Coordination Compounds
Introduction:
We have already seen the
ability of transition elements to form coordination compounds. This property is
however not restricted to transition elements alone. It is also exhibited by
certain other metals although to a lesser extent. In this unit we will study
the different aspects of coordination compounds including their application. We
will also study about organometallic compounds
Double Salt
and a Complex Ion:
Double
salt: Ionizes completely in solution. Example:
Mohr’s salt 
The
complex ion: The complex ion does not dissociate. Example: 
Co-ordination
compounds:
Co-ordination
compounds are compounds in which the central metal atom is linked to a number
of ions or neutral molecules by co-ordinate bonds i.e., donation of electron
pairs. They retain their identities even when dissolved in
water
Example:
Ni(CO)4, [Co(NH3)6]Cl3, K[Ag (CN)2]
Types of Complexes
1. Cationic
complexes- Those in
which the complex ion has a positive charge.
Example: [Co(NH3)6]3+
2. Anionic
complexes - Those
in which the complex carries a negative charge. Example: [Ag (CN)2]-
3. Neutral
complex- The
complex does not carry a charge. Example: [Ni(CO)4]
Ligands
Ligand (ligare is Latin, to bind): A ligand is a
molecule or ion that is directly bonded to a metal ion in a coordination
complex. Molecule or ion having a lone electron pair that can be used to form a
bond to a metal ion. A ligand uses a lone pair of electrons (Lewis base)
to bond to the metal ion (Lewis acid)
The donor
atoms, molecules or anions which donate an electron pair to the metal atom or
the atoms or group of atoms that surround the central metal atom. Example: NH3, H2O and CN-
Classification
of Ligands:
1.
Based on denticity:
Unidentate/ monodentate - If one
donor atom is present. Forms one bond to metal ion
Example: NH3, H2O.
Bidentate
- If two donor atoms are present on the ligand. Forms two bond to metal ion
Example:
NH2 - CH2 - CH2 - NH2 ethylene diamine.
Multidentate/ polydentate –If
more than two atoms are present on the lignd. more than
two bonds to a metal ion possible
Example: Ethylene diamine tetra acetic acid
(EDTA) is hexadentate
Some multidentate ligands
2. Based on charge
|
Name
|
Formula
|
Charge
|
Name
of Ligand
|
|
Ammonia
|
NH3
|
Zero
|
ammine
|
|
Water
|
H2O
|
Zero
|
aqua
|
|
Phosphine
|
PH3
|
Zero
|
Phosphine
|
|
Nitrogen
oxide
|
NO
|
Zero
|
Nitrosyl
|
|
Carbon
monoxide
|
CO
|
Zero
|
Carbonyl
|
|
Ethylene
diammine
|
NH2-CH2-CH2-NH2
or (en)
|
Zero
|
Ethane-1,2-diamine
|
|
Pyridine
|
C5H5N
|
Zero
|
Pyridine(Py)
|
|
Triphenyl
phosphine
|
(C6H5)3P
or P(PH3)3
|
Zero
|
Triphenyl
phosphine
|
|
Thiocarbonyl
|
CS
|
Zero
|
Thiocarbonyl
|
|
Thio
urea
|
H2NCSNH2
|
Zero
|
Thiourea(tu)
|
|
Halide
ion
|
X-(X=Cl,Br,I)
|
-1
|
halido
|
|
Hydroxide
ion
|
:OH -
|
-1
|
hydroxo
|
|
Cyanide
ion
|
:CN -
|
-1
|
Cyanato
|
|
Nitrite
ion
|
NO2-
|
-1
|
Nitrito-N-
|
|
Nitrito
ion
|
ONO-
|
-1
|
Nitrito-O-
|
|
Thiocyanate
ion
|
SCN-
|
- 1
|
Thiocyanato-S-
|
|
Isothiocyanato
|
NCS-
|
-1
|
Thiocyanato-N-
|
|
Acetate
ion
|
CH3COO-
|
- 1
|
Acetato
|
|
Nitrate
ion
|
NO3-
|
-1
|
Nitrato
|
|
Amide
ion
|
NH2-
|
-1
|
Amido
|
|
Oxide
ion
|
O2
-
|
- 2
|
Oxo
|
|
Peroxide
ion
|
O22
-
|
- 2
|
Peroxo
|
|
Carbonate
ion
|
CO32
-
|
- 2
|
Carbonato
|
|
Sulphate
ion
|
SO42
-
|
-2
|
Sulphato
|
|
Sulphide
ion
|
S2
-
|
- 2
|
Sulphido
|
|
Thiosulphate
|
S2O32
-
|
- 2
|
Thiosulphato
|
|
Sulphite
ion
|
SO32-
|
-2
|
Sulphito
|
|
Imide
ion
|
NH2
-
|
- 2
|
Imido
|
|
Nitrosonium
|
NO+
|
+1
|
Nitrosonium
|
|
Nitronium
|
NO2+
|
+1
|
Nitronium
|
3.
Chelate ligands: The
ligands which when attach to central metal atom form a cyclic/ ring structures
are called chelate ligands. It bind through more than one donor atom per ligand.
Example: en, ox, gly
etc
•
The octahedral [Co(en)3]3+ is a typical en complex.
•
Chelate effect: More stable complexes are formed with chelating agents
than the equivalent number of monodentate ligands.
Co-ordination
Number (CN):
The
number of co-ordinate bonds formed with the central metal ion by the ligand or
the no of ligands surrounding the
central metal atom.
Examples:
[Ag(CN)2]-
[Cu(NH3)4]2+
[Cr(H2O)6]3+
C.N = 2 C.N = 4 C.N = 6
Co-ordination Sphere
The
central atom and the ligands which are directly attached are enclosed in square
brackets called the coordination sphere. The ionisable groups are written
outside the brackets.
Example: 
Common
Structures
Octahedral Tetrahedral Square planar
The formation of a coordinate complex is a Lewis
acid-base reaction
Rules for Writing
Chemical Formula of a complex:
Rules for
formula writing:
1)
Formula of the cation must be written first followed by anion.
2) The
Co-ordination sphere is written in square brackets.
3) Within
the co-ordination sphere the sequence is (Metal atom, anionic, neutral,
cationic, ligand).
4)
Polyatomic ligands are enclosed in parentheses.
Rules for
Naming Co-ordination Compounds (IUPAC Nomenclature)
(1) The cation is named first (whether
simple or complex) then the anion.
[Co(NH3)6]Cl3
hexaammine cobalt
(III) chloride.
K2[PtCl6] potassium
hexachloroplatinate (IV).
(2) Name is started with a small
letter and the complex part is written as one word.
(3) For non-ionic or molecular
complexes one word name is given
[Co(NO2)3
(NH3)3] triamminetrinitrocobalt
(III)
name of the
central metal atom along with its oxidation state in Roman numerals within a
bracket.
(5) Naming
of ligands:
Negative ligands end in - O. Example: CN-
(cyano), Cl- (chlorido)
Neutral ligands have no special
ending: NH3 (ammine), H2O (aqua), CO (Carbonyl) ,NO
(Nitrosyl).
Positive ligands end in-ium. Example: NO2+
(nitronium) NO+ (nitrosonium)
(6) In unidentate ligands with more than one
co-ordinating atoms (ambidentate ligands) the point of attachment is
designated.
Example: - SCN thiocyanato-S; - NCS isothiocyanato-N
- NO2 nitrito-N - ONO nitrito-O
(7) If several ligands of the same type
are present the prefix di, tri, tetra, penta and hexa are used.
For complex ligands like ethane-1,2-diamine
the prefix bis, tris and tetrakis are used.
Preference
order
All
ligands are named in alphabetical order. The prefix di, tri are not considered.
Example: [PtCl(NO2)(NH3)4]
SO4
Tetra
ammine chloronitro platinum (IV) sulphate.
Naming
the complex
Ligands
are named first followed by the metal atom. The ending of the name of the metal
depends on the nature of the complex ion.
a) If the complex ion is a cation
or non-ionic, the name of the central metal ion is written as such followed by
its oxidation state, indicated by roman numerals (II, III, IV).
Example:
i) [Cu(NH3)4]SO4
Tetramminecopper (II) sulphate
ii) [Pt
Cl4 (NH3)2] Diammine tetrachloroplatinum (IV).
b) If the complex ion is an anion
the name of the central metal atom is made to end in -ate.
Example:
Na3 [AlF6] sodium hexafluoro aluminate (III)
K2[Co(NH3)2Cl4]- potassium
diamminetetrachlorocobaltate(II)
[Co(NH3)4Cl2]Cl
- tetraamminedichlorocobalt(III)
chloride
[Cr(NH3)6]3+ - hexamine
Chromium (III) ion
[Ru(NH3)4(HSO3)2]- bis(bisulfite)tetraamineruthenium(II)
K3[Fe(CN)6]- potassium hexacyanoferrate(III)
[Cu(NH3)4]SO4
- tetraaminecopper(II) sulfate
[Ti(H2O)6][CoCl6]- hexaaquatitanium(III)hexachlorocobaltate(III)
Isomerism in
Coordination Compounds
Two or
more substances having the same molecular formula but different structural or
spatial arrangement are called isomers.
These are
of two types: a) Structural
isomerism b) Stereo isomerism
Structural Isomerism
Two or
more substances having the same molecular formula but different structural
arrangement are called isomers.
(a) Ionisation isomerism: Compounds, which give different ions in solution although they have same
composition are called ionization isomers. There is an exchange in groups
between the coordination sphere of the metal ion and ions outside the sphere.
Example: 
The first
compound gives a white precipitate with Barium chloride. The second compound
gives a precipitate with AgNO3 and not with BaCl2.
(b) Hydrate Isomerism:Compounds, which have the same
composition but differ in the number of water molecules present as ligands (in
the co-ordination sphere) and as molecules of hydration (present outside the
coordination sphere). Example:
(c) Linkage Isomerism: It occurs when more than one atom in a mono-dentate
ligand may function as a donor. Example:
(d) Coordination
Isomerism: This type
of isomerism is possible when both positive and negative ions of a salt are
complex ions and the two isomers differ in the distribution of ligands in the
cation and anion. Example:
Stereo Isomerism or Space Isomerism
It arises
because of the different positions and arrangements of ligands in space around
the metal ion.
(a) Geometrical Isomerism
or Cis-trans Isomerism: When two identical groups
(ligands) occupy adjacent positions, the isomer is called cis and when arranged
opposite to one another the isomer is
called trans.
It is not possible for co-ordination number of 2
and 3 and for tetrahedral complexes of co-ordination number 4.
It is
common in square planar and octahedral complexes.
Square planar complexes
Example:
[Pt Cl2(NH3)2] diammine dichloro platinum
(II).
Octahedral complexes
Example:
[CoCl2(NH3)4]+- tetraammine dichloro cobalt (III) ion.
[CoCl2
(en)2]+
[Co(NH3)4Cl2]+
•
In case of
complexes of the type Ma3b3
•
fac (facial) => three identical ligands occupying the corners of a
common triangular
•
mer (meridional) => three identical ligands occupying three
consecutive corners of a square plane
(b)
Optical Isomerism: Optical isomers rotate the
plane of polarised light in opposite directions. The two isomers are mirror images of
each other. The two mirror images are non superimposable, do not possess a
plane of symmetry. They are called dextro and laevo (d and l) rotatory depending upon the
direction in which plane polarised light is rotated.
It occurs in only octahedral complexes with
coordination number 6 with 2 or 3 bidentated ligands.
Examples:[CoCl2(en)2 ]+,
[Co(en)3]3+ and [Cr(C2O4)3
]3-
[CoCl2(en)2]+
It can exist in cis and trans forms. The
trans form is symmetrical and optically inactive. Cis form exists in d and e
forms.
Non superimposable mirror images
Bonding in Co-ordination
Compounds
The first
theory was called the Werner's theory of co-ordination compounds.
1) Metals
possess two types of valencies:
a)
Primary valency or ionizable valency. It is also referred to oxidation state.
b)
Secondary valency which a metal atom or cation exercises towards neutral
molecules or negative groups
(ligands) in the formation of complex ions.The secondary valency is also called
the coordination number.
Example:
In [Pt(NH3)6]Cl4 secondary valency of Pt is 6
2)
Primary valencies are satisfied by negative ions, secondary valencies may be
satisfied by negative ions or neutral molecules.
3)
Ligands satisfying secondary valencies are directed towards fixed positions in
space giving a definite geometry to the complex, but the primary valencies are
non-directional.
Six
valencies are directed towards a regular octahedron while four are directed
towards either a tetrahedral manner or square planar.
Complex
Structure using Werner's Theory
Example:
In the complex between CoCl3 and NH3 the ionizable
chloride ions are found by precipitation with AgNO3. The remaining
Cl and NH3 are present around the central Co in such directions so
as to minimize repulsion and are linked by secondary valencies.
a)
Primary valencies - represented by dotted lines.
b)
Secondary valencies - by solid lines.
c)
Groups satisfying both primary and secondary valency - by solid and dotted
lines.
Werner
was given the Nobel Prize in 1913 for his work in complexes.
Valence Bond
(VB) Theory
I) The
metal ligand bond arises by donation of pair of electrons by ligands to the
central metal atom.
II) To
accommodate these electrons the metal ion must possess requisite number of
vacant orbitals of equal energy. These orbitals of the metal atom undergo
hybridisation to give hybrid orbitals.
III)
Sometimes the unpaired (n-1)d orbitals pair up before bond formation making
(n-1)d orbitals vacant. The central metal atom makes available number of
d-orbitals equal to its co-ordination number.
iv) The
metal ligand bonds are thus formed by donation of electron pairs by the ligands
to the empty hybridized orbitals. These bonds are equal in strength and
directional in nature.
v)
Octahadral, square planar and tetrahedral complexes are formed as a result of d2sp3
(or sp3 d2), dsp2 and sp3
hybridization respectively.
Square
Planar Complexes
dsp2 hybridization - square planar
Diamagnetic - No unpaired electrons.
Inner and Outer Orbital Complexes
In the
octahedral structure the central metal atom uses inner (n -1) d - orbitals or
outer (n)d-orbitals for hybridization. This results in
Inner
orbital complex: involving (n-1)d orbitals for d2sp3
hybridization.
The
electrons in the metal pair, so the complex is either diamagnetic or will have
lesser number of unpaired electrons. Example: [V(H2O)6]3+,
[Co(NH3)6]3+
Outer
orbital complex: When the
complex formed involves use of outer nd - orbitals sp3d2.
It is called outer orbital complex. It has a large number of unpaired
electrons.
Example: [CoF6]3-,
[MnF6]3-, [Fe F6]3- because F-
is a weak ligand, it cannot cause force pairing.
Drawbacks of Valence Bond Theory
1) It is
a qualitative approach describing bonding in coordination compounds.
2) The
theory fails to explain the optical absorption spectra and magnetic properties
of coordination compounds.
3) The
theory does not provide an answer to the origin of characteristic colors of
complex ions.
4) It
does not explain why [Co(NH3)6]3+ is an inner
orbital complex and [CoF6]3- is an outer orbital complex
Crystal Field
Theory:
Bonding
in a complex ion is due to electrostatic interactions between the positively
charged nucleus of the central metal ion and electrons in the ligands i.e.,
attractive as well as repulsive interactions.
MAIN
POSTULATESOFCRYSTALSPLITTING FIELD THEORY:
a. It considers the bond to be ionic
arising purely from electrostatic interactions between the metal ion and the ligands.
b. It treats each ligand as a point of
negative charge.
c. Ligands
arranges around metal ion in such a way so as to minimise repulsions
d. In
a free transition metal ion all the five d orbitals have equal energies DEGENERATE
e. The attractive forces will arrive due to the
positive metal ion and the negatively charged ligand.
f. The
repulsive forces arise between the lone pairs on the ligand and the electrons
in the d-orbital of the metal.
d-orbitals
and Crystal Field Splitting
The
five d-orbitals can be divided into two groups depending upon the nature of
their orientation in space.
(i)
The three d-orbitals (dxy, dyz, dzx) which
orient in the regions in between the coordinate axes are designated as t2g
orbitals.
(ii)
The two other orbitals (dx2 - y2 and dz2)
which orient along the axes are labeled as e.g., orbitals.
Orientation of five d orbitals in
octahedral environment
In the case of a free ion,
all the five d-orbitals have the same energy, i.e., they are degenerate
(meaning energetically alike).
When the ligand approaches
the central metal ion, the electrons in d-orbital of the central metal ion will
be repelled by the lone pairs of the ligands. As a result of these interactions
the degeneracy of d-orbitals of metal ions is lost and these split into two set
of orbitals having different energies. This is known as Crystal Field Splitting
(D) and it forms the basis of crystal field theory.
The crystal field splitting
depends upon the number of ligands approaching the central metal ions. That is,
crystal field splitting will be different in different structures with
different coordination numbers.
Crystal Field Splitting in Tetrahedral
Complexes
The
tetrahedral arrangement of four ligands surrounding the metal ions is as shown
in the figure.
Tetrahedral arrangement of four
ligands surrounding the metal ions
It
is clear from the figure that none of the d-orbitals point exactly towards the
ligands. The three d-orbitals dxy, dyz and dzx
are pointing close to the direction in which ligands are approaching.
As
a result of this, the energy of these three orbitals increases much more than
the other two d-orbitals (dx2
- y2 and dz2).
The
d-orbitals will thus split as shown below:
Octahedral
Complex Crystal Field Splitting
The
octahedral arrangement of six ligands surrounding the central metal ion is as
shown in the figure.
Octahedral arrangement of six ligands
surrounding the central metal ion
Looking at the figure, we
observe that as the ligands approach the x,y and z axis, the two d-orbitals
lying along the axis namely dx2 - y2 and dz2
will suffer more electrostatic repulsions and hence their energy will be
greater than the other three d-orbitals dxy, dyz and dzx,
Accordingly,
the d-orbitals will thus split as follows.
fig
10.5
Magnitude of Crystal Field Splitting
Different
ligands differ in their ability to produce a splitting of the d-orbitals.
For e.g.,
strong lewis bases, such as CN- and NH3, produce a strong
field. They repel d electrons most strongly and cause a greater separation of
the two groups (t2g and eg) of d-orbitals than do weak
bases like H2O and F-.
The
crystal field splitting is measured in terms of energy difference between t2g
and eg orbitals and is denoted by symbol D. It is generally measured
in terms of a parameter, Dq.
When a
d-electron enters or resides in one of the t2g orbital, the energy
of that orbital becomes 4Dq or 0.4D less or more than that of
hypothetical degenerate orbitals.
Similarly,
for each electron entering e.g., orbital, the energy of that orbital becomes 6
Dq or 0.6 D less or more, than that of degenerate orbitals.
•
Spectrochemical Series:
•
Ligands can be arranged into a spectrochemical
series according to the magnitude of splitting of the d-orbitals
•
Large splitting is associated with
strong field ligands
•
Small splitting is associated with
weak field ligands
CN- >
en > NH3 > H2O > F- >SCN- >
Cl- > Br- > I-
Co-ordination
Compounds Stability
Although
a complex ion does not dissociate generally, it may sometimes dissociate to a
small extent.
Example:
The
complex is in equilibrium with its ions in solution.
The
instability constant
The
stability constant
For
a general expression
a+,
x- and b+ are the charge on the metal, the ligand and the complex ion
respectively.
(a+)
+ n (x-) = (b+)
Greater
the value of K more is the stability.
Factors Affecting Stability of a Complex Ion
1) Charge
on the central metal ion: Greater the charge, more the stability.
2) Basic
nature of ligand: More the basic nature, more is the stability.
3)
Presence of chelate rings: Formation of chelate rings increases the stability.
Example:
Preparation of Co-ordination Compounds
In this
method a stronger ligand replaces a weaker ligand.
Direct
combination
Redox
method: In this
method, a complex with the metal atom in a higher oxidation state than the
starting material is obtained.
Example:
Pentaamminenitro cobalt (III)
nitrate is prepared by adding NH3 and NH4NO3
to cobalt (II) nitrate solution in the presence of H2O2.
Applications
of Co-ordination Compounds
1)
Estimation of hardness in water: Ca2+
and Mg2+ ions water can be estimated using EDTA
2) Animal
and plant world: Chlorophyll -
Contains Mg complex
Haemoglobin - Contains Fe complex
Vitamin B12 - Contains Co complex
3)
Electroplating of metal: Co-ordination compounds of gold and silver are used in
electroplating both for controlled deposition of metal ions.
Example: K[Ag(CN)2]
4)Extraction
of metals
Au and Ag
are extracted from their ores through formation of cyanide complexes, [Ag(CN)2]-
and [Au(CN)4]-
5)
Estimation of Ni(II) as red glyoxime using dimethyl glyoxime.
Synthesis of
Organometallic Compounds
i)
Organometallic compounds and Grignard reagent.
ii) Tetra
ethyl lead and tetra alkyl lead obtained by reaction of metal halide with
organometallic compound.
(Bu = C4Hg)
(Et = C2H5)
TEL
(Tetra Ethyl Lead) used as an antiknock agent is prepared as follows:
Preparation
of p-Complexes
1) Zeise's salt
2) Ferrocene
3) Dibenzene chromium
Direct
combination
Reaction between Ni and CO is
used in the Mond's process for purification of Ni.
Applications of Organometallic Compounds
Many
reactions in solution are catalyzed by organometallic compounds or transition
metal complexes.
Example:
Wilkinson's catalyst
(Ph3P)3RhCl
- Catalyst in hydrogenation of alkenes.
Heterogenous
Catalysts
Ziegler -
Natta, example trialkyl aluminium + TiCl4 used for low temperature
polymerization of alkenes.
Ethyl mercury chloride - To
prevent infection in small plants.
Coordination Compounds Conclusion
Transition
elements are known for their ability to form many complex compounds. The complex
compounds in which the metallic ion is surrounded by two or more ions or
molecules are called coordination compounds, also known as complex compounds.
The chapter has covered the nomenclature, isomerism, bonding, applications and
stability of coordination compounds and organometallic compounds.
VERY
SHORT ANSWER TYPE QUESTIONS: ( 1 Mark each)
01) What is the coordination number of central
metal ion in [Fe(C2O4)3]3- ?
Ans
= 6
02) Write IUPAC name of [ Pt Cl2 (NH3)4
] 2+.
Ans
= dichloride tetraannine
platinum (IV) ion.
03)
Name the type of isomerism that occurs in complex in which both cation and
anion are complex ions.
Ans
= Co-Ordination isomerism.
04) Write IUPAC name for ionization isomer [Co
(NH3)5 SO4]Br.
Ans =
The ionization isomer of the given complex compound is [Co (NH3)5 Br] SO4.
Its IUPAC name is
pentaamminebromidocobalt.(III) sulphate
05) How many moles of AgCl will be precipitated
when an excess of AgNO3 is
added to a molar solution of [Cr(H2O)5Cl]Cl2.
Ans =
2 moles of AgCl will be precipitated .
06) If the geometry of (PtCl4)2- is square
planar, which orbitals of Pt are involved in the bonding ?
Ans
= dsp2 hybridization is
taking place involving 5d, 6s and 6p.
07) Which of the two, K4 [Fe(CN)6]
or K3 [Fe(CN)6] is more stable ?
Ans
= K3[Fe(CN)6] is
more stable.
08) What is the coordination number of a central
metal ion in an octahedral complex ?
Ans
= 6
09) How many ions are produced from the complex
[Co(NH3)6]Cl2 in solution ?
Ans
= 03 ions are produced.
10) Amongst the following ions which one has the
highest magnetic moment values ?
a)
[Cr(H2O)6]3+ (b) [Fe(H2O)6]2+ (c) [Zn(H2O)6]2+.
Ans =
a) Cr3+ :
[Ar]3d3 µ = √3
(3+2) = √15
=
b) Fe 2+
: [Ar]3d6 µ
= √4 (4+2) =
√24 =
c) Zn2+ :
[Ar] 3d10 µ = 0
SHORT
ANSWER TYPE QUESTIONS ( 02 Marks Each).
01) A
coordination compound has the formula
CoCl3. 4NH3. It does not librated NH3 but
precipitates chloride ion as AgCl. Write the IUPAC name of the complex and
write its formula.
Ans = [CO(NH3)4Cl2]Cl
IUPAC name : Tetraamminedichloridocobalt(III)
chloride.
02) Using
valence bond theory of complexes, explain the geometry and magnetic nature of
[Cr(H2O)6]3+.
Ans = Cr has
oxidation state of +3
Its configuration is [Ar]3d3.
3d 4s 4p
Cr3+
As a weak
ligand, H2O approaches towards it, the pairing against Hund’s rule
does not take place and thus 3d orbitals participate in hybridization.
3d 4s 4p
[Cr(H2O)6]3+
d2sp3 hybridization.
As the
hybridization is d2sp3,
the shape of the ion is octahedral since, it contains 3 unpaired e-s
is paramagnetic in nature.
03) Explain
the following :
a) [Co(NH3)6]3+
is diamagnetic, while [CoF6]3- is paramagnetic.
b) [Ti(H2O)6]3+
is coloured while [Sc(H2O)6]3+ is colourless.
Ans
= a)
[ Co (NH3)6]3+, co exists as Co3+.
Co3+ - [
Ar] 3d6.
3d 4s 4p
Co3+
As strong ligand NH3 approaches,
the e-s tend to pair up against the
Hunds rule, thus.
3d 4s 4p 4d
[Co(NH3)6]3+
d2sp3
But in [CoF6]3-,
a weak ligand F- approaches,
thus pairing against Hund,s
rule does not, take place.
3d 4s 4p 4d
[CoF6]3-
sp3d2.
Thus due to absence of unpaired e-s
in [Co(NH3)6]3+, it is diamagnetic but
dues to presence of unpaired e-s
in [CoF6]3-, it is paramagnetic.
Ans
= b) [Ti (H2O)6]3+,
Ti exists as Ti3+
Ti 3+ [Ar]3d1
While Sc exists as
Sc3+ in [Sc(H2O)6]3+ with no
unpaired e-.
Due to presence
of un paired e- in [Ti (H2O)6]3+,
it undergoes d-d transition & hence appears coloured.
04)
Mention application of coordination compound in
a) metallurgical operations.
b) medicine.
Ans =
a) metals like Ag & Au are
extracted by the formation of their cyanide complexes.
Ag2S +
NaCN
Na [ Ag (CN)2] + Na2S
2Na [ Ag (CN)2]
+ Zn Na2
[Zn (CN)4 + 2 Ag.
Ans =
b) cis – platin i.e, cis [ PtCl2 (NH3)2]
is used in cancer chemotherapy.
05)
Name the isomerism associated with these coordination compounds.
[Cr(CH3)5
Cl]SO4 and [Cr(NH3)5 SO4]Cl. State a chemical
test to distinguish between these two
isomers.
Ans =
They are ionisation isomers. They can be differentiated by adding BaCl2 solution in their
aqueous solutions.
[Cr(NH3)5Cl]SO4 aq
[ Cr(NH3)5Cl]2+ + SO42-
BaCl2
BaSO4
(White ppt)
While [Cr(NH3)5 SO4
] Cl does not form white ppt when ag
BaCl2 is added to its aq. solution.
06)
Draw fac and mer isomers of [Co(NH3)3Cl3].
NH3
NH3
Ans Cl NH3 Cl NH3
CO
CO
Cl NH3 Cl Cl
Cl
NH3
Fac – isomer Mer – isomer
SHORT
ANSWER TYPE QUESTIONS ( 03 Marks )
01) Draw
a figure to show splitting of degenerate d orbitals in an octahedral crystal
field. How does the magnitude of ∆0
decide the configuration of d
orbitals in a complex entity ?
Ans : Refer note
02)
The value of dissociation
constant of [Cu(NH3)4]2+ and [Co(NH3)6]3+ are 1 x 10-12, and
6.2 x 10-36
respectively, which complex would be more stable and why ?
Ans
= [Cu(NH3)6]3+ would be more stable as
its dissociation constant is less and
dissociation constant
is
reciprocal of stability
constant. Thus ß 4 (stability constant ) = 1/Kd (dissociation const)
HIGHER
ORDER THINKING QUESTIONS:
01) A
cationic complex has 2 isomers
(A) and (B) Each one
has one Co3+, five NH3, One Br- and
one SO42-. (A) gives a white ppt with BaCl2
solution while (B) gives a yellow ppt
with AgNO3 solution.
a) What are the possible structures of
(A) and (B) ?
b) Will the two complexes have same
colour ?
Ans a)
(A) gives a white ppt with BaCl2,
SO42- ion is present in its ionization
sphere and thus
its structure is [ Co(NH3)5 Br]SO4.
(B)gives a yellow ppt with AgNO3 solution, … Br- ion must be
present in its ionization
sphere…. Its structure is [Co(NH3)5 SO4]Br.
b)
No, the two complexes will not have the same colour.
02) A
student prepared following coordination complexes containing pt having the following
characteristics.
Formula Cations
present in solution per formula unit.
(A)
Pt Cl4 . 6 NH3 4
(B)
Pt Cl4 . 5 NH3 3
(C)
Pt Cl4 . 4 NH3 2
a)
Write the structures of these comnplexes.
b)
Write IUPAC name of the complex
(B) & (C).
Ans =
a) A – [ Pt
(NH3)6] Cl4.
B – [ Pt
(NH3)5 Cl ]
Cl3.
C – [ Pt
(NH3)4 Cl2 ] Cl2.
b)
IUPAC name of :
B - pentaamminechloridoplatinum (IV) chloride
C – tetraammine dichloridoplatinum (IV) chloride.
03) A metal ion Mn+ having d5
valence electronic configuration combines with 3 didebtateligands to form a
complex compound. Assuming ∆0
> P,
i)
draw the diagram assuming d – orbital
splitting during the complex formation.
ii)
Write the electronic configuration of valence electrons of the metals Mn+ ion in
terms of
t2g & eg.
iii)
What type of hybridization will Mn+ ion have
?
Ans = (i) figure
ii) t2g5eg0 :.
0 > P,
as P ( Pairing energy ) is less
than ∆0 pairing
of e- takes place in lower
energy t2g
level.
iii) Mn+ ion
will have d2sp3
hybridisation as its coordination no. is 6 due to its tendency to
combine with 3 bidentate ligands and also
due to presence of vacant orbitals.
04) Arrange the following in increasing order of
electrical conductivity.
[Co(NH3)3 Cl3], [Co(NH3)5 Cl]Cl2, [Co(NH3)6]Cl3, [Co(NH3)5Cl]Cl.
Ans
= [Co(NH3)3Cl3]
< (Co(NH3)5 Cl
]Cl < [Co(NH3)5Cl]Cl2
< [Co(NH3)6]Cl3.
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