chemistry. coordination compounds and organometallics - 1 session
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Chemistry
COORDINATION COMPOUNDS AND ORGANOMETALLICS - 1
Session
Session Objectives
1. Werner’s coordination theory
2. IUPAC nomenclature
3. Isomerism in Coordination Compounds
Coordination compound, central metal ion, ligand, coordination sphere, coordination number
Coordination compounds
3 6 3Co(NH ) Cl
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.
Werner’s coordination theory
Illustrative Example
An isomer of CrCl3 6H2O reacts with one mole AgNO3 to give one mole of precipitate of AgCl. What is the compound?
[CrCl2(H2O)4]Cl . H2O + AgNO3 AgCl
Solution
Only those ions which are present outside the coordination sphere can be ionised.
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.
[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.
Oxidation number of the central metal ion.
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
d) Numerical prefixes to indicate the number of ligands. –di, tri, etc.
a) Cation is named first.
b) Name the coordination sphere.
c) Ligands are named in the alphabetical orderirrespective of their being neutral, negativelyor positively charged.
[Pt Cl (NO2) (NH3)4] amine, chloro, nitrito – N
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 Romannumerical. (Such as II, III, IV)
g) Bridging of ligand is indicated with a prefix-.
IUPAC nomenclature
[Co(NH3)5(NCS)]Cl2
[Cr(H2O)4Cl2]+
K2HgI4
Na2[CrOF4]
Bis(ethylenediamine)cobalt (III) - - imido - – hydroxo bis(ethylenediammine)cobalt (III) ion.
NH
OH
Co(en)2(en2)Co
3+
IUPAC nomenclature
Tetraaquadichlorochromium (III) ion
Potassium tetraiodomercurate (II)
Sodium tetrafluorooxochromate (IV)
Pentaammineisothiocyanatocobalt (III) chloride
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–
(i) Ag in Tollen’s reagent exists as Ag2O
2x + 1(–2) = 0
x = +1
Solution:
(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) Potassium tetrahydroxozincate(II)(ii) Triamminechlorocyanonitrocobalt(III)(iii) Tetrapyridineplatinum(II) tetrachloroplatinate(II)(iv) 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)
i. [VO(acac)2]
ii. [Co(NH3)5CO3]Cl
iii. Na3[Fe(CN)4NOS]
iv. [Cu{(NH2)2CO}2Cl2]
Solution:
Effective atomic number(EAN)
3+3 6Co (NH )
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
(2) EAN of [Fe(CN)6]4–
26 – 2 + 2 × 6 = 36
(1) EAN of [Fe(CN)6]3-
26 – 3 + 2 × 6 = 35
(3) EAN of [Cu(NH3)4]2–
29 – 2 + 2 × 4 = 35
Solution :
Isomerism in Coordination Compounds
(i) Structural isomerism
(ii) Stereo-isomerism
Structural isomerism
Ionization isomerism
3 4Co(NH )Br SO 2+ 2-
3 5 4Co(NH ) Br SO
[Co(NH3)5SO4] Br + -3 5 4[Co(NH ) SO ] + 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.
Example [Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2
Coordination position isomerism
(NH ) Co43
OH
OH
Co(NH ) Cl23 2
Cl(NH )33 Co
OH
OH
Co(NH ) Cl33 SO 4
SO4 and
- - - 2-2 2 3NO , SCN , CN , S O
Illustrative Example
Write the type of isomerism in the compounds [Co(NH3)4Cl2]NO2 and [Co(NH3)4ClNO2]Cl.
As the two complexes give different ions in solution, they show ionization isomerism.
Solution:
Geometrical isomerism
Complexes with general formula, Ma2b2
M M
a
a a ab
b bTrans-isom er
[M a b ]2 2
C is-isom erb
CN=4
Geometrical isomerism
Complexes with general formula Ma2bc can have cis- and trans-isomers.
M M
c
c
a a
a
a
b bC is
[M a bc]2
Trans
Geometrical isomerism
Geometrical isomerism
Complexes with general formula, Mabcd, can have three isomers.
M
c
(I)
d
a b
M
d
b(II)
a
cM
b
d(III)
a
c
Geometrical isomerism
Octahedral complexes of the type Ma4b2
CN=6
M
ba b
aa
a
M
ba a
aa
b
Cis-isomerTrans-isomer
Geometrical isomerism
Geometrical isomerism
Octahedral complexes of the type Ma3b3
M
ba b
ba
a
M
ba b
aa
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
(Where AA = Symmetrical bidentate ligand)
[ ] , [ ], [ ( ) ] , [ ( ) ]Ma b C Mabcdef M AA M AA an n n2 2 2 3 2 2
[ ( ) [ ( ) ]M AA ab] and M ABn n2 3
(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).
Since A is readily soluble in dilute aqueous ammonia. It must have Cl-
outside the coordination sphere I.e. its formula is [Cr(NH3)4ClBr] Cl
Solution :
Similarly, form B gives pale yellow precipitate of AgBr which are sparingly soluble in NH4OH.
3 3 2 3 2 34 4pale yellow ppt
AgNO Cr NH Cl Br AgBr Cr NH Cl NO
3 24Hence, from B is Cr NH Cl Br
Solution Cont.
4 3 22Pale yellow ppt.
AgBr 2NH OH Ag NH Br 2H O
The state of hybridisation of chromium in both the complexes is d2sp3. Both the forms have chromium in +3 oxidation state.
3d 4s 4p
d sp hybrid isation2 3
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.
3 3 53
2
3 2 32 4
(i) Pt NH Cl (ii) Pt NH Cl
(iii) Pt NH ClNO (iv) Pt NH ClBr
(a) No isomers are possible for a square planar complex of the type MA3B
+
Pt
H N3
Cl
H N3
NH3
NH3
Solution
Solution contd.
(b) No isomers are possible for an octahedral complex of the type of MAB5.
–
Pt
Cl
Cl
Cl
Cl
NH3
Cl
(c). Cis and trans isomers are possible for a square planar complex of the type MA2BC.
Pt
H N3
Cl NH3
NO 2
trans
Pt
Cl
O N2
NH3
NH3
cis
Solution Contd.
(d). Cis and trans isomers are possible for an octahedral complex of the type MA4BC.
H N3
Pt
H N3
H N3
H N3
NH3
NH3
Cl
Brtrans
H N3
Pt
H N3
H N3
H N3
Cl
Br
NH3
NH3
cis
Thank you
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