<<

Coordination compounds

CHAPTER COORDINATION COMPOUNDS 10 LEARNING OBJECTIVES (i) Appreciate the postulates of Werner’s theory of coordination compounds. (ii) Know the meaning of the terms: coordination entity, central atom/, ligand, coordination number, coordination sphere, coordination polyhedron, oxidation number, homoleptic and heteroleptic. (iii) Learn the rules of nomenclature of coordination compounds. (iv) Write the formulas and names of mononuclear coordination compounds. (v) Define different types of isomerism in coordination compounds. (vi) Understand the nature of bonding in coordination compounds in terms of the Valence Bond and Crystal Field theories. (vii) Learn the stability of coordination compounds. (viii) Appreciate the importance and applications of coordination compounds in our day to day life.

INTRODUCTION Coordination compounds are a special class of compounds in which the central metal atom is surrounded by or molecules beyond their normal valency. These compounds are also referred to as complex compounds or simply complexes. In the modem terminology, these compounds are called coordination compounds. These compounds are widely present in the minerals, plants and animals, and play many important functions. Many biologically important compounds are coordination compounds in which complicated organic species are bound to the metal ions. The common examples are: haemoglobin, which is a coordination compound of iron, chlorophyll, which is a coordination compound of magnesium, Vitamin B12 which is a coordination compound of cobalt etc. The coordination compounds are finding extensive applications in metallurgical processes, analytical and medicinal chemistry. Many complex metal oxides and sulphides which constitute minerals are solid-state coordination compounds.

MOLECULAR OR ADDITION COMPOUNDS : When solutions containing two or more simple stable compounds in molecular proportions are allowed to evaporate, crystals of new substances are obtained. These substances are termed molecular or addition compounds. Some common examples are as follows. Simple stable compounds Addition or molecular compound CuSO4 + 4NH3  CuSO4.4NH3 AgCN + KCN  KCN.AgCN 4KCN + Fe(CN)2  Fe(CN2).4KCN K2SO4 + Al2 (SO4)3 + 24H2O  K2SO4. Al2(SO4)3.24H2O Alum The molecular or addition compounds are of two types. Double salts or lattice compounds and coordination or complex compound. (A) Double salts or lattice compounds : The addition compounds which are stable in solid state only but are broken down into individual constituents, when dissolved in water are called double salts or lattice compounds.Their solution have the same properties as the mixture of individual compounds. For example when carnallite (KCl. MgCl2. 6H2O) is dissolved in water it 2+ + exhibit the properties of KCl and MgCl2. Mohr’s salt [FeSO4.(NH4)2SO4.6H2O] when dissolved in water gives Fe , NH4 and 2– SO4 ions in the solution, which gives the tests of all these ions. (B) Coordination or complex compounds : The addition compounds in which some of the constituent ions or molecules lose their identity and when dissolved in water they do not break up completely into individual ions are called coordination compounds. The properties of their solutions are different than those of their constituents. In such compounds there is complex ion which is a central metal ion with lewis bases attached to it through coordinate covalent bonds. On the basis of stability of complex ion, complex ions are further divided as follows. (a) Perfect complexes : Those in which complex ion is fairly stable and is either not dissociated or feebly dissociated in solution + 4– state, Ex. K4 [Fe(CN)6]  4K + [Fe(CN)6]  Fe2+ + 6CN– (feebly dissociated) 4– The ferrocyanide ion [Fe(CN)6] is so in significantly dissociated that it can be considered as practically undissociated and does not give the test of Fe2+ or CN¯ ions. Gyaan Sankalp 1 Coordination compounds (b) Imperfect complexes: Those in which complex ion is less stable and is reversibly dissociated to give enough simple ions and thus imparts their tests, Ex. + 2– K2[Cd(CN)4]  2K + [Cd(CN)4]  Cd + 4CN– (appreciably dissociated) An imperfect complex may be too unstable to exist and may be completely dissociated in solution, it then becomes a double salts. Cl WERNER’S COORDINATION THEORY HN3 NH Alfred Werner (considered as the father of coordination chemistry) studied 3 the structure of coordination complexes such as CoCl3.6NH3 and CuSO4. 4NH3 in 1893. According to him. (i) Each metal in coordination compound possesses two types of valencies : HN3 Co NH3 (a) Primary valency or principal valencies or ionisable valencies. (b) Secondary valency or non ionisable valencies. Cl (ii) Primary valency are satisfied by anions only. The number of primary valencies NH NH3 depends upon the oxidation state of the central metal. It may change from one 3 Cl compound to other. These are represented by dotted lines between central metal atom and anion [Co(NH3 ) 6 ]Cl 3 (iii) Secondary valencies are satisfied only by electron pair donor, the ions or the neutral species. These are represented by thick lines. (iv) Each metal has a fixed number of secondary valencies also referred as coordination number. The coordination number depends mainly on the size and the charge on the central atom. The maximum number of ions or molecules that the central atom can hold by secondary valencies is known as coordination number. (v) The ion attached to primary valencies possess ionising nature whereas the ions attached to secondary valencies do not ionise when the complex is dissolved in a solvent. (vi) Every central ion tends to satisfy its primary as well as secondary valencies. (vii) The secondary valencies are directional and are directed in space about the central metal ion. The primary valencies are non- directional. The presence of secondary valencies given rise to stereoisomerism in complexes. Initially, Werner had pointed out coordination number of a metal atom to be four or six. The six valencies were regarded to be directed to the corners of a regular octahedron circumscribed about the metal ion. For metals having four coordination number, the four valencies are either arranged in a planar or tetrahedral nature. Thus on the basis of werner theory, the CoCl3. 6NH3 is called hexaminecobalt (III) chloride because there are six ammonia ligands and the cobalt is in the +3 oxidation state, i.e. cobalt has three primary valencies and six secondary valencies. Now, it has been proposed that coordination number of a metal may be any whole number between 2 and 9. Definitions of some important terms pertaining to coordination compounds : (A) Central ion : (Centre of coordination) (i) The cation to which one or more neutral molecules or anions are attached is called as centre of coordination. (ii) Since the central ion acts as an acceptor and thus has to accommodate electron pairs donated by the donor atom of the ligands, it must have empty orbitals. (iii)This explains why the transition metals having empty d-orbitals form coordination compounds very readily. 2+ 3– 2+ 3+ (iv) In the complexes [Ni(NH3)6] and [Fe(CN)6] , Ni and Fe respectively are the central ions. (B) Ligand : (i) The neutral molecules, anions or cations which are directly linked with the central metal atom or ion in a complex ion are called 2+ 3+ ligands. For example in [CoCl(NH3)5] ligand is NH3, central ion is Co . (ii) Ligand donate a pair of electrons to the central atom. (iii)Ligands are attached to the central metal ion or atom through coordinate bonds or dative linkage. ..  – – – (iv) Ligands are normally polar molecules like NH3 ; HO:2 or anions much as Cl , OH , CN etc. which contain atleast one ensured pair of valency electrons. (v) With few exceptions free ligands have lone pair of electron that is not engaged in bonding for example ..  ,  :CN: : Cl.. : (vi) Ligands may be termed as Lewis and central ion as Lewis acid. Types of ligands on the basis of number of donor atoms : Mono or unidentate ligands : They have one donor atom, i.e. they supply only one electron pair to central metal atom or ion. Ex. Neutral molecule % H2O , ROH, NH3, CO, NO, R3P, C5H5N etc. – – – –2 – Anion % Cl , OH , CN , SO4 , C6H5 etc. + + Cation % NO , NH2.NH3 etc.

2 Gyaan Sankalp Coordination compounds Ligands are three types : (i) Cation ligands (ii) Anion ligands (iii) Neutral ligands Monodentate Ligands : 1. Negative Ligand 2. Positive Ligand

S. No. Formula Name of ligand S. No. Formula Name of ligand – 01 X Halo  Nitrosonium 01 NO 02 OH – Hydroxo  02 NO Nitronium – 2 03 CN Cyano  NH Ammonium 04 O –2 Oxo 03 4 –2 Peroxo Cl+ Chloronium 05 O2 04 – ..  06 NH Amido Hydrazinium 2 05 NHNH2 3 07 S – – Sulphido – – – 08 N Nitrido 3. Neutral Ligand 09 P – – – Phosphido 10 CH COO – Acetato 3 1 H2O Aqua (Aquo) 11 NO – Nitrato 3 2 NH Ammine – Nitro 3 12 NO2 13 ONO – Nitrito 3 CO Carbonyl –2 Carbonato 14 CO3 4 CS Thio carbonyl 15 SO –2 Sulphato 4 5 NO Nitrosyl –2 Sulphito 16 SO3 – 6 17 CNS Thiocyanato NS Thionitrosyl 18 S O –2 Thiosulphato 2 3 7 C6H5N Pyridine (py)

Bidentate ligands : Ligands which have two donor atoms and have the ability to link with central metal ion at two positions are called bidentate ligands. Some examples are : H H

CH2 N O O N N CH2 N C C H H O O Ethylenediamine (en) Oxalate (ox) 1,10-Phenanthroline (–phen)

H

CH2 N H N N O C O C O O O Glycinato (Gly) 2,2'-Dipyridine (Dipy) Carbonato

Gyaan Sankalp 3 Coordination compounds Tridentate ligands : The ligands having three donor atoms are called tridentate ligands. H H

H N N H

N N HC2 CH2 N HC2 N HC2

H Diethylene triamine (dien) 2,2',2''Terpyridine (terpy) Tetradentate ligands : These ligands possess four donor atoms. Examples are : H H CH2 COO

N (CH2 ) 2 N (C ) 2 CH2 COO C H H 2 2 ) ( 2 N N N H H CH2 COO H H Nitriloacetato Triethylene tetramine (Trien) Pentadentate ligands : They have five donor atoms. For example, ethylenediamine triacetate ion. O C O

O

O C O O C CH2

CH2 N (CH2 ) 2 N CH2 N CH CH N H 2 2 Hexadentate ligands : They have six donor atoms. CH2 CH2 The most important example is ethylenediamine CH2 tetra acetate ion. HC2 O C O O C O

C O O C O O Ethylenediamine tetra acetate ion (EDTA)4–

Chelating ligands : (i) Polydentate ligands whose structures permit the attachment of two or more donor sites to the same metal ion simultaneously, thus closing one or more rings are called chelating ligands and the compounds formed are known as chelate compounds. (ii) A chelate may be defined as a ring structure formed by the combination of a polydentate ligand having two or more donor atoms with a metal ion forming part of the ring. (iii) The process of formation of chelates is known as chelation. (iv) Chelate complexes are more stable than ordinary complexes in which the ligand is a monodenate. (v) This increased stability of a compound due to chelation is called the chelate effect. (vi) In the complex ion given below, 5 membered rings are formed. So all these are called chelate complexes.

CH2 H2 N NH2 CH 2 Cu

CH2 H 2 N NH2 CH 2

4 Gyaan Sankalp Coordination compounds (vii) Generally the chelate complexes with 5 or 6 membered rings are more stable. (viii) Out of these, 5 membered rings are very stable when they involve saturated ligands. (ix) On the other hand 8-membered ring structures acquire maximum stability when they involve unsaturated ligands containing conjugate double bond. This is due to the resonance effect. Involving metal d-orbitals and ligand p-orbital electrons. Bridging ligands : The monodentate ligands which can coordinate simultaneous with more that one metal ion are called bridging – – – – ligands. For example, – OH , – NH2 , NO2 , Cl , CO etc. Ambidentate ligands : The monodentate ligands which can coordinate with the central atom or iron through more than one site are called Ambidentate ligands. For example, – NO2 can coordinate to the central metal atom or ion through N or O. O  MONO   MNO  Nitrito Nitro Similarly – CN can coordinate through C or N and – SCN can coordinate through S or N. (C) Coordination number : (i) The number of atoms of the ligands that are directly bond to the central metal atom or ion by coordinate bonds is known as the coordination number of the metal atom or ion. (ii) It is actually the number of coordinate covalent which the ligands form with the central metal atom or ion. (iii) Some common coordination numbers exhibited by metal ions are 2,4,6.The light transition metals exhibit 4 and 6 (iv) For example, the coordination number Ni in the complex [Ni(NH3)4]Cl2 is 4 and that of Pt in the complex K2[PtCl6] is 6. (v) While calculating the coordination number of any central atom/ion, each bidentate contributes two to the coordination number. Similarly, a tridentate ligand contributes three to the coordination number and so on. For example, in the complex ion, 3– 2– 3+ [Fe(C2O4)3] , there are three bidentate (C2O4 is a bidentate ligand) ligands around Fe . Thus, the coordination number of Fe3+ is 6 (= 3 × 2). (D) Coordination sphere : (i) The central metal atom and the ligands directly attached to it are collectively termed as the coordination sphere. 3+ (ii) Coordination sphere is written inside square bracket, for example [Co(NH3)6] (iii) The central metal atom and the ligands inside the square bracket behave as a single entity. (iv) The species present in the coordination sphere are non ionizable. (v) The species present in the ionization sphere are ionisable. Co-ordination sphere Central metal ion Ligand Ni (NH3 ) 6 Cl2 Ionisation sphere Co-ordination number

(E) Coordination polyhedron : The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. The most common coordination polyhedra are octahedral, square 3+ 2– planar and tetrahedral. For example, [Co(NH3)6] is octahedral, [Ni(CO)4] is tetrahedral and [PtCl4] is square planar. (F) Charge of a complex ion : The charge number carried by a complex ion is the algebraic sum of the charges carried by the central 2+ ion and the ligands coordinated to it. For example, the complex ion [CoCl(NH3)5] , carries a net charge of +2 because it is formed by the coordination of one CO3+ ion with five molecules of ammonia and one Cl– ion. 2+ [CoCl(NH3)5] : Charge = +3 – 1 + 5 (0) = + 2 Similarly, [CoCl3(NH3)3] : Charge = +3 – 3 + 3 (0) = 0 2+ [Ni(NH3)6] : Charge = +2 + 6 × 0 = + 2 (G) Oxidation State : (i) It is a number which represents the electric charge on the central metal atom of a complex ion. 4– 3+ (ii) For example O.N. of Fe, Co and Ni in [Fe(CN)6] , [Co(NH3)6] and Ni(CO)4 is +2, +3 and zero respectively. (iii) The charge of the complex is the sum of the charges of the constituent parts. (iv) For the calculation of the oxidation number of the central metal atom, its oxidation number is assumed to be x. The algebraic sum of the oxidation numbers of the central metal atom and the ligands is equated to the charge number of the complex and the value of x is calculated. For example, 2+ [CoCl(NH3)5] : x – 1 + 5 (0) = +2  x = + 3, O.N. of Co = + 3 2+ [Ni(NH3)6] : x + 6 (0) = +2  x = + 2, O.N. of Ni = + 2

Gyaan Sankalp 5 Coordination compounds (H) Effective atomic number (EAN) : (i) In order to explain the stability of the complex. Sidgwick proposed effective atomic number. (ii)It can be defined as the resultant number of electrons with the metal atom or ion after gaining electron from the donor atoms of the ligands. (iii) The EAN generally coincides with the atomic number of next noble gas in some cases. (iv) EAN is calculated by the following relation EAN = atomic number of the metal–number of electrons lost in formation + number of electrons gained from the donor atoms of the ligands. (v) The EAN values of different metal in their respective complexes are tabulated as follows :

Complex of metal Oxidation state of metal At No. Coordination number Effective atomic number K4[Fe(CN)6] +2 26 6 (26–2) + (6 × 2) = 36 [Kr] [Cu(NH3)4]SO4 +2 29 4 (29–2) + (4 × 2) = 35 [Co(NH3)6]Cl3 +3 27 6 (27–3) + (6 × 2) = 36 [Kr] Ni(CO)4 0 28 4 (28–0) + (4 × 2) = 36 [Kr] K2[Ni(CN)4] +2 28 4 (28–2) + (4 × 2) = 34 K2[PtCl6] +4 78 6 (78–4) + (6 × 2) = 86 [Rn] K3[Cr(C2O4)3] +3 24 6 (24–3) + (6 × 2) = 33 K3[Fe(CN)6] +3 26 6 (26–3) + (6 × 2) = 35 K2[HgI4] +2 80 4 (80–2) + (4 × 2) = 86 [Rn] [Ag(NH3)2]Cl +1 47 2 (47–1) + (2 × 2) = 50 K2[PdCl4] +2 46 4 (46–2) + (4 × 2) = 52

3+ (I) Homoleptic and heteroleptic complexes : Complexes in which a metal is bound to only one kind of donor groups, e.g., [Co(NH3)6] , + are known as homoleptic. Complexes in which a metal is bound to more than one kind of donor groups, e.g., [Co(NH3)4Cl2] , are known as heteroleptic.

IUPAC NAMING OF COORDINATION COMPOUNDS : Due to wide variety of coordination compounds it is essential to use a unified system of nomenclature. Though a lot of compounds are known by their trivial names only, but IUPAC has introduced rules for the naming of coordination compounds as follows : (A) The order of listing the ions : (i) In common salts cation is named first and then the anion. (ii)In the complex ion (Cation or anion) ligands are named first followed by the name of central atom. (iii) The oxidation state of the central metal is indicated by Roman numberal in brackets immediately after its name; (0) indicates zero o.s. (iv) In case the complex is non ionic, it is named as one word e.g. [Ni(CO)4] is called tetracabonyl nickel (0) (v) The suffix - ate is added to the name of central metal forming anionic complex ion. In cationic complex ion, the name of metal (usual name) is followed by the oxidation number in bracket. However in some metals Latin named are preferred in place of English names e.g. ion as ferrate, lead as plumbate and silver as argentate. (B) Name of the Ligand : (i) If there are two or more different kinds of ligands, they are named in alphabetical order without separation by hyphen. (ii)When there are several ligands of same kind, they are listed alphabetically. (iii) Anionic ligands ending with 'ide' are named by replacing 'ide' by suffix ‘O’. Symbol Name as ligand Symbol Name as ligand – 2– Cl Chloro O2 Peroxo – – Br Bromo O2H Perhydroxo CN– Cyano S2– Sulphido 2– – O Oxo NH2 Amido OH– Hydroxo N3– Nitrido

(iv)Ligands whose names end in 'ite' or 'ate' become ‘ito’ i.e. by replacing the ending ‘e’ with ‘o’ as follows : Symbol Name as ligand 2– CO3 Carbonato 2– C2O4 Oxalato 2– SO4 Sulphato

6 Gyaan Sankalp Coordination compounds – NO3 Nitrato 2– SO3 Sulphito – CH3COO Acetato ONO– Nitrite (bonded through oxygen) Nitrito, – NO2 Nitrite (bonded through nitrogen) Nitro

(v) Neutral ligands are given the same names at the neutral molecules. For example. Ethylene diamine as a ligand is named ethylene diamine in the complex. However some exceptions to this rule are H2O Aqua NH3 Ammine CO Carbonyl NO Nitrosyl CS Thiocarbonyl

(vi) Names of positive ligands ends in 'ium' e.g. + NH4 Ammonium NO+ Nitrosonium + NH2NH3 Hydrazinium

(vii) If the number of a particular ligand is more than one in the complex ion, the number is indicated by using Greek numbers such as di, tri, tetra, penta, hexa etc. (viii) However, when the name of the ligand includes a number e.g. dipyridyl, ethylene diamine then bis tris, tetrakis are used in place of di, tri tetra etc. (ix) In case of chelating ligands or ligands having di, tri tetra etc. in their name the prefixes bis, tris, tetrakis are used before ligands placed in paranthesis. (x) In poly nuclear complexes the bridging group is indicated in the formula of the complex by separating it form the rest of complex by hyphens and adding before its name or in poly nuclear complex (a complex with two or more metal atoms) bridging ligands (which links two metal atoms) is denoted by the prefix  before its name.

NH2 (SO ) name is : (en)2 Co Co(en)2 4 2 OH

Bis (ethylenediamine) cobalt(III)--amido--hydroxobis-(ethylenediamine) cobalt(III) sulphate.

Example 1 : Name the compound : [Cr(NH3)3(H2O)3]Cl3 Sol. IUPAC name of the compound is triamminetriaquachromium(III) chloride The complex ion is inside the square bracket, which is a cation. The amine ligands are named before the aqua ligands according to alphabetical order. Since there are three chloride ions in the compound, the charge on the complex ion must be +3 (since the compound is electrically neutral). From the charge on the complex ion and the charge on the ligands, we can calculate the oxidation number of the metal. In this example, all the ligands are neutral molecules. Therefore, the oxidation number of chromium must be the same as the charge of the complex ion, +3.

Example 2 : Name the compound : [Co(H2NCH2CH2NH2)3]2(SO4)3 is named as: Sol. IUPAC name of the compound is tris(ethane-1,2–diammine)cobalt(III) sulphate The sulphate is the counter anion in this molecule. Since it takes 3 sulphates to bond with two complex cations, the charge on each complex cation must be +3. Further, ethane-1,2– diamine is a neutral molecule, so the oxidation number of cobalt in the complex ion must be +3. Remember that you never have to indicate the number of cations and anions in the name of an ionic compound.

Gyaan Sankalp 7 Coordination compounds Nomenclature of some complex ions and coordination compounds :

S.No. Formula Name 1. [Co(NH3)6Cl3 hexaamminecobalt (III) chloride 2. [CrCl2(H2O)4]NO3 tetraaquadichlorochromium (III) nitrate 3. [CoCl(NO2)(NH3)4]NO3 tetraamminechloronitrocobalt (III) nitrate 4. K3[Fe(C2O4)3] potassium trioxalatoferrate (III) 5. K3[Co(CN)5(NO)] potassium pentacyanonitrosylcobaltate (II) 6. K[PtCl3(NH3)] potassium amminetrichloroplatinate (II) 7. Na2[CrOF4] sodium tetrafluorooxochromate (IV) 8. Na2[SiF6] sodium hexafluorosilicate (IV) 9. [Cr(en)3]Cl3 tris (ethylenediamine) chromium (III) chloride. 10. [CoCl2(en)2]SO4 dichlorobis (ethylenediamine) cobalt (IV) sulphate 11. Dextro K3[Ir(C2O4)3] potassium (+) or d-trioxalatoiridate (III). 12. Na3[Fe(C2O4)3 sodium trioxalatoferrate (III) 13. [CoBr(H2O)(NH3)4] (NO3)2 tetraammineaquabromocobalt (III) nitrate 14. [Cr(H2O)6]Cl3 hexaaquachromium (III) chloride 15. Hg[Co(SCN)4] mercuric tetrathiocyanatocobaltate (II) + 16. [CoCl(ONO)(en)2] chlorobis (ethylenediamine) nitritocobalt (III) ion. 17. [Ni(CO)4] tetracarbonylnickel (0) 18. K4[Ni(CN)4] potassium tetracyanonickelate (0) 19. [CrCl2(H2O)4Cl tetraaquadichlorochromium (III) chloride 20. [Co(NO2)3(NH3)3] triamminetrinitrocobalt (III) 21. [PtCl2(NH3)2] diamminedichloroplatinum (II) 22. [Co(CO3)(NH3)5]Cl pentaamminecarbonatocobalt (III) chloride 23. [CoCl2(en)2]SO4 dichlorobis (ethylenediamine) cobalt (IV) sulphate 24. [PtCl2(NH3)4] [PtCl4] tetraamminedichloroplatinum (IV) tetrachloroplatinate (II) 25. Fe4[Fe(CN)6]3 ferric hexacyanoferrate (II) 26. [CoCl2(NH3)4]3 [Cr(CN)6] tetraammilledichlorocobalt (III) hexacyanochromate (III) 27. [Cr(PPh3) (CO)5] pentacarbonyltriphenyl phosphinechromium (0) 28. [Fe(C2H5)2] bis(cyclopentadienyl) iron (II) 29. [Ni(dmg)2] bis(dimethylglyoximato) nickel (II) 30. [Mn3(CO)12] dodecarbonyltrimanganese (O) 31. [(NH3)5Cr – OH – Cr(NH3)5]Cl5 pentaammine chromium {III)-µ-hydroxo pentaammine chromium (III) chloride 32. [Co(NH3)6]ClSO4 hexaammine cobalt (III) chloride sulphate

ISOMERISM IN COORDINATION COMPOUNDS : The compounds having same molecular formula but different physical and chemical properties on account of different structures are called isomers and the phenomenon as isomerism. (A) Classification :

Type of isomerism

Structural isomerism Stereo isomerism

Ionization Hydrate Coordination Linkage Geometrical Optical

Polymerisation

8 Gyaan Sankalp Coordination compounds (i) Ionization Isomerism % (a) This is due to exchange of groups or ion between the coordinating sphere and the ionization sphere. (b) This type of isomerism gives different ions in solution because only the ion from ionisation sphere is releases in solution. –2 – (c) The example is red violet [Co(NH3)5]SO4 and red [Co(NH3)5SO4]Br which give. SO4 and Br respectively. + – [Co(NH3)5SO4]Br  [Co(NH3)5SO4] + Br 2+ –2 [Co(NH3)5Cl]SO4  [Co(NH3)5Cl] + SO4 (ii) Hydrate isomerism : (a) This type of isomerism is due to presence of different number of water molecules inside and outside the coordination sphere. (b) Examples : Violet [Cr(H2O)6]Cl3 Green [Cr(H2O)4Cl2]Cl.2H2O Dark green [Cr(H2O)5Cl]Cl2.H2O (iii) Polymerisation isomerism : (a) This type of isomerism exists in compound having same stoichiometric composition but different molecular compositions. (b) The molecular compositions are simple multiples of the simplest stoichiometric arrangement. (c) Examples : [Pt(NH3)2Cl2], [Pt(NH3)4] [PtCl4]; [Pt(NH3)3Cl]2 [PtCl4] (i) (ii) (iii) In the above three examples (ii) and (iii) compounds are the polymers of the first. (iv) Coordination isomerism : (a) This type of isomerism occurs when both cation and anion are complex. (b) The isomerism is caused by the interchange of ligands between the two complex ions of the same complex. (c) Examples : [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6][Co(CN)6] (v) Linkage isomerism : (a) This is also called as salt isomerism. – – – (b) This type of isomerism arises due to presence of ambidentate ligands like NO2 , CN CO and SCN (c) These ligands have two donor atoms but at a time only one atom is directly linked to the central metal atom of the complex. (d) Such type of isomers are distinguished by infra red (I.R.) spectroscopy. (e) Examples : [Co(NH3)5NO2]Cl2 and [Co(NH3)5ONO]Cl2 (Yellow) (Red) (vi) Geometrical Isomerism : (a) This isomerism is due to ligands occupying different positions around the central metal atom or ion. (b) The ligands occupy position either adjacent or opposite to one another. (c) This type of isomerism is also known as cis-trans isomerism. (d) This is common in coordination number 4 & 6 complexes. (e) Basically in di-substituted complexes, the similar groups may be adjacent or opposite to each other. This gives rise to geometrical isomerism. (f) When similar groups are adjacent to each other it is called a cis isomer whereas when they are opposite to each other, it is called as trans isomer.

CH2 NH2 O CO Example : Pt

CO O NH2 CH2 Trans–isomer Diglycinato platinum (IV) complexes (g) C.N. = 4 Square planar complexes : (i) Complexes with general formula, Ma2b2 (where both a and b are monodentate) can have cis-and trans-isomers. a b a a HN3 Cl HN3 NH3 M M Pt Pt

b a b b Cl NH3 Cl Cl Trans–isomer Cis–isomer Trans Cis

[Ma2 b2 ] [Pt(NH3 )2 Cl 2 ]

Gyaan Sankalp 9 Coordination compounds

(ii) Complexes with general formula, Ma2bc can have cis- and trans-isomers. a a a c HN3 NH3 HN3 NO2 M M Pt Pt

NH b c b a Cl NO2 Cl 3 Cis Trans Cis Trans

[Ma2 bc] [Pt(NH3 ) 2 ClNO 2 ] (iii) Complexes with general formula, Mabcd, can have three isomers. a b a d a b

M M M

d c c b c d (i) (ii) (iii) Ex : [Pt(NH3)(NH2OH)(NO2)(Py)]NO2 Geometrical isomerism is not observed in complexes of coordination number 4 of tetrahedral geometry. (h) Octahedral complexes : (Coordination number = 6). These may be classified into the following different types : + (i) MA4B2 or MA2B4 or MA4BC or MA3B3 type : An example of the type MA4B2 or MA2B4 is [CrCl2(NH3)4] ion or + – [CoCl2(NH3)4] ion or [Fe(CN)4(NH3)2] . In cis-form, similar groups (trio of donor atoms) occupy the corners of one of the octahedral faces and hence is called facial or fac- isomer while in trans-form, the positions of trio of donor atoms are around the meridian of the octahedron. Hence it is called meridional or mer-isomer. Another example of this type is [Co(NH3)3Cl3] or [Co(NO2)3(NH3)3]. (ii) [M(AA)2B2] or [M(AA)2BC] type , i.e. containing one symmetrical bidentate ligand (AA) and the other two monodentate + ligands may be same or different, .e.g., [CoCl2(en)2] ion exists in two geometrical isomers. 2+ (iii) MA2B2C2 type : These types of complexes can exist in five geometrical isomers. The complex ion [PtCl2(NH3)2(py)2] for examples, shows the following five geometrical isomers. (iv) M(AA)3 type : [Cr(gly)3] is a typical example of this type of complex. Octahedral complex of the type [MABCDEF], i.e., having all the six different monodentate ligands forms 15 different geometrical isomers. The only complex of this type prepared so far is [Pt(Cl) (Br) (I) (NO2) (NH3) (py)]. (vii) Optical Isomerism : (a) A coordination compound which can rotate the plane of polarised light is said to be optically active. (b) When the coordination compounds have same formula but differ in their abilities to rotate directions of the plane of polarised light are said to exhibit optical isomerism and the molecules are optical isomers. The optical isomers are pair of molecules which are non-superimposable mirror images of each other. (c) This is due to the absence of elements of symmetry in the complex. (d) Optical isomerism is expected in tetrahedral complexes of the type Mabcd. Ex. : Bis - benzoylacetonato - beryllium (II)

CH CH CH CH3 3 C O O C 3 3 C O O C H C Be C H H C Be C H C O O C C O O C HC CH HC5 6 CH6 5 5 6 6 5 Mirror (e) Optical isomerism is not found in square planar complexes on account of the presence of axis of symmetry. (f) Optical isomerism is very common in octahedral complexes. n± 2+ Ex. : (a) [Ma2b2c2] [Pt(py)2(NH3)2Cl2] 2+ 2+ py py Cl py py Cl

Pt Pt

Cl HN3 Cl NH3 NH3 NH3 Cis–d–isomer Mirror Cis– –isomer

10 Gyaan Sankalp Coordination compounds

(b) [Mabcdef] : [Pt(py)NH3NO2ClBr]

Br Br py NO2 ON2 py Pt Pt Cl HN Cl NH3 3 I I d–isomer Mirror – isomers

n± 3+ (c) [M(AA)3] ; [Co(en)3]

en 3+ en 3+

en Co Co en

en en

d–form Mirror –form

en Co en en

'Meso' or optically inactive form n± + (d) [M(AA)2a2] [Co(en)2Cl2]

en + en + Cl Cl

Co Co

Cl Cl en en Cis–d–isomer Mirror Cis– –isomer

n± Trans form of [M(AA)2a2] does not show optical isomerism. n 2+ (e) [M(AA)2ab] [Co(en)2NH3Cl]

en 2+ en 2+ Cl Cl

Co Co

HN3 HN3 en en Cis–d–isomer Mirror Cis– –isomer

Gyaan Sankalp 11 Coordination compounds

(f) [M(AB)3] [Cr(gly)3]

gly gly

gly Cr Cr gly

gly gly Cis or trans–d–isomer Mirror Cis or trans– –isomer

Some more examples are : 3– 2+ – 4+ [Cr(ox)3] ; [Fe(dipy)3] ; [Cr(ox)2(H2O)2] ; [Pt(en)3]

BONDING IN COORDINATION COMPOUNDS : Valence Bond Theory : The salient features of the theory are : (i) The central metal ion has a number of empty orbitals for accommodating electron donated by the ligands. The number of empty orbitals is equal to the coordination number of the metal ion for the particular complex. (ii) The atomic orbitals (s, p or d) of the metal ion hybridize to form hybrid orbitals with definite directional properties. These hybrid orbitals now overlap with the ligand orbitals to form strong chemical bonds. Number of orbitals and types of hybridisations :

Coordination number Type of hybridisation Distribution of hybrid orbitals in space 4 sp3 Tetrahedral 4 dsp2 Square planar 5 sp3d Trigonal bipyramidal 6 sp3d2 Octahedral 6 d2sp3 Octahedral

(iii) The d-orbitals involved in the hybridization may be either inner (n-1) d orbitals or outer nd-orbitals. The complexes formed in these two ways are referred to as low spin and high spin complexes, respectively. (iv) Each ligand contains a lone pair of electrons. (v) A covalent bond is also sometimes called as a coordinate bond. (vi) If the complex contains unpaired electron, it is paramagnetic in nature, while if it does not contain unpaired electron, it is diamagnetic in nature. (vii) The number of unpaired electrons in the complex, points out the geometry of the complex as well as hybridisation of central metal ion and vice-versa. In practice, the number of unpaired electrons in a complex is found from magnetic moment measurements as illustrated below :  = n n 2 ; where n = no of unpaired electron

Magnetic Moment 0 1.73 2.83 3.87 4.90 5.92 (Bohr Magnetons)

Number of unpaired electrons 0 1 2 3 4 5

Under the influence of a strong ligand the electrons can be forced to pair up against the Hund’s rule of maximum multiplicity. Shapes of complexes basis of VBT : The shape of complexes depends upon hybridization state of central atom, it is described as follows: (A) Octahedral Complexes : On the basis of hybridized orbitals it can be of two type as d2sp3 (inner orbital) or sp3d2 (outer orbital) hybridized. (a) Inner Orbital Complexes : In these type of complexes the d-orbitals used are of lower quantum number i.e. (n-1) various examples are as follow : 12 Gyaan Sankalp Coordination compounds Complexes formed by the use of inner orbitals are diamagnetic or have reduced paramagnetism. These are called as low spin or spin paired complexes. 4– 3d 4s 4p (i) [Fe(CN)6] (Ferrocyanide ion) e– configuration of Fe26 = [Ar] 3d64s2 e– configuration of Fe+2 = [Ar] 3d6 e¯ configuration of Fe+2 after rearrangement.

The above rearrangement is due to presence of cyanide ligand. In this state Fe2+ undergoes d2sp3 hybridisation to from six d2sp3 hybrid orbitals, each of which accepts 2 3 d sp electron pair donated by CN¯ ions. The resultant complex is inner octahedral as shown in figure NH3 and it should be diamagnetic as it has no unpaired electron. NH 3 NH3 Form action of [Co(NH ) ]3+ takes place in the same manner. 3 6 +3 Co

NH3 NH3 NH3 3+ Octahedral shape of [Co(NH3)6] 3– (ii) [Fe(CN)6] (Ferricyanide ion) 3d 4s 4p

Fe26 3d 4s 4p Fe3+ Fe3+ (rearranged due to presence of CN–) 3d 4s 4p

2 3 d sp hybridization The resulting complex is octahedral due to d2sp3 hybridization. Due to presence of unpaired e– it is paramagnetic 3+ (iii) [Cr(NH3)6] 5 1 3d 4s 4p

Cr24

Cr3+ Cr3+, d2sp3 hybridized state

2 3 d sp hybridization

2 3 3+ This d sp hybridisation leads to octahedral geometry, the complex [Cr(NH3)6] will be octahedral in shape. Since the complex ion has 3 unpaired electrons it must be paramagnetic which is founded to be so. 3– 3+ Other complexes of chromium with similar inner structure are [Cr(CN)6] and [Cr(H2O)6] . (b) Outer orbital complexes : In these complexes s, p as well as d orbitals involved in hybridization, belong to the highest quantum number (n). Complexes formed by the use of outer n, d orbitals will be paramagnetic. These complexes are called high-spin free complexes. The outer orbital complexes have greater number of unpaired electrons.

Gyaan Sankalp 13 Coordination compounds 3– [CoF6] 3d 4s 4p 4d

Co27

Co3+ ion Co3+ ion is sp3d2 hybridised state

3 2 sp d 3 2 3– Due to octahedral orientation of six sp d hybridised orbitals shape [CoF6] complex ion is octahedral. 3– Due to presence of four unpaired electrons in 3d orbitals [CoF6] ion has paramagnetic. 3– 2+ 2+ 2+ 3+ Other examples are – [FeF6] , [Fe(NH3)6] , [Ni(NH3)6] , [Cu(NH3)6] , [Cr(H2O)6]

(B) Tetrahedral Complexes : These are formed by sp3 hybridisation. Complexes of Zn2+ are invariable tetrahedral because they involve sp³ hybrid orbitals. 2+ (i) [Zn(NH3)4] 3d 4s 4p e– configuration of Zn30

3d 4s 4p e– configuration of Zn+2

Zn2+ in sp3 hybridised state 3 sp hybridised

Zn [(NH ) ]2+ 3 4 3 sp NH3 NH3 NH3 NH3 Since the complex is formed by sp3 hybridisation, it is tetrahedral. Since all electrons are paired it is diamagnetic. (ii)[Ni(CO)4] 3d 4s 4p

Ni28 3d 4s 4p

Ni after rearrangement 3 sp CO The empty one 4s and three 4p orbitals mix to form four sp³ hybridised orbitals. Each orbital accepts one electron pair from carbon monoxide molecule forming Ni [Ni(CO)4]. The shape of nickel tetracarbonyl is tetrahedral as shown below. It is diamagnetic in nature. CO CO CO

14 Gyaan Sankalp Coordination compounds (C) Square planar complex : These are formed due to dsp2 hybridisation. These complexes tend to be formed when the central ion has only one d orbital available in the inner shell. 2– (i) [Ni(CN)4] 3d 4s 4p e– configuration of Ni

3d 4s 4p e– configuration of Ni2+

2 dsp The rearrangement is due to presence of strong ligand CN–. The four orbitals make dsp2 hybridisation. The shape of resulting complex is square planar. Due to paired electrons it is diamagnetic. Coordination number and shapes of complex :

Coordination Hybridised orbital Geometrical shape of Examples of Number the complex complex

o 180  [Ag(NH3 ) 2 ] 2 sp  L M L [Ag(CN)2 ] Linear

2 – 3 sp [HgI3]

L 3 – 4 sp [FeCl4] 109o 28' 0 [Ni(CO)4] +2 M [Zn(NH3)4] –2 –2 [ZnCl4] , [CuX4] L L where X = CN Cl–, Br–, I–, CNS

Tetrahedral

4 dsp2 L L The d– orbital involved is [Ni(CN) ]–2 90o 4 o d 2 2 o +2 x y orbital of the inner 90 90 [Pt(NH3)4] 90o d shell, i.e. it is (n–1) x2 y 2 orbital L L

L L L 90o 90o

5 dsp3 ; d-orbital is n orbital [SbF ]2–, IF dx2 y2 L L 4 5 Gyaan Sankalp 15 Coordination compounds Coordination Hybridised orbital Geometrical shape of Examples of Number the complex complex

2 3 6 d sp L L L +3 When d-orbitals are (n–1)d [Cr(NH3)6] +3 orbitals(Inner orbital 90° [Ti(H2O)6] 3 2 –2 complexes) or sp d When [Fe(CN)6] 90° M +3 d-orbital are nd orbital (outer [Co(NH3)6] –2 –3 orbital complexes) In both [PtCl6] ,[CoF6] cases d-orbitals are d L L z2 L and d orbitals x2 y 2 L = Ligands indicated by white circles (same or different) M = Central metal indicated by black circles. Properties of complex ions

S. No. Magnetic Low spin Complex ions Form. Metal ion Coordination No. Configuration Hybridisation Geometry U.E. Properties High spin

–2 Ni+2 8 2 Sq. planar 0 Diamagnatic 01 [Ni(CN)4] 4 d dsp L.S. Ni 8 2 3 Tetrahedral 0 Diamagnatic 02 [Ni(CO)4] 4 d s sp L.S. +2 Cu+2 9 2 Sq. planar 1 Paramagnetic 03 [Cu(NH3)4] 4 d dsp L.S. +2 Zn+2 10 3 Tetrahedral 0 Diamagnetic 04 [Zn(NH3)4] 4 d sp L.S. –2 Co+2 7 3 Tetrahedral 3 Paramagnetic 05 [CoCl4] 4 d sp L.S. –2 Ni+2 8 3 Tetrahedral 2 Paramagnetic 06 [NiCl4] 4 d sp L.S. +2 Pt+2 8 3 Tetrahedral 2 Paramagnetic 07 [Pt(NH3)4] 4 d sp L.S. –3 Fe+3 5 2 3 Octahedral 1 Paramagnetic 08 [Fe(CN)6] 6 d d sp L.S. –4 Fe+2 6 2 3 Octahedral 0 Diamagnetic 09 [Fe(CN)6] 6 d d sp L.S. +2 Fe+2 6 3 2 Octahedral 4 Paramagnetic 10 [Fe(H2O)6] 6 d sp d H.S. +3 Fe+3 5 3 2 Octahedral 5 Paramagnetic 11 [Fe(H2O)6] 6 d sp d H.S. –3 –3 Fe+3 5 3 2 Octahedral 5 Paramagnetic 12 [FeF6] , [FeCl6] 6 d sp d H.S. +3 Co+3 6 2 3 Octahedral 0 Diamagnetic 13 [Co(H2O)6] 6 d d sp L.S. –4 Co+2 7 2 3 Octahedral 1 Paramagnetic 14 [Co(CN)6] 6 d d sp L.S. –3 Co+3 6 3 2 Octahedral 4 Paramagnetic 15 [CoF6] 6 d sp d H.S. +2 Cr+2 4 3 2 Octahedral 4 Paramagnetic 16 [Cr(H2O)6] 6 d sp d H.S. +3 Cr+3 3 2 3 Octahedral 3 Paramagnetic 17 [Cr(NH3)6] 6 d d sp L.S. –3 Cr+3 3 2 3 Octahedral 3 Paramagnetic 18 [Cr(Cl)6] 6 d d sp L.S. +3 Ti+3 1 2 3 Octahedral 1 Paramagnetic 19 [Ti(H2O)6] 6 d d sp L.S. +3 V+3 2 2 3 Octahedral 2 Paramagnetic 20 [V(H2O)6] 6 d d sp L.S. +2 Zn+2 10 3 2 Octahedral 0 Diamagnetic 21 [Zn(NH3)6] 6 d sp d H.S. +4 Pt+4 6 2 3 Octahedral 0 Diamagnetic 22 [Pt(NH3)6] 6 d d sp L.S. –4 Mn+2 5 2 3 Octahedral 1 Paramagnetic 23 [Mn(CN)6] 6 d d sp L.S.

16 Gyaan Sankalp Coordination compounds Draw backs of valence bond theory : (i) It describes bonding in coordination compounds only qualitatively. (ii) It does not offer any explanation for the optical absorption spectra of complex. (iii) It does not describe the detailed magnetic properties of coordination compounds. (iv) It does not explain the colour exhibited by coordination compounds. (v) Why should certain ligands form high spin, while other low spin complexes ?

CRYSTAL FIELD THEORY : Crystal Field Theory was developed by Hans Be The and John Van Vleck and gives much more satisfactory explanation for the bonding and the properties of complexes. The crystal field theory (CFT) is an electrostatic model which considers the metal- ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. The main assumptions of this theory are : l. The valence bond theory considers the bonding between the metal ion and the ligands as purely covalent. On the other hand, crystal field theory is based on the assumption that the metal ion and the ligands act as point charges and the interactions between them are purely electrostatic. In case of negative ligands (anions such as Cl–, Br–, CN–), the interactions with metal ions are ion-ion interactions. If the ligands are neutral molecules (such as NH3, H2O, CO), the interactions with the metal ions are ion- 3– 3+ – dipole interactions. For example, in the case of complex ion [CoF6] the interactions are between Co and F ions whereas in + 3+ [Co(NH3)6] , the interactions are between negatively charged end of NH3 molecules (-charge on N atom) and Co ion. 2. In the case of nee metal ion, all the five d-orbitals have the same energy. These orbitals having the same energies are called degenerate orbitals. This means that an electron can occupy anyone of these five d-orbitals with equal ease. However, on the approach of the ligands, the orbital electrons will be repelled by the lone pairs of the ligands. The repulsion will raise the energy of the d-orbitals. If all the ligands approaching the central metal ion are at equal distance from each of the d-orbitals, the energy of each orbital will increase by the same amount, due to spherical field or symmetrical field of ligands. Therefore, these orbitals will still remain degenerate, but they will have higher energy than that of the free ion. 3. We know that d-orbitals have different orientations and, therefore, these orbitals will experience different interactions from the ligands. The orbitals lying in the direction of the ligands, will experience greater repulsion and their energies will be raised relative to their positions in a symmetrical field. (Remember that orbitals represent regions of negative charge; because like charges repel each other and therefore, potential energy increases when they are brought together). On the other hand, the orbitals lying away from the approach of the ligands will have lesser interactions with the negative charge of donor atoms and therefore, their energies will be lower than they would be in a spherical field. 4. Due to the electrical field held by the ligands. The energies of the five d-orbitals will split up. Thus, this conversion of five degenerate d-orbitals of the metal ion into different sets of orbitals having different energies in the presence of electrical field of ligands is called crystal field splitting. Thus, in coordination compounds, the interaction between the ligand and the metal ion causes the five d-orbitals to split up. The number of ligands (coordination number) and their arrangement (geometry) around the central metal ion will have different effect on the relative energies of five d-orbitals. In other words, the crystal field splitting will be different in different structures with different coordination numbers Crystal field splitting energy : The energy difference between the two sets of energy levels is called crystal field splitting energy and is represented as o (the subscript 'o' stands for octahedral). It measures the crystal field strength of the ligands. It may be noted that the crystal field splitting occurs in such a way that the average energy of the d-orbitals does not change. This means that the three orbitals lie at an energy that is (2/5) 0 below the average d-orbital energy and the two d-orbitals lie at an energy (3/5)o above the average energy. Spectrochemical Series : The crystal field theory depends upon the nature of the ligands. The greater the case with which the ligand can approach the metal ion, the greater will be the crystal field splitting caused by it. The ligands which cause only a small degree of crystal field splitting are called weak field ligands while those which cause a large splitting are called strong field ligands. The arrangement of ligands in order of their crystal field splitting energy (C.F.S.E.), 0 values is known as spectrochemical series. Thus, the ligands can be arranged according to the magnitude of the 0 and this arrangement is called spectrochemical series. The spectrochemical series in the increasing order of crystal field splitting is : Weak field ligands Strong ligands –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– – 2– – – – – – 2– 2– I < Br < S < SCN < CI < NO3 < F < OH < ox < O < H2O < NCS < py NH3 –––––––––––––––––––––––––––––––– – < en < dipy < o-phen < NO2 < CN < CO – – – – – From the series it is clear that CO, CN and NO2 are strong field ligands whereas I , Br and Cl are weak field ligands. In general, the ligands lying above H2O are called strong field ligands and they cause greater crystal field splitting (or high o ), On the other hand the ligands water and below it are called weak field ligands and they cause lesser crystal field splitting (or low o).

Gyaan Sankalp 17 Coordination compounds

d 2 2 d Crystal field splitting in octahedral complexes : In octahedral field of splitting , the two d-orbitals; x y and z2 are called

e.g., orbitals (pronounced as “e–g”) and the three orbitals dxy, dyz and dzx are called t2g orbitals (pronounced as “t-two–g”). These names are derived from spectroscopic terms.

d 2 2 d In an octahedral field the five d-orbitals split up into two sets of orbitals, one set consisting of two orbitals x y and z2 of

higher energy and another set of three orbitals (dxy, dyz and dzx) of lower energy. This splitting of d-orbitals is shown in figure. Crystal field splitting energy ( 0 ) dx–2 y2 dz2

y 0.6 0 g e

r g e n E 0 Average energy of orbitals in a spherical 0.4 0 dxy dyz dxz crystal field

t2 g d-orbitals in free ion Figure : Splitting of d-orbitals in octahedral field Crystal field splitting in tetrahedral complexes : The splitting of d-orbital in a tetrahedral complex takes place in the reverse pattern compared to the splitting pattern observed in the octahedral complexes. Two-fold degenerate set ‘e’ has lower energy than three-fold degenerate set, ‘t2’. The subscript ‘g’ is not used to denote the two sets in tetrahedral complexes. The splitting in tetrahedral complexes is smaller than that in case of octahedral complexes. 4 The difference of energy represented by,     . This energy is not able to force the electrons to pair up. As such, t9 0 tetrahedral complexes have high spin configuration. The splitting of d-orbitals in the tetrahedral complexes is shown in figure.

dxy dyz dxz

y 0.4 t g t

r 2 e n

E t Average energy of 0.6 d-orbitals in a spherical t dx–2 y2 dz2 crystal field

e

d-orbitals in free ion Figure : Splitting of d-orbitals in tetrahedral crystal field The actual configuration adopted by the complex is decided by the relative values of  and P, where P represents the energy required for electron pairing in a single orbital and is crystal field splitting energy. High spin and low spin complexes : If the crystal field splitting energy () is less than the energy required for electron pairing in an orbital (P) the complex will be high spin complex. Weak field ligands also tend to form high spin complexes. However, if the crystal field splitting energy () is greater than the energy required for electron pairing in an orbital (P), the complex, will be low spin complex. Strong field ligands also tend to make low spin complexes. If 0< P, it will be high spin complex. 18 Gyaan Sankalp Coordination compounds

If 0 > P, it will be low spin complex. Strong field ligands form – low spin complexes. Weak field ligands form – high spin complexes. Colour in coordination compounds : The colour in the coordination compounds can be readily explained in terms of the crystal 3+ field theory. Consider, for example, the complex [Ti(H2O)6] , which is violet in colour. This is an octahedral complex where the 3+ 1 single electron (Ti is a 3d system) in the metal d orbital is in the t2g level in the ground state of the complex. The next higher state available for the electron is the empty eg level. If light corresponding to the energy of yellow-green region is absorbed by 1 0 0 1 the complex, it would excite the electron from t2g level to the eg level (t2g e g t 2g e g ) . Consequently, the complex appears violet in colour. The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron. It is important to note that in the absence of ligand, crystal field splitting does not occur and hence the substance is colourless. For example, removal of water from [Ti(H2O)6]Cl3 on heating renders it colourless. Similarly, anhydrous CuSO4 is white, but CuSO4.5H2O is blue in colour. Limitations of crystal field theory : The crystal field model is successful in explaining the formation, structures, colour and magnetic properties of coordination compounds to a large extent. However, from the assumptions that the ligands are point charges, it follows that anionic ligands should exert the greatest splitting effect. The anionic ligands actually are found at the low end of the spectrochemical series. Further, it does not take into account the covalent character of bonding between the ligand and the central atom. These are some of the weaknesses of CFT, which are explained by ligand field theory (LFT) and molecular orbital theory.

METAL CARBONYLS : Organometallic compounds in which carbon monoxide (CO) acts as a ligand are called metal carbonyls. The compounds which contain only carbonyl (CO) ligands are called homoleptic carbonyls. In metal carbonyls, CO is bonded to metal through C-atom. The ligand CO possess vacant orbitals in addition to the lone-pair. Therefore, CO acts as a weak donor (or a weak base) and an acceptor ligand. In carbonyls, CO binds itself to the metal atoms through its carbon atom. The lone-pair donation to the metal atom gives rise to sigma () bond between metal and C of CO. The vacant orbitals on CO then accept back the electron-density back into its empty -orbitals is known as -acidity (from Lewis nomenclature), and such ligands are called -acceptors. This characteristic property of back donation of electron density stablises the metal-ligand (CO) bonding. Structures : Carbonyls have simple, well defined structures. Tetracarbonylnickel(0) is tetrahedral, pentacarbonyliron(0) is trigonalbipyramidal while hexacarbonyl chromium(0) is octahedral. Decacarbonyldimanganese(0) is made up of two square pyramidal Mn(CO)5 units joined by a Mn – Mn bond. Octacarbonyldicobalt(0) has a Co – Co bond bridged by two CO groups.

STABILITY OF COMPLEX : (i) A complex is formed in solution by the stepwise addition of ligands to a metal ion. (ii) This can be expressed as follows M + L  ML (iii) The stability constant K for this reaction is as shown. ML K  [M][L] (iv) This metal can again get a ligand. ML + L  ML2 (v) The forthcoming stability constant K1 is [ML ] K  2 1 [ML][L] its value is less than K (vi) The higher the value of stability constant stabler is the complex. (vii) The value of stability constants for some of the complexes are given below : Complex Stability Constant 2+ 11 [Cu(NH3)4] 4.5 × 10 + 7 [Ag(NH3)2] 1.6 × 10 2+ 6 [Co(NH3)6] 1.12 × 10 3+ 33 [Co(NH3)6] 5.0 × 10 – 5 [AgCl2] 1.11 × 10 – 7 [AgBr2] 1.28 × 10 – 22 [Ag(CN)2] 1.0 × 10 2– 27 [Cu(CN)4] 2.0 × 10 3– 43 [Fe(CN)6] 7.69 × 10 Gyaan Sankalp 19 Coordination compounds Factors influencing the stability of complex : Charge  (i) Nature of central ion : The complex will be more stable for higher values of charge density . Radius  The higher the electronegativity of the central ion the greater is the stability of its complexes. The higher the oxidation state of the metal the more stable is the compound. (ii) Nature of ligand : A basic ligand is likely to easily donate its electrons. Thus a more basic ligand will form more stable complex. For example cyano and fluoro complexes of any metal ions are more stable than ammine complexes. Chelating ligands form more stable complexes as compared to monodentate ligands.

USES OF COORDINATION COMPOUNDS : 1. Coordination compounds find use in many qualitative and quantitative chemical analysis. The familiar colour reactions given by metal ions with a number of ligands (especially chelating ligands), as a result of formation of coordination entities, form the basis for their detection and estimation by classical and instrumental methods of analysis. Examples of such reagents include EDTA, DMG (dimethylglyoxime), –nitroso––naphthol, cupron, etc. 2+ 2+ 2. Hardness of water is estimated by simple titration with Na2EDTA. The Ca and Mg ions form stable complexes with EDTA. The selective estimation of these ions can be done due to difference in the stability constants of calcium and magnesium complexes. 3. Some important extraction processes of metals, like those of silver and gold, make use of complex formation. Gold, for example, – combines with cyanide in the presence of oxygen and water to form the coordination entity [Au(CN)2] in aqueous solution. Gold can be separated in metallic form from this solution by the addition of . 4. Purification of metals can be achieved through formation and subsequent decomposition of their coordination compounds. For example, impure nickel is converted to [Ni(CO)4], which is decomposed to yield pure nickel. 5. Coordination compounds are used as catalysts for many industrial processes. Examples include rhodium complex, [(Ph3P)3RhCl], a Wilkinson catalyst, is used for the hydrogenation of alkenes. – 6. Articles can be electroplated with silver and gold much more smoothly and evenly from solutions of the complexes, [Ag(CN)2] – and [Au(CN)2] than from a solution of simple metal ions. 7. In black and white photography, the developed film is fixed by washing with hypo solution which dissolves the undecomposed 3– AgBr to form a complex ion, [Ag(S2O3)2] . Examples 3 : Write the IUPAC names of the following coordination compounds: (i) [Pt(NH3)2Cl(NO2)] (ii) K3[Cr(C2O4)3] (iii) [CoCl2(en)2]Cl Sol. (i) Diamminechloridonitrito-N-platinum(II) (ii) Potassium trioxalatochromate(III) (iii) Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride Example 4 : – Draw structures of geometrical isomers of [Fe(NH3)2(CN)4]

NH3 NH3 NC NH3 NC CN Fe Fe Sol. NC CN NC CN CN NH3 cis trans Example 5 : 2– [NiCl4] is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why? 2– Sol. In Ni(CO)4, Ni is in zero oxidation state whereas in NiCl4 , it is in +2 oxidation state. In the presence of CO ligand, the unpaired d electrons of Ni pair up but Cl– being a weak ligand is unable to pair up the unpaired electrons. Example 6 : 3+ 3– [Fe(H2O)6] is strongly paramagnetic whereas [Fe(CN)6] is weakly paramagnetic. Explain. Sol. In presence of CN–, (a strong ligand) the 3d electrons pair up leaving only one unpaired electron. The hybridisation is d2sp3 3 2 forming inner orbital complex. In the presence of H2O, (a weak ligand), 3d electrons do not pair up. The hybridisation is sp d forming an outer orbital complex containing five unpaired electrons, it is strongly paramagnetic. Example 7 : 2– Predict the number of unpaired electrons in the square planar [Pt(CN)4] ion. Sol. For square planar shape, the hybridisation is dsp2. Hence the unpaired electrons in 5d orbital pair up to make one d orbital empty for dsp2 hybridisation. Thus there is no unpaired electron.

20 Gyaan Sankalp Coordination compounds TRY IT YOURSELF Q.1 How many ions are obtained in the aqueous solution of [Co(H2O)6]Cl2 (A) 3 (B) 2 (C) 4 (D) 6 Q.2 Which one of the following complexes is paramagnetic :– 3+ 3+ (A) [Co(NH3)6] (B) [Co(H2O)6] (C) [CoF3(H2O)3] (D) All the above – 3 Q.3 Hybridization of Fe in the complex [FeF6] is - (A) sp3d2 (B) d2sp3 (C) dsp2 (D) sp2 Q.4 [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br are examples of which type of isomerism :– (A) Linkage (B) Geometrical (C) Ionization (D) Optical Q.5 The complex compound [Co(NH3)3NO2ClCN] is named as :– (A) Chlorocyanonitrotriammine cobalt (III) (B) Nitrochlorocyanotriammine cobalt (III) (C) Cyanonitrochlorotriammine cobalt (III) (D) Triamminenitrochlorocyano cobalt (III) Q.6 IUPAC name of K3Fe(CN)6 is :– (A) Potassium ferrocyanide (II) (B) Potassium hexaferrocyanate (III) (C) Potassium ferrohexacyanate (II) (D) Potassium hexacyanoferrate (III) Q.7 The pair of complex compounds [Cr(H2O)6Cl3] and [Cr(H2O)5Cl] Cl2H2O are an example of (A) Linkage isomerism (B) Ionisation isomerism (C) Coordination isomerism (D) Hydrate isomerism Q.8. What is the co–ordination number of cobalt in [Co(NH3)3Cl3] :– (A) 3 (B) 4 (C) 5 (D) 6 Q.9 The example of dsp2 hybridisation is :– 3– 2– 2+ 3– (A) Fe(CN)6 (B) Ni(CN)4 (C) Zn(NH3)4 (D) FeF6 –3 Q.10 Oxidation number of Co in [Co (CN)6] is - (A) –6 (B) –3 (C) +3 (D) +6 ANSWERS (1) (A) (2) (D) (3) (A) (4) (C) (5) (A) (6) (D) (7) (D) (8) (D) (9) (B) (10) (C)

USEFUL TIPS 1. Coordination isomerism is caused by the interchange of the ligands between the complex cation and complex anion. 2. Linkage isomerism is shown by the coordination compounds which contain ligands that are able to attach themselves to the central ion through more than one donor atoms, (ambidentate ligands). 3. In tetrahedral geometry, all four ligands are symmetrically placed relative to each other. Therefore, cis-trans isomerism is not possible in tetrahedral coordination compounds. 4. The most well-known cases of optical isomerism occur among octahedral complexes of the type [M(aa)3], where ‘aa’ is a bidentate ligand, e.g., trioxalatochromate (III) ion, and tris (ethylenediamine) cobalt (III) ion show optical isomerism. 5. The energy gap 0 between the two groups of d-orbitals is called crystal field splitting. 6. The ligands which cause only a small crystal field splitting are called weak field ligands. The ligands which cause a large crystal splitting are called strong field ligands. 2+ 3– 7. The weak ligands produced high-spin complexes. [Fe(H2O)6] and [CoF6] are high spin complexes. 4– 3+ 8. The strong field ligands produce low-spin complexes. [Fe(CN)6] and [Co(NH3)6] are low spin complexes. 9. The organometallic compounds in which carbon monoxide (CO) acts as a ligand are called metal carbonyls. 10. The ability of CO molecule to accept electron-density back into its empty orbitals is called -acidity (or back donation). 11. The stability of a complex depends upon the following factors. (a) Nature of the central ion – High range, small size favours the stability of the complex. (b) Nature of the ligand – More basic ligand forms more stable complex.

MISCELLANEOUS SOLVED EXAMPLES Example 1 : 3+ 2+ Explain [Co(NH3)6] is an inner orbital complex whereas [Ni(NH3)6] is an outer orbital complex. 2 3 Sol. In the presence of NH3, the 3d electrons pair up leaving two d orbitals empty to be involved in d sp hybridisation forming inner 3+ 2+ 8 orbital complex in case of [Co(NH3)6] . In Ni(NH3)6 , Ni is in +2 oxidation state and has d configuration, the hybridisation involved is sp3d2 forming outer orbital complex. Example 2 : 2+ Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)4 ion, given that 4 for this complex is 2.1 × 1013. –14 Sol. The overall dissociation constant is the reciprocal of overall stability constant i.e. 1/4 = 4.7 × 10 Gyaan Sankalp 21 Coordination compounds Example 3 : (a) Draw the shapes of the d , d , d , d and d orbitals. xy xz yz x2 y 2 z2 (b) Explain why a splitting of the energy levels occurs for the d-orbitals in an octahedral environment. (c) The splitting pattern in a trigonal-bipyramidal environment is shown. Explain the splitting pattern.

Sol. (a)

dxz and dyz orbitals have the same shape but in the x-z and y-z planes

(b) The ligands have either lone pairs or carry negative charges. In an octahedral environment, the ligands approach the metal ion along the + and – X-axis, + and – Y-axis, + and – Z-axis. Thus, all the d-electrons are repelled. However, since the dxy, dxz, and dyz

orbitals do not point exactly towards the ligands they are less repelled than d 2 2 and d orbitals [pointing along the axes]. x y z2

The former orbitals are triply degenerate (t2g) and the latter are doubly degenerate (eg). The splitting pattern is shown below.

(c) In a tbp environment, the repulsion of the two axial ligands raises the d orbitals relative to the d and d orbitals. The z2 xz yz three other ligands in the XY plane raise the energy of the d and the d orbitals. x2 y 2 xy Example 4 : Find out the hybridization, geometry and magnetic moment of the complexes : 3+ 3– (i) [Co(NH3)6] (ii) [Cr(CN)6] 3+ Sol. (i) The oxidation state of cobalt in the complex [Co(NH3)6] is +3 3+ The electronic configuration of Co ion is :

22 Gyaan Sankalp Coordination compounds

3d 4s 4p

3+ : : : : : : [Co(NH3 ) 6 ] Rearrangement d2 sp 3 hybridization

Octahedral, zero magnetic moment. (ii) The oxidation state of chromium in the complex is +3. The electronic configuration of Cr3+ ion is

3d 4s 4p

3+ : : : : : : [Cr(CN6 )] Rearrangement d2 sp 3 hybridization

Octahedral, mag. moment = 3 (3  2) = 15 = 3.87 BM. Example 5 : (a) How does the t2g, eg-orbitals splitting pattern change, when a regular octahedral geometry gets distorted, by the gradual moving away of two trans-ligands from the central metal ion ? Draw the relevant diagram. 3 –1 (b) The visible absorption spectrum of [Ti(H2O)6] shows a maximum at ~ 20000 cm . Calculate the corresponding wavelength and the transition energy corresponding to this. What colour may one expect for the ion in solution ?

Sol. (a) The moving away of the trans-ligands stabilises the d orbital relative to the d 2 2 orbital. Similarly the d and the d z2 x y xy yz orbitals are stabilized

relative to the dxy orbital. (b) Using the relation, E = c/. We have 1/ = 20000 cm–1 1 8    = × 1 cm = 10 Å = 5000 Å. 20000 20000  E = 3 × 1010 × 20000 = 6 × 1014 s–1 ergs = 6 × 107 J = 6 × 104 kJ Since the broad absorption is somewhere in the middle part of the spectrum, the colour is mild pink. Example 6 : Explain the following observations. (a) The blood-red colour obtained by adding KCNS to a ferric salt fades out when NaF is gradually added to it. (b) When NaOH is added to a cupric salt solution, pale bluish green Cu(OH)2 is precipitated. But if Rochelle salt is added to the Cu++ ion solution, addition of NaOH produces no precipitate. A deep blue solution is obtained. (c) The chromium compound with the formula CrCl3 . 6H2O exists in 3 isomeric forms with different colours, violet, dark green and pale green. (d) Addition of excess of Cl– ions to the pink aqueous solutions of cobalt nitrate gives a blue solution. Sol. (a) The colour is attributed to ferric thiocyanate. Addition of NaF leads to the formation of the more stable Na3[FeF6] and the colour of ferric thiocyanate fades out.

Gyaan Sankalp 23 Coordination compounds

(b) Precipitation of Cu(OH)2 is prevented by the formation of the deep blue tartrate complex.

(c) [Cr(H2O)6] Cl3 is violet, trans – [Cr Cl2 (H2O)4] Cl.2H2O is dark green, [Cr Cl (H2O)5] Cl2. H2O is pale green. (d) There is a change from octahedral to tetrahedral to tetrahedral coordination. 2  2 [Co(H2O)6] + 4 Cl  + [CoCl4] + 6H2O pink blue Example 7 : How would you account for the following : 2– Ni(CO)4 possesses tetrahedral geometry while [Ni(CN)4] is square planar. 3 Sol. In the formation of Ni(CO)4, nickel undergoes sp hybridization, hence it is tetrahedral in shape. 3d 4s 4p Ni(0)3d8

: : : : Ni(CO)4 Rearrangement sp3 hybridization 2– 2+ 2 In the formation of [Ni(CN)4] , Ni ion undergoes dsp hybridization, hence it is square planar in shape. 3d 4s 4p Ni2+

2– : : : : [Ni(CN)4] : Rearrangement dsp2 hybridization Example 8 : Square planar complexes with coordination number of 4 exhibit geometrical isomerism whereas tetrahedral complexes do not. Why? Sol. In tetrahedral complexes, the relative position of atoms with respect to each other is same thus these do not show geometrical isomerism. Square planar complexes show cis, trans isomerism. Example 9 : (a) Write the formulae for the following compounds / ions: (i) Potassium trioxalato ferrate (lll) (ii) Tetrakis (Oxalato)- µ dihydroxodichromium (lll) ion (b) Give an example of geometric isomerism in coordination compounds. (c) Nickel salts yield a rose- red precipitate with dimethyl glyoxime in ammoniacal medium. Write the formula of the compound formed. O 4– C O O C OH O O C O Fe (C2 O 4 )2 Cr Cr(C2 O 4 ) 2 C O Sol. (a) (i) O O (ii) OH C O C O O

24 Gyaan Sankalp Coordination compounds

O H Cl O NH3 H3N Cl N N CH3 HC3 C Ni C C C (b) (c) CH3 HC3 N N

Cl NH O H O NH3 Cl 3 (rose-red complex)

Example 10 : Using the EAN rule, predict the molecular formulae of simple carbonyls of Fe (at. number 26) and Cr (at. number 24). The metals have zero oxidation state in the carbonyls. Sol. At. number of iron = 26. At. No. of the next noble gas; Kr (i.e. EAN) = 36 No. of electrons to be provided = 36 – 26 = 10 Since each CO provides 2 electrons, 10 Total number of carbonyl (CO) group = = 5 2  Molecular formula of simple iron carbonyl = Fe(CO)5 Molecular formula of chromium carbonyl At. no. of chromium = 24 EAN = 36  No. of electrons to be provided by CO = 36 – 24 = 12  No. of CO group = 12/2 = 6 Hence the molecular formula of chromium carbonyl will be Cr(CO)6. Example 11 : What is the geometry, hybridisation and magnetic behaviour of the following complex ions. 3+ 3– 3+ 3– 4– (a) [Cr(NH3)6] (b) [Co(CN)6] (c) [Co(NH3)6] (d) [Fe(CN)6] (e) Ni(CO)4 (f) [Fe(CN)4] . Sol. (a) Octahedral, d2sp3, paramagnetic (3 unpaired electrons) (b) Octahedral, d2sp3, diamagnetic (no unpaired electron) (c) Octahedral, d2sp3, diamagnetic (no unpaired electron) (d) Octahedral, d2sp3, paramagnetic (one unpaired electron) 3 (e) Tetrahedral, sp diamagnetic (no unpaired electron) (f) Octahedral, d2sp3, diamagnetic (no unpaired electron). Example 12 : + – In the reaction [CoCl2(NH3)4] + Cl  [CoCl3(NH3)3] + NH3, only one isomer of the complex product is obtained. Is the initial complex cis or trans ? Sol. The original complex must have the two chloride ions in the trans position, since than all the other four positions which could be replaced are equivalent. If the original isomer were cis, the third chlorine could replace an ammonia cis to both or trans to one, and two isomers would be expected. Example 13 : Using IUPAC norms write the systematic names of the following : (a) [Co(NH3)6]Cl3 (b) CoCl(NO2) (NH3)4Cl (c) [Ni(NH3)6]Cl2 (d) [PtCl(NH2CH3) (NH3)2]Cl 2+ 3+ (e) [Mn(H2O)6] (f) [Co(en)3] 3+ 2– (g) [Ti(H2O)6] (h) [NiCl4] (i) [Ni(CO)4] Sol. (a) Hexaamminecobalt (III) chloride (b) Tetraamminechloronitrocobalt (III) chloride (c) Hexaamminenickel (IV) chloride (d) Diamminechloromethylamineplatinum (II) chloride (e) Hexaaquamanganese (II) (f) Tris (ethylenediamine) cobalt (II) (g) Hexaaquatitanium (III) (h) Tetrachloronickelate (II) (i) Tetracarbonylnickel (0)

Gyaan Sankalp 25 Coordination compounds Example 14 : (a) Draw the structure of the geometrical isomers of the following square planar or octahedral complexes + (i) [Pd(NH3)2Cl2] (ii) Ru(NH3)4Br2 (b) A cobalt complex has a composition corresponding to the formula CoBr2Cl. 4NH3. What is the structural formula if the conductance measurements show two ions per formula unit ? AgNO3 solution gives an immediate precipitate of AgCl, but no AgBr. Write the structural formula of an isomer ? Cl HN3 NH3 HN3 Pd Pd Sol. (a) (i)

Cl Cl Cl NH3 cis trans (ii) Square planar

NH3 NH3

Br NH3 Br NH3

Ru Ru

Br HN3 Br NH3

NH3 NH3

(b) Structural formula : [Co(NH3)4Br2]Cl an ionization isomer–the structure is [Co(NH3)4BrCl]Br. Example 15 : Aqueous copper sulphate solution (blue in colour) gives : (a) a green precipitate with aqueous potassium fluoride, and (b) a bright green solution with aqueous potassium chloride. Explain these experimental results. Sol. (a) Copper sulphate solution when treated with potassium fluoride solution gives a green precipitate. So a normal salt of copper (II) having poor solubility in water is formed. The green precipitate is the hydrated copper (II) fluoride. (b) The bright green solution obtained when KCl solution is added to copper sulphate solution is due to the formation of 2– [CuCl4] in the solution. Example 16 : + Explain why NH4 ion does not form complexes ? + Sol. NH4 ion does not act as ligand because nitrogen atom has no lone pair of electrons which it can donate to metal atom. Example 17 : What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate ? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution ? 2 2– Sol. Copper sulphate solution on treating with KCN solution forms the complex entity, [Cu(CN)4] . The complex entity [Cu(CN)4] 2+ is a stable complex, and therefore the concentration of free Cu ions in the solution is very low. So when H2S is passed through such a solution, no precipitate of copper sulphide is formed. Example 18 : Give the oxidation state, d-orbitals occupation and coordination number of the central metal ion in the following complexes : (a) K3[Co(C2O4)3] (b) (NH4)2 [CoF4] (c) cis-[CrCl2 (en)2] Cl (d) [Mn(H2O)6]SO4 Sol. Complex Oxidation state ofd-level Configuration Coordination number central metal ion of metal ion 6 (a) K3[Co(C2O4)3] +3 3d 6 7 (b) (NH4)2[CoF4] +2 3d 4 3 (c) cis-[CrCl2(en)2]Cl +3 3d 6 5 (d) [Mn(H2O)6]SO4 +2 3d 6

26 Gyaan Sankalp Coordination compounds Example 19 : Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex : (a) K|Cr(C2O4)2 (H2O)2|.3H2O (b) CrCl3(py)3 (c) K4|Mn(CN)6| (d) Co(NH3)5Cl|Cl2 (e) Cs|FeCl4| Sol. Complex and Name (IUPAC) Oxidation state Electronic Coordination stereochemistry Magnetic of central metal Configuration number moment ion/atom(outer) (BM) 3 3 (a) K[Cr(C2O4)2(H2O)2].3H2O +3 d or t 2g 6 Octahedral 3.87 Potassium diaqua bis (oxalato) chromate (III) trihydrate 3 3 (b) [CrCl3(py)3] +3 d or t 2g 6 Octahedral 3.87 Trichloridotri-pyridine- chromium (III) 5 5 (c) K4[Mn(CN)6] +2 d or t 2g 6 Octahedral 1.73 Potassium hexacyanidomanganate (II) 6 6 (d) [CoCl(NH3)5]Cl2 +3 d or t 2g 6 Octahedral 0 Pentaammine chlorido-cobalt (III) chloride 5 2 3 (e) Cs[FeCl4] +3 d or e g t 2g 4 Tetrahedral 5.9 Cesium tetrachloridoferrate (III) Example 20 : Amongst the following ions which one has the highest magnetic moment value ? 3+ 2+ 2+ (a) [Cr(H2O)6] (b) [Fe(H2O)6] (c) [Zn(H2O)6] Sol. Complex Central metal ion Electronic configuration No. of upe Magnetic moment 3+ 3+ 3 (a) [Cr(H2O)6] Cr d 3 2+ 2+ 6 (b) [Fe(H2O)6] Fe d 4 Highest 2+ 2+ 10 (c) [Zn(H2O)6] Zn d 0 2+ Thus [Fe(H2O6)] has the highest magnetic moment value. Example 21 : 2+ 2– A solution of [Ni(H2O)6] is green but a solution of [Ni(CN)4] is colourless. Explain. 2+ 2+ 8 Sol. Ni in [Ni(H2O)6] is in +2 slate. Ni is d case. H2O molecules are weak field ligands. So, these 8 electrons occupy the 3d orbitals as follows : As a result it contains 2 unpaired electrons. This permits d-d transition and hence colour in the visible region. CN– is a strong field ligand, and hence forces the 3d-electrons to pair up and therefore no d-d transition can take place. Example 22 : (a) Selecting trise (ethylene diamine) cobalt (lll) ion as example, discuss opticalisomerism in coordination compounds. Write down the structures of both optically active and inactive (meso) forms of 4+

NO2

(en)2 Co Co(en)2

NH2 ion.

(b) Discuss briefly the phenomenon of crystal field splitting in a tetrahedral environment. Sol. (a) In coordination compound , both geometrical and optical isomerism are possible

en en en NO NO en en 2 en NO2 en 2 Co Co en en Co Co NH NH NH en 2 en 2 en 2 en en en en

Optical isomers Non- Optical Mesoform superposable mirror isomers image forms Gyaan Sankalp 27 Coordination compounds (b) In a tetrahedral environment, the ligands approach the central metal ion in the manner shown in the diagram. The splitting pattern is as shown in the Figure. Thus in a strong field the lower orbitals (two in no.) get filled then the upper 3- orbitals

L2

+ L1 L1 L2 M L3 L 4 are Ligands L4

L3

d xy dxz dyz

0.4 {

0.6 unsplit { d-orbitals x 2 2 d z  y z2

d-orbitals in the metal ion

Example 23 : 4– 2+ [Fe(CN)6] and [Fe(H2O)6] are the different colours in dilute solutions. Why ? 4– 2+ Sol. [Fe(CN)6] is a low spin complex. All the six electrons in Fe are paired. 2– 2+ [Fe(H2O)] is a high complex. Fe contains four unpaired electrons. That is why the two complexes give different colours in solutions. Example 24 : (a) Discuss one example for each of the following (i) Use of complexation in metallurgy (ii) Complexation in qualitative analysis. (b) Give one example of each of the following in complexes. (i) Ionization isomerism (ii) Linkage isomerism (iii) Hydrate isomerism. Sol. (a) (i) In the metallurgy of Ag, Ag2S ore (Powdered ) is digested with NaCN ; Ag2S + 4NaCN  2NaAg(CN)2 + Na2S, NaAg(CN)2 is the complex ion, sodium argento cyanide. (ii) Nessler's reagent is K2Hgl4 (Comlex compound) solution containing NaOH. It is used for testing ammonium salts. (b) (i) Ionization isomerism [Co(NH3)5 Br] SO4 and [CO(NH3)5 SO4]Br 2+ 2+ (ii) Linkage isomerism [Co(NH3)5 ONO] and [Co(NH3)5 NO2] (iii)Hydrate isomerism [Cr(H2O)6]Cl3 and Cr(H2O)5Cl]Cl2.H2O.

28 Gyaan Sankalp Coordination compounds QUESTION BANK

EXERCISE - 1 ONLY ONE OPTION IS CORRECT (A) +2, +3, 0 (B) +3, 0, +3 Q.1 Which of the following is not an organo- metallic com- (C) +2, +3, +4 (D) 0, +3, +4 pound ? Q.14 Which of the following sets is/are example(s) of co-ordina- (A) C4H9 Li (B) (C2H5)4 Pb tion isomerism is complex ? (C) [(C5H5)2Fe] (D) C2H5–O–Na (A) [Co(NH3)6] [Cr(CN)6] and [Co(CN6)] [Cr(NH3)6] Q.2 Which of the following complex will show geometrical as (B) [Cr(H2O)5Cl]Cl2.H2O and [Cr(H2O)4Cl2]Cl.2H2O well as optical isomerism. [en = ethylene diamine] (C) [Co(NH3)5Br] SO3 and [Cr(NH3)5SO4]Br (D) [Pt(NH ) Cl ] and [Pt(NH ) ] [PtCl ] (A) [Pt(NH3)2Cl2] (B) [Pt(NH3)Cl4] 3 2 2 3 4 4 4+ Q.15 Which of the following has least conductivity in aqueous (C) [Pt(en3)] (D) [Pt(en)2Cl2] Q.3 What is the name of the isomerism of the compounds ? solution- (A) Co(NH ) Cl (B) Co(NH ) Cl (i) [Co(NH3)6] [CrOx3] (ii) [Cr(NH3)6] [Co Ox3] 3 4 3 3 3 3 (A) Coordination isomerism (B) Linkage isomerism (C) Co(NH3)5Cl3 (D) Co(NH3)6Cl3 (C) Ligand isomerism (D) Geometrical isomerism Q.16 In nitroprusside ion, the iron and NO exist as Fe(II) and Q.4 The number of ions formed in aqueous solution by the NO+ rather than Fe(III) and NO. These forms can be differ- entiated by – compound [Co(NH3)4Cl2]Cl is- (A) 2 (B) 3 (A) estimating the concentration of iron (C) 4 (D) 7 (B) measuring the concentration of CN– 2+ (C) measuring the solid state magnetic moment Q.5 Co-ordination number of Co in [Co(NH3)6] is- (A) 4 (B) 5 (D) thermally decomposing the compound (C) 6 (D) 8 Q.17 When the configuration is d7 in a transition metal, the paramagnetic susceptibility will be equal to- Q.6 The correct IUPAC name of AlCl3(EtOH)4 is- (A) Aluminium (II) chloride - 4-ethanol (A) 3.87 B.M. (B) 2.68 B.M. (B) Aluminium (III) chloride -4-ethanol (C) 5.92 B.M. (D) 6.92 B.M. (C) Aluminium (IV) chloride-4-hydroxy ethane Q.18 Among the following complexes the one which contains a (D) Aluminium chloride -4-ethanol d6 metal ion is- – Q.7 Complex with CN ligands are usually (A) K3[Fe(CN)6] (B) K4[Fe(CN)6] (A) High spin complexes (B) Low spin complexes (C) [Co(NH3)6]Cl2 (D) [Ni(NH3)6] Cl2 (C) Both (D) None of these Q.19 Ni2+ is a system of : Q.8 Of the species given below which has a geometry different (A) d7 type (B) d8 type from the other three ? (C) d6 type (D) d5 type  Q.20 The EAN of nickel in Ni(CO) is- (A) Ni(CO)4 (B) [SbCl4] 4 2 2– (A) 36 (B) 38 (C) [Cd(CN)4] (D) [CoCl4] Q.9 The structure of iron pentacarbonyl is- (C) 28 (D) 54 (A) Square planar (B) Trigonal bipyramid Q.21 For which one of the following would it not be possible to (C) Triangular (D) None of these distinguish high spin from low spin complexes in octahe- dral geometry ? Q.10 A cobaltamine has the formula Co(NH3)6Cl3, with AgNO3 solution, one third of the chloride is precipitated. The com- (A) Co (II) (B) Fe (II) plex shows cis-trans isomerism. It can have the structure – (C) Co (III) (D) Ni (II) Q.22 The neutral ligand is- (A) [Co(NH3)6]Cl3 (B) [Co(NH3)5Cl]Cl2 (A) Chloro (B) Hydroxo (C) [Co(NH3)4Cl2]Cl (D) [Co(NH3)5.H2O]Cl3 Q.11 The formula of dichlorobis (urea) copper (II) is- (C) Ammine (D) Oxalato Q.23 Which one of the following complex ions has the largest d- (A) [CuO = C(NH2)2] Cl2 orbital spliting energy ? (B) [CuCl2 {O = C(NH2)2}] (A) [Co(H O) ]+3 (B) [Fe(H O) ]+3 (C) [Cu {O = C (NH2)2} Cl] Cl 2 6 2 6 (C) [CrCl ]–3 (D) [Cr(NH ) ]+3 (D) [CuCl2] [O = C (NH2)2] H2 6 3 6 Q.24 The shape of the complex Ag(NH ) + is : Q.12 An octahedral complex has the formula [Ma2b2c2]. How 3 2 many isomeric structures are possible for it ? (A) Octahedral (B) Square planar (A) 5 (B) 4 (C) Tetrahedral (D) Linear (C) 6 (D) 8 Q.25 Hexafluoroferrate (III) lon is an outer orbital complex. The 4– Q.13 The oxidation number of Fe in [Fe(CN)6] , Co in number of unpaired electrons present in it is. [Co(NH3)3(NO2)3] and Ni in [Ni(CO)4] are respectively (A) 1 (B) 5 (C) 4 (D) Unpredictable

Gyaan Sankalp 29 Coordination compounds +3 3– Q.26 The crystal field splitting energy for Cr ion in an octahe- Q.37 The charge on cobalt in [Co(CN)6] is- – – dral field increases for the ligands I , H2O, NH3, CN and (A) –6 (B) +3 the order is such that (C) –3 (D) +6 – – – – (A) I < H2O < NH3 < CN (B) CN < I < H2O < NH3 Q.38 Which of the following is not a simple salt ? – – – – (C) CN < NH3 < H2O < I (D) NH3 < H2O < I < CN (A) FeSO4 (B) KCl.MgCl26H2O 3+ Q.27 An aqueous solution of [Ti(H2O)6] ion has a mild violet (C) Al2(CO3)3 (D) NH4Br colour of low intensity. Which of the following statements Q.39 When FeCl3 reacts with K3[Fe(CN)6] no blue colour is is incorrect ? observed. K3[Fe(CN)6] is used as indicator in estimation 2+ 2– (A) The ion absorbs visible light in the region of ~ 5000 Å. of Fe by Cr2O7 ion in acidic medium. End point will be (B) The colour results from an electronic transition of one when : electron from the t2g to an eg orbital ? (A) Solution of iron salt does not give blue colour with (C) The low colour-intensity is because of a low probabil- indicator. ity of transition. (B) Solution of iron salt gives blue colour with indicator (D) The transition is the result of metal-ligand back bond- (C) No colour before and after with indicator ing. (D) Always blue colour before and after Q.28 The EAN of platinum in potassium hexachloroplatinate (IV) Q.40 The oxidation number of cobalt in K[Co(CO)4] is- is : (a) +1 (b) +3 (A) 46 (B) 86 (c) –1 (d) –3 (C) 36 (D) 84 Q.41 Point out the wrong statement: In an octahedral geometry, Q.29 Which one of the following complexes has geometrical (A) a central metal ion with d8 configuration has two un- isomers ? paired electrons. + 6 (A) [Co(NH3)3Cl] Cl2 (B) [Co(en)2Cl2] (B) an ion with d configuration is diamagnetic in a strong +2 + (C) [Cr(H2O)5Cl] (D) [Pt(NH3)3Cl] field. Q.30 From the stability constants (hypothetical values), given (C) an ion with d5 configuration has one unpaired electron below, predict which one is the strongest ligand ? both in weak and strong fields. 2+ 2+ 11 4 5 6 7 (A) Cu + 4NH3 [Cu(NH3)4] , K = 4.5 × 10 (D) in d , d , d and d configurations, weak and strong 2+ – 2– 27 field complexes have different numbers of unpaired elec- (B) Cu + 4CN [Cu(CN)4] , K = 2.0 × 10 (C) Cu2+ + 2en [Cu(en) ]2+, K = 3.0 × 1015 trons. 2 Q.42 Amongst the following, the most stable complex is- (D) Cu2+ + 4H O [Cu(H O) ]2+, K = 9.5 × 108 2 2 4 (A) [Fe(H O) ]3+ (B) [Fe(NH ) ]3+ Q.31 Mohr’s salt is a : 2 6 3 6 (C) [Fe(C O ) ]3– (D) [FeCl ]3– (A) Simple salt (B) Complex salt 2 4 3 6 Q.43 The compound [Co(NO )(NH ) ]Cl and (C) Complex ion (D) Double salt 2 3 5 2 [Co(ONO)(NH ) ]Cl are examples of – Q.32 The number of unpaired electrons in [Cr(CN) ]–4 and 3 5 2 6 (A) Geometrical isomers (B) Linkage isomers [Cr(H O) ]+2 respectively are- 2 6 (C) Ligand isomers (D) Ionisation isomers (A) 0 and 4 (B) 2 and 4 Q.44 Among the following complex ion which one exhibits high- (C) 0 and 2 (D) 2 and 2 est paramagnetism Q.33 In lithium tetrahydroaluminate, the ligand is : (A) [Cr(H O )]+3 (B) [Fe(H O) ]+2 (A) Al+ (B) H 2 6 2 6 (C) [Cu(H O) ]+2 (D)[Zn(H O) ]+2 (C) H– (D) None 2 6 2 6 Q.45 Each of the compounds Pt(NH ) Cl . Cr(NH ) Cl , Q.34 The complex species which does not obey the EAN rule is 3 6 4 3 6 3 Co(NH ) Cl and K Pt. Cl were dissolved in water to make (A) Ni(CO) (B)[Cr(NH ) ]+3 3 4 3 2 6 4 3 6 its 0.001 M solution. The correct order of their increasing (C) [Fe(CN) ]–3 (D) [Cu(CN) ]–3 6 4 conductivity in solution is : Q.35 Which has a complex anion ? (A) K Pt.Cl < Co(NH ) .Cl < Cr(NH ) Cl < Pt(NH ) Cl (A) Na AlF (B) K SiF 2 6 3 4 3 3 6 3 3 6 4 3 6 2 6 (B) Cr(NH ) Cl < Co(NH ) Cl < K Pt.Cl < Pt(NH ) Cl (C) K PtCl (D) All 3 6 3 3 4 3 2 6 3 6 4 2 6 (C) Co(NH ) Cl < K Pt.Cl < Cr(NH ) Cl < Pt(NH ) Cl Q.36 A co-ordination complex compound of cobalt has molecu- 3 4 3 2 6 3 6 3 3 6 4 (D) Pt(NH ) Cl < Co(NH ) Cl < Cr(NH ) Cl < K Pt.Cl lar formula containing five ammonia molecules, one nitro 3 6 4 3 4 3 3 6 3 2 6 Q.46 Consider the following complexes : group and two chlorine atoms for one cobalt atom. One (I) K PtCl (II) PtCl .2NH mole of this compound produces three mole ions in an 2 6 4 3 (III) PtCl .3NH (IV) PtCl .5NH aqueous solution. On reacting this solution with excess of 4 3 4 3 Their electrical conductances in aq. solutions are- silver nitrate solution, two molecules of AgCl get precipi- (A) 256, 0, 97, 404 (B) 404, 0, 97, 256 tated. The ionic formula of this compound would be – (C) 256, 97, 0, 404 (D) 404, 97, 256, 0 (A) [Co(NH ) NO Cl] [(NH )Cl] 3 4 2 3 Q.47 Which of the following pairs exhibits linkage isomerism – (B) [Co(NH3)5Cl] [Cl(NO2)] (A) [Pt(NH3)Cl3]Br2 and [Pt(NH3)4Br2]Cl2 (C) [Co(NH3)5NO2]Cl2 (B) [Co(NH3)5NO3]SO4 and [Co(NH3)5SO4]NO3 (D) [Co(NH3)5] [(NO2)2Cl2] 2+ 2+ (C) [Cr(H2O)5SCN] and [Cr(H2O)5NCS] 30 Gyaan Sankalp Coordination compounds

(D) [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl2]Cl2.H2O Q.54 [CO(NH3)5Br]SO4and [Co(NH3)5SO4] Br are examples of Q.48 K3CoF6 is high spin complex. What is the hydbrid state of (A) Linkage isomerism (B) Ionization isomerism Co atom in this complex – (C) Geometric isomerism (D) Optical isomerism 3 3 2 (A) sp d (B) sp d Q.55 Which of the following is most likely structure of CrO3.6H2O (C) d2sp3 (D) dsp2 if 1/3 of total chlorine of the compound is precipitated by Q.49 Hexafluoro ferrate (III) ion is an outer orbital complex. The adding AgNO3 to its aqueous solution – number of unpaired electrons present in it is – (A) CrCl3.6H2O (B) [Cr(H2O)3Cl3].(H2O)3 (A) 1 (B) 3 (C) [CrCl2(H2O)4]Cl.2H2O (D) [CrCl(H2O)5]Cl2.H2O (C) 4 (D) 5 Q.56 The oxidation state of Ag in Tollen’s reagent is : Q.50 Which complex has a square planar structure ? (A) 0 (B) +1 –2 (A) Ni(CO)4 (B) [NiCl4] (C) +2 (D) +1.5 –2 +2 1 10 (C) [Ni(CN)4] (D) [Cu(NH3)4] Q.57 Consider transition metal ions which have d to d elec- Q.51 The hybridisation of Mn in hexacyano-manganate (II) is trons. Which of these can form high spin and low spin sp3d2. The number of unpaired electrons in it is – complexes in octahedral geometry. (A) 1 (B) 2 (A) d3, d5, d6, d9 (B) d4, d5, d6, d7 (C) 5 (D) 3 (C) d2, d5, d7, d9 (D) d4, d6, d7, d10 –3 2– 2– Q.52 The oxidation state of chromium in [Cr(C2O4)3] is Q.58 Among [Ni(CO)4], [Ni(CN)4] and [Ni(Cl)4] : 2– (A) +6 (B) +4 (A) [Ni(CO)4] and [Ni(Cl)4] are diamagnetic and 2– (C) +2 (D) +3 [Ni(CN)4] is paramagnetic 2– 2– Q.53 A certain complex ion, with octahedral geometry has six (B) [Ni(Cl)4] and [Ni(CN)4] are diamagnetic and n+ different ligands : [ML1L2L3L4L5L6] . How many iso- [Ni(CO)4] is paramagnetic 2– meric structures are possible, counting optical isomers (C) [Ni(CO)4] and [Ni(CN)4] are diamagnetic and 2– separately? [Ni(Cl)4] is paramagnetic 2– (A) 120 (B) 30 (D) [Ni(CO)4] is diamagnetic and [Ni(CN)4] and 2– (C) 24 (D) 15 [Ni(Cl)4] are paramagnetic.

EXERCISE - 2 ONE OR MORE THAN ONE CHOICE MAY Q.4 Which of the following statements is/are correct – BE CORRECT (A) Formation of co-ordinate bond between a ligand and a Q .1 Which of the following statements is/are correct – central metal ion with a vacant orbital is acid-base reaction (A) Cyanides are very toxic in nature according to Lewis theory (B) The cyanide ion often brings out the maximum co-ordi- (B) The acidity of a hydrated metal ion depends on the nation number of a metal strength of the bond between cation and oxygen (C) Many metal ions, which are too unstable to exist in (C) Highly charged cations are expected to form most solution, are quite stable when complexed with cyanide strongly acidic hydrated cations ions (D) None is correct (D) None is correct Q.5 Which of the following ligands form generally the high Q.2 Which of the following is (are) correct statement(s) (as- spin complexes – – – suming oxidation number of metal does not affect crystal (A) I (B) Br – – field energy) (C) Cl (D) F Q.6 Which statements is/are correct for the magnitude of crys- (A) Considering H2O to be a weak ligand then on the basis 2+ tal field splitting for octahedral complexes of the same of CFSE only, we can say that [Co(H2O)6] is more stable 3+ ligand with metal ions of the same group in the same oxida- than [Co(H2O)6] 2+ tion state, but in different rows of the periodic table – (B) On the basis of CFSE only [Fe(NH3)6] is more stable +3 (A)  increases by about 25-50% on going from first tran- than [Fe(NH3)6] 0 (C) All octahedral complexes of Ni(ll) are bound to be outer sition series to the second d-complex (B) 0 increases by another 25-50% on going from the (D) The type of d-orbital involved in the hybridisation for second transition series to the third a square planar complex (CN = 4) is d . (C) 0 decreases by about 50% on going from first transi- z2 tion series to the second Q.3 Which of the following complex ions do not obey EAN (D) 0 decreases by about 25-50% on going from the sec- rule – ond transition series to the third 2– – (A) Zn(OH)4 (B) Ag(CN)2 2– 2+ (C) HgI4 (D) Cu(NH3)4

Gyaan Sankalp 31 Coordination compounds

Q.7 Identify the complexes which are expected to be coloured 3 + – Statement 2 : Size of [Al(H2 O) 6 ] is smaller than (A) Ti(NO3)4 (B) [Cu(NCCH3)4] BF4 3+ – (C) [Cr(NH3)6] 3Cl (D) K3 [VF6] [Mg(H O) ]2 and possesses more effective nuclear Q.8 Wilkinson’s catalyst react with H to form an octahedral 2 6 2 charge. complex in which Rh(Z = 45) has the following electronic 3+ 3+ configuration in the ligand field t 2,2,2 , e 0,0. Then which Q.17 Statement- 1 : [Ti(H2O)6] is coloured, while [Sc(H2O)6] 2g g is colourless. of the following is (are) correct about this new complex ? 3+ – Statement - 2 : Ti has one e in one of its 3d orbitals (A) It is diamagnetic – 3+ (B) Its IUPAC name is chlorodihydridotris while there is no e in 3d orbitals of Sc . (triphenylphosphine) rhodium (III) (C) Hybridisation of Rh(l) is d2sp3 MATCH THE COLUMN TYPE QUESTIONS (D) It is a paramagnetic complex (Q.18-Q.20) Each question contains statements given in two columns ASSERTION AND REASON QUESTIONS which have to be matched. Statements (Q.9-Q.17) (A, B, C, D) in column I have to be matched with state- Note : Each question contains STATEMENT-1 (Assertion) ments (p, q, r, s) in column II. and STATEMENT-2 (Reason). Each question has 5 choices Q.18 Match the column correctly – (A), (B), (C), (D) and (E) out of which ONLY ONE is cor- Column I Column II rect. (A) V (CO)6 (p) Paramagnetic with 1 (A) Statement-1 is True, Statement-2 is True; Statement-2 unpaired electron is a correct explanation for Statement-1. (B) [NiCl2 (PPh3)2] (q) Paramagnetic with 2 unpaired electrons (B) Statement-1 is True, Statement-2 is True; Statement-2 2+ 3 is NOT a correct explanation for Statement-1. (C) [Ni(NH3)4] (r) sp hybridization (C) Statement -1 is True, Statement-2 is False. (D) Ni (CO)4 (s) Diamagnetic (D) Statement -1 is False, Statement-2 is True. Q.19 Match the column correctly – (E) Statement -1 is False, Statement-2 is False. Column I Column II (A) [Pt(NH3)2Cl2] (p) 15 geometrical, 30 optical Q.9 Statement 1 : The solution of [PtCl2(NH3)2] is a nonconductor. (B) [Pt (NH3)(Py)(I)(SH)] (q) no geometrical, no optical (C) [Pt(Py) (OH) (Br) (I) (Cl) (r) 3 geometrical, 0 optical Statement 2 : The solution of [Pt(NH3)4]Cl4 is a conductor. (NH3)] Q.10 Statement - 1 : K3[Fe(CN)6] is a low spin complex. – 2+ 3 2 (D) [Pt(NH3)Cl5] (s) 2 geometrical, 0 optical Statement - 2 : Fe ion in this complex undergoes sp d hybridization. Q.20 Match the column correctly – Column I Column II Q.11 Statement - 1 : Ratio of no. of geometrical isomers to optical isomers of a metal complex of coordination no. 6 having 3 Complex Hybridization / Magnetic different types of monoatomic monodentate ligands is property 2 greater than one. (A) Fe NH  (p) Octahedral  3 6  Statement- 2 : More than 50% of the geometrical isomers Na Fe CN NO  2 3 of the complex are optically inactive. (B) 2  5  (q) d sp Q.12 Statement -1 : [Ni(CN) ]2– is diamagnetic. 3 4 (C) Fe CN  (r) Paramagnetic Statement - 2 : Ni2+ is dsp2 hybridised. 6  4 (D) Fe CN  (s) diamagnetic Q.13 Statement - 1 : K4[Fe(CN)6] is a low spin complex.  6  2+ 3 2 Statement - 2 : Fe ion in this complex undergoes sp d hybridization. PASSAGE BASED QUESTIONS Q.14 Statement-1:Tetrahedral complexes do not show Passage 1- (Q.21-Q.23) geometrical isomerism. In coordination chemistry there are a variety of methods Statement - 2 : The relative positions of the ligands in the applied to find out the structures of complexes. One method tetrahedral complex are the same w.r.t. each other. involves treating the complex with known reagents and 2+ Q.15 Statement 1 : Ammonia forms the complex ion [Cu(NH3)4] from the nature of reaction, the formula of the complex can with copper ions in alkaline solution but not in acidic be predicted. An isomer of the complex Co(en)2(H2O)Cl2Br, solutions. on reaction with concentrated H2SO4 (dehydrating agent) Statement 2 : In acidic solutions protons coordinate with it suffers loss in weight and on reaction with AgNO + 3 ammonia molecules forming NH4 ions and NH3 molecules solution it gives a white precipitate which is soluble in are not available. NH3 (aq). 3 Q.21 The correct formula of the complex is : Q.16 Statement 1 : [Al(H2 O) 6 ] is stronger acid then (A) [CoClBr(en)2] H2O (B) [CoCl(en)2 (H2O)] BrCl 2 (C) [CoBr(en) (H O)]Cl (D) [CoBrCl(en) ]Cl.H O [Mg(H2 O) 6 ] . 2 2 2 2 2 32 Gyaan Sankalp Coordination compounds Q.22 If all the ligands in the coordination sphere of the above (A) The C-O bond length increases complex be replaced by SCN–, then the paramagnetic mo- (B) The bond order of CO increases ment of the complex ion (due to spin only) will be : (C) The C-O bond length does not change (A) 2.8 BM (B) 5.9 BM (D) None of these (C) 4.9 BM (D) 1.73 BM Q.23 Similarly if all the ligands in the coordination sphere be Directions : Q. 27-Q.29 replaced by NCS– then the paramagnetic moment of the The cationic and neutral complex ions are named by writ- complex ion (due to spin only) will be : ing, “the number and names of the ligands followed by the (A) 1.73 BM (B) 0.0 BM name of the central metal ion with its oxidation number in (C) 4.9 BM (D) 5.9 BM Roman numerals within the parenthesis. The name of the complex cation followed by the name of Passage 2- (Q.24-Q.26) the simple anion is used to name coordination compounds Compounds containing M-C bonds are called organome- having complex cation and simple anion. tallic compounds of which carbonyls constitute an impor- Q.27 Name of [CoBr(NH3)4(NO2)] is – tant part. Synergic effect in bonding in the metal carbonyls (A) tetraamminebromonitrocobalt(II) makes them stable species, metal carbonyls abide by the (B) triamminetri(thiocyanato-S)cobalt(III) effective atomic number (EAN) rule. (C) dichlorobis(ethylenediamine)cobalt(II) Q.24 Which of the following is not an organ metallic compound (D) tetraamminedichlorocobalt(III) chloride Q.28 Name of [Co(NH ) (SCN) ] is – (A) Al C H O (B) Fr C H 3 3 3  2 5 3  5 5 2 (A) tetraamminebromonitrocobalt(II)  (B) triamminetri(thiocyanato-S)cobalt(III) (c) PtCl C H  (D) Cr Co 3 2 2    6 (C) dichlorobis(ethylenediamine)cobalt(II) x (D) tetraamminedichlorocobalt(III) chloride Q.25 The value of x in Mn Co  is  5  Q.29 Name of [CoCl2(en)2] is – (A) 3 (B) 2 (A) tetraamminebromonitrocobalt(II) (C) 1 (D) 0 (B) triamminetri(thiocyanato-S)cobalt(III) Q.26 Which of the following is true for the synergic bonding in (C) dichlorobis(ethylenediamine)cobalt(II) metal carbonyls ? (D) tetraamminedichlorocobalt(III) chloride EXERCISE - 3

SUBJECTIVE QUESTIONS Q.8 Using IUPAC norms, write the formulae for the following : Q.1 Determine the oxidation state of metal in the complex ion, (a) Tetrahydroxozincate (II) 2– [PtCl6] . (b) Hexaamminecobalt (III) sulphate Q.2 Write the formulae of the following coordination com- (c) Potassium tetrachloropalladate (II) pounds (d) Potassium tri(oxalato) chromate (III) (i) Potassium hexanitritoferrate (III) (e) Diamminedichloroplatinum (II) (ii) Dichloroplatinum (IV) tetrachloroplatinate (II) (f) Hexaammineplatinum (IV) (iii) Bis (acetylacetonato) oxovanadium (IV) (g) Potassium tetracyanonickelate (II) (iv) Potassium tetrahydroxozincate (II) (h) Tetrabromocuprate (II) (v) Sodium pentacyanonitrosylsulphidoferrate (III) (i) Pentaamminentrio-O-cobalt (III) Q.3 What will be the correct order for the wavelengths of ab- (j) Pentaamminenitrio-N-cobalt (III) sorption in the visible region for the following : Q.9 Assign the charge on the following ions : [Ni(NO ) ]4–, [Ni(NH ) ]2+, [Ni(H O )]2+ x 2 6 3 6 2 6 [Co(NH3)2Cl4] Q.4 Why NH3 forms co-ordinate complexes, while the isoelec- Q.10 Which of the following names are not correct ? Point out tronic species CH4 does not ? the mistakes. Q.5 The EAN of each Mn (Z = 25) in its carbonyl is 36. What is (i) [Cu(H2O) (NH3)] Br2; Amminoaquodibromocopper (I) the structure of the carbonyl with molecular formula Mn 2 (ii) Na3 [Al(C2O4)3] Trisodium trioxalatoaluminate (III) (CO) ? 10 (iii) Na2 [Ni(EDTA)]; Sodium ethylenediaminetetracetato Q.6 Give examples of (a) Sigma bonded, and (b) pi bonded nickel (III) compounds. (iv) [Co(NH3)5 ONO] SO4; Pentaamminenitrocobalt (III) Q.7 FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 mo- sulphate. lar ratio gives the test of Fe2+ ion but CuSO solution 3+ 3+ 4 Q.11 [Ti(H2O)6] is coloured while [Sc(H2O)6] is colourless. mixed with aqueous ammonia in 1:4 molar ratio does not Explain. give the test of Cu2+ ion. Explain why ?

Gyaan Sankalp 33 Coordination compounds Q.12 (a) Give the IUPAC names of the following coordination Q.14 (a) Each of the following complex ions is either tetrahedral compounds or square planar. On the basis of number of unpaired elec- tron (given in brackets) decide which is the correct geom- Pt NH Cl  Pt NH Cl  (i)  34 2  Cl2 (ii)  32 2  etry ? (i) Pt(NH ) ++(0) (ii) Co (en) ++ (1) (iii) K2[PtCl6] 3 4 2 – – – (b)Write the formulae of the given coordination compounds (iii) FeCl4 (5) (iv) Co(NCS)4 (3) (i) Hexa aqua iron (II) chloride (b) Obtain the distribution of d-electrons in the complex (ii) Potassium tetra fluoro argentate ions listed below, using the crystal field theory. ++ ++ (iii) Penta chlorotitanate (II) ion (i) Pt(NH3)4 (0) (ii) Co (en)2 (1) – – – Q.13 Give the name of the following : (iii) FeCl4 (5) (iv) Co(NCS)4 (3) 2+ + Q.15 [Cu(CN) ]2– is more stable complex than [Cu(NH ) ]2+. (a) [Cu(NH3)2(H2O)2] (b) [Co(NH3)4Cl2] 4 3 4 + Explain. (c) [Pt(NH3)3CN] EXERCISE - 4 + PREVIOUS YEAR IIT-JEE QUESTIONS (A) [Mn(CO)6] (B) [Fe(CO)5] – Q.1 The complex ion which has no ‘d’ electron in the central (C) [Cr(CO)6] (D) [V(CO)6] metal atom is – [2001] Q.7 Match the complexes in Column-I with their properties listed – 3+ in Column-II. [2007] (A) [MnO4] (B) [Co(NH3)6] 3– 3+ Column-I Column-II (C) [Fe(CN)6] (D) [Cr(H2O)6] Q.2 The species having tetrahedral shape is – [2004] (A) [Co(NH3)4(H2O)2]Cl2 (p) geometrical isomers 2– 2– (B) [Pt(NH ) Cl ] (q) paramagnetic (A) [PdCl4] (B) [Ni(CN)4] 3 2 2 2– 2– (C) [Co(H O) Cl]Cl (r) diamagnetic (C) [Pd(CN)4] (D) [NiCl4] 2 5 Q.3 The spin magnetic moment of cobalt in the compound (D) [Ni(H2O)6]Cl2 (s) metal ion with +2 oxidation state Hg[Co(SCN)4] is – [2004] Q.8 The IUPAC name of [Ni(NH ) ][NiCl ] is [2008] (A) (B) 3 4 4 3 8 (A) Tetrachloronickel (II) – tetraamminenickel (II) (C) 15 (D) 24 (B) Tetraamminenickel (II) – tetrachloronickel (II) Q.4 Which kind of isomerism is exhibited by octahedral (C) Tetraamminenickel (II) – tetrachloronickelate (II) (D) Tetrachloronickel (II) – tetraamminenickelate (0) Co(NH3)4Br2Cl – [2005] (A) Geometrical and ionization Q.9 Statement 1 : The geometrical isomers of the complex (B) Geometrical and optical [M(NH3)4Cl2] are optically inactive. [2008] (C) Optical and ionization and (D) Geometrical only Statement 2 : Both geometrical isomers of the complex [M(NH ) Cl ] possess axis of symmetry. Q.5 CuSO4 decolourises an addition of KCN, the product formed 3 4 2 is – [2006] (A) Statement 1 : is True, statement 2 is True; statement 2 is 2+ 3– correct explanation for statement 1 (A) Cu get reduced to form [Cu(CN)4] (B) [Cu(CN) ]2– (C) CuCN (B)Statement 1 : is True, statement 2 is True; statement 2 is 4 NOT a correct explanation forstatement 1 : (D) Cu(CN)2 Q.6 Among the following metal carbonyls, the C – O bond (C) Statement 1 : is True, statement 2 is False order is lowest in – [2007] (D) Statement 1 : is False, statement 2 is True

34 Gyaan Sankalp Coordination compounds HINTS & SOLUTIONS

EXERCISE - 1 Q 1 2 3 4 5 6 7 8 9 10 11 A D D A A C B B B B C B Q 12 13 14 15 16 17 18 19 20 21 22 A C A A B C A B B A D C Q 23 24 25 26 27 28 29 30 31 32 33 A D D B A D B B B D B C Q 34 35 36 37 38 39 40 41 42 43 44 A C C C B B A C C C B B Q 45 46 47 48 49 50 51 52 53 54 55 A C A C B D C A D B B C Q 56 57 58 A B B C (1) (D) Compounds in which the carbon or organic groups are have been changed within the co-ordination sphere. directly boned to metal atoms are known as organometallic Choices (B), (C) and (D) are examples of hydrate isomers, compounds. D is sodium ethoxide and alkoxide. ionisation isomers and polymerisation isomers, respectively. (2) (D). [Pt(en)2Cl2] will exhibit geometrical as well as optical (15) (B). Werner’s theory isomerism. Geometrical isomerism is found in compounds [Co(NH3)3Cl3] is non electrolyte. having co-ordination number 4 (having square planar struc- (16) (C). Fe (II) contains 4 unpaired electrons, ture) and co-ordination number 6. So  4(4  2)BM  4.9BM (4) (A). The dissociation of given compound is as follows + – Fe (III) contains 5 unpaired electrons, [Co(NH3)4Cl2]Cl [Co(NH3)4Cl2] + Cl (5) (C). Co-ordination number is equal to total number of So  5(5  2) BM  5.9 BM . ligands in a complex. (17) (A). The magnetic moment values in Bohr magneton for 1 (7) (B). A strong field ligand produces low spin complexes. lone electron is 1.73, for 2 is 2.83, for 3 is 3.87 for 4 is 4.90, (9) (B) for 5 is 5.92, for 6 is 6.92 and so on d7 configuration has xx xx xx xx xx three lone electrons, hence,  = 3.87 BM. (19) (B). Ni2+ has eight electrons in 3d orbital. dsp3 (20) (A). Effective atomic no. (EAN) = at. no. of central atom - oxidation state xx electron pair donated by CO. + 2 × (no. of ligands) = 28 – 0 + 2 × 4 = 36. (10) (C). This compound has only one ionisable Cl– out of given (22) (C). :NH3 has no charge but acts as ligand. three. + + – (24) (D). Ag(NH3)2 has sp-hybridisation and linear complex. [Co(NH3)4Cl2]Cl [Co(NH3)4Cl2] + Cl (25) (B). Electronic configuration of [FeF ]3– is : It also exhibits cis-trans isotherm. 6 (11) (B). Follow IUPAC rules. 3d 4– (13) (A). [Fe(CN)6] x + { – 6} = – 4 c = + 2 4s 4p 4d [Co(NH ) (NO ) ] 3 3 2 3 xx xx x + (3 x 0) + (–3) = 0 xx xx xx xx

x – 3 = 0 x = + 3 3 2 sp d [Ni(CO)4] x + (4 × 0) = 0 (26) (A). Weak field ligands produce small degree of splitting of x = 0 ‘d’ orbitals and strong field ligands cause large splitting. I– (14) (A). When both positive and negative ions are complex, is the weakest ligand, CN– is the strongest. co-ordination isomerism may occur due to the interchange (27) (D). Ti(H O )3 has one unpaired electron in its d-subshell of ligands within the co-ordination sphere itself e.g. 2 6 which gives rise to d-d transition to impart colour. [Co(NH3)6] [Cr(CN)6] and [Co(CN)6] [Cr(NH3)6] are two co-ordination isomers, where ammonia and cyanide ligands Gyaan Sankalp 35 Coordination compounds + 3– (28) (B). EAN = atomic no. of the metal - no. of electron lost in (48) (B). K3CoF6 3K + CoF6 3+ 3– ion formation + no. of electrons gained from the donor The electronic configuration of Co in CoF6 is atoms of the ligand = 78 – 4 + 2 × 6 = 86. 3d 4s 4p 4d (30) (B). Greater the value of stability constant, stronger is the ligand. (31) (D). The union of two or more molecules of simple salts, 3– (49) (D). Hexafluoroferrate (III) is [FeF6] . The O.N. of Fe in it when combining through electrostatic forces of attraction, is x + (–6) = – 3 or x – 6 = –3 or x = + 3. combine with each other without loosing their chemical Thus in it Fe is present as Fe3+, which contains 5 unpaired identity, generally combining by simple physical methods, electrons. the resulting products are known as double salts; e.g., (51) (A). The configuration of Mn is [Ar]d5. Thus it is expected – carnallite (KCl.MgCl2.6H2O), chromealum to have 5 unpaired electrons. But the strong ligand CN (Cr2(SO4)3.K2SO4.12H2O), Mohr’s salt causes the pairing of electrons as (FeSO4.(NH4)2.6H2O), etc. – (33) (C). LiH + AlH3  Li[AlH4]; H of LiH acts as ligand. (35) (C). Na3AlF6 and K2SiF6 actually represent double salt Thus there is only one unpaired electron left. – since presence of F is detected as 3NaF.AlF3 and 2KF.SiF4. (53) (B). 5 × 3 × 2 ways of accommodating the ligands. True complex is K2PtCl6. (55) (C). [CrCl2(H2O)4]Cl.2H2O has only one ionisable Cl out (36) (C). Because only (C) gives two moles of AgCl with AgNO3 of three Cl atoms present in the compound. Two Cl atoms and yields three ions (including two ionisable Cl– ions) in which are non-ionisable are present in co-ordination aqueous solution. sphere. 2+ – [Co(NH3)5NO2]Cl2]  [Co(NH3)5NO2] + 2Cl (56) (B). Ag in Tollens reagent exists as Ag2O (37) (B). The algebraic sum of oxidation no. of all atoms in a 2 × a + 1 × (–2) = 0 a = + 1. complex ion is equal to charge on that ion, (57) (B). Metal ions with d4, d5, d6 and d7 configuration can i.e., a + 6 × (–1) = – 3  a = + 3. form high spin and low spin complexes. For transition metal (38) (B). A simple salt is a compound which has been prepared ions with configuration d1, d2, d3, d8, d9 or d10 all the by the combination of two simple ions like Na+ and Cl– octahedral complexes of a given metal ion have approxi- ions to form NaCl. Similarly KBr, K2S, MgCl2 CaC2 etc., mately the same magnetic moment and there is no high simple ionic salts. Another set of simple salts can be Zn3P2, spin versus low spin complexes. V2O5 etc., which are bonded covalently. Salts like FeSO4, (58) (C). [NiCO)4] : Here, nickel is present in zero oxidation Al2(CO3)3, NH4Br etc., are ionic, but formed by the combi- state, i.e. in the form of metallic nickel. 2 nation of a simple ion and a compound ion such as SO4 , Ni (Atomic number = 28) : 2– + CO3 , NH4 etc. 3d 4s 4p (39) (A). At the end point, there is no Fe2+, all iron in Fe3+ hence no blue colour with K3[Fe(CN)6] hence (A). (40) (C). The correct answer is (c) In the presence of strong CO ligand, rearrangement takes +1 x 0 K [Co (CO )4] place and electrons are paired up against the Hund’s rule. So x + 1 = 0 or x = – 1 i.e. the two 4s electrons go to the 3d orbitals in order to 5 (41) (C). A d -ion has 5 unpaired electrons in a weak field. vacant the 4s orbital for electrons donated by the ligand 2– (42) (C). It is due to chelation effect of C2O4 ligand. CO. (43) (B). The ligand NO2 has two types of linkage with central Ni after rearrangement : atom. In NO , N donates electron pair, while in O – NO the 2 3d O atom donates the electron pair. 4s 4p (45) (C). In aqueous solution (0.001 M), the complexes will dis- sociate to give the ions : The empty one 4s and three 4p orbitals mix (sp3 Pt(NH ) Cl [Pt(NH ) ]4+ + 4Cl–; 5 ions. 3 6 4 3 6 hybridisation) to form four equivalent hybrid orbitals each 3+ – Cr(NH3)6Cl3 [Cr(NH3)6] + 3Cl ; 4 ions. of which accepts an electron pair from CO molecule form- Co(NH ) Cl [Co(NH ) Cl ]+ + Cl–; 2 ions. 3 4 3 3 4 2 ing [Ni(CO)4]; which is, thus, tetrahedral and diamagnetic 2– + K2PtCl6 [PtCl6] + 2K ; 3 ions. in nature. 2– (46) (A). C.N. of Pt is 6 hence Similarly in [Ni(CN)4] : 2+ I is K2PtCl6 - Three ions Ni ion : II [Pt(NH3)2Cl4] - One molecule 3d 4s 4p III [Pt(NH3)3Cl3] Cl - Two ions IV [Pt(NH3)5Cl]Cl3 - Four ions Conductivity  no. of ions. The 3-d electrons rearrange against Hund’s rule in pres- 2+ 2+ – (47) (C). [Cr(H2O)5SCN] and [Cr(H2O)5NCS] exhibits link- ence of strong CN ligand as follow : age isomerism 36 Gyaan Sankalp Coordination compounds 2+ Ni ion : 3d 4s 4p 3d 4s 4p

Since, chloride ion is a weak ligand and can not force the The empty one 3d, one 4s and two 4p orbitals mix (dsp2 pairing up of electrons against Hund’s rule, none of 3d hybridisation) to form four dsp2 orbitals, each of which orbital is involved in hybridisation. The presence of two – 2– unpaired electrons corresponds to the paramagnetic accepts an electron pair from a CN ion forming [Ni(CN)4] 2– ion. The complex is square planar and is diamagnetic. behaviour of [Ni(Cl)4] ion. 2– Here, one 4s and 3p orbitals mix to give four equivalent sp3 In [Ni(Cl4)] Ni2+ ion : hybrid orbitals which can accept four lone pairs donated by four ligands i.e. Cl– ions.

EXERCISE - 2 (1) (ABC). (B) When cation oxygen bond is strong, the bond between (A) Toxicity of cyanide is due to the fact that CN– ion oxygen and hydrogen is weakened and the proton can be forms complexes with metals present in enzymes and hae- denoted to a base. If cation-oxygen bond is weak, the oxy- moglobin in the body and prevents normal metabolism. gen-hydrogen bond is difficult to break, and only the en- 3+ – – tire water molecule can be split from the cation. (B) For example, Fe gives [FeCl4] with Cl ions but 3– (C) Trivalent cations, as a group, are more acidic than biva- [Fe(CN)6] with cyanide ions. (C) Cu+, Ni+, Mn+, Au+ and Mn3+ etc. are too unstable to lent cations, and most monovalent cations have negligible exist in solution. In the extraction of Ag and Au, the metal acidic character. When the cation is a Lewis acid and has dissolves in a solution of AgCN in the presence of air and vacant atomic orbitals, the bond between cation and oxy- form sodium argentocyanide, from which the metal is re- gen has covalent character. covered by reduction with Zn. (5) (ABCD). The halide ligands produce a small crystal field splitting (a weak ligand field). Therefore, they usually form 2Ag + 4NaCN + H2O + O 2Na[Ag(CN)2] + 2NaOH (2) (ABC). high spin complexes. (6) (AB). For the third transition series the valence electrons (A) [Co(H O) ]2+  t 2, 2, 1 e 1, 1 2 6 2g g are 5d electrons, which extend farther out from the nucleus  CFSE = – 0.8  0 than the 4d and 3d, and therefore, interact more strongly 3+ 2, 1, 1 1, 1 [Co(H2O)6]  t2g eg with the ligands.  CFSE = – 0.4  3+ 3+ 3+ 0 (7) (CD). Are coloured because Cr in [Cr(NH3)6] and V 2+ 2, 2, 2 0, 0 in [VF ]3– has unpaired electron in d-subshell. (B) [Fe(NH3)6]  t2g eg 6 (8) (AB).  CFSE = – 2.4 0 3+ 2, 2, 1 0, 0 I III [Fe(NH ) ]  t e [Rh Cl(PPh3)3] + H2  [Rh ClH2 (PPh3)3] 3 6 2g g III 2,  CFSE = – 2.0  [Rh Cl H2 (PPh3)3] gives electronic configuration of t2g 0 2, 2 0, 0 (C) For co-ordination no six, two empty d-orbitals are not , eg , inner orbital complex and thus to have two d- 2 3 available for d2sp3 as Ni(ll) has 3d8 4s0 configuration. orbitals empty, electrons will pair up and give d sp hybridisation. RhI with C.N = 4 has dsp2 hybridisation (D) d-orbital involved is d not d 2 . x2 y 2 z (4d series element) (3) (BD). Z of Ag = 47. Ag+ ion contributes 46 electrons and (9) (B). The solution of [PtCl2(NH3)2] is a nonconductor. This 2CN– ligands contribute 4 electrons. is because this complex does not furnish any ion solution. Thus EAN = 46 + 4 = 50. The solution of [Pt(NH3)4]Cl4 is a conductor. This is There is no rare gas with Z = 50. Similarly Z of Cu = 29. because this complex compound dissociates into ions 2+ when dissolved in water. Thus Cu ion contributes 27 electrons and 4 NH3 ligands contribute 8 electrons. Thus EAN = 27 + 8 = 35. There is no (10) (C) (11) (A) rare gas with Z = 35. Since Cu2+ has an odd number of (12) (B) (13) (C) electrons, there is no co-ordination number that could (14) (A) achieve the EAN of a rare gas. It should be noted that if the (15) (A). : NH3 molecules is a base due to the presence of a lone bonding between metal and ligand is largely covalent, EAN pair of electrons on it. + rule is expected to apply. The EAN rule is less likely to In acidic medium, it accepts a proton to form NH4 and apply if the metal-ligand bond has a large amount of ionic there is no free NH3 molecule. character. : NH H  NH  (4) (ABC). 3 4 almost no free NH3 it does not act (A) A ligand acts as Lewis base (electron pair donor) and remains in the solution as a ligand metal ion acts as Lewis acid (electron pair acceptor) (16) (A) (17) (A) Gyaan Sankalp 37 Coordination compounds (18) (A)  p, (B)  q, r ; (C)  s (D) rs no. of unpaired electrons = 4 ;  = 4(4 2) = 4.9 B.M. (19) (A)  s, (B)  r ; (C)  p (D) q – (20) (A)  p, r, (B)  p, q, s ; (C)  p, q, r (D) p, q, s (23) (B). NCS is a strong field ligand and, therefore, all elec- 3+ 6 (21) (D). As it suffer loss in wt with H SO , ionization sphere trons will pair up in Co ion giving  = 0, (3d configura- 2 4 tion) contains H2O molecule. As complex also give white ppt. with AgNO , the ionisation sphere should contain Cl– ion. (24) (A) (25) (C) (26) (A) 3 (27) (A) (28) (B) (29) (C) So formula may be [CoBrCl(en)2]Cl.H2O. (22) (C). SCN– is weak field ligand [CoBr(NH3)4(NO2)] is tetraamminebromonitrocobalt(II) [Co(NH3)3(SCN)3] is triamminetri(thiocyanato-S)cobalt(III) [CoCl2(en)2] dichlorobis(ethylenediamine)cobalt(II) 3d6

EXERCISE - 3 SUBJECTIVE QUESTIONS like simple salts dissociate into the constituent ions when 2+ (1) Charge on the complex ion = Oxidation state of metal + dissolved in water. So, it gives the test of Fe ion. charge on ligands CuSO4 with aqueous ammonia forms a complex 2+ – 2 = x + 6 × (– 1) or x = + 4 [Cu(NH3)4] . This complex does not dissociate into free 2+ The oxidation state of Pt in the complex ion is +4. Cu ions and NH3 molecules. As a result it does not give 2+ (2) (i) K [Fe(ONO) ] (ii) [PtCl ] [PtCl] (iii) [VO(acac) ] test for Cu ions. 3 6 2 2 2– (8) (a) [Zn(OH)4] (b) [Co(NH3)6]SO4 (iv) K2 [Zn(OH)4] (v) Na4[Fe(CN)5NOS] (3) The strength of ligands follows the order (c) K2[PdCl4] (d) K3[Cr(ox)3] 4+ NO – > NH > H O (e) [PbCl2(NH3)2] (f) [Pt(NH3)6] 2 3 2 2– So, the splitting of energy levels follow the order (g) K2[Ni(CN)4] (h) [CuBr4] 2+ 2+ 4– 2+ 2+ (i) [Co(ONO) (NH3)5] (j) [Co(NO2) (NH3)5] [Ni(NO2)6] > [NI(NH3)6] > [Ni(H2O)6] Therefore, the order of wavelength of absorption in visible (9) The oxidation state of Co in the complex is +3 – region is [Ni(H O) ]2+ > [Ni(NH ) ]2+ > [Ni(NO ) ]4– So x = + 3 + 2 × 0 + 4 × (–1) = – 1, i.e., [Co(NH3)2Cl4] 2 6 3 6 2 6 (10) (i) Ammine aqua copper (II) bromide (4) NH3 has lone pair of electrons on N unlike CH4 (5) Electron from each Mn = 25 (ii) Sodium trioxalato aluminate (III) Electrons from five CO ligands = 2 × 5 = 10 (iii) It is correct Electron from (Mn – Mn) bond = 1 (iv) Pentammine nitrito cobalt (III) sulphate 3+ Thus EAN = 25 + 10 + 1 = 36 (11) Sc(H2O)6 has no unpaired electron in its d subshell and 3+ Thus in the complex five CO (ligands) are coordinated to thus d-d transition is not possible whereas Ti(H2O)6 each Mn atom and sixth coordination number is attained has one unpaired electron in its d subshell which gives by other Mn atom of (Mn – Mn) bond. Thus it can have rise to d-d transition, to impart colour. structure. (12) (a) (i) Tetramine dichloro platinum (IV) chloride (ii) Diamine dichloro platinum CO CO CO (iii) Potassium hexachloro platinate CO 3– (b) (i) [Fe(H2O)6] Cl2 (ii) K[AgF4] (iii) TiCl5 CO Mn Mn CO (13) (a) Diammine diaquo copper (II) ion. CO CO (b) Tetrammine dichloro cobalt (III) ion. CO (c) Triammine cyano platinum (II) ion. (6) Grignard reagents (RMgX), dimethyl aluminium, tetraethyl (14) (a) (i) Square planar (ii) Square planar (iii) Tetrahedral (iv) Tetrahedral lead, Pd(C2H5)4, n-butyl lithium etc. are sigma bonded com- pounds in which metal and carbon atom are linked with  (b) (i) 8 electrons in square planar field (low spin) bonds. Dibenzene chromium, ferrocene, Ziesse’s salt, (ii) 7 electrons in square planar field (low spin) (iii) 5 electrons in tetrahedral field (high spin) (PH3P)3 RhCl etc are bonded compounds in which  electrons of ligands take part in the formation of bonds (iv) 7 electrons in tetrahedral field (high spin) between metal and ligand. (15) The higher stability constant (K = 2 × 1027) for Cu2+ + 4CN–  [Cu(CN) ]2– than for [Cu(NH ) ]2+ which (7) FeSO4 and (NH4)2SO4 solutions when mixed in 1:1 molar 4 3 4 ratio forms a double salt called Mohr’s salt. Double salt is 4.5 × 1011 explains stability. EXERCISE - 4 (1) (B) (2) (B) (7) A  p, q, s (B)  p, r, s , (C)  q, s (D)  q, s (3) (C) (4) (A) (8) (C) (5) (A) (6) (D) (9) (B). The molecule should not posses alternate axis of sym- metry to be optically active. 38 Gyaan Sankalp