
twice the expected depression for a single particle. If three ions are formed this will give three times the
expected depression. Thus:

LiCl → Li + + Cl − (2 particles) ⎡(2 charges)⎤

MgCl 2 → Mg 2 + + 2Cl − (3 particles) ⎢⎢(4 charges)⎥⎥
LaCl 3 → La 3+ + 3Cl − (4 particles) ⎢⎣(6 charges)⎥⎦

The number of particles formed from a complex molecule determines the size of the depression of
freezing point. Note that the number of particles formed may be different from the total number of charges
which can be obtained from conductivity measurements. These two types of information can be used
together to establish the structure (Table 5.4).

Table 5.4 Establishing the structure of complexes

Formula Cryoscopic measurement Molar conductivity Structure

CoCl3 ∙ 6NH3 4 particles 6 charges [Co(NH3)6]3+ 3Cl−

CoCl3 ∙ 5NH3 3 particles 4 charges [Co(NH3)5Cl]2+ 2Cl−
CoCl3 ∙ 4NH3 2 particles 2 charges [Co(NH3)4Cl2] + Cl−
CoCl3 ∙ 3NH3 1 particle 0 charge [Co(NH3)3Cl3]
Co(NO2)3 ∙ KNO2 ∙ 2NH3 2 particles 2 charges K+ [Co(NH3)2(NO2)4]−
Co(NO2)3 ∙ 2KNO2 ∙ NH3 3 particles 4 charges 2K+ [Co(NH3)(NO2)5]2−
Co(NO2)3 ∙ 3KNO2 4 particles 6 charges 3K+ [Co(NO2)6]3−

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 5.4 Classification of Ligands 175

The magnetic moment can be measured and this provides information about the number of unpaired
electron spins present in a complex. From this it is possible to decide how the electrons are arranged and
which orbitals are occupied. Sometimes the structure of the complex can be deduced from this. For exam-
ple, the compound NiII(NH3)4(NO3)2 ∙ 2H2O might contain four ammonia molecules coordinated to Ni in
a square planar [Ni(NH3)4]2+ ion and two molecules of water of crystallization and have no unpaired elec-
trons. Alternatively the water might be coordinated to the metal, giving an octahedral [Ni(H2O)2(NH3)4]2+
complex with two unpaired electrons. Both these complex ions exist and their structures can be deduced
from magnetic measurements.
Dipole moments may also yield structural information but only for non-ionic complexes. For example,
the complex [Pt(NH3)2Cl2] is square planar, and can exist as cis or trans forms. The dipole moments from
the various metal–ligand bonds cancel out in the trans configuration. However, a finite dipole moment is
given by the cis arrangement (Figure 5.4).
Electronic spectra (UV and visible) also provide valuable information on the energy of the orbitals, and
on the shape of the complex. By this means it is possible to distinguish between tetrahedral and octahedral
complexes, and whether the shape is distorted or regular.
The most powerful method, however, is the X-ray determination of the crystal structure. This provides
details of the exact shape and the bond lengths and angles of the atoms in the structure.

5.4 | CLASSIFICATION OF LIGANDS

There are several ways to classify ligands and these are discussed as follows:
1. Based upon charges
a. Neutral ligands: H 2O, NO, CO, C 6 H 6, etc.
+ +

b. Positive ligands: NO and NH - NH 2 3
– – – –
c. Negative ligands: Cl , NO , CN , OH
2

2. Based upon denticity of the ligand

The number of donations accepted by a central atom from a particular ligand is known as the denticity
of the ligand. Ligands may be classified as follows based on their denticity:
a. Monodentate: Only one donation is accepted from the ligand. For example, H 2O, NO, CO, NH 3 ,
CO23 − , Cl − , etc.
b. Bidentate: Two donations are accepted from the ligand. For example,
(i) en: ethylenediamine (ii) pn: propylenediamine
CH3 CH CH2
CH2 CH2
NH2 NH2
H2N NH2

(iii) tn: trimethylenediamine (iv) bn: butylenediamine

CH2 CH2 CH2 CH3 CH CH CH3

NH2 NH2 H2N NH2

(v) ox 2− : oxalate (vi) acac −: acetylacetonate

−
O O
C O O−

C C C
O O
− H3C CH CH3

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176 Chapter 5 Coordination Compounds

(vii) dmg −: dimethylglyoximate (viii) gly − : glycinate

−
CH3 O
C N O
CH2 C
C N
NH2 O−
H3C O H

(ix) dipy: dipyridyl (x) o-phen: ortho- phenanthroline

N N N N

(xi) N, N′- Diethylthiocarbamate ion (xii) Salicylaldehyde anion

H
S
C
Et2N C O

S O−

(xiii) 8-Hydroxyquinolinol ion (oxine) (xiv) o-Phenylenebisdimethylarsine (diarsine)

As(CH3)2
N
O−
As(CH3)2

c. Tridentate: Three donations are accepted from the ligand. For example,
(i) dien: diethylenetriamine (ii) imda 2−: iminodiacetate
NH
CH2 CH2 CH2 CH2

H2C NH CH2 C C O
O

H2N NH2 O− O −

d. Tetradentate: Four donations are accepted from the ligand. For example,
(i) trien: triethylenetetraamine (ii) NTA 3− : nitrilotriacetate
CH2 CH2 O CH2
CH2
CH2 NH CH2 C N C O
CH2
H2N NH O− O−

CH2 C
H2N
CH2 O O
−

e. Pentadenate: Five donations are accepted from the ligand. For example,
EDTA 3−: ethylenediamine triacetate
CH2 CH2 CH2
CH2
N N C O
O C
CH2 H
O− O−

C
O O−

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 5.4 Classification of Ligands 177

f. Hexadentate: Six donations are accepted from the ligand. For example,
EDTA 4− : ethylenediamine tetracetate
CH2 CH2 CH2
CH2
N N C O
O C
CH2 CH2
O O−
−
C
C
O O−
O− O

Note:
1. Flexidentate : A ligand which shows variable denticity is called a flexidentate ligand.
For example,
(i)
O O O O

S and S
O O O O
Bidentate ion Monodentate ion

(ii) [Co(NH3)4CO3]Br and [Co(NH3)5CO3]Br (iii) NO−3 , CH3COO−, S2O2− 3

CO2−
3 is bidentate CO2−3 is monodentate
may also act as flexidentate
ligand
2. Chelating : A ligand that can form a ring structure with the central atom is called a chelating ligand.
All polydentate ligands are the example of chelating ligands.
Chelated complexes are more stable than similar complexes with monodentate ligands as dissociation
of the complex involves breaking two bonds rather than one.
However, it must be noted that NH 2 NH 2 and N ( CH 2 − CH 2 ) 3 N cannot act as chelating ligands due
to the formation of three membered ring and locked structure, respectively.
3. Ambidentate ligand : A ligand that may have more than one kind of donor sites but at a time only one
kind of donor site is utilized for donation is called as ambidentate ligand. Ambidentate ligand may be
of two types.
a. Monodentate and ambidentate:
C N O C N S C N O
or or
:

or .N
C N S C N
:

O:
O C N
or
O
O O .
O S :N
:

N or N N O O:
S
S− S or
or O O N O
− or
O
O S +
N :C O:
S or
S
+
O O :C O:

b. Bidentate and ambidentate:

S O− S− O
C C
or
C C
−
S O− S O
dithiooxalate

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178 Chapter 5 Coordination Compounds

3. Based upon bonding interaction between the ligand and the central atom
a. Classical or simple donor ligand: These ligands only donate the lone pair of electrons to the central
atom. For example, O2 − , OH − , F − , NH 2− , NH 3 , N 3−, etc.
b. Non-classical or p - acid or p - acceptor ligand: These ligands not only donate the lone pair of elec-
trons to the central atom but also accept the electron cloud from the central atom in their low-lying
vacant orbitals. This kind of back donation is known as ‘synergic effect’ or ‘synergic bonding’. For
− +
example, CO, C N, NO, PF3 , PR 3 (R = H, Et, Ph…), C 2 H 4 , C 2 H 2 , CO2, etc.
(i) In case of CO, the back donation to the p * orbital of central atom may be depicted as:

p∗

M C O M C O

p∗

By valence bond or molecular orbital theory, it is well understood that the bond order of C − O
bond decreases but the C − O bond length must increase due to synergic effect. Similarly, as
− +
C N and NO are isoelectronic with CO, so back donation takes place in these species also in
the p * orbitals and the same conclusion can be drawn for the bond order and bond lengths.
(ii) In case of PR3, the back donation may be depicted as:

vacant 3d orbital
accepts the back
donation

M P R3

(iii) In case of C 2 H 4 , the back donation may be depicted using the example of Zeise’s salt.

H
Cl Cl p∗
H
Pt C

Cl
C H
p∗

Here the back donation is accepted in the t * orbital of C − C bond. Hence, the bond order of
C − C bond decreases and the bond length increases as compared to free C 2 H 4 molecule. Due
to backbonding, C2H4 molecule loses its planarity. Similarly, C2H2 molecule loses its linearity
not the planarity.

5.5 | EFFECTIVE ATOMIC NUMBER (EAN)

Physically EAN signifies the number of electrons available around a central atom within a complex, includ-
ing the electrons accepted from ligands. It is mathematically defined as follows:
EAN of a central atom in a complex = Atomic number of the central atom (Z)
− (oxidation number of the central atom with sign)
+ (number of electrons collected from the ligands).

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 5.5 Effective Atomic Number (EAN) 179

The following points need to be noted with regard to EAN.

1. All donations contribute two electrons, while NO is considered as a 3-electron donor.
2. For p-donors, the number of p-electrons involved in donation from a particular ligand are to be
considered. For example,

−
is a 6-electron donor.

− is a 4-electron donor.

is a 2-electron donor.
−
CH 2 = CH − C H 2 is a 2-electron donor.

3. For the compounds having d bond, for example, Mn 2 (Co)10, the EAN of each Mn atom is calculated as:
1
EAN of Mn = [2 × 25 − 0 + 10 × 2 + 2* ] = 36
2
*These two electrons are considered for d -bond.
The EAN of some metal atoms in different complexes are given in Table 5.5.

Table 5.5 Effective atomic numbers of some metals in complexes

Atom Atomic Complex Electrons lost Electrons gained EAN
number in ion formation by coordination
Cr 24 [Cr(CO)6] 0 12 36
⎫
Fe 26 [Fe(CN)6] 4−
2 12 36 ⎪
Fe 26 [Fe(CO)5] 0 10 36 ⎪
⎪
⎬ (Kr)
Co 27 [Co(NH3)6]3+ 3 12 36 ⎪
Ni 28 [Ni(CO)4] 0 8 36 ⎪
⎪⎭
Cu 29 [Cu(CN)4]3− 1 8 36
Pd 46 [Pd(NH3)6]4+ 4 12 54 (Xe)
2−
Pt 78 [PtCl6] 4 12 86 (Rn)
Fe 26 [Fe(CN)6]3− 3 12 35
Ni 28 [Ni(NH3)6]2+ 2 12 38
2−
Pd 46 [PdCl4] 2 8 52
2+
Pt 78 [Pt(NH3)4] 2 8 84
Ti 22 [Ti(s −C5 H5)2 (p −C5 H5)2] 0
4 16 34
Fe 26 [Fe(p −C5 H5)2] 0
2 12 36
0
Fe 26 [Fe(CO)2 (NO)2] 0 10 36
Co 27 [Co(CO)4]− −1 8 36
−
V 23 [V(CO)6] −1 12 36

Sidgwick EAN rule

In 1927, Sidgwick suggested that electron pairs from ligands were added to the central metal atom until
the central atom was surrounded by the same number of electrons as the next noble gas. The stability of
the resulting state can be explained on the basis of the molecular orbital theory. However, this rule fails in
many cases and works best for metals in low oxidation state.

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180 Chapter 5 Coordination Compounds

Metal carbonyls exhibit a strong tendency to achieve Sidgwick EAN values and as a result of this:
1. The number of CO molecule attached in mononuclear carbonyls can be predicted. For example, in

Fe(CO)x : x = 5, Ni(CO) y : y = 4, and Cr(CO)z : z = 6.

2. [Mn(CO)6 ]0 can act as a reducing agent. The complex loses an electron to attain the noble gas configu-
ration and hence obey Sidgwick EAN rule.

[Mn(CO)6 ] ⎯−⎯
e
→ [Mn(CO)6 ]+
EAN = 37 EAN = 36

3. [V(CO)6 ]0 can act as oxidizing agent. The complex gains an electron to attain the noble gas configura-
tion and hence obey Sidgwick EAN rule.

[V(CO)6 ]0 ⎯+⎯
e
→ [V(CO)6 ] − .
EAN = 35 EAN = 36

4. [Mn(CO)5 ]0 undergoes dimerization to attain the noble gas configuration and hence obey Sidgwick
EAN rule.

2[Mn(CO)5 ]0 ⎯⎯
→ [Mn 2 (CO)10 ]
EAN = 35 EAN = 36

5.6 | SHAPES OF d ORBITALS

Since d orbitals are often used in coordination complexes it is important to study their shapes and
distribution in space. The five d orbitals are not identical and the orbitals may be divided into two sets. The
three t2g orbitals have identical shape and point between the axes, x, y and z. The two eg orbitals have
different shapes and point along the axes (Figure 5.5). Alternative names for t2g and eg are de and dg
respectively.

x x y

t2g
orbitals y z z
(de)

dxy dxz z dyz

x

eg
orbitals y
(dg)

dx 2−y2 dz2

Figure 5.5 Shapes of d orbitals.

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 5.8 Valence Bond Theory 181

5.7 | BONDING IN TRANSITION METAL COMPLEXES

There are three theories of metal to ligand bonding in complexes, all dating back to the 1930s.

Valence bond theory

This theory was developed by Pauling. Coordination compounds contain complex ions, in which ligands
form coordinate bonds to the metal. Thus the ligand must have a lone pair of electrons, and the metal
must have an empty orbital of suitable energy available for bonding. The theory considers which atomic
orbitals on the metal are used for bonding. From this the shape and stability of the complex are pre-
dicted. The theory has two main limitations. Most transition metal complexes are coloured, but the
theory provides no explanation for their electronic spectra. Further, the theory does not explain why
the magnetic properties vary with temperature. For these reasons it has largely been superseded by the
crystal field theory. However, it is of interest for study as it shows the continuity of the development of
modern ideas from Werner’s theory.

Crystal field theory

This theory was proposed by Bethe and van Vleck. The attraction between the central metal and ligands in
the complex is considered to be purely electrostatic. Thus bonding in the complex may be ion–ion attrac-
tion (between positive and negative ions such as Co3+ and Cl−). Alternatively, ion–dipole attractions may
give rise to bonding (if the ligand is a neutral molecule such as NH3 or CO). NH3 has a dipole moment with
a d - charge on N and d + charges on H. Thus in [Co(NH3)6]3+ the d - charge on the N atom of each NH3
points towards the Co3+. This theory is simple. It has been remarkably successful in explaining the elec-
tronic spectra and magnetism of transition metal complexes, particularly when allowance is made for the
possibility of some covalent interaction between the orbitals on the metal and ligand. When some allow-
ance is made for covalency, the theory is often renamed as the ligand field theory. Three types of interaction
are possible: s overlap of orbitals, p overlap of orbitals, or dp−pp bonding (back bonding) due to p overlap
of full d orbitals on the metal with empty p orbitals on the ligands.

Molecular orbital theory

Both covalent and ionic contributions are fully allowed for in this theory. Though this theory is probably
the most important approach to chemical bonding, it has not displaced the other theories. This is because
the quantitative calculations involved are difficult and lengthy, involving the use of extensive computer
time. Much of the qualitative description can be obtained by other approaches using symmetry and group
theory.

5.8 | VALENCE BOND THEORY

The formation of a complex may be considered as a series of hypothetical steps. First the appropriate metal
ion is taken, e.g. Co3+. A Co atom has the outer electronic structure 3d74s2. Thus a Co3+ ion will have the
structure 3d6, and the electrons will be arranged:
3d 4s 4p 4d
full
inner
shell

If this ion forms a complex with six ligands, then six empty atomic orbitals are required on the metal ion
to receive the coordinated lone pairs of electrons. The orbitals used are the 4s, three 4p and two 4d. These
are hybridized to give a set of six equivalent sp3d2 hybrid orbitals. A ligand orbital containing a lone pair
of electrons forms a coordinate bond by overlapping with an empty hybrid orbital on the metal ion. In this
way a s bond is formed with each ligand. The d orbitals used are the 4dx2 − y2 and 4dz2 . In the diagrams below,
electron pairs from the ligands are shown as .

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182 Chapter 5 Coordination Compounds

3d 4p 4d
full
inner
shell
(sp3d 2 hybridization)
octahedral shape
outer orbital complex
high-spin complex

Since the outer 4d orbitals are used for bonding this is called an outer orbital complex. The energy of these
orbitals is quite high, so that the complex will be reactive or labile. The magnetic moment depends on the
number of unpaired electrons. The 3d level contains the maximum number of unpaired electrons for a
d6 arrangement, so this is sometimes called a high-spin or a spin-free complex. An alternative octahedral
arrangement is possible when the electrons on the metal ion are rearranged as shown below. As before,
lone pairs from the ligands are shown as .

Since low energy inner d orbitals are used this is called an inner orbital complex. Such complexes are more
stable than the outer orbital complexes. The unpaired electrons in the metal ion have been forced to pair
up, and so this is now a low-spin complex. In this particular case all the electrons are paired, so the complex
will be diamagnetic.
The metal ion could also form four-coordinate complexes, and two different arrangements are possible.
It must be remembered that hybrid orbitals do not actually exist. Hybridization is a mathematical manipula-
tion of the wave equations for the atomic orbitals involved.

3d 4s 4p 4d
full
inner
shell
(sp 3 hybridization)
tetrahedral shape

3d 4s 4p 4d
full
inner
shell
(dsp 2 hybridization)
square planar shape

The theory does not explain the colour and spectra of complexes. The theory shows the number of
unpaired electrons and from this the magnetic moment can be calculated. However, it does not explain why
the magnetic moment varies with temperature.

5.9 | CRYSTAL FIELD THEORY

The crystal field theory is now much more widely accepted than the valence bond theory. It assumes that
the attraction between the central metal and the ligands in a complex is purely electrostatic. The transi-
tion metal which forms the central atom in the complex is regarded as a positive ion of charge equal to
the oxidation state. This is surrounded by negative ligands or neutral molecules which have a lone pair of
electrons. If the ligand is a neutral molecule such as NH3, the negative end of the dipole in the molecule is
directed towards the metal ion. The electrons on the central metal are under repulsive forces from those

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 5.9 Crystal Field Theory 183

on the ligands. Thus the electrons occupy the d orbitals furthest away from the direction of approach of
ligands. In the crystal field theory the following assumptions are made.
1. Ligands are treated as point charges.
2. There is no interaction between metal orbitals and ligand orbitals.
3. The d orbitals on the metal all have the same energy (that is degenerate) in the free atom. However,
when a complex is formed the ligands destroy the degeneracy of these orbitals, i.e. the orbitals now
have different energies. In an isolated gaseous metal ion, the five d orbitals do all have the same energy,
and are termed degenerate. If a spherically symmetrical field of negative charges surrounds the metal
ion, the d orbitals remain degenerate. However, the energy of the orbitals is raised because of repul-
sion between the field and the electrons on the metal. In most transition metal complexes, either six
or four ligands surround the metal, giving octahedral or tetrahedral structures. In both of these cases
the field produced by the ligands is not spherically symmetrical. Thus the d orbitals are not all affected
equally by the ligand field.

Octahedral complexes
In an octahedral complex, the metal is at the centre of the octahedron, and the ligands are at the six corners.
The directions x, y and z point to three adjacent corners of the octahedron as shown in Figure 5.6.
The lobes of the eg orbitals (dx2 − y2 and dz2 ) point along the axes x, y and z. The lobes of the t2g orbitals
(dxy, dxz and dyz) point in between the axes. It follows that the approach of six ligands along the x, y, z,
−x, −y and −z directions will increase the energy of the dx2 − y2 and dz2 orbitals (which point along the axes)
much more than it increases the energy of the dxy, dxz and dyz orbitals (which point between the axes).
Thus under the influence of an octahedral ligand field the d orbitals split into two groups of different
energies (Figure 5.7).

eg

d orbitals are
split into two
groups

z
Energy

t2g

Free metal ion Metal ion

(five degenerate in octahedral
x d orbitals) field

Figure 5.6 The directions in an Figure 5.7 Crystal field splitting of energy
octahedral complex. levels in an octahedral field.

Rather than referring to the energy level of an isolated metal atom, the weighted mean of these two
sets of perturbed orbitals is taken as the zero: this is sometimes called the Bari centre. The difference in
energy between the two d levels is given either of the symbols Δо or 10 Dq. It follows that the eg orbitals are
+0.6Δо above the average level, and the t2g orbitals are −0.4Δо below the average (Figure 5.8).
The size of the energy gap Δо between the t2g and eg levels can be measured easily by recording the
UV–visible spectrum of the complex. Consider a complex like [Ti(H2O)6]3+. The Ti3+ ion has one d electron.

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184 Chapter 5 Coordination Compounds

eg

+0.6Δo

Δo

Energy
Average energy level
(Bari centre)
−0.4Δo

t2g
Average energy Metal ion
of metal ion in octahedral
in spherical field
field

Figure 5.8 Diagram of the energy levels of d orbitals in an octahedral field.

In the complex this will occupy the orbital with the lowest energy, that is one of the t2g orbitals (Figure 5.9a).
The complex absorbs light of the correct wavelength (energy) to promote the electron from the t2g level to
the eg level (Figure 5.9b).

eg

t2g

(a) (b)

Figure 5.9 d1 configuration: (a) ground state, (b) excited state.

The electronic spectrum for [Ti(H2O)6]3+ is given in Figure 5.10.

The steep part of the curve from 27 000 to 30 000 cm−1 (in the UV
region) is due to charge transfer. The d−d transition is the single
Wavelength (Å)
broad peak with a maximum at 20 300 cm−1. Since 1 kJ mol−1 = 3000 4000 5000 7000
83.7 cm−1, the value of Δо for [Ti(H2O)6]3+ is 20 300/83.7 = 243 kJ mol−1. 10
This is much the same as the energy of many normal single bonds
(see Appendix F).
The above method is the most convenient way of measuring
Molar absorbance

Δо values. However, Δо values can also be obtained from values

of observed lattice energies and those calculated using the Born–
5
Landé equation (see Chapter 3).
Solutions containing the hydrated Ti3+ ion are reddish violet
coloured. This is because yellow and green light are absorbed to
excite the electron. Thus the transmitted light is the complemen-
tary colour red–violet (Table 5.6).
Because of the crystal field splitting of d orbitals, the single d 0
30000 20000 10000
electron in [Ti(H2O)6]3+ occupies an energy level 2/5Δо below the
Frequency (cm−1)
average energy of the d orbitals. As a result the complex is more
stable. The crystal field stabilization energy (CFSE) is in this case Figure 5.10 Ultraviolet and visible
2/5 × 243 = 97 kJ mol−1. absorption spectrum of [Ti(H2O)6]3+.

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 5.9 Crystal Field Theory 185

Table 5.6 Colours absorbed and colours observed

Colour absorbed Colour observed Wavenumber observed (cm−1)
Yellow–green Red–violet 24 000–26 000
Yellow Indigo 23 000–24 000
Orange Blue 21 000–23 000
Red Blue–green 20 000–21 000
Purple Green 18 000–20 000
Red–violet Yellow–green 17 300–18 000
Indigo Yellow 16 400–17 300
Blue Orange 15 300–16 400
Blue–green Red 12 800–15 300

The magnitude of Δо depends on three factors:

1. The nature of the ligands.
2. The charge on the metal ion.
3. Whether the metal is in the first, second or third row of transitions elements.
Examination of the spectra of a series of complexes of the same metal with different ligands shows that
the position of the absorption band (and hence the value of Δо) varies depending on the ligands which are
attached (Table 5.7).

Table 5.7 Crystal field splittings by various ligands

Complex Absorption peak

(cm−1) (kJ mol−1)

[CrIIICl6]3− 13 640 163
[CrIII(H2O)6]3+ 17 830 213
[CrIII(NH3)6]3+ 21 680 259
[CrIII(CN)6]3− 26 280 314

Ligands which cause only a small degree of crystal field splitting are termed weak field ligands. Ligands
which cause a large splitting are called strong field ligands. Most Δ values are in the range 7,000 cm−1 to
30,000 cm−1. The common ligands can be arranged in ascending order of crystal field splitting Δ. The order
remains practically constant for different metals, and this series is called the spectrochemical series.

Spectrochemical series
weak field ligands

I − < Br − < S 2 − < Cl − < NO3− < F − < OH − < EtOH < oxalate < H2O
< EDTA < (NH3 and pyridine) < ethylenediamine < dipyridyl
< o-phenanthroline < NO2− < CN −< CO
strong field ligands
The spectrochemical series is an experimentally determined series. It is difficult to explain the order as it
incorporates both the effects of s and p bonding. The halides are in the order expected from electrostatic
effects. In other cases we must consider covalent bonding to explain the order. A pattern of increasing s
donation is followed:
halide donors < O donors < N donors < C donors
The crystal field splitting produced by the strong field CN − ligand is about double that for weak field ligands
like the halide ions. This is attributed to p bonding in which the metal donates electrons from a filled t2g

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186 Chapter 5 Coordination Compounds

orbital into a vacant orbital on the ligand. In a similar way, many unsaturated N donors and C donors may
also act as p acceptors.
The magnitude of Δо increases as the charge on the metal ion increases.

Table 5.8 Crystal field splittings for hexa-aqua complexes of M2+ and M3+
Oxidation state Ti V Cr Mn Fe Co Ni Cu

Electronic d2 d3 d4 d5 d6 d7 d8 d9
(+II)
configuration
Δo in cm−1 − 12 600 13 900 7 800 10 400 9 300 8500 12 600
−1
Δo in kJ mol − 151 (166) 93 124 111 102 (151)
Electronic d1 d2 d3 d4 d5 d6 d7 d8
(+ III)
configuration
Δo in cm−1 20 300 18 900 17 830 21 000 13 700 18 600 − −
Δo in kJ mol−1 243 226 213 (251) 164 222 − −

Note: Values in paranthesis for d4 and d9 are approximate because of tetragonal distortion.

Table 5.9 Δo crystal field splittings in one group

cm−1 kJ mol−1
[Co(NH3)6]3+ 24 800 296
3+
[Rh(NH3)6] 34 000 406
[Ir(NH3)6]3+ 41 000 490

For first row transition metal ions, the values of Δо for M3+ complexes are roughly 50% larger than the values
for M2+ complexes (Table 5.8).
The value of Δо also increases by about 30% between adjacent members down a group of transition
elements (Table 5.9). The crystal field stabilization energy in [Ti(H2O)6]3+, which has a d1 configuration,
has previously been shown to be −0.4Δо. In a similar way, complexes containing a metal ion with a d2
configuration will have a CFSE of 2 × −0.4Δо = −0.8Δо, by singly filling two of the t2g orbitals. (This is in
agreement with Hund’s rule that the arrangement with the maximum number of unpaired electrons is the
most stable.) Complexes of d3 metal ions have a CFSE of 3 × −0.4Δо = −1.2Δо.
Complexes with a metal ion with a d4 configuration would be expected to have an electronic arrange-
ment in accordance with Hund’s rule (Figure 5.11a) with four unpaired electrons, and the CFSE will
be (3 × −0.4Δо) + (0.6Δо) = −0.6Δо. An alternative arrangement of electrons which does not comply with
Hund’s rule is shown in Figure 5.11b. This arrangement has two unpaired electrons, and the CFSE is

eg

eg

Large Δo
Energy

Small Δo
value value

t2g
t2g

(a) (b)
4
Figure 5.11 High- and low-spin complexes: (a) d high-spin arrangement
(weak ligand field); (b) d 4 low-spin arrangement (strong ligand field).

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 5.9 Crystal Field Theory 187

Table 5.10 CFSE and pairing energy for some complexes

Complex Configuration Δo (cm−1) P (cm−1) Predicted Found
6
[FeII(H2O)6]2+ d 10 400 17 600 high spin High spin
II 4− 6
[Fe (CN)6] d 32 850 17 600 low spin Low spin
III 3− 6
[Co F6] d 13 000 21 000 high spin High spin
[CoIII(NH3)6]3+ d6 23 000 21 000 low spin Low spin

(4 × −0.4Δо) = −1.6Δо. The CFSE is larger than in the previous case. However, the energy P used to pair
the electrons must be allowed for, so the total stabilization energy is −1.6Δо + P. These two arrangements
differ in the number of unpaired electrons. The one with the most unpaired electrons is called ‘high-
spin’ or ‘spin-free’, and the other one the ‘low-spin’ or ‘spin-paired’ arrangement. Both arrangements
have been found to exist. Which arrangement occurs for any particular complex depends on whether
the energy to promote an electron to the upper eg level (that is the crystal field splitting Δо) is greater
than the energy to pair electrons (that is P) in the lower t2g level. For a given metal ion P is constant.
Thus the amount of crystal field splitting is determined by the strength of the ligand field. A weak field
ligand such as Cl− will only cause a small splitting of energy levels Δо Thus it will be more favourable
energetically for electrons to occupy the upper eg level and have a high-spin complex, rather than to
pair electrons. In a similar way, strong field ligands such as CN− cause a large splitting Δо. In this case it
requires less energy to pair the electrons and form a low-spin complex. (see Table 5.10.).
Similar arguments apply to high- and low-spin complexes of metal ions with d5, d6 and d7 configura-
tions. These are summarized in Table 5.11.
Table 5.11 CFSE and electronic arrangements in octahedral complexes
Number Arrangement in weak ligand field Arrangement in strong ligand field
of d
t2g eg CFSE Spin only t2g eg CFSE Spin only
electrons
Δо magnetic Δо magnetic
moment ms(D) moment ms(D)

d1 − 0.4 1.73 − 0.4 1.73

d2 − 0.8 2.83 − 0.8 2.83

d3 − 1.2 3.87 − 1.2 3.87

− 1.2
d4 + 0.6 4.90 − 1.6 2.83
= − 0.6
− 1.2
d5 + 1.2 5.92 − 2.0 1.73
= − 0.0
− 1.6
d6 + 1.2 4.90 − 2.4 0.00
= − 0.4
− 2.0 − 2.4
d7 + 1.2 3.87 + 0.6 1.73
= − 0.8 = − 1.8
− 2.4 − 2.4
d8 + 1.2 2.83 + 1.2 2.83
= − 1.2 = − 1.2
− 2.4 − 2.4
d9 + 1.8 1.73 + 1.8 1.73
= − 0.6 = − 0.6
− 2.4 − 2.4
d 10 + 2.4 0.00 + 2.4 0.00
= − 0.0 = − 0.0

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188 Chapter 5 Coordination Compounds

5.10 | EFFECTS OF CRYSTAL FIELD SPLITTING

In octahedral complexes, the filling of t2g orbitals decreases the energy of a complex, that is makes it more
stable by −0.4 Δo per electron. Filling eg orbitals increases the energy by +0.6Δо per electron. The total crystal
field stabilization energy is given by

CFSE(octahedral) = − 0.4 n( t2 g ) + 0.6 n( eg )

where n( t2 g ) and n( eg ) are the number of electrons occupying the t2g and eg orbitals respectively. The CFSE
is zero for ions with d0 and d10 configurations in both strong and weak ligand fields. The CFSE is also zero
for d5 configurations in a weak field. All the other arrangements have some CFSE, which increases the
thermodynamic stability of the complexes. Thus many transition metal compounds have a higher mea-
sured lattice energy (obtained by calculations using the terms in the Born–Haber cycle) than is calculated
using the Born–Landé, Born–Meyer or Kapustinskii equations. In contrast, the measured (Born–Haber)
and calculated values for compounds of the main groups (which have no CFSE) are in close agreement
(Table 5.12). There is also close agreement in MnF2 which has a d5 configuration and a weak field ligand:
hence there is no CFSE.

Table 5.12 Measured and calculated lattice energies

Compound Structure Measured lattice Calculated lattice Difference (measured –
energy (kJ mol−1) energy (kJ mol−1) calculated) (kJ mol−1)

NaCl Sodium chloride −764 −764 0

AgCl Sodium chloride −916 −784 −132
AgBr Sodium chloride −908 −759 −149
MgF2 Rutile −2908 −2915 +7
MnF2 Rutile −2770 −2746 −24
FeF2 Rutile −2912 −2752 −160
NiF2 Rutile −3046 −2917 −129
CuF2 Rutile −3042 −2885 −157

A plot of the lattice energies of the halides of the first row transition elements in the divalent state is
given in Figure 5.12. In the solid, the coordination number of these metals is 6, and so the structures are
analogous to octahedral complexes. The graphs for each halide show a minimum at Mn2+, which has a
d5configuration. In a weak field this has a high-spin arrangement with zero CFSE. The configurations d 0
and d10 also have zero CFSE. The broken line through Ca2+, Mn2+ and Zn2+ represents zero stabilization. The
heights of other points above this line are the crystal field stabilization energies.
The hydration energies of the M2+ ions of the first row transition elements are plotted in Figure 5.13a.
2+
M (g) + excess H 2 O → [M(H 2 O)6 ]2 +
The ions Ca2+, Mn2+ and Zn2+ have d 0, d5 and d10 configurations, and have zero CFSE. An almost straight
line can be drawn through these points. The distance of the other points above this line corresponds to the
CFSE. Values obtained in this way agree with those obtained spectroscopically. A similar graph of the M3+
ions is shown in Figure 5.13b: here the d 0, d5 and d10 species are Sc3+, Fe3+ and Ga3+.
The ionic radii for M2+ ions might be expected to decrease smoothly from Ca2+ to Zn2+ because of the
increasing nuclear charge, and the poor shielding by d electrons. A plot of these radii is given in Figure 5.14.
The change in size is not regular.
A smooth (broken) line is drawn through Ca2+, Mn2+ and Zn2+. These have d 0, d5 and d10 configurations
as the d orbitals are empty, half full or full. These arrangements constitute an almost spherical field round
the nucleus. In Ti2+ the d electrons occupy orbitals away from the ligands, providing little or no shielding
of the nuclear charge. Thus the ligands are drawn closer to the nucleus. The increased nuclear charge has
an even greater effect in the case of V2+. At Cr2+ the eg level contains one electron. This is concentrated

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 5.11 Tetragonal Distortion of Octahedral Complexes (Jahn-Teller Distortion) 189

kJmol−1

3000 F 2200

2000
2900
1800
2800
1600
Cl
2700 Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga
Br
Lattice energy

(a) M2+
2600 I

2500 4800

2400 4600

2300 4400

4200
2200

4000
2100
Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga
Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn (b) M3+

Figure 5.12 CFSE of dihalides of the first transition Figure 5.13 Enthalpies of hydration
series. (After T.C. Waddington. Lattice energies and for (a) M2+ and (b) M3+, in kJ mol−1.
their significance in inorganic chemistry. Advances in
Inorganic Chemistry and Radiochemistry, 1,
Academic Press, New York, 1959.)
1.0

0.9
in the direction of the ligands, thus providing very good
A°
shielding. Thus the ligands can no longer approach so 0.8
closely and the ionic radius increases. This increase in
size is continued with the filling of the second eg orbital 0.7
at Mn2+. The screening by the eg orbitals is so good that
the radius of Mn2+ is slightly smaller than it would be if 0.6
it were in a truly spherical field. The same sequence of Ca2+ Sc2+ Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+
size changes is repeated in the second half of the series.
Figure 5.14 Octahedral ionic radii of M2+
for first row transition elements.

5.11 | TETRAGONAL DISTORTION OF OCTAHEDRAL COMPLEXES

(JAHN-TELLER DISTORTION)
The shape of transition metal complexes is determined by the tendency of electron pairs to occupy posi-
tions as far away from each other as possible. This is the same as for the main group compounds and
complexes. In addition, the shapes of transition metal complexes are affected by whether the d orbitals are
symmetrically or asymmetrically filled.
Repulsion by six ligands in an octahedral complex splits the d orbitals on the central metal into t2g and eg
levels. It follows that there is a corresponding repulsion between the d electrons and the ligands. If the d
electrons are symmetrically arranged, they will repel all six ligands equally. Thus the structure will be a
completely regular octahedron. The symmetrical arrangements of d electrons are shown in Table 5.13.

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190 Chapter 5 Coordination Compounds

Table 5.13 Symmetrical electronic arrangements

Electronic configuration t2g eg Nature of ligand field Examples

Ti IVO2 .[Ti IV F6 ]2 −
d0 Strong or weak
[Ti IVCI 6 ]2 −

[Cr III (oxalate)3 ]3−

d3 Strong or weak
[Cr III (H 2O)6 ]3+

[Mn II F6 ]4 −
d5 Weak
[FeIII F6 ]3−

[FeII (CN)6 ]4 −
d6 Strong
[CoIII (NH 3 )6 ]3+

[Ni II F6 ]4 −
d8 Weak
[Ni II (H 2O)6 ]2 +

[Zn II (NH 6 )6 ]2 +
d10 Strong or weak
[Zn II (H 2O)6 ]2 +

All other arrangements have an asymmetrical arrangement of d electrons. If the d electrons are asym-
metrically arranged, they will repel some ligands in the complex more than others. Thus the structure is dis-
torted because some ligands are prevented from approaching the metal as closely as others. The eg orbitals
point directly at the ligands. Thus asymmetric filling of the eg orbitals results in some ligands being repelled
more than others. This causes a significant distortion of the octahedral shape. In contrast the t2g orbitals do
not point directly at the ligands, but point in between the ligand directions. Thus asymmetric filling of the t2g
orbitals has only a very small effect on the stereochemistry. Distortion caused by asymmetric filling of the
t2g orbitals is usually too small to measure. The electronic arrangements which will produce a large distortion
are shown in Table 5.14.

Table 5.14 Asymmetrical electronic arrangements

Electronic configuration t2g eg Nature of ligand field Examples

d4 Weak field (high-spin complex) Cr (+ II), Mn (+ III)

d7 Strong field (low-spin complex) Co (+ II), Ni (+ III)

d9 Either strong or weak Cu (+ II)

The two eg orbitals dx2 − y2 and dz2 are normally degenerate. However, if they are asymmetrically filled
then this degeneracy is destroyed, and the two orbitals are no longer equal in energy. If the dz2 orbital con-
tains one more electron than the dx2 − y2 orbital then the ligands approaching along +z and −z will encounter
greater repulsion than the other four ligands. The repulsion and distortion result in elongation of the octa-
hedron along the z axis. This is called tetragonal distortion. Strictly it should be called tetragonal elonga-
tion. This form of distortion is commonly observed.
If the dx2 − y2 orbital contains the extra electron, then elongation will occur along the x and y axes. This
means that the ligands approach more closely along the z axis. Thus there will be four long bonds and two
short bonds. This is equivalent to compressing the octahedron along the z axis and is called tetragonal com-
pression. Tetragonal elongation is much more common than tetragonal compression, and it is not possible
to predict which will occur.
For example, the crystal structure of CrF2 is a distorted rutile (TiO2) structure. Cr2+ is octahedrally sur-
rounded by six F−, and there are four Cr—F bonds of length 1.98−2.01 Å, and two longer bonds of length

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 5.12 Square Planar Arrangements 191

2.43 Å. The octahedron is said to be tetragonally distorted. The electronic arrangement in Cr2+ is d4. F− is
a weak field ligand, and so the t2g level contains three electrons and the eg level contains one electron. The
dx2 − y2 orbital has four lobes whilst the dz2 orbital has only two lobes pointing at the ligands. To minimize
repulsion with the ligands, the single eg electron will occupy the dz2 orbital. This is equivalent to splitting
the degeneracy of the eg level so that dz2 is of lower energy, i.e. more stable, and dx2 − y2 is of higher energy, i.e.
less stable. Thus the two ligands approaching along the +z and −z directions are subjected to greater repul-
sion than the four ligands along +x, −x, +y and −y. This causes tetragonal distortion with four short bonds
and two long bonds. In the same way MnF3 contains Mn3+ with a d4 configuration, and forms a tetragonally
distorted octahedral structure.
Many Cu(+II) salts and complexes also show tetragonally distorted octahedral structures. Cu2+ has a
9
d configuration:
t 2g eg

To minimize repulsion with the ligands, two electrons occupy the dz2 orbital and one electron occupies
the dx2 − y2 orbital. Thus the two ligands along −z and −z are repelled more strongly than are the other four
ligands.
The examples above show that whenever the dz2 and dx2 − y2 orbitals are unequally occupied, distor-
tion occurs. This is known as Jahn–Teller distortion. The Jahn–Teller theorem states that ‘Any non-linear
molecular system in a degenerate electronic state will be unstable, and will undergo some sort of distortion
to lower its symmetry and remove the degeneracy.’ More simply, molecules or complexes (of any shape
except linear), which have an unequally filled set of orbitals (either t2g or eg), will be distorted. In octahedral
complexes distortions from the t2g level are too small to be detected. However, distortions resulting from
uneven filling of the eg orbitals are very important.

5.12 | SQUARE PLANAR ARRANGEMENTS

If the central metal ion in a complex has a d8 configuration,
six electrons will occupy the t2g orbitals and two electrons will eg
occupy the eg orbitals. The arrangement is the same in a complex
with weak field ligands. The electrons are arranged as shown in
Figure 5.15. The orbitals are symmetrically filled, and a regular
octahedral complex is formed, for example by [NiII(H2O)6]2+ and
Energy

[NiII(NH3)6]2+.
The single electron in the dx2 − y2 orbital is being repelled by
four ligands, whilst the electron in the dz2 orbital is only being
repelled by two ligands. Thus the energy of the dx2 − y2 increases
t2g
relative to that of dz2 . If the ligand field is sufficiently strong,
the difference in energy between these two orbitals becomes
larger than the energy needed to pair the electrons. Under these Figure 5.15 d 8 arrangement in weak
conditions, a more stable arrangement arises when both the eg elec- octahedral field.
trons pair up and occupy the lower energy dz2 orbital. This leaves
the dx2 − y2 orbital empty (Figure 5.16). Thus four ligands can now
approach along the +x, −x, +y and −y directions without any difficulty, as the dx2 − y2 orbital is empty. However,
ligands approaching along the +z and −z directions meet very strong repulsive forces from the filled dz2
orbital (Figure 5.17). Thus only four ligands succeed in bonding to the metal. A square planar complex is
formed, the attempt to form an octahedral complex being unsuccessful.
The amount of tetragonal distortion that occurs depends on the particular metal ion and ligands.
Sometimes the tetragonal distortion may become so large that the dz2 orbital is lower in energy than
the dxy orbital as shown in Figure 5.18. In square planar complexes of CoII, NiII and CuII the dz2 orbital has
nearly the same energy as the dxz and dyz orbitals. In [PtCl4]2− the dz2 orbital is lower in energy than the dxz
and dyz orbitals.

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192 Chapter 5 Coordination Compounds

dx 2−y 2 dx 2−y 2

dz2 dz2

Energy

dxy
t2g

dxz and dyz

(a) (b)
8
Figure 5.16 d arrangement in very strong octahedral field. Tetragonal distortion splits (a) the eg level;
and (b) also splits the t2g level. The dxy orbital is higher in energy than the dxz or dyz. (For simplicity this
is sometimes ignored.)

dx 2−y 2

eg

z dxy
y
Energy

dz2

t2g

dxz and dyz

x

Figure 5.17 d 8 arrangement, Figure 5.18 Tetragonal distortion.

strong field. (The dz2 orbital is
full, the dx2 − y2 empty.)

Square planar complexes are formed by d8 ions with strong field ligands, for example [NiII(CN)4]2−.
The crystal field splitting Δо is larger for second and third row transition elements, and for more highly
charged species. All the complexes of Pt(+II) and Au(+III) are square planar – including those with weak
field ligands such as halide ions.
Square planar structures can also arise from d4 ions in a weak ligand field. In this case the dz2 orbital
only contains one electron. Some ions that form square planar complexes are given in Table 5.15.

Table 5.15 Ions that form square planar complexes

Electronic configuration Ions Type of field Number of unpaired electrons
4
d Cr(+II) Weak 4
6
d Fe(+II) (Haem) 2
d7 Co(+II) Strong 1
8
d Ni(+II), Rh(+I), Ir(+I) Strong 0
Pd(+II), Pt(+II), Au(+III) Strong and weak 0
9
d Cu(+II), Ag(+II)) Strong and weak 1

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 5.13 Tetrahedral Complexes 193

5.13 | TETRAHEDRAL COMPLEXES

A regular tetrahedron is related to a cube. One atom is at the centre of the cube, and four of the eight
corners of the cube are occupied by ligands as shown in Figure 5.19.
The directions x, y and z point to the centres of the faces of the cube. The eg orbitals point along x, y
and z (that is to the centres of the faces). The t2g orbitals point between x, y and z (that is towards the centres
of the edges of the cube) (Figure 5.20).
z

dx 2−y 2 dz 2

x
y

dxy dyz dxz

Figure 5.19 Relation of a tetrahedron
to a cube. Figure 5.20 Orientation of d orbitals relative to a cube.

The direction of approach of the ligands does not coincide exactly with either the eg or the t2g orbit-
als. The angle between an eg orbital, the central metal and the ligand is half the tetrahedral angle =
109°28 ′ / 2 = 54°44 ′. The angle between a t2g orbital, the central metal and the ligand is 35°16′. Thus the t2g
orbitals are nearer to the direction of the ligands than the eg orbitals. (Alternatively the t2g orbitals are half
the side of the cube away from the approach of the ligands, whilst the eg orbitals are half the diagonal of
the cube away.) The approach of the ligands raises the energy of both sets of orbitals. The energy of the t2g
orbitals is raised most because they are closest to the ligands. This crystal field splitting is the opposite way
round to that in octahedral complexes (Figure 5.21).
The t2g orbitals are 0.4Δt above the weighted average energy of the two groups (the Bari centre) and the
eg orbitals are 0.6Δt below the average (Figure 5.22).

t 2g
t2g

d orbitals are
split into two +0.4 Dt
groups Average
energy level
Energy

Dt (Bari centre
Energy

eg

−0.6 Dt

eg
Average energy Metal ion
Free metal ion Metal ion of metal ion in tetrahedral
(five degenerate in tetrahedral in spherical field
d orbitals) field field

Figure 5.21 Crystal field splitting of energy Figure 5.22 Energy levels for d orbitals in
levels in a tetrahedral field. a tetrahedral field.

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194 Chapter 5 Coordination Compounds

The magnitude of the crystal field splitting Δt in tetrahedral complexes is considerably less than in octa-
hedral fields. There are two reasons for this:
1. There are only four ligands instead of six, so the ligand field is only two thirds the size: hence the ligand
field splitting is also two thirds the size
2. The direction of the orbitals does not coincide with the direction of the ligands. This reduces the crystal
field splitting by roughly a further two thirds.
Thus the tetrahedral crystal field splitting Δt roughly 2/3 × 2/3 = 4/9 of the octahedral crystal field splitting
Δo. Strong field ligands cause a bigger energy difference between t2g and eg than weak field ligands.
However, the tetrahedral splitting Δt is always much smaller than the octahedral splitting Δ0. Thus it is
never energetically favourable to pair electrons, and all tetrahedral complexes are high-spin.

Table 5.16 CFSE and electronic arrangements in tetrahedral complexes

Number Arrangement Spin only Tetrahedral Tetrahedral Octahedral

of d of electrons magnetic CFSE CFSE scaled CFSE Δo
electrons moment for comparison
with octahedral Weak Strong
values, assuming field field
4
eg t2g m (e) Δt Δt = Δo
9

d1 1.73 − 0.6 − 0.27 − 0.4 − 0.4

d2 2.83 − 1.2 − 0.53 − 0.8 − 0.8

d3 3.87 − 1.2 + 0.4 = − 0.8 − 0.36 − 1.2 − 1.1

d4 4.90 − 1.2 + 0.8 = − 0.4 − 0.18 − 0.6 − 1.6

d5 5.92 − 1.2 + 1.2 = 0.0 − 0.00 0.0 − 2.0

d6 4.90 − 1.8 + 1.2 = − 0.6 − 0.27 − 0.4 − 2.4

d7 3.87 − 2.4 + 1.2 = − 1.2 − 0.53 − 0.8 − 1.8

d8 2.83 − 2.4 + 1.6 = − 0.8 − 0.36 − 1.2 − 1.2

d9 1.73 − 2.4 + 2.0 = − 0.4 − 0.18 − 0.6 − 0.6

d10 0.00 − 2.4 + 2.4 = 0.0 0.00 0.0 0.0

The CFSE in both octahedral and tetrahedral environments is given in Table 5.16. This shows that for
d0, d5 and d10 arrangements the CFSE is zero in both octahedral and tetrahedral complexes. For all other
electronic arrangements there is some CFSE, and the octahedral CFSE is greater than the tetrahedral
CFSE. It follows that octahedral complexes are generally more stable and more common than tetrahedral
complexes. This is partly because there are six bond energy terms rather than four, and partly because there
is a larger CFSE term. Despite this some tetrahedral complexes are formed, and are stable. Tetrahedral
complexes are favoured:
1. Where the ligands are large and bulky and could cause crowding in an octahedral complex.
2. Where attainment of a regular shape is important. For tetrahedral structures d 0, d2, d5, d7 and d10
IV 0 0 VII −
configurations are regular. Some tetrahedral complexes which are regular are: Ti Cl 4 (eg , t2 g ),[Mn O4 ]
(eg0 , t20g ),[FeVI O4 ]2 − (eg2 , t20g ),[FeIII Cl 4 ]− (eg2 , t23g ),[CoII Cl 4 ]2 − (eg4 , t23g ) and [Zn II Cl 4 ]2− (eg4 , t26g ).
3. When the ligands are weak field, and the loss in CFSE is thus less important.
4. Where the central metal has a low oxidation state. This reduces the magnitude of Δ. www.mediit.in
 5.14 Magnetism 195

5. Where the electronic configuration of the central metal is d 0, d 5 or d 10 as there is no CFSE.

6. Where the loss of CFSE is small, e.g. d1 and d 6 where the loss in CFSE is 0.13Δo or d2 and d7 where
the loss is 0.27 Δo.
Many transition metal chlorides, bromides and iodides form tetrahedral structures.

5.14 | MAGNETISM
The magnetic moment can be measured using a Gouy balance. If we assume that the magnetic moment
arises entirely from unpaired electron spins then the ‘spin only’ formula can be used to estimate n, the
number of unpaired electrons. This gives reasonable agreement for complexes of the first row of transition
metals.
m s = n(n + 2)

Once the number of unpaired electrons is known, either the valence bond or the crystal field theory
can be used to work out the shape of the complex, the oxidation state of the metal, and, for octahedral
complexes, whether inner or outer d orbitals are used. For example, Co (+III) forms many complexes, all
of which are octahedral. Most of them are diamagnetic, but [CoF6]3− is paramagnetic with an observed
magnetic moment of 5.3 BM. Crystal field theory explains this (Figure 5.23).
Co(+II) forms both tetrahedral and square planar four-coordinate complexes.These can be distinguished
by magnetic measurements (Figure 5.24).
However, orbital angular momentum also contributes to a greater or lesser degree to the magnetic
moment. For the second and third row transition elements not only is this contribution significant, but
spin orbit coupling may occur. Because of this, the ‘spin only’ approximation is no longer valid, and there
is extensive temperature-dependent paramagnetism. Thus the simple interpretation of magnetic moments
in terms of the number of unpaired electrons cannot be extended from the first row of transition elements
to the second and third rows. The temperature dependence is explained by the spin orbit coupling.
This removes the degeneracy from the lowest energy level in the ground state. Thermal energy then allows
a variety of levels to be populated.

Co2+ in a tetrahedral field

t2g
Co3+ octahedral complex with strong field ligands
Energy

eg
eg
3 unpaired electrons, m = 3(3 + 2) = 15 = 3.87 BM
Energy

t 2g
Strong field ligands, e.g [Co(NH3)6]3+ Co2+ in square planar complex
alternatively no unpaired electrons
dx2−y2
hence diamagnetic

Co3+ octahedral complex with weak field ligands

Energy

dxy
dz2
Energy

Weak field ligands, e.g [CoF6]3−

four unpaired electrons dxz and dyz
hence paramagnetic 1 unpaired electron, m = 1(1 + 2) = 3 = 1.73 BM
assuming ms = n(n + 2) = 4(4 + 2) = 4.90 BM
Figure 5.24 Co2+ in tetrahedral and square
3+
Figure 5.23 Co in high-spin and low-spin complexes. planar complexes.

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5.15 | EXTENSION OF THE CRYSTAL FIELD THEORY

TO ALLOW FOR SOME COVALENCY
The crystal field theory is based on purely electrostatic attraction. At first sight this seems to be a most
improbable assumption. Nevertheless, the theory is remarkably successful in explaining the shapes of com-
plexes, their spectra and their magnetic properties. Calculations can be carried out quite simply. The dis-
advantage of the theory is that it ignores evidence that some covalent bonding does occur in at least some
transition metal complexes:
1. Compounds in the zero oxidation state such as nickel carbonyl [Ni0(CO)4] have no electrostatic attrac-
tion between the metal and the ligands. Thus the bonding must be covalent.
2. The order of ligands in the spectrochemical series cannot be explained solely on electrostatic grounds.
3. There is some evidence from nuclear magnetic resonance and electron spin resonance that there is some
unpaired electron density on the ligands. This suggests the sharing of electrons, and hence some covalency.
The Racah interelectron repulsion parameter B is introduced into the interpretation of spectra. This makes
allowance for covalency arising from the delocalization of d electrons from the metal onto the ligand. If B
is reduced below the value for a free metal ion, the d electrons are delocalized onto the ligand. The more
B is reduced the greater the delocalization and the greater the amount of covalency. In a similar way an
electron delocalization factor k can be used in interpreting magnetic measurements.

5.16 | NOMENCLATURE OF COORDINATION COMPOUNDS

The rules for systematic naming of coordination compounds are discussed as follows:
1. The name of the cationic part is written first followed by the anionic part.
2. The name of the ligand is listed before the name(s) of the central atom(s).
3. No space is left between the names that refer to the same coordination entity.
4. Ligand names are listed in the alphabetical order (multiplicative prefixes indicating the number of
ligands are not considered in determining that order).
5. a. When there are several ligands of the same kind, prefixes like di, tri, tetra, penta, hexa, etc. are used
before the name of the ligand to indicate their number.
b. When the name of the ligand includes terms like di, tri, tetra, etc. then prefixes like bis, tris, tetrakis,
pentakis, etc. are written before the name of the ligand, and the ligand name is kept within brackets
to avoid ambiguity. For example,
(i) (NH 3 )2: diammine
(ii) Me2 NH: dimethylamine
(iii) (en)2: bis (ethylenediamine)
(iv) (NH 2 Me)2: bis (methylamine)
(to make it distinct from dimethylammine)
c. There is no deletion of vowels or use of a hyphen. For example,
(NH 3 )4: tetraammine.
6. The names of cationic ligands are ended with ‘ium’. For example,
+
(i) NO: nitrosylium
+
(ii) NH 2 − N H 3: hydrazinium
7. The neutral ligand names have no special endings. For example,
(i) H2O: aqua
(ii) NH3: ammine
(iii) CO: carbonyl
(iii) NO: nitrosyl

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 5.16 Nomenclature of Coordination Compounds 197

(iv) MeCO NH 2: acetamide

(v) Me NH 2: methanamine
(vi) Me PH 2 : methylphosphane
(vii) Me O S(= O)OH : methyl hydrogen sulphite
(ix) Me2O: dimethyl ether
(x) Me2S: dimethyl thioether
(xi) C 6 H 6: benzene
8. The names of anionic ligands are ended with ‘o’. For example,
a. Ligand names ending with ‘ate’ are changed to ‘ato’.
(i) NO3−: nitrato
(ii) CO2−
3 : carbonato

(iii) MeCO2− : acetato or ethanoato

(iv) acac − : acetylacetonato
b. Ligand names ending with ‘ite’ are changed to ‘ito’.
(i) SO2−
3 : sulphito

(ii) ClO2−: chlorito

−
(iii) OCl: hypochlorito
(iv) NO2−: nitrito
c. Ligand names ending with ‘ide’ are changed to ‘ido’.
(i) F−: fluorido
(ii) Cl − : chlorido
(iii) CN − : cyanido
(iv) H − : hydrido
(v) OH − : hydroxido
(vi) NH 2−: amido
(vii) D− : deuterido or [ 2 H] hydrido
d. The anionic organic ligands where the C atom is the donor atom, are not ended with ‘o’. For example,
(i) CH 3− : methyl (not methylo)
−
(ii) C H 2CH = CH 2 : allyl
(iii) Ph − : phenyl
(iv) Z- : cyclopropyl
−
(v) CH 2 = C H: vinyl
e. For the p - donors, the prefix like hx is to be used, where h indicates p - electron donation and x is known
as the hapticity of the ligand, i.e. the number of atoms involved in the p - donation. For example,
(i) p - C 5 H -5 : h 5 - cyclopentadienyl or pentahaptocyclopentadienyl
(ii) π − C 6 H 6 : η6 − benzene or hexahaptobenzene
(iii) π − C 3 H 5− : η 3 − allyl or trihaptoallyl
f. Ambidentate ligands are named as follows:
-
(i) ¬: C N : cyanido or cyanido − C
−
←: N C : isocyanido or cyanido − N
(ii) ←: NO 2− : nitro or nitrito − N
−
← ON O : nitrito or nitrito − O
−
(iii) ←: OCN : cyanato or cyanato − O
−
←: NC O : isocyanato or cyanato − N

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198 Chapter 5 Coordination Compounds

−
(iv) ←: SCN : thiocyanato or thiocyanato − S
−
←: NC S : isothiocyanato or thiocyanato − N
O S
C

C
(v) O S : dithiooxalato (O, O′)

S O
C

C
(iv) S O : dithiooxalato (S, S′)
9. For anionic complexes, metal names are ended with ‘ate’, i.e. ‘ate’ is to replace ‘ium’, ‘um’, or ‘enum’
in the metal’s name. For example,
(i) Aluminium : Aluminate
(ii) Platinum : Platinate
(iii) Molybdenum : Molybdate
(iv) Zinc : Zincate
(v) Nickel : Nickelate
(vi) Cobalt : Cobaltate
(vii) Iron : Ferrate
(viii) Manganese : Manganate
10. The oxidation number of the central atom is written in Roman numerals within brackets, after the
name of the central atom. For example, as. (II), (−I), (0), etc.
Alternatively, the charge on a coordination entity may be indicated. The net charge is written in Arabic
numerals, with the number preceding the charge sign and enclosed in parenthesis. It follows the name
of the central atom (including the ending ‘ate’ if applicable) without the intervention of a space.
For example,
(i) K 4 [ Fe(CN)6 ]: potassium hexacyanidoferrate (II)
or potassium hexacyanidoferrate (4−)
or tetrapotassium hexacyanidoferrate
(ii) K 2 [OsCl 5 N]: potassium pentachloridonitridoosmate (2−)
or potassium pentachloridonitridoosmate (VI)
(iii) [CuCl 2 {O = C(NH 2 )2 }2 ]0 : dichloridobis (urea) copper (II)
(iv) [CoCl(NH 3 )5 ]Cl 2 : pentamminechloridocobalt (III) chloride
or pentaamminechloridocobalt (2+) chloride
11. If there is any water of crystallization, it is to included in the name. For example,
[Cr(H 2O)4 Cl 2 ]Cl ⋅ 2 H 2O: tetraaquadichlorido chromium (III) chloride - 2 - water or
tetraaquadichlorido chromium (III) chloride dihydrate
These rules are illustrated by the following examples:

[Co(NH3)6]Cl3 Hexaamminecobalt(III) chloride

[CoCl(NH3)5]2+ Pentaamminechloridocobalt(III) ion
[CoSO4(NH3)4]NO3 Tetraamminesulphatocobalt(III) nitrate
[Co(NO2)3(NH3)3] Triamminetrinitrocobalt(III)
[CoCl ∙ CN ∙ NO2 ∙ (NH3)3] Triamminechloridocyanonitrocobalt(III)
[Zn(NCS)4]2− Tetrathiocyanato-N-zincate(II) ion

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 5.17 Isomerism 199

[Cd(SCN)4]2− Tetrathiocyanato-S-cadmiate(II) ion

Li[AlH4] Lithium tetrahydridoaluminate(III)

(lithium aluminium hydride)
Na2[ZnCl4] Sodium tetrachloridozincate(II)
K2[Cr(CN)2O2(O2)NH3] Potassium amminedicyanodioxoperoxo chromate(VI)

[Pt(py)4][PtCl4] Tetrapyridineplatinum(II) tetrachloridoplatinate(II)

[Cr(en)3]Cl3 d or l Tris(ethylenediamine)chromium(III) chloride
[CuCl2(CH3NH2)2] Dichloridobis(methylamine)copper(II)

5.17 | ISOMERISM
Compounds that have the same chemical formula but different structural arrangements are called isomers.
Because of the complicated formulae of many coordination compounds, the variety of bond types and the
number of shapes possible, many different types of isomerism occur. Werner’s classification into polymer-
ization, ionization, hydrate linkage, coordination, coordination position, and geometric and optical isomerism
is still generally accepted.

Polymerization isomerism
This is not true isomerism because it occurs between compounds having the same empirical formula,but different
molecular weights. Thus [Pt(NH3)2Cl2], [Pt(NH3)4][PtCl4], [Pt(NH3)4][Pt(NH3)Cl3]2 and [Pt(NH3)3Cl]2[PtCl4]

3+ 6+
OH OH
(NH3)3 Co OH Co(NH3)3 and Co Co(NH3)4
OH OH
3

Figure 5.25 Polymerization isomers.

all have the same empirical formula. Polymerization isomerism may be due to a different number of nuclei in
the complex, as shown in Figure 5.25.

Ionization isomerism
This type of isomerism is due to the exchange of groups between the complex ion and the ions outside it.
[Co(NH3)5Br]SO4 is red–violet. An aqueous solution gives a white precipitate of BaSO4 with BaCl2 solution,
thus confirming the presence of free SO2− 4 ions. In contrast [Co(NH3)5SO4]Br is red. A solution of this complex
does not give a positive sulphate test with BaCl2. It does give a cream-coloured precipitate of AgBr with AgNO3,
thus confirming the presence of free Br − ions. Note that the sulphate ion occupies only one coordination posi-
tion even though it has two negative charges. Other examples of ionization isomerism are [Pt(NH3)4Cl2]Br2
and [Pt(NH3)4Br2]Cl2, and [Co(en)2NO2 ∙ Cl]SCN, [Co(en)2NO2 ∙ SCN]Cl and [Co(en)2Cl ∙ SCN]NO2.

Hydrate isomerism
This type of isomerism is shown by compounds having the same formula but differing only in the number
of water molecules of crystallization. More broadly, it can also be called solvent isomerism to include
other solvents also (like NH 3 or other ligands) present in a similar manner. Some examples of hydrate
isomerism are as follows
1. CrCl 3 ⋅ 6 H 2 O can exist in following forms:
[Cr(H 2 O)6 ]Cl 3: violet (three chloride ions)

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200 Chapter 5 Coordination Compounds

[Cr(H 2 O)5 Cl]Cl 2 ⋅ H 2 O : bright blue green (two chloride ions)

[Cr(H 2 O)4 Cl 2 ]Cl ⋅ 2 H 2 O : dark green (one chloride ion)
[Cr(H 2 O)3 Cl 3 ] ⋅ 3H 2 O : dark green (no chloride ion)
2. [Co(NH 3 )4 (H 2 O)Cl]Br2 and [Co(NH 3 )4 Br2 ]Cl ⋅ H 2 O .
3. [Co(NH 3 )5 (H 2 O)](NO3 )3 and [Co(NH 3 )5 NO3 ](NO3 )2 ⋅ H 2 O .
Note that hydrate isomerism is also a kind of ionization isomerism.

Linkage isomerism
Certain ligands contain more than one atom which could donate an electron pair. Thus this type of isom-
erism is shown by ambidentate ligands. For example, in the NO2− ion, either N or O atoms can act as the
electron pair donor. Thus there is the possibility of isomerism. Two different complexes [Co(NH3)5NO2]Cl2
have been prepared, each containing the NO2− group in the complex ion. One is red and is easily decom-
posed by acids to give nitrous acid. It contains Co—ONO and is a nitrito complex. The other complex is
yellow and is stable to acids. It contains the Co—NO2 group and is a nitro compound. The two materials
are represented in Figure 5.26.

NH3 2+ NH3 2+
H3N ONO H3N NO2
Co and Co
H3N NH3 H3N NH3
NH3 NH3
Red Yellow
Nitritopentamminecobalt(III) Nitropentamminecobalt(III)
ion ion

Figure 5.26 Nitrito and nitro complexes.

This type of isomerism also occurs with other ligands such as SCN− , OCN−, S2O32−, CN−, NOS−, NO and
C2O2S22− (dithiooxalate ion). For example,
CH2 CH2

Me2N CH2

S C N Pd P(Ph)2 [O − N − Ru(NO)4 (OH)]2−  [N − O − Ru(NO)4 OH]2−

Stable form Metastable form
S

C
N

2− 2−
S O O S O S S O
C C C C
Pd2+ and Pd2+
C C C C
S O O S O S S O

Coordination isomerism
When both the positive and negative ions are complex ions, isomerism may be caused by the interchange
of ligands between the anion and cation, for example [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6].
Intermediate types between these extremes are also possible. For example,

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 5.17 Isomerism 201

Formula Possible number of isomers

3+ 3−
[Co(NH 3 )6 ] [Cr(NO2 )6 ] 6
[Co(NH 3 )6 ]3 + [Co(NO2 )6 ]3 − 3
2+ 2−
[Pt(NH 3 )4 ] [Pt Cl 4 ] 2
[Pt(NH 3 )4 ]2 + [CuCl 4 ]2 − 4
3+ 3−
[Ni(NH 3 )6 ] [ Fe(C 2O4 )3 ] 4
[Ni(en)3 ]3+ [ Fe(C 2O4 )3 ]3− 4

Coordination position isomerism

In polynuclear complexes an interchange of ligands between the different metal nuclei gives rise to
positional isomerism. An example of this type of isomerism is given in Figure 5.27.

NH2
(NH3)4Co Co(NH3)2Cl2 Cl2 Cl Cl Cl
Pd Pd
O2
Ph3P Cl PPh3
and and
Ph3P Cl Cl
NH2
Pd Pd
Cl(NH3)3Co Co(NH3)3Cl Cl2 Ph3P Cl Cl
O2

Figure 5.27 Coordination position isomers.

Geometrical isomerism and optical isomerism

In disubstituted complexes, the substituted groups may be adjacent or opposite to each other. This gives
rise to geometric isomerism. This type of isomerism is observed in complexes with coordination number
four, five and six.
If a molecule is asymmetric, it cannot be superimposed on its mirror image. The molecule and its mirror
image have the type of symmetry as shown by the left and right hands and are called an enantiomorphic
pair. The two forms are optical isomers. They are called either dextro or laevo (often shortened to d or l),
depending on the direction in which they rotate the plane of polarized light in a polarimeter. (d rotates to
the right, l to the left.) Optical isomerism is common in octahedral complexes involving bidentate groups.
The geometrical and optical isomerism for complexes having coordination numbers four, five and six
is discussed as follows.
1. For coordination number 6
The complexes with coordination number 6 can exist in octahedral geometry. The following cases may
arise in complexes with octahedral geometry.
a. All ligands are monodentate, having no chiral centre: The various structures possible are listed
below. Here M is the metal ion and a, b, c, d, e and f represent monodentate ligands.

Formula Possible number Possible number of Possible number of

of stereoisomers enantiomer pairs geometrical isomers
Ma6 1 0 1
Ma5 b 1 0 1

(Continued)

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202 Chapter 5 Coordination Compounds

(Continued)

Formula Possible number Possible number of Possible number of

of stereoisomers enantiomer pairs geometrical isomers
Ma4 b2 2 0 2
Ma4 bc 2 0 2
Ma3 b3 2 0 2
Ma3 b2c 3 0 3
Ma3 bcd 5 1 4
Ma2 b2c2 6 1 5
Ma2 b2cd 8 2 6
Ma2 bcde 15 6 9
Mabcdef 30 15 15

b. When ligands are bidentate symmetrical and monodentate type, having no chiral centre: The various
structures possible are listed below. Here M is the metal ion, AA is a bidentate symmetrical ligand
and a, b, c, and d represent monodentate ligands.

Formula Possible number Possible number of Possible number of

of stereoisomers enantiomer pairs geometrical isomers
[M(AA)3] 2 1 1
[M(AA)2a2] 3 1 2
[M(AA)2ab] 3 1 2
[M(AA)a4] 1 0 1
[M(AA)a3b] 2 0 2
[M(AA)a2b2] 4 1 3
[M(AA)a2bc] 6 2 4
[M(AA)abcd] 12 6 6

c. When ligands are bidentate unsymmetrical and monodentate type having no chiral centre: The vari-
ous structures possible are listed below. Here M is the metal ion, AB is a bidentate unsymmetrical
ligand and a, b, c, and d represent monodentate ligands.

Formula Possible number of Possible number of Possible number of

stereoisomers enantiomer pairs geometrical isomers
[M(AB)3] 4 2 2
[M(AB)2a2] 8 3 5
[M(AB)2ab] 11 5 6
[M(AB)a4] 1 0 1
[M(AB)a3b] 4 1 3
[M(AB)a2b2] 6 2 4
[M(AB)a2bc] 12 5 7
[M(AB)abcd] 24 12 12

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 5.17 Isomerism 203

Some examples of each kind are illustrated as follows:

(i) For [Ma 2 b 2 c 2 ]n± : The various structures possible are shown in Figure 5.28.
a a a b c
c a c b b c a c a b

M M M M M

c b b c b c a c a b
b a a b c
Optically active Optically inactive Optically inactive Optically inactive Optically inactive
(cis w.r.t. all) (trans w.r.t all) (trans w.r.t a) (trans w.r.t b) (trans w.r.t c)

Figure 5.28 Possible structures for octahedral complexes of the type [Ma2b2c2]n±.

(ii) For [M(AA)a 2 bc]n± : The various structures possible are shown in Figure 5.29.
a a a b
A b A a A a A a

M M M M

A c A b A c A a
a c b c
Optically inactive Optically active Optically active Optically inactive

Figure 5.29 Possible structures for octahedral complexes of the type [M(AA)a2bc]n± .

(iii) For [M(AB)a 2 b 2 ]n± : The various structures possible are shown in Figure 5.30.
a a a b
A b A a A b A a

M M M M

B b B b B a B a
a b b b
Optically inactive Optically active Optically active Optically inactive

Figure 5.30 Possible structures for octahedral complexes of the type [M(AB)a2b2]n± .

(iv) For Ma 3 b 3: The various structures possible are shown in Figure 5.31.
a a
a a b a

M and M

b b a b
b b
Facial (Fac) Meridional (Mer)

Figure 5.31 Possible structures for octahedral complexes of the type Ma3b3.

2. For coordination number 5

a. The formula Ma 5 can exist in two possible geometries, i.e. trigonal bipyramidal and square pyramidal.
These two are also called allogon isomers (Figure 5.32).
a
a a a a

M a M

a a a a
Trigonal bipyramidal Square pyramidal

Figure 5.32 Possible structures for complexes of the type Ma5.

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204 Chapter 5 Coordination Compounds

b. The formula Ma 2 b 3 can exist in six possible geometries as shown in Figure 5.33.
a b b
b a b a b b b a b a
a b

M b M a M a M M M

b a b b a b b b a b a b
(a) (b) (c) (d) (e) (f)

Figure 5.33 Possible structures for complexes of the type Ma2b3. Here (a), (b), (c) are geometrical
isomers of each other but any one of (a), (b), (c) and any one of (d), (e), (f) are allogon isomers
of each other.
3. For coordination number 4
The complexes with coordination number four (e.g. Ma 4 ) can have two possible geometries, i.e. tetra-
hedral or square planar (Figure 5.34).
a
a a

M M
a
a a
a a
Tetrahedral Square planar

Figure 5.34 Possible structures for complexes with coordination number four.
a. For tetrahedral geometry:
(i) If the ligands are different, then no geometrical isomerism is observed in tetrahedral geometry,
i.e. complexes of the type Mabcd exist as two optical isomers only (Figure 5.35).
a a

M d M
d

b c c b
dl - pair

Figure 5.35 Possible structures for complexes of the type Mabcd.

(ii) If symmetrical bidentate ligands having no chiral center are present, then the complexes are
optically inactive. For example, the following compounds are optically inactive: [Zn(acac)2 ]0 ,
o -
[Be(C 2O4 )2 ]2− and B C6H4 2
o
(iii) If unsymmetrical bidentate ligands are present, then the compound will be optically active. For
example, the following compounds are optically active:
o -
B C
C6 H 4 2 and [Zn(gly)2 ]0
o

O
b. For square planar geometry:
(i) The complexes with formula [Ma 4 ]n ± , [Ma 3 b]n ± , [M(AA)2 ]n ± , [M(AA)a 2 ]n ± and [M(AB)a 2 ]n ±
do not show geometrical isomerism since possible geometry is only one. Here, a, b are simple
monodentate ligands; AA is a symmetrical bidentate ligand and AB is an unsymmetrical
bidentate ligand, all having no chiral centre.
(ii) The complexes with formulae [Ma 2 b 2 ]n± and [Ma 2 bc] can have two geometrical isomers, as
shown in Figure 5.36 a and 5.36 b respectively.

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 5.17 Isomerism 205

a b a a

M M

c/b a b b/c
trans cis
(a) (b)

Figure 5.36 Geometrical isomers for complexes of the type [Ma2b2]n± and [Ma2bc] .
(iii) The complexes with formula [Mabcd]n± can have three geometrical isomers (Figure 5.37).
For these isomers, cis and trans terminology is not applicable or it is to be mentioned with
respect to a, b or b, c, etc.
a b a c a c

M M M

d c d b b d

Figure 5.37 Geometrical isomers for complexes of the type [Mabcd]n±.

(iv) For complex with the type [M(AB)2 ]n± , two geometrical isomers are possible (Figure 5.38).
A A A B

M M

B B B A

Figure 5.38 Geometrical isomers for complexes of the type [M(AB)2]n± .

(v) In complexes of the type [M(AB)ab]n±, two geometrical isomers are possible (Figure 5.39).
A a A b

M M

B b B a

Figure 5.39 Geometrical isomers for complexes of the type [M(AB)ab]n±.

(vi) The complexes of the type [M(AA)2 ]n± can show geometrical isomerism if the ligand AA has a
chiral centre. For example, consider the geometrical isomers of [Pt(bn)2 ]2+ (Figure 5.40), where
∗ ∗
bn : CH 3 − C H(NH 2 ) − C H(NH 2 ) − CH 3 .
Me N N Me Me N N Me

Pt2+ Pt2+

Me N N Me Me N N Me
Optically inactive Optically inactive

Me N N Me Me N N Me

Pt2+ Pt2+

Me N N Me Me N N Me
Optically active Optically active

Me N N Me

Pt2+

Me N N Me
Optically inactive

Figure 5.40 Geometrical isomers of [Pt(bn)2]2+.

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206 Chapter 5 Coordination Compounds

(vii) The complexes of the type [M(AB)2 ]n± can show geometrical as well as∗optical isomerism if the
ligand AB has a chiral centre. For example, the compound [Pd(NH 2 − C H(CH 3 ) − CO2− )2 ]0 can
have four geometrical isomers (Figure 5.41).

Me Me H H2 N O
H2N NH2 O
C C C C
H H Me H
Pd2+ Pd2+
C C C C
O O O NH2 Me
O O O
Optically inactive (Plane of symmetry) Optically active
(cis-cis) (trans-cis)

H Me Me
H2N NH2 H 2N O O
C C C C
H H H
Me Pd2+ Pd2+
C C C C
O O Me
O O O O NH2
Optically active Optically inactive (Centre of symmetry)
(cis-trans) (trans-trans)

Figure 5.41 Geometrical isomers of [Pd(NH2 − CH(CH3) − CO−2 )2]0.

(viii) Geometrical isomers are possible for square planar binuclear complexes of the type,
[Pd 2Cl 4 (PPh 3 )2 ] (Figure 5.42)

Cl Cl Cl Cl Cl PPh3
Pd Pd Pd Pd

Ph3P Cl PPh3 Ph3P Cl Cl

cis trans

Figure 5.42 Geometrical isomers of [Pd2Cl4(PPh3)2].

| SINGLE CORRECT CHOICE TYPE QUESTIONS

1. Which of the following is the correct match between (C) [RhF6]3− : Δo > P : Diamagnetic
IUPAC name and complex? (D) [Fe(NH3)6]2+ : Δo < P : Paramagnetic
(A) Hexa-m-acetato(O,O′)-m4-oxido-tetra- 5. What is the difference in the EAN value of metal in
beryllium(II) : [Be4O(CH3COO)6] the complexes sodium pentacynidonitrosyliumfer-
(B) m-amido-decaamminecobalt(III) nitrate: rate (II) and pentaaquanitrosyliron (II) ion?
[(NH3)5CoNH2Co(NH3)5] (NO3)5 (A) 1
(C) Pentaamminechloridoplatinum (IV) ammine- (B) 2
pentachloridoplatinate (IV): [Pt (NH3)5Cl] (C) Cannot be predicted
[Pt Cl5(NH3)] (D) None of these
(D) Hydrogen hexfluoridosilcate (IV) acid: H2[SiF6]
6. Which of the following statements is incorrect?
2. Which of the following options represents the oxi- (A) According to Werner’s coordination theory, sec-
dation state of Co and Cr in the given complex? ondary valencies of central metal atom is fixed
[CO(NH3)4 (NO2)2] [Cr(NH3)3(NO2)3] towards a particular ligand.
(A) 2, 3 (B) 3, 2 (B) C2H4 and NO act as non-classical ligand.
(C) 3, 3 (D) 2, 2 (C) The bond order of O2 is 2 and it is colourless gas
3. Which of the following complexes does not show and paramagnetic in nature.
optical activity? (D) In PF3, the P–F bond order decreases when PF3
(A) [Co(EDTA)]− (B) [Pt(bn)2]2+ forms complex with a metal and shows synergic
(C) [Pt(pn)2]2+ (D) [Pt(en)2]2+ bonding.
4. Which of the following is not correctly matched? 7. Which of the following statements is incorrect?
(P : Pairing energy) (A) [Fe(EDTA)]− is diamagnetic and optically active.
(A) [Co(H2O)6]3+ : Δo < P : Paramagnetic (B) Nitro triacetate(nta3-) is tetradentate ligand.
(B) [NiF6]2− : Δo < P : Diamagnetic (C) [Mn(CO)5] gets stabilized through dimerization.
(D) [Mn(CO)5 gets stabilized by reduction.

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 Single Correct Choice Type Questions 207

8. Which of the following complex has low spin? 17. Select correct statement about this complex [Pt(ox)
(A) K2 [Fe(CN)6] (py)2(O2)(H2O)]
(B) K3 [Fe(CN)6] (A) Oxidation state of O2 is −1.
(C) K3 [Co(OX)3] (B) EAN of Pt is 86.
(D) All of these (C) Complex will show geometrical as well as optical
9. Which of the following complex is optically inactive isomerism.
but its other geometrical isomer is optically active? (D) Both (B) and (C).
(A) trans-tetraaquadichloridocobalt (III) nitrate 18. Select the incorrect statement about metal carbonyl
(B) mer-triaquatrifluoridocobalt (III) complex compounds.
(C) trans-diamminebis (ethane 1, 2 diamine) cobalt (A) Metal-carbon bonds in metal carbonyls possess
(III) chloride both s and p character.
(D) trans-diamminedichloridoplatinum (II) (B) Due to synergic, bonding metal-carbon bond
10. Which of the following complexes is colourless? becomes weak.
(A) MnO4− (C) Due to synergic bonding carbon-oxygen bond
(B) [Ni(H2O)6]2+ strength decreases.
(C) [Cu(CH3CN)4]+ (D) In metal carbonyls the extent of synergic bond-
(D) [Cr(H2O)6]3+ ing will increase with increase in negative charge
on central metal ion.
11. The correct statement is
(A) [NiCl4]2− is sp3 hybridized and paramagnetic in 19. In which of the following complexes, spin only mag-
nature. netic moment is independent of the nature of ligand?
(B) [PtCl4]2− is dsp2 hybridized and paramagnetic in (L = monodentate ligand)
nature. II II
(C) [Ni(CO)4] is dsp2 hybridized and diamagnetic in (A) [NiL4] (B) [CoL6]
nature.
(D) [Co(H2O)6]3+ is sp3d2 hybridized and diamagnetic III III
in nature. (C) [FeL6] (D) [CrL6]
12. Which of the following complex is most stable? 20. Match the complexes with their characteristics.
(A) [Co(H2O)6]Cl3
(B) K3[Co(CN)6] Column-I Column-II
2−
(C) K3 [Co(OX)3] (P) [Ni(CN)2] (1) Spin magnetic moment
(D) K3 [CoF6] = 8 BM
13. The geometries of [Co(CO)4]− and [Cd (CN)4]2− are (Q) Fe(CO)5 (2) sp3 hybridized state of
(A) both square planar. central metal
(B) both tetrahedral. (R) [MnBr4]2− (3) dsp2 hybridized state
(C) tetrahedral and square planar, respectively. of central metal
(D) square planar and tetrahedral, respectively.
(S) [Ni(NH3)6]2+ (4) Trigonal bipyramidal
14. Which of the following species exists as optically complex
inactive form?
(A) [Fe(en)3]Cl3 Code:
(B) [RhCl(PPh3)(CO)(H2O)] P Q R S
(C) [Pd(en)2Cl2] (A) 3 4 1 2
(D) O (B) 3 4 2 1
=

C (C) 1 4 3 2
O (D) 2 4 3 1
Na B 21. Which of the following statements is correct for the
complex [ Fe(H 2O)5 NO] SO4 ?
O 2
(A) The EAN value of Fe in this complex depends on
15. Which of the following ligands can act as chelating the charge of NO ligand.
agent but does not have a chiral centre? (B) The EAN value of Fe in this complex does not
(A) nta3− (B) bn depend on the charge of NO ligand.
(C) pn (D) None of these (C) The hybridization of the central atom is d 2 sp3.
16. Which type of isomerism may be shown by the com- (D) It is paramagnetic with μ = 1.73 B.M.
plex [Ru(NH3)4 )(H2O)(S2O3)]NO3? 22. The EAN value of [Ti(σ − C 5 H 5 )2 (π − C 5 H 5 )2 ]0 is
(A) Ionization isomerism (A) 32
(B) Linkage isomerism (B) 33
(C) Geometrical isomersim (C) 34
(D) All of these (D) 35

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208 Chapter 5 Coordination Compounds

23. Which of the following ligands can show linkage

isomerism? (C) [ Fe(H 2O)6 ]2 [ Fe(CN)6 ]
(A) NO (D) None of these.
(B) NH3 28. Which of the following statements is true for the
(C) NO3− compounds: [Co F6 ]3− (I) and [Ni F6 ]2− (II)?
(D) None of these. (A) Both I and II are paramagnetic.
24. Which of the splitting patterns is correct for (B) Both I and II are diamagnetic.
[ Fe(H 2O)6 ]3+ and [ Fe(CN)6 ]3− respectively? (C) I is paramagnetic while II is diamagnetic.
(D) I is diamagnetic while II is paramagnetic.
(A) eg
29. The correct order of C − C bond length in the follow-
eg
ing compounds is:
t 2g (I) C 2 F4 (II) C 2 H 4 (III) [PtCl 3 (C 2 H 4 )]−
E t 2g (A) I >II >III
(B) I < II < III
[Fe(H2O)6]3+ [Fe(CN)6]3− (C) I >III >II
(D) I >II = III
(B)
eg eg 30. Which of the following statements is incorrect?
(A) CN− is a weaker p-acceptor than CO
E
t2g t2g (B) The Fe − C bond length is more in [ Fe(CN)6 ]4− as
compared to that in [ Fe(CN)6 ]3−
(C) The Fe − C bond length is less in [ Fe(CN)6 ]4− as
(C) compared to that in [ Fe(CN)6 ]3−
eg
eg (D) The d Fe −O {in [ Fe(H 2O)6 ]3+ } <
E t 2g
t 2g d Fe −O { in [ Fe(H 2O)6 ]2 + }

31. The V−C distances in V(CO)6 and [V(CO)6 ]− are

(D) respectively (in pm)
eg eg (A) 200, 200 (B) 193, 200
a a (C) 200, 193 (D) 193, 193
E
t 2g t 2g
32. Which of the following complexes contains a cationic
ligand?
(A) [Ni(η 5 − C 5 H 5 )2 ]+
25. In [isothiocyanatothiocynato (1-dipheny lphosphino-3-
(B) [V(η6 − C 7 H 8 )(η 7 − C 7 H 7 )]+
dimethylaminopropane) palladium (II)], the Pd−
NCS combination is linear while the Pd−SCN (C) [ Fe(CO)2 (NO)2 ]0
combination is bent with the ∠Pd − S − C of 107.3°. (D) None of these.
Predict the structure adopted for SCN − group in this 33. Which of the following compounds can show cis-trans
complex. isomerism?
−
(A) [Ni(η 3 − C 3 H 5 )2 ]0 (B) [Zn(gly)2 ]0
(A) S − C ≡ N:
− 0
(C) [Pd(acac)2 ] (D) None of these.
(B) S = C = N
(C) Hybrid of (A) and (B). 34. Which among the following statements is correct
(D) Cannot be predicted. regarding the bonding of Pt and C 2 H 4 in Zeise’s salt?
(I) It involves s-donation from the p-orbital of
26. Choose the correct order for Δo for the following alkene into vacant metal hybrid orbital.
complexes. (II) It involves p-donation from the p-orbital of
I: [Co(H 2O)6 ]2+ II: [Co(H 2O)6 ]3+ alkene into vacant metal hybrid orbital.
III: [ Fe(H 2O)6 ]3+ IV: [ Fe(CN)6 ]3− (III) It involves p-back donation from the filled metal
d-orbital (or hybrid) into the vacant antibonding
(A) I < II < III < IV
orbital of alkene.
(B) I < III < II < IV
(A) I, II and III (B) I and III
(C) I < II = III < IV
(C) II and III (D) III only
(D) I < II < IV < III
35. Which of the following compounds is resolvable into
27. [ Fe (H 2O)4 (CN)2 ] is the empirical formula of a com- d or l forms?
pound which has a magnetic moment corresponding o
(A) [Zn(acac)2 ]0 (B) B C6H4
to 2 23 unpaired electrons per iron. The best possible o 2

formula of the compound is o 0

(C) B C6H3 CH3 (D) [PdCl 2 (en)] .
(A) [ Fe(H 2O)4 (CN)2 ]2 [ Fe(H 2O)4 (CN)2 ] o 2

(B) [ Fe(H 2O)6 ][ Fe(H 2O)2 (CN)4 ]

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 Multiple Correct Choice Type Questions 209

| MULTIPLE CORRECT CHOICE TYPE QUESTIONS

1. Which of the following bidentate ligand(s) has simi- 9. Which of the following types of isomerism is/are possible
lar donor atoms? for the compound [Co(Cl)(en)2 (NH 2 − C 6 H 4 − Me)]
(A) py (B) bn Cl 2?
(C) en (D) gly− (A) Geometrical isomerism
(B) Linkage isomerism
2. Consider the formation of the following metal car-
(C) Optical isomerism
bonyl complex.
(D) Ligand isomerism
Ni + 4CO prNi (CO)4s
carbon (Y) 10. Which of the following pairs of compounds has the
monooxide
same EAN value but does not obey Sidgwick EAN
Select the correct statement(s) about compound Y.
rule?
(A) Compound Y has no counter ion.
(A) Fe(CO)5 , Ni(CO)4
(B) Four monodentate uninegative ligands are con-
nected with central metal ion. (B) [Fe(NH 3 )6 ]2 + , [Cr(C 2O4 )3 ]3 −
(C) This compound produces two ions in its aqueous (C) [Fe(H 2O)5 NO]2 + , [Mn(CO)6 ]
solution. (D) [Ti(CO)6 ], [Mn(C 2O4 )3 ]3−
(D) Total number of electrons remains the same
during the formation of compound Y. 11. Which of the following statements is/are correct for
ferrocene?
3. Which of the following statement(s) is/are correct?
(A) The dipole moment of the eclipsed form of
(A) Ni(CO)4 is diamagnetic and sp3 hybridized.
ferrocene is zero.
(B) Complex [Cu(CN)4]3− is paramagnetic and Cu2+
(B) The dipole moment of the staggered form of
ion is dsp2 hybridized.
ferrocene is non-zero.
(C) [MnBr4]2− ion is paramagnetic and sp3 hybridized.
(C) All C atoms are equidistant from Fe2+ ion.
(D) Vaska’s catalyst, that is, [IrCl(CO)(PPh3)2] is dia-
(D) Synergic bonding takes place in the π * orbital of
magnetic and Ir+ ion is dsp2 hybridized.
C atoms.
4. Which of the following chemical species can act a 12. Which of following IUPAC names is/are correct for
non-classical ligand(s)? the complex [CO(NH 3 )2 (tn) (s − C 3 H 5 )2 ]NO3 ?
(A) CO (B) C2H4 (A) Diallyldiamminetrimethylenediaminecobalt
(C) NO+ (D) PR3 (III) nitrate.
5. Which of the following is/are characteristic of (B) 1,3-Diaminopropanediamminediallylcobalt
ferrocene? (III) nitrate
(A) Cyclopentadienyl act as a p-donor ligand. (C) Diammine-1, 3-diaminopropanedicyclopropyl
(B) Fe is in (+2) oxidation state. cobalt (III) nitrate.
(C) It is organometallic compound. (D) Diallyldiammine-1, 3-diaminopropanecobalt
(D) It has a metal–carbon bond. (III) nitrate.
6. Which of the following chemical species is/are dia- 13. A complex compound consists of 1 mole of Co 3+
magnetic as well as coloured? ion, 6 moles of NH 3, 6 moles of NO2− and 1 mole of
(A) Hexafluoridoferrate(III) ion Cr 3+ ion. The complex has neither the highest value
(B) Permangnate ion nor the lowest value of electrical conductivity. The
(C) Brown ring complex possible formulae for the complex is/are
(D) Chromate ion (A) [Cr(NH 3 )5 (NO2 )][Co(NH 3 )(NO2 )5 ]
7. In case of [Mn(NH3)6] 2+ (B) [Co(NH 3 )6 ][Cr(NO2 )6 ]
(A) relation in between 'q and P is 'q < P. (C) [Cr(NH 3 )4 (NO2 )2 ][Co(NH 3 )2 (NO2 )4 ]
(B) two unpaired electrons are present in axial (D) [Co(NH 3 )5 (NO2 )][Cr(NH 3 )(NO2 )5 ]
d-orbital of Mn2+.
14. Which of the following complexes is/are square planar?
(C) hybridization of central metal atom of complex
is d2sp3. (A) [Ag F4 ]−
(D) d-orbitals involved in hybridization of central (B) [AuCl 4 ]−
atom are dx and dxy.
2 (C) [NiCl 2 (PPh 3 )2 ]
(D) [NiCl 2 (PMe3 )2 ]
8. [Pt(NH 3 )4 ][PtCl 3 (NH 3 )]2 is the polymerization
isomer of which of the following compounds? 15. Choose the correct statement(s) from the following:
(A) Monovalent silver complexes (coordination
(A) cis -[PtCl 2 (NH 3 )2 ] number 2) are diamagnetic.
(B) trans - [PtCl 2 (NH 3 )2 ] (B) Bivalent silver complexes (coordination number
(C) [Pt(NH 3 )3 Cl]2 [PtCl 4 ] 4 and 6) are paramagnetic with μ = 1.73 BM .
(D) None of these.

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210 Chapter 5 Coordination Compounds

(C) Trivalent silver complexes (coordination number A

(B) n±
4) are diamagnetic. A
(D) AgO is diamagnetic. A

16. Which of the following complexes have geometrical M

isomers that are optically inactive due to the pres- A
A
ence of centre of symmetry?
A
(A) [Pt(bn)2 ]2+
(C) A n±
(B) [ Fe(gly)3 ]0 B A
(C) [Cr(H 2O)3 Cl 3 ]0 M
(D) [Pd{NH 2 − CH(CH 3 ) − CO2 }2 ]0 A B
17. Which of the following complexes is/are optically B
inactive due to the presence of plane of symmetry?
(D)
(A) a n± Me
Me H2N NH2
a a
H
M Pt2+ H
H
H
b c
H2N NH2
b Me Me

| COMPREHENSION TYPE QUESTIONS

Passage 1: For Questions 1–2 3. The number of water molecules of crystallization are
respectively
Isomers in coordination chemistry include many types. (A) zero, two.
In structural isomers hydrate or solvent isomers, ioniza- (B) zero, zero.
tion isomers and coordination isomers have same overall (C) two, zero.
formula but have different ligands attached to the central (D) two, two.
atom or ion.
The terms linkage isomerism or ambidentate isomer- 4. The correct formula and geometry of the first complex is
ism are used for cases of bonding through different atoms (A) [Ni(H 2O)2 (NO3 )2 ] ⋅ 4 NH 3 ; tetrahedral.
of the same ligand. (B) [Ni(NH 3 )4 ](NO3 )2 ⋅ 2 H 2O; tetrahedral.
1. Which of the following is not correctly matched
(C) [Ni(NH 3 )4 ](NO3 )2 ⋅ 2 H 2O; square planar.
against indicated isomerism?
(A) [Co(NH3)4(H2O)Cl]Br2 : Ionization isomerism (D) [Ni(NH 3 )4 (H 2O)2 ](NO3 )2 ; octahedral.
(B) [Rh(PPh3)2(CO)(NCS)2] : Linkage isomerism 5. Which of the following statements are true for the
(C) [Pt(NH3)4][(PtCl4)] : Coordination isomersim second complex?
(D) [Zn(gly)2] tetrahedral : Geometrical isomerism (A) It has the EAN value of 36.
2. Ma3b3 complex has two geometrical forms: facial and (B) It can show optical isomerism.
meridional. Then which of the following statement is (C) It cannot show geometrical isomerism.
incorrect? (D) It produces three-fold freezing point depression.
(A) In facial isomers, three same ligands occupy adja- Passage 3: For Questions 6–8
cent positions on octahedron face.
(B) In meridional isomers, same ligands are present Some ligands not only donate their lone pair to the cen-
at 90° and 180° angles. tral metal atom but also accept the electron cloud from
(C) In facial isomers, same ligands are present only the central metal atom. This is known as synergic bonding.
at 90° angle. 6. In which of the following cases is the bond energy of
(D) Both isomers (facial and merridional) are opti- C − O bond minimum?
cally active. (A) Free CO molecule.
(B) Terminal CO group in a complex.
Passage 2: For Questions 3–5 (C) Doubly bridging CO in a complex.
The magnetic moment for two complexes of empirical (D) Triply bridging CO in a complex.
formula Ni(NH 3 )4 (NO3 )2 ⋅ 2 H 2O is zero and 2.84 BM 7. Select the correct order for the stretching frequencies
respectively. The second complex is not a neutral complex. of C – O bond the following compounds.

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 Assertion–Reasoning Type Questions 211

(I) fac-Mo(CO)3 (PF3 )3 (II) fac-Mo(CO)3 (PCl 3 )3 13. Which of the following has higher multiple bond
(III) fac-Mo(CO)3 (PMe3 )3 character in M−C bond?
(A) I < II < III. (B) I >II >III. (A) [Ni(CO)4]
(C) I = II = III. (D) II >I >III. (B) [Co(CO)4]–
(C) [Fe(CO)4]2–
8. In which of the following complexes is the bond
(D) (B) and (C) both have equal bond character in
order of C – O bond minimum?
(A) [Mn(CO)6 ]+ (B) [Cr(CO)6 ] M−C bond.

(C) [Ti(CO)6 ]2− (D) [V(CO)6 ]− Passage 6: For Questions 14–16

Coordination number of a central atom is its inherent
Passage 4: For Questions 9–11 property and has no role on its isomerism.
+ −
The series [CoCl n (NH 3 )6 − n ]Cl 3− n gives rise to 4, 3, 2, and 14. [CoCl 2 (NH 3 )4 ] + Cl → [CoCl 3 (NH 3 )3 ] + NH 3
zero ions in solution respectively, for n = 0, 1, 2, and 3. If in the above reaction two isomers of the product
are obtained, which is true for the initial (reactant)
9. For which value of n is the electrical conductivity of
complex?
the complex maximum?
(A) compound is in cis form.
(A) 0 (B) 1
(B) compound is in trans form.
(C) 2 (D) 3
(C) compound is in both (cis and trans) forms.
10. For which value of n will the complex not show (D) cannot be predicted.
optical isomerism?
(A) 1 (B) 2 15. When 0.1 mol of CoCl3(NH3)5 is treated with excess
(C) 3 (D) All of these of AgNO3, 0.2 mol of AgCl are obtained. The conduc-
tivity of solution will correspond to (cation : anion)
11. For which value of n, the complex will show mini- (A) 1 : 3 electrolyte
mum freezing point depression? (B) 1 : 2 electrolyte
(A) 0 (B) 3 (C) 1 : 1 electrolyte
(C) 2 (D) 1 (D) 3 : 1 electrolyte

Passage 5: For Questions 12-13 16. Due to the presence of ambidentate ligands coor-
dination compounds show isomerism, palladium
Synergic bond is a type of back bonding. complexes of the type [Pd(C6H5)2(SCN)2] and
12. Which of the following has higher stretching fre- [Pd(C6H5)2(NCS)2] are
(A) linkage isomers.
quency for C−O bond?
(B) coordination isomers.
(A) [Ni(CO)3PF3]
(C) ionisation isomers.
(B) [Ni(CO)3(PMe3)]
(D) geometrical isomers.
(C) Both have equal stretching frequency.
(D) None of these.

| ASSERTION–REASONING TYPE QUESTIONS

In the following set of questions, a Statement I is given 2. Statement I: Under the strong field ligand only, the
and a corresponding Statement II is given below it. Mark degeneracy of the d-orbital is lost.
the correct answer as:
Statement II: After splitting of the d-orbitals also,
(A) If both Statement I and Statement II are true
Hunds rule is not violated anywhere.
and Statement II is the correct explanation of
Statement I. 3. Statement I: NO is 3e-donor.
(B) If both Statement I and Statement II are true but Statement II: The antibonding electron of NO is very
Statement II is not the correct explanation for much susceptible to donate apart from its lone pair.
Statement I.
4. Statement I: Dithiooxalate is bidentate ambidentate
(C) If Statement I is true but Statement II is false.
ligand.
(D) If Statement I is false but Statement II is true.
Statement II: At a time either two S atoms or two O
1. Statement I: Only cis Pt(NH 3 )2 Cl 2 reacts with oxalic atoms act as donor atoms.
acid (H 2C 2O4 ) to form [PtCl 2 (ox )]2 − not the trans
isomer. 5. Statement I: PF3 is weakest donor as well as weakest
acceptor compared to PMe3 in synergic bonding.
Statement II: The oxalate ion is a bidentate ligand
which occupies adjacent positions only. Statement II: Me group is having +I effect while
F atom is having −I effect.

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212 Chapter 5 Coordination Compounds

6. Statement I: SnCl 2 does not act as ligand but SnCl 3− Statement II: t26g eg1 is the electronic configuration for
acts as good ligand. both cases.
Statement II: On addition of Cl − to SnCl 2, the orbital 9. Statement I: CdS is yellow coloured.
containing the lone pair acquires less s-character.
Statement II: The d − d transition does not take place
7. Statement I: [Cr(H 2O)6 ]3+ is less acidic compared to in this compound.
[ Fe(H 2O)6 ]3+ .
10. Statement I: [Co(en)2(NH3)Br]SO4 has lower electri-
Statement II: Both are inner orbital complexes. cal conductivity as compared to [Co(en)2(NH3)SO4]
8. Statement I: Under the influence of a strong field Br.
ligand, d7-system will have only one unpaired elec-
Statement II: Both the ions produce the same
tron either in coordination number six or four.
number of ions in solution.

| INTEGER ANSWER TYPE QUESTIONS

The answer to each of the following questions is a 6. The number of stereoisomers, optically active iso-
non-negative integer. mers and geometrical isomers for the complex
[Co(en)(pn)(NO2 )2 ]+ is ___________ , ___________
1. Find the number of diamagnetic complexes which
show geometrical isomerism. and ___________ respectively.
[NiF6]2−, [Pt(NH3)2Cl2], [CoF3(H2O)3], 7. If CFSE increases by 30% and 40% respectively
[Fe(en)2Cl2]Cl, [Ni(NH3)2Cl2], [Ni(PPh3)2Cl2] for Co 3+ to Rh 3+ and for Rh 3+ to Ir 3+, then the total
increase in CFSE for Ir 3+ with respect to Co 3+ is
2. Find the number of complexes in which stability con- ___________%.
stant value is greater than the stability constant value
of [FeF6]3−. 8. The hapticity of the organic ligand in the following
[Fe(CN)6]3−, [Fe(ox)3]3−, [Fe(NH3)6]3+, complex is ___________.
[Fe(H2O)6]3+, [Fe(en)3]3+
3. Suppose we replace all the fluoride and water ligands
from [CoF3(H2O)3] by oxalate ligand without chang-
ing the oxidation state and coordination number of
central metal atom or ion. Predict the number of
parameters among following which increase in the
Mo(CO)3
newly formed complex.
(I) Δ0 9. The coordination number of the central atom in
(II) Number of t2g electrons [Co(NH 3 )4 SO4 ]NO3 is ___________ .
(III) Number of eg electrons 10. The number of optically active isomers for the com-
(IV) Number of stereoisomers plex of formula [Ma 2 b 2 cd] is___________.
(V) Number of geometrical isomers
(VI) EAN value 11. The number of unpaired electrons in the t2g set of d
orbitals in the complex [Co(H 2O)3 F3 ] is __________.
4. The number of coordination isomers possible for:
12. The number of unpaired electrons present in [NiF6]2–
(a) [Pt(NH 3 )4 Cl 2 ][PtCl 4 ] is ___________.
is ___________.
(b) [Pt(NH 3 )4 Cl 2 ][Pt(CN)4 ] is ___________.
13. The sum of stereoisomers of complex-A, complex-B
(c) [ Fe(NH 3 )6 ]3+ [Cr(C 2O4 )3 ]3− is ___________. and complex-C in following reaction is ___________.
(d) [Pt(NH 3 )4 ]2 + [CuCl 4 ]2 − is ___________. +2(pyridine) + NH
[PtCl 4 ] 2 − ⎯⎯⎯⎯⎯⎯→[Complex-A] ⎯⎯⎯
−→
3
(e) [Pt(NH 3 )3 Cl]2 [Pt(SCN)4 ] is ___________. −2Cl − −Cl
+ Br −
5. The possible number of stereoisomers for the for- [Complex-B] ⎯⎯⎯⎯⎯→[Complex-C ]
−(pyridine)
mula [Ma 3 b 2 ]n± is ___________.

| MATRIX–MATCH TYPE QUESTIONS

In each of the following questions, statements are given while those in Column II are labelled as (P), (Q), (R), (S)
in two columns, which have to be matched. The state- and (T). Any given statement in Column I can have cor-
ments in Column I are labelled as (A), (B), (C) and (D), rect matching with one or more statements in Column II.

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 Matrix–Match Type Questions 213

1. Match the processes with their respective characteristics.

Column-I Column-II Column I Column II

(A) [CoCl6] → 3−
(P) Change in number of (C) [M(AB)3 ] n±
(R) The number of geometrical
[Co(en)2Cl2]+ geometrical isomers isomers is four.
(B) [Fe(EDTA)]− → (Q) Change in number (D) [Ma 2 b 2 cd]n± (S) The number of optically
[Fe(en)3]+3 of stereoisomers active isomers is four.
(C) [Pt(NH3)4]+2 → (R) EAN value does not
3. Match the complex compounds with their properties.
[Pt(ONO)4]2− change
(D) [Co(H2O)6]Cl3 → (S) Change in number of Column I Column II
[Co(H2O)3Cl3]·3H2O structural isomers (A) [Co F3 (H 2O)3 ]0 (P) Diamagnetic and low spin
(T) Change in magni- complex.
tude of Δ (B) [Cr (C 2O4 )3 ]3−
(Q) Paramagnetic and outer
orbital complex.
2. Match the metal complex type with the properties. −
(C) [AuCl 4 ] (R) Paramagnetic but inner
Column I Column II orbital complex.
(A) [M(AA)abcd] n±
(P) All geometrical isomers (D) [ Fe(CN)6 ]3− (S) Diamagnetic and high spin
are optically active. complex.
(B) [M(AB)a 2 b 2 ]n± (Q) The number of (T) Paramagnetic and low spin
stereoisomers is four. complex.

4. Match the processes carried out with the changes observed.

Column I Column II
n± −a / +b n±
(A) [M(AB)a 3 b] ⎯⎯⎯→[M(AB)a 2 b 2 ] (P) The number of pairs of enantiomers is increased by one.
n± −c / +b n±
(B) [M(AB)a 2 bc] ⎯⎯⎯→[M(AB)a 2 b 2 ] (Q) The number of geometrical isomers in the final
product is four.
(C) [Ma 3 b 2 c]n ± ⎯−⎯⎯
b / +d
→[Ma 3 bcd]n ± (R) The number of stereoisomers is increased to double of
the original or decreased to half of the original.
(D) [M(AA)a 3 b]n ± ⎯−⎯⎯
a /+b
→[M(AA)a 2 b 2 ]n ±

5. Match the complex compounds with the properties 7. Match the pair of complex compounds with the
not depicted by them. properties that are same in them.

Column I Column II Column I Column II

2− 2
(A) [ Fe(CO)4 ] (P) Hybridization is dsp . (A) [Ti(H 2O)6 ]3+ (P) Hybridization.
and [Mn(NH 3 )6 ]2+
(B) [Pt(NH 3 )2 Cl 2 ] (Q) Hybridization is sp3.
(C) [Pt(bn)2 ]2+ (R) Exhibits geometrical (B) [AuCl 4 ]− (Q) Magnetic moment value.
isomerism. and [PdCl 4 ]2−
(D) [Zn(gly)2 ]0 (S) Low spin complex. (C) [ Fe(DMG)2 ]0 (R) EAN value.
(T) Exhibits optical 0
and Ni(DMG)2 ]
isomerism.
6. Match the pair of complex compounds with the proper- (D) [ Fe(CO)5 ]0 and (S) Shape.
ties that are different in them. [Ni(CO)4 ]0

Column I Column II 8. Match the coordination complex with the number of

(A) [Mn(NH 3 )6 ]SO4 (P) Hybridization. isomers formed.
and [Co(H 2O)6 ]Cl 3
Column I Column II
(B) (NH 4 )2 [Pt Cl 4 ] (Q) Magnetic moment n±
value. (A) [M(AB)3] (P) Complex having two
and K 2 [NiCl 4 ]
optically active isomers.
(C) K 4 [ Fe(CN)5 O2 ] (R) Magnetic behaviour. (B) [Ma2b2c2]n± (Q) Complex having four
and K 4 [ Fe(CN)6 ] stereoisomers.
(S) Electrical conductivity
with significant change.
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214 Chapter 5 Coordination Compounds

8. Match the coordination complex with the number of 9. Match the complex with its property.
isomers formed.
Column I Column II
Column I Column II 3–
(A) [Co(C2O4)3] (P) High-spin complex.
(A) [M(AB)3]n± (P) Complex having two (B) [CoF6]3– (Q) Low-spin complex.
optically active isomers. 2+
(C) [Ni(NH3)6] (R) Zero electron in eg set of
(B) [Ma2b2c2]n± (Q) Complex having four orbital.
stereoisomers.
(D) [Cr(CN)6]3– (S) Paramagnetic behaviour.
(C) [M(AA)a2b2]n± (R) Complex having odd number
(T) d2sp3 hybridization.
of geometrical isomers.
(D) [Ma3bcd]n± (S) Complex having even
number of geometrical
isomers.
(T) All geometrical isomerism
are not optically active.

| ANSWERS
Single Correct Choice Type Questions
1. (A) 8. (D) 15. (A) 22. (C) 29. (B)
2. (B) 9. (C) 16. (D) 23. (A) 30. (B)
3. (D) 10. (C) 17. (D) 24. (A) 31. (C)
4. (A) 11. (A) 18. (B) 25. (A) 32. (B)
5. (A) 12. (B) 19. (D) 26. (B) 33. (A)
6. (D) 13. (B) 20. (B) 27. (C) 34. (B)
7. (A) 14. (B) 21. (B) 28. (C) 35. (C)

Multiple Correct Choice Type Questions

1. (B), (C) 5. (A), (B), (C), (D) 9. (A), (C), (D) 13. (A), (D) 17. (A), (D)
2. (A), (D) 6. (B), (D) 10. (C), (D) 14. (A), (B), (D)
3. (A), (C), (D) 7. (A), (B) 11. (A), (C), (D) 15. (A), (B), (C), (D)
4. (A), (B), (C), (D) 8. (A), (B), (C) 12. (A), (C), (D) 16. (A), (D)

Comprehension Type Questions

1. (D) 5. (D) 9. (A) 13. (C)
2. (D) 6. (D) 10. (D) 14. (A)
3. (C) 7. (B) 11. (B) 15. (B)
4. (C) 8. (C) 12. (A) 16. (A)

Assertion–Reasoning Type Questions

1. (A) 3. (A) 5. (D) 7. (C) 9. (B)
2. (D) 4. (A) 6. (A) 8. (C) 10. (D)

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 Answers 215

Matrix–Match Type Questions

1. (A) →(P), (Q), (R), (T) (C) → (P) 6. (A) → (P), (Q), (R), (S) (D) → (P), (S), (T)
(B) → (R), (T) (D) → (T), (R) (B) → (P), (Q), (R)
9. (A) → (Q), (R), (T)
(C) → (R), (S), (T) (C) → (Q), (R)
4. (A) → (P), (Q) (B) → (P), (S)
(D) → (P), (Q), (R), (T)
(B) → (Q), (R) 7. (A) → (S) (C) → (S)
2. (A) → (P) (C) → (P), (Q) (B) → (P), (Q), (S) (D) → (R), (S), (T)
(B) → (R), (S) (D) → (P), (R) (C) → (P), (Q), (S)
(C) → (P), (Q), (S) (D) → (Q), (R)
5. (A)→ (P), (R), (S), (T)
(D) → (S)
(B) → (Q), (T) 8. (A) → (Q), (S)
3. (A) → (Q) (C) → (Q) (B) → (P), (R), (T)
(B) → (R) (D) → (P), (R), (S) (C) → (P), (Q), (R), (T)

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