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What are coordination compounds?

Specific type of compounds formed by certain elements

Metal

Anion/neutral atom

Complex or coordination compound in biological system

Chlorophyll - Mg metal complex

Hemoglobin - Fe metal complex

Cisplatin (Anticancer drug) - Pt metal complex

Applications

Hardness of water can be removed by Ca – EDTA complex

Extraction of metal like Ag, Au

Vitamin B12 - Co metal complex

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12C09.1��Introduction to coordination compounds�

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��12C09.1 Introduction to coordination compounds��

Learning Objectives

Werner’s theory

Terminology pertaining to coordination compounds

Ligands

Nomenclature of coordination compounds

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12C09.1

CV1

Werner’s Theory

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��Werner’s Theory��

Experiment that led to Werner’s theory

AgNO₃

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��Werner’s Theory��

Experiment that led to Werner’s theory

AgNO₃

1 : 3 solution conductivity

1 : 2 solution conductivity

1 : 1 solution conductivity

1 C ⁺ and 3 A ^–

1 C ⁺ and 2 A ^–

1 C ⁺ and 1 A ^–

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��Werner’s Theory��

Conclusion of Warner’s experiment

Proposing the structures of the compounds

Gave 3 mole of AgCl

1:3 electrolyte

[Co(NH₃)₆] ³⁺ 3Cl^–

Gave 2 mole of AgCl

1:2 electrolyte

[CoCl(NH₃)₅] ²⁺ 2Cl^–

Gave 1 mole of AgCl

1:1 electrolyte

Gave 1 mole of AgCl

1:1 electrolyte

[CoCl₂ (NH₃)₂] ⁺ Cl^–

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��Werner’s Theory��

Postulates of Warner’s theory

Coordination compounds

Primary Valency

Secondary Valency

Ionisable and normally satisfied by negative ions

Non ionisable and are satisfied by negative ions or neutral atom

Relates to oxidation state

Different spatial arrangement for different coordination number

Relates to coordination number & is fixed for a metal

[CoCl(NH₃)₅] ²⁺ 2Cl^–

Coordination number =

Secondary valency = 6

Primary valency = 2

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��Werner’s Theory��

Postulates of Warner’s theory

Different spatial arrangements

[CoCl₂ (NH₃)₂] ⁺ Cl^–

GREEN

VIOLET

Due to different arrangement of molecules or anions around metal

Thus are ISOMERS

[CoCl₂ (NH₃)₂] ⁺ Cl^–

Coordination entity

Counter ion

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��Werner’s Theory��

Spatial arrangement

Geometrical shape of coordination compounds

Tetrahedral

Octahedral

Square planar

[ Co(NH₃)₆] ³⁺ 3Cl^–

[ CoCl(NH₃)₅] ²⁺ 2Cl^–

[ CoCl₂(NH₃)₄] ⁺ Cl^–

[ PtCl₄] ^2–

[ Ni(CO)₄]

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��Difference between a double salt and a complex��

Double salt

Complex

Salt completely dissociate into ions and each ion give its confirmatory test

Compound incompletely dissociate into ions and each ion does not give its confirmatory test

[ Ni(CO)₄]

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12C09.1

PSV 1

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Q. FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1:4 molar ratio does not give test of Cu²⁺ ion. Explain why?

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Time duration - 2 min

NCERT, Exercise question – 9.2

Page no. 264

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Sol.

FeSO₄.(NH₄)₂SO₄. 6H₂O

dissolved in water

Fe⁺² , NH₄⁺, SO₄^2- in water

FeSO₄

(NH₄)₂SO₄

FeSO₄.(NH₄)₂SO₄.6H₂O

Double salt

Dissociates into its constituent ions

Fe²⁺ is tested

CuSO₄

4 NH₄OH

[Cu(NH₃)₄ ]²⁺

Complex compound

Doesn’t dissociates into its

constituent ions

Cu²⁺ is not tested

[Cu(NH₃)₄]²⁺

Dissolved in water

[Cu(NH₃)₄]²⁺ as it is in water

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12C09.1

CV2

Terminology pertaining to coordination compounds

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Terminology pertaining to coordination compounds

No. of bonded ions or molecules

Coordination entity

Metal atom

Coordination entity

[ CoCl₃(NH₃)₃]

Metal atom surrounded by 3 Cl^- ions and 3 NH₃ molecules

Central atom

[ Co Cl(NH₃)₅]²⁺

[ Fe (CN)₆]^3-

[ Ni Cl₂(H₂O)₄]

Co³⁺

Fe³⁺

Ni²⁺

Also called lewis acid

Central atom

Because accept electron from bounded anions or molecules

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Terminology pertaining to coordination compounds

Ligands

No. of bonded ions or molecules

Metal atom

Ligands

Coordination Sphere

2 –

Coordination sphere

K₄[Fe(CN)₆]

Coordination sphere

Counter ions

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Terminology pertaining to coordination compounds

Coordiation Number (CN)

CN - 4

CN - 6

CN is defined by only sigma bond formed by the ligand with metal. Pi bond formed is not counted for coordination number

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Terminology pertaining to coordination compounds

Coordination polyhedron

Octahedral

Tetrahedral

Trigonal bipyramidal

Square pyramidal

Spatial arrangement of ligand around central atom

Square planar

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Terminology pertaining to coordination compounds

Oxidation number of central atom

3 –

Remove 4 CN ^–

Cu⁺

Charge carried by central metal after removing all ligands

Represented by roman numeral I, II, etc

Homoleptic and heteroleptic complex

Homoleptic complex – Only one type of ligand are in the complex

Heteroleptic complex – More than one kind of ligand are in the complex

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Q. What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S is passed through solution?

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Time duration - 2 min

NCERT, Exercise question – 9.14

Page no. 265

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Sol.

[Cu(H₂O)₄] SO₄ + 4 KCN K₂[Cu(CN)₄]

Coordination entity -

[Cu(CN)₄]^{2 –}

Complex compound

Doesn’t dissociates into its

constituent ions

Cu²⁺ is not dissociated

No combination with H₂S

No Cu₂S ppt

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12C09.1

CV3

Ligands

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Types of Ligands

Ligands are Lewis bases which give their electrons to the Lewis acids which are metal ions

Ligand (Lewis base)

Anionic

Classification of ligands on the basis of charge

Metal (Lewis acid)

Cationic

Neutral

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Anionic Ligand

Ligands carrying negative charge

Symbol of ligand	Ligand name

F^–	Fluorido

Cl^–	Chlorido

Br^–	Bromido

I^–	Iodo

CN^–	Cyano/Cyanido

Symbol of ligand	Ligand name

O^2–	Oxo

OH^–	Hydroxo

	Sulphato

	Nitro/nitrito

	Acetato

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Neutral Ligand

Ligands carrying no charge

Symbol of ligand	Ligand name

	Ammine

	Carbonyl

	Aqua

	Nitrosyl

	Methylammine

Symbol of ligand	Ligand name

	Dioxygen

	Ethylene diamine

	Triphenylphosphine

	Nitrosyl

	Methylammine

Cationic Ligand

Ligands carrying positive charge. Less cases of cationic ligand . NO⁺ is an example

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Denticity of ligands

Single donor atom , thus NH₃ is Unidentate ligand

Number of donor atoms in a single ligand molecule to which it bonds with central atom

Metal

NH₂

CH₂

NH₂

Ethylene diamine (en)

2 ‘N’ donor atom

Oxalate C₂O₄^2-

2 ‘O’ donor atom

Two donor atom , thus ‘en’ and oxalate are Bidentate ligand

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Denticity of ligands

When several donor atoms are present in one ligand, it is Polydentate ligand

EDTA^{4 -} (Ethylenediaminetetraacetate anion)

CH₂

CH₂COO^–

Hexadentate ligand because

6 donor atoms

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Ambidentate ligands

When there are two donor atoms in a unidendate ligand, then either of the two ligand can donate electrons to central atom

Nitrito – N

Nitrito – O

Thiocyanato – S

Thiocyanato – N

Cyano

Isocyano

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12C09.1

CV4

Nomenclature of Coordination compounds

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Nomenclature of coordination compounds

Writing formulas of mononuclear coordination entities

1.) Central atom is listed first

K₄[Fe(CN)₆]

2.) Ligands are listed then according to alphabetical order

[Co(NH₃)₅Cl] ²⁺

3.) Polydentate ligand are also listed alphabetically and their abbreviation or formula should always be enclosed in parenthesis.

[Co(NH₃)₄(en)] ³⁺

4.) There should be no spaces between metal and ligands in coordination sphere.

[Co(NH₃)₅Cl] ²⁺

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Nomenclature of coordination compounds

Writing formulas of mononuclear coordination entities

5.) When counter ion is not written then charge should be mentioned outside the brackets as right superscript with number before the sign.

K₄[Fe(CN)₆]

[Fe(CN)₆] ^{4 –}

6.) Charge of the cation should be balanced by the charge of anions.

K₄[Fe(CN)₆] ^{4 –}

Charge on cations = + 4

Anionic charge = - 4

Cationic charge = anionic charge

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Nomenclature of coordination compounds

Naming the mononuclear coordination entities

1.) Cation is named first in both positive and negative charged coordination entities

K₄[Fe(CN)₆]

Potassium hexacyanoferrate(II)

[Co(NH₃)₅Cl]Cl₂

Pentaamminechloridocobalt(III) chloride

2.) Ligands are named according to alphabetical order and before the name of central atom/ion

[Co(NH₃)₅Cl]Cl₂

Pentaamminechloridocobalt(III) chloride

3.) Names of anionic ligand ending in ‘o’ , those of neutral & cationic are the same except for aqua for H₂O, ammine for NH₃ , carbonyl for CO, nitrosyl for NO

[Cu(H₂O)₆]²⁺

Hexaaquacopper(II)

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Nomenclature of coordination compounds

Naming the mononuclear coordination entities

4.) Prefixes of monodentate ligands – mono, di, tri, tetra, penta, hexa etc.

When name of ligand already include numerical prefix then bis, tris, tetrakis are used

K₄[Fe(CN)₆]

Potassium hexacyanidoferrate(II)

[NiCl₂(PPh₃)₂]

Dichloridobis(tripheylphosphine)nickel(II)

[Fe(en)₃]³⁺

tris(ethane – 1,2 – diamine)iron(III)

5.) Oxidation state of central atom should be written in roman numeral in paranthesis

K₄[Fe(CN)₆]

Potassium hexacyanidoferrate(II)

6.) If coordination sphere is cationic or neutral then metal atom is written as it is otherwise ‘ate’ suffix is added to metal name.

[NiCl₂(PPh₃)₂]

Dichloridobis(tripheylphosphine)nickel(II)

K₄[Fe(CN)₆]

Potassium hexacyanidoferrate(II)

Ag – Argentate ,Au – Aurate, Fe – Ferrate

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Q. Write the formula for the following coordination compounds :

(a) Tetraamminediaquacobalt(III) chloride (b) Potassium tetracyanidonickelate(II)

(c) Tris(ethane-1,2-diamine)chromium(III) chloride (d) Amminebromidochloridonitrito – N – palatinate(II) (e) Dichloridobis(ethane – 1,2 – diamine)platinum(V) nitrate (f) Iron(III) hexacyanoferrate(II)

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Time duration - 2 min

NCERT, Intext question – 9.1

Page no. 251

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Sol.

(a) Tetraamminediaquacobalt(III) chloride

[ Co(NH₃)₄(H₂O)₂]³⁺ Cl₃

Charge is balanced

(b) Potassium tetracyanidonickelate(II)

K₂[ Ni(CN)₄]^{2 –}

Charge is balanced

[ Cr(en)₃]³⁺ Cl₃

Charge on coordination sphere is -1

(d) Amminebromidochloridonitrito – N – palatinate(II)

[ PtNH₃BrClNO₂]^1–

Charge is balanced

(e) Dichloridobis(ethane – 1,2 – diamine)platinum(V) nitrate

[ PtCl₂(en)₂]³⁺ (NO₃)₃

Charge is balanced

(f) Iron(III) hexacyanidoferrate(II)

Fe [ Fe(CN)₆]^{3 –}

Charge is balanced

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Summary

Primary Valency – Valencies which are ionizable and which are satisfied by negative ions

Secondary Valency – Valencies which are not ionizable and which are satisfied by negative/ neutral atom

Coordination entity – Metal + No. of donor atoms

Central atom – Lewis acid which accept electrons from donor atom

Ligands – Lewis bases which donate pair of electrons to the central atom

Oxidation number – Charge on central atom after removing all ligands

Coordination sphere – when metal and ligands are enclosed in square bracket

Ligands

Anionic

Cationic

Neutral

Ambidentate

Ligands carrying negative charge

Ligands carrying Positive charge

Ligands carrying no charge

Monodentate ligand which can donate from two different atoms

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Reference Questions

Example Question, NCERT : 9.1, Page no. – 246

9.2, Page no. – 250

9.3, Page no. – 250

Exercise Question, NCERT : 9.1, Page no. – 264

9.3, Page no. – 264

9.4, Page no. – 264

9.6, Page no. – 264

9.7, Page no. – 264

Workbook Questions: 1, 2, 5, 6, 7, 8, 9, 10, 12, 17

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ConcepTest

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Pause the video

Time duration - 2 min

NCERT, Intext question – 9.2

Page no. 251

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Sol.

a.) [Co(NH₃)₆]Cl₃

Hexaamminecobalt(III) chloride

b.) [Co(NH₃)₅Cl]Cl₂

Pentaamminechloridocobalt(III) chloride

c.) K₃[Fe(CN)₆]

Potassium hexacyanidoferrate(III)

Ferrate is written because

coordination entity is anionic

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d.) K₃[Fe(C₂O₄)₃]

Potassium trioxalatoferrate(III)

Ferrate is written because

coordination entity is anionic

e.) K₂[PdCl₄]

Potassium tetrachloridopalladate(II)

Palladate is written because

coordination entity is anionic

f.) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl

diamminechloridomethylamineplatinum(II) chloride

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12C09.2��Isomerism in coordination compounds�

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12C09.2 Isomerism in coordination compounds

Learning Objectives

Structural Isomerism

Stereoisomerism

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Structural isomerism

Linkage

Coordination

Ionisation

Solvate

Structural isomers – Different chemical formula and different bond arrangement around central atom

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Structural isomerism

Linkage Isomerism

Arises in a coordination compound containing ambidentate ligand

Thiocyanato – S

Thiocyanato – N

Jorgenson experiment

[Co(NH₃)₅(NO₂)]Cl₂

[Co(NH₃)₅(ONO)]Cl₂

Nitrito - N

Nitrito - O

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Structural isomerism

Coordination Isomerism

Arises from the interchange of ligands between cationic and anionic entities of different metal ions

[Cr(NH₃)₆]

[Co(CN)₆]

6 NH₃ is attached with Cr

6 CN is attached with Co

[Co(NH₃)₆]

[Cr(CN)₆]

6 NH₃ is attached with Co

6 CN is attached with Cr

NH₃ & CN exchange their position within entities

[Cr(NH₃)₆]

[Co(CN)₆]

COORDINATION ISOMERS

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Structural isomerism

Ionisation Isomerism

When counter ion in a complex is itself a potential ligand and can displace a ligand which then become counter ion

[Co(NH₃)₅(SO₄)] Br

[Co(NH₃)₅Br] (SO₄)

Displacing

Can act as

potential ligand

Ligand

Becomes counter ion

After displacement

[Co(NH₃)₅(SO₄)] Br

[Co(NH₃)₅Br] (SO₄)

IONISATION

ISOMERS

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Structural isomerism

Solvate Isomerism

When solvent molecule in a complex is itself a potential ligand a ligand which then become counter ion

Also called hydrate isomerism when water is solvent

[Cr(H₂O)₅Cl] Cl₂. H₂O

[Cr(H₂O)₆] Cl₃

Can act as

potential ligand

Ligand

Becomes counter ion

after displacement

[Cr(H₂O)₅Cl] Cl₂. H₂O

[Cr(H₂O)₆] Cl₃

SOLVATE/HYDRATE ISOMERS

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NCERT, Intext question – 9.4

Page no. 254

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Sol.

[Co(NH₃)₅Cl]²⁺

SO₄^2-

BaCl₂

AgNO₃

BaCl₂

AgNO₃

[Co(NH₃)₅SO₄]⁺

Cl ^–

Ionisation isomers

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12C09.2

CV2

Stereoisomerism

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Stereoisomerism

Stereoisomers – Isomers that have same chemical formula but different spatial arrangement of ligands around central metal.

Stereoisomerism

Geometrical

Optical

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Geometrical isomerism

Geometrical Isomerism

Arises in heteroleptic complex with coordination number 4 and 6

Two angles should be different

For coordination number – 6 (Octahedral complexes)

M is metal, L is ligand Y is a different ligand

Geometrical isomerism will not exist in these two because atleast two different ligand should be there

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For coordination number – 6 (Octahedral complexes)

NH₃

Trans

Cis

Trans

Cis

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For coordination number – 6 (Octahedral complexes)

fac –

Meridonial – if three donor atoms occupy position around meridian of octahedron

mer –

Facial – if three donor atoms occupy adjacent position at corners

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For coordination number – 4 (Square planar complexes)

NH₃

Trans

Cis

For coordination number – 4 (Tetrahedral complexes)

Relative position of all ligand is same

Geometrical isomerism is not possible because all angles in a tetrahedral complex are equal of 109^o

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Optical isomerism

Optical Isomerism

Optical isomers which are mirror images that can’t be superimposed on one another

Observed mostly in octahedral complexes

Enantiomers

Chiral molecules

Molecules or ions that can’t be superimposed on one another

Optically active compounds

Compounds which rotate the plane polarized light either towards left or right

LEFT

leavo

RIGHT

dextro

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Optical isomerism

dextro

leavo

dextro

leavo

Mirror

Optical isomers (d & l) of

[Co(en)₃] ³⁺

Optical isomers (d & l) of

[Pt(en)₂Cl₂] ²⁺

Also shows cis and trans

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Summary

When the ligand

can donate electrons

through more than one site.

Structural isomerism

Linkage

Coordination

Ionization

Solvate

Arises from the interchange of ligands between cationic and anionic entities of different metal ions

When counter ion in a complex is itself a potential ligand and can displace a ligand which then become counter ion

When solvent molecule in a complex is itself a potential ligand a ligand which then become counter ion

Stereoisomerism

Geometrical

Optical

Cis/Trans – [ML₄Y₂] / [ML₂Y₂]

Fac/mer – [ML₃Y₃]

Dextro/leavo – Should not have

Any plane of symmetry

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Reference Questions

Example Question, NCERT : 9.4, Page no. – 252

9.5, Page no. – 253

9.6, Page no. – 253

Exercise Question, NCERT : 9.8, Page no. – 264

9.9, Page no. – 264

Intext Question, NCERT : 9.3, Page no. – 254

Workbook Questions: 4, 18

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NCERT, Exercise question – 9.10

Page no. 265

Q. Draw the structures of optical isomers of:

(i) [Co(C₂O₄)₃]^3- (ii) [PtCl₂(en)₂]²⁺ (iii) [Cr(NH₃)₂Cl₂(en)]⁺

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Q. Draw the structures of optical isomers of:

(i) [Co(C₂O₄)₃]^3- (ii) [PtCl₂(en)₂]²⁺ (iii) [Cr(NH₃)₂Cl₂(en)]⁺

Sol.

(i) [Co(C₂O₄)₃]^3-

Dextro

Leavo

Mirror

(ii) [PtCl₂(en)₂]²⁺

Dextro

Leavo

Mirror

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(iii) [Cr(NH₃)₂Cl₂(en)]⁺

NH₃

Dextro

NH₃

Leavo

Mirror

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12C09.2

PSV 1

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Q. Draw all the isomers (geometrical & optical) of :

(i) [CoCl₂(en)₂]⁺ (ii) [Co(NH₃)Cl(en)₂]²⁺ (iii) [Co(NH₃)₂Cl₂(en)]⁺

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Time duration - 2 min

NCERT, Exercise question – 9.11

Page no. 265

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Sol.

(i) [CoCl₂(en)₂]⁺

Trans

Cis

Mirror

Dextro

Leavo

Trans

No optical activity due to symmetry

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(ii) [Co(NH₃)Cl(en)₂]²⁺

NH₃

Cis

NH₃

Stands for en

trans

NH₃

Dextro

NH₃

Leavo

Mirror

NH₃

Not optically active

Plane of symmetry is present

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(iii) [Co(NH₃)₂Cl₂(en)]⁺

NH₃

Trans Cl

Cis

NH₃

Trans NH₃

NH₃

Dextro

NH₃

Leavo

Mirror

Optically inactive due to plane of symmetry in plane of molecule

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12C09.3��Valence Bond Theory (VBT)�

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12C09.2 Isomerism in coordination compounds

Learning Objectives

Introduction to Valence Bond Theory

Tetrahedral and square planar complexes

Magnetic properties and limitations of VBT

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12C09.3

CV1

Introduction to Valence bond theory

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Types of hybridization

Introduction – It is based on the hybridization of orbitals. Hybridization is decided on the basis of magnetic moment.

Hybridized orbital

formed

Ligands give electron pairs to

the hybridized orbital

Complex compound is formed

Types of hybridization

Tetrahedral – sp³

Square planar – dsp²

CN - 4

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Types of hybridization

CN - 5

CN - 6

Trigonal bipyramidal – sp³d

Octahedral – sp³d²

Octahedral –d²sp³

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Octahedral complex

Inner orbital complex

Outer orbital complex

Inner orbital complex

CN - 6

Octahedral – d²sp³

2 x (n – 1) d

Orbitals used for hybridization

3 x np

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Inner orbital complex

[Co(NH₃)₆] ³⁺

Diamagnetic

Oxidation state of Co = +3

Orbitals of Co

Orbitals of Co⁺³

As this complex is diamagnetic so there should be no unpaired electron

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Inner orbital complex

d²sp³

NH₃

d²sp³

3d⁶

No unpaired electron

Inner orbital complex

Also called low spin or spin paired complex because

Low magnetic moment

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Outer orbital complex

CN - 6

Octahedral – sp³d²

2 x nd

Orbitals used for hybridization

3 x np

[CoF₆] ^3–

Paramagnetic

Oxidation state of Co = +3

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Outer orbital complex

Orbitals of Co

Orbitals of Co⁺³

As this complex is paramagnetic so there should be unpaired electrons

sp³d²

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Outer orbital complex

sp³d²

F^–

sp³d²

F^–

4 unpaired electron

Outer orbital complex

Also called high spin or spin free complex because

High magnetic moment

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12C09.3

PSV 1

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Q. Discuss nature of bonding according to the VBT :

(i) [Fe(CN)₆]^4- (ii) [FeF₆]^3- (iii) [Co(C₂O₄)₃]^3- (iv) [CoF₆]^3-

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NCERT, Exercise question – 9.15

Page no. 265

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Q. Discuss nature of bonding according to the VBT :

(i) [Fe(CN)₆]^4- (ii) [FeF₆]^3- (iii) [Co(C₂O₄)₃]^3- (iv) [CoF₆]^3-

Sol.

(i) [Fe(CN)₆]^4-

Orbitals of Fe

Orbitals of Fe²⁺

d²sp³

Inner orbital complex

Octahedral

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(ii) [FeF₆]^3-

Orbitals of Fe

Orbitals of Fe⁺³

sp³d²

Outer orbital complex

Octahedral

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(iii) [Co(C₂O₄)₃]^3-

Orbitals of Co

Orbitals of Co⁺³

d²sp³

Inner orbital complex

Octahedral

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(iv) [CoF₆]^3-

Orbitals of Co

Orbitals of Co⁺³

sp³d²

Outer orbital complex

Octahedral

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12C09.3

CV2

Tetrahedral and square planar complex

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Tetrahedral complex

Tetrahedral – sp³

CN - 4

Orbitals used for hybridization

3 x np

[NiCl₄] ^2–

Paramagnetic

Oxidation state of Ni = +2

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Orbitals of Ni

Orbitals of Ni²⁺

Tetrahedral complex

As this complex is paramagnetic so there should be unpaired electrons

sp³

Cl^–

sp³

2 unpaired electron

High spin or outer orbital complex

High magnetic moment

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[Ni(CO)₄]

Diamagnetic

Orbitals of Ni (0)

Because s orbitals are always needed for any type of hybridization

sp³

Outer orbital complex

Low magnetic moment due to low O.S of Ni

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Square planar complex

Square planar – dsp²

CN - 4

1 x (n – 1) d

Orbitals used for hybridization

2 x np

[Ni(CN)₄] ^2–

Dimagnetic

Oxidation state of Ni = +2

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Square planar complex

Orbitals of Ni

Orbitals of Ni²⁺

As this complex is dimagnetic so there should be no unpaired electrons

dsp²

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dsp²

CN^–

Square planar complex

dsp²

No unpaired electron

Low spin or inner orbital complex

Low magnetic moment

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Q. Predict the number of unpaired electrons in square planar [Pt(CN)₄]^2- ion

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NCERT, Intext question – 9.9

Page no. 261

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Sol.

[Pt(CN)₄]^2-

Orbitals of Pt

Orbitals of Pt²⁺

Zero unpaired electrons

Q. Predict the number of unpaired electrons in square planar [Pt(CN)₄]^2- ion

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Q. [NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic even though both are tetrahedral. Why?

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NCERT, Intext question – 9.6

Page no. 261

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Sol.

[NiCl₄]^2-

Orbitals of Ni

Orbitals of Ni²⁺

sp³

e ^– given by 4 Cl^–

2 unpaired e ^–

[Ni(CO)₄]

Orbitals of Ni (0)

sp³

e ^– given by 4CO

0 unpaired e ^–

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12C09.3

CV3

Magnetic properties and limitation of VBT

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Magnetic properties of coordination compounds

Octahedral Complexes

Orbitals of Ti³⁺

For d¹, d², d³ complexes

Orbitals of V³⁺

Orbitals of Cr³⁺

Inner orbital complexes are more stable because 3d orbital is less diffused and have less energy than 4d orbitals

Thus for d¹, d², d³ electronic configuration, inner orbital complexes are formed

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Magnetic properties of coordination compounds

For d⁴, d⁵, d⁶ complexes

Orbitals of d⁴ - Cr²⁺ or Mn³⁺

Orbitals of d⁵ - Mn²⁺ or Fe³⁺

Orbitals of d⁶ - Fe²⁺ or Co³⁺

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Magnetic properties of coordination compounds

For d⁴, d⁵, d⁶ complexes

For, d⁶ complexes, magnetic data agree with maximum spin pairing in most of the cases

For d⁴ complexes

sp³d² in case of [MnCl₆]^3–

Outer orbital complex

d²sp³ in case of [Mn(CN)₆]³^–

Inner orbital complex

For d⁵ complexes

sp³d² in case of [FeF₆]^3–

Outer orbital complex

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Magnetic properties of coordination compounds

For d⁶ complexes

d²sp³ in case of [Fe(CN)₆]^3–

Inner orbital complex

sp³d² in case of [CoF₆]^3–

Outer orbital complex

d²sp³ in case of [Co(C₂O₄)₃]^3–

Inner orbital complex

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Limitations of Valence Bond Theory (VBT)

It involves a number of assumptions
It does not give quantitative interpretation of magnetic data of coordination compounds
It does not explain the colour exhibited by coordination compounds
It does not give a quantitative interpretation of thermodynamic or kinetic stabilities of coordination compounds

e.g. why [Cu(NH₃)₆]²⁺ is a stable compound whereas it forms an outer orbital

complex ?

It does not make exact prediction regarding the tetrahedral and square planar structure of 4-coordinate compounds
It does not make any distinction between strong and weak ligands

e.g why [Mn(CN)₆]³⁺ form inner orbital complex whereas [Mn(Cl)₆]³⁺ form outer orbital

complex ?

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Summary

VBT

Octahedral

Square planar

Tetrahedral

Inner orbital complex – [Fe(CN)₆]^4-

d²sp³

sp³d²

Outer orbital complex – [Fe(H₂O)₆]²⁺

Inner orbital complex – [Pt(Cl)₄]^2-

dsp²

sp³

Outer orbital complex – [Ni(Cl)₄]^2-

d¹, d², d³ – Inner orbital octahedral complex

d⁴, d⁵, d⁶ – Inner orbital complex/ Outer orbital complex

d⁸, d⁹ – Outer orbital octahedral complex/ Inner orbital square planar complex/Tetrahedral complex

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Reference Questions

Example Question, NCERT : 9.7, Page no. – 257

Exercise Question, NCERT : 9.5, Page no. – 264

9.19, Page no. – 265

9.23, Page no. – 265

9.24, Page no. – 265

Intext Question, NCERT : 9.5, Page no. – 261

9.7, Page no. – 261

Workbook Questions: 11, 13

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12C09.3

PSV 2

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Q. Explain [Co(NH₃)₆]³⁺ is an inner orbital complex while [Ni(NH₃)₆]²⁺ is an outer orbital

complex.

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NCERT, Intext question – 9.8

Page no. 261

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Sol.

Orbitals of Co

Orbitals of Co⁺³

Orbitals of Ni

Orbitals of Ni⁺²

[Ni(NH₃)₆]²⁺

[Co(NH₃)₆]³⁺

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12C09.3

PSV 3

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Q. Amongst the following ions which one has the highest magnetic moment value

(i) [Cr(H₂O)₆]³⁺ (ii) [Fe(H₂O)₆]²⁺ (iii) [Zn(H₂O)₆]²⁺

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NCERT, Exercise question – 9.29

Page no. 266

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Sol.

[Cr(H₂O)₆]³⁺

Orbitals of Cr³⁺

Orbitals of Cr

[Fe(H₂O)₆]²⁺

Orbitals of Fe²⁺

Orbitals of Fe

[Zn(H₂O)₆]²⁺

Orbitals of Zn²⁺

Orbitals of Zn

[Fe(H₂O)₆]²⁺ has highest magnetic moment

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12C09.4��Crystal Field Theory (CFT)�

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12C09.2 Isomerism in coordination compounds

Learning Objectives

Postulates of Crystal Field Theory

Octahedral Splitting

Examples of Octahedral Splitting

Tetrahedral Splitting

Color in coordination compounds

Bonding in Metal Carbonyls

Applications of Coordination compounds

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12C09.4

CV 1

Postulates of Crystal Field Theory (CFT)

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Shapes of d orbitals

All these three d orbitals are between the three axis

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Shapes of d orbitals

These two orbitals

are on the axis

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Postulates of CFT

Electrostatic Model

Ligand

Ionic Bond

Due to electrostatic attraction

Ligands can be of two types

δ -

δ +

Point Charge

Point Dipole

Metal is considered as

M⁺

Test Charge

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Postulates of CFT

Ligand

Electrostatic attraction

due to positive and

negative charge

Electrostatic Repulsion

due to e ^– present in metal and

ligand

Due to ligand field

energy of d orbitals is increased

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M⁺

Postulates of CFT

Free metal atom/ion

All d orbitals are of same energy

Degenerate

M⁺

Symmetrical field of ligands

Energy of all the d orbitals is raised

Degenerate Orbitals

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Postulates of CFT

M⁺

Tetrahedral

Octahedral

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12C09.4

CV 2

Octahedral Splitting

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Octahedral Splitting

Electrostatic Repulsion due to e ^– present in metal’s d orbitals and electrons of ligand

In Octahedral complexes ligand approach metal through the 3 axis

Only these two orbitals

are on the axis

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Octahedral complex

Metal atom

Metal atom surrounded by spherical crystal field

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d orbital Splitting in octahedral complex

3/5 Δ_o

2/5 Δ_o

Δ_o

Free metal atom

Metal atom surrounded by spherical crystal field

Splitting of d – orbital in

octahedral crystal field

Crystal Field Splitting - Δ_o

Barycenter

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Spectrochemical series

Weak Ligand field

Strong Ligand field

Less attraction

More attraction

Less repulsion

More repulsion

More rise in energy of e_g and thus more crystal field splitting

Less rise in energy of e_g and thus less crystal field splitting

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Spectrochemical series

Ligand Strength

Strong ligand field

Creates more crystal field splitting

Weak ligand field

Creates less crystal field splitting

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12C09.4

CV 3

Examples of Octahedral Splitting

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Examples of octahedral splitting

3/5 Δ_o

2/5 Δ_o

Δ_o

3/5 Δ_o

2/5 Δ_o

Δ_o

For d¹, d², d³ complexes

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Examples of octahedral splitting

For d¹, d², d³ complexes

3/5 Δ_o

2/5 Δ_o

Δ_o

According to Aufbau’s principle, in d¹, d², d³ metal atom, electrons will occupy lower t_2g orbital

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Examples of octahedral splitting

For d⁴ complexes

3/5 Δ_o

2/5 Δ_o

Δ_o

3/5 Δ_o

2/5 Δ_o

Δ_o

Two possibilities

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Examples of octahedral splitting

For d⁴ complexes

Δ_o

Crystal Field Splitting energy, Δ_o – Energy separation between t_2g and e_g

Pairing Energy, P – Energy required for electron pairing in single orbital

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Weak field and strong field ligands

Weak Field Ligands

Creates less crystal field splitting

Δ_o < P, then electron will enter in e_g orbital

Δ_o

Strong Field Ligands

Creates less crystal field splitting

Δ_o > P, then electron will enter in t_2g orbital

Δ_o

Ligands

High Spin Complex

Low Spin Complex

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12C09.4

CV 4

Tetrahedral Splitting

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Metal atom

Metal atom surrounded by spherical crystal field

Tetrahedral complexes

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d – orbital splitting in Tetrahedral complexes

2/5 Δ_t

3/5 Δ_t

Δ_t

Free metal atom

Metal atom surrounded by spherical crystal field

Splitting of d – orbital in tetrahedral crystal field

Crystal Field Splitting - Δ_t

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Tetrahedral complexes

For tetrahedral complexes “g” subscript is not used because there is no center of symmetry

Δ_t = 4/9 Δ_o

Δ_tis very small thus pairing does not happen and hence all complexes are high spin complex regardless of ligand strength

Δ_t < Δ_o

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12C09.4

CV5

Colour in Coordination compounds

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Colour in Coordination compounds

Transition Metals

Form coloured complex compounds

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Colour in Coordination compounds

Complementary Colour

Coordination entity	Wavelength of light absorbed (nm)	Colour of light absorbed	Colour of coordination entity
	535	Yellow	Violet
	500	Blue Green	Red
	475	Blue	Yellow orange
	310	Ultraviolet	Pale yellow
	600	Red	Blue
	498	Blue green	Violet

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Colour in Coordination compounds

Violet

Absorbing light

Colour explained by CFT

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Colour in Coordination compounds

Colour explained by CFT

Violet

Blue – green region

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Colour in Coordination compounds

Colour explained by CFT

No ligand

No d orbital splitting

No colour

NO d-d charge transfer

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Colour in Coordination compounds

Influence of Ligand on colour of complex

Nickel(II) chloride

H₂O

Green

Pale blue

Purple

Violet

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Colour in Coordination compounds

Colour of Some Gem Stones

Ruby is Al₂O₃ containing about 0.5-1% Cr³⁺ ions (d³), which are randomly distributed in positions normally occupied by Al³⁺

In emerald Cr³⁺ ions occupy octahedral sites in the mineral beryl (Be₃Al₂Si₆O₁₈)

d – d transitions

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Limitations of CFT

Ligands are point charges

Anionic ligands should exert the greatest splitting effect

Anionic ligands actually are found at the low end of the spectrochemical series

Does not take into account the covalent character of bonding between ligand and central atom

Later explained by LFT and MOT

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ConcepTest

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Q. Explain the change in colour of blue copper sulphate on heating.

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Q. Explain the change in colour of blue copper sulphate on heating.

Sol.

Heat

Water of crystallization evaporated

No ligand

No d orbital splitting

No colour

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12C09.4

CV 6

Bonding in metal carbonyls

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Structures of some important carbonyl complex

Metal Carbonyls – Complexes which have Carbonyl molecule as ligand

Metal Carbonyls

Homoleptic Complex

Heteroleptic Complex

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Structures of some important carbonyl complex

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Synergic bonding

Donation of electrons from CO to metal by σ bond

Increase in formal charge of metal due to electrons

Thus electrons are given by π overlap to π^*

to decrease formal charge

Leading to decrease in bond order of C – O triple bond and increase in bond order of M – C bond.

Thus, Synergic effect strengthen M – CO bond

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12C09.4

CV 7

Applications of Coordination Compounds

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Applications of Coordination Compounds

EDTA

DMG

Cupron

Estimation and detection of metal ions

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Applications of Coordination Compounds

Estimating hardness of water

Mg²⁺ & Ca²⁺

present in water

Titration with Na₂EDTA

Stable complexes of Ca²⁺ & Mg²⁺ are formed

Na₂EDTA

Weaker complex

Mg(EDTA) / Ca(EDTA)

Stable complex

Extraction processes of Ag & Au

Metal can be further extracted by dissolving complex solution with zinc

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Applications of Coordination Compounds

Purification of metal

[Ni(CO)₄]

Impure Nickel

Pure Nickel

Decomposition

Biological Importance

Chlorophyll – Mg complex

Haemoglobin – Fe complex

Vitamin B12 – Co complex

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Enzyme

Applications of Coordination Compounds

Carboxypeptidase A enzyme hydrolyses peptide bonds

Carbonic anhydrase enzyme

Hydrolyses CO₂ to form H₂CO₃

Both enzymes are Zn complexes

Industrial Importance

[Rh(PPh₃)₃Cl]

Wilkinson catalyst

Used in hydrogenation of alkenes

catalyst

Wilkinson

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Applications of Coordination Compounds

Electroplating with Ag & Au

Electroplating is done with complex solutions which makes the coating even

Black and white polarography

Undecomposed AgBr on film

Washed with hypo solution

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Applications of Coordination Compounds

Chelate Therapy

Excess of Cu

Removed by D - penicillamine

By formation of complex

Excess of Fe

Removed by desferrioxime B

By formation of complex

Lead poisoning

Removed by EDTA by Lead – EDTA complex

Inhibit the growth of tumors

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Q. The hexaquo manganese(II) ion contains five unpaired electrons while the hexacyanoion contains only one unpaired electron. Explains using CFT.

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NCERT, Intext question – 9.10

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Sol.

3/5 Δ_o

2/5 Δ_o

Δ_o

3/5 Δ_o

2/5 Δ_o

Δ’_o

Δ’_o > Δ_o

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Q. What are t_2g and e_g orbitals?

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Workbook Q – 14

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Sol.

Q. What are t_2g and e_g orbitals?

t _2g orbitals

t stands for triply

degenerate orbitals

g stands for gerade

Gerade – center of symmetry

is present in molecule

3/5 Δ_o

2/5 Δ_o

Δ_o

e _g orbitals

e stands for doubly

degenerate orbitals