Coordination compounds-1.pptx ppt containing all notes for class 12 neet jee

Coordination compounds

1. Coordination compounds

2. Double salts and coordination compounds: Double salts : Double salts are formed by mixing saturated solutions of two normal salts. These compounds exist only in crystalline state and loose their identity in polar solvents or melted. E.g . Potash alum : K 2 SO 4 .Al 2 (SO 4 ) 3 .24 H 2 O Mohr’s salt : FeSO 4 .(NH 4 ) 2 SO 4 .6H 2 O Ferric alum : K 2 SO 4 .Fe 2 (SO 4 ) 3 .24H 2 O

3. Coordination compounds/complex compounds : When solution of two normal salts are mixed and evaporated to crystalline then a new salt is formed which does not show the properties of its components radicals and does not loose its properties when dissolved in solvent or melted ; these compounds are called complex compounds or coordination compounds. Examples: CuSO 4 + 2 NH 4 OH ----  Cu(OH) 2 + (NH 4 ) 2 SO 4 Cu(OH) 2 + NH 4 OH -----> [Cu(NH 3 ) 4 ] 2+ + 2 OH - + 4 H 2 O (in this solution we can not identify the presence of Cu 2+ ions)

4. When aqueous solution of KCN is added to the aqueous solution of silver nitrate , a white precipitate of silver cynide is first formed which dissolve in excess KCN to form soluble comlex . AgNO 3 ( aq ) + KCN( aq ) ----  AgCN (s) + KNO 3 ( aq ) AgCN + KCN --- K[Ag(CN) 2 ] ( potassium argentocynide ) Solution of thee above complex will give test for K + and [Ag(CN) 2 ] - ions. So a coordination compound may be defined as “ a compound which is formed by the combination of two or more stable salts and which retains its identity I solid as well as in solution.” e.g. : K 4 [Fe(CN) 6 ] , [Fe(H 2 O) 6 ]Cl 3 , K[Ag(CN) 2 ] etc.

5. On the basis of nature, addition (or) molecular compounds are divided into two categories. They are double salts and coordination (or) complex compounds. Mohr’s salt: FeSO 4 .(NH 4 ) 2 SO 4 .6H 2 O double salt.

6. Complex ion : a complex ion may be defined as an electrically charged radical formed by the combination of a central metal atom or ion surrounded by ions or neutral molecules. Complexes may have cation complex ; anion complex or neutral complex. For example: K 3 [Fe(CN) 6 ] cation anion complex [Co(NH 3 ) 6 ]Cl 3 cation complex anion [Ni(CO) 4 ] neutral complex

7. Important terms used in coordination chemistry : 1.: Ligands : The molecules or ions which can donate at least one electron pair to central metal to form coordinate bonds are called ligands or donor groups. The word ligand is originated from Greek word , ligane meaning to bind. e.g.: K 3 [Fe(CN) 6 ] ; in this complex Fe 2+ is coordinated to six cyanide ( - CN) ions.

8. Classification of Ligands : Monodentate or unidentate ligands : The ligands which form only one coordinate bond with the central metal are called monodentate or unidentate ligands. Examples : Common name of ligand Formula Donor atom IUPAC name of ligand Fluoride ion F - F fluorido Chloride ion Cl - Cl Chlorido Bromide ion Br - Br bromido

9. Iodide ion I- I iodido Water H 2 O O Aqua Ammonia NH 3 N ammine Phosphine PH 3 P phosphine Carbon monoxide CO O Carbonyl Oxide O 2- O oxo Peroxide O 2 2- O Peroxo Cynide - CN C Cyanido Isocynide NC - N Isocyanido Nitrosyl NO N nitrosyl Pyridine C 5 H 5 N N Pyridine Acetate ion CH 3 COO - O - acetato Nitro NO 2 - N nitro Nitrite ion - ONO O nitrito Hydroxide ion OH - O Hydroxo Nitride ion N 3- N nitrido

10. Polydentate ligands: These ligands have two or more donor atoms or point of attatchment and can be linked to the central metal using two or more donating sites. Polydentate ligands can be further classified into : Bidentate Tridentate Tetradentate Pentadentate Hexadentate Depending upon the point of attatchment or donor groups.

11. Bidentate ligands :

12. Ambident ligands : These ligands have two or more donor atoms but in complex formation , only one of them is coordinated to the central metal. Examples : - CN cyanido M - CN M - NC isocyanido Nitro M NO 2 - M - ONO nitrito Thiocyanato M - SCN M - NCS isothiocyanato

13. Chelates : These are the complexes in which the donor atoms are attached to each other as well as to the central metal , so that metal becomes a part of heterocyclic ring. Such ligands are called chelate ligands and the resulting complexes are called chelates. Characteristics of chelates : 1. the stability of the complexes increases by chelation. This is called chelate effect. E.g.: porphyrine ring present in chlorophyll and heamoglobin makes them extremely stable. 2. chelate complexes have low melting point. 3. these complexes are soluble in the organic solvents. 4.four membered ring is the smallest ring but it is highly strained hence rare occurs. But five and six membered rings are more stable and are common.

14. Homoleptic and heteroleptic complexes : Homoleptic complexes : Complexes in which the central metal is attached to one kind of ligands are called Homoleptic complexes. E.g. : [Co(NH 3 ) 6 ] 3+ , [Fe(CN) 6 ] 4- , [Ni(CO) 4 ] Heteroleptic complexes : complexes in which the central metal is attached two or more kind of ligands are called heteroleptic complexes. E.g.: [Cr(NH 3 ) 5 Br] 2+ , [Co(NH 3 ) 4 Cl 2 ] +

15. Coordination sphere :The central metal atom or ion and coordinated ligands to it inside the square bracket form coordination sphere. The ionisable ions attached outside the bracket are called counter ions. K 4 [Fe(CN) 6 ] counter ion coordination sphere [Cu(NH 3 ) 4 ] SO 4 coordination sphere counter ion The components present in coordination sphere i.e. central metal ion and ligands are collectively called as coordination entity.

16. Coordination number : The total number of coordinate bonds formed by the central metal in a complex is called its coordination number. Coordination number = ∑ number of ligands x dentate character of ligand ( denticity ) Complex Coordination number [Cu(NH 3 ) 4 ]SO 4 4 x 1 = 4 [Cr(en) 3 ]SO 4 3 x 2 = 6 [Co(NH 3 ) 6 ] 3+ 6 X 1 = 6 [PtCl 6 ] 2- 6 x 1 = 6

17. Oxidation state of central metal atom : Oxidation state is defined as the charge which the central metal would carry if all the ligands are removed along with the electron pairs shared with the central metal. It is represented by Roman numerals I , II , III , IV and so on in the parenthesis after the name of central atom. Charge on the complex ion = [oxidation number of metal ion ] + [ charge on the ligands] For example : [Fe(CN) 6 ] 4- : let the oxidation state of Fe is (x) -4 = x + 6(-1) x = +2 or Fe(II)

18. Coordination polyhedron : The three dimensional arrangement of ligands bonded to central metal is known as coordination polyhedron. Examples : tetrahedral , square planar , sqyare pyramidal , trigonal bipyramidal , octahedral

19. Question 1.: FeSO 4 solution mixed with (NH 4 ) 2 SO 4 solution in 1:1 molar ratio gives the test of Fe 2+ but CuSO 4 solution mixed with ammonia in 1:4 ratio does not give the test of Cu 2+ ions. Explain. Answer : When FeSO 4 ( aq ) is mixed with (NH 4 ) 2 SO 4 ( aq ) in 1:1 ratio by mole , formation of a double salt called Mohr’s salt takes place. FeSO 4 ( aq ) + (NH 4 ) 2 SO 4 ( aq ) + 6 H 2 O ---  FeSO 4 .(NH 4 ) 2 SO 4 .6H 2 O Mohr’s salt This salt will completely ionize in solution. When CuSO 4 ( aq ) is mixed with NH 4 OH in 1:4 ratio by mole ; formation of a comples compound [Cu(NH 3 ) 4 ]SO 4 takes place. CuSO 4 ( aq ) + 4 NH 4 OH -- [Cu(NH 3 ) 4 ]SO 4 + 4 H 2 O complex Above complex in solution will give [ Cu(NH 3 ) 4 ] 2+ ion but will not give Cu 2+ ions , hence Cu 2+ ions can not be detected in this solution by analytical methods.

20. Question 2.: 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. Answer : [Cu(H 2 O) 4 ] 2+ + 4 F - ----  [CuF 4 ] 2- + 4 H 2 O tetrafloridocuprate (II) (green colour ) [Cu(H 2 O) 4 ] 2+ + 4 Cl - ----  [CuCl 4 ] 2- + 4 H 2 O tetrachloridocuprate (II) (bright green)

21. Q.3: 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 ppt of copper sulphide is obtained when H 2 S gas is passed through the solution ? Ans : 2 2 CuSO 4 + 10 KCN -------  2 K 3 [Cu(CN) 4 ] + 2 K 2 SO 4 + CN-CN Solution of complex K 3 [Cu(CN) 4 ] will give [Cu(CN) 4 ] 3- and K + but not Cu 2+ ions; hence this solution will not give ppt of CuS on passing H 2 S gas through it.

22. Q.4: Specify the oxidation numbers of the metals in the following coordination entities : (a) [Co(H 2 O)(CN)(en) 2 ] 2+ (b)[Cr(NH 3 ) 3 Cl 3 ] (c) K 3 [Fe(CN) 6 ] (d)[CoBr 2 (en) 2 ] + Ans. : (a) x + 0 + (-1) + 2x0 = +2 ; x = 3 (b) x + 3(0) + 3(-1) = 0 ; x = 3 (c) 3 K + + [Fe(CN) 6 ] 3- = x + 6(-1) = -3 ; x = +3 (d) x + 2(-1) + 2(0) = +1 ; x = +3

23. IUPAC Nomenclature of Coordination compounds : 1.Cation is named first followed by anion. 2.Ligands are written in the alphabetical order irrespective of their charge . 3. Within the coordination sphere , the ligands are named first followed by the central metal. 4.The oxidation state of central metal is written in Roman numeral in bracket , just after the name of central metal. 5. when the complex ion is anionic then the name of central metal ends in – ate , however , for cationic , neutral or non-ionic complexes , the metal is called with usual names. E.g.: Cobalt = cobaltate ; nickel = nickelate , chromium = chromate , iron = ferrate , copper = cuprate , silver = argentate , lead = plumbate , gold = aurate , aluminium = aluminate . Zinc = zincate , platinum = palatinate , mercury = murcurate

24. 6. If the number of a ligand is two or more then while naming the complex , the number is indicated by Greek prefix i.g . di , tri , tetra , penta , hexa etc. However when the numerals di , tri , tetra is present in the name of ligand e.g. ethane-1,2-diamine , dipyridyl then bis , tris , tetrakis etc. are written as prefix at the place of di , tri , tetra etc. 7. The name of the coordination sphere is written in continuum. K 4 [Fe(CN) 6 ] potassium hexacynoferrate (II) 8. The name of neutral or non-ionic complex is written in one word only. [Ni(CO) 4 ] tetracarbonylnickel (0)

25. Anionic ligands with their name : Anion Symbol Name as ligand Fluoride F - Fluorido Chloride Cl - Chlorido Bromide Br - Bromido Cyanide CN - Cyanido Oxide O 2- Oxo Hydroxide OH - Hydroxo Sulphide S 2- Sulphido Amide NH 2 - Amido Nitride N 3- Nitrido Phosphide P 3- Phosphido Imide NH 2- Imido Azide N 3 - Azido

26. Peroxide O 2 2- peroxo Hydride H - Hydrido Iodide I - Iodido Carbonate CO 3 2- Carbonato Oxalate C 2 O 4 2- Oxalato Sulphate SO 4 2- Sulphato Sulphite SO 3 2- Sulphito Nitrate NO 3 - Nitrato Acetate CH 3 COO - Acetato Chlorate ClO 3 - Chlorato Perchlorate ClO 4 - Perchlorato Thiosulphate S 2 O 3 2- Thiosulphato Water H 2 O Aqua or aquo Ammonia NH 3 ammine Carbon monooxide CO carbonyl

27. CS Thiocarbonyl NO Nitrosyl NS Thionitrosyl C 6 H 5 N Pyridine ( py ) PH 3 Phosphine (C 6 H 5 ) 3 P Triphenylphosphine H 2 NCSNH 2 Thiourea H 2 N-CH 2 -CH 2 -NH 2 Ethane-1,2-diamine (en) NH 4 + Ammonium NH 2 NH 3 + Hydrazinium NO + nitrosonium NO 2 + Nitronium

28. Coordination compounds containing complex anionic ion : Example : K 4 [Fe(CN) 6 ] in this complex K + ion are cations and [Fe(CN) 6 ] 4- ion is anion First calculate the oxidation number of Fe metal in complex ion . x + 6(-1) = -4 ; x = 2 Now write the name of cation first and than anion. Outside the complex ion no di , tri , tetra etc. if the ions are more than 1. When we write the name of complex ion , first write the name of ligands alphabetically than metal name without space and than oxidation number in bracket. If the complex ion is anion , than metal name ends with –ate. So name is : potassium hexacyanidoferrate (II)

29. K 3 [Fe(CN) 6 ] Potassium hexacyanidoferrate (III) K 3 [Cr(C 2 O 4 ) 3 ] Potassium trisoxalatochromate (III ) K 3 [Co(C 2 O 4 ) 2 Cl 2 ] Potassium dichloridodioxalatocobaltate (III) K 2 [HgI 4 ] Potassium tetraiodidomecurate (II) K 2 [PtCl 6 ] Potassium hexachloridoplatinate (IV) Na[Ag(CN) 2 ] Sodium dicyanidoargentate (I) [Ni(CN) 4 ] 2- Tetracyanidonickelate (II) ion K 3 [Fe(CN) 5 NO] Potassium pentacyanidonitrosylferrate (II) Li[AlH 4 ] Lithium tetrahydridoaluminate (III) Na 2 [Zn(CN) 4 ] Sodium tetracyanidozincate (II) K 3 [Al(C 2 O 4 ) 3 Potassium trisoxalatoaluminate (III ) K 2 [Zn(OH) 4 ] Potassium tetrahydroxidozincate (II)

30. K 3 [Fe(CN) 6 ] Potassium hexacyanidoferrate (III) K 3 [Cr(C 2 O 4 ) 3 ] K 3 [Co(C 2 O 4 ) 2 Cl 2 ] K 2 [HgI 4 ] K 2 [PtCl 6 ] Na[Ag(CN) 2 ] [Ni(CN) 4 ] 2- K 3 [Fe(CN) 5 NO] Li[AlH 4 ] Na 2 [Zn(CN) 4 ] K 3 [Al(C 2 O 4 ) 3 K 2 [Zn(OH) 4 ]

31. Coordination compounds containing complex cation : Example : [Co(NH 3 ) 4 (H 2 O) Cl ]Cl 2 In this compound complex ion is cation . So name will be : tetraammineaquachloridocobalt (III)chloride

32. [Cu(en)2]SO4 Bis (ethane-1,2-diamine)copper(II) sulphate [Cr(H 2 O) 4 Cl 2 ] + Tetraaquodichloridochromium (III) ion [Fe(H 2 O) 4 (C 2 O 4 ) 2 ]SO 4 Tetraaquodioxalatoiron (II) sulphate [Ag(NH 3 ) 2 ] Cl Diamminesilver (I) chloride [ Pt (NH 3 ) 6 ]Cl 4 Hexaammineplatinum (IV) chloride [Cr(en) 3 ]Cl 3 Tris (ethane-1,2-diamine) chromim (III) chloride [CoCl 2 (en) 2 ] Cl Dichloridobis (ethane-1,2-diamine)cobalt(III) chloride

33. Coordination compounds containing both cation and anion : [Cr(NH 3 ) 6 ][Co(CN) 6 ] In this compound cation and anion both are complex ions . So the name will be : Hexaamminechromium (III) hexacyanidocobaltate (III) [CoCl 2 (NH 3 ) 4 ] 3 [Cr(CN) 6 ] Tetraamminedichloridocobalt (III) hexacyanidochromate (III)

34. Non-ionic complex : [Ni(CO)4] tetracarbonylnickel (0) [Cr(CO)6] hexacarbonylchromium (0) [Fe(CO)5] pentacarbonyliron (0)

35. Formula of a complex compound from its IUPAC name : Rules : 1. formula of the cation is written first. 2. coordination entity is written in the square bracket. 3.components of the coordination sphere are written within the bracket in the following sequence : ( i ) central metal (ii) anionic ligands (iii) neutral ligands (iv) positively charged ligands. 4. If a number of ligands of same type are present in the complex then they are written in alphabetical sequence of the first alphabet of their sequence.

36. 5. if two ligands have same key atom ( SO 4 2- , S 2 O 3 2- ) then the ligand with fewer key atom will be written. Thus SO 4 2- will be written in the formula before S 2 O 3 2- . 6. If two ligands have same key atom ( with same number of key atoms) then subsequent symbol decides sequence of formula. E.g. : NH 2 - will be written before NO 2 - 7. polyatomic ligands are written in parenthesis. 8. all ligands are formulated without giving any space in between. 9. Total number of cations and anions be written based on the principle that total positive charge on the cation must be balanced by total negative charge on the anionic ligands. The complex as a whole must be electrically neutral. Complex ligands are to be written as abbreviations in parenthesis. E.g. ethane-1,2-diammine (en) Dipyridyl ( dipy ) Dimethyl glyoxime ( dmg ) Pyridine ( py ) Glycinato ( gly ) Diethylene triamine ( trien )

37. Write the formulas for the following coordination compounds : (a) tetraammineaquachloridocobalt (III) chloride as the complex ion is written first so it is cation . Cation : [Co(NH 3 ) 4 (H 2 O) Cl ] Anion is Cl - Oxidation number of Cobalt is +3 So charge on the complex is : 3 + 4(0) + 0 + 1(-1) = +2 Since there is -1 charge on the Chloride ion , so to make the coordination compound as a whole neutral 2 Cl - ions will be outside the coordination entity. Hence the formula of the compound will be : [Co(NH 3 ) 4 (H 2 O) Cl ]Cl 2

38. (b) potassium trioxalatoaluminate (III) K 3 [Al(C 2 O 4 ) 3 ] (c) dichloridobis (ethane-1,2-diamine)cobalt(III) ion [CoCl 2 (en) 2 ] + (d) amminebromidochloridonitrito -N palatinate(II) [ Pt (NH 3 ) BrCl (NO 2 )] -

39. Werner’s theory of Coordination Alfred Werner (1893) put forward “ Werner’s theory of coordination”. Main postulates: Central metal possess two types of valencies : ( i ) primary (ii) secondary 2. Primary valencies corresponds to the oxidation state of the central metal atom and secondary corresponds to the coordination number. 3 . Central metal atom satisfies both types of valencies . 4. Primary valencies are satisfied by the negatively charged ligands only. 5. Secondary valencies are satisfied by all types of ligands e.g. negatively charged as well as neutral ligands. 6. Secondary valencies are directional and the ligands satisfying this valency are directed towards the vertices of polyhedral e.g. tetrahedral ,square planar , octahedral etc.

40. Postulates: The central metal atom (or) ion in a coordination compound exhibits two types of valencies - primary and secondary. Primary valencies are ionisable and correspond to the number of charges on the complex ion. Primary valencies apply equally well to simple salts and to complexes and are satisfied by negative ions.

41. Secondary valencies correspond to the valencies that a metal atom (or) ion towards neutral molecules (or) negative ions in the formation of its complex ions. Secondary valencies are directional and so a complex has a particular shape. The number and arrangement of ligands in space determines the stereochemistry of a complex. Postulates:

42. The postulates of Werner's coordination theory were actually based on experimental evidence rather than theoretical. Although Werner's theory successfully explains the bonding features in coordination compounds, it has drawbacks. Drawbacks : It doesn't explain why only certain elements form coordination compounds. It does not explain why the bonds in coordination compounds have directional properties. It does not explain the colour, and the magnetic and optical properties of complexes. Werner's Theory:

43. Experimental verifications of Werner’s theory : Molecular formula Ionic formulation Colour No. of ions in the molecule Ionisable Cl - ions precipitate d with AgNO3 Primary valency Secondary valency Molar conductance ( mho mol -1 ) CoCl 3 .6NH 3 [Co(NH 3 ) 6 ]Cl 3 Orange 4 3 +3 6 430 CoCl 3 .5NH 3 [Co(NH 3 ) 5 Cl]Cl 2 Violet 3 2 +3 6 250 CoCl3.4NH3 [Co(NH 3 ) 4 Cl 2 ] Cl Green 2 1 +3 6 100 CoCl3.3NH3 [Co(NH 3 ) 3 Cl 3 ] Blue-green +3 6

44. Q .: How many ions are produced from the complex [Co(NH 3 ) 6 ]Cl 2 ? Ans : [Co(NH 3 ) 6 ] 2+ + 2 Cl - i.e. : 3 Q.:Arrange the following complexes in increasing order of conductivity of their solutions : [Co(NH 3 ) 3 Cl 3 ] , [Co(NH 3 ) 4 Cl 2 ] Cl , [Co(NH 3 ) 6 ]Cl 3 , [Co(NH 3 ) 5 Cl]Cl 2 Ans : greater is the ionic concentration in the solution , more will be the conductivity. More is no. of ions produced from one formula unit of compound , more will be ionic concentration. Hence [Co(NH 3 ) 3 Cl 3 ] < [Co(NH 3 ) 4 Cl 2 ] Cl < [Co(NH 3 ) 5 Cl]Cl 2 < [Co(NH 3 ) 6 ]Cl 3 increasing conductivity

45. Valence bond theory : given by Pauling in 1931 Main points : The central metal atom may loose some electrons to form central metal cation . The number of electrons lost represent the valency of metal (primary valency ). In some complexes , metal has zero oxidation state hence no electrons are lost. 2. A central metal atom or ion present in a complex provides vacant orbitals s , p , d equal to the coordination number. These vacant orbitals undergo mixing to form mixed or hybrid orbitals of equivalent energy. Bonds are not formed by pure orbitals but by hybrid orbitals. 3. Each ligand has at least one orbital containing a lone pair of electrons. 4. The filled orbital of ligand overlaps with empty orbital of metal to form coordinate bond , it is represented by an arrow from ligand to metal (L M). Here ligand act as lewis base and central metal acts as lewis acid. 5.The non-bonding electrons of the central metal occupying the inner orbitals obey Hund’s rule but under the influence of some strong ligands like NH3 , -CN , CO etc. there may be some rearrangement of electrons against Hund’s rule. 6.If the complex does not contain unpaired electrons , it is diamagnetic , while the complex having one or more unpaired electrons is paramagnetic.

46. If (n-1)d orbitals are used for hybridization, the complexes is called inner complexes and nd called outer complexes.

47. When strong field ligands like NH 3 and CN - are involved in the formation of complexes, they causes pairing electrons present in metal ions. This process is called as spin pairing.

48. Octahedral complexex : [Cr(NH 3 ) 6 ] 3+ Hexaamminechromium (III) ion In this compound Chromium has +3 oxidation number. 24 Cr = [ Ar ]3d 5 4s 1 and 24 Cr 3+ = [ Ar ]3d 3 4s 24 Cr 3+ 3d 4s 4p octahedral ; d 2 sp 3 hybridization (inner orbital complex) as there are 3 unpaired electrons , so paramagnetic in nature. ↑ ↑ ↑ × × × × × ×

49. [Fe(CN) 6 ] 3- : hexacyanidoferrate (III) ion 26 Fe = [ Ar ] 3d 6 4s 2 and 26 Fe 3+ = [ Ar ] 3d 5 4s 26 Fe 3+ 3d 4s 4p As Cyanide ( - CN) ligand is a strong ligand , so in the presence of strong ligand field elctrons will pair up. 3d 4s 4p Paramagnetic d 2 sp 3 (inner orbital complex) , octahedral ↑ ↑ ↑ ↑ ↑ × × × × ↑ ↓ ↑ ↓ ↑ × × × × × ×

50. [Fe(CN) 6 ] 4- : hexacyanidoferrate (II) ion 26 Fe = [ Ar ] 3d 6 4s 2 and 26 Fe 2+ = [ Ar ] 3d 6 4s 26 Fe 2+ 3d 4s 4p As Cyanide ( - CN) ligand is a strong ligand , so in the presence of strong ligand field elctrons will pair up. 3d 4s 4p diamagnetic d 2 sp 3 (inner orbital complex) , octahedral ↑ ↓ ↑ ↑ ↑ ↑ × × × × ↑ ↓ ↑ ↓ ↑ ↓ × × × × × ×

51. [Fe(H 2 O) 6 ] 3+ ; hexaaquairon (III) ion 26 Fe = [ Ar ] 3d 6 4s 2 and 26 Fe 3+ = [ Ar ] 3d 5 4s 26 Fe 3+ 3d 4s 4p 4d As H 2 O ligand is a weak ligand , so in the presence of weak ligand field elctrons will not pair up. Paramagnetic sp 3 d 2 (outer orbital complex) , octahedral ↑ ↑ ↑ ↑ ↑ × × × × × ×

52. [NiCl 4 ] 2- : tetrachloridonickelate (II) 28 Ni = [ Ar ] 3d 8 4s 2 ; 28 Ni 2+ = [ Ar ] 3d 8 4s 28 Ni 2+ 3d 4s 4p As Cl - is a weak ligand so pairing of electrons will not take place. Paramagnetic ; sp3 hybridization ; tetrahedral ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↑ × × × ×

53. [Ni(CN) 4 ] 2- : tetracyanidonickelate (II) 28 Ni = [ Ar ] 3d 8 4s 2 ; 28 Ni 2+ = [ Ar ] 3d 8 4s 28 Ni 2+ 3d 4s 4p As CN- is a strong ligand so pairing of electrons will take place. 3d 4s 4p dsp2 diamagnetic ; dsp 2 hybridization ; square planar ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↑ × × × × ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ × × × ×

54. [Ni(CO) 4 ] : tetracarbonylnickel (0) 28 Ni = [ Ar ] 3d 8 4s 2 ; 28 Ni 3d 4s 4p As CO is a strong ligand so pairing of electrons will take place. 3d 4s 4p sp3 diamagnetic ; sp3 hybridization ; tetrahedral ↑ ↓ ↑ ↓ ↑↓ ↑ ↑ ↑↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓ × × × ×

55. [Co(NH 3 ) 6 ] 3+ : hexaamminecobalt (III) ion 27 Co = [ Ar ] 3d 7 4s 2 ; 27 Co 3+ = [ Ar ] 3d 6 4s 27 Co 3+ 3d 4s 4p In this compound NH 3 ligand act as strong ligand , hence pairing of electrons takes place. 3d 4s 4p Hence DIAMAGNETIC , d 2 sp 3 , octahedral ↑ ↓ ↑ ↑ ↑ ↑ X X X X ↑ ↓ ↑ ↓ ↑ ↓ x x x x x x

56. Limitations of Valence Bond theory : 1. It gives spin only magnetic moment. It does not explain the magnetic moment arising from orbital motion of electrons. 2.it does not distinguish weak field and strong field ligands. 3. It can not explain the colour of the complex compounds. 4. it can not always correlate between geometry and magnetic behavior of the complex. 5. it is unable to explain why some complexes are inner orbital complex and some other are outer orbital for a metal ion in same oxidation state.

57. Crystal Field Theory : Main points : 1. In a complex the central metal atom/ion is surrounded by ligands. 2. Ligands are either negative ions or neutral molecules with lone pair of electrons. 3. central metal and ligands are considered as point mass and point charges. 4.The interaction between metal and ligands is purely electrostatic or ionic. It is of two types: (a) the attractive forces between central metal cation and the negatively charged ligands or negative pole of neutral molecule. (b) The repulsive forces acts between the lone pairs of ligands and electrons present in the d-orbitals of central metal. This repulsive force is the focus point in the crystal field theory. This repulsive force destroy the five fold degeneracy of d-orbitals and they split into two groups t 2g ( d xy , d yz , d xz ) and e g (d x2-y2 , dz 2 ).

58. e g or axial orbitals : this set of orbitals consists of d x2-y2 and d z2 . These orbitals have their lobes along the axis hence they are called axial orbitals. Here ‘e’ stands for doubly degenerate. t 2g or non-axial orbitals : this set of orbitals consists of d xy , d yz , d xz orbitals. Lobes of these orbitals lie between the axis hence called non-axial orbitals. Here ‘t’ stands for triply degenerate. 5.The electrons of the central metal occupy the splitted d-orbitals as per Hund’s rule. The electrons first occupy those orbitals which suffers less repulsion ( the orbitals farthest from approaching ligands) i.e. the orbitals which have less energy. 6. This theory does not consider any overlapping between the orbitals of ligands central atom. Hence formation of covalent or coordinate bond is ruled out in this theory.

59. The stability of a complex is quantitatively measured by the magnitude of CFSE. The magnitude of CFSE depends on the nature of ligand and the geometry of complex. the zero value predicts that the complex is unstable. Greater the negative value of CFSE : more is the stability of the complex.

60. Application of CFT to tetrahedral complexes In the tetrahedral complex, [MX 4 ] n , the metal atom or ion is placed at centre of the regular tetrahedron and the 4 ligands, are placed at four corners of the tetrahedron. Ligand approach the central Metal atom in between 3 cordinate x, y, z.

61. Lower repulssion In case of strong field ligands, the electrons prefer to pair up in eg orbital giving low spin complexes while in case of weak field ligands, the electrons prefer to enter higher energy t2g orbitals giving more unpaird electrons and hence form high spin complexes. Greater repulsion High energy Bari centre

62. Application of Crystal field theory to octahedral complexes [MX 6 ] n the metal atom or ion is placed at the centre of regular octahedron while 6 ligands Occupy the positions at 6 vertices of the octahedron. Two orbital d x2-y2 and d z2 are Axial greater repulsion and d xy , d yz , d xz less repulsion.

63. Five d orbital lose degenracy and split into two point group. The group t 2g lower energy while e g group have higher energy.

64. Splitting of d-orbitals in octahedral complexes : d x2-y2 d z2 energy e g 3/5∆ =6Dq ------- Bary’s centre ---------- ∆ or 10Dq average energy of d-orbitals in a spherical field 2/5 ∆ = 4Dq degenerate d-orbitals d xy d yz d xz t 2g (splitting of d-orbitals in octahedral complexes) ∆ = 10 Dq ( Dq = differentials of quanta )

65. Splitting of d-orbitals in tetrahedral complexes :

66. Spectrochemical series : The magnitude of CFSE depends on the nature of ligands. The strong ligands have higher splitting power of d-orbitals of central metal atom or ion. Whereas weak ligands have relatively lower splitting power of d – orbitals of central metal atom/ion. A series representing the order of strength of ligands is acalled spectrochemical series. CO > - CN > NO 2 - > en >NH 3 = py > EDTA > - NCS >H 2 O >C 2 O 4 2- >OH - >F - >NO 3 - >S 2- > Cl - > - SCN > Br - > I - strong field ligands weak field ligands C-Donor atom ligands > N-Donor atom ligands > O-donor atoms ligands > Halogens , sulphur -donor atoms ligands strong ligands weak ligands In addition to nature of ligands the crystal field splitting energy depends on some other factors given below: ( i ) CFSE for tetrahedral complexes is significantly less than that of octahedral complexes. ∆ t = 4/9 ∆ (ii) CFSE of a metal in higher oxidation state is greater than that of a metal in lower oxidation state. e.g.: Co 2+ complexes have low value of CFSE than those of Co 3+

67. Pairing energy (P) : The energy required to pair two electrons in the same orbital. In octahedral complexes : when ∆ > P ; pairing of electrons occurs in t 2g set of orbitals. when ∆ < P ; electrons will be singly filled in t 2g and than e g than pairing of electrons takes place. For example : d 4 : when ∆ > P ; t 2g 4 e g when ∆ < P : t 2g 3 e g 1

68. d 5 : when ∆ > P ; t 2g 5 e g when ∆ < P : t 2g 3 e g 2 d 6 : when ∆ > P ; t 2g 6 e g when ∆ < P : t 2g 4 e g 2 d 7 : when ∆ > P ; t 2g 6 e g 1 when ∆ < P : t 2g 6 e g 1 d 8 : when ∆ > P ; t 2g 6 e g 2 when ∆ < P : t 2g 6 e g 2 Same is in d 9

69. Magnetic properties of complexes : Magnetic character of complexes are successfully explained by the Crystal Field Theory. A complex having one or more unpaired electrons is always paramagnetic and its spin only magnetic moment will be : µ = n(n+2) BM Unpaired electron : 1 magnetic moment = 1.73 BM 2 = 2.83 BM 3 = 3.87 BM 4 = 4.90 BM 5 = 5.92 BM  

70. Colours in coordination compounds : Complexes having 1 or more unpaired electrons will be coloured due to d-d transition. Colour of transition metal is complementary to that which is absorbed. Wavelength absorbed (nm) VIOLET BLUE BLUE GREEN YELLOW GREEN ORANGE RED Complementary colour observed GREENISH YELLOW YELLOW RED VIOLET/purple BLUE BLUE GREEN

71. Limitations of crystal field theory: 1. this theory is unable to explain the bonding in metal carbonyls. 2. in spectrochemical series , H 2 O is stronger ligand than OH- .it is not explained. 3 . This theory does not consider the covalent nature of bond between centralmetal and ligand. 4.this theory does not consider s and p electrons of metals in any explanation.

72. Factors affecting stability of complexes : ( i ) Charge density of central metal ion : Charge density = ionic charge/surface area of ion Higher the charge density of metal ion forms strong coordinate bond with the ligand. Higher oxidation state of a metal shows greater stability of complex with identical ligands. If charge of metal ions is same then smaller is the size of ion , greater will be the stability of complex with identical ligand. (ii) nature of ligands : Greater the basic character of the ligand , more will be the stability of complex. Chelate ligands form more stable complex than non-chelated complex.hence greater is the extent of chelation , more is the stability of complex.

73. Q.1: On the basis of the following observations made with aq. solutions , assign secondary valencies to metals in the following compounds : FORMULA Moles of AgCl precipitated per mole of the compound with excess of AgNO 3 ( i ) PdCl 2 .4NH 3 2 (ii) NiCl 2 .6H 2 O 2 (iii) PtCl 4 .2HCl (iv) CoCl 3 .4NH 3 1 (v) PtCl 2 .2NH 3 ANSWERS : FORMULA OF THE COMPOUND SECONDARY VALENCIES ( i ) [Pd(NH 3 ) 4 ]Cl 2 4 (ii) [Ni(H 2 O) 6 ]Cl 2 6 (iii) H 2 [PtCl 6 ] 6 (iv) [CoCl 2 (NH 3 ) 4 ] Cl 6 (v) [PtCl 2 (NH 3 )2] 4

74. Q.2:The spin only moment of [MnBr 4 ] 2- is 5.9 B.M. Predict the geometry of the complex ion ? Ans.: As the coordination number of Mn 2+ is 4 , hence the geometry will be either tetrahedral or square planar. Mn 2+ = [ Ar ]3d 5 4s As the spin only moment is 5.9 B.M. , hence no. of unpaired electrons are 5 i.e. all 5 electrons are present in 3d subshell i.e. 3d 5 . hence one 4s and three 4p orbitals are vacant and available for hybridisation i.e. sp3 hybridisation Hence geometry will be : tetrahedral

75. Q.3:What is spectrochemical series ? Explain the difference between a weak field ligand and strong field ligand . Ans.: Spectrochemical series : The arrangement of ligands in order of their increasing field strength i.e. increasing crystal field splitting energy (CFSE) values is called spectrochemical series. The ligands with small value of CFSE are called weak field ligands whereas those with large value of CFSE are called strong field ligands .

76. Q.4: What is crystal field splitting ? How does the magnitude of ∆ decide the actual configuration of d- orbitals in a coordination entity ? Ans.: Crystal field splitting : When ligands approches a transition metal ion , the degenerate d- orbitals split up into two sets , one set is having higher energy than other set .The difference of energy between two sets of d- orbitals is called crystal field splitting energy (CFSE). ∆ < P (pairing energy) , the 4 th electron enters one of the eg orbitals giving the configuration t 2g 3 e g 1 , thus forming high spin complexes. Such ligands for which ∆ < P , are called weak field ligands . ∆ > P , the 4 th electron pair up in one of the t 2g orbitals giving the configuration t 2g 4 e g , thereby forming low spin complexes . Such ligands for which ∆ > P are called strong field ligand .

77. Q.5:Discuss the nature of bonding in the following coordination entities on the basis of Valence Bond Theory : ( i ) [Fe(CN) 6 ] 4- (ii) [FeF 6 ] 3- (iii) [Co(C 2 O 4 ) 3 ] 3- (iv) [CoF 6 ] 3- (v) [Co(NH 3 ) 6 ] 3+ (vi) [Ni(CN) 4 ] 2- (vii) [NiCl 4 ] 2- (viii) [Ni(CO) 4 ] Ans : ( i ) [Fe(CN) 6 ] 4- : d 2 sp 3 , Octahedral , diamagnetic (ii) [FeF 6 ] 3- : sp 3 d 2 , octahedral , paramagnetic (iii) [Co(C 2 O 4 ) 3 ] 3- : d 2 sp 3 , octahedral , diamagnetic (iv) [CoF 6 ] 3- : sp 3 d 2 , octahedral , paramagnetic (v) [Co(NH 3 ) 6 ] 3+ : d 2 sp 3 , octahedral , diamagnetic (vi) [Ni(CN) 4 ] 2- : dsp 2 , square planar , diamagnetic (vii) [NiCl 4 ] 2- : sp 3 , tetrahedral , paramagnetic (viii) [Ni(CO) 4 ] : sp 3 , tetrahedral , diamagnetic

78. Q.6: Draw figure to show splitting of degenerated orbitals in an octahedral crystal field. Ans : d x2-y2 d z2 energy e g 3/5∆ =6Dq ------- Bary’s centre ---------- ∆ or 10Dq average energy of d- orbitals in a spherical field 2/5 ∆ = 4Dq degenerate d- orbitals d xy d yz d xz t 2g (splitting of d- orbitals in octahedral complexes) ∆ = 10 Dq ( Dq = differentials of quanta )

79. Q.7: What is meant by chelate effect ? Give an example. Ans : When a bidentate or polydentate ligand contains donor atoms positioned in such a way that they coordinate with the central metal ion , a five or six membered ring is formed , the effect is called chelate effect. As a result , the stability of the comlex increases. E.g. : complex of Ni 2+ with (en) is more stable than NH 3 . [Ni(NH 3 ) 6 ] 2+ ; log β = 8.61 [Ni(en) 3 ] 2+ ; log β = 18.28

80. Q.8:How many ions are produced from the complex , Co(NH 3 ) 6 Cl 2 in solution ? Ans : Coordination number of Cobalt = 6 Hence the complex is [Co(NH 3 ) 6 ]Cl 2 . It ionises in the solution as : [Co(NH 3 ) 6 ]Cl 2 + aq. -------  [Co(NH 3 ) 6 ] 2+ + 2 Cl - Hence 3 ions are produced.

81. Q.9:The oxidation number of cobalt in K[Co(CO) 4 ] is : Ans : K + [Co(CO) 4 ] - X + 0 = -1 X = -1

82. Q.10:Among the following , the most stable complex is : ( i ) [Fe(H 2 O) 6 ] 3+ (ii) [Fe(NH 3 ) 6 ] 3+ (iii) [Fe(C 2 O 4 ) 3 ] 3+ (iv) [FeCl 6 ] 3- Ans : In each of these complexes , Fe possess +3 oxidation state . As C 2 O 4 2- is bidentate chelating ligand , it forms chelate rings and hence is the most stable complex.

83. Q.11:Among the following ions , which one has the highest magnetic moment ? ( i ) Fe(H 2 O) 6 ] 2+ (ii) Cr(H 2 O) 6 ] 3+ (iii) Zn(H 2 O) 6 ] 2+ Ans : oxidation states are : Cr(III) , Fe(II) , Zn(II) Electronic configurations: Cr 3+ = 3d 3 ; unpaired electrons = 3 , inner orbital complex Fe 2+ = 3d 6 ; unpaired electrons = 4 , outer orbital complex Zn 2+ = 3d 10 ; unpaired electrons = 0 , outer orbital complex Magnetic moment is directly proportional to unpaired electrons. Hence ( i ) ion possess highest magnetic moment

84. Assertion –Reason type questions : In each of the following questions , a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it . Of the statements , mark the correct answer as : If both A and R are true and R is the true explanation of A. If both A and R are true , but R is not the true explanation of A. If A is true , but R is false . If both A and R are false.

85. Q.12 : Assertion (A): Toxic metal ions are removed by the chelating ligands . Reason (R): Chelate complexes tend to be more stable. Ans : Assertion (A) and reason (R) both are true and (R) is the correct explanation of (A ). Hence the answer is (a) Q.13: Assertion (A) : [Cr(H 2 O) 6 ]Cl 2 and [Fe(H 2 O) 6 ]Cl 2 are reducing in nature . Reason (R) : unpaired electrons are present in their d- orbitals . Answer : Cr 2+ = 3d 4 , so in aq. Solution Cr(II) will lose 1 electron to acquire 3d3 (t 2g ) exactly half filled electronic configuration and hence reducing in nature. ; (TRUE) Fe 2+ = 3d 6 , so Fe(II) lose 1 electron to give 3d 5 exactly half filled electronic configuration and hence reducing in nature .(TRUE) Both Cr(II) and Fe(II) both have unpaired electrons .(TRUE) So both (A) and (R) are true but (R) is not the exact explanation of (A). Hence answer is (b)

86. Q.14: Assertion (A) : [Fe(CN) 6 ] 3- ion shows magnetic moment corresponding to two unpaired electrons. Reason (R) : Because it has d 2 sp3 type hybridisation . Answer : Fe 3+ = 3d 5 ; under the influence of strong ligand field of CN- ( cynide ion) elctrons will pair up in 3d – orbitals and hence only 1 unpaired electron is present. So Assertion (A) is False. Fe(III) undergo d2sp3 hybridisation and will form inner orbital complex. Hence Reason (R) is TRUE. So (A) is false and (R) is TRUE

87. Q.15: Assertion (A) : The complex [Co(NH 3 ) 3 Cl 3 ] does not give precipitate with AgNO 3 . Reason (R) : The given complex does not contain counter ions. Answer : Assertion (A) is TRUE and Reason (R) is also TRUE. (R) is the correct explanation of (A). Hence answer is (a) .

88. Q.16: Assertion (A) : Both [Ni(CN) 4 ] 2- and [NiCl 4 ] 2- has same shape and same magnetic behaviour . Reason : Both are square planar and diamagnetic. Answer : [Ni(CN) 4 ] 2- = dsp 2 hybridisation and square planar [NiCl 4 ] 2- = sp 3 and tetrahedral Hence (A) is FALSE Reason (R) is also FALSE. Answer is (d) i.e. both (A) and (R) are FALSE

89. Effective atomic number (EAN) EAN = Z – X + Y Where, Z = atomic no. of the metal. X = no. of electron lost during the formation of the metal ion from its atom. Y = no. of electrons donated by the ligands. Eg. [Fe(CN) 6 ] -4 , Fe oxidation = +2 & Z = 26 [Fe(CN) 6 ] -4 = 26 – 2 + 6(2) = 36

90. Effective atomic numbers are calculated for the central metal atoms of coordination compounds. They give information regarding the stability of complexes and also help in understanding if the complex has reducing or oxidising properties. For eg : • EAN of Ni in [Ni(CO)4] can be calculated as follows- 28 + 4×2=36 which is equal to the no: of e-s present in krypton,an inert gas. Thus the complex is highly stable. • Similarly, EAN of Fe in [Fe(CN)6]^3- can be found out as: 26- 3+ 6×2=35. Thus the metal can accomodate an e- to get the EAN to be 36. And thus the complex has an oxidising nature.