Noo Xii Ch09 Coordination Compounds
Noo Xii Ch09 Coordination Compounds
Noo Xii Ch09 Coordination Compounds
COORDINATION COMPOUNDS
c/3 After studying this unit thoroughly, the reader should be able to
• have an idea of the nature of coordination compounds and the basic concepts used in coordination chemistry;
P • learn the nomenclature of different types of coordination compounds;
Q • learn the different types of isomerism exhibited by coordination compounds;
£> • learn the nature of bonding in coordination compounds;
• discuss the stability of coordination compounds;
• have an idea of the applications of coordination compounds.
ransition metals possess a characteristic property important biological functions. For example, photosynthesis
T of undergoing complex formation. Due to in plants involves chlorophyll, which is a coordination
this remarkable tendency, they form a large number of
compound of magnesium. The red pigment haemoglobin
coordination compounds. Coordination compounds are not present in blood, which is responsible for the regulation of
only formed by transition elements but are also formed by respiration process in the animal world, is a coordination
several other non-transition metals although the tendency compound of iron. A variety of metallurgical processes,
is much less. The branch of chemistry which deals with the catalysts and analytical reagents make use of coordination
study of coordination compounds or complex compounds compounds. Therefore, the study of coordination compounds
is known as Coordination Chemistry. is highly significant from biological, industrial and analytical
Coordination compounds are widely distributed in points of view. In the forthcoming sections, \ye shall briefly
minerals, plants and animals and are known to play many study the chemistry of coordination compounds.
nh3, h2o :, — nh2, h2n—ch2—ch2—nh2, :f~: cN~etc. (ii) Anionic complexes : The complexes in which the
(ethylene diamine, en) complex ions carry a net negative charge are called anionic
complexes. For example, K4[Fe(CN)6], K[Ag(CN)2],
3. Complex ion (Coordination entity) : The electrically
K2[HgI4], etc. These complexes possess anionic complex
charged speciesformed by the union ofa central metal atom orion
ions as is clear from the following.
with one or more ligands is called a complex ion. For example, K4[Fe(CN)6] 4K++ [Fe(CN)6]4“
[Fe (CN)6]4~ ion is a complex ion. In this complex ion, the
K2[HgI4] 2K++[HgI4]2"
central Fe ’’ ion is attached to six CN ligands. Similarly, (iii) Neutral complexes : The complexes which carry
oi oi
[Cu (NH3)4] is a complex ion in which Cu ’r ion is linked no net charge are called neutral complexes. For example,
to four NH3 ligands. [Ni(CO)4], [Pt(NH3)2Cl2], etc.
A complex ion may have positive or negative charge 7. Coordination number: The maximum number of ligands
on it. A positively charged complex ion is called cationic which can be coordinated to a central metal atom or ion is
complex ion, while the negatively charged complex ion is known as the coordination number of that central atom
termed as anionic complex ion. For example, or ion. For example, in K4[Fe(CN)6], six CN- ligands are
9i
Cationic complex ions : [Ag(NH3)2]+, [Cu(NH3)4]2+,
coordinated to central Fe ion. Therefore, its coordination
[Co(NH3)6]3+, etc.
number is six. Similarly, in [Cu(NH3)4]SO4, the coordination
Anionic complex ions .'[Fe(CN)6]4-, [Cu(Cl)4]2-,
number of Cu is four. Coordination numbers of metals vary
[Ag(CN)2]- etc.
from 2 to 10 but the most common coordination numbers are
While writing the formula of a complex ion, the
4 and 6. The coordination number of a metal ion depends
ligands are usually written inside the parenthesis, and the
entire complex ion in a square bracket. The net charge mainly upon its nature, its oxidation state and on the nature
present on the complex ion is written at the right hand top of ligands surrounding it. However, coordination number
corner of the square bracket. may be influenced by temperature, pressure and nature of
4. Coordination sphere and ionic sphere : The part solvent. A particular metal may exhibit different coordination
consisting of the- central metal atom and ligands directly numbers in different complexes. For example, platinum has
coordination number 4 in [PtCl4]2-, while coordination
attached to it (enclosed in square brackets) is called the
coordination sphere of the complex compound, while number 6 in complex ion [PtClg]4-.
the part which gets ionised in solution (written outside The geometry of a complex depends upon the
square brackets) is called the ionic sphere of the complex coordination number of its central metal atom. Central
compound. For example, the complex, [Cu(NH3)4jSO4 metals with coordination number 6 usually form octahedral
ionises in solution as complexes, while those with coordination number 4 usually
[Cu(NH3)4]SO4 [Cti(NH3)4]2+ + SO2- form tetrahedral or square planar complexes.
Complex compound (coordination sphere) (ionic sphere) 8. Charge number of a complex ion : The net charge
-X 2+ carried by a complex ion is called its charge number. It
Obviously, the part [Cu(NH3)4] constitutes the
is equal to the algebraic sum of the charges carried by
coordination sphere while SO2- the ionic sphere of this
central ion and the ligands attached to it. For example, the
complex. charge number of [Fe(CN)6]4- is - 4. It may be calculated
5. Coordination polyhedron : The spatial arrangement as shown below :
of the ligands directly attached to the central atom or Charge number of
ion constitute a polyhedron about the central atom. This [Fe(CN)6]4 = (charge on Fe2+) + (6 x charge on CN- ion)
polyhedron is termed as coordination polyhedron. The = (+2) + 6 x (-1) = - 4.
coordination polyhedron may be tetrahedral, square
9. Oxidation number or oxidation state of the central metal
planar, octahedral, trigonal bipyramidal, etc., in shape.
atom : The number which represent the electrical charge which
6. Cationic, anionic and neutral complexes: Coordination
the central atom actually has or appears to have when combined
compounds or complexes may be of following three types.
with other atoms or groups of atoms is called the oxidation
(i) Cationic complexes : The complexes in which the
number or the oxidation state of the central metal atom.
complex ions carry a net positive charge are called cationic
For example, the oxidation state of iron in the complex,
complexes. For example, [Fe(H2O)6]Cl3, [Co(NH3)6]C13,
K4[Fe(CN)6] is +2.
[Ni(NH3)6]Cl2, etc. The complex ions present in these
complexes carry a net positive charge as is clear from the The oxidation number or oxidation state of the central
following. metal atom may be calculated according to the common
[Fe(H2O)6]Cl3 [Fe(H2O)6]3+ + 3CF rules used for the purpose which we have already studied
[Co(NH3)6]C13 [Co(NH3)6]3++ 3C1 in class XI. Following examples illustrate the procedure.
Coordination Compounds ___________ ________________________________ 485
(i) Oxidation number of Cu in [Cu(NH3)4] SO4 : [B] Classification on the Basis of the Number of
Suppose the oxidation number of Cu in this complex is x.
Donor Atoms
Since it is a neutral molecule, the sum of oxidation numbers The mode of attachment of a ligand to the central metal
of all the species in it should be equal to zero. Hence, atom or ion is decided by the number of donor atoms present
x + 4x(0)-2 = 0 or, x = + 2. in it. The number of such ligating groups (donor atoms) is
Therefore, the oxidation number of Cu in this complex called denticity of the ligand. Depending upon the number
is+ 2. of donor atoms, ligands can be classified as follows.
o
(ii) Oxidation number of Fe in [Fe(CN)6] ion : 1. Unidentate ligands : The ligands which possess only
Suppose the oxidation number of iron is x. The sum of one donor atom and can form only one coordinate bond to the
oxidation numbers of all the species present in it should be central metal atom or ion are called unidentate ligands or
equal to the charge present on the ion. Hence, monodentate ligands. Such ligands can attach themselves
x + 6 x (-1) = - 3 or, x = + 3. to the central atom through one point only. They may be
neutral as well as anionic. Some important monodentate
Thus, the oxidation number of Fe in this complex ion
ligands, their names and donor atoms present in them are
is +3.
listed in Table 9.1.
(iii) Oxidation number of Au in H[AuC14] : If the Table 9.1 Some Important Unidentate Ligands
oxidation number of gold in this complex is x, we have
Donor Donor
+1 +x + 4 x (-1) = 0 or, x = + 3. Ligand Name Ligand Name
atom atom
Thus, the oxidation number of Au in this complex is
H2O aqua 0 r iodido I
+ 3.
(iv) Oxidation number of Ni in [Ni(CO)4] : If the nh3 ammine N CN” cyanido C or N
oxidation number of nickel in this complex is x, we have CO v. carbonyl 0 OH” hydroxo 0
x + 4 x (0) = 0 or, x - 0.
NO nitrosyl N no2 nitro N
Thus, the oxidation number of Ni in this complex is 0.
10. Homoleptic and heteroleptic complexes : The C5H5N pyridine N —ONO” nitrito 0
complexes in which the central metal atom or ion is linked (py)
to only one kind of donor groups (ligands) are called F" fluorido F CH3COO” acetato 0
homoleptic complexes, e.g, [Fe(CN)6]4~, [Co(NH3)6]3+, cr chlorido Cl SCN” thiocyanato S
etc. Complexes in which the central metal is bound to more Br” bromido Br NCS” isothiocyan ato N
than one kind of donor groups are termed as heteroleptic NH2" imido N NH2 amido N
complexes, e.g, [Co(NH3)4Cl2]+ [Co(NH3)5(SO4)]+, etc.
9.1.3 Classification of Ligands 2. Didentate ligands : The ligands which possess two
donor atoms and can form two coordinate bonds with the
As discussed /earlier, ligands are those atoms or
central metal atom or ion are called didentate ligands.
groups of atoms which have a tendency to donate a lone
Such ligands can get attached to the central metal atom or
pair of electrons to the central metal atom or ion. The
ion through two points. Some important didenate ligands
ligands can be classified as follows.
are listed in Table 9.2.
|A| Classification on the Basis of Charge Table 9.2 Some Important Didentate Ligands
Depending upon the nature of charge (electrical) Ligand Name Donor atoms
Present, ligands can be classified as follows. oxalato (ox) O and O
COO2”
1. Neutral ligands : Such ligands possess no electrical
charge present on them. They are usually molecular
COO
species having one or more lone pairs of electrons. For
NH2CH2COO" glycino (gly) N and O
example, H2OI (aqua), NH3 (ammine), CO (carbonyl),
ch2nh2 ethylenediamine or N and N
h'O (nitrosyl), C5H5N (pyridine, py), etc.
ethane-1, 2-diamine
2. Anionic ligands : Such ligands carry negative charge (en)
ch2nh2
them. These are anionic species containing one or
■"ore lone pairs of electrons. For example, F- (fluorido), CH3 — C = N — OH dimethylglyoximato N and O
-i (chlorido), Br” (bromido), I” (iodido), CN” (cyanido), (DMG)
(hydroxo), CH3COO” (acetato), etc. CH3 — C = N — 0“
3. Cationic ligands : Such ligands carry positive charge 2, 2'-dipyridyl N and N
(pUo)
occur very rarely in complexes. For example, NO+ (dipy)
(’litrosylium), NH2NH3 (hydrazinium), etc. •• ••
486 Nootan ISC Chemistry-XII
3. Tridentate ligands : The ligands which possess three Some common chelating ligands are ethylene-diamine
donor atoms and can form three coordinate bonds with the (en), oxalato (ox), diethylenetriamine (dien), EDTA, etc.
central metal atom or ion are called tridentate ligands. For For example, the complexes formed by the bidendate
example, ligand ethylenediamine (en) with Cu (II) and Pt (II) ions
are chelates. These are shown below.
NH2 —CH2 —CH2 —NH—CH2 —CH2 —nh2
Diethylenetriamine (dien) H2C—H2N NH2— ch2
I Cu |
ooc—ch2— ch —cop/ H2C—H2N NH2 — CH2
nh2 or, [Cu(en)2]2+
Aspartate ion (asp)
H2C—H2N NH2 —CH2'
4. Tetradentate ligands : The ligands which possess I Pt I
four donor atoms are called tetradentate ligands. h2c—h2n r nh2-ch2
For example,
or, [Pt(en)2]2+
NH2 —CH2 —CH2 —NH—CH2 —CH2 — NH—
Chelating ligands stabilise a complex. This is why
chelates are more stable than the ordinary complexes
CH2— ch2— nh2
Triethylenetetramine (trien)
containing monodentate ligands which are unable to form
chelates.
5. Pentadentate ligands : The ligands which contain five
Chelates find a wide range of applications in the
donor atoms are called pentadentate ligands. For example,
laboratory as well as in industry. They are used for
.. /CH2COO“ the detection and estimation of several ions such as
~OOCCH2 —NH—CH2 —CH2 —N < nickel (II), copper (II), magnesium (II), etc., in qualitative
’• xCH2COO“
and quantitative analysis. They are also used for softening
Ethylenediaminetriacetato of hard water and in the separation of lanthanoids and
6. Hexadentate ligands : The ligands which contain six actinoids.
donor atoms are termed as hexadentate ligands. 9.1.5 Werner's Theory of Coordination
For example,
Compounds
OOCCH2 <b .. ch2coo~
Alfred Werner (1892) prepared and isolated several
)N—CH2— CH2— N<
~oocch2z xCH2COO~ complex compounds by the action of ammonia on cobalt
chloride. He characterised these compounds as follows.
Ethylenediaminetetraacetato (EDTA)
Compound Colour Name according to colour
7. Bridging ligands : The monodentate ligands which can
simultaneously attach themselves to more than one metal ion CoCl3 ■ 6NH3 Yellow Luteo complex
are called bridging ligands. The ligands OH-, NH2, NO2, C0CI3 ■ 5NH3 Purple Purpureo complex
Cl-, CO, etc., act as bridging ligands. C0CI3 ■ 4NH3 Green Praseo complex
C0CI3 • 4NH3 Violet Video complex
9.1.4 Chelating Ligands and Chelates
The ligands containing two or more donor atoms are He studied the properties of these compounds in detail.
usually referred to as polydentate ligands. These ligands In order to explain the properties of these compounds, he
may contain donor atoms positioned in such a way that proposed a theory commonly known as Werner’s theory of
they may form a five or six membered ring with the metal coordination compounds. This theory is briefly described
below.
ion. Such ligands are termed as chelating ligands and the
Postulates of Werner’s theory : The main postulates of
rings thus formed are called chelates. They may be defined
Werner’s theory are as follows :
as follows.
(i) The central metal atom or ion present in a complex
When a polydentate ligand attaches itself to a central exhibits two types of valencies-primary valency and
ion through two or more donor atoms in such a way that it secondary valency.
forms a five or six membered ring with the central ion, the (a) Primary valency : This type of valency (linkage)
is ionisable and corresponds to the oxidation state
ligand is called chelating ligand and the ring thus formed is
of the central metal ion. It is always satisfied by
called a chelate.
a negative ion. The primary valencies are shown
Coordination Compounds ___________________________________________ ' 487 |
by dotted lines. The ions attached to the central of cobalt is six, it must have six secondary valencies.
metal ion through primary valencies can ionise In order to satisfy all the six secondary valencies,
in solution. one Cl atom (already attached by a primary valency) must
(b) Secondary valency : This type of valency also be linked by a secondary valency and the complex
(linkage) is non-ionisable and corresponds to the should have the structure as shown in Fig. 9.2. The Cl
coordination number of the central metal atom or atom linked by both primary and secondary valencies does
ion. Every central metal atom or ion has a fixed not ionise in the solution. Therefore, this complex should
number of secondary valencies (this number, ionise in the solution as follows :
infact, refers to the coordination number of the
CoC13-5NH3 ----- > [CoCi • 5NH3]2+ + 2CF
central metal ion). The secondary valencies are
satisfied either by negative ions or by neutral
molecules. They are represented by thick lines
and the groups attached to the central metal ion
through them are unable to ionise in the solution.
(ii) The central metal atom or ion has a tendency to satisfy
all of its primary and secondary valencies. In order to
meet this requirement, the central metal atom or ion
may attach one or more negative ions through both
primary and secondary valencies. Thus, some of the
negative ions may play a dual role. Such ions do not Fig. 9.2 Werner’s structure of C0CI3 • 5NH3.
ionise in solution. This has been confirmed conductometrically. One
(iii) The secondary valencies are directional, i.e., they point mole of the complex is found to give two moles of CF ions
in definite directions in space. The primary valencies are and onexmole of [CoC1-5NH3]2+ ion.
non-directional. 3. C0CI3 4NH3 and C0CI3 3NH3 : Following the
Structures of Some Coordination Compounds on arguments as given above, the structures of these complexes
can be written as shown in Fig. 9.3.
the Basis of Werner's Theory
The structures of some complexes derived on the basis
of Werner’s theory are discussed below.
1. C0CI3 • 6NH3: In this complex, cobalt is in +3 oxidation
state. Therefore, its primary valency is 3. Experimentally, it
has been observed that one mole of the complex gives three
moles of CF ions in solution. Therefore, the three Cl atoms
(negative ions) present in it should be linked to the central
cobalt by primary valencies. The coordination number of
cobalt is six. Therefore, it must have six secondary valencies.
Obviously, the six NH3 molecules (neutral molecules) must
Fig. 9.3 Werner’s structure of (a) CoC13-4NH3 and
be attached to cobalt by these secondary valencies. Thus,
(b) CoC13-3NH3.
on the basis of Werner’s theory, the complex C0CI3 • 6NH3
should be represented as shown in Fig. 9.1. On the basis of the Werner’s structures of the above
mentioned complexes of cobalt, the ionisation behaviour of
these complexes can be summarised as given in Table 9.3.
Table 9.3. Ionisation Behaviour of Some Cobalt Complexes
Number
of Total
Complex Ionisation behaviour ionisable number
chlorine of ions
atoms
nh3
CoC13-6NH3
Fig. 9.1 Werner’s structure of C0CI3 • 6NH3. CoC13-6NH3 3 4
----- > [Co-6NH3]3+ + 3CF
2. C0CI3 5NH3 : The oxidation state of cobalt in
this complex is +3. Therefore, it should have three
CoCl3 • 5NH3
primary valencies and all the three Cl atoms should
CoC13-5NH3 ----- > [CoCi • 5NH3]2+ + 2CF 2 3
be linked to cobalt by these valencies. The remaining
five NH3 molecules should be attached to cobalt by
secondary valencies. Since the coordination number
Nootan ISC Chemistry-XII
C0CI3 • 4NH3 ionisable atoms or groups out side the square brackets. If
CoC13-4NH3 1 2 ligands attached to the central metal atom or ions are of
—> [coci2- 4nh3]+ + cr different types, they are represented in the order-anionic
CoC13-3NH3 Does not ionise ligands, neutral ligands, cationic ligands. If there are several
- -
ligands of the same kind, they are listed alphabetically
Werner’s theory was the first attempt to explain the according to the first symbol of their formulae. Thus, the
nature of bonding in coordination compounds and the modem notation of a coordination compound is in the
theory successfully explained the ionisation behaviour of following form :
several complexes. However, the bonding in complexes is For example,
more complicated. Therefore, the theory was vindicated Werner’s Modern Werner’s Modern
and several advanced theories were given to explain the Formula Notation Formula Notation
nature of bonding in complexes. Some of these theories CoC13-6NH3 [Co(NH3)6]C13 CoC13-5NH3- [Co(H2O)
will be discussed later on in this unit. h2o (NH3)5]C13
Modern Notation for Representing Coordination CoC13-5NH3 [CoC1(NH3)5] CoSO4-5NH3-Br [Co(SO4)
Compounds Cl2 (NH3)5]Br
Now-a-days, coordination compounds are represented CoC13-4NH3 [CoCl2(NH3)4] PtClBrNH3py [PtBrCl(NH3)
Cl py]
by enclosing complex ions (central metal atom or ion and
unionisable groups, i.e., ligands) in square brackets with CoCl3 • 3NH3 [C0CI3 (NH3) 3]
REVIEW EXERCISES
9.1 What are coordination compounds and how do they 9.8 What is the oxidation state of platinum in the
differ from double salts? Explain with examples. complex, [Pt(NH3)2CI2]C12?
9.2 Explain the following terms with respect to [Ans. + 4]
coordination compounds: 9.9 What is the oxidation state of cobalt in the complex,
(i) Ligands [Co(NH3)4(H2O)Br] (NO3)2?
(ii) Coordination sphere [Ans. + 31
(iii) Coordination number. 9.10 What is the coordination number of Fe in the
9.3 What is the oxidation state of Ni in Ni(CO)4? complex, K3[Fe(C2O4)3]?
|Ans. 0] [Ans. six]
9.4 Calculate the oxidation number of the central metal 9.11 What is the coordination number of cobalt in
[Co(en)(H20)(Br)(Cl)2]?
atom (underlined) in each of the following complex
[Ans. six]
species:
9.12 What is meant by a hexadentate ligand? Give an
(i) K4[Fe(CN)6] (ii) [Fe(H2O)6]CI3 example.
9.13 What do you understand by chelating ligands and
(iii) [Co(NH3)6]' (iv) [Fe (C2O4)3]'
what are chelates? Give one example of each.
[Ans. (i) +2, (ii) +3, (iii) +3, (iv) + 3] 9.14 Give one example of each of the following :
9.5 In the complex ion [Co(NH3)3(H2O)2CI]+: (i) a neutral ligand
(a) identify the ligands' formulae and charge on each (ii) an anionic ligand
of them; (iii) a cationic complex ion
(b) write the geometry of the complex ion. (iv) an anionic complex.
[Ans. (a) (i) NH3, 0 (ii) H2O, 0 (iii) CP, - 1 9.15 Calculatethechargenumberofthefollowingcomplex
(b) octahedral] ions :
9.6 Define coordination number. Find the coordination (i) [Co(NH3)4Cl2l+ (ii) [Pt(NH3)2CI2].
number of the central metal atom in each of the |Ans. (i) + 1, (ii) 0]
following : 9.16 What is the coordination number of central metal
ion in [Fe (C2O4)3]3 ?
(i) [Co(NH3)5CI] Cl2 (ii) K2[FeCI4]
[Ans. (i) six, (ii) four] [Ans. six]
9.7 What is the oxidation state of Co in the complex, 9.17 Give an example of chelate complexes.
9.18 Why is ammonia molecule a good ligand?
[Co(NH3)2(N02)CI] [Au(CN)2]?
9.19 Name a ligand which is bidentate and give an
[Ans. + 3] example of the complex formed by this ligand.
Coordination Compounds 489
K3[Fe(CN)6] potassium hexacyanidoferrate ligand by symbol p (mu) separated by hyphens. The symbol
(III) p is repeated before the name of each different kind of
bridging ligand present in the complex. Following examples
K2[HgI4] potassium f etraiodidomercurate
illustrate the procedure of nomenclature of the bridged
(ID polynuclear complexes.
K3[A1(C2O4)3] potassium trioxalatoaluminate
(III) z0H\
K3[Co(CN)5NO] potassium (H2O)4Fe< )Fe(H2O)4 (SO4)2
XOHZ
pentacyanidonitrosylcobaltate
(ID tetraaquairon (Ill)-p-dihydroxotetraaquairon (III) sulphate.
K[PtCl3(NH3)] potassium
amminetrichloridoplatinate (II) NH2
(NH3)4Co( XCo(NH3)4 (NO3)4
Dextro K3[Ir(C2O4)3] potassium-d-trioxalatoiridate xNO2z
(HI)
ammonium diamminetetra tetraamminecobalt (Ill)-p-amido-p-nitrotetraamminecobalt (III)
NH4[Cr(NCS)4
nitrate
(NH3)2] (thiocyanato - N) chromate (III)
Na2[Ni(EDTA)] sodium ethylenediaminetetraac- /NH\
etatonickelate (II) (en)2Co<f yCo(en)2
It is to be noted that the name of the cation present in an XOHZ
anionic complex is not prefixed by numerical prefixes such as bis (ethylenediamine)
di, tri, etc., even when there are more than one cation present cobalt(III)-p- hydroxo-p- imido-
bis (ethylenediamine) cobalt (III) ion
in the complex.
9.2.3 Nomenclature of Complexes z°H\
(C2O4)2Crv /Cr(C2O4)2
Containing Complex Cationic and XOHZ
Complex Anionic Species dioxalatochromium (III) -p-dihydroxodioxalatochromium (III) ion
The coordination compounds containing complex (C6H5)3P Cl Cl
cations and complex anions are also quite common. Such
coordination compounds are named by writing the name /pd\
of the complex cation followed by the name of the complex Cl Cl 'P(C6H5)3
anion. The complex cation and complex anion are named chloridotriphenylphosphinepalladium(II)-p-di
exactly in the same way as described earlier. Following chloridochloridotriphenylphosphinepalladium (II).
examples illustrate the procedure.
9.2.5 Writing Formula of a Simple
[Cr(NH3)6] [Co(CNJ6] hexaamminechromium (III)
hexacyanidocobaltate (III) Mononuclear Complex from its
[PtCl2(NH3)4] [PtCl4] tetraamminedichloridoplati- IUPAC Name
num (IV) tetrachloridoplati- The formula of a simple mononuclear complex can
nate (II) easily be written if its IUPAC name is known. The rules
involved and the procedure are as follows.
[CoC12(NH3)4]3 [Cr(CN)6] tetraamminedichloridocobalt (i) Identify the central metal and the ligands attached
(III) hexacyanidochromate to it. Write their formulae in the following order :
(III) [Metal atom; anionic, neutral, cationic ligands]
[Pt(py)4] [PtCl4] tetrapyridineplatinum (II) This constitutes the coordination sphere (complex
tetrachloridoplatinate (II) species) of the given complex. Enclose it in square
[Co(NH3)6] [CdCl5] hexaamminecobalt (III) brackets.
pentachloridocadmate (II) (ii) Calculate the charge on the complex species as
follows :
9.2.4 Nomenclature of Bridged Charge on a complex species = Oxidation state of the
Polynuclear Complexes central metal atom/ion + total charge on the ligands
attached.
The complexes having two or more acceptor metal
atoms are called polynuclear complexes. In such complexes, It is to be noted that the oxidation state of a
two or more acceptor atoms are bridged together through central metal atom/ion is always positive in a
certain ligands called bridging ligands. As mentioned earlier, complex. Consider it while carrying out the above
a bridging ligand is represented by prefixing the name of the calculation.
Coordination Compounds ___________________________________________ 493
(iii) If the charge calculated as above on the given central metal in this complex is iron (Fe) and it is attached
complex species is positive, the complex under to six cyanide CN- ligands. The oxidation state of iron is +2.
consideration is a cationic complex and would Charge on the complex species = + 2 + 6 x (-1) = -4.
contain anions in the ionic sphere. On the other Thus, the complex species should be represented as
hand, if the charge is negative, it is an anionic
[Fe(CN)6]4~. Further, the ionic sphere contains potassium
complex and would contain cations in the ionic
sphere. (K+) ions. In order to balance the charge, there should be
(iv) The number of cations or anions present in the four K+ ions in the ionic sphere. Hence, the given complex
ionic sphere of the given complex can be calculated should have the formula K4[Fe(CN)6].
by multiplying the formula of the cation or anion Example 9.2 Give the chemical formula of the compound
(mentioned in the name) by a suitable factor in nitropentaamminecobalt (III) nitrate.
such a way that the total charge of the ionic sphere
Ans. The name nitropentaamminecobalt (III) implies that
becomes equal to that of the coordination sphere.
the given complex contains cobalt in the +3 oxidation state
(v) In case of a cationic complex, write the ionic
and it is attached to one nitro and five NH3 ligands.
sphere immediately after the coordination sphere;
while in case of an anionic complex, write the .'. Charge on the complex species
ionic sphere immediately before the coordination = +3 + (-1) + 5 x (0) = +2
sphere. This will give the structure of the given Thus, the complex species should be represented as
9i
simple mononuclear complex. Following examples [Co(NOH2)(NH3)5] . Further, the ionic sphere contains
illustrate the procedure. NO3 ions. In order to balance the charge, there should be
Example 9.1 Give the chemical formula of potassium
hexacyanidoferrate (II). two NO3 ions. Hence, the given complex should have the
Ans. As the name hexacyanidoferrate (II) implies, the formula [Co(NO2)(NH3)5] (NO3)2.
REVIEW EXERCISES
9.20 Name the following complex ions in IUPAC system : [Ans. (i) potassium trioxalatoaluminate (III)
(i) [Co(CN)6]3~ (ii) pentaamminechloridocobalt (III) chloride
(ii) [Hgl4]2- (iii) potassium hexacyanidoferrate'(III) (iv) sodium
(iii) [Fe(H2O)6]3+ hexa nitrocobaltate (III)
(iv) [Cu(NH3)4]2+ (v) tris (ethylenediamine) chromium (III) chloride
(v) [CoCI(en)2NH3]2+ (vi) triamminetrichloridoplatinum
(vi) [CoCI2(NH3)4]+ (IV) chloride (vii) tris (ethylenediamine) cobalt
(vii) [Co,j(ONO) (NH3)5]2+ (III) hexacyanidochromate (III)
[Ans. (i) hexacyanidocobaltate (III) ion (viii) sodium ethylenediaminetetraacetatochromate
(ii) tetraiodidomercurate (II) ion (iii) hexaaquairon (II) (ix) pentaamminenitrocobalt
(III) ion (iv) tetraamminecopper (II) ion (III) chloride (x) Ammonium dramminetetra
(v) amminechloridobis (ethylenediamine) cobalt (III) (thiocyanato-N) chromate (III)
ion (vi) tetraamminedichloridocobalt (III) ion (xi) tetraamminediaquacopper (II) sulphate (xii)
(vii) pentaamminenitritocobalt (III) ion] tetrapyridineplatinum (II) tetrachloridoplatinate (II)
9.21 Give the IUPAC names of the following coordination (xiii) pentaamminenitritocobalt (III) chloride
compounds : (xiv) sodium hexafluoridoaluminate (III)]
(i) K3[AI(C2O4)3] 9.22 Give the chemical formula for each of the following
(ii) [CoCl(NH3)5]CI2
compounds:
(iii) K3[Fe(CN)6]
(i) copper (II) hexacyanidoferrate (II)
(iv) Na3[Co(NQ2)6]
(ii) potassium hexacyanidocobaltate (III)
(v) [Cr(en)3] Cl3
(iii) hexaammineplatinum (IV) chloride
(vi) [PtCl3(NH3)3] Cl
(vii) [Co(en)3][Cr(CN)6] (iv) potassium hexacyanidoferrate (III)
(viii) Na2[Cr(EDTA)] (v) pentaamminenitrocobalt (III) chloride
(ix) [Co(N02)(NH3)5] Cl2 (vi) tetrachloridocuprate (II) ion
(x) NH4[Cr(NH3)2 (NCS)4] (vii) chloridobis (ethylenediamine) nitrocobalt (III)
(xi) [Cu(H2O)2(NH3)4] SO4 ion
(xii) [Pt (py)4] [PtCI4] (viii) hexaaquamanganese (II) ion
(xiii) [Co(ONO)(NH3)5] Cl2 (ix) potassium hexachloridoplatinate (IV)
(xiv) Na3[A!F6] (x) triamminebromidonitrocobalt (II)
494 Nootan ISC Chemistry-XII
[Ans. (i) Cu2[Fe(CN)6] (ii) K3[Co(CN)6] 9.25 Write the chemical formulae of the following
(iii) [Pt(NH3)6] Cl4 (iv) K3[Fe(CN)6] complexes :
(v) [Co (NH3)5 (NO2)1 Cl2 (vi) [CuCI4]2- (i) Hexaammineplatinum (IV) chloride
(vii) [CoCI(N02) (en)2]+ (viii) [Mn(H2O)6]2+ (ii) Tetraamminedichloridocobalt (III) chloride
lAns. (i) |Pt(NH3)6]CI4 (ii) [Co(NH3)4CI2]CI]
(ix) K2[PtCI6] (x) [CoBr(NO2)(NH3)3] 9.26 Write IUPAC name of the complex Na3[Cr(OH)2F4].
9.23 Write the names of the following complexes using [Ans. sodium tetrafluoridodihydroxochromate (III)]
IUPAC norms : 9.27 Name the following complex using IUPAC norms :
(i) [Cu(NH3)4] SO4 [Co (en)2 (ONO) Cl] Cl.
(ii) [Pt(NH3)2CI2] [Ans. chloridobis (ethylene diamine) nitritocobalt
(iii) [Ag(NH3)2CI] (III) chloride]
(iv) Na3[AlF6] 9.28 A coordination compound has the formula
[Ans. (i) tetraamminecopper (II) sulphate CoCI3 ■ 4NH3. It does not liberate ammonia but
(ii) diamminedichloridoplatinum (II) forms a precipitate with AgNO3. Write the structure
(iii) diamminechloridoargentate (I) (iv) sodium and IUPAC name of the complex compound.
hexafluoridoaluminate (III)] [Ans. [Co(NH3)4CI2]CI,
9.24 Write the IUPAC names of the complexes tetraamminedichloridocobalt (III) chloride]
[CoBr(NH3)5]S04 and [Co(SO4)(NH3)5]Br and give a 9.29 Give the IUPAC name of [PtCI(NH2CH3)(NH3)2]CI.
simple test to distinguish them. [Ans. diamminechlorido(methylamine)platinum (II)
[Ans. pentaamminebromidocobalt (III) sulphate, chloride]
pentaamminesulphatocobalt (III) bromide; the 9.30 Write the formula of the following complexes:
former gives a white precipitate with BaCI2 while (i) Hexaammineplatinum chloride
the latter does not] (ii) Dichloridotetraammine cobalt ion.
[Ans. (i) [Pt(NH3)6]Cl4, (ii) [Co(NH3)4CI2]+]
now. Theoretically, octahedral complexes containing only The d- and Z-forms along with the optically inactive trans
monodentate ligands are optically active and should exist form are shown in Fig. 9.13.
in d- and Z-forms, but the paucity of adequate experimental
techniques to resolve them makes their optical isomerism
of little value.
2. Octahedral complexes containing one or more
symmetrical bidentate chelating ligands : Octahedral
complexes containing all the monodentate ligands could not
be resolved. However, if an octahedral complex contains one
or more bidentate chelating ligands, it is possible to resolve
it into its optically active forms. Some examples of such
complexes are given below.
(i) Complexes of the type [M(AA)3]n± : In the
Optically active forms (cis)
complexes of this type, three symmetrical bidentate
chelating ligands AA are coordinated to the central metal
atom M. Such complexes do not possess any element
of symmetry and are optically active. Moreover, these
complexes can be resolved into optical isomers.
An example of this type of complexes is [Cr(C2O4)3]3-.
It is optically active and has been resolved into d-and
Z-forms (Fig. 9.11).
Optically inactive
trans form
Fig. 9.13 Optically active (cis) and optically inactive
(trans) forms of the complex [CoCl2(en)2]+.
(iii) Complexes of the type [M(AA)2ab]n± : In this
case AA are symmetrical bidentate chelating ligands, while
a and b are monodentate ligands. Such complexes exist in
three forms, two are optically active (d- and Z-forms) and
the third one is inactive meso form. An example of this type
(OX refers to bidentate oxalato ligand.) of complexes is [CoCl(en)2(NH3)]2+. Its three forms are
shown in Fig. 9.14.
Other examples of this type are [Co(en)3]3+,
[Co(pn)3]3+, [Pt(en)3] + and [Cd(pn)3]2+. The optical
(d)
L CN J
Fig. 9.16 Formation of [Fe(CN)6]3“ complex ion by valence
bond method.
(iii) Hexaamminecobalt (III) ion, [Co (NH3)6]3+ :
This is also an octahedral complex ion formed by
d sp hybridisation. The central Co nucleus has
27 - 3 + 12 = 36 electrons in the complex and the net charge
on it is +3 + 0 = +3 because neutral ammonia ligands do not
contribute any charge to the central metal ion.
fig. 9.15 Formation of [Fe(CN)6l4 complex ion by The formation of this complex ion is shown in
valence bond method. Fig. 9.17.
Nootan ISC Chemistry-XII
3d 4s 4p
(a) Cr atom iTpr111 ? I T j e iiii
3d 4s 4p
(b) Cr3+ ion I T I T 111 1 J 1—1 IIII
1_—
__ . 1
▼ d2sp 3 hybridisation
(c) cPsp3 hybrid
orbitals of 1 T | T Ltl I'T 1 1 IT'I
Cr3+ ion vacant d2sp3 hybrid orbitals
2. Tetrahedral Complexes
Tetrahedral complexes are formed by sp hybridisation.
A few examples of such complexes are [Ni(CO)4], [NiCl4]2“,
[MnCl4]2-, [CuCl4]3-, etc. Formation of [Ni(CO)4] complex
is discussed below.
Nickel carbonyl, In this complex,
[Ni(C0)4] :
Fig. 9.19 Formation of [Fe(H2O)6]2j complex ion by central nickel atom is attached to four CO ligands. The
valence bond method. oxidation state of nickel is zero. Nickel atom has the
configuration 3d84s2 (a in Fig. 9.21). In the presence
(vi) Hexafluoridocobaltate (III) ion, [CoF6]3- : In
this complex ion, central Co3+ ion is surrounded by six of strong CO ligands, rearrangement takes place
and the 4s-electrons are forced to go into 3d-orbitals
fluoride ligands. Each fluoride ligand contributes two
(b in Fig. 9.21). Thus, Ni can now provide 4s- and 4p-orbitals
electrons and one negative charge to the central ion. Thus o
for hybridisation. One 4s- and three 4p-orbitals undergo sp
the total number of electrons possessed by Co3+ ion in the
hybridisation and form four hybrid orbitals of equal energy
complex is 27 - 3 + 12 = 36 and the net charge possessed
which are directed towards the four corners of a regular
by the complex ion is +3 - 6 = -3. Fluoride is also a weak
ligand and can not pair up the electrons of Co3+ ion against tetrahedron (c in Fig. 9.21). These orbitals overlap with the
orbitals of CO ligands (d in Fig. 9.21). Thus, a tetrahedral
Hund’s rule. Therefore, this complex is also formed by
[Ni(CO)4] complex is formed (e in Fig. 9.21).
3d 4s 4p
(a) Co atom
(a)
Ni atom |T>l|T4jT4<| T | T |
IT4-I
(b) Co3’’ ion 3d 4s 4p
(b) Ni atom after the |T4<|T4<|T>L-|T4<|T>1<|
11-— 1
---- 1 1 11
(c) sp3cP hybrid rearrangement in
orbitals of the presence of strong sp3 hybridisation
Co3* ion CO ligands
vacant sp3 d2 hybrid orbitals
(c) sp3 hybrid |T4<|T4<|T>l|T4<|TJ<| I I I I I
d High No. of
No. of
electrons
Low energy energy eg unpaired CFSE OOOOQ/L . L
t2g orbitals orbitals electrons / degenerate
in the '■■Too l10Dq
d-orbitals of
metal ion the metal ion in ^x2-y2’ ^z2 (eg)
1 -4Dq OOOOO the crystal field splitting of d orbitals
Five degenerate (Hypothetical) in the tetrahedral
2 -8 Dq d-orbitals of the
d2 T central metal ion
ligand field
3 - 12 Dq
T in the absence of
d3 ligands
T 4 - 6 Dq
d* tT t Fig. 9.27 Splitting of d-orbitals in a tetrahedral crystal
T 5 0 Dq field. Here At = 10 Dq, 0.4 A( = 4 Dq and 0.6 At = 6 Dq.
ds t
t 4 -4Dq +P Calculations have shown that t2g orbitals are raised by
d6 f
T 3 - 8 Dq + 2P 4 Dq, whereas eg orbitals are lowered by 6 Dq as compared
d7 n Ti t t T 2 -12 Dq + 3P
to the energy of hypothetical degenerate d orbitals in
d8 u U Tl n T 1 - 6 Dq + 4P
ligand field. The energy gap, i.e., the crystal field splitting 1
d? u
U between the two sets is At or 10 Dq (the subscript t in Af I
s
Q Dq + 5P
T4 n n 0 refers to tetrahedral geometry).
e
d10 T4
Coordination Compounds
Calculations have also shown that each electron yellow green region, the electron present in the lower
entering into low energy eg orbitals stabilises the complex t2g level gets excited to the higher eg level. Thus, a d-d
by 6 Dq, whereas each electron entering into high energy transition takes place as shown in Fig. 9.28.
t2g orbitals destabilises it by 4 Dq.
— — eg
The distribution of electrons in these orbitals depends
on whether the ligand field is strong or weak. In a strong
ligand field, electrons prefer to get paired up in low energy — — ----- kg - ----- ----- kg
orbitals in place of going to high enregy t2g orbitals. Ground state Excited state
Thus, we shall have a low spin complex. But if the ligand f2g eg p1
'2g eg
field is weak, the electrons would like to go to high energy
Fig. 9.28 d-d transition in [Ti(H2O)6]3+.
t2g orbitals in place of being paired up in low energy eg
orbitals and we shall have a high spin complex. On account of this transition, the complex appears
violet in colour.
|C| Colour in Coordination Compounds In the absence of ligands, there is no crystal field
Coordination compounds exhibit a wide range of splitting and hence the substance is colourless. For
colours. The sensation of colour is produced when light example, if water is removed from [Ti(H2O)6]3+ by
having a wavelength within the visible region (400 nm - heating, it becomes colourless. Similarly CuSO4 • 5H2O is
750 nm) of electromagnetic spectrum strikes the retina blue in colour, whereas anhydrous CuSO4 is white.
of the eye. A complex shows colour when the white light
passing through it is partly absorbed and partly reflected. [D] Limitations of Crystal Field Theory
The colour shown by the complex is the complementary 1. Crystal field theory assumes that metal-ligand
colour generated from the wavelengths present in the interaction is purely electrostatic. This assumption
partly reflected light. For example, if complex absorbs can not be said to be very realistic.
green light, it shows red colour. The relationship between 2. The theory takes into account only the d-orbitals
the wavelength of the light absorbed and the colour shown of central metal atom or ion and their splitting is
studied; the other orbitals, e.g., s and p orbitals of
by some complexes is given in Table 9.7.
central metal are not taken into.account.
Table 9.7 Relationship between the Wavelength of Light
Absorbed and the Colour Shown by Some Com 3. The theory is unable to explain why a certain
plexes ligand causes a large splitting while the other a
small, e.g., the theory has no explanation why H2O
Complementary is a stronger ligand than OH-.
Wavelength Colour of
colour shown 4. This theory rules out the possibility of any
Complex species of light the light
by the complex 7i-bonding in complexes. This is a serious
absorbed absorbed
species drawback of the theory because 7t-bonding is
[Co(CN)6]3" 310 nm Ultraviolet Pale yellow found to exist in complexes.
[Co(NH3)6]3+ 475 nm Blue Yellow orange 5. The theory does not give any weightage to the
orbitals of ligands. Therefore, all properties
[Ti(H2O)6]3+ 498 nm Yellow Purple related to ligand orbitals and their interaction
green with metal orbitals remain unexplained.
[Co(NH3)5(H20)]3+ 500 nm Blue green Red
[CoC1(NH3)5]2+ [E] Superiority of Crystal Field Theory Over
535 nm Yellow Violet
Valence Bond Theory
[Cu(H2O)4]2+ 600 nm Red Blue
Inspite of all the above limitations, crystal field theory
The colour shown by a complex can easily be explained has an edge over valence bond theory. The crystal field theory
on the basis of crystal field theory. According to the crystal is supposed to be superior to valence bond theory due to
field theory, colour of complexes is due to d-d transitions following facts.
of electrons. For example, the violet colour of [Ti(H2O)6]3+ 1. Crystal field theory provides the fine details of
can be explained as follows. magnetic properties of the complexes and gives a
[Ti(H2O)6]3+ is an octahedral complex. The central suitable explanation of the variation of magnetic
metal ion Ti3+ constitutes a 3d1 system. The only electron moments with temperature. These fine details are
present in the 3d-subshell lies in the t2g level in the ground not provided by VBT.
state of the complex. The next higher state available is 2. Crystal field theory provides a quantitative
empty eg level. When the complex absorbs light from the measure of the stability of a complex. With this
measure, the geometry acquired by a particular
510 ‘_____________________________________ Nootan ISC Chemistry-XII
complex can be predicted. VBT does not provide 4. The observed d-d transitions in complexes and
any such measure. their colour can be explained only on the basis
3. Crystal field theory gives an explanation for of crystal field theory. The VBT neither explains
certain thermodynamic and kinetic properties spectroscopic properties nor gives an adequate
of complexes. VBT does not provide such an explanation for the colour of complexes.
explanation.
precipitate containing AgCl, Hg2Cl2 and pbcl2 is treated Ag2S + 4NaCN^=± 2Na [Ag (CN)2] + Na2S
Sodium dicyanoargentate (I)
with aqueous ammonia solution to separate Ag+ ions from
2+ 2+ 2Na[Ag(CN)2] + Zn ----- > Na2 [Zn (CN)4] + 2Ag
Hg2 and Pb ions. AgCl forms a complex with ammonia Sodium tetracyanozincate (II)
and passes into the solution as given below.
AgCl + 2NH3 [Ag (NH3)2] Cl (iv) In electroplating : Since complexes are quite stable
Hg2Cl2 and PbCl2 remain undissolved as they do not and dissociate in solution to a very small extent, they give a
form complexes with ammonia. This helps in the detection controlled supply of metal ions in the solution. This property
of Ag+ ions. of complexes has been used in the electroplating of noble
(b) In the ILA group of qualitative analysis, Cd2+ are metals like silver and gold. The complex, K [Ag(CN)2] is
9-4- widely used as an electrolyte in the electroplating of inferior
tested in the presence of Cu ions by complex formation.
metals by silver. Similarly, the complex, K[Au(CN)2] is used
The solution containing both Cu2+ and Cd2+ ions is treated
for gold plating.
with excess of KCN, when both the ions form their cyanido
(v) In medicine : Complexes have also been used
complexes.
in medicine. For example, the platinum complex,
Cu2+ + 4CN“ [Cu (CN)4]2-
cis-[PtCl2(NH3)2] known as cisplatin has been found
Cd2+ + 4CN“ [Cd (CN)4]2“
useful in the treatment of cancer.
Since [Cu (CN)4]2- is more stable than [Cd(CN)4]2-,
(vi) In modifying the redox behaviour of metal ions :
the solution contains a higher equilibrium concentration
The redox behaviour of a metal ion may be modified by
of Cd2+ ions. Therefore, on passing H2S, only CdS gets
complexing it with a particular type of ligand. For example,
Precipitated. Thus, Cd2+ ions can easily be detected in the cobalt is easily oxidised from +2 to +3 state in the presence
9i
Presence of Cu ions. of some ligands. This property is utilised by many enzymes
(c) Ni ions can easily be detected and estimated by containing metal ions (e.g., Mo, Cu, Fe) to perform catalysed
complexing them with dimethylglyoxime. biochemical reactions.
(ii) For the estimation of hardness of water: The hardness (vii) Biological importance : Several coordination
°fwater is due to the presence of Ca2+ and Mg2+ ions in it. The compounds play significant roles in a number of processes
^xadentate ligand EDTA (ethylenediaminetetraacetato) occurring in plants and animals. Several biologically
forms stable complexes with Ca2+ and Mg2+ ions, and important natural compounds are complexes. For example,
can be used for the estimation of hardness of water by chlorophyll present in plants is a complex of magnesium;
simple titration method. Further, Ca2+ and Mg2+ ions can haemoglobin present in blood is a complex of iron; vitamin
k selectively estimated by this method because of the B12 is a complex of cobalt.
■
512 Nootan ISC Chemistry-XII
REVIEW EXERCISES
9.57 Using valence bond theory of complexes explain 9.69 Explain the square planar shape of the
the geometry and diamagnetic nature of ion tetracyanidonickelate (II) ion and account for its
[Co(NH3)6]3+. magnetic property.
9.58 Describe the hybridisation scheme, the resultant 9.70 Using the valence bond approach, predict the shape
geometry and the magnetic behaviour of and magnetic character of [Ni(CO)4].
[Co(NH3)6]3+. 9.71 Deduce the shape and magnetic behaviour of the
9.59 How would you account for the following? complex ion [Co(NH3)5NO2]2+.
(i) [Ti(H2O)6]3+ is coloured, while [Sc(H2O)6]3+ is 9.72 Explain the following observation :
colourless. Tetrahedral Ni(ll) complexes are usually
(ii) [Fe(CN)6] is weakly paramagnetic, while paramagnetic but square planar Ni(ll) complexes are
|Fe(CN)6]4‘ is diamagnetic. diamagnetic.
(iii) [Ni(CO)4] possesses tetrahedral geometry, while 9.73 What magnetic behaviours are expected for
n_
[Ni(CN)4] is square planar. [Ni(CO)4] and [NiCI4]2- and why?
9.60 Using valence bond approach, predict the shape and 9.74 Giving an example, describe how the formation
magnetism (i.e., paramagnetic or diamagnetic) of of coordination compounds is useful in analytical
[Co(CN)4]“. chemistry.
9.61 Using valence bond approach, predict the shape 9.75 Using the valence bond approach, deduce the shape
and magnetism (paramagnetic or diamagnetic) of and magnetic character of [Cr(CO)6],
[Ni(CN)4]2-. 9.76 Using the valence bond approach, predict the shape
9.62 Following valence bond scheme, explain the bonding and magnetic character of [Fe(CN)6] ion.
in [Cr(H2O)6]3+. 9.77 Write the neutral molecule in which the central
9.63 What do you understand by the term stability atom is sp d hybridised.
constant, K of a complex? Knowing that the value of 9.78 Using valence bond theory, deduce the geometry
K for [Cu(NH3)4]2+ is 4.5 x 1011 and for [Cu(CN)4]2- and magnetic nature of [Ni(CN)4]2~.
is 2.0 x 1027, suggest (a) which complex species will 9.79 Draw a figure to show splitting of degenerate
furnish less Cu2+ ions in solution, and (b) which out d-orbitals in an octahedral crystal field. How does
of NH3 and CN~ is a stronger ligand. the magnitude of Ao decide the actual configuration
9.64 Select a complex formation reaction and write an of d-orbitals in a complex entity?
expression for the stability constant of the complex 9.80 Compare the magnetic behaviour of the complex
species. What information is conveyed regarding the entities [Fe(CN)6]4- and [FeF6]3-.
strengths of ligands from the stability values of their 9.81 (a) Thevaluesofdissociationconstantof[Cu(NH3)4]2’r
complexes with a metal ion? Illustrate your answer and [Co(NH3)6]3+ are 1.0 x ,10-12 and
with examples of monodentate ligands. 6.2 x 10~34 respectively. Which complex would
9.65 Using the valence bond approach, explain the shape be more stable and why?
and magnetic behaviour of [Ni(NH3)6]2+. (b) Using valence bond theory of complexes,
9.66 On passing H2S through a solution containing Cu2+ explain the geometry and magnetic nature of
and Cd2+ ion and excess of KCN solution, only Cd [Ni(NH3)6]2+.
gets precipitated. Explain. 9.82 (a) Name two properties of the central metal ion
9.67 Draw the structure and write the hybridisation state of which enable it to form stable complex entities.
the central atom of each of the following species: (b) The formation of complex compounds, finds
(i) |Co(NH3)6]3+ application in the extraction of some metals.
(ii) [NiCI4]2". Furnish one example to support the above
9.68 Describe the following : statement.
(i) Linkage isomerism with an example. 9.83 Draw a sketch to show the splitting of d-orbitals in
(ii) Magnetic behaviour of [Ni(CO)4] ion. an octahedral crystal field.
Coordination Compounds 513
16. The complexes involving d sp hybridisation are called 17. According to the Crystal field theory, the interaction
inner orbital complexes. These complexes usually contain lesser between the metal ion and the ligands is purely electrostatic. On
number of unpaired electrons. Therefore, they are also called
the approach of ligands, the five degenerate d-orbitals of the
low spin complexes or spin paired complexes. Some examples of
this type of complexes are [Fe(CN)6]4- [Co(NH3)6]3', [Co(CN)6]4-, central metal atom get split into two point groups-t2g (dxy, dyz, dzx)
(Fe(CN)6]3-, etc. 2
ande„(d2 2, dz ). The splitting depends upon the geometry of
The complexes involving spa^ hybridisation are called outer y x —y
orbital complexes. They possess comparatively higher number of the complex. The energy separation between these groups is
unpaired electrons. Therefore, they are also referred to as high spin referred to as 10 Dq or A. The occupation of electrons in splitted
complexes or spin free complexes. Some examples of this type of
complexes are [Fe (H2O)6]2+, [CoF6]3~, [Cr(H2O)6]2+, [Zn(NH3)6]2+, d-orbitals is in accordance to Hund's rule.
etc.
CH3—C = NOH
CH3— C= NOH
Dimethylglyoxime
OH 0
T
ch3-—C=N N=C—CH3
1 1
ch3 —C=N XN = C—CH3
0 OH
Nickel dimethylglyoxime (Red)
Mer-triamminetrichloridocobalt (III)
42. (a) [Zn(OH)4]1
2~ (b) [Co(NH3)6]2(SO4)3
(A.I.E.E.E., 2002) 44. The IUPAC name for the complex [Co(NO2)(NH3)5] Cl2 is
35. One mole of the complex compound Co(NH3)5CI3, gives 3 (a) nitrito-N-pentaaminecobalt (III) chloride
moles of ions on dissolution in water. One mole of the same (b) nitrito-N-pentaamminecobalt (II) chloride
complex reacts with two moles of AgN03 solution to yield (c) pentaamminenitrito-N-cobalt (II) chloride
(d) pentaamminenitrito-N-cobalt (III) chloride.
two moles of AgCl (s). The structure of the complex is
(A.I.E.E.E., 2006)
(a) [Co(NH3)5CI]CI2
45. Nickel (Z = 28) combines with a uninegative monodentate
(b) [Co(NH3)3 Cl3] -2NH3 ligand X“ to form a paramagnetic complex [NiX4]2-. The
(c) [Co(NH3)4CI2]CI ■ NH3
number of unpaired electron(s) in the nickel and geometry of
(d) [Co(NH3)4CI]CI2 • NH3. . (A.I.E.E.E., 2003)
this complex ion are respectively
36. The coordination number of a central metal atom in a complex
(a) one, tetrahedral (b) two, tetrahedral
is determined by
(a) the number of ligands around a metal ion bonded by (c) one, square planar (d) two, square planar.
sigma bonds (A.I.E.E.E., 2006)
(b) the number of ligands around a metal ion bonded by pi- 46. The coordination number and the oxidation state of the
bonds element ‘E’ in the complex [E(en)2(C2O4)]NO2 (where (en) is
(c) the number of ligands around a metal ion bonded by ethylenediamine) are, respectively
sigma and pi-bonds both
(a) 6 and 2 (b) 4 and 2
(d) the number of only anionic ligands bonded to the metal
(c) 4 and 3 (d) 6 and 3.
ion. (A.I.E.E.E., 2004)
(A.I.E.E.E., 2008)
37. Which one of the following complexes is an outer orbital
47. In which of the following octahedral complexes of Co (at.
complex?
(a) [Fe(CN)6]4’ (b) [Mn(CN)6]4~ no. 27), will the magnitude of Ao be the highest?
(c) [Co(NH3)6]3+ (d) [Ni(NH3)6]2+. (a) [Co(CN)6]3’ (b) [Co(C2O4)3]3’
(c) [Co(H2O)5]3+ (d) [Co(NH3)6]3+.
(A.I.E.E.E., 2004)
38. Coordination compounds have great importance in biological (A.I.E.E.E., 2008)
systems. In this context which of the following statements is 48. Which of the following complex ions is not expected to absorb
incorrect? visible light?
(a) Chlorophylls are green pigments in plants and contain (a) [Ni(CN)4]2- (b) [Cr(NH3)6]3+
calcium. (c) [Fe(H2O)6]2" (d) [Ni(H20)6]2r.
(b) Haemoglobin is the red pigment of blood and contains (A.I.P.M.T., 2010)
iron. 49. The existence of two different coloured complexes with the
(c) Cyanidocobalamin is vitamin B12 and contains cobalt. composition of [Co(NH3)4CI2F is due to
(d) Carboxypeptidase-A is an enzyme and contains zinc.
(a) linkage isomerism
(A.I.E.E.E., 2004)
(b) geometrical isomerism
39. Which one of the following has largest number of isomers?
(a) [Ru(NH3)4CI2]+ (b) [Co(NH3)5CI]2+ (c) coordination isomerism
(c) [lr(PR3)2H(CO)]2+ (d) [Co(en)2CI2]+. (d) ionisation isomerism. (A.I.P.M.T., 2010)
50. The d-electron configurations of Cr2+, Mn2+, Fe2+ and Co2'
(/? = alkyl group, en = ethylenediamine) (A.I.E.E.E., 2004)
are d4, d5, d6 and d7 respectively. Which one of the following
40. The correct order of magnetic moments (spin only values in
BM) among the following is will exhibit minimum paramagnetic behaviour?
(a) [MnCI4]2- > [CoCI4r > [Fe(CN)6]4“ (At. no. Cr = 24, Mn = 25, Fe = 26, Co = 27)
(b) [MnCI4]2" > [Fe(CN)6]4^ > [CoCI4]2’ (a) [Fe(H2O)6]2+ (b) [Co(H2O)6]2+
(c) [Fe(CN)6]4^ > [MnCI4]2~ > [CoCI4]2’ (c) [Cr(H2O)6]2+ (d) [Mn(H2O)6]2+.
(d) [Fe(CN)6]4- > [CoCI4]2- > [MnCI4]2-. (A.I.E.E.E., 2004)
(A.I.P.M.T., 2011)
Coordination Compounds 521
51. The complexes [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6] 62. Which one of the following complexes shows optical
[Co(CN)6] are the examples of which type of isomerism? isomerism?
(a) Ionisation isomerism (b) Coordination isomerism (a) [Co(NH3)3CI3] (b) c/s[Co(en)2CI2]CI
(c) Geometrical isomerism (d) Linkage isomerism. (c) trans[Co(en)2CI2]CI (d) [Co(NH3)4CI2]CI
(A.I.P.M.T., 2011) (en = ethylenediamine) (J.E.E. Main, 2016)
52. The complex, [Pt(Py)(NH3)BrCI] will have how many 63. Among [Ni(C0)4], [NiCI4]2-, [Co(NH3)4CI2]CI, Na3[CoF6],
geometrical isomers? Na202 and CsO2, the total number of paramagnetic
(a) 4 (b) 0 (c) 2 (d)3. compounds is
(A.I.P.M.T., 2011) (a) 2 (b) 3
53. Which one of the following is an outer orbital complex and (c) 4 (d) 5 (J.E.E. Ad., 2016)
exhibits paramagnetic behaviour? 64. Which of the following has longest C—0 bond length? (Free
(a) [Ni(NH3)6]2+ (b) [Zn(NH3)6]2+ C—0 bond length in CO is 1.128 A.)
(c)[Cr(NH3)6]3+ (d) [Co(NH3)6]3+. (a) Ni(C0)4 (b) [Co(C0)4]
(A.I.P.M.T., 2012) (c) [Fe(CO)4]2" (d) [Mn(C0)6]+
54. Among the following complexes, the one which shows zero (N.E.E.T., 2016)
crystal field stabilisation energy (CFSE) is 65. The correct order of the stoichiometries of AgCI formed
(a) [Mn(H2O)6]3+ (b) [Fe(H2O)6]3+ when AgNO3 in excess is treated with the complexes :
(c) [Co(H2O)6]2+ (d) [Co(H2O)6]3+. CoCI3 • 6NH3, CoCI3 • 5NH3, CoCI3 ■ 4NH3 respectively is
(A.I.P.M.T., 2014) (a) 1 AgCI, 3 AgCI, 2 AgCI (b) 3 AgCI, 1 AgCI, 2 AgCI
55. Which of the following complexes is used to be as an (c) 3 AgCI, 2 AgCI, 1 AgCI (d) 2 AgCI, 3 AgCI, 1 AgCI
anticancer agent? (N.E.E.T., (UG)-2017)
(a) mer-[Co(NH3)3CI3] (b) cis-[PtCI2(NH3)2] 66. Correct increasing order for the wavelengths of absorption in
the visible region for the complexes of Co3+ is
(c) cis-/<2[PtCI2Br2] (d) Na2CoCI4.
(A.I.P.M.T., 2014) (a) [Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+
56. Which of the following complex species is not expected to (b) [Co(H2O)6]3+, [Co(en)3]3+, [Co(NH3)6]3+
exhibit optical isomerism? (c) [Co(H2O)6]3+, [Co(NH3)6]3+, [Co(en)3]3+
(a) [Co(en)3]3+ (b) [Co(en)2CI2]+ (d) [Co(NH3)6]3+, [Co(en)3]3+, [Co(H2O)6]3+
(c) [Co(NH3)3CI3] (d) [Co(en)(NH3)CI2]+. (N.E.E.T., (UGJ-2017)
□_
(J.E.E. Main, 2013) 67. Pick out the correct statement with respect [Mn(CN)6]
57. The octahedral complex of a metal ion M3+ with four 3 2.
(a) It is sp d hybridised and octahedral
monodentate ligands Llt L2, L3 and L4 absorb wavelengths 3 2
(b) It is sp d hybridised and tetrahedral
in the region of red, green, yellow and blue respectively. The 2 3
(c) It is d sp hybridised and octahedral
increasing order of ligand strength of the four ligands is
(a) L4 < L3 < L2 < /-i (b) Ly < L3 < L2 < t4 (d) It is dsp hybridised and square planar
(N.E.E.T., (UG)-2017)
(c) L3 < L2 < L4 < Z-i (d) < L2 < L4 < L3.
68. On treatment of 100 mL of 0.1 M solution of CoCI3 • 6H?O
(J.E.E. Main, 2014)
o_ with excess AgNO3; 1.2 x 10 ions are precipitated. The
58. The complex ion [Ni(CN)4] is
complex is
(a) square planar and diamagnetic
(a) [Co(H2O)6]CI3 (b) [Co(H2O)5CI]CI2 • H2O
(a) tetrahedral and paramagnetic
(c) [Co(H2O)4CI2]CI • 2H20 (d) [Co(H2O)3CI3] • 3H2O
(a) square planar and paramagnetic (J.E.E. Main, 2017)
(a) tetrahedral and diamagnetic. (I.S.C., 2016) 69. Consider the following reaction and statements :
59. The number of geometric isomers that can exist for square
planar [Pt (Cl) (py) (NH3) (NH20H)]+ is (py = pyridine) [Co(NH3)4Br2] + Br ----- > [Co(NH3)3Br3] + NH3
(a) 2 (b) 3 (I) Two isomers are produced if the reactant complex ion is
(c) 4 (d) 6 (J.E.E. Main, 2015) a cis-isomer.
60. Which of the following compounds is not coloured yellow ? (II) Two isomers are produced if the reactant complex ion is
(a) Zn2[Fe(CN)6] (b) K3[Co(NO2)6] a trans-isomer.
(c) (NH4)3 [As(Mo3O10)4] (d) BaCrO4 (III) Only one isomer is produced if the reactant complex ion
(J.E.E. Main, 2015) is a trans-isomer.
61. The pair having the same magnetic moment is (IV) Only one isomer is produced if the reactant complex ions
[At. No. : Cr = 24, Mn = 25, Fe = 26, Co = 27] is a cis-isomer.
(a) [Cr(H2O)6]2 + and [CoCI4]2" The correct statements are :
(b) [Cr(H2O)6]2+ and [Fe(H2O)6]2+ (a) (I) and (II) (b) (I) and (III)
(c) [Mn(H2O)6]2+ and [Cr(H2O)6]2+ (c) (III) and (IV) (d) (II) and (IV)
(d) [CoCI4]2- and [Fe(H2O)6]2+ (J.E.E. Main, 2016) (I.I.T., J.E.E. 2018)
Nootan ISC Chemistry-XII
70. The type of isomerism shown by the complex [CoC^fen^l is 71. The geometry and magnetic behaviour of the complex
[Ni(CO)4] are
(a) ionization isomerism
(a) square planar geometry and paramagnetic
(b) coordination isomerism
(b) tetrahedral geometry and diamagnetic
(c) geometrical isomerism
(c) square planar geometry and diamagnetic
(d) linkage isomerism (N.E.E.T., 2018)
(d) tetrahedral geometry and paramagnetic (N.E.E.T., 2018)
Answers
1. (d) 2. (a) 3. (a) 4. (0 5. (c) 6. (0 7. (c) 8. (d) 9. (0 10. (a)
11. (d) 12. (d) 13. (b) 14. (b) 15. (b) 16. (b) 17. (a) 18. (c) 19. (b) 20. (0
21. (0 22. (a) 23. (c) 24. (a) 25. (d) 26. (b) 27. (d) 28. (b) 29. (d) 30. (b)
31. (b) 32. (b) 33. (a) 34. (0 35. (a) 36. (a) 37. (d) 38. (a) 39. (d) 40. (a)
41. (a) 42. (a) 43. (d) 44. (d) 45. (b) 46. (d) 47. (a) 48. (a) 49. (b) 50. (b)
51. (b) 52. (d) 53. (a) 54. (b) 55. (b) 56. (c) 57. (b) 58. (a) 59. (b) 60. (a)
61. (b) 62. (b) 63. (b) 64. (0 65. (0 66. (a) 67. (0 68. (b). 69. (b) 70. (0
71. (b)
Hints
26. Higher the number of unpaired electrons in the central 3d 4s 4p
Mn3+ (3tf4) : |U| ? | ? | |~| □ iiii
metal atom, greater is the paramagnetism of the complex.
2_
28. C204 is a bidentate ligand. 3d 4s 4p
Cr3+ (3b3) : | T| T | T | | ~1 □ IIII
29. Cl” is a weak ligand and is unable to pair up 3d electrons
of Ni2+ ion. Therefore, the complex is formed by sp3 hybridisation (CN- ligands are strong and pairup electrons)
_ o_
and acquires a tetrahedral shape. The complex [Co(CN)6] has no unpaired electron and
34. Cyanido complexes are more stable because CN” is a therefore possesses the lowest value of paramagnetic behaviour.
strong ligand and stabilises the complex to a greater extent. It is actually diamagnetic in nature.
35. [Co(NH3)5CI]CI2 [Co (NH3)5CI]2+ + 2CF 43. gs = Jri (n + 2) BM or 2.84= Jn(n + 2)
1 mole 3 moles of ions in water
which gives n= 2.
2AgNO3 + [Co(NH3)5 CI]CI2 —>
In strong ligand field, cr configuration becomes
2 moles 1 mole
m| T | T | | | and there are only two unpaired electrons left.
[Co(NH3)5CI](NO3)2 + 2AgCI(s)
2 moles 2—
2 4. 3 2 45. The formation and geometry of [NiX4] is as follows :
37. [Ni(NH3)fi] is formed by sp d hybridisation because 3d 4s 4p
configuration of Ni2+ is Is2 2s2 2p6 3s2 3p° 3d8 and NH3 is a (a) Ni atom |TT|t4|T4| T I T I ITT] | | | I
weak ligand, i.e., it is unable to pair up d-electrons.
38. Chlorophyll contains magnesium and not calcium.
(b) Ni2+ ion |T4<|T4<|T4<| T | T 11 11 | | |
39. [Co(en)2CI2]+ shows both geometrical as well as optical
isomerism. ,, sp3 hybridisation
40. The number of unpaired electrons possessed by Mn2+,
(c) sp3 hybrid orbitals |T4jT4<|T4-| T | T | | | | | ~|
Co2+ and Fe2+ in the given complexes are respectively 5, 3 and
0. Hence, [MnCI4]2- has the highest and the [Fe(CN)6F the least ofNi2+ion Vacant sp3
hybrid orbitals
magnetic moment.
41. The optical isomers of [Cr2(C2O4)3]3- are as follows (d) sp3 hybrid orbitals |TT|T4|T4<| T | T | |T4<|T4<|T4<|TT|
in [NiX4]2- Four pairs of electrons
Z X-r x U I FM-v i2-• from four X" ligands
(e) Tetrahedral [NiX4] ion :
46. Both en and C2O2" are bidentate ligands. Therefore, < OH" < C2O2" < H2O < NCS"
coordination number of E in the given complex is 6. The oxidation < CH3CN < NH3 < en < NO2 < CN" < CO.
state of E is +3. Thus, among the ligands present in the given complexes,
47. The arrangement of ligands in the order of increasing CN" is the strongest ligand and causes maximum splitting in the
field strength is called spectrochemical series. The series is as cf-orbitals of the central metal ion. Hence, for [Co(CN)6] , the
follows : magnitude of Ao is the highest.
I" < Br" < S2" < SCN" < CF < NO3 < F"
Answers
1. T 2. F 3. T 4. T 5. T 6. F 7. F 8. F 9. T 10. F
11. T 12. F 13. T 14. F 15. T 16. F 17. F 18. T 19. T.
3. The coordination number of platinum in [PtCI4]2" is.............. 13. The complex [Ni(C0)4]° is............. in shape and is...............
4. Ethylenediamine is a............. dentate ligand. magnetic.
5. The IUPAC name of the complex K3[AgF4] is................ 14. Tetracyanocuprate(ll) ion involves............. hybridisation and
6. The isomerism arising due to the interchange of ligands is............. in shape.
between the coordination spheres of positive and negative 15. Greater the basic strength of the ligand, ............. is the
parts is called............. isomerism. stability of the complex.
7. When all the electrons in a complex are paired, it is............. 16. The ligand used in the estimation of the hardness of water
I in nature. is.................
8. Outer orbital complexes involve ............. hybridisation and 17. Vitamin B12 is a complex of................
are............. spin complexes. 18. Cisplatin found useful in the treatment of cancer has the
9. The complexes involving ............. hybridisation are called formula................
inner orbital complexes. They are............. spin complexes.
Ewers
l.five 2. complex ion 3. 4 4. di
5. potassium tetrafluoroargentate (I) 6. coordination 7. diamagnetic
8. sp3d2, high 9. d2sp3, low 10.+3 11. octahedral, dia
0 12. 3 13. tetrahedral,dia 14. dsp , square planar 15. greater
16. EDTA 17. cobalt 18. czs-[PtCI2(NH3)2]
524 Nootan ISC Chemistry-XII I (
Assertion Reason
1. Mohr’s salt, FeSO4 • (NH4)2S04 ■ 6H2O is not a coordination The aqueous solution of this salt gives the test of Fe , NH4 and
compound. SO 4 ions.
2. The complexes, [Co(pn)2CI2]+ and [Co(tn)2CI2]+ are ligand isomers. The ligand 1,2-diaminopropane can exist both as
1,2-diaminopropane (pn) and 1, 3-diaminopropane (t„).
3. The tetrahedral complexes can show geometrical isomerism. This is because all the four ligands lie at the same distance from
central metal atom in a tetrahedral geometry.
Answers
1. (b) 2. (a) 3. (d) 4. (a) 5. (c)
6NCERT TEXT-BOOK'
I Exercises (with Hints and Solutions)
9.1 Explain the bonding in coordination compounds in terms of 9.5 Specify the oxidation numbers of the metals in the
Werner’s postulates. coordination entities :
Hint: See § 9.1.5. (i) [Co(H2O)(CN)(en)2]2+
9.2 FeS04 solution mixed with (NH4)2SO4 solution in 1 : 1 molar (ii) [CoBr2 (en)2]+
ratio gives the test of Fe2+ ion but CuS04 solution mixed with (iii) [PtCI4]2”
aqueous ammonia in 1 : 4 molar ratio does not give the test (iv) K3[Fe(CN)6] I 94
of Cu2 ' ion. Explain, why? (v) [Cr(NH3)3CI3]
Ans. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 Ans. Suppose the oxidation number of the metal present in
ratio forms a double salt, FeSO4 ■ (NH4)2SO4 • 6H2O known the given complex isx. I 9.S
as Mohr's salt. It ionises as follows : (i) x+(0) + (-l) + 2 x (0) = + 2;
FeS04 ■ (NH4)2S04 • 6H2O ----- > x=+3
(ii) x + 2 x (-l) + 2 x (0) = + 1;
Fe2+ + 2NH4 + 2SO4~ + 6H20 .-. x=+3
(iii) x + 4 x (-1) = -2;
Due to the presence of Fe2+ ions in solution, it gives the test x=+2
of Fe2+ ion. (iv) 3 x (+l)+x + 6 x (—1) = 0;
x=+3
When CuS04 solution is mixed with aqueous ammonia in
(V) X + 3x(0) + 3x(-l) = 0;
1 : 4 molar ratio, a complex salt. [Cu(NH3)4]SO4 is obtained. x= + 3 19.10
It ionises in solution as follows. 9.6 Using IUPAC norms write the formulas for the following :
[Cu(NH3)4] S04 ----- > [Cu(NH3)4]2+ + SO2- (i) Tetrahydroxozincate(ll)
The [Cu(NH3)4]2+ ion does not ionise further to give Cu2+ (ii) Potassium tetrachloridopalladate(ll)
ions. Therefore, its solution does not give the test of Cu2+ ion. (iii) Diamminedichloridoplatinum(ll)
(iv) Potassium tetracyanidonickelate(ll)
9.3 Explain with two examples each of the following: coordination
(v) Pentaamminenitrito-O-cobalt(lll)
entity, ligand, coordination number, coordination polyhedron,
(vi) Hexaamminecobalt (III) sulphate
homoleptic and heteroleptic.
(vii) Potassium tri(oxalato) chromate(lll)
Hint: See § 9.1.2.
(viii) Hexaammineplatinum(IV)
9.4 What is meant by unidentate, didentate and ambidentate
ligands? Give two examples for each. (ix) Tetrabromidocuprate(ll)
(ii) K2[PdCI4]
(iii) [Pt(NH3)2CI2]
(iv) K2[Ni(CN)4]
(v) [Co(NH3)5(ONO)]2+
(vi) [Co(NH3)6]2(SO4)3
(vii) K3[Cr(C2O4)3]
(viii) [Pt(NH3)6]4+
(ix) [CuBr4]2"
(x) [Co(NH3)5(NO2)]2+ (i) [CoCI2(en)2r
(ii) [Co(NH3)CI(en)2]2+
9.7 Using IUPAC norms write the systematic names of the
(iii) [Co(NH3)2CI2(en)]+
following :
(i) [Co(NH3)6]CI3
(ii) [Pt(NH3)2CI(NH2CH3)]CI
(iii) [Ti(H2O)6]3+
(iv) [Co(NH3)4CI(NO2)]CI
(v) [Mn(H2O)6]2+
(vi) [NiCI4]2“
(vii) [Ni(NH3)6]CI2
(viii)
[Co(en) 3]3+
(ix) Ni(C0)4]
Ans. (i) Hexaamminecobalt(lll) chloride
(ii) Diamminechlorido (methylamine) platinum(ll) chloride
(iii) Hexaaquatitanium(lll) ion
(iv) Tetraamminechloridonitrito-N-cobalt(lll) chloride
(v) Hexaaquamanganese(ll) ion
(vi) TetrachloridonickelateGI) ion
(vii) Hexaamminenickel(ll) chloride
(viii) Tris (ethane-1, 2-diamine) cobalt(lll) ion
Geometrical isomers
(ix) Tetracarbonylnickel(O). Optical isomers
9.8 List various types of isomerism possible for coordination
9.12 Write all the geometrical isomers of [Pt(NH3)(Br)(CI)(py)] and
compounds, giving an example of each.
how many of these will exhibit optical isomerism?
Hint: See § 9.3.
9.9 How many-geometrical isomers are possible in the following Hint: See Fig. 9.6.
coordination entities?
None of these isomers will show optical isomerism.
(i) [Cr(C2O4)3]3' 9.13 Aqueous copper sulphate solution (blue in colour) gives :
(ii) [Co(NH3)3CI3]
(i) a green precipitate with aqueous potassium fluoride and
Ans. (i) It is a complex of the type [M(AA)3]n± and is unable to (ii) a bright green solution with aqueous potassium chloride.
show geometrical isomerism. Hence, no geometrical isomers Explain these experimental results.
9.14 What is the coordination entity formed when excess of get paired up in the presence of weak H2O ligands. The
aqueous KCN is added to an aqueous solution of copper unpaired electrons undergo d-d transition due to sunlight.
sulphate? Why is it that no precipitate of copper sulphide is The transition absorbs red light. Therefore, the complex entity
obtained when H2S(g) is passed through this solution? shows complimentary green colour.
In [Ni(CN)4]2-, nickel again exists as Ni24- with 3d8
Ans. On addition of aqueous KCN to an aqueous solution
of copper sulphate, the coordination entity [Cu(CN)4] is configuration and has two unpaired electrons. But in the
formed as shown below. presence of strong CN- ligands, these electrons get paired
[Cu(H2O)4]2+ +4CN“ ----- > [Cu(CN)4]2- +4H2O up. In the absence of any unpaired electron, no d-d transition
9.19 [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2“ is and indicate the oxidation state, electronic configuration and
coordination number. Also give stereochemistry and magnetic
diamagnetic. Explain, why?
moment of the complex :
Hint: See § 9.4.1 [B] 1 (octahedral complexes)(iv) and
§ 9.4.1 [B] 3 (square planar complexes)(i). (i) K[Cr(H20)2(0204)2] • 3H2O
(ii) [Co(NH3)5CI]CI2
9.20 A solution of [Ni(H20)6] is green but a solution of
[Ni(CN)4]2- is colourless. Explain. (iii) CrCI3(py)3
Ans. In [Ni(H2O)g]2+ nickel exists as Ni2+ with 3d8 (iv) Cs[FeCI4]
configuration. It has two unpaired electrons which do not (v) K4[Mn(CN)6]
Coordination Compounds 527
Ans.
Cr3+ OzV3 x 3 .0 n = 3,
(i) Potassium + 1 +x + 2x(0) 6 Octahedral 3c/ , ttgCg
diaquadioxalatochromate + 2 x (-2) = 0 73(3 + 2)
(III) hydrate x=+3
= 3.87 BM
Co3+ 3d , t2geg n = 0,
(ii) Pentaamminechl- x + 5 x (0) + 6 Octahedral
oridocobalt (III) 3 x(-l)=0 70(0 + 2)
chloride .'. x = + 3
= 0 BM
9.25 What is meant by stability of a coordination compound Ans. (ii). [Fe(H2O)6]2+ has the highest magnetic moment.
in solution? State the factors which govern stability of (This is because in (i) Cr3+ has 3 unpaired electrons, whereas
complexes. in (iii) Zn2+ has no unpaired electron. Fe2+ in (ii) has 4
Hint: See § 9.5. unpaired electrons).
9.26 What is meant by the chelate effect? Give an example. 9.30 The oxidation number of cobalt in K[Co(CO)4)] is
Ans. When a polydentate ligand attaches itself to a central (i) + 1 (ii) + 3
metal ion through two or more donor atoms in such a way (iii)-l (iv) - 3
that it forms a five or six membered ring with the central
ion, the effect is called chelate effect. Chelates stabilise a Ans. (iii) -1.
complex. For example, [+1 + x + 4 x (0) = 0, x =- 1]
H2C^H2N> NH2—CH2']2+ 9.31 Amongst the following, the most stable complex is
(i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6]3+
(iii) [Fe(C2O4)3]3" (iv) [FeCI5]3'
h2o—h2n nh^-ch2
Ans. (iii) [Fe(C2O4)3]3-
i.e., [Cu(en)2]2+
(C2O24 is a chelating ligand and stabilises the complex).
9.27 Discuss briefly giving an example in each case the role of
coordination compounds in : 9.32 What will be the correct order for the wavelengths of
(i) biological systems (ii) medicinal chemistry absorption in the visible region for the following :
[Ni(N02)6]4~, [Ni(NH3)6]2+, [Ni(H2O)6]2+?
(iii) analytical chemistry and
(iv) extraction/metallurgy of metals. Ans. The order of ligands present in the given complexes in
Hint: See § 9.6. the spectrochemical series is as follows :
9.28 How many ions are produced from the complex Co(NH3)6CI2 H2O < NH3 < NO2
in solution? Hence, the wavelengths of the light observed will be in the
K (i) 6 (ii) 4
order
(iii) 3 (iv) 2 [Ni(H20)6]2+ < [Ni(NH3)6]2+ < [Ni(N02)6]4-
Ans. (iii) 3 ions are produced. Since the wavelength observed is complementary to that
[Co(NH3)6]CI2 ----- > [Co(NH3)6]2+ + 2CF 7 he
absorbed, the wavelengths of the light absorbed E = —
k A 7
9.29 Amongst the following ions which one has the highest
magnetic moment value? will be in the opposite order, i.e.,
[Ni(NO2)6]4” < [Ni(NH3)6]2+ < [Ni(H2O)6]2+.
(i) [Cr(H2O)6]^ (ii) [Fe(H2O)6]2+ (iii) [Zn(H2O)6]2+
528 Nootan ISC Chemistry-XII
(a) [Co(NH3)5NO2]CI2 and [Co(NH3)5ONO]CI2. ion according to the valence bond theory. (2015)
(b) [Cr(H2O)5CI]CI2 • H2O and [Cr(H2O)4CI2]CI • 2H2O. (2009) 9. What type of isomers are [Co(NH3)5Br]SO4 and
[Co(NH3)5SO4]Br? Give a chemical test to distinguish
3. (a) Give the IUPAC names for the following :
between them. (2015)
(i) Na3[AIF6] (ii) [Co(NH3)5]CI3
10. Write the structures of optical isomers of the complex ion
(b) For the complex ion, [Fe(CN)6] :
[Co(en)2CI2]+. (2015)
(i) Show the hybridisation diagrammatically.
11. (a) Write the IUPAC names of the following :
(ii) Is it an inner orbital complex or an outer orbital complex?
(i) [Co(NH3)4SO4]NO3
(iii) State its magenetic property. (2011)
(ii) K[Pt(NH3)CI3]
4. (a) Write the formulae of the following coordination
compounds : (b) What type of isomerism is exhibited by the following pairs
of compounds :
(i) potassium tetracyanidonickel (0)
(i) [PtCI2(NH3)4]Br2 and [PtBr2(NH3)4]CI2,
(ii) triamminetrinitrocobalt (III)
(ii) [Cr(SCN)(H2O)5]2+ and [Cr(NCS)(H2O)5]2+?
(b) [CoF6] is a coordination complex ion.
(i) What is the oxidation number of cobalt in the complex? (c) How does K2[PtCI4] get ionised when dissolved in water?
(ii) How many unpaired electrons are there in the Will it form precipitate when AgNO3 solution is added to
complex? it? Give a reason for your answer. (2016)
(iii) State the magnetic behaviour of the complex. 12. (a) Write the formula of the following compounds :
(iv) Give the IUPAC name of the complex. (i) Potassium trioxalatoaluminate (III)
(ii) Hexaaquairon(ll) sulphate.
(c) Draw the structural isomer of [Co(NH3)5NO2]CI2 and
name the type of isomerism. (2012) (b) Name the types of isomerism shown by the following
pairs of compounds :
5. (a) (i) State the geometry and magnetic property of
(i) [Cu(NH3)4] [PtCI4] and [Pt(NH3)4][CuCI4]
tetracarbonyl nickel according to the valence bond
(ii) [Co(Pn)2CI2]+ and [Co(tn)2CI2]+
theory.
(c) For the coordination complex ion [Co(NH3)6]3+:
(ii) What type of structural isomers are
(i) Give the IUPAC name of the complex ion.
[Pt(OH^(NH3)4]SO4 and [PtSO4(NH3)4](OH)2? How
(ii) What is the oxidation number of cobalt in the complex
will you identify the isomers with a chemical test?
ion?
(b) Name the coordination compound used for the following :
(iii) State the type of hybridisation of the complex ion.
(i) Treatment of cancer. (iv) State the magnetic behaviour of the complex ion.
(ii) Treatment of lead poisoning. (Sample Paper, 2013)
(2017)
6. (a) Write the formula of the following compounds : 13. What type of isomerism is shown by the following coordination
(i) Triamminetriaquachromium (III) chloride compounds :
(ii) Potassium hexacyanidoferrate (III) [Pt Cl2 (NH3)4] Br2 and [Pt Br2 (NH3)4] Cl2.
(b) Name the types of isomerism shown by the following Write their IUPAC names. (2018)
pairs of compounds : 14. For the complex ion [Fe(CN)6]3-, state :
(i) [CoCI(H2O)(NH3)4]CI2 and [CoCI2(NH3)4]CI • H2O (a) the type of hybridisation.
(ii) [Pt(NH3)4][PtCI6] and [Pt(NH3)4CI2][PtCI4] (b) the magnetic behaviour.
(c) For the complex ion of [Co(NH3)6]3+: (c) the oxidation number of the central metal atom. (2018)
(i) State the hybridisation of the complex. 15. Write the IUPAC name of [Co(en)2CI2]+ ion and draw the
(ii) State the magnetic nature of the complex. (2014) structures of its geometrical isomers. (2018)
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