Download 57. Coordination Compounds and Organometallics

Chemistry
Session
COORDINATION COMPOUNDS AND
ORGANOMETALLICS - 1
Session Objectives
1. Werner’s coordination theory
2. IUPAC nomenclature
3. Isomerism in Coordination Compounds
Coordination compounds
Coordination compound, central metal ion, ligand,
coordination sphere, coordination number
Co(NH3 )6  Cl3
Werner’s coordination theory
Metal has two types of valencies.
(a) 1o valency - ionisable, non-directional, equal to the
valency of the metal.
(b) 2° valency - non-ionisable, directional, equal to the
coordination number
- decides the hybridization and geometry of molecule.
Illustrative Example
An isomer of CrCl3 6H2O reacts with one mole AgNO3 to give
one mole of precipitate of AgCl. What is the compound?
Solution
Only those ions which are present outside the coordination
sphere can be ionised.
[CrCl2(H2O)4]Cl . H2O + AgNO3  AgCl
Therefore the compound is [CrCl2(H2O)4]Cl.H2O.
Basic terms used in coordination
chemistry
Coordination entity: it constitutes a central atom/ion, usually of a
metal, to which are attached a fixed number of other atoms or
groups each of which is called a ligand.
For example [PtCl4]2- and [Fe(CN)6]3-.
Central atom/ion: in a coordination entity the atom/ion to which
are bound a fixed number of ligands in a definite geometrical
arrangement around it.
For example: Ni2+ in [NiCl2(OH2)4], Fe3+ in [Fe(CN)6]3-.
Coordination number: it is determined by the sigma bonds between
the ligands and the central atom/ion.
For example coordination number in [Fe(CN)6]3- is six.
Oxidation number of the
central metal ion.
[Co(NH3)4 Cl (NO2)]
K3[Fe(CN)6]
x + 4(o) + 1 (–1) + 1 (–1) = 0
3(1) + x + 6(–1) = 0
x=+2
x = + 3.
Ligands
The ligands are the ions or molecules bound to the central atom/ion
in the coordination entity.
Ligands are Lewis base because they works as electron donor.
Ambidentate ligands:
Ligands which can ligate through two different atoms present in it.
For example SCN- ion can coordinate through the sulphur or
nitrogen atom.
Denticity and chelation: when coordination of more than one sigma
electron pair donor group from the ligand to the same central
atom/ion takes place, it is called chelation.
The number of such ligating groups indicate the denticity of the ligand,
e.g., ethylenediamine, EDTA For example unidentate, didentate,
terdentate tetradentate etc.
Some common ligands
IUPAC nomenclature
a) Cation is named first.
b) Name the coordination sphere.
c)
Ligands are named in the alphabetical order
irrespective of their being neutral, negatively
or positively charged.
[Pt Cl (NO2) (NH3)4] amine, chloro, nitrito – N
d) Numerical prefixes to indicate the number of
ligands. –di, tri, etc.
IUPAC nomenclature
e) Ending form: For anionic complex,
metal ion ends with ‘ate’.
K[Pt Cl5 (NH3)] potassium ammine
pentachloro-platinate (IV)
Latin name of the metals are commonly used.
cuperate for Cu, ferrate for Fe, argentate for Ag.
f)
Oxidation state of metal is designated by a Roman
numerical. (Such as II, III, IV)
g) Bridging of ligand is indicated
with a prefix-.
IUPAC nomenclature
[Co(NH3)5(NCS)]Cl2
Pentaammineisothiocyanatocobalt (III) chloride
[Cr(H2O)4Cl2]+
Tetraaquadichlorochromium (III) ion
Potassium tetraiodomercurate (II)
K2HgI4
Sodium tetrafluorooxochromate (IV)
Na2[CrOF4]
3+
NH
Co(en)2
(en2)Co
OH
Bis(ethylenediamine)cobalt (III)
-  - imido -  – hydroxo bis
(ethylenediammine)cobalt (III) ion.
Rules for writing formula of
complexes
•
complexes are enclosed in square brackets.
•
first the name of the central atom is given.
•
followed by first the anionic ligand and then the neutral
ligands; within each group they are alphabetically
ordered according to the first character of their formula.
Examples:
[PtCl2(C2H4)(NH3)]
K2[PdCl4]
[Co(en)3]Cl3
Illustrative example
Write the oxidation state of central metal atom in
the following compounds.
(i)Ag in Tollen’s reagent is
(ii)[Mo2O4(H2O)2]2–
Solution:
(i) Ag in Tollen’s reagent exists as Ag2O
2x + 1(–2) = 0
x = +1
(ii) 2x + 4(–2) + 2 × 0 = –2
x = +3
Illustrative Example
Write IUPAC name of the following compounds:
(i) K2[Zn(OH)4]
(ii) [CoCl.CN.NO2.(NH3)3]
(iii) [Pt(py)4][PtCl4]
(iv) K2[Cr(CN)2O2(O2)NH3]
Solution :
Follow IUPAC rules.
(i)
(ii)
(iii)
(iv)
Potassium tetrahydroxozincate(II)
Triamminechlorocyanonitrocobalt(III)
Tetrapyridineplatinum(II) tetrachloroplatinate(II)
Potassium amminedicyanodioxoperoxochromate(VI)
Illustrative Example
Write the formula of the following coordination compounds.
i. Bis(acetylacetonato) oxovanadium (IV)
ii. Pentaammine carbonatocobalt (III) chloride
iii. Sodium tetratacyanonitrosylsulphidoferrate (III)
iv. Dichlorobis (urea) copper (II)
Solution:
i.
[VO(acac)2]
ii.
[Co(NH3)5CO3]Cl
iii. Na3[Fe(CN)4NOS]
iv. [Cu{(NH2)2CO}2Cl2]
Effective atomic number(EAN)
Co (NH3)6  3+
Atomic number of cobalt is 27.
As such Co3+ has 24 electrons.
Each of the six NH3 will donate
one pair of electron to cobalt.
So EAN of Co3+ is 24 + 12 = 36.
Illustrative Example
What is the EAN of the central atom in
following compounds?
(1)K3[Fe(CN)6]
(2)K4[Fe(CN)6]
(3)[Cu(NH3)4]SO4
Solution :
(1) EAN of [Fe(CN)6]326 – 3 + 2 × 6 = 35
(2) EAN of [Fe(CN)6]4–
26 – 2 + 2 × 6 = 36
(3) EAN of [Cu(NH3)4]2–
29 – 2 + 2 × 4 = 35
Isomerism in Coordination Compounds
(i) Structural isomerism
(ii) Stereo-isomerism
Structural isomerism
Ionization isomerism
2+
SO24
Co(NH3)Br  SO4
Co(NH3)5Br 
[Co(NH3)5SO4] Br
[Co(NH3)5SO4]+ + Br-
Isomerism
Hydrate isomerism
[Cr(H2O)6]Cl3 (Violet)
[Cr(H2O)5Cl]Cl2.H2O (Green)
[Cr(H2O)4Cl2]Cl.2H2O (Green)
Coordination isomerism
[Co(NH3)6] [Cr (CN)6]
and
[Cr(NH3)6] [Co (CN)6]
Isomerism
Linkage isomerism
This type of isomerism occurs in
ambidentate ligands like and CO.
NO-2 , SCN- , CN- , S2O23
Example
[Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2
Coordination position isomerism
OH
(NH3)4Co
Co(NH3)2Cl2
SO4 and
OH
OH
Cl(NH3)3Co
Co(NH3)3Cl SO4
OH
Illustrative Example
Write the type of isomerism in the compounds
[Co(NH3)4Cl2]NO2 and [Co(NH3)4ClNO2]Cl.
Solution:
As the two complexes give different
ions in solution, they show ionization
isomerism.
Geometrical isomerism
CN=4
Complexes with general formula, Ma2b2
b
a
a
a
M
M
a
b
Trans-isomer
[Mab]
2 2
b
b
Cis-isomer
Geometrical isomerism
Complexes with general formula Ma2bc
can have cis- and trans-isomers.
a
a
M
M
c
b
c
a
a
b
Trans
Cis
[Mab
]
2 c
Geometrical isomerism
Geometrical isomerism
Complexes with general formula,
Mabcd, can have three isomers.
a
b
d
a
M
b
a
d
M
c
M
(I)
c
(II)
b
c
(III)
d
Geometrical isomerism
CN=6
Octahedral complexes of the type Ma4b2
b
b
a
a
b
a
M
M
a
a
a
Cis-isomer
a
a
b
Trans-isomer
Geometrical isomerism
Geometrical isomerism
Octahedral complexes of the type
Ma3b3
b
b
a
a
b
b
M
M
b
a
a
a
a
b
Geometrical isomerism
three NH3 molecules can be situated (1) on the same face
of the octahedron (fac isomer) or (2) around a perimeter of
the octahedron (mer isomer
Optical isomerism
Optical isomerism
[Ma2b2C 2 ]n  , [Mabcdef ], [M( AA ) 3 ]n  , [M( AA ) 2 a2 ]n 
(Where AA = Symmetrical bidentate ligand)
[M( AA ) 2 ab]n  and [M( AB) 3 ]n 
(Where AB = Unsymmetrical ligands)
Illustrative Example
A metal complex having composition Cr(NH3)4Cl2Br has been isolated in
two forms (A) and (B). The form (A) reacts with AgNO3 to give a white
precipitate readily soluble in dilute aqueous ammonia whereas (B) gives
a pale yellow precipitate soluble in concentrated ammonia. Write the
formula of (A) and (B) and state the hybridisation of chromium in each.
Calculate the magnetic moment (spin only value).
Solution :
Since A is readily soluble in dilute aqueous ammonia. It must have Cloutside the coordination sphere I.e. its formula is [Cr(NH3)4ClBr] Cl
Similarly, form B gives pale yellow precipitate of AgBr which are
sparingly soluble in NH4OH.
Hence, from B is Cr NH3 4 Cl2  Br
AgNO3  Cr NH3 4 Cl2  Br 
AgBr 
pale yellow ppt
 Cr NH3 4 Cl2  NO3
Solution Cont.
AgBr
 2NH4OH  Ag NH3 2 Br   2H2O
Pale yellow ppt.
The state of hybridisation of chromium in both the complexes is
d2sp3. Both the forms have chromium in +3 oxidation state.
3
d
p
4
s4
2
3
d
s
p
h
y
b
r
i
d
i
s
a
t
i
o
n
As there are three unpaired electrons so
magnetic moment
()  n(n  2)
 3(3  2)
 3.87 BM
Illustrative Example
Platinum II forms square planar complexes and platinum
IV gives octahedral complexes. How many geometrical
isomers are possible for each of the following complexes.
Describe their structures.
(i) Pt NH3 3 Cl

(ii) Pt NH3  Cl5 
(iii) Pt NH3 2 ClNO2 

(iv) Pt NH3 4 ClBr 
2
Solution
(a) No isomers are possible for a square planar
complex of the type MA3B
+
Cl
NH3
Pt
HN
3
NH3
Solution contd.
(b) No isomers are possible for an octahedral
complex of the type of MAB5.
–
NH3
Cl
Cl
Pt
Cl
Cl
Cl
(c). Cis and trans isomers are possible for a square
planar complex of the type MA2BC.
C
l
N
H
3
C
l
P
t
O
N
2
N
H
3
P
t
N
H
N
3 H
3
c
is
N
O
2
tr
a
n
s
Solution Contd.
(d). Cis and trans isomers are possible for an
octahedral complex of the type MA4BC.
N
H
3
H
N
3
C
l
C
l
H
N
3
P
t
H
N
3
P
t
B
r
N
H
3
c
is
N
H
3
H
N
3
N
H
3
B
r
tra
n
s
Thank you