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