By:-
DR. VIKRAM SINGHTANUSHREE SINGH
YEAR OF PUBLICATION-2010All rights reserved. No part of this publication may be reproduced,
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SAVANT INSTITUTE
TM
CLASS XIICHEMISTRY
Chemistry Coordination Compounds 95
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7
COORDINATION COMPOUNDS
Slide 1
§ Coordination chemistry is an important and challenging area of modern inorganic chemistry.
§ Coordination compounds find use in analytical chemistry, medicial chemistry, metallurgical processes and industry.
§ Modern concepts of chemical bonding and structure make it possible to understand the formation and functioning of such compounds.
§ Their nomenclature, properties and other structural aspects would be taken up in this chapter.
_____________________ Slide 2 ______________________
Double salts and complexes
(a) Those which lose their identity in solution (double salt) (b) Those which retain their identity in solution (complexes) When crystals of carnallite are dissolved in water, the solution shows the properties of K+, Mg2+ and Cl– ions.
KCl. MgCl2. 6H2O à K+ + Mg2+ + 3Cl– + 6H2O Carnallite is a double salt.
_____________________ Slide 3 ______________________
§ Similarly potash alum and Mohr’s salt are double salts ( )
( )( )
3 2–2 4 2 4 2 43
2 2–4 4 4 2 4 42
K S O . A l SO .24HO 2K 2Al 4SO
Potash alum
FeSO . NH SO .6H O Fe 2NH 2SO
Mohr'ssatt
+ +
+ +
→ + +
→ + +
§ Potassium ferrocyanide, on the other hand is a complex
compound. § It dissociates as:
( ) ( )4
4 6 6K Fe CN 4K Fe CN
−+ + ���⇀↽���
_____________________ Slide 4 ______________________
Types of complexes
§ A complex in which the ion carries a net positive charge is called cationic complex e.g. [Co (NH3)6]3+, [Ni(NH3)6]2+ etc.
§ A complex in which the ion carries a net negative charge is called an anionic complex, e.g. [Ag (CN)2]–, [Fe (CN)6]4–
§ A complex carrying no net charge is called a neutral complex or simply a complex, e.g.
[Ni (CO)4], [CoCl3 (NH3)3] etc.
Slide 5
Some important points
(i) Central ion or Centre of co-ordination– The cation to which one or more neutral molecules or ions are attached is called the centre of co-ordination.
(ii) Ligands– The neutral molecules or ions which are attached with the central metal ion in coordination entity are called ligands. In most of complexes a ligand acts as a donor partner i.e. it donates one (or more) electron pair to the central metal ion. The common donor atoms in ligands are nitrogen, oxygen and less common are arsenic and phosphorous etc.
_____________________ Slide 6 ______________________
(iii) Co-ordination number– It is the total number of the atoms of the ligands that can co-ordinate to the central metal ion. For example, in the complex ions [Ag(CN)2]–, [Cu(NH3)4]2+ and [Cr(H2O)6]3+ the co-ordination number of Ag, Cu and Cr are 2, 4 and 6 respectively. Similarly in the complex ion, [Fe(C2O4)3]3–, the co-ordination number of Fe is 6 because C2O4
2- is a bidentate ligand.
_____________________ Slide 7 ______________________
(iv) Co-ordination sphere – The central metal ion and the ligands that are directly attached to it are enclosed in a first sphere of attraction and written in a square bracket. It is called as coordination sphere. The ionizable groups are written outside the brackets. The ions being outside the square bracket form the second sphere of attraction.
(v) Spatial arrangement of ligand atoms attached to central atom form a coordination polyhedra which can be octahedral
[Co(NH3)6]3+ i.e., or tetrahedral [Ni(CO)4]
i.e., or square planar [Pt Cl4]2– i.e., etc.
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Slide 8
(vi) Oxidation number or oxidation state – It is a number that represents the charge which the central atom or ion actually has or appears to have when combined with other atom, e.g., oxidation number of copper in [Cu(NH3)4]2+ is +2 but co-ordination number is 4.
(vii) Charge on the complex ion – The charge carried by a complex ion is the algebric sum of the charges carried by central metal ion and the ligands co-ordinated to the central metal ion. For example, the net charge on the complex ion [Ag(CN)2]- is +1 – 2 = –1.
_____________________ Slide 9 ______________________
Classification of ligands
Depending on number of donor atoms present in the ligands. (i) Monodentate and unidentate ligands – The ligands which
have only one donor atom. These ligands may be neutral molecules, negatively charged ion (anions) or positively charged ion (cations).
_____________________ Slide 10 _____________________
(ii) Bidentate , tridentate , polydentate ligands: Ligands having two, three, four, five or six donor atoms are called bi, tri (or ter-) tetra (or quadri) Penta and Hexa (or sex–) dentate ligands respectively. When a ligand can bind through two donor atoms as in C2O4
2– (oxalate) it is said to be didentate. When several donor atoms are present in a single ligand the ligand is said to be polydentate.
_____________________ Slide 11 _____________________
Ambi dentate ligands:
§ A ligand which can ligate through two different atoms is ambidentate ligand.
§ Examples are NO2–, SCN– etc.
_____________________ Slide 12 _____________________
Actual Name
Formula
Ch
arg
e
Don
or
Ato
m/
Ato
ms Name Given in
the Complex
(a) Monodentate
Cyanide ion
: CN – – 1 C cyano
Halide ion
X – (F–, Cl–, Br–, I–)
– 1 X halo (fluoro, chloro, bromo, iodo)
Hydride ion
H– – 1 H hydrido
Nitro NO2– – 1 N nitro
Slide 13
Nitrite ion ONO– –1 O nitrito
Nitrate ion NO3– –1 N nitrato
Hydroxide ion OH– –
1 O hydroxo
Amide ion NH2– –1 N amido
Thiocyanate ion SCN– –1 S thiocyanato
Isothiocyanate
ion NCS– –1 N isothiocyanato
Acetate ion CH3COO– –1 O acetato
Oxide ion O2– –2 O oxo
Peroxide ion O22– –2 O peroxo
Sulphide ion S2– –2 S sulphido
_____________________ Slide 14 _____________________
Sulphite ion SO32– –2 S sulphito
Sulphate ion SO42– –2 S sulphato
Thiosulphate
ion S2O3
2– –2 S thiousulph
ato
Carbonate ion CO32– –2 O carbonato
Imide ion NH2– –2 N imido
(b) Bidentate
Oxalate ion(ox)
COO–
|
COO–
–2 Two O–
atoms oxalato
_____________________ Slide 15 _____________________
NEUTRAL LIGANDS
(a) Monodentate
Ammonia NH3 zero N ammine
Water H2O zero O aqua or aq
Nitric oxide NO zero N nitrosyl
Carbon monoxide CO zero O carbonyl
Thiocarbonyl CS zero S thiocarbonyl
Phosphine PH3 zero P phosphine
Triphenyl
phosphine
(C6
H5)3 P zero P
Triphenyl
Phoshphine
Thiourea (tu) H2N C
S NH2 zero S Thiourea (tu)
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Slide 16
Pyridine (py)
zero
N pyridine
(b) Bidentate
Ethylenedi-
amine (en)
CH2 – NH2
|
CH2 – NH2 zero
N ethylene
diamine
2,2-Dipyridyl
(dipy) ze
ro N dipyridyl
(c) Tridentate
Diethylene
triamine
(dien) zero
N diethylen triamine
_____________________ Slide 17 _____________________
(d) Tetradentate
Triethylene tetramine
(trien)
.. ..
CH2 NH (CH2)2 NH2
| .. CH2 NH (CH2)2 NH2 ze
ro
Fou
r N
ato
ms triethylene
tetramine
POSITIVE LIGANDS
Nitrosonium ion NO+ + 1 N nitrosonium
Nitronium ion 2NO+ + 1 N nitronium
_____________________ Slide 18 _____________________
Chelation
§ Polydentate ligands whose structures permit the attachment
of their two or more donor atoms (or sites) to the same metal ion simultaneously and produce one or more rings are
called chelate or chelating ligands or chelating groups. The
formation of such rings is termed as chelation and the resulting ring structures are called chelate rings or simply
chelates.
Slide 19
Coordination Compounds
Bidentate and polydentate ligands are called chelating agents
_____________________ Slide 20 _____________________
EDTA Complex of Lead
_____________________ Slide 21 _____________________
Characteristics of Chelates
(i) Chelating ligands form more stable complex than the monodentate analogs. This is called chelating effect.
(ii) Chelating ligands which do not contain double bonds e.g. ethylenediamine form five membered stable rings. The chelating ligands such as acetylacetonate form six membered stable ring complexes.
(iii) Ligands with large group, form unstable ring than the ligands with smaller groups due to steric hinderance.
_____________________ Slide 22 _____________________
Use of Chelates
(i) In analytical chemistry (ii) In water softening (iii) In the elimination of harmful radioactive metal, from the body (iv) In solvent extraction (v) In food preservation
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Slide 23
Nomenclature of coordination compounds
The basic rules for systematic naming the complex compounds are summarized here. (i) The positive ion is named first followed by the negative ion. (ii) When writing the name of a complex, the ligands are quoted
in alphabetical order, regardless of their charge (followed by the metal).
(iii) When writing the formula of complexes, the complex ion should be enclosed by square brackets. The metal is named first, then the coordinated groups are listed in the order: negative ligands, neutral ligands, positive ligands (and alphabetically according to the first symbol within each group).
_____________________ Slide 24 _____________________
(a) The names of negative ligands end in -o, for example:
F– fluoro H– hydrido HS– mercapto Cl– chloro OH– hydroxo S2– thio
Br– bromo O2– oxo CN– cyano
I – iodo peroxo NO2– nitro
(b) Neutral groups have no special endings. Examples
include (NH3) ammine, (H2O) aqua, (CO) carbonyl and
(NO) nitrosyl. The ligands N2 and O2 are called dinitrogen and dioxygen. Organic ligands are usually
given their common names, for example phenyl, methyl,
ethylenediamine, pyridine, triphenylphosphine. (c) Positive groups end in - ium, e.g. NH2-NH2 hydrazinium.
_____________________ Slide 25 _____________________
(iv) When there are several ligands of the same kind, we normally use the prefixes di, tri, tetra, penta and hexa to show the number of ligands of that type.
(v) An exception occurs when the name of the ligand includes a number, e.g. dipyridyl or ethylenediamine. To avoid confusion in such cases, bis, tris and tetrakis are used instead of di, tri and tetra and the name of the ligand is placed in brackets.
(vi) The oxidation state of the central metal is shown by a Roman numeral in brackets immediately following its name (e.g. titanium (III).
_____________________ Slide 26 _____________________
(vii) Complex positive ions and neutral molecules have no special ending but complex negative ions end with -ate.
(viii) If the complex contains two or more metal atoms, it is
termed polynuclear. The bridging ligands which link the two
metal atoms together are indicated by the prefix µ-. If there are two or more bridging groups of the same kind, this is
indicated by di -µ-, tri-µ- etc.
(ix) Bridging groups are listed alphabetically with the other groups unless the symmetry of the molecule allows a
simpler name. If a bridging group bridges more than two
metal atoms it is shown as µ3, µ4, µ5 or µ6 to indicate how many atoms it is bonded to.
_____________________ Slide 27 _____________________
(x) Sometimes a ligand may be attached through different
atoms. Thus M-NO2 is called nitro and M-ONO is called
nitrito. Similarly the SCN group may bond M-SCN thiocyanato or M-NCS isothiocyanato. These may be named
systematically thiocyanato-S or thiocyanato-N to indicate
which atom is bonded to the metal. This convention may be extended to other cases where the mode of linkage is
ambiguous.
(xi) If any lattice components such as water or solvent of crystallization are present, these follow the name, and are
preceded by the number of these groups in Arabic
numerals.
_____________________ Slide 28 _____________________
These rules are illustrated by the following examples:
Complex cations
[Co (NH3)6]Cl3 Hexa-amminecobalt (III) chloride [Co Cl (NH3)5]2+ Penta-amminechlorocobalt (III)
[Co SO4 (NH3)4] NO3 Tetra-amminesulphatocobalt (III)
nitrate [Co (NO2)3 (NH3)3] Tri-amminetrinitrocobalt (III)
[CoCl CN NO2 (NH3)3] Tri-amminechlorocyanonitrocobalt (III)
[Zn (NCS)4]2+ Tetra-thiocyanato-N-zinc(II) [Cd (SCN)4]2+ Tetra-thiocyanato-S-cadmium(II)
_____________________ Slide 29 _____________________
Complex anions
Li [AlH4] Lithium tetrahydridoaluminate(III)
(lithium aluminium hydride)
Na2 [ZnCl4] Sodium tetrachlorozincate (II) K4 [Fe(CN)6] Potassium hexacyanoferrate (II)
K3 [Fe(CN)5NO] Potassium pentacyanonitrosoferrate (II)
K2 [OsCl5N] Potassium pentachloronitridoosmate(VI) Na3 [Ag(S2O3)2] Sodium bis(thiosulphato)argentate(I)
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Slide 30
Organic groups
[Pt (Py)4] [Pt Cl4] Tetrapyridineplatinum(II)
tetrachloroplatinate(II) [Cr(en)3]Cl3 Tris(ethylenediamine)chromium(III)
chloride
[CuCl2 {(CH2CH2)3(NH2)4}] Dichlorotriethylenetetraaminecopper(II)
Fe(C5H5)2 Bis(cyclopentadienyl)iron(II)
[Cr(C6H6)2] Bis (benzene) chromium (0)
Bridging groups
[(NH3)5 Co.NH2 Co (NH3)5] (NO3)5
µ-amidobis [pentaamminecobalt (III)]
nitrate
_____________________ Slide 31 _____________________
Isomerism in co-ordination compound
Structural isomerism
Compounds which have the same chemical formula but different structural arrangements are called isomers, and the phenomenon is known as isomerism. Werner classified all the possible type of isomers of co-ordination compounds in the following classes.
_____________________ Slide 32 _____________________
1. Ionisation Isomerism:–
This type of isomerism is due to the exchange of groups
between the complex ion and the ions outside it. [Co (NH3)5
Br] SO4 is red-violet. An aqueous solution gives a white precipitate of BaSO4 with BaCl2 solution, thus confirming the
presence of free SO42- ions. In contrast [Co (NH3)5SO4]Br is
red. A solution of this complex does not give a positive test with BaCl2. It does give a cream – coloured precipitate of
AgBr with AgNO3, thus confirming the presence of free Br–
ions.
_____________________ Slide 33 _____________________
2. Hydrate Isomerism:
This type of isomerism is due to different number of water molecules present in the co-ordination sphere, three
isomers of CrCl3. 6H2O are known
[Cr (H2O)6] Cl3 violet (three ionic chlorines) [Cr (H2O)5Cl] Cl2. H2O green (two ionic chlorines)
[Cr (H2O)4Cl2] Cl. 2H2O dark green (one ionic chlorine)
Other examples of hydrate isomerism are [Co (NH3)4 (H2O) Cl] Cl2 and [Co (NH3)4Cl2] Cl. H2O
[Co (NH3)3 (H2O)2 Cl] Br2 and [Co (NH3)3 (H2O) Cl Br] Br.
H2O
_____________________ Slide 34 _____________________
§ Certain ligands contain more than one atom which could donate an electron pair.
§ In the NO2– ions, either N or O atoms could act as the
electron pair donor. Thus there is the possibility of isomerism. Two different complexes [Co (NH3)5 NO2] Cl2 have been prepared, each containing the NO2
– group in the complex ion.
_____________________ Slide 35 _____________________
3. Co-ordination isomerism:
This type of isomerism occurs when both cation and anion are complex. The isomerism is caused by the interchange of ligands between the two complex ions of the same complex. Examples are, (i) [Co (NH3)6] [Cr (CN)6] and [Cr (NH3)6] [Co (CN)6] (ii) [Co (NH3)6] [Cr (C2O4)3] and [Cr (NH3)6] [Co (C2O4)3]
_____________________ Slide 36 _____________________
4. Coordination position isomerism:
In polynuclear complexes an interchange of ligands between
the different metal nucleus gives rise to positional isomerism.
_____________________ Slide 37 _____________________
Stereoisomerism or space isomerism:
Space isomerism arises on account of the different position and
arrangements of ligands (atoms or groups) in space around the
metal ion. It is of two types (i) Geometrical isomerism
(ii) Optical isomerism
_____________________ Slide 38 _____________________
(i) Geometrical isomerism § Geometric isomers differ in the arrangement of the
attached ligands, forming either cis- (same side) or trans- (opposite sides) compounds.
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_____________________ Slide 39 _____________________
§ When two identical groups (ligands) occupy adjacent positions, the isomer is called cis and when these are opposite to one another the isomer is called trans.
§ This isomerism is not possible for complexes with coordination number 2, 3 and tetrahedral complexes with coordination number 4.
§ However, cis-trans isomerism is quite common in square planar and octahedral complexes.
_____________________ Slide 40 _____________________
The square planar
These complexes may be classified into the following different types. (i) MA2B2 type e.g. PtCl2 (NH3)2 i.e. diammine dichloroplatinum
(II). § A few other examples of this type include [Pd (NO2)2
(NH3)2], [PtCl2(Py)2] § Usually X-ray diffraction methods are used to
distinguish between cis and trans - isomers.
_____________________ Slide 41 _____________________
(ii) MA2BC type, e.g. [Pt (py)2 (NH3) Cl]– [Co(NH3)2Cl(Br)]
_____________________ Slide 42 _____________________
(iii) Square planar complexes of the type MABCD form three isomers. Their structures may be obtained by fixing the
position of one ligand e.g. A and placing at the trans position anyone of the remaining three ligands one by one.
_____________________ Slide 43 _____________________
Octahedral complexes (Coordination number = 6)
These may be classified into the following different types: MA4B2 or MA2B4 , MA4BC, MA3B3 type. (i) MA4B2 Type : [CrCl2(NH3)4]+ or [CoCl2(NH3)4]+n
_____________________ Slide 44 _____________________
(ii) An example of the type MA3B3 is [RhCl3 (py)3]. Its cis and trans isomers may be represented as follows:
Another example of this type is [Co (NH3)3Cl3]
_____________________ Slide 45 _____________________
(iii) [M (AA)2B2] or [M (AA)2BC] type i.e. containing one
symmetrical bidentate ligand (AA) and the other two monodentate ligands may be same or different e.g. [CoCl2
(en)2]+ ion exists in two geometrical isomers as shown
below.
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_____________________ Slide 46 _____________________
Notes:-
(i) Octahedral complexes of the type MA6 or MA5B would not
show geometrical isomerism for obvious reasons.
(ii) Octahedral complex of the type [MABCDEF) i.e. having all the six different monodentate ligands form 15 different
geometrical isomers. The only complex of this type prepared
so far is [Pt (Cl) (Br) (I) (NO2) (NH3) (py)].
_____________________ Slide 47 _____________________
Optical isomerism
§ Optical isomers differ in their ability to rotate the plane of
polarized light. § Molecules that are nonsuperimposable mirror images of one
another (like right and left hands) are called enantiomers
and each enantiomer rotates polarized light in opposite directions.
_____________________ Slide 48 _____________________
§ The optical isomers are called dextro and laevo (d and l) depending upon the direction in which plane of the polarised
light is rotated (i.e. towards right or left).
§ The d and l isomers of a compound are called enantiomers or enantiomorphs.
§ Optical isomerism is common in octahedral complexes with
coordination number six involving 1, 2 or 3 symmetrical
bidentate ligands (i.e. a ligand attached to the central atom by two coordinate bonds).
_____________________ Slide 49 _____________________
(i) [M (AA) 3] type i.e. containing three symmetrical bidentate
ligands e.g. [Co (en)3]3+ and [Cr (ox)3]3-. Then dextro and
laevo forms are represented below:
_____________________ Slide 50 _____________________
_____________________ Slide 51 _____________________
(ii) [M (AA)2 B2] or [M(AA)2BC] type
However, we have seen earlier that this complex forms
geometrical isomers (cis and trans). Trans isomer does not show optical isomerism since it is
symmetrical while cis isomer shows optical isomerism as it
is unsymmetrical.
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Slide 52
_____________________ Slide 53 _____________________
(iii) [M (AA) B2C2] type i.e. containing one symmetrical bidentate ligand e.g. [Co (en) (NH3)2Cl2]+. Its two optical isomers may be represented as follows:
_____________________ Slide 54 _____________________
Optical isomers with co-ordination number - 4
Square planar complexes do not show optical isomerism because they contain a plane of symmetry but tetrahedral complexes containing unsymmetrical bidentate ligands e.g. [Ni(NH3CH2COO)2] i.e., bis(glycinato) nickel (II) shows optical isomerism.
_____________________ Slide 55 _____________________
Valence bond Theory (VBT) :-
This theory is mainly due to Pauling. It deals with the electronic structure of central metal ion in its ground state, kind of bonding, geometry (i.e., shape) and magnetic properties of the complexes. this is based on the following assumptions: (i) The central metal atom or ion (as the case may be) makes
available a number of empty s, p and d atomic orbitals equal to its coordination number. These vacant orbitals hybridise together to form hybrid orbitals which are the same in number as the atomic orbitals hybridizing together. These are vacant, equivalent in energy and have definite geometry.
Slide 56
(ii) The ligands have at least one σ-orbital containing a lone pair
of electrons. Vacant hybrid orbitals of the metal atom or ion overlap with
the filled (containing lone pair of electrons) σ-orbitals of the
ligands to form ligand à metal σ - bond (represented as Là M). This bond, which is generally known as coordinate bond
is special type of covalent bond and shows the
characteristics of both the overlapping orbitals. However, it also possesses a considerable amount of polarity because
of the mode of its formation.
_____________________ Slide 57 _____________________
SNo. Type of Hybridiz
ations
Geometry of the hybrid
orbitals or complex
Example of
complexes
1. sp
(4s 4p)
+3 2
-2
[Ag(NH ) ] ,
[Ag(CN) ]
2. sp2
(4s 4p)
[HgI3]–
_____________________ Slide 58 _____________________
3. sp3
(4s 4p3)
[Ni (CO)4]0,
[ ]2+3 4
2-4
2-4
2-4
2-4
- - -
Zn (NH ) ,
[ZnCl ] ,
[CuX ] ,
[MnX ] ,
[NiX ] ,
X = Cl, Br, I .
4.
dsp2 ( 2 2x – y3d
4s
4Px
4Py)
[Ni(CN)4]2–, [Ni(NH3)4]2+,
[Cu(NH3)4]2+
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Slide 59
5.
dsp3
(3dz2 4s 4p3)
[CuCl5]3–,
[Fe(CO)5]0
6.
sp3d
(4s 4p3 4dx2 –
y2)
[Sb F5]2–
_____________________ Slide 60 _____________________
7.
d2sp3
(3dx2–y2
3dz2 4s 4p3)
sp3d2
(4s 4p3 4dx2–y2
4dz2)
[FeF6]3–
[Co(H2O)6]2+,
[Ni(NH3)6]2+ etc.
_____________________ Slide 61 _____________________
A few examples are given below:
(a) Octahedral complexes
§ Formation of [Cr (NH3)6]3+, The oxidation state of chromium in [Cr (NH3)6]3+ ion is + 3. The electronic
configuration of Cr is [Ar] 3d5 4s1. Hence we have Cr3+
ion provides six empty orbitals to accommodate six pairs of electrons from six molecules of ammonia
(shown as crosses). The resulting complex [Cr (NH3)6]3+
involves d2sp3 hybridization and is thus octahedral. The presence of these unpaired electrons in the complex
explains its paramagnetic character.
Slide 62
_____________________ Slide 63 _____________________
§ Formation of ferrocyanide ion, [Fe(CN)6]4-. Oxidation state of Fe here is + 2. Hence we have
_____________________ Slide 64 _____________________
The resulting complex ion (involving d2Sp3 hybridization) is octahedrhal and diamagnetic as it does not contain any unpaired electron.
§ Formation of [Co F6]3- ion. Oxidation state of cobalt in this complex is +3. Hence we have,
_____________________ Slide 65 _____________________
_____________________ Slide 66 _____________________
It may be pointed out that F- ion provides a weak ligand field and is unable to pair up the unpaired electrons of the 3d orbitals. Hence six equivalent hybrid orbitals are obtained by mixing up of one 4s, three 4p and two 4d orbitals. The highly paramagnetic nature of the complex further confirms the presence of four unpaired electrons.
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Slide 67
Inner and Outer Orbital Complexes:
We have seen that in octahedral structures, the central metal atom either uses inner (n -1) d-orbitals or outer n d-orbitals for hybridization. Accordingly, the complexes are classified as follows:
_____________________ Slide 68 _____________________
(a) Inner orbital complex:
When the complex formed involves the inner (n -1) d-orbitals for hybridization (d2sp3), the complex is called inner orbital complex. In this case, the electrons of the metal are made to pair up, so the complex will be either diamagnetic or will have lesser number of unpaired electrons. This type of complex is also known as low spin complex. The first three examples discussed above are inner orbital complexes. Some other examples of inner orbital complexes are [V(H2O)6]3+, [Co(NH3)6]3+, [Mn(CN)6]3-, [Co(CN)6]4-, [Co(CN)6]3-, [Fe(H2O)6]2+.
_____________________ Slide 69 _____________________
(b) Outer orbital complex:
When the complex formed involves the use of outer nd-orbitals for hybridization, (sp3 d2), the complex is called outer orbital complex. The complex will have large number of unpaired electrons as the configuration of the metal remains unchanged. This type of complex is also called high spin complex. Examples:- [CoF6]3- [MnF6]3-, [FeF6]3-, [Ni (NH3)6]2+, [Fe (H2O)6]3+.
_____________________ Slide 70 _____________________
(c) Tetrahedral complexes
(i) Formation of [Ni (CO)4]0. Oxidation state of nickel in this complex is '0'. Its electronic configuration is [Ar] 3d84s2. Hence we have sp3 hybrid orbitals accommodate four pair of electrons from four CO molecules and the resulting tetrahedral complex is diamagnetic due to absence of unpaired electrons.
Slide 71
(ii) Formation of [Zn (NH3)4]2+. Complexes of Zn2+ are
invariably tetrahedral because they involve sp3
hybridisation as shown below:
_____________________ Slide 72 _____________________
Some other examples of tetrahedral complexes are [NiCI4]2-, [FeCI4]2-
(d) Square planar complexes.
Formation or [Ni (CN)4]2-. It may be noted that from the
configuration of Ni2+, it appears at the first sight that the hybridisation should be sp3 but actually it is dsp2 i.e. the two
unpaired 3d electrons pair up against Hund's rule
before hybridisation. This is on account of the following two reasons:
_____________________ Slide 73 _____________________
(i) If no pairing up occurs, the complex should be paramagnetic but actually it is diamagnetic.
(ii) CN- is a strong ligand and as it approaches the metal
ion, the electrons must pair up.
Note:-
In [NiCI4]2-, Cl– provides a weak ligand field. It is, therefore, unable to pair up the unpaired electrons of
the 3d orbital. Hence the hybridisation involved is sp3
(involving one 4s and three 4p orbitals) and the shape is tetrahedral. Moreover it will be paramagnetic.
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Slide 74
Sidgwick concept of Effective Atomic Number – EAN concept (also called Noble Gas Rule)
Sidgwick suggested that after the ligands have donated a certain
number of electrons to the central metal ion through bonding, the total number of electrons on the central atom, including those
gained from ligands in the bonding is called the effective atomic
number (EAN) of the central metal ion and in many cases this total number of electrons (i.e. EAN) surrounding the coordinated
metal ion is equal to the atomic number of the inert gas.
_____________________ Slide 75 _____________________
Example – EAN of Co(III) in [Co (NH3)6]3+ can be calculated as
follows, Electrons in Co0 atom = atomic number of Co = 27 electrons
Electrons in Co3+ ion = 27 – 3 = 24 electrons
Electrons donated by 6 (NH3) = 2 × 6 = 12 electrons EAN of Co III in [Co (NH3)6]3+ = 24 + 12 = 36 electrons
EAN (=36) of Co III is evidently equal to the atomic number of Kr.
_____________________ Slide 76 _____________________
Exception of EAN rule
Complexes of Ni (II), Co (II), Ag(I) etc. which have more than one co-ordination number depending on the nature of the ligand, generally do not follow the EAN rule. Some metal atoms such as Fe (III) which has its co-ordination number equal to 4 in [FeCl4]– and equal to 6 in [Fe(CN)6]3+ never follow this rule.
_____________________ Slide 77 _____________________
§ The V.B. theory does not explain the colour and spectra of complex.
§ The theory shows the number of unpaired electrons. § From this the magnetic moment can be calculated. § However., it does not explain why the magnetic moment
varies with temperature . § It does not give a quantitative interpretation of the
thermodynamic stability of coordination compounds. § It involves a number of assumptions.
_____________________ Slide 78 _____________________
Crystal field theory
§ The crystal field theory is now much more widely accepted than the valence bond theory.
§ It assumes that the attraction between the central metal and the ligands in a complex is purely electrostatic.
§ The transition metal which forms the central atom in the complex is regarded as a positive ion of charge equal to the oxidation state.
§ This is surrounded by negative ligands or neutral molecules which have lone pair of electrons.
_____________________ Slide 79 _____________________
§ If the ligand is a neutral molecule such as NH3, the negative end of the dipole in the molecule is directed towards the metal ion.
§ The electrons on the central metal are under repulsive forces from those on the ligands.
§ Thus the electrons occupy the d orbitals farthest away from the direction of approach of ligands.
§ In the crystal field theory the following assumptions are made: (i) Ligands are treated as point charges. (ii) There is no interaction between metal orbitals and
ligand orbitals.
_____________________ Slide 80 _____________________
§ The d orbitals on the metal all have the same energy (that is degenerate) in the free atom.
§ However, when a complex is formed the ligands destroy the degeneracy of these orbitals, i.e., the orbitals now have different energies.
§ In an isolated gaseous metal ion, the five d orbitals have the same energy, and are termed degenerate.
_____________________ Slide 81 _____________________
§ If a spherically symmetrical field of negative charges
surrounds the metal ion, the d orbitals remain degenerate. § However, the energy of the orbitals is raised because of
repulsion between the field and the electrons on the metal.
§ In most transition metal com plexes, either six or four ligands surround the metal, giving octahedral or tetrahedral
structures.
§ In both of these cases the field produced by the ligands is not spherically symmetrical.
§ Thus the d orbitals are not all affected equally by the ligand
filed. § These points are further elaborated in the following
diagrams:
_____________________ Slide 82 _____________________
Bonding in Complexes: Crystal Field Theory
§ Attractions between a central atom or ion and its ligands are largely electrostatic.
§ Ligands can distort the d-orbitals of the central species, leading to a splitting of energy levels of those orbitals.
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Slide 83
_____________________ Slide 84 _____________________
Octahedral complexes
§ In an octahedral complex, the metal is at the centre of the octahedron, and the ligands are at the six corners.
§ The directions x, y and z point to three adjacent corners of the octahedron.
§ The lobes of the eg orbitals point along the axes x, y and z. § The lobes of the t2g orbitals (dxy, dxz and dyz) point in
between the axes.
_____________________ Slide 85 _____________________
§ It follows that the approach of six ligands along the x, y, z, –x, –y and –z directions will increase the energy of the orbitals which point along the axes much more than it increases the energy of the dxy, dxz and dyz orbitals (which point between the axes).
§ Thus under the influence of an octahedral ligand field the d orbitals split into two groups of different energies as shown below (figure).
_____________________ Slide 86 _____________________
Slide 87
_____________________ Slide 88 _____________________
Orbital Diagrams for High Spin and Low Spin Octahedral
Complexes
_____________________ Slide 89 _____________________
§ Rather than referring to the energy level of an isolated metal atom, the weighted mean of these two sets of perturbed orbitals is taken as the zero: this is sometimes called the Bari centre.
§ The difference in energy between the two d levels is given either of the symbols ∆o or 10 Dq.
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§ It follows that the eg orbitals are +0.6∆o above the average level, and the t2g orbitals are –0.4∆o below the average value as shown in figure.
_____________________ Slide 90 _____________________
The size of the energy gap ∆o between the t2g and eg levels can be measured easily.
_____________________ Slide 91 _____________________
§ The magnitude of ∆o depends on three factors: (i) The nature of the ligands. (ii) The charge on the metal ion. (iii) Whether the metal is in the first, second or third row of
transition elements. § Ligands which cause only a small degree of crystal field
splitting are termed weak field ligands. § Ligands which cause a large splitting are called strong field
ligands. § Most ∆ values are in the range 7000 cm–1 to 30000 cm –1.
_____________________ Slide 92 _____________________
§ The common ligands can be arranged in ascending order of crystal field splitting ∆.
§ The order remains practically constant for different metals, and this series is called the spectrochemical series.
Spectrochemical series
Weak field ligands 2
3 2
3
2
I Br S Cl NO F OH EtOH oxalate H OEDTA (NH andpyridine) ethylenediamine dipyridyl
o phenanthroline NO CN CO
− − − − − −
− −
< < < < < < < < << < < <
< − < < < Strong field ligands
_____________________ Slide 93 _____________________
The spectrochemical series shows the relative abilities of some common ligands to split the d-orbital energy levels.
_____________________ Slide 94 _____________________
Crystal field splitting in tetrahedral coordination entities
_____________________ Slide 95 _____________________
§ In tetrahedral coordination entity formation, the d orbital splitting (see figure) is inverted and is smaller as compared to the octahedral field splitting.
§ For the same metal, the same ligands and metal-ligand distances, it can be shown that ∆t = (4 / 9) ∆0.
§ Consequently, the orbital splitting energies are not sufficiently large for forcing pairing and, therefore, low spin configuration are rarely observed.
_____________________ Slide 96 _____________________
Energy-Level Diagram for a Square-Planar Complex
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CURRICULUM BASED WORKSHEET
Topics for Worksheet – I
§ IMPORTANT TERMS
§ NOMENCLATURE
§ ISOMERISM
Worksheet – I
1. Explain the following terms (a) Coordination number (b) Counter ions
2. What is a complex ion? 3. What do you understand by the term stability constant K
for a complex? 4. There are two compounds [Cr(NH3)4]2+ and [Cr(CN)4]2–
having 5.2 × 1011 and 3.2 × 1022 respectively as their K values, which complex is more stable and which ligand NH3 or CN– is a stronger ligand?
5. Write the chemical formula for the following coordinate compounds — (a) Potassium trioxalato ferrate (III). (b) Dichloro bis (ethylenediamine) copper (II). (c) Penta aquanitrosonium ion (I) sulphate. (d) Potassium hexacyanoferrate (III). (e) Tetracyanonickelate (II) ion.
6. Write IUPAC name of the following complexes: (i) [Fe (H2O)6 SO4] (ii) [Cu (en)2] (NO3)2
7. Explain the facial and meridional isomerism. 8. Explain the optical isomerism in coordination
compounds. 9. What is macrocyclic effect? 10. Calculate the oxidation state of the central metal atom
in the following coordinate compounds? (a) Na[AlH4] (b) [Ni(CO)4] (c) [PtCl2(NH3)2]2+ (d) [Co(en)2
(ONO)Cl]Cl (e) K3 [Co(ox) 3]
11. Give IUPAC name of the following complexes: (i) Na2[Fe(CN)5(NO) (ii) K2[Zn(CN)4]
12. Provide systematic names to the following complexes: (i) [Cr(NH3)6]3+ (ii) [Pt(NH3)2Cl2] (iii) [NiCl4]2–
13. Write the chemical formula of (i) Sodium tetrahydridoborate (III) (ii) Tetrahydroxozincate (II) ion
14. Account for the fact that both [Ni(CO)4] and [Ni(CN)4]2–
are diamagnetic. 15. Briefly explain the shortcomings of valence bond theory. 16. What are ambident ligands? What are their importance?
Topics for Worksheet – II
§ VALENCE BOND CONCEPT
§ CRYSTAL FIELD THEORY
Worksheet – II
1. Explain the hybridization and magnetic behavior of [Ni(CN)4]2–.
2. [NiCl4]2– is paramagnetic in nature, explain. 3. Draw figure to show splitting of degenerate d-orbitals in
an octanhedral crystal field? 4. How does the metal carbonyls gain stability although
CO is a weak donor? 5. Discuss the concept of backbonding in metal carbonyls. 1. What are the main features of the valence bond theory?
Explain briefly. 2. Briefly explain the concept of crystal field theory. 3. What are inner orbital and outer orbital coordination
entities? Explain them with examples. 4. Explain the crystal field effects and splitting up of d-
orbitals in octahedral complezes.
CURRICULUM BASED CHAPTER ASSIGNMENT
2 Mark Questions
1. What do you mean by ambidentate ligands? 2. Give two examples of coordination isomerism. 3. Giving a suitable example describe the importance of
the formation of complex compounds in: (i) The estimation of hardness of water. (ii) The extraction of a particular metal from its natural
source.
3 Marks Questions
4. Why geometrical isomerism is not found in tetrahedral complexes?
5. Write the main assumption of valence bond theory. 6. What do you mean by coordination number? Explain
with example. 7. What is an ionization isomerism? Explain it with an
example. 8. What are hydrate isomers? 9. What is the significance of ∆0?
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5 Marks Questions
10. How would you purify the impure sample of Ni? 11. Explain with examples Cis -trans geometrical isomerism. 12. Define cis and trans isomerism in complexes giving
suitable examples. 13. How would you account for the following:
(a) [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless,
(b) [Fe(CN)6]3– is weakly paramagnetic while [Fe(CN)6]4– is diamagnetic,
(c) [Ni(CO)4] possesses tetrahedral geometry while [Pt(NH3)Cl2] is square planner.
OR Using the valence bond approach predict the shape and magnetic character of [FeCN6)]3–. At no. of Fe = 26.
14. Briefly describe the nature of bonding in metal carbonyls.
QUESTION BANK FOR COMPETITIONS
1. Which one of the following has largest number of isomers? (a) [Ir(PR3)2H(CO)]2+ (b) [Co(NH3)5CI]2+ (c) [Ru(NH3)4Cl2]+ (d) [Co(en)2Cl2]+
2. In the compound lithium tetrahydrido-aluminate, the l igand is (a) AI+ (b) H (c) H– (d) None of these
3. The EAN of iron in [Fe(CN)6]3– is (a) 34 (b) 36 (c) 37 (d) 35
4. Coordination number of Ni in [Ni(C2O4)3]4– is (a) 3 (b) 6 (c) 4 (d) 5
5. The IUPAC name of K3[Ir(C2O4)3] is (a) Potassium trioxalatoiridium (III) (b) Potassium trioxalatoiridate (III) (c) Potassium tris (oxalato) iridium (III) (d) Potassium tris (oxalato) iridate (III)
6. The chemical formula for iron (III) hexacyanoferrate (II) is (a) Fe[Fe(CN)6] (b) Fe3[Fe(CN)6] (c) Fe3[Fe(CN)6]4 (d) Fe4[Fe(CN)6]3
7. Which one is the most likely structure of CrCl3.6H2O if 1/3 of total chlorine of the compound is precipitated by adding AgNO3? (a) CrCl3.6H2O (b) [Cr(H2O)3Cl3]. (H2O)3 (c) [CrCl2(H2O)4]Cl.2H2O (d) [CrCl(H2O)5]Cl2.H2O
8. Which one of the following will not show geometrical isomerism? (a) [Cr(NH3)4Cl2]Cl (b) [Co(en)2Cl2]Cl (c) [Co(NH3)5NO2]Cl2 (d) [Pt(NH3)2Cl2]
9. Which of the following species represent the example dsp2 – hybridization? (a) [Fe(CN)6]3– (b) [Ni(CN)4]2– (c) [Zn(NH3)4]2+
(d) [FeF6]3– 10. Which one of the following is an example of octahedral
complex? (a) [FeF6]3– (b) [Zn(NH3)4]2+ (c) [Ni(CN)4]2– (d) [Cu(NH3)4]2+
11. Consider the following complex [Co(NH3)5CO]ClO4. The coordination number, oxidation number, number of d-electrons and number of unpaired d-electrons on the metal are respectively (a) 6,3,6,0 (b) 7,2,7,1 (c) 7,1,6,4 (d) 6,2,7,3
12. Atomic numbers of Cr and Fe are respectively 24 and 26. Which of the following is paramagnetic with the spin of the electron? (a) [Cr(CO)6] (b) [Fe(CO)5] (c) [Fe(CN)6]4– (d) [Cr(NH3)6]3+
13. Which of the following is organo-metallic compound? (a) Ti(C2H4)4 (d) Ti(OC2H5)4 (c) Ti(OCOCH3)4 (d) Ti(OC6H5)4
14. Which of the following organometallic compound is σ and π-bonded? (a) [Fe(η5 – C 5H5)2] (b) [PtCl3(η2 – C2H4)] (c) [Co(CO)5NH3]2+ (d) Al(CH3)3
15. Which of the following will exhibit maximum ionic conductivity? (a) K4[Fe(CN)6] (b) [Co(NH3)6]Cl3 (c) [Cu(NH3)4Cl2] (d) [Ni(CO)4]
16. Ziegler-Natta catalyst is (a) K[PtCl3(C2H4)] (b) (Ph3P)3RhCl (c) Al2(C2H5)6 + TiCl4 (d) Fe(C5H5)2
17. Hybridization of Ag in the linear complex [Ag(NH3)2]+ is : (a) dsp2 (b) sp (c) sp2 (d) sp3
18. Which of the following have square planar structure? (a) [Ni(CN)4]2+ (b) [Ni(CO)4]2+ (c) [Ni(Cl)4]2+ (d) All of above
19. Which of the following is wrong statement? (a) Ni(CO)4 has oxidation number +4 for Ni (b) Ni(CO)4 has zero oxidation number for Ni (c) Ni is metal (d) CO is gas
20. IUPAC name of Na3[Co(ONO2)6] is (a) Sodium cobaltinitrite (b) Sodium hexanitrito cobaltate (III) (c) Sodium hexanitrocobalt (III) (d) Sodium hexanitritocobaltate (II)
21. Which has highest paramagnetism? (a) [Cr(H2O)6]3+ (b) [Fe(H2O)6]2+ (c) [Cu(H2O)6]2+ (d) [Zn(H2O)6]2+
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22. The number of geometrical isomers of [CO(NH3)3(NO2)3] (a) Zero (b) 2 (c) 3 (d) 4
23. In solid CuSO4.5H2O copper is coordinated to (a) 4 water molecules (b) 5 water molecules (c) One sulphate molecule (d) One water molecule
24. One mole of the complex compound Co(NH3)5Cl3, gives 3 moles of ions on dissolution in water. One mole of the same complex reacts with two moles of AgNO3 solution to yield two moles of AgCl (s). The structure of the complex is (a) [Co(NH3)3Cl3].2NH3 (b) [Co(NH3)4Cl2]Cl. NH3 (c) [Co(NH3)4Cl]Cl2. NH3 (d) [Co(NH3)5Cl]Cl2
25. Which one of the following complexes is an outer orbital complex? (Atomic nos : Mn = 25; Fe = 26; Co = 27; Ni = 28) (a) [Co(NH3)6]3+ (b) [Mn(CN)6]4– (c) [Fe(CN)6]4– (d) [Ni(NH3)6]2+