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Absolute

CHEMISTRY Vol. 2.1

NEET – UG & JEE (Main)

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Target’s “Absolute Chemistry Vol - 2.1” is compiled according to the notified Std. XI syllabus for NEET-UG & JEE (Main). The content of this book is framed after reviewing various state syllabi as well as the ones prepared by CBSE, NCERT and COBSE. The sections of Theory, Quick Review, Formulae, MCQs and Topic Test form the backbone of every chapter and ensure adequate revision. These MCQs are framed considering the importance given to every topic as per the NEET-UG & JEE (Main) exam. They are a healthy mix of theoretical, numerical, multi-step reactions and graphical based questions. The level of difficulty of these questions is at par with that of various competitive examinations like AIIMS, CPMT, JEE (Main), NEET-UG, TS-EAMCET (Med. and Engg.), BCECE, Assam CEE, AP EAMCET (Med. and Engg.) & the likes. Also to keep students updated, questions from most recent examinations such as NEET-UG, MHT CET, KCET, WB JEE, JEE (Main), of years 2016 to 2018 are exclusively covered. We are confident that this book will cater to needs of students across a varied background and effectively assist them to achieve their goal. We welcome readers’ comments and suggestions which will enable us to refine and enrich this book further. Please write to us at: [email protected]

All the best to all Aspirants! Yours faithfully, Authors Edition: Second

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No. Topic Name Page No. 1 Solid State 1 2 Solutions 55 3 Electrochemistry 127 4 Chemical Kinetics 208 5 Surface Chemistry 267 6 General Principles and Processes of Isolation of Elements 320 7 p-Block Elements 371 8 d and f-Block Elements 497 9 Coordination Compounds 546

'Chapters 10 to 16 are a part of Absolute Chemistry Vol - 2.2'

Index

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546

Absolute Chemistry Vol - 2.1 (Med. and Engg.)

Coordination compounds: i. Coordination chemistry is the branch of chemistry dealing with the study of coordination (or complex) compounds. Coordinate bond: A coordinate bond is a coverlent bond in which both electrons come from the same atom.

It is also called as dative bond. ii. Coordination compounds contain central metal atom/ion surrounded by oppositely charged ions or neutral

molecules (called as ligands). iii. These ions/molecules are attached to central metal atom through coordinate bond. eg. [Cu(NH3)4]2+ is a coordination compound that contains central Cu2+ ion attached to four neutral

ammonia molecules through coordinate bonds. iv. Coordination compounds retain their identity in aqueous solutions. The properties of coordination

compounds are entirely different from those of the constituent ions. eg. When a solution of ferrous sulphate is mixed with a solution of KCN, a precipitate of ferrous cyanide

is obtained which dissolves in excess KCN to form potassium ferrocyanide. Potassium ferrocyanide dissociates in solution to form K+ ions and [Fe(CN)6]4 ion. K4[Fe(CN)6] 4K+ + [Fe(CN)6]4 Potassium ferrocyanide Potassium ion Ferrocyanide ion Ferrocyanide ion retains its identity in solution. It will not dissociate to Fe2+ and CN ions. Double salts: i. Double salts are molecular or addition compounds which are formed by two apparently saturated

compounds, but they lose their identity when dissolved in water. eg. Mohr’s salt: (NH4)2 Fe(SO4)2.6H2O Potash alum: K2SO4.Al2(SO4)3.24H2O Carnallite: KCl.MgCl2.6H2O ii. When the saturated solutions of ferrous sulphate and ammonium sulphate are mixed and the resultant

mixture is cooled, Mohr’s salt is obtained.

Introduction9.0

09 Coordination Compounds

FeSO4 + 2KCN Fe(CN)2 + K2SO4Ferrous cyanide

precipitate

ferrous sulphate

Potassiumsulphate

potassium cyanide

Ferrous cyanide

Potassium ferrocyanide

Potassium cyanide

Fe(CN)2 + 4KCN K4[Fe(CN)6]

Waterferrous sulphate Mohr’s salt FeSO4 + (NH4)2SO4 + 6H2O FeSO4.(NH4)2SO4.6H2O

Ammonium sulphate

9.0 Introduction 9.1 Werner’s theory of coordination compounds 9.2 Ligands, chelation and denticity 9.3 Coordination number, oxidation number

and charge number 9.4 IUPAC nomenclature of mononuclear

coordination compounds 9.5 Isomerism in coordination compounds 9.6 Sidgwick’s electronic theory

9.7 Effective atomic number (EAN) 9.8 Valence bond theory (VBT) 9.9 Crystal field theory (CFT) 9.10 Colour of coordination compounds 9.11 Magnetic properties of coordination compounds 9.12 Bonding in metal carbonyls 9.13 Stability of coordination compounds 9.14 Importance of coordination compounds

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Chapter 09: Coordination Compounds

iii. When Mohr’s salt is dissolved in water, it dissociates to give Fe2+, 4NH and 24SO ions.

iv. The properties of double salts are same as those of the constituents. v. The metal ions in double salts show their normal valency.

Double salts Coordination compounds i. Double salts are the molecular or addition

compounds that exist only in solid state and dissociate into ions in aqueous solution or in any other solvent.

Coordination compounds are the molecular or addition compounds that exist in the solid state as well as when dissolved in water or any other solvent.

ii. They lose their identity in aqueous solution. They do not completely lose their identity in aqueous solution.

iii. The properties of the double salts are essentially the same as those of constituent ions.

The properties of coordination compounds are different from the constituent ions or atoms.

iv. In a double salt, metal ions exhibit their normal valency.

In coordination compound, metal ion is surrounded by a number of oppositely charged ions or neutral molecules more than its normal valency.

Homoleptic and heteroleptic complexes: i. In homoleptic complexes, a metal is attached to one kind of donor atoms or groups. eg. [Co(NH3)6]3+ is a homoleptic complex with six ammonia molecules attached to central cobalt atom. ii. In heteroleptic complexes, a metal is attached to more than one kind of donor atoms or groups. eg. [Co(NH3)4Cl2]+ is a heteroleptic complex with four ammonia molecules and two chloride ions

attached to the central cobalt atom. Complex ion, coordination entity, coordination sphere and coordination polyhedron: i. A complex ion is more or less stable, charged aggregate formed when an ion, mostly of a metal is directly

linked to a group of neutral molecules and/or ions. eg. a. [Cu(NH3)4]2+ is a complex ion in which Cu(II) ion is directly linked to four neutral ammonia molecules. b. [Fe(CN)6]3 is a complex ion in which Fe(III) ion is directly linked to six negatively charged cyanide ions. ii. A coordination entity constitutes central metal atom or ion bonded to fixed number of ions or molecules. eg. [Co(NH3)3Cl3], [Fe(CN)6]4, [Co(NH3)6]3+ and [Ni(CO)4] are coordination entities. iii. The central metal atom/ion and the coordinating groups (ligands) attached to it are written inside the

square bracket and are together called as coordination sphere. The ionisable groups are written outside the bracket and termed as counter ions.

eg. The coordination sphere in K3[Fe(CN)6] is [Fe(CN)6]3 and the counter ion is K+. iv. The spatial arrangement of the ligand atoms, which are directly attached to the central atom/ion, is called

coordination polyhedron around the central atom/ion. The generally observed coordination polyhedra are tetrahedral, square planar and octahedral. eg. [Ni(CO)4] has tetrahedral geometry, [PtCl4]2 has square planar geometry while [Co(NH3)6]3+ and

[Co(NH3)4Cl2]+ have octahedral geometry.

L

L

LL

M

TetrahedralShapes of different coordination polyhedra (M = Central metal atom/ion, L = Ligand)

M

LL

L

L

L

LOctahedral

M

L

L L

L

Square planar

M

L

L

L

L

LTrigonal bipyramidal

L

M

L

L

L

L

Square pyramidal

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Cationic, anionic and neutral complexes: i. The complexes in which the complex ion carries a net positive charge are called cationic complexes. eg. [Co(NH3)6]Cl3 and [Ni(NH3)6]Cl2 are cationic complexes. This can be seen from the following

chemical equations.

[Co(NH3)6]Cl3 [Co(NH3)6]3+ + 3Cl

[Ni(NH3)6]Cl2 [Ni(NH3)6]2+ + 2Cl ii. The complexes in which the complex ion carries net negative charge are called anionic complexes. eg. K4[Fe(CN)6] and K2[HgI4] are anionic complexes. This can be seen from the following chemical equations.

K4[Fe(CN)6] 4K+ + [Fe(CN)6]4

K2[HgI4] 2K+ + [HgI4]2–

iii. The complexes which carry no net charge are called neutral complexes. eg. [Ni(CO4)], [Pt(NH3)2Cl2] are neutral complexes. Basic postulates of Werner’s theory: i. Two types of valences are shown by most metallic elements: a. Primary valence or principal valence or ionisable valence. b. Secondary valence or non-ionisable valance. ii. Every metal tends to satisty both primary and secondary valence. iii. The number of secondary valence shown by each metal is fixed. iv. The secondary valence directs towards fixed positions in space. Salient features of Werner’s theory: i. The primary valence or principal valence is also known as ionisable valence. It is designated by solid

line () and it corresponds to the oxidation state of metal ion. According to Werner, it forms outer sphere (or second sphere) also known as ionisation sphere. The groups present in the ionization sphere can be separated easily by dissolution in suitable solvent as they are loosely attached to metal ion.

ii. The secondary valence is also known as auxiliary valence or subsidiary valence or residual valence or non-ionisable valence. It is designated by dotted line () and it is equivalent to the coordination number of metal ion. According to Werner, it forms inner sphere (or first sphere) also known as the coordination sphere. It is difficult to separate groups present in the coordination sphere as they are firmly attached to metal ion.

iii. The metal ion exercises primary valency towards the negative groups to satisfy its normal charge by the formation of simple salts.

eg. Pt, Co, Cu, Ag have primary valences 4, 3, 2 and 1 respectively where they form simple salts PtCl4, CoCl2, CuCl2 and AgCl respectively.

iv. The metal ion exercises secondary valence towards the negative ions, neutral molecules or both to form a coordination sphere.

eg. The secondary valency of Co3+ is six. It forms complexes such as [Co(NH3)6]3+, [Co(NH3)5Cl]2+, [Co(NH3)4Cl2]+, [Co(NH3)3Cl3].

v. The secondary valence is fixed for every element. vi. The primary valences are non-rigid and non-directional while the secondary valences are directional. The

number and position of the secondary valences in space determines the geometry of the complex. This leads to the stereoisomerism in complexes.

vii. Square planar or tetrahedral complexes have secondary valence of 4. Octahedral complexes have secondary valence of 6.

Werner’s theory of coordination compounds9.1

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Werner’s theory in the light of modern electronic theory of valence: i. According to Werner, two spheres of attraction are present around the metal. The inner zone is coordination

sphere and the outer zone is ionisation sphere. ii. In terms of modern electronic theory of valence, the

coordination sphere is equivalent to the coordination number of the metal ion and the ionisation sphere is equivalent to the ionisable valence or the oxidation state of the metal ion.

iii. Werner introduced square brackets [ ] to enclose the central metal ion and the ligands. This represents the coordination sphere. The ions of the ionisation sphere are placed outside the square bracket.

eg. In [Co(NH3)6]Cl3 complex, the coordination sphere includes six ammonia ligands around central cobalt ion and the ionisation sphere includes three chloride ions. This is schematically represented in the figure.

Application of Werner’s theory to Co(III) amines: Co(III) amines are the complexes of Co(III) with ammonia. They have compositions CoCl3.6NH3,

CoCl3.5NH3, CoCl3.4NH3, CoCl3.3NH3. The designation and formulation of cobalt (III) amines are shown in the following figure:

Experimental observations of Werner’s theory: i. Cobalt has primary valence (or oxidation state) 3 and secondary valence (or coordination number) 6. ii. Ammonia is not liberated during heating cobalt amines with hydrochloric acid at 373 K. Thus ammonia is

firmly attached to cobalt. iii. The electrical properties of Co(III) amines are different from each other. iv. The reactivity towards silver nitrate is different for all these complexes. a. CoCl3.6NH3: 1. It reacts with AgNO3 to give AgCl precipitate corresponding to three chloride ions. CoCl3.6NH3 + Ag+ 3AgCl (excess) Thus, three chlorides should be present in ionization sphere and satisfy the primary valence of

cobalt ion. 2. Ammonia is not liberated during heating CoCl3.6NH3 with hydrochloric acid at 373 K. Thus, all

six ammonia molecules are firmly attached to cobalt and they satisfy secondary valence of cobalt. 3. As per cryoscopic study of CoCl3.6NH3, four particles are present. 4. As per molar conductivity study of CoCl3.6NH3 in nitrobenzene, six charges are present. Thus,

CoCl3.6NH3 is formulated as [Co(NH3)6]Cl3. [Co(NH3)6]Cl3 [Co(NH3)6]3+ + 3Cl

Cobalt (III) ammines

Cl

Cl

Cl

Co

(A) CoCl36NH3 or [Co(NH3)6]Cl3

Cl

Cl

Cl

Co

(B) CoCl35NH3 or [Co(NH3)5Cl]Cl2

Cl

Cl Cl

Co

(D) CoCl33NH3 or [Co(NH3)3Cl3]

Cl

Cl

Cl

Co

(C) CoCl34NH3 or [Co(NH3)4Cl2]Cl

Cl

Cl

Cl

Co

[Co(NH3)6]Cl3 Outer sphere or ionisation sphere

(field of primary valence)

Inner sphere or coordination sphere(field of secondary valence)

Two spheres of attraction of metal

M

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b. CoCl3.5NH3: 1. It reacts with AgNO3 to give AgCl precipitate corresponding to two chloride ions. CoCl3.5NH3 + Ag+ 2AgCl (excess) Thus, two chlorides should be present in ionization sphere and satisfy the primary valence of

cobalt ion. 2. Ammonia is not liberated during heating CoCl3.5NH3 with hydrochloric acid at

373 K. Thus all five ammonia molecules are firmly attached to cobalt and they satisfy secondary valence of cobalt.

3. Cryoscopic and conductivity study of CoCl3.5NH3 indicates three ions and four charges. Thus CoCl3.5NH3 is formulated as [Co(NH3)5Cl]Cl2. Here chloride serves the dual role of satisfying both a primary and a secondary valence. The bond between this chloride and cobalt is shown by combined representation of dotted line and solid line ( ).

[Co(NH3)5Cl]Cl2 [Co(NH3)5Cl]2+ + 2Cl The structure and bonding in CoCl3.4NH3 and CoCl3.3NH3 can be explained on similar grounds. The properties (features) and results of conductivity measurements for cobalt (III) complexes are summarized as follows:

Composition Colour No. of Cl ions precipitated by

AgNO3

No. of ions/ particles

Total charge Formula

Solution conductivity

corresponds to CoCl3.6NH3 Yellow 3 4 6 [Co(NH3)6]3+3Cl 1:3 electrolyte CoCl3.5NH3 Purple 2 3 4 [Co(NH3)5Cl]2+2Cl 1:2 electrolyte CoCl3.4NH3 Green 1 2 2 [Co(NH3)4Cl2]+Cl 1:1 electrolyte CoCl3.4NH3 Violet 1 2 2 [Co(NH3)4Cl2]+Cl 1:1 electrolyte CoCl3.3NH3 0 0 0 [Co(NH3)3Cl3] -

In above table, CoCl3.4NH3 shows two different complexes having identical formula, [Co(NH3)4Cl2]+Cl, but different properties. They are called as isomers (cis and trans). Limitations of Werner's theory: Werner's theory failed to explain why, a. only certain elements form coordination compounds. b. the coordination sphere/entity has a definite geometry. c. the coordination compounds possess definite magnetic and optical properties. i. Ligand is an atom or molecule or ion which is capable of donating a pair of electrons to the central metal

atom or ion and forms a coordinate bond with it. OR The molecules or ions which are coordinated to the metal atom or ion in a coordination compound are

called ligands. eg. In [Cu(NH3)4]2+, ammonia molecules act as ligands. ii. Ligands are also called donor groups. Ligands (negative ions, positive ions and neutral molecules):

Negative ions Positive ions Neutral molecules CN (cyanide) NO+ (nitrosonium) CO (carbon monoxide)

2NO (nitrite) 2 5N H (hydrazinium) NH3 (ammonia)

OH (hydroxide) NO2+ (nitronium) H2O (water)

3NO (nitrate) NH2OH (hydroxylamine)

F (fluoride) CH3NH2 (methylamine) Cl (chloride) C5H5N (pyridine) Br (bromide) NO (nitric oxide)

Ligands, chelation and denticity9.2

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O2 (oxide) PH3 (phosphine) CH3COO– (acetate) NH2CSNH2 (thiourea) SCN– (thiocyanate) CS (thiocarbonyl) (C6H5)3P - Triphenyl phosphine

iii. Classification of ligands: Depending on the number of donor atoms present, ligands are classified as monodentate ligands, polydentate ligands and ambidentate ligands.

a. Monodentate (Unidentate) ligands: 1. Monodentate (Unidentate) ligands are the ligands which have only one donor atom which acts

as point of attachment and coordinates with metal ion. eg. NH3, Cl, OH, H2O, etc. (Each of these ligands have only one donor atom). b. Polydentate (Multidentate) ligands: 1. Polydentate (Multidentate) ligands are the ligands which have two or more donor atoms which

act as point of attachment and coordinate with metal ion. 2. Thus, multiple sites of ligands are used in the coordination with metal. 3. These multiple sites of ligands alongwith metal form ring structure. Such coordination

compounds are called metal chelates. 4. The polydentate ligands are called chelate ligands and the process of formation of such

compounds is called chelation. 5. Based on the number of donor atoms, polydentate ligands are further classified as bidentate,

tridentate, tetradentate, etc. i. Bidentate (didentate) ligands have two donor atoms. eg. Ethylenediamine or ethane-1,2-diamine (en) with two nitrogen atoms as donor

atoms. A pair of electrons is donated by each nitrogen atom thereby forming coordinate bonds.

ii. Tridentate ligands have three donor atoms. eg. Diethylenetriamine (dien) with three nitrogen atoms as donor atoms. iii. Tetradentate ligands have four donor atoms. eg. Triethylenetetramine (trien) with four nitrogen atoms as donor atoms. iv. Hexadentate ligands have six donor atoms. eg. Ethylenediaminetetraacetate ion (EDTA4)

Name of ligand Structure Donor atom Abbreviation

Bidentate ligands Ethylene diamine

N en

Oxalate ion

O ox

NH2

NH2 NH2

NH2

2

3 5

4

1 Cu2+ + Cu2+

Central metal ion

ComplexEthylenediamine (en)

CH2 CH2

H2N NH2

COO–

COO–

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Glycinate ion

N and O gly

Acetyl acetonate

O acac

Dimethyl glyoximato

N and O dmg

2, 2-Dipyridyl

N dipy

1, 10-phenanthroline or ortho phenanthroline

N phen

Tridentate ligands Diethylene triamine

N dien

Tetradentate ligands Triethylene tetraamine

N trien

Pentadentate ligand Ethylenediamine triacetato

N and O

NH2

COO–

CH2

N N

N N

H2C NH CH2 | | H2C CH2

NH2 NH2

H2C |

NH2

H2C

CH2 | CH2

NH2

NH NHC H2

C H2

O = C

H2C

HN

CH2 CH2

CH2CH2

N

C = O

C = OO–

O–O–

H3C – C

H3 C – C = N

OH

O–N

H3C C CH = C CH3 || O

| O–

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Hexadentate ligand Ethylenediamine tetraacetate ion

N and O EDTA

c. Ambidentate ligands: Ambidentate ligands are the ligands which have two or more donor atoms

capable of forming coordinate bonds; however, only one donor atom is utilized during complex formation.

eg. 1. 2NO group can form complexes by utilizing either N or O as donor atom but not both. This results in formation of either MNO2 or MONO complex respectively.

2. Similarly, SCN group can form complexes by utilizing either S or N as donor atom. This results in formation of either MSCN or MNCS complex respectively.

3. CN group can form complexes by utilizing either C or N as donor atom. This results in formation of either MCN or MNC complex respectively.

Note: H2O has 2 lone pairs of electrons on oxygen atom but it is a monodentate ligand (not bidentate). One atom can donate only one pair of electrons.

Denticity: The denticity of the ligand is the number of ligating groups or coordinating groups present in a ligand.

Based on denticity, the chelation is classified into didentate chelation, tridentate chelation, tetradentate chelation, and so on.

i. Didentate chelation: In didentate chelation, the chelating ligand coordinates through two sigma electron pair donor groups. The ligand is called didentate ligand and its denticity is 2.

eg. In the complex [PtCl2(en)], the denticity of ethylenediamine (en) ligand is 2. ii. Tridentate chelation: In tridentate chelation, the chelating ligand coordinates through three sigma electron

pair donor groups. The ligand is called tridentate ligand and its denticity is 3. eg. In the complex [PtCl(dien)]+, the denticity of diethylenetriamine (dien) ligand is 3. iii. Tetradentate chelation: In tetradentate chelation, the chelating ligand coordinates through four sigma

electron pair donor groups. The ligand is called tetradentate ligand and its denticity is 4. eg. In the complex [Pt(trien)]2+, the denticity of triethylenetetramine (trien) ligand is 4. Coordination number: i. Coordination number (CN) or ligancy is the total number of coordinate bonds formed between ligands and

central metal atom/ion. ii. It is the property which is characteristic of metal. iii. The range of CN is from 2 to 10. Common CN are 4, 6 and are usually observed in lighter transition metals

whereas CN 8 is usually observed in heavier transition metals. eg. 6VF , 3

8NbF and 38TaF

Coordination number, oxidation number and charge number 9.3

M NCS Isothiocyanato

Ambidentate ligands

M SCN Thiocyanato

M NO

ONitritoN

M O N = O NitritoO

CH2 CH2

H2C CH2

O

O C = O

Ethylenediaminetetraacetate ion (EDTA4)

N N

CH2 CH2

O = C C = O

O–O = C

O

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Absolute Chemistry Vol - 2.1 (Med. and Engg.) Oxidation state and CN of metal ions:

Metal ion Oxidation state CN Example Ag+ +1 2 [Ag(NH3)2]+ Cu2+, Ni2+, Zn2+,Cd2+, Hg2+, Pt2+ +2 4 [Cu(NH3)4]2+ Fe3+, Co3+, Cr3+ +3 6 [Co(NH3)6]3+ Sn4+, Pt4+ +4 6 [Pt(NH3)6]4+ Mo4+ +4 8 [Mo(CN)8]4

iv. In case of metal chelates, CN is the number of pair of electrons involved in formation of bonds between metal and ligands.

eg. [Cu(en)2]2+ has CN 4, as 4 pairs of electrons are involved in bond formation. v. Factors influencing the CN of the metal ion are: a. Charge on metal ion b. Charge on ligand c. Relative size of metal ion and the legands. d. Inter-ligand repulsion vi. Geometry and shape of the coordination compounds is governed by CN. Oxidation number: The oxidation number of the central metal atom is the charge it would carry if all the ligands are removed

alongwith the electron pairs that are shared with the central metal atom. eg. Let the oxidation number of Ni in [Ni(CO)4] be x. [Ni(CO)4] is a neutral complex; the sum of oxidation numbers of all the species in it is zero. x 0 [Ni(CO)4] x + 4(0) = 0 x = 0 Therefore, the oxidation number of Ni in [Ni(CO)]4 is 0. Charge number: i. The charge number of a complex ion is the net charge carried by the complex ion. ii. Charge number is the algebraic sum of the charges carried out by the central metal ion and ligands attached to it. eg. Charge number of [Fe(CN)6]3 = (Charge on Fe3+) + (6 charge on CN ion) = (+3) + 6(1) = +3 6 = 3 IUPAC is the acronym for International Union of Pure and Applied Chemistry. IUPAC rules (1990) are the

recommendations for the systematic naming of coordination compounds. Formulae of mononuclear coordination entities: The complex containing a single central metal atom is known as mononuclear coordination entity. The rules

for writing their formulae are as follows: i. The central metal atom is written first. ii. The ligands are then written in alphabetical order irrespective of the charge on them. iii. The abbreviated ligands are written according to the first letter of the abbreviation in alphabetical order.

Polydentate ligands are also written alphabetically. iv. The formula of entire coordination entity is enclosed in square brackets independent of the presence or

absence of charge on it. v. There should not be any space between the ligands and metal within a coordination sphere. vi. The formulae of polyatomic ligands and ligand abbreviations are enclosed in parentheses. e.g. [Ni(CO)4], [CoCl2(en)2]Cl vii. In case of a charged coordination entity without the counter ion, the charge is written outside the square

bracket as a right superscript with the number before the sign. The cationic counter ion is written before the square bracket, where as the anionic counter ion is written after.

eg. [Co(CN)6]3, [Cr(H2O)6]3+, [Cr(H2O6)]Cl3 and K3[Co(CN)6] viii. The charge of the cation(s) is balanced by that of the anion(s).

IUPAC nomenclature of mononuclear coordination compounds 9. 4

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Naming of mononuclear coordination compounds: The latest IUPAC rules for naming the coordination compounds are are follows: i. Naming complex compounds: First, the name of cation is written, followed by the name of anion. eg. In the complex K2[Pt(Cl)6] the cation K(potassium) is named first. K2[Pt(Cl)6] – Potassium hexachloridoplatinate (IV) ii. For anionic ligands, the name ends in O. eg. Chlorido for Cl, bromido for Br, cyanido for CN Note: As per new IUPAC rules, if the name of negative ligand ends in ‘–ide’, the last letter ‘e’ is replaced

by ‘O’. Similarly for ‘ite’ and ‘ate’ it becomes ‘ito’ and ‘ato’. However, the old name cyano is retained for –CN (cyanide).

The names of neutral and cationic ligands are same. eg. Ethylenediamine or ethane-1,2-diamine for en Following are the exceptions: a. H2O is written as aqua b. NH3 is written as ammine c. CO is written as carbonyl d. NO is written as nitrosyl Organic neutral molecules retain their names. eg. CH3NH2 (methylamine), C5H5N (pyridine or py) iii. Alphabetical order: For two or more different ligands, the names of ligands are arranged in alphabetical order. eg. diamminedicarbonylnickel (0) for [Ni(NH3)2(CO)2] iv. Prefixes: For two or more identical ligands; prefixes di, tri, tetra, penta, hexa, etc. are used. For chelate ligands containing prefixes in name itself, the number of chelate ligands are indicated by

prefixes bis, tris, tetrakis, etc. v. Naming of complex ions: For complex cation, first the name of ligand is written followed by the name of

metal along with the oxidation state which is written in Roman number within parentheses. eg. Tetraamminecopper (II) for [Cu(NH3)4]2+, Tetraamminedichloridoplatinum (IV) ion for [Pt(NH3)4Cl2]2+ For complex anion, first the name of ligand is written followed by the name of metal, ending with suffix

‘ate’. eg. hexacyanoferrate (II) for [Fe(CN)6]4 For non-ionic complex, the name is written as one word. eg. Tetracarbonylnickel (0) for Ni(CO)4 Names of common ligands in coordination compounds:

Ligands Name of ligand in coordination compounds Ligands Name of ligand in

coordination compounds Negative ligands

F– Fluorido O2– Oxo Cl– Chlorido N3– Nitrido Br– Bromido P3– Phosphido NO2

– Nitrito-N N3– Azido

ONO– Nitrito-O NCS– Isothiocyanato SO4

2– Sulphato H– Hydrido OH– Hydroxo CO3

2– Carbonato COO–

COO–

Oxalato NO3– Nitrato

CN– Cyanido NH2– Amido

CH3COO– Acetato NH2– Imido SCN– Thiocyanato ClO3

– Chlorato O2

2– Peroxo S2O32– Thiosuplhato

SAMPLE C

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Positive ligands NO+ Nitrosonium NH2NH3

+ Hydrazinium NO2

+ Nitronium Neutral ligands

CH3NH2 Methylamine NH2CH2CH2NH2 Ethane-1,2-diamine or Ethylenediamine

C5H5N Pyridine (py) (C6H5)3P Triphenylphosphine PH3 Phosphine H2NCSNH2 Thiourea (tu)

Dipyridyl (dipy)

Names of the metal in an anionic complex:

Ligands Name of metal in anionic complex Metal Name of metal in anionic

complex Aluminium Aluminate Manganese Manganate Chromium Chromate Molybdenum Molybdate Cobalt Cobaltate Nickel Nickelate Copper Cuprate Silver Argentate Gold Aurate Tin Stannate Iron Ferrate Zinc Zincate Lead Plumbate — —

Note: The name ethane-1,2-diamine should be used instead of ethylenediamine Names of some complex compounds:

Formula Name 1. Li[AIH4] Lithium tetrahydridoaluminate (III) 2. Na3[Co(NO2)6] Sodium hexanitrito-N-cobaltate (III) 3. Na[Au(CN)2] Sodium dicyanoaurate (I) 4. Na2[ZnCl4] Sodium tetrachloridozincate (II) 5. Na3[Fe(C2O4)3] Sodium trioxalatoferrate (III) 6. K4[Fe(CN)6] Potassium hexacyanoferrate (II) 7. K3[Al(C2O4)3] Potassium trioxalatoaluminate (III) 8. Ni(CO)4 Tetracarbonylnickel (0) 9. [Ni(dmg)2] Bis (dimethylglyoximato)nickel (II) 10. [Ag(NH3)2]Cl Diamminesilver (I) chloride 11. [Pt (NH3)4Br2]Br2 Tetraamminedibromidoplatinum (IV) bromide 12. (NH4)2[Pt(SCN)6] Ammonium hexathiocyanatoS-platinate (IV) 13. [Co(NH3)6]Cl3 Hexaamminecobalt (III) chloride 14. [Co(NH3)5Cl]2+ Pentaamminechloridocobalt (III) ion 15. [Co(NH3)3(NO2)3] Triamminetrinitrito-N-cobalt (III) 16. [Co(H2O)(NH3)5]I3 Pentaammineaquacobalt (III) iodide 17. [Co(en)3]Cl3 Tris (ethane-1, 2-diamine)cobalt (III) chloride 18. [Co(en)2Cl2]+ Dichloridobis(ethane-1,2-diamine)cobalt (III) ion 19. [Ag(NH3)2]NO3 Diamminesilver (I) nitrate 20. K[Au(CN)4] Potassium tetracyanoaurate (III) ion 21. [Co(C2O4)3]3 Trioxalatocobaltate (III) ion. 22. [Ni(CN)4]2 Tetracyanonickelate (II) ion 23. [Cr(H2O)4Cl2]Cl Tetraaquadichloridochromium (III) chloride 24. [Cu(H2O)2 (NH3)5]Cl2 Pentaamminediaquacopper (II) chloride 25. Ba[CuCl4] Barium tetrachloridocuprate (II)

N N

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26. [Pt(NH3)4Cl2]Cl2 Tetraamminedichloridoplatinum (IV) chloride 27. [Ni(en)3]3+ Tris(ethane-1,2-diamine)nickel (III) ion 28. [Cu(CN)4]2 Tetracyanocuprate (II) ion 29. Cu2[Fe(CN)6] Copper hexacyanoferrate (II) 30. [Fe(CO)5] Pentacarbonyliron (0) or Ironpentacarbonyl 31. [Co(NH3)5(CO3)]Cl Pentaamminecarbonatocobalt (III) chloride 32. Hg[Co(SCN)4] Mercury tetrathiocyanato-S-cobaltate (III) 33. [CoCl2(en)2]Cl Dichloridobis(ethane-1,2diamine)cobalt (III) chloride 34. [Ag(NH3)2][Ag(CN)2] Diamminesilver (I) dicyanoargentate (I) 35.

Trans-bis(ethane-1,2-diamine)dinitrito-N-cobalt (III) chloride

Note: For ligands which act as a bridge between two metal atoms, the greek letter and hyphen is written before their names.

Isomerism: i. Two or more substances having the same molecular formula but different structural or spatial arrangements

are called isomers. ii. The term isomerism was used for the first time by Berzelius. It is derived from Greek words ‘isos’ meaning

equal and ‘meros’ meaning parts. iii. Isomerism is classified into: stereoisomerism and structural isomerism. Stereoisomerism: Directional nature of coordinate bonds results in stereoisomerism. It is due to difference in the spatial

arrangement of atoms or groups of atoms around the central metal atom/ion. Stereoisomerism is classified into: geometrical isomerism and optical isomerism. i. Geometrical isomerism/Cis-trans isomerism: a. Geometrical isomerism is due to difference in the spatial arrangement of ligands around the central

metal atom/ion in heteroleptic compunds. b. It is present in square planar complexes with coordination number 4 and octahedral complexes with

coordination number 6. Tetrahedral complexes does not show geometrical isomerism because all the four positions are equivalent.

c. In cis isomer, similar ligands are arranged adjacent to each other while in trans isomer, similar ligands are arranged opposite to each other.

Geometrical isomerism in complexes with coordination number 4: a. Complexes of the type [Ma2b2]n: 1. Square planar complexes of the type [Ma2b2]n

(where, a and b are monodentate ligands) show cis-trans isomerism.

eg. [Pt(NH3)2Cl2] 2. In the cis isomer of this complex, two Cl ligands

are arranged adjacent to each other. Also two NH3 ligands are arranged adjacent to each other. In the trans isomer, two Cl ligands are arranged opposite to each other.

Note: Complexes of the type MA4, MA3B, MAB3 do not show geometrical isomerism because all possible spatial arrangements are equivalent.

Isomerism in coordination compounds 9.5

Geometrical isomers of [Pt(NH3)2Cl2]

Pt

cis

Cl NH3

NH3Cl trans

Pt

H3N Cl

NH3Cl

NO2

NO2

en

en Cl Co

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b. Complexes of the type [Ma2bc]n: 1. Square planar complexes of the type [Ma2bc]n,

eg. [Pt(NH3)2ClBr] also shows cis-trans isomerism.

2. In the cis isomer of this complex, two NH3 ligands are arranged adjacent to each other. In the trans isomer, two NH3 ligands are arranged opposite to each other.

c. Complexes of the type [Mabcd]n: 1. Square planar complexes of the type [Mabcd]n, eg. [Pt(NH3)BrCl(Py)] with all the four monodentate

ligands different, has three geometrical isomers. 2. The ligand arranged opposite to Br is different in all three isomers. In one isomer, it is NH3, in the

other isomer it is Cl and in third isomer it is Py. This is true not only for Br ligand but also for remaining three ligands.

d. Complexes of the type [M(AB)2]n: 1. Square planar complexes of the type

[M(AB)2]n where (AB) is a bidentate and unsymmetrical ligand, eg. [Pt(gly)2] has two geometrical isomers, cis and trans. Here gly is glycinate ligand (NH2CH2COO).

2. In the cis isomer of this complex, two NH2 groups are arranged adjacent to each other. Also two O atoms of the two groups are arranged adjacent to each other.

3. In the trans isomer, two NH2 groups are arranged opposite to each other. Also two O atoms of the

two groups are arranged opposite to each other. Geometrical isomerism in complexes of coordination number 6 a. Complexes of the type [Ma4b2]n: 1. Octahedral complexes of the type [Ma4b2]n where a and b are monodentate ligands, eg. [Co(NH3)4Cl2]+ has two geometrical isomers. 2. In the cis isomer, chloride ligands are adjacent and occupy i.e., 1, 2 positions. In the trans isomers,

chloride ligands are opposite and occupy i.e., 1, 6 positions.

C O O

C O O

Geometrical isomers of [Pt(NH3)2ClBr]

Pt

cis

H3N Cl

H3N Br

Pt

trans

Cl

NH3Br

H3N

Geometrical isomers of [Pt(NH3)BrCl(Py)]

Cl Br

Py NH3

Pt

ClBr

H3N Py

Pt

Py Br

H3N Cl

Pt

Geometrical isomers of [Co(NH3)4Cl2]+

+

H3N

NH3 H3N

Co

Cl

Cl

NH3

cis

+

H3N

NH3H3N

Co

Cl

Cl

NH3

trans

(1)

(2)

(1)

(6)

Geometrical isomers of [Pt(gly)2]

Pt

H2N

CH2

CO

O NH2

CH2

COO

trans

Pt

H2N

CH2

COO

NH2

CH2

COO

cis

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b. Complexes of the type [Ma3b3]n: 1. Octahedral complexes of the type [Ma3b3]n

where a and b are monodentate ligands, eg. [Cr(NH3)3Cl3] has two geometrical

isomers. 2. In cis isomer, three Cl ligands form one

triangular face and occupy positions 1, 2 and 3. Three NH3 ligands form opposite triangular face and occupy positions 4, 5 and 6. This isomer is also known as facial isomer or fac isomer.

3. In trans isomer, three Cl ligands are present on edges of the octahedron and occupy positions 1, 2 and 6. Three NH3 ligands are present on opposite edges of the octahedron and occupy positions 3, 4 and 5. This isomer is also known as meridional isomer or mer isomer or peripheral isomer.

c. Complexes of the type [M(AA)2a2]n: 1. Octahedral complexes of the type [M(AA)2a2]n where a is a monodentate ligand and (AA) is a

symmetrical bidentate ligand eg. [CoCl2(en)2]+ has two geometrical isomers. 2. In cis isomer, two Cl ligands are adjacent to each other while in trans isomer, two Cl ligands are

opposite to each other. Note: Octahedral complexes of the type Ma6 and Ma5b do not show geometrical isomerism, as different

spatial arrangement of ligands is not possible in these complexes. ii. Optical isomerism: a. Isomers that are non-superimpossible mirror image of each other and which rotate the plane of

polarised light in opposite directions are called optical isomers. b. Optically active isomers have chirality and do not have any element of symmetry. c. As they rotate the plane of polarised light they are also called as enantiomers or enantiomorphs. d. The isomer which rotates the plane of polarized light in the clockwise direction is known as dextro

rotatary or d-form. The isomer which rotates the plane of polarized light in the anti clockwise direction is known as laevo rotatory or l-form.

e. Optical isomerism in different complexes can be explained as follows: 1. Four coordination compounds: i. Tetrahedral complexes: When four different atoms or groups of atoms are attached to central

metal atom or ion in a tetrahedral complex, optical activity is expected. However in general, ligands are labile. Therefore, d and l forms of such complexes are not resolved. Optical activity is observed for tetrahedral complexes containing bidentate unsymmetrical ligands.

Geometrical isomers of [Cr(NH3)3Cl3]

(6)

H3N

NH3 H3N

Cr

Cl

Cl

Cl

trans/meridional/mer

(2)

(1) (1)

(2)

(3)

H3N

ClH3N

Cr

Cl

Cl

NH3

cis/facial/fac

Geometrical isomers of [CoCl2(en)2]+

Cl

Cl

Co

N

N

en +

cis

N

N

en

Cl

Cl

N N

+

trans

N N

en en Co

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e.g. [Ni (CH2NH2COO2)] ii. Square planar complexes: Square planar complexes contain plane or axis of symmetry as all

the four ligands and central metal atom are present in one plane. Thus, they are not chiral. In very rare cases, square planar complexes show optical activity.

2. Six coordination compounds: i. Octahedral complexes with monodentate ligands: [Ma2b2c2], [Ma2b2cd], [Ma2bcde] and [Mabcdef] are different types of octahedral compounds

containing monodentate ligands and they do not possess any element of symmetry. They are expected to be optically active, however, they are not resolved so far.

ii. Octahedral complexes with one or more symmetrical bidentate chelating ligands: The resolution of these complexes in optical isomers is possible. Given below are some of the

examples of such type of compounds: a. Complexes of the type [M(AA)3]n: They contain central metal atom/ion coordinated to

three symmetrical bidentate chelating ligands (AA). They are optically active as they do not contain any element of symmetry.

eg. [Cr(C2O4)3]3 and [Co(en)3]3+

O O O

O

N N N N

Ni Ni

Mirror

O

N

represents glycinate ion

Mirror

O

O

Cr

O

O

ox 3

d-form

O ox

ox

O

O

O

Cr

O O

l- form

O O

ox

ox

ox

3

Optical isomers of [Cr(C2O4)3]3 and [Co(en)3]3+

N

N

Co

N N

l- form

N N

en

en

en

3+

Mirror

N

N

Co

N

N

en 3+

d-form

N

N

en

en

SAMPLE C

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b. Complexes of the type [M(AA)2a2]n: They contain central metal atom/ion coordinated to two symmetrical bidentate chelating ligands (AA) and two monodentate ligands (a). They are optically active as they do not contain any element of symmetry.

eg. [CoCl2(en)2]+ is optically active in cis isomer. Trans isomer contains a plane of symmetry and is optically inactive.

c. Complexes of the type [M(AA)2ab]n: They contain central metal atom/ion coordinated

to two symmetrical bidentate chelating ligands (AA) and two monodentate ligands a and b. Out of three isomers, two are optically active and one is optically inactive meso form.

eg. [CoCl(en)2(NH3)]2+ d. Complexes of the type [M(AA)B2C2]: They contain one symmetrical didentate ligand,

e.g. [Co(en)(NH3)2Cl2]. Its two optical isomers may be represented as follows:

ClCo

2+

l- form

Mirror

Cl

Co

2+

trans isomer

enen

en

en

NH3 NH3

Optically active forms and meso form of the complex [Co(NH3)Cl(en)2]2+

optically active(cis forms)

(optically inactive meso form)

2+

d-form

Cl NH3

Co

en

en

Mirror

Optically active forms and meso form of the complex [CoCl2(en)2]+

Cl

Cl Co

+

l-form en

en

optically active(cis forms)

Cl

Cl

Co

+

trans isomer (optically inactive

meso form)

enen

Cl

ClCo

en+

d-form

en

Mirror

dextro

Cl

NH3NH3

NH3

Co Co

Cl ClCl

N3H

en en

laevo

SAMPLE C

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Structural isomerism: The compounds showing structural isomerism have same molecular formula but different structural

arrangement of ligands around central metal atom/ion. Structural isomerism is classified into ionisation isomerism, linkage isomerism, coordination isomerism and solvate isomerism.

i. Ionisation isomerism (ion-ion exchange isomerism): a. It involves exchange of ions inside and outside the coordination sphere. b. Compounds showing ionisation isomerism have the same formula but they give different ions in solution. eg. 1. [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4 are ionisation isomers. Br and 2

4SO ions are exchanged inside and outside the coordination sphere. The aqueous solution of [Co(NH3)5SO4]Br is red violet in colour and gives Br. This is confirmed by the precipitate of AgBr formed by reaction with AgNO3. The aqueous solution of [Co(NH3)5Br]SO4 is red in colour and gives 2

4SO ion. This is confirmed by the precipitate of BaSO4 formed by reaction with BaCl2.

2. [Co(NH3)5NO2]SO4 and [Co(NH3)5SO4]NO2 3. [Pt(NH3)4Cl2]Br2 and [Pt(NH3)4Br2]Cl2 ii. Linkage isomerism: a. It involves ambidentate ligands attached to central metal atom/ion through different donor atoms. b. Compounds showing linkage isomerism have same molecular formula and they differ in the linkage

between ligand and central metal atom/ion. eg. 1. Yellow coloured [Co(NH3)5NO2]Cl2 and red coloured [Co(NH3)5ONO]Cl2 show linkage

isomerism. NO2 group is an ambidentate ligand and can form complexes by utilizing either N or O as donor atom but not both. This results in formation of either [Co(NH3)5NO2]Cl2 or [Co(NH3)5ONO]Cl2. NO2 is nitro and ONO is nitrito.

2. Cyano CN and thiocyanate SCN are other ambidentate ligands. iii. Coordination isomerism: When compounds containing both cationic and anionic entities differ in the distribution of ligands in cationic

and anionic parts, then the isomers are called as coordination isomers. eg. 1. In the complex [Co(NH3)6] [Cr(CN)6], NH3 ligands are attached to Co3+ and CN ligands are

attached to Cr3+. In its coordination isomer [Cr(NH3)6] [Co(CN)6], NH3 ligands are attached to Cr3+ and CN ligands are attached to Co3+.

2. [Cu(NH3)4][PtCl4] and [Pt(NH3)4][CuCl4] 3. [Cr(NH3)6][Cr(CN)6] and [Cr(NH3)4(CN)2][Cr(NH3)2(CN)4] 4. [Cr(NH3)6][Cr(SCN)6] and [Cr(SCN)4(NH3)2][Cr(NH3)4(SCN)2] iv. Hydrate isomerism (solvate isomerism): a. It involves exchange of H2O molecules inside and outside the coordination sphere. b. Compounds showing hydrate isomerism have the same formula but different number of molecules of

H2O inside and outside the coordination sphere. eg. 1. CrCl3.6H2O has three hydrate isomers viz. violet coloured [Cr(H2O)6]Cl3, bluish green coloured

[Cr(H2O)5Cl]Cl2.H2O and green coloured [Cr(H2O)4Cl2]Cl.2H2O The number of H2O molecules inside coordination sphere in these three hydrate isomers is 6, 5

and 4 respectively. The number of H2O molecules outside coordination sphere in these three hydrate isomers is 0, 1 and 2 respectively. When aqueous solutions of these three hydrate isomers are treated with excess of AgNO3, the ratio of precipitates obtained in moles is 3:2:1 respectively.

2. [Co(H2O)6]Cl3, [Co(H2O)5Cl]Cl2.H2O, [Co(H2O)4Cl2]Cl.2H2O and [Co(H2O)3Cl3].3H2O i. The Sidgwick’s electronic theory is an extension of the electronic theory of Lewis. It explains the formation

of coordination compounds. ii. According to this theory, a number of coordinate bonds are formed when the ligands donate the electron

pairs to the central metal ion.

Sidgwick’s electronic theory9.6

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eg. a. Four CO ligands donate four electron pairs to Ni to form the complex [Ni(CO)4] which contains four coordinate bonds.

b. Six CN ligands donate six electron pairs to Fe3+ to form the complex [Fe(CN)6]3. i. Effective atomic number is the total number of electrons around the central metal ion present in a complex. ii. It is calculated as the sum of electrons on the metal ion and the number of electrons donated by the ligands. EAN = Z X + Y where, Z = atomic number of metal, X = number of electrons lost during oxidation of metal to metal ion

and Y = number of electrons donated by ligands. eg. For [Co(NH3)6]3+, Z = atomic number of Co = 27 X = 3 (because Co Co3+ + 3e) Y = 6 2 = 12 electrons (because each NH3 donates 2 electrons) EAN = Z X + Y = 27 3 + 12 = 36 iii. EAN rule states that “a metal ion continues to accept electron pairs from the ligands till the total number

of electrons present around the metal ion in the complex becomes equal to the atomic number of the next rare gas atom.”

eg. a. In [Fe(CN)6]4, the Fe2+ ion continues to accept electron pairs from CN ligands till the total number of electrons present around Fe2+ ion in [Fe(CN)6]4 becomes equal to 36 which is the atomic number of the next rare gas, krypton.

b. In [Pt(NH3)6]Cl4, the Pt4+ ion continues to accept electron pairs from NH3 ligands till the total number of electrons present around Pt4+ ion in [Pt(NH3)6]Cl4 becomes equal to 86 which is the atomic number of the next rare gas, radon.

iv. Effective atomic number of few metal ions: Metal Complex Z X Y EAN

Ni [Ni(CO)4] 28 0 8 36 Fe [Fe(NH3)6]2+ 26 2 12 36 Co [Co(NH3)6]3+ 27 3 12 36 Zn [Zn(NH3)4]2+ 30 2 8 36 Pt [Pt(NH3)6]4+ 78 4 12 86

v. Few exceptions for EAN rule: In some cases, the EAN is different from the atomic number of the next rare gas.

eg. Metal Complex Z X Y EAN

Fe [Fe(CN)6]3 26 3 12 35 Cu [Cu(NH3)4]2+ 29 2 8 35 Pt [Pt(NH3)4]2+ 78 2 8 84

Effective atomic number (EAN) 9.7

COOC

Ni

CO

CO

Ni + 4CO

Fe3+ + 6CN

NC

CNNC

Fe3+

CN

CN

CN

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Absolute Chemistry Vol - 2.1 (Med. and Engg.) Valence bond theory (VBT): i. Linus Pauling developed the valence bond theory in 1931. ii. It is based on the assumption that the bonds formed between metal ion and ligand are purely covalent coordinate. iii. Each ligand orbital contains a lone pair of electrons. iv. Vacant hybrid orbitals of the central metal atom/ion overlaps with the filled ligand orbitals to form co-

ordinate bonds. Salient features of valence bond theory: i. Definite number of vacant metal orbitals such as s, p and d are utilized for the formation of coordinate

bonds with the ligands. ii. The number of vacant metal orbitals utilized for the formation of coordinate bonds is equal to the

coordination number of metal ion. iii. Hybridization of vacant metal orbitals results in formation of equal number of hybrid orbitals. iv. Each ligand provides at least one orbital which contains a lone pair of electrons. v. Vacant hybrid orbital of the central metal atom/ion overlaps with ligand orbitals to form coordinate bond. vi. The hybridisation of metal orbitals determines the geometrical shape of the complex. Hybridisation and Geometry of Complexes:

Type of hybridisation Geometrical shape sp Linear sp2 Triangular sp3 Tetrahedral

dsp2 Square planar sp3d Trigonal bipyramidal

d2sp3 or sp3d2 Octahedral d3sp3 Pentagonal bipyramidal

vii. Strength of the coordinate bond is directly proportional to extent of the orbital overlap. viii. Inner complexes are formed by hybridization of (n1) d orbitals. Outer complexes are formed by

hybridization of nd orbitals. ix. Paramagnetic complexes contain one or more unpaired electrons while diamagnetic complexes contain no

unpaired electron. x. Presence of strong field ligands like NH3 and CN result in spin pairing i.e., pairing of electrons present in

metal ions. Structures of Complex Compounds based on valence bond theory: i. Structure of nickel tetracarbonyl [Ni(CO)4]: a. Nickel tetracarbonyl is a neutral compound containing Ni as central metal atom in ground state and

four CO ligands. The ground state outer electronic configuration of Ni is 3d84s2. b. Presence of CO ligands results in transfer of two electrons of Ni from 4s orbital to 3d orbital as CO is

a strong ligand. This results in spin pairing (electron pairing) and the configuration now becomes 3d104s0.

c. One 4s and three 4p orbitals of Ni undergo sp3 hybridisation to form four hybrid orbitals with tetrahedral geometry.

d. These four sp3 hybrid metal orbitals overlap with four CO ligand orbitals to form four coordinate bonds between Ni and four CO groups. Metal hybrid orbitals are empty whereas ligand orbitals contains 2 electrons each. Total number of electrons donated by four CO ligands is 4 2 = 8.

Valence bond theory (VBT) 9.8

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e. [Ni(CO)4] is diamagnetic as it contains no

unpaired electrons. Its magnetic moment is

zero. ii. Structure of [NiCl4]2: a. The oxidation state of Ni is +2. The outer electronic configuration of Ni+2 is 3d8. b. sp3 hybridisation of one 4s and three 4p orbitals result in tetrahedral geometry. c. Coordinate bonds are formed by overlap of metal sp3 hybrid orbitals with ligand orbitals. d. [NiCl4]2 is paramagnetic, due to the presence of two unpaired electrons. Note: Cl is a weak ligand so does not cause pairing of 3d electrons.

Structure of [Ni(CO)4]

CO

Ni

COCOOC

CO

Ni

CO

COOC

Ni (ground state)

Transfer of 2 electronsfrom 4s to 3d in presenceof 4CO ligands and sp3

hybridisation of empty 4sand 4p orbitals

(10 electrons of Nickel)

3d8 4s2 4p

3d10 4s 4p

[Ni(CO)4]

sp3 hybrid orbitals (8 electrons of CO ligands)

sp3 hybridisation


3d8 4s 4p

[NiCl4]2

sp3 hybrid orbitals (8 electrons of Cl ligands)

Ni+2 3d8 4s 4p

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iii. Structure of [Ni(CN)4]2: a. The oxidation state of Ni is +2. The outer electronic configuration of Ni+2 is 3d8. b. dsp2 hybridisation of one d, one s and two p orbitals results in square planar geometry. c. [Ni(CN)4]2 is diamagnetic, due to absence of unpaired electrons. iv. Structure of [Cu(NH3)4]2+: a. The oxidation state of Cu in [Cu(NH3)4]2+ is +2. The outer electronic configuration of Cu+2 is 3d94s0. b. dsp2 hybridisation of one d, one s and two p orbitals results in square planar geometry which has been

proved by X-ray analysis. c. Due to presence of one 4p electron, [Cu(NH3)4]2+ is paramagnetic. The magnetic moment, is 1.73 B.M v. Structure of [Co(NH3)6]3+: a. The oxidation state of Co is +3. The outer electronic configuration of Co3+ is 3d6. b. d2sp3 hybridisation of two d, one s and three p orbitals results in octahedral geometry.

Spin pairing and dsp2 hybridisation

[Ni(CN)4]2

Ni+2


3d8 4s 4p

dsp2 hybrid orbitals (8 electrons of CN ligands)

3d8 4s 4p

Transfer of one electronfrom 3d to 4p and dsp2 hybridisation

Cu+2

dsp2 hybrid orbitals (8 electrons of NH3 ligands)

3d9 4s 4p

[Cu(NH3)4]2+

(8 electrons of Cu)

electron from Cu2+ ion

Spin pairing and d2sp3

hybridisation

Co3+

[Co(NH3)6]3+

d2sp3 hybrid orbitals (12 electrons of six NH3 ligands)

3d6 4s 4p

4s 4p

3d

3d

(6 electrons of Co)

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c. Due to absence of unpaired electron, [Co(NH3)6]3+ is diamagnetic. d. [Co(NH3)6]3+ is also known as inner orbital or low spin or spin paired complex due to utilization of

inner 3d orbitals in d2sp3 hybridisation. vi. Structure of [CoF6]3: a. The oxidation state of Co is +3. The outer electronic configuration of Co3+ is 3d6. b. sp3d2 hybridisation of one s, three p and two d orbitals results in octahedral geometry. c. Due to presence of unpaired 3d electrons, [CoF6]3 is paramagnetic. Magnetic moments is 4.89 B.M. d. [CoF6]3 is also known as outer orbital or high spin or spin free complex due to utilization of outer

4d orbitals in sp3d2 hybridisation. Magnetic properties of coordination compounds based on Valence Bond Theory: i. Metal ions like Ti3+ (d1), V3+(d2) and Cr3+(d3) contain one, two and three electrons in d-orbitals respectively.

So, octahedral hybridisation is possible by utilizing two vacant d orbitals along with one s and three p orbitals. The magnetic properties of free ions and coordination compounds are similar.

ii. Low spin complexes utilize inner d-orbitals for d2sp3 hybridization. For more than three 3d electrons, the pairing of electrons occurs as per Hund's rule. Thus for d4(Cr2+, Mn3+), d5(Mn2+, Fe3+), d6(Fe2+, Co3+), undergo spin pairing which give two empty d-orbitals each and two, one and zero valence electrons, respectively.

iii. The magnetic data and spin pairing do not always match for d4 and d5 conglexes. eg. [Mn(CN)6]3– has magnetic moment of two unpaired electrons while [Mn(Cl)6]3– has that of four

unpaired electrons. This difference in data is explained by valance bond theory using formation of inner orbital and outer

orbital complexes. iv. Hence, [Mn(CN)6]3– is a low complex involving d2sp3 hybridisation. [Mn(Cl)6]3– is a high spin compelx

involving sp3d2 hybridisation. v. Similarly, [Fe(CN)6]3– and [Co(C2O4)]3– are inner orbital complexes where [Fe(CN)6]3– is paramagnetic. vi. [Fe(F6)]3– and [CoF6]3– are outer orbital complexes and paramagnetic with five and four unpaired electrons

respectively. Limitations of valence bond theory: i. VBT is unable to explain the spectral properties (colours) of complexes. ii. VBT explains spin only contribution to magnetic moment which depends on the number of unpaired

electrons. There is no explanation in VBT for orbital contribution to magnetic moment which is the magnetic moment arising due to orbital motion of electrons.

iii. There is no explanation in VBT about why a metal ion with a particular oxidation state can have low spin and high spin complexes.

iv. In some cases, the correlation between geometry and magnetic property cannot be explained on the basis of VBT. v. VBT does not provide quantitative interpretations of thermodynamic or kinetic stabilities of coordination

compounds. vi. There is no distinction between weak field and strong field ligands. vii. VBT cannot predict if a 4 coordinate complex will have tetrahedral or square planar geometry. viii. The order of reactivity of inner orbital inert complexes of d3, d4, d5 and d6 ions cannot be explained by VBT.

sp3d2 hybridisation

Co3+

sp3d2 hybrid orbitals (12 electrons of six F ligands)

3d 4p 4d

4s 4p 3d

4s

4d

6 electrons of cobalt

[CoF6]3

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Introduction to crystal field theory i. According to crystal field theory, only electrostatic interaction is present between metal ions and ligands.

This means there is only ionic bond between metal and ligands. ii. Ligands are considered as point charge or dipoles, depending on anions and neutral molecules respectively. iii. As the ligand approaches the central metal ion, the degeneracy is disturbed by the negative field of ligand

(caused by anions or negative ends of dipolar molecules like NH3, H2O.) iv. Due to this electrostatic interaction, degenerate d orbitals of metals split into t2g and eg. This splitting is

known as crystal field splitting (). Application of crystal field theory to octahedral complexes: i. In octahedral complexes, the central metal atom is at the

centre of octahedron. Six ligands are present at six vertices of octahedron.

ii. 2zd and 2 2x y

d

are along the areas in direction of the ligands.

They have maximum electron density along the axes and they experience maximum electrostatic repulsion due to ligands and will be raised in energy.

iii. dxy, dyz and dxz are planar orbitals. They have maximum electron density in planes and in between axes, they experience less electrostatic repulsion due to ligands and will be lower in energy

iv. The five degenerate d orbitals of metal split into two levels. The lower energy level is t2g and contains dxy, dyz and dxz

orbitals. The upper level is eg level and contains 2 2x yd

and 2z

d orbitals. The energy difference between t2g

and eg level is 10 Dq or o. The subscript ‘o’ in o represents octahedral geometry.

v. As per the calculations, the difference in energy levels of eg orbitals in octahedral crystal field and

hypothetical five degenerate metal d orbitals in octahedral crystal field is equal to 0.6 o or 6 Dq. vi. As per calculations, the difference in energy levels of t2g orbitals in octahedral crystal field and hypothetical

five degenerate metal d orbitals in octahedral crystal field is equal to 0.4 o or 4 Dq. vii. When an electron enters t2g orbital, the complex is stabilized by 0.4 o or 4Dq. When an electron enters eg

orbital, the complex is destabilized by 0.6 o or 6 Dq. Crystal field stabilization energy (CFSE) is the total gain of energy obtained by filling electrons in d-orbitals.

Crystal field theory (CFT) 9.9

Barycentre

t2g

Splitting of metal d-orbitals in an octahedral crystal field

Ener

gy Degenerate

metal d-orbitals (in crystal field)

Degenerate metal d-orbitals (in absence of ligands)

dxy dyz dzx

Splitting of metal d-orbitals in octahedralcrystal field.

∆o

0.6∆o

0.4∆o

eg

2 2x yd

2zd

Octahedral complex having metal (M) at the centre and six ligands (L) at

the vertices of the octahedron.

Z L

L

L

L

L

M

Y

X

L

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viii. Strength of ligands determine the magnitude of o or 10 Dq. Strong ligands result in large splitting and high value of o. Weak ligands result in small splitting and low value of o. Thus, the strength of ligands determine the filling of electrons in the d-shell.

Note: For octahedral complexes, crystal field stabilization energy (CFSE) = 4 Dq (number of electrons in t2g orbital) + 6 Dq (number of electrons in eg orbital) Here, the negative sign indicates lowering in energy (i.e., gain in stability)

Metal ion configuration

Low spin state (Strong ligand field) CFSE (0)

High spin state (Weak ligand field) CFSE (0)

d4 eg t2g

4 × (–0.4) = – 1.6 3 × (–0.4) + 0.6 = –0.6

d5 eg t2g

5 × (–0.4) = –2.0 3 × (–0.4) + 2 (+0.6) = 0.0

d6 eg t2g

6 × (–0.4) = –2.4 4 × (–0.4) + 2(+0.6) = –0.4

d7 eg t2g

6 × (–0.4) + 0.6 = –1.8

5 × (–0.4) + 2 (0.6) = –0.8

d8 to d10 configurations can have only one possible arrangement and hence fixed CFSE value. d8 = 6 2

2g gt e , CFSE = 6(–0.4) + 2 (+0.6) = –1.2

d9 = 6 32g gt e , CFSE = 6(–0.4) + 3(+0.6) = –0.6

d10 = 6 42g gt e , CFSE = 6(–0.4) + 4(–0.6) = 0.0

Spectrochemical series: i. In spectrochemical series, common ligands are arranged in a series in the order of decreasing field strength,

as follows: I– < Br– < S2– < SCN– < Cl– < F– < OH– < C2O4

2– < O2– < H2O < NCS– < NH3 < en < NO2– < CN– < CO

Weak field ligands Increasing order of 0 Strong field ligands The order of field strength of common ligands is independant of geometry of complex and nature of central

metal ion. ii. For d1, d2 and d3 ions electrons occupy lower t2g energy level. iii. First three electrons of d4 occupy lower t2g energy level. For the fourth electron there are two possibilities: a. When o > P, fourth electron occupies t2g level. This results in electron pairing. The energy required

to overcome electron electron repulsion is the pairing energy ‘P’. This results in formation of low spin (LS), strong field or spin paired octahedral complex. Ligands that cause this effect are called strong field ligands.

b. When P > o, fourth electron occupies eg level and it is unpaired. This results in formation of high spin (HS), weak field or spin free octahedral complex. Ligands that cause this effect are called weak field ligands.

iv. The values of o and P are obtained from spectroscopy. Therefore, the name spectrochemical series. d configuration Complex Value of P(cm1) Value of o (cm1) Spin state

d4 [Mn(H2O)6]3+ 28,800 21,000 HS (o < P) 36Mn(CN) 28,800 38,500 LS (o > P) [Cr(H2O)6]2+ 23,500 13,900 HS (o < P)

d5 [Mn(H2O)6]2+ 25,500 7,800 HS (o < P) [Fe(H2O)6]3+ 30,000 13,700 HS (o < P)

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Application of crystal field theory to tetrahedral complexes: i. In tetrahedral complexes, the central metal atom is at the centre

of tetrahedron. Four ligands are present at four vertices of tetrahedron. Ligands approach the central metal atom in between three coordinate axes.

ii. dxy, dyz and dxz are now pointed towards ligands. Hence these orbitals experience more repulsion and move to higher energy levels.

iii. Axial obitals (dx2

–y2, dz

2) lie in between metal ligand bond axes. Hence these orbitals remain at lower energy levels.

iv. The five degenerate d orbitals of metal split into two levels. The lower energy level is eg and contains 2 2x y

d

and 2zd orbitals.

The upper level is t2g and contains dxy, dyz and dxz orbitals. The energy difference between eg and t2g level is considered as 10 Dq or t. The subscript ‘t’ in t represents tetrahedral geometry.

v. For the same metal and same ligands, the difference of energy between tetrahedral and octahedral complexes is

represented by t 49

o. This is because splitting is much smaller as compared to octahedral complexes. Limitations of Crystal Field Theory: i. In crystal field theory, only d orbitals of central metal are considered. There is no explanation for s and p orbitals. ii. bonding in complexes is not accounted for. iii. Satisfactory explanation is not provided for the fact that water is a stronger ligand than OH. iv. Partial covalent nature of metal ligand bond is not explained. Crystal field theory only explains ionic bond.

It cannot explain all the properties of complexes. Superiority of Crystal Field Theory over Valence Bond Theory: i. Magnetic properties of complexes and variation with temperature are explained by crystal field theory.

Valence bond theory cannot explain these. ii. Crystal field theory gives the quantitative measure of the stability of complexes. It predicts the geometry of

complexes. It is not possible by valence bond theory. iii. Kinetic and thermodynamic properties of some complexes are explained by crystal field theory but not by

valence bond theory. iv. Crystal field theory explains d-d transitions and colour of complexes. This is not explained by valence bond theory. i. The crystal field theory explains the origin of colour of coordination compounds which is due to d-d transition. ii. Consider an octahedral complex [Ti(H2O)6]3+ with d1 configuration with one electron in lower t2g level.

Tetrahedral complex having metal (M) at the centre and four ligands (L) at the

four corners of the tetrahedron

L

X

Y

M

L

L Z

L

Colour of coordination compounds 9.10

Splitting of metal d-orbitals in a tetrahedral crystal field

Ener

gy

Degenerate metal d-orbitals (in crystal field)

Splitting of metal d-orbitals in tetrahedralcrystal field.

∆t

0.6∆t

0.4∆t

eg

Degenerate metal d-orbitals (in absenceof ligands)

t2g

dzxdyz dxy

2 2x yd

2zd

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Complex absorbs light in the blue green region. The electron is excited from lower t2g level to higher eg level. 1 0

2g gt e hv

light absorbed 0 12g gt e

iii. The colour observed is violet colour. iv. Colour of some coordination compounds:

Coordination entity

Wavelength of light absorbed (nm)

Colour of light absorbed

Colour of coordination entity (transmitted)

[Ti(H2O)6]3+ 498 blue green violet [Co(CN)6]3 310 ultraviolet pale yellow [Co(NH3)5Cl]2+ 535 yellow violet [Co(NH3)5(H2O)]3+ 500 blue green red [Co(NH3)6]3+ 475 blue yellow orange [Cu(H2O)4]2+ 600 red blue

v. Spectroscopy enables to determine the wavelength absorbed and quantifies crystal field splitting. eg. The wavelength corresponding to [Ti(H2O)6]3+ ion is 498 nm which is in the visible region. The crystal field splitting energy is calculated from the wavelength absorbed. where, E is the crystal field splitting energy, h is Planck’s constant (6.63 1034 J s), is the frequency of light, c is the velocity of light (3.00 108 m/s) and is the wavelength of light. The calculation for the wavelength 498 nm for [Ti(H2O)6]3+ ion is shown below:

E = hc

= 34 8

9

(6.63 10 Js)(3.00 10 m/s)(498 nm)(1 10 m/1 nm)

= 3.99 1019 J/ion

E calculated above, can be converted from J/ion to J/mol by multiplying with Avogadro’s number (6.02 1023 ions/mol).

= (3.99 1019 J/ion) (6.02 1023 ions/mol) = 240,437 J/mol = 240 kJ/mol vi. Crystal field splitting is not observed in the absence of ligands due to which the substance is colourless. eg. a. [Ti(H2O)6]Cl3 becomes colourless on removal of water by heating. b. Anhydrous CuSO4 is white, but CuSO4.5H2O is blue in colour. vii. The influence of ligand on the colour of a complex can be explained in detail by considering [Ni(H2O)6]2+

complex that is formed by the dissolution of nickel (II) chloride in water. When the complex is treated with didentate ligand, ethane-1,2-diamine (en) in various molar ratios (en: Ni = 1:1, 2:1, 3:1), the colour changes occur as follows:

Note: Colour of some gemstones: Many gemstones found in nature are coloured due to electronic transitions within the d orbitals of transition metal ion present in it: i. Ruby is aluminium oxide (Al2O3) containing 0.5 – 1% Cr3+ ions (d3). These Cr3+ ions are randomly distributed in

positions normally occupied by Al3+ ions. The chromium (III) species are in the form of octahedral complex incorporated into the alumina lattice. The d-d transitions at these centres give ruby its beautiful red colour.

ii. Emerald, contains Cr3+ ions which occupy octahedral sites in the mineral beryl (Be3Al2Si6O18). The absorption bands observed in ruby are shifted to longer wavelength in emerald due to which it transmits light in the green region.

E = hc

E = h

= c

2

2 6 aqNi H O

+ en(aq)

2

2 4 aqNi H O en

+ 2H2O

pale blue green

2

2 4 aqNi H O en

+ en(aq)

2

2 2 2 aqNi H O en

+ 2H2O

blue/purple

2

2 2 2 aqNi H O en

+ en(aq)

2

3 aqNi en

[Ni(en)3]2+(aq) + 2H2O

violet

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i. Magnetic properties of coordination compounds depends on the number of unpaired electrons. ii. The number of unpaired electrons depends on a. the electronic configuration of central metal ion b. the magnitude of crystal field splitting. iii. Compounds with d1 electronic configuration such as [Ti(H2O)6]3+ are always paramagnetic due to presence

of unpaired electron. iv. For compounds with d5 electronic configuration such as [FeF6]3 and [Fe(CN)6]3, the number of unpaired

electrons depends on the magnitude of crystal field splitting. a. For high spin [FeF6]3 ion with low crystal field splitting, the difference between t2g and eg levels is

small. The electron distribution is as per Hund’s rule. Five electrons are placed in five different orbitals and have parallel spins. This results in maximum stability. However, two electrons are in the higher eg level. This arrangement requires energy corresponding to the promotion of two of the five electrons to the higher 2 2x y

d

and 2zd orbitals.

b. For low spin [Fe(CN)6]3 ion with high crystal field splitting, the difference between t2g and eg levels is large. All the five electrons are placed in lower t2g energy level. There is only one unpaired electron in accordance with Pauli’s exclusion principle.

Thus, high spin complexes are more paramagnetic than low spin complexes. i. In metal carbonyls, the ligand is CO. Since they contain carbonyl ligand only, they are called homoleptic

carbonyls. ii. Most of the transition metals form metal carbonyls. iii. They have simple well defined structure. Structures of metal carbonyls:

Magnetic properties of coordination compounds 9.11

Bonding in metal carbonyls9.12

Fe3+ [FeF6]3

(high spin)

t2g

eg

[Fe(CN)6]3 (low spin)

t2g

eg

Cr(CO)6 (Octahedral)

CO

COCOOC

Cr OC CO

CO

Fe

CO

COOCOC

Fe(CO)5 (Trigonal

bipyramidal)Ni(CO)4

(Tetrahedral)

CO

Ni

CO COOC

Structures of some metal carbonyls

[Mn2(CO)10]

Mn

CO CO

OC Mn

CO CO

CO

CO CO CO CO[Co2(CO)8]

Co Co

OC

C O

CO

COCO

OCOCOC

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iv. In decacarbonyldimanganese (0) [Mn2(CO)10], two square pyramidal Mn(CO)5 units are joined by MnMn bond. v. Octacarbonyldicobalt (0) [Co2(CO)8] contains Co Co bond bridged by two CO ligands. vi. The bonding in metal carbonyls involves a sigma bond and a pi bond. a. The carbonyl carbon donates a lone pair of electrons to the

vacant orbital of the metal. This results in formation of M C sigma bond.

b. The filled d orbital of metal donates a pair of electrons to the vacant antibonding * orbital of CO ligand. This results in the formation of M C pi bond. CO is known as -acid ligand as it can accept electron pairs from metal ion. This is also called as back donation or back bonding.

vii. Presence of both sigma and pi bonds in metal carbonyls result in synergic effect due to which the bond between metal and CO strengthens.

Note: Ligands like CO, CN– and NO+ are called -acceptor or -acid ligands. These are strong field ligands as they increase the value of 0.

Stability of coordination compounds: i. The stability of coordination compounds depends on the interaction between the metal ion and ligands. For

thermodynamically stable compounds, this interaction is strong. ii. The thermodynamic stability is quantitatively expressed in terms of the stability constant or formation

constant which is an equilibrium constant for the equilibrium between metal ion and ligand. Ma+ + nLx [MLn]b+ The equilibrium constant for this equilibrium (i.e., stability constant) is expressed as,

K = b

na x n

[ML ][M ][L ]

iii. High value of stability constant indicates greater thermodynamic stability. Stability constants for complex formation:

System Stability constant (K) Cd2+ + 4NH3 [Cd(NH3)4]2+ 1.3 107

Ag+ + 2NH3 [Ag (NH3)2]+ 1.6 107

Cu2+ + 4NH3 [Cu(NH3)4]2+ 4.5 1011

Ag+ + 2CN [Ag(CN)2] 5.5 1018

Cu2+ + 4CN [Cu(CN)4]2 2.0 1027

Co3+ + 6NH3 [Co(NH3)6]3+ 5.0 1033 iv. The values of the stability constants of the cyano complexes are higher than those of amine complexes. This

indicates greater thermodynamic stability of cyano complexes. Hence CN is a stronger ligand than NH3. v. Consider the reaction, M + 4L ML4. Here, larger the stability constant, the higher the proportion of ML4

in the solution. Free metal ions hardly exist in the solution as they are surrounded by solvent molecules which will compete with the ligand molecules. We generally ignore these solvent molecules for simplicity.

M + L ML ; K1 = [ML] / [M] [L] ML + L ML2 ; K2 = [ML2] / [ML] [L] ML2 + L ML3 ; K3 = [ML3] / [ML2] [L] ML3 + L ML4 ; K4 = [ML4] / [ML3] [L] where, K1, K2, K3, K4 are stepwise stability constants.

Synergic bonding interactionsin a metal carbonyl

M

C O

*

Stability of coordination compounds9.13

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The overall stability constant () is given by, M + 4L ML4 ; 4 = [ML4] / [M] [L]4

The stepwise and overall stability constants can be related as follows: 4 = K1 K2 K3 K4

n = K1 K2 K3 ………….. Kn. eg. Formation of cuprammonium ion involves following steps: Step 1: Cu2+ + NH3 Cu(NH3)2+ ; K1 = [Cu(NH3)2+] / [Cu2+][NH3]

Step 2: Cu(NH3)2+ + NH3 23 2Cu NH ; K2 = 2

3 2Cu NH / [Cu(NH3)2+][NH3] and so on.

where K1 and K2 are stepwise stability constants. The overall stability constant is given by, 4 = 2

3 4Cu NH / [Cu2+][NH3]4

The addition of four NH3 groups to copper shows the decrease in the successive stability constants. In this case, the four constants can be given as:

log K1 = 4.0, log K2 = 3.2, log K3 = 2.7, log K4 = 2.0, or log 4 = 11.9 The reciprocal of the formation constant is known as the instability constant or the dissociation constant

of coordination compounds. Factors affecting stability of complexes: The thermodynamic stability of complexes depends on the following two factors. i. Charge density of the central metal ion: a. The charge density of the central metal ion depends on the charge/radius ratio. The charge density is

higher when the magnitude of charge is higher and the size of the ion is smaller. b. Higher is the charge density of metal ion, greater is the stability of complex. eg. The ionic radii of Cu2+ and Cd2+ ions are 69 pm and 97 pm respectively. Thus the size of Cu2+

ion is smaller than the size of Cd2+ ion. But the charge is same for both ions. Hence the charge density of Cu2+ is higher than the charge density of Cd2+ ion. Hence, Cu2+ complexes are more stable than Cd2+ complexes. This can be seen from the stability constant values. For [Cu(NH3)4]2+, K = 4.5 1011 and for [Cd(NH3)4]2+, K = 1.3 107.

c. Irving William order of stability indicates the order of stability of complexes of divalent metal ions with same ligand. This order is:

Cu2+ > Ni2+ > Co2+ > Fe2+ > Mn2+ > Cd2+ Since the charge is same for all the metal ions, the order of stability of complexes increases with

decrease in the ionic radii of metal ions. Ion Cu2+ Ni2+ Co2+ Fe2+ Mn2+ Cd2+ Ionic radius (pm) 69 78 82 83 91 97

ii. Nature of ligands: a. Ligands donate electrons to form complexes. Hence, they act as lewis bases. b. The stability of complex depends on the basic strength of ligands. Ligands with higher basicity can easily

donate lone pairs of electrons to the central metal ion, resulting in the formation of stable complexes. eg. The basicity of CN is greater than that of NH3. Hence, cyano complexes are more stable than

ammine complexes. K for [Ag(CN)2] = 5.5 1018 , K for [Ag(NH3)2]+ = 1.6 107

i. Extraction of metals: a. Metals like silver and gold are extracted by using complex formation technique. b. Treatment of finely powdered silver ore with dilute NaCN results in formation of cyano complex of silver

which then passes into solution. Silver from the complex is then displaced with more electropositive zinc.

Importance of coordination compounds 9.14

Ag2S + 4NaCN 2Na[Ag(CN)2] + Na2S Silver (I) Sodium Sodium Sodium Sulphide cyanide dicyanoargentate(I) sulphide

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ii. Qualitative and quantitative analysis in analytical chemistry: a. During detection of Cd2+ ions in the presence of Cu2+ ions of group II A, cyano complexes are formed

by treatment with KCN.

Cu2+ + 4CN [Cu(CN)4]2, Cd2+ + 4CN [Cd(CN)4]2 Treatment with H2S gives precipitate of CdS as [Cu(CN)4]2 is more stable than [Cd(CN)4]2. b. During separation of group I Ag+ ions from 2

2Hg and Pb2+ ions, reaction of aqueous ammonia with a precipitate containing AgCl gives water soluble [Ag(NH3)2]Cl complex. Hg2Cl2 and PbCl2 remains undissolved as they do not form complexes.

AgCl + 2NH3 [Ag(NH3)2]Cl c. Detection and estimation of Ni2+ ions, is done by complexing with DMG (dimethylglyoxime) to

obtain red precipitate. d. Some other reagents used in the detection and estimation of metal ions are EDTA, cupron, -nitroso-

-naphthol, etc. iii. Biological importance: Coordination compounds play vital role in many processes occuring in plants and animals. eg. a. Chlorophyll is present in plants. It is a complex of magnesium. b. Haemoglobin acts as oxygen carrier and is a complex of iron. c. Vitamin B12 (cyanocobalamine) is a complex of cobalt. d. The enzymes like, carboxypeptidase-A and carbonic anhydrase (catalysts of biological systems)

are also coordination compounds with biological importance. iv. Use in medicine: a. Cisplatin [Pt(NH3)2Cl2] is useful in treatment of cancer. b. EDTA is useful in treatment of poisoning by lead. c. The excess of copper and iron in toxic proportions in plant/animal systems are removed by using the

chelating ligands like D-penicillamine and desferrioxime-B via formation of coordination compounds. v. In electroplating: Stable complexes with very small dissociation in solution are used for electroplating

with silver and gold. a. The complex K[Ag(CN)2] is used for electroplating with silver. b. The complex K[Au(CN)2] is used for electroplating with gold. vi. Estimation of hardness of water: The presence of Ca2+ and Mg2+ ions in water results in hardness. EDTA

titrations are used to selectively estimate Ca2+ and Mg2+ ions. These ions form stable complexes with EDTA and can be estimated selectively due to different stability constants of calcium and magnesium complexes.

vii. Modification of the redox behaviour of metal ions: a. It is possible to modify the redox behaviour of a metal ion by complex formation. eg. Complexation to oxidize Co2+ to Co3+. It is useful in enzymes and catalysts. b. Many coordination compounds are used as catalysts in chemical reactions. eg. 1. Wilkinson catalyst, [(Ph3P)3RhCl] for hydrogenation of alkenes. 2. Co2(CO)8 catalyst for conversion of olefins into alcohols. viii. Purification of metals is carried out by formation and subsequent decomposition of their coordination

compounds. eg. Impure nickel gets converted into [Ni(CO)4], which on decomposition gives pure nickel. ix. In black and white photography, the developed film is fixed by washing with hypo solution. This process

dissolves the undecomposed silver bromide (AgBr) to form a complex ion, Na3[Ag(S2O3)2].

2Na[Ag(CN)2] + Zn Na2[Zn(CN)4] + 2Ag Sodium dicyanidoargentate (I) Zinc Sodium Silver tetracyanidozincate (II)

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Quick Review

Three donor atoms eg. diethylenetriamine (dien)

Six donor atoms eg. Ethylenediaminetetraacetate ion (EDTA4–)

Ligands

Polydentate or multidentate ligands or chelate ligands

Ligands which have two ormore donor atoms.

Ambidentate ligands

Ligands which have two or more donor atomsbut only one is used to form coordinate bond. eg. In 2NO , both N and O are donor atoms. InSCN–, S and N are donor atoms.

Monodentate or unidentate ligands

Ligands which have only onedonor atom. eg. NH3, Cl–, OH–, H2O, etc.

Bidentate or didentate ligands

Two donor atoms eg. ethylenediamine(en)

Tridentate ligands Tetradentate ligands

Four donor atoms eg. Triethylenetetraamine (trien)

Hexadentate ligands

Stereoisomerism Difference in the spatial arrangement of atoms or

groups of atoms around the central metal atom/ion.

Isomerism (Same molecular formula but different physical or chemical properties)

Structural isomerism Same molecular formula, but different structural

arrangement of ligands around central metal atom/ion.

Geometrical isomerism (Cis-trans isomerism)

Ionisation isomerism Exchange of ions inside and outside the

coordination sphere. eg. [Co(NH3)5NO2]SO4 and [Co(NH3)5SO4]NO2

Linkage isomerism Ambidentate ligands attached through different

donor atoms. eg. [Co(NH3)5NO2]Cl2 and [Co(NH3)5 ONO]Cl2

Coordination isomerism Different complex ions with same molecular

formula and interchange of ligands between cationsand anions.

eg. [Co(NH3)6] [Cr(CN)6] and [Co(CN)6][Cr(NH3)6]

Optical isomerism (Non-superimposable mirror images)

(enantiomers)

Hydrate isomerism Exchange of solvent (H2O) molecules inside and

outside the coordination sphere. eg. [Cr(H2O)6]Cl3,

[Cr(H2O)5Cl]Cl2.H2O and[Cr(H2O)4Cl2]Cl.2H2O

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Summary of stereoisomerism in coordination compounds: Geometry

of the complex

Geometrical isomerism Optical isomerism (d and l form)

Tetrahedral geometry

Does not exhibit geometrical isomerism

Not been possible to resolve.

Square planar geometry

[Ma2b2]n

Not common and is very rarely observed.

[Ma2bc] n [Mabcd]n [M(AB)2]n

Octahedral geometry

[Ma4b2]n

[Ma2b2c2]n, [Ma2b2cd]n, [Ma2bcde], [Mabcdef] are optically active but these complexes have not been resolved till now. [M(AA)3]n

[Ma3b3]n

cis trans b

b b

b

a a

a a M M

cis trans

c c

b b

M a

a a

a

M

cis

A

trans

A

A

A

M M

B

B B B

b c c

b

b c

M

a a a

d

d d

MM

a trans/meridional/mer cis/facial/fac

b

b

b b

b

b

a

a a a

a

MM

b cis trans

b

b b

a

a a

a a a

a a

M M

d-form l-form mirror

A

A

A

A

A

A

A

A

A

A

A

AMM

SAMPLE C

ONTENT

578


[M(AA)2a2]n

[M(AA)2a2]n

[M(AA)2ab]n

where, a, b, c, d, e, f monodentate ligands AA symmetrical bidentate chelating ligand AB unsymmetrical bidentate chelating ligand

mirror

Optically active (cis forms)

d-form l-form

A

AA

A

b b

A

A

A

A

a a

M M

a Optically inactive

meso form (trans isomer)

b

A A

AAM

cis trans

A

A

A

A

A A

AAa

a

a

a

M MOptically active

(cis forms)

d-form l-form mirror Optically inactive meso form

(trans isomer)

A

A

A

A

A

A

A

A

A

AA

Aa

a a

a

a

a

M M M

SAMPLE C

ONTENT

579


1. Coordination compounds contain ligands attached

to central metal atom/ion through _______. (A) covalent bond (B) ionic bond (C) coordinate bond (D) metallic bond 2. Potassium ferrocyanide is a _______.

[AFMC 2000] (A) normal salt (B) complex salt (C) double salt (D) all of these 3. An example for a double salt is _______.

[KCET 2002] (A) potassium ferricyanide (B) cobalthexammine chloride (C) cuprous sulphate (D) Mohr’s salt 4. How many ions per molecule are produced in

the solution when Mohr’s salt is dissolved in excess of water? [KCET 2015]

(A) 4 (B) 5 (C) 6 (D) 10 5. An aqueous solution of potash alum gives

_______. [UPSEAT 2004] (A) two types of ions (B) only one type of ion (C) four types of ions (D) three types of ions 6. Carnallite in aqueous solution gives _______

ions. (A) K+, Mg2+, Cl (B) K+, Cl, 2

4SO , Br (C) K+, Mg2+, 2

3CO (D) K+, Mg2+, Cl, Al3+ 7. According to Werner’s theory _______.

[MP PMT 2000, 02] (A) primary valency can be ionized (B) secondary valency can be ionized (C) primary and secondary valences both

cannot be ionized (D) only primary valency cannot be ionized 8. Which one of the following statements is

INCORRECT about Werner’s theory? (A) Primary valence is the same thing as

oxidation state. (B) Secondary valence is the same thing as

coordination number.

(C) Primary valences are satisfied by neutral molecules.

(D) Secondary valences are directional whereas primary valences are non-directional.

9. Which one of the following will give a white precipitate with 3AgNO in aqueous medium?

[MP PMT 1994] (A) 3 25 2Co NH Cl NO

(B) 3 22Pt NH Cl

(C) 2PtCl (en)

(D) 3 24Pt NH Cl 10. The number of moles of AgCl precipitated

when excess of AgNO3 is added to one mole of 3 24[Cr NH Cl ]Cl is _______.

[EAMCET 1998] (A) zero (B) one (C) two (D) three 11. When l mol CrCl3.6H2O is treated with excess

of AgNO3, 3 mol of AgCl are obtained. The formula of the complex is _______ .

[NCERT Exemplar] (A) [CrCl3(H2O)3].3H2O (B) [CrCl2(H2O)4]Cl.2H2O (C) [CrCl(H2O)5]Cl2.H2O (D) [Cr(H2O)6]Cl3 12. Which of the following complexs will give

white precipitate with aqueous BaCl2? [JIPMER 1997]

(A) 3 4 4 2[Co(NH ) SO ]NO (B) 3 45[Cr NH SO ]Cl (C) 3 5 4[Cr(NH ) Cl]SO (D) [Co(NH3)4SO4]CN 13. The number of ions formed when

cuprammonium sulphate is dissolved in water is _______. [KCET 1993]

(A) 1 (B) 2 (C) 4 (D) 0 14. Which of the following exhibits highest molar

conductivity? [MP PET 1994] (A) 3 36Co NH Cl

(B) 3 25Co NH Cl Cl

(C) 3 24Co NH Cl Cl

(D) 3 33Co NH Cl

Multiple Choice Questions

Introduction 9.0

Werner’s theory of coordinationcompounds

9.1

SAMPLE C

ONTENT

580


15. The aqueous solution of which of the following complex has the least conductivity under identical conditions. [GUJ CET 2018]

(A) Pentaaquachlorido chromium (III) chloride

(B) Tetraaquadichlorido chromium (III) chloride

(C) Hexaaquachromium (III) chloride (D) Triaquatrichlorido chromium (III) 16. Which of the following complex compounds

is a poor electrolytic conductor in solution? [MP PMT 1994]

(A) K2[PtCl6] (B) [Co(NH3)3(NO2)3] (C) K4[Fe(CN)6] (D) [Cu(NH3)4]SO4 17. The solution of this compound will show

maximum ionic conductivity: [BCECE 2015] (A) K4[Fe(CN)6] (B) [Co(NH3)6]Cl3 (C) [Cu(NH3)4]Cl2 (D) [Ni(CO)4] 18. A complex compound of cobalt contains five NH3

molecules, one nitro group and two chlorine atoms for one Co atom. One mole of this compound produce three moles of ions in aqueous solution. On reacting with excess of AgNO3 solution, two moles of AgCl get precipitated. The molecular formula of the compound is _______.

(A) [Co(NH3)5Cl](NO2)Cl (B) [Co(NH3)5 Cl(NO2)]Cl (C) [Co(NH3)5(NO2)]Cl2 (D) [Co(NH3)5(NO2)Cl2]

19. Cobalt(III) chloride forms several octahedral complexes with ammonia. Which of the following will NOT give test for chloride ions with silver nitrate at 25 °C? [AIPMT 2015]

(A) CoCl3.3NH3 (B) CoCl3.4NH3 (C) CoCl3.5NH3 (D) CoCl3.6NH3 20. The CORRECT order of the stoichiometries of

AgCl formed when AgNO3 in excess is treated with the complex: CoCl3.6NH3, CoCl3.5NH3, CoCl3.4NH3, respectively is _______.

[NEET (UG) 2017] (A) 3 AgCl, 1 AgCl, 2 AgCl (B) 3 AgCl, 2 AgCl, 1 AgCl (C) 2 AgCl, 3 AgCl, 1 AgCl (D) 1 AgCl, 3 AgCl, 2 AgCl 21. When 0.1mol CoCl3(NH3)5 is treated with

excess of AgNO3, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to _______ . [NCERT Exemplar]

(A) 1:3 electrolyte (B) 1:2 electrolyte (C) 1:1 electrolyte (D) 3:1 electrolyte

22. An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichloridotetraaquachromium(III) chloride. The number of moles of AgCl precipitated would be _______. [NEET 2013]

(A) 0.001 (B) 0.002 (C) 0.003 (D) 0.01 23. On treatment of 100 mL of 0.1 M solution of

CoCl3.6H2O with excess AgNO3; 1.2 1022 ions are precipitated. The complex is _______.

[JEE (Main) 2017] (A) [Co(H2O)4Cl2]Cl.2H2O (B) [Co(H2O)3Cl3].3H2O (C) [Co(H2O)6]Cl3 (D) [Co(H2O)5Cl]Cl2.H2O 24. Which of the following is a homoleptic

complex? (A) [Co(NH3)4Cl2]+ (B) [Pt(NH3)2BrCl] (C) [Co(NH3)6]3+ (D) [CrCl2(en)2]+ 25. The counter ion in K3[Fe(CN)6] is _______. (A) K+ (B) Fe2+ (C) Fe3+ (D) CN 26. The charge on [Ag(CN)2] complex is

_______. [AIIMS 2001] (A) 1 (B) +1 (C) +2 (D) +3 27. The cationic complex is _______. (A) [Ni(NH3)6]Cl2 (B) K[Ag(CN)2] (C) K2[HgI4] (D) [Pt(NH3)2Cl2] 28. Which of the following is a neutral complex?

[MHT CET 2016] (A) [Pt(NH3)2Cl2] (B) [Co(NH3)6]Cl3 (C) [Ni(NH3)6]Cl2 (D) K4[Fe(CN)6] 29. The ion or molecule which forms a complex

compound with transition metal ion is called _______.

(A) complex ion (B) ligand (C) coordinate ion (D) counter ion 30. Ligands in a complex salt are _______.

[KCET 1992] (A) anions linked by coordinate bonds to a

central metal atom or ion (B) cations linked by coordinate bonds to a

central metal atom or ion (C) molecules linked by coordinate bonds to

a central metal atom or ion (D) all of these

Ligands, chelation and denticity9.2

SAMPLE C

ONTENT

581


31. Which of the following species is NOT expected to be a ligand? [NCERT Exemplar]

(A) NO (B) 4NH (C) NH2CH2CH2NH2 (D) CO 32. _______ is a neutral ligand. (A) chloride (B) hydroxide (C) ammonia (D) oxide 33. _______ is NOT a neutral ligand. (A) CO (B) 2NO

(C) H2O (D) NH2OH 34. The ligand in potassium ferricyanide is/are

_______. (A) K+ (B) CN (C) Fe3+ (D) both K+ and CN

35. A monodentate ligand has _______. (A) one coordination site (B) two coordination sites (C) three coordination sites (D) many coordination sites 36. Which of the following represents a chelating

ligand? [JIPMER 2002] (A) H2O (B) OH (C) DMG (D) Cl

37. A chelating agent has two or more than two

donor atoms to bind to a single metal ion. Which of the following is NOT a chelating agent? [NCERT Exemplar]

(A) thiosulphato (B) oxalato (C) glycinato (D) ethane-1,2-diamine 38. Which of the following ligands forms a

chelate? [MP PET/PMT 1998] (A) Acetate (B) Oxalate (C) Cyanide (D) Ammonia 39. Which of the following ligands is bidentate? (A) Br (B) en (C) CH3NH2 (D) OH 40. What is the number of donor atom/s in

dimethylglyoximato ligand? [MHT CET 2018]

(A) 1 (B) 2 (C) 3 (D) 4 41. Which of the following is an example of

hexadentate ligand? (A) Ethylenediamine (B) Glycinate ion (C) Acetyl acetonate (D) Ethylenediaminetetraacetate ion

42. EDTA combines with metal ions to form _______. (A) ion-exchange resins (B) chelates (C) dimer (D) polymers 43. An ambidentate ligand is the one which _______. (A) is linked to the metal atom at two points (B) has two or more donor atoms but only

one donor atom is utilized to form a co-ordinate bond

(C) forms chelate rings (D) all of these 44. Which of the following is not an ambidentate

ligand? (A) NO2 (B) SCN (C) H2O (D) CN 45. Match the following.

Column - I Column - II i. en a. monodentate ii. trien b. ambidentate iii. NCS c. tetradentate iv. PH3 d. bidentate

(A) i - c, ii - b, iii - a, iv - d (B) i - d, ii - c, iii - b, iv - a (C) i - c, ii - d, iii - a, iv - b (D) i - b, ii - d, iii - a, iv - c 46. Coordination number of Fe in the complexes

4

6Fe CN ,

3

6Fe CN

and 4FeCl would be respectively _______. [MP PET 2003]

(A) 2, 3, 3 (B) 6, 6, 4 (C) 6, 3,3 (D) 6, 4, 6 47. What is the coordination number of cobalt in

[Co(NH3)3Cl3]? [MP PET 1994] (A) 3 (B) 4 (C) 5 (D) 6 48. The coordination number EDTA is _____.

[AFMC 2004] (A) 3 (B) 4 (C) 5 (D) 6 49. Factor(s) influencing the coordination number

of metal ion is/are _______. (A) charge of metal ion and ligand (B) size of metal ion and ligand (C) inter-ligand repulsion (D) all of these

Coordination number, oxidation number and charge number

9.3

SAMPLE C

ONTENT

582


50. The coordination and oxidation number of X in the compound [X(SO4)(NH3)5]Cl will be _______ respectively.

[JIPMER 1997; DCE 2004] (A) 6 and 4 (B) 10 and 3 (C) 2 and 6 (D) 6 and 3 51. In 3 44Ni NH SO , the oxidation number

and coordination number of Ni will be respectively _______.

(A) 3 and 6 (B) 2 and 4 (C) 4 and 2 (D) 4 and 4 52. The oxidation number of Ni in [Ni(C2O4)3]4 is

_______. [BCECE 2015] (A) +3 (B) +4 (C) +2 (D) +6 53. The oxidation number of cobalt in

4K[Co(CO) ] is _______. [MP PMT 2001; J & K CET 2005]

(A) +1 (B) –1 (C) +3 (D) –3 54. In which of the following, the central atom

does NOT exhibit an oxidation state of +2? (A) K2[Ni(CN)4] (B) K4[Fe(CN)6] (C) [Fe(C2O4)3]3 (D) [Cu(NH3)4]2+

55. The oxidation states of Cr in [Cr(H2O)6]Cl3,

[Cr(C6H6)2], and K2[Cr(CN)2(O)2(O2)(NH3)] respectively are _______.

[JEE (Main) 2018] (A) +3, +4, and +6 (B) +3, +2, and +4 (C) +3, 0, and +6 (D) +3, 0, and +4 56. The CORRECT charge and co-ordination

number of ʻFe’ in K3[Fe(CN)6] is _______. [MHT CET 2017]

(A) + 2, 4 (B) + 3, 6 (C) + 2, 6 (D) + 3, 3 57. The coordination number and oxidation state

of Cr in 3 2 4 3K Cr C O are, respectively _______. [CBSE PMT 1995]

(A) 4 and +2 (B) 6 and +3 (C) 3 and +3 (D) 3 and 0 58. The sum of coordination number and

oxidation number of the metal M in the complex [M(en)2(C2O4)]Cl (where en is ethylenediamine) is _______.

[AIPMT RE-TEST 2015] (A) 7 (B) 8 (C) 9 (D) 6 59. The value of charge number x which appears in

the Ni2+ complex i.e., [Ni(CN)4]x is _______. (A) + 2 (B) – 2 (C) 0 (D) 4

60. Among the properties (a) reducing, (b) oxidising and (c) complexing; the set of properties shown by CN ion towards metal species is _______. [AIEEE 2004]

(A) c, a (B) b, c (C) a, b (D) a, b, c 61. The CORRECT formula of potassium

hexacyanoferrate (II) is _______. [CBSE PMT 1988]

(A) 4 6K Fe CN

(B) 2 26K Fe CN H O

(C) 3 6K Fe CN

(D) 3 2 6K Fe CN 62. The formula of amminebromidochloridonitrito-

N-platinate (II) is _______. (A) [Pt(NH3)BrCl(ONO)] (B) [Pt(NH3)BrCl(NO2)] (C) [Pt(NH3)2BrCl(NO2)]+ (D) [Pt(NH3)BrCl(NO2)]+ 63. IUPAC name of 3 2 4 3K Al C O is _______. [MP PMT 1993, 02, 03] (A) potassium aluminooxalato (B) potassium aluminium (III) trioxalate (C) potassium trioxalatoaluminate (III) (D) potassium trioxalatoaluminate (IV) 64. The IUPAC name of 3 36Co NH Cl is

_______. [IIT-JEE 1994] (A) hexaamminecobalt (III) chloride (B) hexaamminecobalt (II) chloride (C) triamminecobalt (III) trichloride (D) hexaamminecobalt (III) trichloride 65. The CORRECT IUPAC name of

[Pt(NH3)2Cl2] is _______ . [NCERT Exemplar]

(A) diamminedichloridoplatinum (II) (B) diamminedichloridoplatinum (IV) (C) diamminedichloridoplatinum (0) (D) dichloridodiammineplatinum (IV) 66. IUPAC name of [Pt(NH3)2Cl(NO2)] is

_______ . [NCERT Exemplar] (A) platinum diamminechloridonitrite (B) chloridonitrito-N-ammineplatinum (II) (C) diamminechloridonitrito-N-platinum (II) (D) diamminechloridonitrito-N-platinate (II)

IUPAC nomenclature of mononuclear coordination compounds

9.4

SAMPLE C

ONTENT

583


67. The name of the complex ion, [Fe(CN)6]3 is _______. [AIPMT RE-TEST 2015]

(A) tricyanidoferrate (III) ion (B) hexacyanidoferrate (III) ion (C) hexacyanoiron (III) ion (D) hexacyanitoferrate (III) ion 68. The CORRECT IUPAC name of

[Co(NH3)3(NO2)3] is _____. [MH CET 2015] (A) triamminetrinitrito - N-cobalt (III) (B) triamminetrinitrito - N-cobalt (II) (C) triamminecobalt (III) nitrite (D) triamminetrinitrito - N-cobaltate (III) 69. The IUPAC name of

2 3 2 2 2K [Cr(NH )(CN) O (O) ] is _______. [DCE 2003]

(A) potassium amminedicyanidodioxoperoxo-chromate (VI)

(B) potassium amminecyanidoperoxodioxo-chromium (VI)

(C) potassium amminecyanidoperoxodioxo-chromium (IV)

(D) potassium amminedicyanidoperoxodioxo-chromate (IV)

70. Select the CORRECT IUPAC name of

[Co(NH3)5(CO3)]Cl. [AP EAMCET (Engg.) 2016]

(A) Pentaammoniacarbonatecobalt (III) chloride

(B) Pentamminecarbonatecobalt chloride (C) Pentamminecarbonatocobalt (III)

chloride (D) Cobalt (III) pentamminecarbonate

chloride 71. As per IUPAC norms, the name of the

complex [Co(en)2(ONO)Cl]Cl is ______ . [KCET 2016]

(A) Chloridobis(ethane-1,2-diamine)nitrito-O-cobalt (III) chloride

(B) Chlorobis(ethylenediamine)nitro-N-cobalt (III) chloride

(C) Chloridodi(ethylenediamine)nitro-O-cobalt (III) chloride

(D) Chloroethylenediaminenitro-O-cobalt (III) chloride

72. The IUPAC name of [Co(NH3)4Cl(NO2)]Cl is _______. [KCET 2018]

(A) Tetraamminechloridonitrito-N-cobalt(III) chloride

(B) Tetraamminechloridonitrocobalt(II) chloride

(C) Tetraamminechloridonitrocobalt(I) chloride

(D) Tetraamminechloridodinitrocobalt(III) chloride

73. The compounds [Co(SO4)(NH3)5]Br and

[Co(SO4)(NH3)5]Cl represent _______ . [NCERT Exemplar]

(A) linkage isomerism (B) ionization isomerism (C) coordination isomerism (D) no isomerism 74. Indicate the complex ion which shows

geometrical isomerism. [NCERT Exemplar] (A) [Cr(H2O)4Cl2]+ (B) [Pt(NH3)3Cl] (C) [Co(NH3)6]3+ (D) [Co(CN)5(NC)]3

75. The number of geometrical isomers of the

complex 3 2 2 2Co(NH ) (NO ) is _______. [CBSE PMT 1997]

(A) 2 (B) 3 (C) 4 (D) 0 76. How many geometrical isomers are there for

[Pt(py)(NH3)BrCl] ? [CBSE PMT (Prelims) 2011; BCECE 2014]

(A) 2 (B) 3 (C) 4 (D) 0 77. The number of geometric isomers that can exist

for square planar [Pt(Cl)(py)(NH3)(NH2OH)]+ is _______. (py = pyridine)

[JEE (Main) 2015] (A) 2 (B) 3 (C) 4 (D) 6 78. Which of the following will be able to show

geometrical isomerism? [KCET 2015] (A) MA3B – Square planar (B) MA2B2 Tetrahedral (C) MABCD Square planar (D) MABCD – Tetrahedral 79. Square planar complex of the type MAXBL

(where A, B, X and L are unidentate ligands) shows following set of isomers _______.

(A) two cis and one trans (B) two trans and one cis (C) two cis and two trans (D) three cis and one trans 80. Which one of the following octahedral complexes

will NOT show geometrical isomerism (a and b are monodentate ligands)? [CBSE PMT 2003]

(A) [Ma5b] (B) [Ma2b4]

Isomerism in coordination compounds9.5

SAMPLE C

ONTENT

584


(C) [Ma3b3] (D) [Ma4b2] 81. The type of isomerism shown by the complex

[CoCl2(en)2] is _______. [NEET (UG) 2018] (A) geometrical isomerism (B) coordination isomerism (C) ionization isomerism (D) linkage isomerism 82. Consider the following reaction and statements: [Co(NH3)4Br2]+ +Br− [Co(NH3)3Br3] + NH3 (I) Two isomers are produced if the

reactant complex ion is a cis-isomer. (II) Two isomers are produced if the

reactant complex ion is a trans-isomer. (III) Only one isomer is produced if the

reactant complex ion is a trans-isomer. (IV) Only one isomer is produced if the

reactant complex ion is a cis-isomer. The CORRECT statements are _______.

[JEE (Main) 2018] (A) (I) and (II) (B) (I) and (III) (C) (III) and (IV) (D) (II) and (IV) 83. Which of the following complexes will show

geometrical isomerism? (A) Aquachlorobis(ethylenediamine)

cobalt(II) chloride (B) Pentaaquachlorochromium(III) chloride (C) Potassium amminetrichloroplatinate(II) (D) Potassium tris(oxalato)chromate(III) 84. Fac-Mer isomerism is associated with which

of the following general formula? (A) M (AA)2 (B) M (AA)3 (C) Mabcd (D) Ma3b3 85. Which of the following complexes shows

facial isomer? [GUJ CET 2018] (A) K[Fe(NH3)2(CN)4] (B) [Co(NH3)3(NO2)3] (C) [Co(NH3)4CO3]Cl (D) [Ni(H2O)4(NH3)2]SO4 86. Assertion: Octahedral complexes of the type

Ma6 and Ma5b do not show geometrical isomerism (a and b are monodentate ligands).

Reason: They have coordination number of 6. (A) Assertion and Reason both are true. Reason

is the correct explanation of Assertion. (B) Assertion and Reason both are true. Reason

is not the correct explanation of Assertion. (C) Assertion is true. Reason is false. (D) Assertion is false. Reason is true.

87. Tetrahedral complexes of the types Ma4 and Ma3b (where M stands for a metal, and a and b are achiral ligands) do NOT show optical isomerism because they have _______.

(A) a centre of symmetry (B) a plane of symmetry (C) non-superimposable mirror images (D) all of these 88. Which of the following does NOT have

optical isomer? (A) 3 3 3[Co(NH ) Cl ] (B) 3 3[Cr(en) ]Cl (C) 2 2[CoCl (en) ]Cl (D) 3 2[Co(NH )Cl(en) ] 89. Which of the following compounds shows

optical isomerism? [AIEEE 2005; CBSE PMT 2005]

(A) 23 4[Cu(NH ) ]

(B) 24[ZnCl ]

(C) 32 4 3[Cr(C O ) ] (D) 3

6[Co(CN) ]

90. Which one of the following complexes shows

optical isomerism? (en = ethylenediamine) [JEE (Main) 2016]

(A) [Co(NH3)3Cl3] (B) cis[Co(en)2Cl2]Cl (C) trans[Co(en)2Cl2]Cl (D) [Co(NH3)4Cl2]Cl 91. Which of the following coordination

compounds would exhibit optical isomerism? (A) trans-Dicyanidobis(ethylene-1,2-diamine) chromium (III) chloride (B) Tris-(ethylene-1,2-diamine)cobalt (III)

bromide (C) Pentaamminenitrocobalt (III) iodide (D) Diamminedichloridoplatinum (II) 92. Which of the following shows geometrical as

well as optical isomerism?[Assam CEE 2017] (A) [PtCl2 (en)2]2+ (B) [Pt(NH3)2

Cl2] (C) [Pt(NH3)2

Cl4] (D) [Pt (en)3]4+ 93. The number of possible isomers for the

complex [Co(en)2Cl2]Cl will be _______. (en = ethylenediamine)

[AIPMT RE-TEST 2015] (A) 3 (B) 4 (C) 2 (D) 1 94. Which one of the following has the largest

number of isomers ? [AIEEE 2004]

SAMPLE C

ONTENT

585


(A) 23 2[Ir(PR ) H(CO)] (B) 2

3 5[Co(NH ) Cl] (C) 3 4 2[Ru(NH ) Cl ] (D) 2 2[CoCl (en) ] (R = alkyl group) 95. 3 2 24Pt NH Cl Br and 3 4 2 2Pt(NH ) Br Cl

are related to each other as _______. [MP PET 1996; AFMC 2000;

CBSE PMT 2001] (A) optical isomers (B) coordinate isomers (C) ionisation isomers (D) linkage isomers 96. Which of the following compounds exhibits

linkage isomerism? [MP PMT 2001] (A) 3 3[Co(en) ]Cl (B) 3 6 6[Co(NH ) ][Cr(CN) ] (C) 2 2[CoCl(en) NO ]Br (D) 3 5 2[Co(NH ) Cl]Br 97. Due to the presence of ambidentate ligands

coordination compounds show isomerism. Palladium complexes of the type [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] are _______ . [NCERT Exemplar]

(A) linkage isomers (B) coordination isomers (C) ionisation isomers (D) geometrical isomers 98. [Fe(NO2)3Cl3]3 and [Fe(O – NO)3Cl3]3 show

_______. [KCET 2018] (A) linkage isomerism (B) geometrical isomerism (C) optical isomerism (D) hydrate isomerism 99. Coordination isomerism is caused by the

interchange of ligands between the _______. [UPSEAT 2002]

(A) cis and trans structure (B) complex cation and complex anion (C) inner sphere and outer sphere (D) lower oxidation and higher oxidation states 100. [Cr(NH3)6] [Cr(SCN)6] and [Cr(NH3)2(SCN)4]

[Cr(NH3)4(SCN)2] are the examples of what type of isomerism? [MH CET 2015]

(A) Ionisation isomerism (B) Linkage isomerism (C) Coordination isomerism (D) Solvate isomerism 101. Which of the following would exhibit

coordination isomerism?

(A) 3 6 6Cr(NH ) Co(CN)

(B) 3 4 2 2Pt (NH ) Cl Br

(C) 3 36Cr NH Cl

(D) 2 2CrCl (en)

102. The total number of possible isomers for the

complex compound II II3 4 4Cu (NH ) Pt Cl

are _______. [CBSE PMT 1998; DPMT 2004; J & K CET 2005]

(A) 3 (B) 4 (C) 5 (D) 6 103. Isomers which differ in the number of water

molecules inside and outside the coordination sphere are called _______ isomers.

(A) ionization (B) ligand (C) hydrate (D) geometrical 104. What kind of isomerism exists between

[Cr(H2O)6]Cl3 (violet) and [Cr(H2O)5Cl]Cl2H2O (greyish-green)?

[MP PMT 1997; NCERT Exemplar] (A) Linkage isomerism (B) Solvate isomerism (C) Ionisation isomerism (D) Coordination isomerism 105. The total number of possible isomers for

square-planar [Pt(Cl)(NO2)(NO3)(SCN)]2− is _______.

(A) 8 (B) 12 (C) 16 (D) 24 106. As per Sidgwick’s electronic theory, ligands

in complex compounds _______. (A) accept electron pair(s) (B) donate electron pair(s) (C) neither accept nor donate electron

pair(s) (D) either accept or donate electron pair(s) 107. In Ni(CO)4, the carbonyl ligands donate

_______ electron pairs to Ni. (A) 1 (B) 2 (C) 3 (D) 4 108. The EAN of Zn in [Zn(NH3)4]2+ is _______. (A) 33 (B) 34 (C) 35 (D) 36 109. The EAN of Cu in [Cu(CN)6]3– is _______. (A) 33 (B) 36 (C) 35 (D) 38 110. Which of following coordinate complexes is

an exception to EAN rule?

Sidgwick’s electronic theory9.6

Effective atomic number (EAN) 9.7

SAMPLE C

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(Given At. No. Pt = 78, Fe = 26, Zn = 30, Cu = 29) [MHT CET 2017]

(A) [Pt(NH3)6]4+ (B) [Fe(CN)6]4- (C) [Zn(NH3)4]2+ (D) [Cu(NH3)4]2+ 111. Valence bond theory is based on the

assumption that the bonds formed between metal ion and ligands are _______.

(A) coordinate covalent (B) ionic (C) metallic (D) both (A) and (B) 112. Which of the following is INCORRECT

statement about valence bond theory? (A) A definite number of vacant metal

orbitals such as s, p and d are utilized for the formation of coordinate bonds with the ligands.

(B) The number of vacant metal orbitals utilized for the formation of coordinate bonds is equal to the coordination number of metal ion.

(C) The number of hybrid orbitals formed, is independent of the number of participating vacant metal orbitals.

(D) The hybridisation of metal orbitals determines the geometrical shape of the complex.

113. A tetrahedral complex ion is formed due to

______ hybridization. (A) sp2 (B) sp3 (C) dsp2 (D) d2sp3 114. The shape of 2

3 4Cu NH

is square planar, 2Cu in this complex is _______.

[NCERT 1989; RPET 1999] (A) 3sp hybridised (B) 2dsp hybridised (C) 3sp d hybridised (D) 3 2sp d hybridised 115. In the complex 2 3

5[SbF ] , sp d hydridisation is present. Geometry of the complex is _______.

[Pb. PMT 2000] (A) trigonal bipyramidal (B) square bipyramidal (C) tetrahedral (D) square planar

116. Match the types of hybridisation given in column I with the geometrical shapes given in column II.

Column I Column II

i. sp2 a. Octahedral ii. dsp2 b. Tetrahedral iii. dsp3 c. Triangular iv. sp3d2 d. Linear

e. Trigonal bipyramidal f. Square planar

(A) i - c, ii - f, iii - e, iv - a (B) i - a, ii - f, iii - f, iv - b (C) i - d, ii - b, iii - a, iv - e (D) i - c, ii - b, iii - a, iv - e 117. Pick out the CORRECT statement with

respect to [Mn(CN)6]3. [NEET (UG) 2017] (A) It is sp3d2 hybridized and tetrahedral. (B) It is d2sp3 hybridized and octahedral. (C) It is dsp2 hybridized and square planar. (D) It is sp3d2 hybridized and octahedral. 118. In Wilkinson’s catalyst, the hybridization of

central metal ion and its shape are respectively _______.

(A) sp3d, trigonal bipyramidal (B) sp3, tetrahedral (C) dsp2, square planar (D) d2sp3, octahedral 119. The CORRECT combination is _______. (A) [Ni(CN)4]2 tetrahedral ; [Ni(CO)4]

paramagnetic (B) [NiCl4]2 paramagnetic ; [Ni(CO)4]

tetrahedral (C) [NiCl4]2 square-planar ; [Ni(CN)4]2

paramagnetic (D) [NiCl4]2 diamagnetic ; [Ni(CO)4]

square-planar 120. The species having tetrahedral shape is

_______. [IIT-JEE (Screening) 2004] (A) 2

4Cu Cl (B) [Zn(NH3)4]2+

(C) 24[NiCl ] (D) all of these

121. In 2

4 4[Ni(CO) ] and [NiCl ]

species, the hybridization states of the Ni atom are respectively _______. (At. no. of Ni = 28)

[CBSE PMT 2004; MP PMT 1992; BHU 1995; AFMC 1997]

(A) 3 2sp , dsp (B) 2 3dsp ,sp (C) 2 2dsp ,dsp (D) 3 3sp ,sp

Valence bond theory (VBT) 9.8

SAMPLE C

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122. The hybridization states of the central atoms in the complexes [Fe(CN)6]4 and [Co(NH3)6]3+, are _______.

(A) d2sp3 and sp2 respectively (B) d2sp3 and sp3d2 respectively (C) d2sp3 and dsp2 respectively (D) both d2sp3 123. Chromium hexacarbonyl is an octahedral

compound involving _______ hybridization. (A) sp3d2 (B) dsp2

(C) d2sp3 (D) sp3d2 124. Hybridization involved in the complex

[Ni(CN)4]2 is ________. (At. No. of Ni = 28) [AIPMT RE-TEST 2015]

(A) d2sp2 (B) d2sp3 (C) dsp2 (D) sp3

125. 3 24[Pt NH ]Cl is _______. [DCE 2001] (A) square planar (B) tetrahedral (C) linear (D) pentagonal bipyramidal 126. Which one of the following complexes is an

outer orbital complex? [AIEEE 2004] (A) 3

3 6[Co(NH ) ] (B) 46[Mn(CN) ]

(C) 46[Fe(CN) ] (D) 2

3 6[Ni(NH ) ] Atomic numbers : Mn 25,Fe 26,Co 27, Ni 28

127. The number of unpaired electrons in [NiCl4]2–, Ni(CO)4 and [Cu(NH3)4]2+ respectively are _______. [WB JEE 2017]

(A) 2, 2, 1 (B) 2, 0, 1 (C) 0, 2, 1 (D) 2, 2, 0 128. The numbers of unpaired electrons in low spin

octahedral complexes of Cr2+(d4) and Fe3+ (d5) are _______ respectively.

(A) 0 and 1 (B) 2 and 1 (C) 0 and 2 (D) 1 and 0 129. The number of unpaired electrons in high spin

octahedral complexes of Co3+ (d6) is _______. (A) 1 (B) 2 (C) 3 (D) 4 130. Which of the following is INCORRECT about

valence bond theory? (A) There is no distinction between weak

field and strong field ligands. (B) In some cases, the correlation between

geometry and magnetic property cannot be explained on the basis of valence bond theory.

(C) The order of reactivity of inner orbital complexes of d3, d4, d5 and d6 ions cannot be explained by valence bond theory.

(D) Valence bond theory can explain the spectral properties of complexes.

131. According to crystal field theory, the

interaction(s) present between metal ion and ligands is/are _______.

(A) electrostatic (B) covalent (C) van der Waals forces (D) both (A) and (B) 132. The energy difference between t2g and eg level

in an octahedral crystal field is _______. (A) 4Dq (B) 6Dq (C) 8Dq (D) 10Dq 133. The magnitude of CFSE depends upon _______. (A) nature of ligands (B) number of ligands (C) geometry of complex (D) all of these 134. Among the following complexes the one

which shows zero crystal field stabilization energy (CFSE) is ________. [AIPMT 2014]

(A) [Mn(H2O)6]3+ (B) [Fe(H2O)6]3+ (C) [Co(H2O)6]2+ (D) [Co(H2O)6]3+ 135. Which of the following spectrochemical series

is TRUE? [GUJ CET 2015] (A) SCN < NH3 < F < en < CO (B) SCN < F < NH3 < en < CO (C) SCN < F < en < NH3 < CO (D) SCN < F < en < CO < NH3 136. Which of the following sequences is

CORRECT regarding field strength of ligands as per spectrochemical series? [KCET 2016]

(A) SCN < F < CN < CO (B) F < SCN < CN < CO (C) CN < F < CO < SCN

(D) SCN < CO < F < CN

137. The strongest ligand in the following is

_______. [MP PET 1995] (A) CN (B) Br (C) OH (D) F 138. A strong ligand gives a complex which is

generally called _______ complex. (A) high spin (B) spin free (C) low spin (D) weak field

Crystal field theory (CFT)9.9

SAMPLE C

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139. The value of the ‘spin only’ magnetic moment for one of the following configurations is 2.84 BM. The CORRECT one is _______. [AIEEE 2005]

(A) d4 (in strong ligand field) (B) d4 (in weak ligand field) (C) d3 (in weak as well as in strong fields) (D) d5 (in strong ligand field) 140. Considering H2O as a weak field ligand, the

number of unpaired electrons in 2

2 6[Mn(H O) ] will be _______. (Atomic number of Mn = 25)

[CBSE PMT 2004] (A) two (B) four (C) three (D) five 141. The complex ion having minimum magnitude

of o (CFSE) is _______ . [KCET 2015] (A) [Cr(CN)6]3 (B) [Co(NH3)6]3+

(C) [CoCl6]3 (D) [Cr(H2O)6]3

142. Which of these statements about [Co(CN)6]3–

is CORRECT? [AIPMT 2015] (A) It has no unpaired electrons and has

low-spin configuration. (B) It has four unpaired electrons and has

low-spin configuration. (C) It has four unpaired electrons and has

high-spin configuration. (D) It has no unpaired electrons and has

high-spin configuration. 143. The CORRECT order of spin-only magnetic

moments among the following is _______. (Atomic number : Mn=25, Co=27, Ni=28,

Zn=30) [JEE (Main) 2018] (A) [ZnCl4]2− > [NiCl4]2− > [CoCl4]2− >

[MnCl4]2− (B) [CoCl4]2− > [MnCl4]2− > [NiCl4]2− >

[ZnCl4]2− (C) [NiCl4]2− > [CoCl4]2− > [MnCl4]2− >

[ZnCl4]2− (D) [MnCl4]2− > [CoCl4]2− > [NiCl4]2− >

[ZnCl4]2− 144. Which one of the following statement is

CORRECT for d4 ions [P = pairing energy]? [TS-EAMCET 2017]

(A) When Δ0 > P, low-spin complex form (B) When Δ0 < P, low-spin complex form (C) When Δ0 > P, high-spin complex form (D) When Δ0 < P, both high-spin and low-

spin complexes form

145. The CFSE for octahedral [CoCl6]4 is 18,000 cm1. The CFSE for tetrahedral [CoCl4]2 will be _______ . [NCERT Exemplar]

(A) 18,000 cm1 (B) 16,000 cm1

(C) 8,000 cm1 (D) 20,000 cm1 146. Which of the following is INCORRECT about

the application of crystal field theory to tetrahedral complexes?

(A) 2zd and 2 2x y

d

are in the direction of

metal-ligand axes. (B) Ligands point directly towards dxy, dyz

and dxz. (C) eg level is lower in energy than t2g level. (D) Splitting is more as compared to

octahedral complex 147. Which of the following is NOT TRUE about

crystal field theory? (A) bonding in complexes is not

accounted for. (B) Satisfactory explanation is not provided

for the fact that water is a stronger ligand than OH.

(C) Crystal field theory considers s, p and d orbitals of central metal.

(D) Partial covalent nature of metal ligand bond is not explained.

148. The wavelength of light absorbed by

[Cu(H2O)4]2+ is 600 nm. The crystal field splitting energy is _______.

(A) 150 kJ/mol (B) 180 kJ/mol (C) 200 kJ/mol (D) 220 kJ/mol 149. CORRECT increasing order for the

wavelengths of absorption in the visible region of the complexes of Co3+ is _______.

[NEET (UG) 2017] (A) [Co(H2O)6]3+, [Co(en)3]3+, [Co(NH3)6]3+ (B) [Co(H2O)6]3+, [Co(NH3)6]3+, [Co(en)3]3+ (C) [Co(NH3)6]3+, [Co(en)3]3+, [Co(H2O)6]3+ (D) [Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+ 150. The octahedral complex of a metal ion M3+

with four monodentate ligands L1, L2, L3 and L4 absorbs wavelengths in the region of red, green, yellow and blue respectively. The increasing order of ligand strength of the four ligands is _______. [JEE (Main) 2014]

(A) L4 < L3 < L2 < L1 (B) L1 < L3 < L2 < L4 (C) L3 < L2 < L4 < L1 (D) L1 < L2 < L4 < L3

Colour of coordination compounds 9.10

SAMPLE C

ONTENT

589


151. How many out of 2 36 6TiF , CoF , Cu2Cl2

and 24NiCl are colourless?

(Ti = 22, Co = 27, Cu = 29, Ni = 28) [BCECE 2014]

(A) 0 (B) 2 (C) 3 (D) 1 152. Assertion : CuSO4.5H2O is blue in colour, but

anhydrous CuSO4 is white. Reason : Crystal field splitting is not observed

in the absence of ligands. (A) Assertion and Reason both are true.

Reason is the correct explanation of Assertion.

(B) Assertion and Reason both are true. Reason is not the correct explanation of Assertion.

(C) Assertion is true. Reason is false. (D) Assertion is false. Reason is true. 153. Which one of the following compounds is

NOT coloured? [AIIMS 1997] (A) Na2[CuCl4] (B) Na2[CdCl4] (C) K4[Fe(CN)6] (D) K3[Fe(CN)6] 154. Which one of the following complexes is

paramagnetic? [RPMT 1997] (A) 3

6[CoF ] (B) 32 6[Co(H O) ]

(C) 2 3 3[Co(H O) F ] (D) All of these 155. Which one of the following is expected to be a

paramagnetic complex? [MP PMT 1991, 2000]

(A) 2

2 6Ni H O

(B) 4Ni CO

(C) 2

3 4Zn NH

(D) 3

3 6Co NH

156. Which of the following complexes is

paramagnetic? [GUJ CET 2015] (A) [Ni(CO)4] (B) [Co(NH3)6]3+ (C) [Ni(CN)4]2– (D) [NiCl4]2–

157. 3

2 6[Ti(H O) ] is paramagnetic in nature due to the presence of _______. [RPMT 2002]

(A) one unpaired electron (B) two unpaired electrons (C) three unpaired electrons (D) no unpaired electron

158. What is true for [Fe(CN)6]3– and [FeF6]3–? [RPET 1999]

(A) Both are paramagnetic. (B) Only 3

6[Fe(CN) ] is paramagnetic. (C) Only 3

6[FeF ] is paramagnetic. (D) Both are diamagnetic. 159. Among the following ions, which one has the

highest paramagnetism? [IIT 1993; UPSEAT 2002]

(A) 32 6Cr(H O) (B) 22 6Fe(H O)

(C) 2

2 6Cu H O

(D) 2

2 6Zn H O

160. Which one of the following complexes has the

highest magnetic moment value? [TS EAMCET (Med.) 2015]

(A) [Fe(CN)6]3 (B) [Co(CN)6]3 (C) [Fe(H2O)6]3+ (D) [CoF6]3

161. A magnetic moment of 1.73 BM will be shown

by which of the following? [NEET 2013] (A) [Cu(NH3)4]2+ (B) [Ni(CN)4]2– (C) TiCl4 (D) [CoCl6]4– 162. What is the magnetic moment of 3 6K [FeF ]?

[Orissa JEE 2005] (A) 5.91 BM (B) 4.89 BM (C) 3.87 BM (D) 6.92 BM 163. Assertion: [CoF6]3 ion shows magnetic

moment = 1.73 BM. Reason: It has 4 unpaired electrons. (A) Assertion and Reason both are true.

Reason is the correct explanation of Assertion.


(C) Assertion is true. Reason is false. (D) Assertion is false. Reason is true. 164. The magnetic moment of a complex ion is

3.87 BM. The ion is _______. (A) [Co(H2O)6]2+ (B) [MnF6]3

(C) [Ni(CN)4]2 (D) [Mn(CN)6]4

165. The pair having the same magnetic moment is

_______ . [JEE (Main) 2016] [At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27] (A) [Cr(H2O)6]2+ and [CoCl4]2

(B) [Cr(H2O)6]2+ and [Fe(H2O)6]2+ (C) [Mn(H2O)6]2+ and [Cr(H2O)6]2+ (D) [CoCl4]2 and [Fe(H2O)6]2+

Magnetic properties of coordination compounds

9.11

SAMPLE C

ONTENT

590

Absolute Chemistry Vol - 2.1 (Med. and Engg.) 166. What is the shape of Fe(CO)5?

[CBSE PMT 2000] (A) Square pyramidal (B) Octahedral (C) Linear (D) Trigonal bipyramidal 167. CO acts as bridging ligand in case of _______. (A) Cr(CO)6 (B) [Co2(CO)8] (C) [Mn2(CO)10] (D) Fe(CO)5 168. Which of the following is TRUE about metal

carbonyl bond? (A) Both M C sigma pi bonds are present. (B) Only M C sigma bond is present. (C) Only M C sigma bond is present. (D) Only M C pi bond is present. 169. Study the following diagram. It explains _______ . (A) Synergic effect in metal carbonyls (B) Pauli’s exclusion principle (C) formation of M C sigma bond (D) formation of C M pi bond 170. Iron carbonyl, Fe(CO)5 is _______.

[NEET 2018] (A) tetranuclear (B) mononuclear (C) trinuclear (D) dinuclear 171. [Co2(CO)8] displays _______. (A) one Co − Co bond, six terminal CO and

two bridging CO (B) one Co − Co bond, four terminal CO

and four bridging CO (C) no Co − Co bond, six terminal CO and

two bridging CO (D) no Co − Co bond, four terminal CO and

four bridging CO 172. Which of the following has the longest C – O

bond length? (Free C – O bond length in CO is 1.128 Å) [NEET P-I 2016]

(A) [Fe(CO)4]2 (B) [Mn(CO)6]+

(C) Ni(CO)4 (D) [Co(CO)4] 173. Which of the following is a acid ligand?

[KCET 1996; AIIMS 2003] (A) NH3 (B) CO (C) F (D) Ethylenediamine

174. Which of the following complexes formed by

Cu2+ ions is most stable? [NCERT Exemplar] (A) Cu2+ + 4NH3 [Cu(NH3)4]2+, log K = 11.6 (B) Cu2+ + 4CN [Cu(CN)4]2, log K = 27.3 (C) Cu2+ + 2en [Cu(en)2]2+, log K = 15.4 (D) Cu2+ + 4H2O [Cu(H2O)4]2+, log K = 8.9 175. The stability of coordination compound

depends on _______. (A) nature of ligand (B) charge of the central metal ion (C) radius of the central metal ion (D) all of these 176. The complexes formed by Cu2+ ion are more

stable than those formed by Cd2+ ion because _______.

(A) the value of the stability constant for Cd2+ complexes is greater than that of Cu2+ ions

(B) the charge density on Cu2+ is greater than that of Cd2+ ion

(C) the ionic radius of Cu2+ ion is more than that of Cd2+ ion

(D) Cu2+ ion forms chelate compounds 177. Irving William order of stability of complexes

of divalent metal ions with same ligands is _______.

(A) Cu2+ > Ni2+ > Co2+ > Fe2+ > Mn2+ > Cd2+ (B) Ni2+ < Co2+ < Fe2+ < Mn2+ < Cd2+ < Cu2+ (C) Fe2+ > Mn2+ > Cd2+ > Cu2+ > Ni2+ > Co2+ (D) Fe2+ < Mn2+ < Cd2+ < Cu2+ < Ni2+ < Co2+

178. Assertion: Higher the basicity of ligands,

higher the stability of complex. Reason: Ligands with higher basicity can

easily donate lone pairs of electrons to the central metal ion form stable complexes.

(A) Assertion and Reason both are true. Reason is the correct explanation of Assertion.


(C) Assertion is true. Reason is false. (D) Assertion is false. Reason is true.

M

C O*

Stability of coordination compounds9.13 Bonding in metal carbonyls9.12

SAMPLE C

ONTENT

591


179. The stabilization of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species? [NCERT Exemplar]

(A) [Fe(CO)5] (B) [Fe(CN)6]3 (C) [Fe(C2O4)3] 3 (D) [Fe(H2O)6]3+

180. During detection of Cd2+ and Cu2+, which of

the following complexes gives precipitate on treatment with H2S?

(A) [Cd (CN)4]2– (B) [Cu (CN)4]2– (C) [Cd (CN)6]3– (D) [Cu (CN)6]3– 181. _______ reacts with excess of CN to form a

complex with coordination number 2. [AIIMS 2004]

(A) Cu+ (B) Ag+ (C) Ni2+ (D) Fe2+ 182. AgCl precipitate dissolves in ammonia due to

the formation of _______. [AIIMS 1991; MP PET 1993; CBSE PMT 1998]

(A) 4 2[Ag(NH ) ]OH (B) 4 2[Ag(NH ) ]Cl (C) 3 2[Ag(NH ) ]OH (D) 3 2[Ag(NH ) ]Cl 183. CuSO4 solution with excess of KCN gives

_______. [MP PMT 1992;IIT 1996; UPSEAT 2001, 02; BCECE 2014]

(A) Cu(CN)2 (B) CuCN (C) K2[Cu(CN)4] (D) K3[Cu(CN)4] 184. Which of the following reagents can be used

to identify nickel ion? (A) Resorcinol (B) Dimethylglyoxime (C) Diphenylbenzidine (D) Potassium ferrocyanide 185. The metal atoms present in haemoglobin,

chlorophyll and vitamin B12 respectively are _______. [Assam CEE 2015]

(A) Fe, Mg, Co (B) Fe, Co, Mg (C) Fe, Mg, Zn (D) Mg, Fe, Co 186. Which of the following complexes is used to

be as an anticancer agent? [AIPMT 2014] (A) mer-[Co(NH3)3Cl3] (B) cis-[Pt(NH3)2Cl2] (C) cis-K2[PtBr2Cl2] (D) Na2CoCl4 187. The chelating ligand which is used in the

treatment of lead poisoning is ________. [Assam CEE 2015]

(A) ethane-1,2-diamine (B) EDTA

(C) dimethylglyoxime (D) None of these 188. Assertion: The excess of copper in toxic

proportions in plants is removed by using D-penicillamine.

Reason: D-penicillamine is a monodentate ligand which forms a stable complex with copper.

(A) Assertion and Reason both are true. Reason is the correct explanation of Assertion.


(C) Assertion is true. Reason is false. (D) Assertion is false. Reason is true. 189. The complex used for the electroplating of

silver is _______. (A) [Ag(NH3)2]Cl (B) K[Ag(CN)2] (C) [Ag(NH3)2]NO3 (D) [Ag(NH3)(CN)] 190. The hardness of water is estimated by a simple

titration method using _______. (A) en (B) EDTA (C) NH3 (D) CN 191. The molecular formula of Wilkinson catalyst,

used in hydrogenation of alkenes is _______ . [MHT CET 2016]

(A) Co(CO)8 (B) (Ph3P)3RhCl (C) [Pt(NH3)2Cl2] (D) K[Ag(CN)2] 192. Wilkinson’s catalyst is used as homogeneous

catalyst in _______. (A) polymerization (B) condensation (C) halogentaion (D) hydrogenation 193. Match the coordination compounds given in

column I with their applications given in column II.

Column I Column II i. cis-[Pt(NH3)2Cl2] a. Treatment of lead

poisoning ii. K[Au(CN)2] b. Conversion of olefins

into alcohols iii. [RhCl(PPh3)3] c. Electroplating with

gold iv. Co2(CO)8 d. Treatment of cancer

e. Hydrogenation of alkenes

(A) i - a, ii - d, iii - b, iv - e (B) i - e, ii - b, iii - a, iv - d (C) i - d, ii - c, iii - e, iv - b (D) i - e, ii - a, iii - c, iv - d

Importance of coordination compounds9.14

SAMPLE C

ONTENT

592


194. Which compound is zero valent metal

complex? [KCET 2005] (A) [Cu(NH3)4]SO4 (B) [Pt(NH3)2Cl2] (C) [Ni(CO)4] (D) K3[Fe(CN)6] 195. The IUPAC name of the complex ion formed

when gold dissolves in aqua regia is _______. [KCET 2014]

(A) tetrachloridoaurate (III) (B) tetrachloridoaurate (I) (C) tetrachloridoaurate (II) (D) dichloridoaurate (III) 196. Which of the following complexes will show

geometrical as well as optical isomerism (en = ethylene diamine)? [KCET 1996]

(A) 3 22Pt NH Cl (B) 3 42Pt NH Cl

(C) 4

3Pt en

(D) 2 2PtCl en 197. For 1 molal aqueous solution of the following

compounds, which one will show the highest freezing point? [JEE (Main) 2018]

(A) [Co(H2O)6]Cl3

(B) [Co(H2O)5Cl]Cl2.H2O (C) [Co(H2O)4Cl2]Cl.2H2O (D) [Co(H2O)3Cl3].3H2O 198. Mixture X = 0.02 mol of [Co(NH3)5SO4]Br

and 0.02 mol of [Co(NH3)5Br]SO4 was prepared in 2 litres of solution.

1 litre of mixture X + excess AgNO3 Y. 1 litre of mixture X + excess BaCl2 Z Number of moles of Y and Z are _______.

[IIT JEE 2003] (A) 0.01, 0.01 (B) 0.02, 0.01 (C) 0.01, 0.02 (D) 0.02, 0.02 199. The complex salt can be made by the

combination of [CoIII(NH3)5Cl]x with _______. [Pb. CET 2001]

(A) 34PO (B) Cl

(C) 2Cl (D) 2K+

200. If the magnetic moment of 1

2Ag CN

is zero, then the number of unpaired electrons will be _______. [MP PET 1995]

(A) 1 (B) 2 (C) 3 (D) Zero 201. Both [Ni(CO)4] and [Ni(CN)4]2 are

diamagnetic. The types of hybridisation of Ni in these complexes are _______ and _______ respectively. [GUJ CET 2015]

(A) sp3, sp3 (B) sp3, dsp2 (C) dsp2, sp3 (D) dsp2, dsp2 202. The geometry and magnetic behaviour of the

complex [Ni(CO)4] are _______. [NEET (UG) 2018]

(A) square planar geometry and diamagnetic (B) tetrahedral geometry and diamagnetic (C) square planar geometry and paramagnetic (D) tetrahedral geometry and paramagnetic 203. Which of the following statements is

INCORRECT? [GUJ CET 2018] (A) K4[Ni(CN)4] and K2[Ni(CN)4] both have

same magnetic moment. (B) K2[Ni(CN)4] is diamagnetic while

K2[NiCl4] is paramagnetic. (C) K4[Ni(CN)4] is square planar while

K2[Ni(CN)4] is tetrahedral. (D) K2[NiCl4] and K4[Ni(CN)4] both have

same geometrical shapes. 204. Which one of the following is an inner orbital

complex as well as diamagnetic in behaviour (Atomic number : Zn = 30, Cr = 24, Co = 27, Ni = 28)?

(A) [Co(H2O)6]2+ (B) 33 6[Cr(NH ) ]

(C) 33 6[Co(NH ) ] (D) 2

3 6[Ni(NH ) ]

205. [Sc(H2O)6]3+ ion is _______. [Pb. CET 2004] (A) colourless and diamagnetic (B) coloured and octahedral (C) colourless and paramagnetic (D) coloured and paramagnetic

1. (C) 2. (B) 3. (D) 4. (B) 5. (D) 6. (A) 7. (A) 8. (C) 9. (D) 10. (B) 11. (D) 12. (C) 13. (B) 14. (A) 15. (D) 16. (B) 17. (A) 18. (C) 19. (A) 20. (B) 21. (B) 22. (A) 23. (D) 24. (C) 25. (A) 26. (A) 27. (A) 28. (A) 29. (B) 30. (D) 31. (B) 32. (C) 33. (B) 34. (B) 35. (A) 36. (C) 37. (A) 38. (B) 39. (B) 40. (B) 41. (D) 42. (B) 43. (B) 44. (C) 45. (B) 46. (B) 47. (D) 48. (D) 49. (D) 50. (D) 51. (B) 52. (C) 53. (B) 54. (C) 55. (C) 56. (B) 57. (B) 58. (C) 59. (B) 60. (A) 61. (A) 62. (B) 63. (C) 64. (A) 65. (A) 66. (C) 67. (B) 68. (A) 69. (A) 70. (C) 71. (A) 72. (A) 73. (D) 74. (A) 75. (A) 76. (B) 77. (B) 78. (C) 79. (A) 80. (A)

Miscellaneous

Answers to MCQs

SAMPLE C

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81. (A) 82. (B) 83. (A) 84. (D) 85. (B) 86. (B) 87. (B) 88. (A) 89. (C) 90. (B) 91. (B) 92. (A) 93. (A) 94. (D) 95. (C) 96. (C) 97. (A) 98. (A) 99. (B) 100. (C) 101. (A) 102. (D) 103. (C) 104. (B) 105. (B) 106. (B) 107. (D) 108. (D) 109. (D) 110. (D) 111. (A) 112. (C) 113. (B) 114. (B) 115. (A) 116. (A) 117. (B) 118. (C) 119. (B) 120. (D) 121. (D) 122. (D) 123. (C) 124. (C) 125. (A) 126. (D) 127. (B) 128. (B) 129. (D) 130. (D) 131. (A) 132. (D) 133. (D) 134. (B) 135. (B) 136. (A) 137. (A) 138. (C) 139. (A) 140. (D) 141. (C) 142. (A) 143. (D) 144. (A) 145. (C) 146. (D) 147. (C) 148. (C) 149. (D) 150. (B) 151. (B) 152. (A) 153. (B) 154. (D) 155. (A) 156. (D) 157. (A) 158. (A) 159. (B) 160. (C) 161. (A) 162. (A) 163. (D) 164. (A) 165. (B) 166. (D) 167. (B) 168. (A) 169. (A) 170. (B) 171. (A) 172. (A) 173. (B) 174. (B) 175. (D) 176. (B) 177. (A) 178. (A) 179. (C) 180. (A) 181. (B) 182. (D) 183. (C) 184. (B) 185. (A) 186. (B) 187. (B) 188. (C) 189. (B) 190. (B) 191. (B) 192. (D) 193. (C) 194. (C) 195. (A) 196. (D) 197. (D) 198. (A) 199. (C) 200. (D) 201. (B) 202. (B) 203. (C) 204. (C) 205. (A) 2. In potassium ferrocyanide, K4[Fe(CN)6], the species [Fe(CN)6]4 retains its identity in solid as well as in

solution (aqueous) state. Hence, it is a complex salt. 3. Mohr’s salt [FeSO4.(NH4)2SO4.6H2O] is a double salt as it loses its identity when dissolved in water. 4. Mohr’s salt is FeSO4.(NH4)2SO4.6H2O. When it is dissolved in excess of water, one molecule forms one Fe2+ ion, two NH4

+ ions and two 24SO ions

(i.e., total five ions per molecule). 5. Potash alum is a double salt of K2SO4 and Al2(SO4)3 and on dissolving, it gives all three ions Al3+, K+ and

24SO , of which it is made.

6. Carnallite is a double salt with molecular formula KCl.MgCl2.6H2O. It gives K+, Mg2+ and Cl ions in solution. 7. Groups which satisfy the primary valence are loosely bound and can be separated as ions on dissolving the

complex in suitable solvent. Hence, primary valences are ionizable. Groups which satisfy the secondary valence are firmly attached and cannot be easily separated. Hence, secondary valences are non ionizable.

8. The metal ion exercises primary valences towards the negative groups to satisfy its normal charge by the formation of simple salts. It exercises secondary valences towards the negative ions, neutral molecules or both to form a coordination sphere.

9. [Pt(NH3)4]Cl2 will ionize to give two Cl ions; 23 4 2 3 4[Pt (NH ) ]Cl [Pt (NH ) ] 2Cl

As it gives Cl ions in solution so it will give white precipitate of AgCl with 3AgNO . 10. One chloride is present in the ionisation sphere and satisfies the primary valence of chromium ion. Chloride

will react with one mole of AgNO3 to give one mole of AgCl. 11. 1 mol CrCl36H2O reacts with excess of AgNO3 to give 3mol of AgCl. This shows that one molecule of

complex gives three chloride ions in the solution. Hence, the formula of the complex is: [Cr(H2O)6]Cl3 12. [Cr(NH3)5Cl]SO4 will ionize to give 2

4SO ions which then react with BaCl2 to form white precipitate of BaSO4. 2 2

3 5 4 3 5 4[Cr(NH ) Cl]SO [Cr(NH ) Cl] SO 2 2

3 5 4 2[Cr (NH ) Cl] SO BaCl 3 5 2 4[Cr (NH ) Cl]Cl BaSO . 13. Cuprammonium sulphate 3 4 4[Cu (NH ) ]SO dissociates to give [Cu(NH3)4]2+ and 2

4SO ions

3 4 4[Cu (NH ) ]SO 2 23 4 4[Cu(NH ) ] SO

So, it will give two ions in water.

Hints to MCQs

SAMPLE C

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14. On ionisation, [Co(NH3)6]Cl3 gives maximum number (four) of ions. Hence, it exhibits highest molar conductivity.

15. Since Triaquatrichlorido chromium (III) produces no ions in the solution, it has the least electrical conductivity.

Complex Number of ions

(A) [Cr(H2O)5Cl]Cl2 3 (B) [Cr(H2O)4Cl2]Cl 2 (C) [Cr(H2O)6]Cl3 4 (D) [Cr(H2O)3Cl3] 0

16. [Co(NH3)3(NO2)3] will not give any ions in solution. Hence it is a poor electrolytic conductor in solution. 17. The given compounds will ionize as follows: K4[Fe(CN)6] 4K+ + [Fe(CN)6]4

[Co(NH3)6]Cl3 [Co(NH3)6]3+ + 3Cl

[Cu(NH3)4]Cl2 [Cu(NH3)4]2+ + 2Cl [Ni(CO)4] No ions Since, K4[Fe(CN)6] gives maximum number of ions (i.e., 5 ions per molecule) in the solution, it will show

maximum ionic conductivity among the given compounds. 18. On reacting with excess of AgNO3 solution, two moles of AgCl get precipitated. Thus, both chloride ions

are present in the ionization sphere. One mole of complex produces three mole ions in aqueous solution, out of which two mole ions are chloride ions present in the ionization sphere. Thus, third ion is the complex ion. Hence, the formula of the compound is [Co(NH3)5(NO2)]Cl2

[Co(NH3)5(NO2)]Cl2 [Co(NH3)5(NO2)]+2 + 2Cl

19.

Composition Formula of the complex No. of Cl ions precipitated by AgNO3 CoCl3.3NH3 [CoCl3(NH3)3] 0 CoCl3.4NH3 [CoCl2(NH3)4]Cl 1 CoCl3.5NH3 [CoCl(NH3)5]Cl2 2 CoCl3.6NH3 [Co(NH3)6]Cl3 3

Therefore, CoCl3.3NH3 will not give test for chloride ions with silver nitrate. 20. CoCl3.6NH3 contains 3 Cl– ions in ionization sphere. CoCl3.5NH3 contains 2 Cl– ions in ionization sphere. CoCl3.4NH3 contains 1 Cl– ion in ionization sphere. 21. In the given reaction, 0.1 mol CoCl3(NH3)5 reacts with excess of AgNO3 to give 0.2 mol of AgCl. This

indicates that there must be two free chloride ions in the solution of electrolyte CoCl3(NH3)5. Thus, the molecular formula of complex is [Co(NH3)5Cl]Cl2 which is a 1:2 electrolyte.

[Co(NH3)5Cl]Cl2 [Co(NH3)5Cl]2+(aq) + (aq)2Cl

22. 100 mL 0.01 M solution means 0.001 mole 100 0.01 0.0011000

[Cr(H2O)4Cl2]Cl has one ionizable Cl. 0.001 moles of the complex precipitates the same amount of AgCl. 23. Reaction with excers AgNO3. Gives AgCl as precipilate. So, 1.2 × 1022 moles AgCl means 1.2 × 1022 moles

of Cl– ions are produced

SAMPLE C

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595


Mol of Cl– ions = 22

231.2 10

6.022 10

= 0.02 mol

Mol of COCl3.6H2O = volume molarity1000 = 100 0.1

1000 = 0.01 mol

1 mole complex produces = 0.020.01

= 2 mol Cl– ions

There are two Cl– ions as primary valence in the given complex. Hence, the formula of the given complex will be [Co(H2O)5Cl]Cl2.H2O.

27. K[Ag(CN)2] and K2[HgI4] are anionic complexes while [Pt(NH3)2Cl2] is a neutral complex. 31. 4NH ion has no lone pair of electrons to donate to the metal ion. 32. Ammonia (NH3) is a neutral ligand, while Cl, OH and O2 are negative ions. 33. 2NO is a negatively charged (ion) ligand. 34. In K3[Fe(CN)6], CN donates lone pair of electron and forms coordinate bond. Hence CN acts as ligand. K

is a counter ion where as Fe is central metal atoms. 36. DMG is a bidentate ligand which binds the central atom and forms 5 membered ring. Hence, it is a

chelating ligand while H2O, OH and Cl are monodentate ligands. 37. Thiosulphato 2

2 3S O is an unidentate ligand. 38. Acetate, cyanide and ammonia are monodentate ligands. They do not form chelates. Oxalate is a bidentate

ligand and thus it forms a chelate. 40. Dimethylglyoximato is a bidentate ligand. 41. Ethylenediamine, glycinate ion and acetyl acetonate are bidentate ligands. (EDTA)4 is hexadentate ligand.

2 2

2 2

2 24(EDTA)

OOCH C CH COO

N CH CH N

OOCH C CH COO

. . ..

46. In the complexes [Fe(CN)6]4, [Fe(CN)6]3 and [FeCl4], the central metal ion is surrounded by 6,6 and 4

ligands respectively. 47. In [Co(NH3)3Cl3], the central cobalt ion is attached to three NH3 ligands and three chloride ligands. Hence,

the coordination number is 6. 48. EDTA forms six bonds with metal. Hence, EDTA has coordination number of 6. 50. Co-ordination number is 6 because one SO4 and five NH3 ligands are coordinated to X. Let the oxidation number of X in [X(SO4)(NH3)5]Cl be x.

N

CH2

O = C – O–

C – O–

H2C

— CH2 — CH2 — N

CH2

O– – C = O CH2 M

O

O– – C

O

SAMPLE C

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Oxidation state of X in 2 0 1

345

X SO NH Clx

is x 2 + 0 = +1

x – 2 + 0 – 1 = 0 x = +3 51. Let the oxidation number of Ni in [Ni(NH3)4]SO4 be x.

0 2

3 44

Ni NH SOx

x + 0 + (2) = 0 x = +2 is oxidation number and coordination number of Ni is 4 as it is surrounded by 4 NH3 ligands. 52. Charge on the ligand 2

2 4C O is 2 and charge on the complex ion is 4. Let the oxidation number of Ni be x. x + 3 (2) = 4 x = +2 Oxidation number of Ni in [Ni(C2O4)3]4– = +2

53. Let the oxidation number of cobalt in 1 0

4

K Co COx

be x.

1 (+1) + x + 4 (0) = 0 1 + x = 0 ; x = 1 Oxidation number of Co = – 1

54. Let the oxidation state of Fe in 3

2

2 43

Fe C Ox

be x.

x + 3(2) = 3 x 6 = 3 x = 3 Thus, iron in [Fe(C2O4)3]3 has +3 oxidation state while in K2[Ni(CN)4], K4[Fe(CN)6] and [Cu(NH3)4]2+; the

central metal atom exhibits an oxidation state of +2. 55. Oxidation state of Cr in [Cr(H2O)6]Cl3 x + (0) (6) + (1) (3) = 0 x = + 3 [Cr(C6H6)2] y + (2 0) = 0 y = 0 K2[Cr(CN)2(O)2 (O2) (NH3)] (2) (1) + z + (2) (1) + (2) (2) + 2 + 0 = 0 z = + 6 56. K3[Fe(CN)6] 3K+ + [Fe (CN)6]3- Oxidation state [Fe(CN)6]3-

x + 6 (–1) = 3 x 6 = 3 x = + 3 Coordination number of Fe is 6 as 6 CN ligands are attached to metal ion Fe3+. 57. The coordination number of K3[Cr(C2O4)3] will be 6 because it is bonded with three bidentate ligands

(oxalate).

SAMPLE C

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Oxidation number (x) of Cr in 1 2

3 2 43

K Cr C Ox

is x + 3(+1) + 3(2) = 0

x = +3 58. [M(en)2(C2O4)]Cl The complex has two molecules of en and one C2O4

2 ion. Thus the complex has three bidentate ligands in all. Coordination number = 2 number of bidentate ligands = 2 3 = 6 Let the oxidation number of metal ‘M’ in the complex be ‘x’. The charge on complex ion, en and C2O4

2 are +1, 0 and 2 respectively. Hence, x + 0 2 = +1 x = 3 The sum of coordination number and oxidation number of the metal M = 6 + 3 = 9 59. [Ni(CN)4]x , (Ni = +2) (CN = 1) x = 2 + 4(1) x = 2 60. CN ions act both as reducing agent as well as good complexing agent. 69. The complex ion is the anion and contains chromium metal with +6 oxidation state written as chromate

(VI). One NH3, two CN, two O and one O2 ligands are present which are written in alphabetical order as amminedicyanidodioxoperoxo. The name of the cation (potassium) is written first followed by the name of the anion.

71. The names of Cl, en and ONO ligands are written in alphabetical order as chlorido, ethane-1,2-diamine and nitrito-O. Prefix bis is used as two en ligands are present. Co is written as cobalt (III).

73. Isomers are the compounds having same molecular formula but different structural formulae. [Co(SO4)(NH3)5]Br and [Co(SO4)(NH3)5]Cl do not have same molecular formula.

74. Complex [Cr(H2O)4Cl2]+ is of [Ma4b2]n+ type and hence shows cis-trans geometrical isomerism. 75. The complex 3 2 2 2Co(NH ) (NO ) has two geometrical isomers i.e., cis and trans. 76. 77. For square planar complexes of the type [Mabcd]n± where all the four ligands are different, three

geometrical isomers are possible. So, for the given square planar complex [Pt(Cl)(py)(NH3)(NH2OH)]+, following three geometrical isomers are possible:

where a = Cl, b = py, c = NH3 and d = NH2OH. 78. For square planar complexes of the type MABCD where all the four ligands are different, three geometrical

isomers are possible.

Trans Cis

Co

H3N NH3

O2N NO2

Co

H3N NO2

NH3O2N

Pt ; NH3

py

Cl

Br Pt

NH3

Br

py

Cl; Pt

NH3

Cl

py

Br

Pt c

a

d

bPt

b

a

d

cPt

d

a

c

b +++

SAMPLE C

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79. Square planar complexes of type, [MAXBL] has three geometrical isomers i.e. two cis and one trans. 80. Octahedral complexes of the type [Ma4b2], [Ma2b4], [Ma3b3] exhibit geometrical isomerism while

octahedral complexes of the type [Ma5b] will not show geometrical isomerism. 81. Complex [CoCl2(en)2] shows geometrical isomerisms. 82. [Co(NH3)4Br2]+ +Br− [Co(NH3)3Br3] + NH3

Br 83. Complex (B): Pentaaquachlorochromium(III) chloride, [Cr(H2O)5Cl]Cl2

does not show geometrical isomerism.

Complex (C): Potassium amminetrichloroplatinate(II), K[Pt(NH3)Cl3] does not show geometrical isomerism.

Complex (D): Potassium tris(oxalato)chromate(III), K3[Cr(C2O4)3] does not show geometrical isomerism. Complex (A): Aquachlorobis(ethylenediamine)cobalt(II) chloride, [Co(H2O)(Cl)(en)2]Cl forms two

geometrical isomers i.e. cis and trans. 84. Compounds with general formula Ma3b3 (eg. [Cr(NH3)3Cl3] and [Rh Cl3(Py)3]) exhibit fac-mer isomerism.

A

MX B

C A

M B C

X A

M

B X

C

Geometrical isomers of [CoCl2(en)2]+

Cl

Cl

N N

trans

N N

en en Co

Cl

Cl

Co

N

N

en

cis

N

N

en

Co

NH3 Br

Br

NH3

NH3 NH3

(Trans)

Co

NH3 Br

Br

Br

NH3 NH3

(Only meridional isomer)

Co

NH3 NH3

Br

NH3 Br

(Cis)

NH3

Co

NH3 Br

NH3 Br

(Facial isomer)

NH3

Br

Co

NH3 Br

NH3

(Meridional isomer)

Br

Br

NH3

Br

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85. Octahedral complex of type [Ma3b3]n exists in two geometrical isomers, e.g. [Co(NH3)3(NO2)3]. 86. The octahedral complexes of the type Ma6 and Ma5b do not show geometrical isomerism due to the

presence of plane of symmetry. 88. 3 3 3[Co(NH ) Cl ] does not have optical isomers because it has plane of symmetry. 89. [Cr(C2O4)3]3 shows optical isomerism. The two optical isomers possible are: 90. Cis isomer of [Co(en)2Cl2]+ is optically active whereas trans isomer contains a plane of symmetry and is

optically inactive.

mer or meridional

NH3

NH3

NO2

NH3 O2N

O2N Co

fac or facial

NH3

NH3

NH3

NO2 O2N

Co O2N

Ma5b

Plane of symmetry

M

b

a

a

aa

a

a

a

Ma6

Plane of symmetry

a

a a

a

M

Cr

3ox

ox

ox

ox

Cr

3–

ox

ox

l-formMirror d-form

Co

Cl

en

en

cis[CO(en)2Cl2]+

Cl

SAMPLE C

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91. 33[Co(en) ] exhibits optical isomerism.

92. Octahedral complex [PtCl2 (en)2]2+ has two geometrical isomers. It is optically active in cis-isomer. 93. [Co(en)2Cl2]+ is optically active in cis isomer. Trans isomer contains a plane of symmetry and is

therefore optically inactive. The resolution of cis isomer in d and l-form is possible. 94. 2 2[CoCl (en) ] shows geometrical as well as optical isomerism. Hence, it has the largest number of isomers. 95. Both 3 2 24Pt NH Cl Br and 3 4 2 2Pt(NH ) Br Cl have same formula but give different ions in the solution.

Hence, they are ionisation isomers. 96. Two linkage isomers are 2 2[CoCl(en) NO ]Br and 2[CoCl(en) ONO]Br because the ambidentate ligand NO2

is attached by either N or O as a donor atom. 97. SCN is an ambidentate ligand and can be linked to a metal atom(M) either by using S atom as MSCN or

N atom as MNCS. Such complexes show linkage isomerism. 98. Linkage isomerism involves ambidentate ligands attached to central metal atom/ion through different donor

atoms. NO2 group is an ambidentate ligand. 102. Following six isomers are possible for the given complex compound II II

3 4 4Cu (NH ) Pt Cl .

1. 3 4 4[Cu (NH ) ][Pt Cl ] 2. 3 3 3 3[Cu (NH ) Cl] [Pt(NH )Cl ] 3. 3 2 2 3 2 2[Cu (NH ) Cl ][Pt (NH ) Cl ]cis 4. 3 2 2 3 2 2[Cu (NH ) Cl ][Pt (NH ) Cl ] trans 5. 3 3 3 3[Cu (NH )Cl ][Pt(NH ) (Cl)] 6. 3 4 4[ Pt (NH ) ] [Cu Cl ]

Cl

ClCo

+

l-form en

en

Mirror

+

Optically inactivetrans-form

Cl

ClCo

en +

d- form

en

Optically activecis-forms

Cl

Cl

Co en en

Co

en

en

en

3+

Co

en

en 3+

en

d-form Mirror l-form

SAMPLE C

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104. Solvate (Hydrate) isomers have same molecular formula but different number of molecules of H2O inside and outside the coordination sphere.

105. In square planar complex [Pt(Cl) (NO2) (NO3) (SCN)2–], since all the four ligands are different, three geometrical isomers are possible. If one of the ligands is ambidentate, then 2 3 = 6 geometrical isomers are possible. Here two ligands 2NO and SCN– are ambidentate, hence total 12 isomers are possible.

106. As per Sidgwick’s electronic theory, the ligands donate the electron pair(s) to the central metal atom/ion

and thus form a number of coordinate bonds. 107. Each of the four carbonyl ligands donate one electron pair. Therefore four carbonyl ligands donate four

electron pairs to Ni. 108. Z = atomic number of (Zn) metal = 30 X = number of electrons lost during oxidation of metal to metal ion = 2 Y = number of electrons donated by ligands = (4 2) = 8 EAN = Z X + Y= 30 2 + 8 = 36 109. Z = atomic number of (Cu) metal = 29 X = number of electrons lost during oxidation of metal to metal ion = 3 Y = number of electrons donated by ligands = (6 2) = 12 EAN = Z X + Y = 29 3 + 12 = 38 110. By EAN rule, EAN = Z X + Y

Metal Complex Z X Y EAN Pt [Pt(NH3)6] 4+ 78 2 8 86 Fe [Fe(CN)6]4- 26 2 12 36 Zn [Zn(NH3)4]2+ 30 2 8 36 Cu [Cu(NH3)4]2+ 29 2 8 35

The EAN of [Cu(NH3)4]2+ is different from the atomic number of next rare gas, i.e. Kr. 112. Hybridization of vacant metal orbitals results in formation of equal number of hybrid orbitals. 115. Complexes with sp3d hybridisation show trigonal bipyramidal geometry. 117. 119.

Complex Geometry Magnetic Property

[Ni(CN)4]2 square planar Diamagnetic [NiCl4]2 Tetrahedral Paramagnetic [Ni(CO)4] Tetrahedral Diamagnetic

Ni + 4 CO COOC

Ni

CO

CO

Mn3+(3d4)

In strong field: [Mn(CN)6]3–

(Octahedral) d2sp3

3d 4s 4p

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121. 122. 124. In [Ni(CN)4]2, Ni has +2 oxidation state. Ni2+ [Ar] 3d8 4s0

125. [Pt(NH3)4]Cl2 forms a square planar complex with dsp2 hyrbidisation. Pt2+ [Pt(NH3)4]Cl2 126. Complex ion Hybridization of central metal ion 4

6[Fe(CN) ] 2 3d sp (inner) 4

6[Mn(CN) ] 2 3d sp (inner) 3

3 6[Co(NH ) ] 2 3d sp (inner) 2

3 6[Ni(NH ) ] 3 2sp d (outer) 127. Ni2+ : [NiCl4]2 : Cl is a weak field ligand. So, there is no pairing of electrons. Number of unpaired electrons = 2 [Ni(CO)4] :

[Ni(CO)4] = ↿⇂ ↿⇂↿⇂↿⇂ ↿⇂ xx

sp3 hybridization

xx xx xx

[NiCl4]2– = ↿⇂ ↿⇂ ↿⇂ xx

sp3 hybridization

xx xx xx

Co3+

d2sp3 hybridization

↿⇂ ↿⇂ ↿⇂ xx xx xx xx xx xx

4p4s3d ↿⇂

[Co(NH3)6]3+

Fe2+

d2sp3 hybridization


4p 4s3d

↿⇂

[Fe(CN)6]4

[Ni(CN)4]2– =

dsp2 hybridization

4d 5s ↿⇂ ↿⇂ ↿⇂

×× ↿⇂ ↿⇂ ↿⇂ ↿⇂ 4d

×× 5s 5p

dsp2 hybridisation

×× ××

4s 3d

3d (sp3)

4s:

CO

:

CO

:

CO

4p :

CO

3d (sp3)

4s :

Cl

:

Cl

:

Cl

4p :

Cl

SAMPLE C

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CO is a strong field ligand. So there is a pairing of electrons Number of unpaired electrons = 0 [Cu(NH3)4]2+ : NH3 is a strong field ligand. So there is a pairing of electrons. Number of unpaired electrons = 1 128. 129. 130. Valence bond theory is unable to explain the spectral properties of complexes. 134. Crystal field stabilization energy (CFSE) = 4 Dq (number of electrons in t2g orbital) + 6 Dq (number of electrons in eg orbital)

Complex Outer electronic configuration CFSE [Mn(H2O)6]3+ 3d4 3 1

2g g(t e ) 4Dq(3) + 6Dq(1) = 6 Dq

[Fe(H2O)6]3+ 3d5 3 22g g(t e ) 4Dq(3) + 6Dq(2) = 0

[Co(H2O)6]2+ 3d7 5 22g g(t e ) 4Dq(5) + 6Dq(2) = 8 Dq

[Co(H2O)6]3+ 3d6 4 22g g(t e ) 4Dq(4) + 6Dq(2) = 4 Dq

137. CN > OH > F > Br is the order of field strength of the given ligands. 139. Magnetic moment of 2.84 BM corresponds to two unpaired electrons. = n(n 2) = 2(2 2) = 2.84 BM d4 in strong ligand field contains two unpaired electrons. 140. 5 2Mn 3d 4s 2 5Mn 3d In presence of weak field ligand, there will be no pairing of electrons. So it will form a high spin complex.

i.e., the number of unpaired electrons = 5. 141. According to the spectrochemical series, the weakest ligand among the given is Cl. Weak field ligands

cause lesser crystal field splitting. Hence, [CoCl6]3 has minimum magnitude of o (CFSE).

4s:

NH3

3d :

NH3

(dsp2) :

NH3

4p :

NH3

low spin d4

p sd

No. of unpaired electrons = 2

xx xxxx xx xx xx

d2sp3 hybridization

↿⇂

↿⇂ ↿⇂d2sp3 hybridization

xx xx p s

xx xx xx xx

No. of unpairedelectron = 1

d low spin d5

high spin d6

p d

(No. of unpairedelectrons = 4)

xx xx s

xx xxxx xx

sp3d2 hybridization

d ↿⇂

SAMPLE C

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142. In [Co(CN)6]3–, Co has +3 oxidation state. Co3+ [Ar] 3d6 CN being strong field ligand causes pairing up of d-electrons. Thus, [Co(CN)6]3 has no unpaired electrons and will be in a low-spin configuration (i.e., it forms low spin

complex). 143.

Complex Number of unpaired electrons Oxidation state of metal [MnCl4]2− 5 + 2 [CoCl4]2− 3 + 2 [NiCl4]2− 2 + 2 [ZnCl4]2− 0 + 2

144. When Δ0 > P (Pairing energy), fourth electron occupies t2g level. This results in electron pairing. This

results in formation of low spin or strong field octahedral complex. 145. CFSE for octahedral (o) and tetrahedral (t) complexes are related to each other by formula:

t = 49o

t = 49

18,000 cm1

t = 8,000 cm1 147. In crystal field theory, only d orbitals of central metal are considered. There is no explanation for s and p

orbitals. 148. Crystal field splitting energy (E) = hc

h = Planck’s constant = 6.63 1034 Js c = speed of light = 3.00 108 m/s = wavelength = 600 nm

E = 34 8

9(6.63 10 Js)(3.00 10 m/s)

(600nm)(1 10 m/ 1nm)

= 3.32 1019 J/ion = (3.32 1019 J/ion) (6.02 1023 ions/mol) = 199,864 J/mol = 199.9 kJ/mol 200 kJ/mol 149. The strength of ligands is in order en > NH3 > H2O. Therefore, the wavelength of absorption will be

[Co(en)3]3+ < [Co(NH3)6]3+ < [Co(H2O)6]3+.

Co3+ion En

ergy

(3d6)

Large value of o

Low spin complex (Strong field ligand)

dx2 y

2 dz2

dxy dyz dxz

[Co(CN)6]3

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150. The increasing order of energies of light is red < yellow < green < blue. Higher the energy of light absorbed, higher is the ligand strength. Hence, the increasing order of ligand strength is L1 < L3 < L2 < L4. 151. In 26TiF , Ti4+ : 3d0, colourless; as there is no unpaired electron.

In 36CoF , Co3+ : 3d6, coloured; as there are 4 unpaired electrons. In Cu2Cl2, Cu+ : 3d10, colourless; as there is no unpaired electron. In 2

4NiCl , Ni2+ : 3d8, coloured; as there are 2 unpaired electrons.

Thus, among the given options, two species viz. 26TiF and Cu2Cl2 are colourless. 153. Na2[CdCl4] does not contain any unpaired electrons. 154. [CoF6]3, [Co(H2O)6]3+ and [Co(H2O)3F3] are d6 high spin complexes in which Co3+ contains four unpaired

electrons. Hence, they are paramagnetic. 155. The configuration of 2Ni has two unpaired electrons. So [Ni(H2O)6]2+ is paramagnetic. [Ni(CO)4], and

[Zn(NH3)4]2+ do not have unpaired electrons. Hence, they are diamagnetic. [Co(NH3)6]+3 is d6 low spin complex. Hence, it is diamagnetic.

156. CO, NH3 and CN are strong ligands, thus they induce pairing of electrons and their complexes are diamagnetic.

Cl is a weak ligand, thus it does not induce the pairing of electrons and its complex is paramagnetic. 158. Due to the presence of one and five unpaired electrons in [Fe(CN)6]3 and [FeF6]3 respectively, both are

paramagnetic. 159. 2

2 6[Fe(H O) ] , 32 6[Cr(H O) ] , 2

2 6[Cu(H O) ] and 22 6[Zn(H O) ] have 4, 3, 1, 0 unpaired electrons

respectively. Since, [Fe(H2O)6]2+ has maximum number of unpaired electrons, it has the highest paramagnetism.

160. The magnetic moment of a substance increases with increase in the number of unpaired electrons. For the complexes [Fe(CN)6]3 and [Fe(H2O)6]3+, the central metal ion is Fe3+ with outer electronic

configuration 3d5. [Fe(CN)6]3 has magnetic moment of a single unpaired electron while [Fe(H2O)6]3+ has magnetic moment of five unpaired electrons.

For the complexes [Co(CN)6]3 and [CoF6]3, the central metal ion is Co3+ with outer electronic configuration 3d6. [Co(CN)6]3 is diamagnetic (no unpaired electrons) while [CoF6]3 has magnetic moment of four unpaired electrons.

Hence, among the given complexes, [Fe(H2O)6]3+ has the highest magnetic moment value.

5s0

Cd2+

4d10

(6 electrons of Cobalt)

d2sp3 hybrid orbitals (12 electrons of NH3 ligands)

(B) Co3+ in [Co(NH3)6]3+


sp3 hybrid orbitals (8 electrons of CO ligands)

(A) Ni in [Ni(CO)4]


sp3 hybrid orbitals (8 electrons of Cl ligands)

(D) Ni2+ in [NiCl4]2–


dsp2 hybrid orbitals (8 electrons of CN ligands)

(C) Ni2+ in [Ni(CN)4]2–

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161. Magnetic moment () = n (n 2) (n = number of unpaired electrons)

1.73 = n (n 2) n = 1 So, compound must contain one unpaired electron. The compound [Cu(NH3)4]2+ contains one unpaired

electron. 162. [FeF6]3 is an outer orbital complex involving sp3d2 hybridisation. It has paramagnetic moment of 5

unpaired electrons. [FeF6]3 = Number of unpaired electrons (n) = 5 Magnetic moment n(n 2) 5(5 2)

35 = 5.91 BM. 163. [CoF6]3 involves sp3d2 hybridisation. Thus, [CoF6]3 ion shows magnetic moment corresponding to four unpaired electrons, and it can be

calculated by using the ‘spin – only’ formula, i.e.,

= n n + 2 where, n = number of unpaired electrons

= 4(4 2) = 24 = 4.89 BM 164. The magnetic moment of 3.87 BM corresponds to 3 unpaired electrons. = n(n 2) = 3(3 2) = 15 = 3.87 BM. In [Co(H2O)6]2+, Co(II) has three unpaired electrons. 165. Magnetic moment () = n n 2 , (n = number of unpaired electrons) Therefore, complexes having same number of unpaired electrons will have same magnetic moment. By

considering H2O and Cl to be weak field ligands:

Complex Central metal ion No. of unpaired electrons [Cr(H2O)6]2+ Cr2+ (3d4) 4 [CoCl4]2 Co2+ (3d7) 3 [Fe(H2O)6]2+ Fe2+ (3d6) 4 [Mn(H2O)6]2+ Mn2+ (3d5) 5

[Cr(H2O)6]2+ and [Fe(H2O)6]2+ have the same magnetic moment. 166. 5Fe (CO) has dsp3-hybridisation; so it shows trigonal bipyramidal geometry.

[CoF6]3

(Co3+ : 3d6)

4s 4d

3d 4p

sp3d2 hybridisation

4p 3d

4s 4d

sp3d2 hybridisation

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169. The given diagram explains synergic effect in metal carbonyls which is due to the formation of C M sigma bond and M C pi bond between metal and CO ligands.

170. [Fe(CO)5] EAN = 26 0 + 10 EAN = 36 In given complex only one central metal atom is present and it follows EAN rule, so it is mononuclear. 171. 172. In metal carbonyls, the donation of a pair of electrons from the filled d-orbital of metal to vacant

antibonding * orbital of CO ligand (back bonding) reduces the bond order of CO bond from the triple bond to double bond. This results in increase in CO bond length from 1.128 Å to 1.15 Å. Among these four options, the central metal ion Fe2 in [Fe(CO)4]2 has maximum electron density and back bonding will be more pronounced in this complex. Thus, CO bond length in [Fe(CO)4]2 is the longest.

173. CO has strong M (metal) L (ligand) bonding ability. Since, it can accept electron pairs from metal ion, it is a acid ligand.

174. Stability of complex compound formed will be higher for the higher value of log K. 179. The complexes formed by chelation are more stable than the similar complexes formed by using

monodentate ligands. This stabilization is known as chelate effect. In the given examples of complex species, the complex containing chelate ligand (C2O4

2 i.e, oxalate) is [Fe(C2O4)3]3. 180. [Cd (CN)4]2– gives precipitate with H2S as it is less stable than [Cu (CN)4]2–. 181. Ag+ reacts with excess of CN to form complex [Ag(CN)2] with coordination number 2. 182. AgCl dissolves in ammonia to form [Ag(NH3)2]Cl complex. AgCl + 2NH3 [Ag(NH3)2]Cl Diammine silver (I) chloride (water soluble) 183. CuSO4 reacts with KCN solution to form K2[Cu(CN)4] and K2SO4. 4KCN + CuSO4 K2[Cu(CN)4] + K2SO4 Potassium tetracyanidocuprate (II) 188. D-penicillamine is a chelating ligand and it is used in the removal of excess copper by formation of stable

complex with the copper. 190. The hexadentate ligand EDTA forms stable complexes with Ca2+ and Mg2+ ions which are responsible for

the hardness of water. Hence, it is used in the simple titration for estimating the hardness of water. 192. Wilkinson’s catalyst is used as homogeneous catalyst in the hydrogenation of alkenes. 194. [Ni(CO)4] is a zero valent metal complex as the oxidation state of Ni is 0. The oxidation state of Cu, Pt and

Fe in [Cu(NH3)4]SO4, [Pt(NH3)2Cl2] and K3[Fe(CN)6] are +2, +2 and +3 respectively. 195. A mixture of three parts of concentrated hydrochloric acid and one part of concentrated HNO3 is known as

aqua regia.

Co Co

C CO

COCO

OCOCOC

O

CO

C = C + H2 (Ph P) RhCl3 3 C C

H H

SAMPLE C

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Gold (Au) dissolves in aqua regia to form the complex ion, [AuCl4]. 196. 2 2[Pt Cl (en) ] shows geometrical as well as optical isomerism. 197. [Co(H2O)5Cl]Cl2.H2O i = 3 [Co(H2O)4Cl2]Cl.2H2O i = 2 [Co(H2O)3Cl3].3H2O i = 1 [Co(H2O)6]Cl3 i = 4 ΔTf ∝ i Depression in freezing point is maximum for [Co(H2O)6]Cl3 and minimum for [Co(H2O)3Cl3].3H2O. Hence, maximum freezing point is shown by complex [Co(H2O)3Cl3].3H2O. 198. 1 litre of mixture X contains 0.01 mol of [Co(NH3)5SO4]Br and 0.01 [Co(NH3)5Br]SO4. On using one litre solution, we will get 0.01 mole Y and 0.01 mole Z. 3 5 4 3

0.01mole[Co(NH ) SO ]Br AgNO 3 5 4 3

0.01mole (Y)[Co (NH ) .SO ]NO AgBr

3 5 2 4 2

0.01mole[Co (NH ) Br ]SO BaCl 3 5 2 4

0.01mole (Z)[Co (NH ) Br]Cl BaSO

199. In the complex ion 3 0 1III

35

Co N H Cl

x, charge on the complex ion:

x = 3 + (0 5) + (1) x = 3 1 = 2 So it will combine with 2Cl to produce [Co(NH3)5Cl]Cl2 complex. 200. = n(n 2) = 0

n = 0 201. In [Ni(CO)4], Ni has 0 oxidation state

Cl

ClCl

en

Cl

Pt

Cl

Optical isomers

Pt en

en

en

Cl

Pt

ClCl

Geometrical isomers

en en en Pt

en Cl

Nitrateion

Au + 4H+ + 3NO + 4Cl 4AuCl

+ NO + 2H2OGold Chloride

ion Nitricoxide

Water Tetrachlorido aurate (III)

ion

Ni =

3d

4s 4p

SAMPLE C

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In [Ni(CN)4]2–, Ni has +2 oxidation state Ni2+ 1s2 2s2 2p6 3s2 3p6 3d8 4s0

202. [Ni(CO)4] is tetrahedral due to sp3 hybridization. It contains no unpaired electrons. Hence, it is diamagnetic. 203.

Complex Oxidation state of Ni Hybridisation Geometry Magnetic property K4[Ni(CN)4] 0 sp3 tetrahedral diamagnetic

K2[Ni(CN)4] + 2 dsp2 square planar diamagnetic

K2[Ni(Cl)4] + 2 sp3 tetrahedral paramagnetic

204. [Co(NH3)6]3+ forms octahedral inner orbital complex with d2sp3 hybridisation.Oxidation number of Co 3

3d 4s 4p 3 6Co (d )

33 6[Co(NH ) ]

Due to paired electrons, it is diamagnetic. [Co(H2O)6]2+ and [Ni(NH3)6]2+ are outer orbital complexes with sp3d2 hybridisation. [Cr(NH3)6]3+ formsinner orbital complex with d2sp3 hybridisation but it is paramagnetic, as it contains 3 unpaired electrons.

205. 1 221Sc [Ar] 3d 4s ; 3 0 0Sc [Ar] 3d 4s . Since there are no unpaired electrons in d subshell, it is diamagnetic and colourless.

1. How many geometrical isomers are possiblefor the square-planar complex[Pt(NO2)(Py)(NH3)(NH2OH)]NO2?(A) Four (B) Five(C) Two (D) Three

2. The coordination number of Fe in[Fe(C2O4)3]3 is _______.(A) 2 (B) 3 (C) 4 (D) 6

3. Which of the following is an example of dsp2

hybridisation?(A) 2

2 6[Fe(H O) ] (B) 24[Ni CN ]

(C) 36CoF (D) [Fe(CN)6]3

4. Two complexes PtCl4.2NH3 and PtCl4.2KCldo not give precipitate of AgCl with AgNO3solution. The conductance studies indicate

presence of zero and 3 moles of ions respectively in their aqueous solutions. The structures of these complexes are _______. (A) [Pt(NH3)2Cl2]Cl2, K2[PtCl6](B) [Pt(NH3)2]Cl4, K2[PtCl6](C) [Pt(NH3)2Cl4], K2[PtCl6](D) [Pt(NH3)2Cl4], K2[PtCl5]Cl

5. Match the coordination compounds given incolumn I with the central metal atoms given incolumn II.

Column I Column II i. Cisplatin a. Rhodium ii. Chlorophyll b. Platinum iii. Haemoglobin c. Cobalt iv. Cyanocobalamine d. Magnesium

e. Iron

Topic Test

[Ni(CN)4]2– =

dsp2 hybridization

[Ni(CO)4]

sp3 hybridization

d2sp3 (Inner)


↿⇂ ↿↿↿ ↿

SAMPLE C

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(A) i - a, ii - c, iii - d, iv - b (B) i - d, ii - c, iii - e, iv - a (C) i - c, ii - b, iii - e, iv - d (D) i - b, ii - d, iii - e, iv - c 6. The formula of the complex

triamminetriaquachromium (III) chloride is _______ .

(A) [Cr(NH3)3(H2O)3]Cl3 (B) [Cr(NH3)3(H2O)]Cl3 (C) [Cr(NH3)3(H2O)3]Cl2 (D) [Cr(NH3)3(H2O)3]Cl 7. Which of the following will NOT show

geometrical isomerism? (A) [Ma2b2]n square planar (B) [Ma2b4]n octahedral (C) [Ma2bc]n square planar (D) [Ma2b2]n tetrahedral 8. Which of the following is homoleptic complex? (A) KClMgCl26H2O (B) [Co(NH3)4Cl2]+ (C) [Ni(CN)4]2 (D) [Ni(NH3)4Cl2] 9. Identify the INCORRECT statement for the

behaviour of ethane-1,2-diamine as a ligand. (A) It is a neutral ligand. (B) It is a didentate ligand. (C) It is a chelating ligand. (D) It is an unidentate ligand. 10. Which of the following is/are diamagnetic? (A) [NiCl4]2 (B) [Ni(CO)4] (C) [MnCl6]3

(D) [FeF6] 11. Which of the following complexes does NOT

show linkage isomerism? (A) [Co(NH3)5 (NO2)]2+ (B) [Co Cl(en)2NO2]+ (C) [Cr(NH3)5SCN]2+ (D) [Fe(en)2Cl2]+

12. Which of the following is INCORRECT? (A) [CoF6]3 is an outer orbital complex. (B) [Mn(CN)6]3 is an inner orbital complex. (C) [Co(C2O4)3]3 is an inner orbital complex. (D) [Co(NH3)6]3+ is an outer orbital complex. 13. When aqueous solution of nickel (II) chloride

is treated with ethylenediamine in the molar ratio en:Ni = 3:1, green colour of the solution changes to violet due to formation of _______.

(A) [Ni(H2O)6] 2+ (B) [Ni(en)3]2+ (C) [Ni(H2O)4Cl2]4+ (D) [NiCl2(en)2]2+

14. In which of the following complex ions, the oxidation number of the central metal ion is +3?

(A) [CoCl(NH3)5]2+ (B) [Ni(NH3)4]2+ (C) [Cu(CN)4]3 (D) [Zn(OH)4]2

15. Which of the following compound shows 3

different types of structural isomerism? (A) [Co(NH3)3 (H2O) Cl(NO2)] Br

(B) [Co(NH3)4 (H2O) (Cl)] Br (C) [Co(NH3)4 (Cl) (NO2)] Br (D) [Co(NH3)2 (H2O)2 (Cl2)] Br 16. Which of the following systems has maximum

number of unpaired electrons? (A) d5 (high spin octahedral) (B) d9 (octahedral) (C) d7 (octahedral) (D) d6 (low spin octahedral) 17. CORRECT IUPAC name of the compound

K2[Cu(ox)2 H2O] is _______. (A) Dipotassium aquadioxalatocopper (II) (B) Potassium aquabis (oxalate)cuprate

(III). (C) Aquabis (oxalato)copper (II)

potassium (I) (D) Potassium aquadioxalatocuprate (II) 18. Find the least stable complex from given data.

ion Cu2+ Ni2+ Fe2+ Cd2+

ionic radius (pm) 69 78 83 97 (A) [Cu(H2O)6]2+ (B) [Ni(H2O)6]2+ (C) [Fe(H2O)6]2+ (D) [Cd(H2O)6]2+ 19. In which of the following pairs, both the

complexes show optical isomerism? [AIIMS 2005]

(A) cis- 32 4 2 2[Cr(C O ) Cl ] , cis- 3 4 2[Co(NH ) Cl ]

(B) 3 3[Co(en) ]Cl , cis- 2 2[CoCl (en) ]Cl (C) [PtCl(dien)]Cl , 2

2 2[NiBr Cl ]

(D) 3 3 3 3[Co(NH ) (NO ) ] , cis- 2 2[PtCl (en) ] 20. Which of the following has a magnetic

moment of 3.87 BM? (A) [CoF6]3 (B) [Fe(CN)6]3

(C) [Ti(H2O)6]3+ (D) [Cr(H2O)6]3+

1. (D) 2. (D) 3. (B) 4. (C) 5. (D) 6. (A) 7. (D) 8. (C) 9. (D) 10. (B) 11. (D) 12. (D) 13. (B) 14. (A) 15. (A) 16. (A) 17. (D) 18. (D) 19. (B) 20. (D)

Answers to Topic Test

SAMPLE C

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