C1-Allens Made Chemistry Theory {PART-1}

EFFECTIVE ATOMIC NUMBER1 5 .Calculate EAN of following complexes ?[Co(NH ) ]Cl3 63 ; K [Fe(CN) ] ; K [Pt Cl ] ; K [Fe(CN) ] ; [Ni (NH ) ]Cl46263 63 62Ans. Complexoxidation stateAtomic number Co-ordinationEffective atomic of the metalof the metalNumbernumber (EAN)[Co(NH ) ]Cl3 63+ 32 76 (27 – 3) + (6 × 2) = 36 (Kr)K [Fe(CN) ]46+ 22 66 (26 – 2) + ( 6 × 2) = 36 (Kr)K [Pt Cl ]26+ 47 86 (78 – 4) + (6 × 2) = 86 (Rn)K [Fe(CN) ]3 6+ 32 66 (26 – 3) + ( 6 × 2) = 35*[Ni (NH ) ]Cl3 62+ 22 86 (28 – 2) + (6 × 2) = 38*DO YOURSELF - V 1.Calculate EAN of the following complexes ?(i)[Ni(CO) ]4(ii)K [Ni(CN) ]24(iii)K [Hgl ]24(iv)[Ag(NH ) ]Cl3 2VALENCE BOND THEORY1 6 .Write hybridisation, geometry, magnetic nature of K [Fe(CN) ] ; [Zn(NH ) ]SO ; [Ni(CN) ]463 444–2 usingVBT ?Ans. (i)K [Fe(CN) ]4 6Fe  [Ar] 3d 4s62Fe +2 [Ar] 3d 4s°6Fe +2 [Ar] 3d4s4pIn presence of ligandFe +2 [Ar] 3dCN CN CN–xx xx xx ––xx xx CN CN––xxCN –4s4pHybridisation = d sp23geometry - octahedralmagnetic nature - diamagnetic(ii)[Zn(NH ) ]SO3 44Zn  [Ar] 3d 4s102Zn +2 [Ar] 3d 4s°10Zn +2 [Ar] 4s3d

In presence of ligandZn +2 [Ar] 4sxxNH NH NH3xx xx xx334p3dNH 3Hybridisation = sp3geometry - tetrahedralmagnetic nature - diamagnetic(iii)[Ni(CN) ]4–2Ni  [Ar] 3d 4s82Ni +2 [Ar] 3d 4s°8Ni +2 [Ar] 3d4sIn presence of ligandNi +2 [Ar] 3dCN CN–xx xx –xx CN –xxCN –4s4pHybridisation = dsp2geometry - Square planarmagnetic nature - diamagneticDO YOURSELF - VI 1.Write Hybridisation, geometry, magenatic moment of following complexes ?(i)[MnCl ]4–2(ii)[Mn(CN) ]6–4(iii)[Mn(H O) ]26+2 2.Discuss the nature of bonding in the following coordination entities on the basis of valence bondtheory :(i)[Fe(CN) ]64–(ii) [FeF ]63–(iii)[Co(C O ) ]24 33–(iv) [CoF ]63–CRYSTAL FIELD THEORY1 7 .Draw shape of d-orbitals ?Ans.In d subshell their are five d-orbitals d , d , d , dxyyzzxx – y22 and d their geometry are :z2zyx3dyz++–––3dxyyzx++–zyx3dzx–++–Electron density between the axis

zyx++–3d z 2yx3dx – y2 2 ++––zElectron density along the axis1 8 .Explain the term Degenerate orbitals :Ans.Orbitals which have same energy in a subshell are known as degenerate orbitals.1 9 .What is crystal field splitting ?Ans.According to CFT the interaction between a transition metal and ligands arises from the attraction betweenthe positively charged metal cation and negative charge of ligand.As a ligand approaches the metal ion, the electrons of ligand will be closer to some of the d-orbitals andfarther away from other causing a loss of degenercy.The electrons in the d-orbitals and those in the ligand repel each other due to repulsion between likecharges. Thus the d-electrons closer to the ligands will have a higher energy than those further away asa results in the d-orbitals splitting in energy.This loss of degeneracy of d-orbital is known as crystal field splitting.• (splitting energy)The state I represents degeneracy of all the five d-orbitals in the isolated central ion. The state II representshypothetical degeneracy of all the orbitals at a higher energy level if the negative charge of all the ligandsis assumed to be uniformly affecting the electrons in the d-orbitals of the metal ion. The state III representscrystal field splitting.2 0 .Explain crystal field splitting into octahedral complex ?Ans.In a octahedral complex, the co-ordination number is 6. The metal ion is at the centre and the ligandsoccupy the six corners of the octahedron as shown in figure.

We know that two orbitals, 22xydand 2z d are oriented along the axis while the remaining three orbitals,viz., d , d and d are oriented in between the axis.xyyzzxthe two orbtials 22xydand 2z d are designated as e orbitals while the three gYZXLLLLLLorbitals d , d and d are designated as t orbitals. As the six ligandsxyyzzx2gapproach the central ion along the axis, e orbitals, is repelled more bygthe ligand than in the t orbitals.2gIn other words, the energy of the 2z dand 22xydorbitals increasesmuch more than the energy of the d , d and d orbitals.xyyzzxThus, in octahedral complexes, the five d-orbitals split up into two sets : one set consistingof two orbitals (22x -y d and 2z d)of higher energy (e orbitals) and the other set consisting ofgthree orbitals (d , d and d ) of lower energy (t orbitals).xyyzzx2g.6  –.4   = crystal field splitting t2g eg energy (C.F.S.E.)The state I represents degeneracy of all the five d-orbitals in the isolated central ion. The state II representshypothetical degeneracy of all the orbitals at a higher energy level if the negative charge of all the ligandsis assumed to be uniformly affecting the electrons in the d-orbitals of the metal ion. The state III representscrystal field splitting discussed above.2 1 .Explain crystal field splitting into tetrahedral complex ?Ans.The co-ordination number for tetrahedral complexes is 4. The tetrahedralarrangement of four ligands surrounding a metal ion may be visualized byplacing ligands at the alternate corners of a cube, as shown in figure.It can be shown that in a tetrahedral structure, none of the d-orbitals points M +XYTetrahedral arrangement of four lignadsZLLLLexactly towards the ligands.When ligand approaches it is more close d , d , d in comparision ofxyyzxzdx –y 22 and d because d , d , d are between the axis and d and dz2xyyzzxz2x –y 22are along the. So d , d , d feels more repulsion as compare to d andxyyzzxz2dx – y22 .Thus, the d orbitals are also split into two groups but in a reverse order. The three orbitals, d , d andxyyzd , designated as t orbitals, now have higher energy than the two orbitals xz2222xyzdand ddesignated ase-orbitals.–0.6 t0.4  t tt2e

2 2 .Compare the energy of d-orbitals in square planar complex with respect to crystal field splitting ?Ans.The splitting of d-orbitals in square planar complexes can be understood by gradually withdrawing twoligands lying along the Z axis from an octahedral complex. As the ligands lying on the Z axis are movedaway, the ligands in the XY plane come more closely to metal. As a result of this, the electrons in d-orbitals in the XY plane experience greater repulsion from the electrons of ligands in a square planarcomplex than in an octahedral complex. This causes an increase in the energy of d-orbitals in XY plane.i.e., an increase in the energy of 22xydand dxy orbitals in square planar complexes compared to theirenergies in octahderal complexes, as illustrated in figure.Further, since the ligands lying on the Z axis have been moved away, the electrons in the d orbitals alongthe Z axis as well as in the XZ and YZ planes experience relatively smaller repulsions from the electronsof the ligands. This results in appreciable fall in the energy of2z dorbital as well as d and d orbitals.xzyzThese changes are shown in figure.MMLLLLLLLLLLt2g eg dx — y 22d xydxzdyzdz 2STATE-ISTATE-IISQUARE PLANAR STRUCTURE Crystal field splitting in spuare planar complexes2 3 .Define (a) crystal field stablisation energy (b) Pairing energy ?Ans.(a) Crystal field stablisation energy : The lowering in the energy of a transition metal ion in a givenligand environment due to crystal field effects.(b) Pairing energy : The energy required to pair the electrons.2 4 .Explain the term( a ) Inner orbital complex and outer orbital complexes ?(b) Low spin and high spin complexes ?Ans.(a)The empty 'd' orbitals involved in hybridisation may be inner (n-1)d or outer \"nd\" orbitals and thesecomplexes are called as Inner orbital complexes and outer orbital complexes respectively.For example in d sp hybridisation (n–1)d, ns and np orbitals are mixed it forms inner orbital complex23and in sp d ns, np and nd orbitals are mixed so it forms outer orbitals complex.3 2

(b)When the strong field ligand approaches to metal ion value to splitting energy () is greater than,pairing energy, so it is unfavourable to put electron into high energy orbitals. Therefore, the lowerenergy orbitals are completely filled before population of the upper sets starts according to the Aufbauprinciple. Such type of complexes are called low spin complex.For low spin complexessplitting energy ( ) > pairing energy (P.E.)Weak field ligand causes a small splitting of the d-orbitals where splitting energy is less than pairingenergy. It is easier to put electrons into the higher energy set of orbitals than to pair up in the samelow energy orbitals because two electrons in the same orbitals repel each other. So one electron isput into each of the five d-orbitals before any pairing occur in accordance with hund rule.Such complexes are known as high spin complex.For high spin complexes splitting energy ( ) < pairing energy (P.E.)2 5 .In octahedral complex if central metal have configuration d , d ,d always make inner orbtial1 23 complex, Why ?Ans.Central metal ion which have d , d ,d configuration have at least two vacant orbitals in any ligand environment1 23 or with any central metal ion with any oxidation state. So always make inner orbital complex.2 6 .How to calculate the crystal field stablising energy (C.F.S.E.) for octahedral and tetrahedral complex?Ans.(i)For octahedral CFSE =  2 g teg0–0.4 n0.6 n + Paring energy (P.E.)where 2 g t n= number of electron in t orbitals2g n = number of electron in eg orbitalseg 0 = crystal field splitting energy(ii)For tetrahedral CFSE = –0.6 n  2 tte0.4 n +Paring energy (P.E.)where n = number of electron in t orbitals2 t2 n = number of electron in e orbitalset = crystal field splitting energy2 7 .Explain the formation of Na [Fe(CN) ] and Na [FeF ] ? Show which is low spin and which is4646high spin complex and also calculate the Crystal field stablisation energy (CFSE) ?Ans.In given compounds CN is strong field ligands and F is weak field ligand and in both compounds ions––is in +2 oxidation state d configuration.6So in Na [Fe(CN) ]46 Na [FeF ]46 ot (d , d , d )2gxyyzzx e (d, d )gx – y z 2223dIn presence of ligande (d, d )gx – y z  o222t (d , d , d )2gxyyzzx

For octahedral CFSE =  2 g teg0–0.4 n0.6 n+ P.E.where 2g t n= number of electron in t orbitals2g n = number of electron in e orbitalseggNa [Fe(CN) ] CFSE =–2.4 46 o+ 3P.[2g t n= 6, n = 0]egNa [Fe(F) ] CFSE =–0.4 46 o + P[2g t n= 4, n = 2]egwhere P = pairing energy to pair up electron.2 8 .What are the factors which affect the splitting in C.F.T ?Ans.factor affecting splitting(i)Strength of ligand [C.F.S.E. is more in case of S.F.L. as compare to W.F.L.](ii)Oxidation state of central metal ion[C.F.S.E. oxidation state](iii) Transition series (d-series)[C.F.S.E. , 5d > 4d > 3d](iv) Geometry (number of ligands).[ sq >  0 >  t] sq = 43 0 t = 49 02 9 .Which factors affect strength of ligands?Ans.Strength of ligand depends upon :(i) good donor (ii) good acceptor (iii) high negative charge (iv) Small in size3 0 .What is spectro-chemical series for ligands ?Ans.Series which shows the relative strength of ligandsI (weakest) < Br < SCN < Cl < S –– – – 2– < F < OH < C O– – 242– < H O < NCS2–< edta4– < NH < en < CN < CO(strongest)3 – 3 1 .What is the relation between splitting energy of octahedral ( 0) and tetrahedral ( ) ? tAns. t49 03 2 .Compare the splitting energy ( 0) into the following compound and give appropritate reason ?[Co(NH ) ] ,3 63+[Rh(NH ) ] ,3 63+[Ir(NH ) ]3 63+Ans.In given compounds number of ligands, types of ligands and oxidation state is same for central atom belongsto same group but different transition series 3d, 4d and 5d respectively. We know that as move top tobottom size of d-orbital(3d4d 5d) is increases so ligand approches to d-orbitals more closely so therepulsion between d-orbital of metal and ligand is high and splitting energy increases.order of splitting energy [Co(NH ) ] < [Rh(NH ) ] < [Ir(NH ) ]3 63+3 63+ 3 63+3 3 .Compare the splitting energy ( 0) in the following compound and give appropritate reason ?[CrCl ] ,63–[Cr(H O) ] ,263+[Cr(NH ) ] ,3 63+[Cr(CN) ]63–

Ans.In above compound oxidation state, central metal ion and number of ligand is same so compound on thebasis of nature of ligand.According to spectro chemical series strength of given lignadsCl < H O < NH < CN–23–We know that as strength of ligand increases splitting energy increases. So the order is[CrCl ] < [Cr(H O) ] < [Cr(NH ) ] < [Cr(CN) ]63–263+3 63+ 63–3 4 .Compare the splitting energy ( 0) in the following compound and give appropritate reason ?[Fe(H O) ] ,262+[Fe(H O) ]263+Ans.As the oxidation state of central metal ion increases ligand approches more closely to the central metalion where the d-orbital exprience the greater repulsion.[Fe(H O) ] < [Fe(H O) ]262+263+DO YOUR SELF-VII 1.Calculate the crystal field stablization energy (CFSE) for(i)d low spin octahedral 5(ii) d high spin octahedral5(iii) d high spin octahedral4(iv) d low spin octahedral6 2.Why the spliting energy on tetrahedral complexes  t is less than splitting energy of octahedral( 0). Give suitable reason ? 3.Discuss the structure of the following compounds on the basis of the crystal field theory[Co(NH ) ] , [CoF ] , [Fe(H O) ] , [Fe(CN) ]3 63+63–262+63–APPLICATION OF CRYSTAL FIELD THEORY3 5 .What are the applications of crystal field theory (C.F.T) ?Ans.Applications of C.F.T(i)To predict the geometry that the compound is either inner orbital or outer orbital complex.(ii)To calculate the magnitude of paramagnetism.(iii) To show the colour property.(I)PARAMAGNETISM :3 6 .How to calculate the magnitude of paramagnetism of compound ?Ans.Paramagnetism µ = n(n 2) B.M.Where n is the number of unpaired electrons present in the metal ion.3 7 .Calculate the paramagnetism into following compound ?[Cr(H O) ]263+[Fe(H O) ]263+[Zn(H O) ]262+Ans.In all compound H O is a weak field ligand so pairing of electron will not occur in2

(I) [Cr(H O) ]263+(II) [Fe(H O) ]263+(III) [Zn(H O) ]263+ eg0e g2 e g4 t 2g 3t 2g 3 t 2g 6n = 3 n = 5 n = 0µ = 15 µ = 35 µ = 03 8 .Why the d configuration always shows paramagnetism 2.83 B.M in octahedral complex ?8Ans.In given d configuration for octahedral complex for both strong field lignad and weak field ligand is always8have two unpaired electron.Strong field ligand Weak field lignad n = 2 n = 2µ= n(n2) B.M. = 2(2 2) = 2.83 B.M.3 9 .Why metal ion with d ,d ,d ,d ,d ,d configuration show fix paramagnetism in octahedral123 8910complex. Give suitable reason ? also give the value of paramagnetism for given configurations?Ans.For d ,d ,d ,d ,d ,d configuration they have always fix number of unpaired electrons in octahedral geometry123 8910in any lignad enviroment either their is strong field lignad or weak field field lignad.In the given configuration their is no effect of crystal field splitting :Eg : d 3Strong fieldd 3Weak fieldd 1n =1  µ =1(1 2) = 1.73 B.M.d 2n =2  µ =2(2 2) = 2.83 B.M.d 3n =3  µ =3(3 2) = 3.87 B.M.d 8n =2  µ =2(2 2) = 2.83 B.M.d 9n =1  µ =1(1 2) = 1.73 B.M.d 10n =0  µ = 0

DO YOUR SELF-VIII1.Calculate the paramagnetism of following configuration ?(i)d high spin octahedral 4(ii) d low spin octahedral4(iii) d high spin octahderal5(iv) d tetrahedral5(v) d tetrahedral6(vi) d low spin octahedral8(vii) d tetrahderal7(viii) d high spin octahedral.7(II)COLOUR PROPERTY :4 0 .Why the complex compound show colour?Ans.Due to d-d transition of electrons.4 1 .[Ti(H O) ]26+3 is violet in colour explain using CFT.Ans.In [Ti(H O) ] d-robitals of Ti lost their degeneracy in the presence of octahedral ligand field and produce263+3+12 g t & eg . orbital of different energy complex absorbed visible light for excitation of electron from002 g t to eg (d-d transition) and show complimentary violet colour.1.2senergyt2g 1eg 0ground stateexcited statet2g 0eg 1d-d transition4 2 .How the complex compounds show the colour?Ans.When d-electrons absrobs energy from visible region they will get excited. Absorbed energy is related toa particular wavelength.absorbedabsorbedhcE( )when electrons fall into lower energy level it will show colour whose wavelength ( ) is the complimentaryof absorbed wavelength (absorbed).4 3 .Write down the complementry colour relationship between colour spectrum ?Ans.VBROYGFor example complementry colour of red is green.4 4 .Why violet coloured [Ti(H O) ]Cl becomes colourless when heated ?263Ans.When [Ti(H O) ]Cl is heated water molecules are removed and in the absence of ligand crystal field splitting263 does not occur and hence the substance is colourless.

DO YOUR SELF-IX1.[Fe(CN) ] and [Fe(H O) ] are of different colours in dilute solutions. Why ?64–262+2.What will be the correct order for the wavelengths of absorption in the visible region for thefollowing :[Ni(NO ) ]2 64– , [Ni(NH ) ]3 62+ , [Ni(H O) ]262+ ?ISOMERISM4 5 .What is isomerism ?Ans.The compounds having same molecular formula but different physical and chemical properties on accountof different structures are called isomer and the phenomenon is known as isomerism.4 6 .What do you mean by structural isomerism ?Ans.It arises due to the difference in the type of chemical linkage and distribution of ligands within and outsidethe co-ordination sphere.4 7 .What is Ionisation isomerism ? Give example.Ans.This type of isomerism which is due to the exchange of groups or ion between the coordinating sphereand the ionisation sphere. Ex.(i)Co(NH ) Br SO can represent3 424[Co(NH ) Br ] SO (red violet) and [Co (NH ) SO ]Br (red)3 4243 442 (ii)[Pt(NH ) Cl ] Br and [Pt (NH ) Br ]Cl3 4223 422(iii) [Co(NH ) (NO ) ]SO and [Co(NH ) .SO ] (NO )3 43 243 443 24 8 .How can you differentiate ?(i)[Co(NH ) SO ]Br3 44(ii) [Co(NH ) Br]SO3 54Ans.[Co(NH ) SO ]Br give does not white ppt. of BaSO with BaCl solution whereas isomer [Co(NH ) Br]SO3 44423 54does form a precipitate.4 9 .What is the Hydrate isomerism ? Give example.Ans.The isomerism in which different number of water molecules are present inside the coordination sphere.Example Cr(H O) Cl has three possible structures.263(i)[Cr(H O) ] Cl violet263(ii)[Cr(H O) Cl]Cl .H O green2522(iii) [Cr(H O) Cl ]Cl. 2H O dark green24225 0 .One mole of which hydrated isomer of CrCl .6H O gives maximum moles of AgCl when treated32with excess of AgNO ?3Ans.[Cr(H O) ]Cl263

5 1 .What is coordination isomerism ? Give Examples.Ans.This type of isomerism is observed in the coordination compounds having both cationic and anionic complexions. The ligands are interchanged in both the cationic and anionic ions to form isomers.Ex. [Co(NH ) ] [Cr(CN) ] and [Co(NH ) ] [Cr(C O ) ]3 663 624 3[Cr(NH ) ] [Co(CN) ] and [Cr(NH ) ] [Co(C O ) ]3 663 624 35 2 .What do you mean by linkage isomerim ? Give examples.Ans.This type of isomerism occurs in complex compounds which contain ambidanate ligands like NO ,2–SCN , CN , S O . These ligands have two donor atoms but at a time only one atom is directly linked––232–to the central metal atom of the complex. These type of isomer are distinguished by infra-red (I.R./UV/Visible) spectroscopy.Ex. [Co(NH ) NO ] Cl and [Co(NH ) ONO]Cl3 5223 525 3 .What do you mean by ligand isomerism ? Give example.Ans.This type of isomerism occurs in complexes which have same molecular formula, but differ with respectto their ligands are called ligand isomers.Ex. [Fe(H O) C H (NH ) Cl ] has two different structures.22362 22Fe (H O) CH –CH––CH Cl22 322NH NH22 and Fe (H O) CH –CH ––CH Cl22 2222NH 2NH 25 4 .What do you mean by stereo isomerism ?Ans.Compounds which contains the same ligands in their co-ordination sphere but differ in the way that theseligands are arranged in space are known as stereo isomers and this phenomenon is known as stereo isomerism.Stereo-isomerism is of two types, viz. geometrical isomerism and optical isomerism.5 5 .What do you mean by geometrical isomerism. How can you divide in two parts.Ans.This isomerism is due to ligands occupying different positions around the central metal atom or ion. Theligand occupy positions either adjacent or opposite to one another. This type of isomerism is also knownas cis-trans isomerism•When two identical ligands are coordinated to the metal ion from same side, the it is cis isomer. (latin,cis means same).•If the two identical ligands are coordinated to the metal ion from opposite side then it is trans isomer.(in latin, trans means across).5 6 .Why geometrical isomerism cannot arise in a tetrahedral complex ?Ans.Because this geometry contains all the ligands in cis (i.e. adjacent) position with respect to eachother i.e. each ligand is equidistant from the other three ligands and all bond angles are the same(= 109.5°). This isomerism is, however found in many square planar (C.N. =4) and octahedral(C.N. =6) complex.

5 7 .Why [Ma ] , [Ma b] , [Mab ]4n±3n± 3n± type square planar complex do not show geometrical isomerism ?Ans.Because every conceivable spatial arrangement of ligands around the metal ion is exactly same.5 8 .Write the example of geometrical isomers with co-ordination number 4 (square planar complex) ?Ans.Geometrical isomers with Coordination number = 4(i)Complexes with general formula, Ma b (where both a and b are monodentate) can have Cis-and2 2trans isomers.Cis-isomerM aabbTrans-isomerM aabb[Pt (NH ) Cl ]3 22PtHN3NH 3ClClCis(Cis-platin) (use as anti cancer)PtNH 3ClClHN3Trans(ii)Complexes with general formula Ma bc can have Cis - and trans-isomers.2CisM acbatransM acba[Pt(NH ) ClBr]3 2PtNH 3ClCisNH 3BrPtHN3NH 3ClBrTrans(iii)Complexes with general formula, Mabcd can have three isomers.M acdb(i)M adcb (ii)M adbc(iii)5 9 .Write the example of geometrical isomer with coordination number 6 ?Ans.Geometrical isomers with Coordination number = 6(i)[Fe(NH ) Cl ]3 42FeHN3NH 3TransHN3NH 3ClClFeHN3CisHN3NH 3NH 3ClCl

(iii)Facial and Meridional isomerism (Ma b )33aaabbbFacial (fac)aaabbbMeridional (Mer)6 0 .Give the isomeric form of [Cr (NH ) Cl ]° ?III33Ans.[Cr (NH ) Cl ]° which exist in two isomeric forms in one isomer, the three Cl ions are on one triangularIII3 33–face and the three NH molecules are on the opposite triangular face of the regular octahedron. This3isomer is called 1, 2, 3 or facial isomer. In the other isomer the Cl ions are around an edge of the octahedron–and the NH molecules are around the opposite edge.3H N3H N3H N3H N3Cl(2)Cl(2)Cl(3)NH 3CrCrCl(1)Cl(1)NH 31,2,3- Isomer1,2,6- IsomerCl(6)6 1 .Why following pairs are not geometrical isomers ?(i)MMbbccaadd(a,c) are at trans(a,c) are at trans(b,d) are at trans(b,d) are at trans(ii) CoCoClClClClNNNNNNNN(Cl, Cl) are at trans (Cl, Cl) are at trans(iii)MMaaccbbcaacbbAns.In pair (i), (ii) and (iii) all the ligands have identical space orientation but represented different side so thatthe pairs have two Identical complex.

6 2 .What is optical isomerism ?Ans.A coordination compound which can rotate the plane of polarised light is said to be optically active. Whenthe coordination compounds have same formula but differ in their abilities to rotate directions of the planeof polarised light are said to exhibit optical isomerism and the molecules are optical isomers.• optically active complexes are those which are non superimposable over the mirror image structure.• If molecule does not have palen of symmetry then it is optically active.6 3 .What do you mean by d and -form ?Ans.The complex which rotates plane polarised light to left hand side is laevo rotatory i.e. ' ' or '–' and if thecomplex rotates the plane polarised light to right hand side then it is dextro rotatory 'd' or '+'.6 4 .Define the optically active & optically inactive forms ?Ans.When d and forms are capable of rotating the plane of polarised light, these are said to be opticallyactive forms or optical isomer and this phenomena is called optical activity or optical isomerism.One which is not capable of rotating the plane of polarised light is called optically inactive.6 5 .What do you mean by Enantiomorphs?Ans.The 'd' and ' ' isomers of a compound are called as enantiomers or enantiomorphs of each other.6 6 .Which of the molecule show optical isomerism ?Ans.Asymmetric molecule show optical isomerism.6 7 .Write the properties of asymmetric molecule.Ans.(i)Asymmetric molecule never has a plane of symmetry.(ii)An asymmetric molecule cannot be superimposed on its mirror image.6 8 .Why [Ma ], [Ma b] and [Ma b ] type complexes do not show optical isomerism ?4322Ans.Because these complexes have plane of symmetry.6 9 .Why Cis form of [Co(en) Cl ] ion, shows optical isomerism but trans form of this ions not22+shows optical isomerism.Ans.CoClClenen++(a)Mirror planeenenCoClCld-cis forml-cis formunsymmetrical and hence optically active forms

CoenenClCl+(b)trans-meso form (symmetrical and hence optically inactive)The cis-isomer of [Co(en) Cl ] ion shown in fig. (a) can be resolved into two optically active isomers, since22+it has no plane of symmetry. Its trans isomer shown at (b) cannot be resolved into two forms, since nomirror-image of this ion is possible i.e. it has a plane of symmetry. Thus trans isomer is an optically inactiveforms (meso-forms)7 0 .Write the example of optical isomers with coordination number 6 ?Ans.Optical isomers with Coordination number = 6(i)[Ma b c ]222n+[Pt(py) (NH ) Cl ]23 222+2+ClpyNH 3NH 3ClPtCis-d-isomer2+ClClpyNH 3HN3pyPtCis- -isomerpyMirror(ii)[Mabcdef] [Pt(py) (NH ) (NO ) ClBrI]32ClBrIHN3Pt-isomerpyCld-isomerBrINH 3NO 2PtpyNO 2Mirror(iii)[M(AA) ] 3n+[Co(en) ]33+3+end-formenen3+enCo-formenenCoMirror

DO YOURSELF-X 1.Give evidence that [Co(NH ) Cl]SO and [Co(NH ) SO ]Cl are ionisation isomers ?3 543 54 2.How many geometrical isomers possible in the following coordination entities ?(i) [Cr(C O ) ]24 33–(ii) [Co(NH ) Cl ]3 33 3.Draw the structures of optical isomers of :(i) [Cr(C O ) ]24 33–(ii) [PtCl (en) ]222+(iii) [Cr(NH ) Cl (en)]3 22+ 4.Draw all the isomers (geometrical and optical) of :(i) [CoCl (en) ]22+(ii) [Co(NH )Cl(en) ]322+(iii) [Co(NH ) Cl (en)]3 22+ 5.Write all the geometrical isomers of [Pt(NH )(Br)(Cl)(py)] and how many of these will exhibit optical3isomers ? 6.Indicate the types of isomerism exhibited by the following complexes and draw the structures forthese isomers :(i) K[Cr(H O) (C O ) ]2224 2(ii) [Co(en) ]Cl33(iii) [Co(NH ) (NO )](NO )3 523 2(iv) [Pt(NH )(H O)Cl ]322BONDING IN METAL CARBONYL7 1 .What is synergic bonding ?Ans.The electronic configuration of CO molecule shows that it has lone pair of electrons on carbon and oxygenatom each. Carbon atom can donate its electron pair of a transition metal atom (M), forming OC Mcoordinate bond.Since the metal atom in metal carbonyl is in zero oxidation state, the formation of M  CO bond accumulatesa negative charge on the metal atom. The accumulation of negative charge on the metal atom can becounter balanced by transferring some negative charge from the metal atom to CO molecule (ligand). Thistransfer can be done by making a M  CO bond by the overlap between an appropriate filled orbitalon the metal atom and empty * or * molecular orbital on CO molecule. This type of bonding betweenyzM and CO is called synergic bonding.– M– M+C O  C O bond+M++––C OC O++++––––MSchematic of orbital overlaps in metal carbonyls.

7 2 .What is the effect of synergic bonding ?Ans.The fillling or partial filling of the antibonding orbital on C reduce the bond order of C–O bond fromthe triple bond in CO towards a double bond. This shown by the increase in C–O bond length from1.128 Å in CO to about 1.15 Å in many carbonyls. As decrease in (C–O) bond order their will be increasein (M–C) bond order and (M–C) bond order increases from one to towards two.7 3 .Which bond is formed in Zeises salt ?Ans.Zeises salt K [Pt Cl ( -C H )]324The bonding of alkenes to a transition metal to form complexes has two components. First, the -electrondensity of the alkene overlaps with a -type vacant orbital or the metal atom. Second is the back bondingformed by the flow of electron density from a filled d-orbital on the metal into the vacant  *-antibondingmolecular orbital on the carbon atom as shown below:– M– M+++CCCC++++––––-overlapMMCCCC+

d-BLOCK ELEMENTS1 .INTRODUCTION(a)The element lying between s- and p-block element of the periodic table are collectively known as transitionor transitional elements. (T.E'.S.)(b)Their properties are transitional between the highly electropositive s- block element to least electropositivep-block element.(c)In d- block elements, the last differentiating electron is accommodated to the penultimate shell.(d)The general electronic configuration of transition element is (n-1)d1-10 ns0, 1 or 2(e)These elements either in their atomic state or in any of their common oxidation state have partly filled(n-1)d orbitals of (n-1) main shell.th(f)The transition elements have an incompletely filled d-level. Since Zn, Cd, Hg elements have d10configuration and are not considered as transition elements but they are d-block elements.ELECTRONIC CONFIGURATIONIst Transation SeriesSymbolScTiVC rMnFeCoNiC uZnAtomic No.2122232 4252627282 9303d electrons123556781 0104s electrons2221222212Irregular electronic configuration Cr, CuIInd Transation SeriesSymbolYZrN bM oTcR uR hP dA gCdAtomic No.39404 14 2434 44 54 64 7484d electrons12455781 01 0105s electrons2211211012Irregular electronic configuration Nb, Mo, Ru, Rh, Pd, AgIII Transation SeriesrdSymbolLaHfTaWReOsIrP tA uHgAtomic No.5772737 47576777 87 9805d electrons123456791 0106s electrons2222222112Irregular electronic configuration W, Pt, Au The irregularities in the observed configuration of Cr (3d 4s instead of 3d 4s5 1 4 2 ), Cu (3d10 4s ), Mo (4d1 55s ), Pd ([Kr] 4d 5s ), Au ( [Xe] 4f1100 14 5d10 6s ), Ag ([Kr] 4d1 10 5s ) are explained on the basis of the concept1 that half-filled and completely filled d-orbitals are relatively more stable than other d-orbitals.2 .GENERAL PROPERTIES OF d-BLOCK ELEMENTS(a)The properties of d-block elements of any given period are not so much different from one another asthose of the same period of non transtion elements.

(b)It is due to the fact that, in transition series, there is no change in number of electrons of outermost shelland only change occur in (n-1)d electron from member to member in a period.3 .METALLIC CHARACTER(a)All the d-block elements are metals as the numbers of electrons in the outer most shell are one or two.(b)They are hard, malleable and ductile (except Hg). IB group elements Cu, Ag and Au are most ductile and soft.(c)These are good conducter of heat and electricity (due to free e ) Elements of IB group are most conductive—in nature. Their order of conductivity is Ag > Cu > Au > Al(d)Covalent and metallic bonding both exist in the atom of transition metals.(e)The presence of partially filled d-subshell favour covalent bonding and metallic bonding. These bondingare favourable also due to possession of one or two electron in outermost energy shell.4 .REDUCING POWER(a)Reducing power of d-block elements depends on their electrode potential.(b)Standard oxidation potential (SOP) of Cu is minimum in the 3d series so it is least reducing elements in 3dseries.(c)Au is the least reducing element in the d-block because of highest +ve value of Standard reduction potential.(d)The poor reducing capacity of the transition metal is due to high heats of vaporization, high ionizationpotential and low heat of hydration of their ions, because reduction potential depends upon all thesethree factors.5 .DENSITY(a)The atomic volume of the transition elements are low, compared with s-block, so their density iscomparatively high (D = M/V)(b)Os (22.57 gm cm—3) and Ir (22.61 gm cm—3) have highest density.(c)In all the groups (except IIIB) there is normal increase in density from 3d to 4d series, and from 4d to 5d,it increases just double. Due to lanthanide contraction Ex.Ti < Zr << Hf(d)In 3d series Cu ZnDensity increasesDensity decreasesScT iVC rM nF eC oN i(e)In 3d series highest density – Culowest density – Sc(f)Some important orders of densityFe < Ni < CuFe < Cu < AuFe < Hg < Au6 .MELTING AND BOILING POINTS(a)Melting and boiling point of d-block > s-blockReason : Stronger metallic bond and presence of covalent bond formed by unpaired d-electrons.)(b)In Zn, Cd, and Hg there is no unpaired electron present in d-orbital, hence due to absence of covalentbond melting and boiling point are very low in series. (Volatile metals Zn, Cd, Hg)(c)In 3d series Sc to Cr melting and boiling point increases then Mn to Zn melting and boiling point decreases(d)As the number of d-electron increases, the number of covalent bond between the atoms are expected toincrease up to Cr-Mo-W family where each of the d-orbital has only unpaired electrons and the opportunityfor covalent sharing is greatest.(e)Mn and Tc have comparatively low melting point, due to weak metallic bond because of stable Half filled(d ) configuration5

(f)Lowest melting point Hg (– 38°C),Highest melting point W ( ~ 3400°C)20001500 Sc10005000IIIBIVBV BVIBVIIBVIIII BIIB13971672TiV17101900CrMn12441530FeCo1495Ni14551083Cu420 ZnMGraphic representation m.p. of 3d-series elementsCharacteristic properties of transition elements :(a) Variable oxidation state(b) Coloured ions(c) Paramagnetic properties(d) Catalytic properties(e) Formation of alloys(f) Formation of interstitial compounds(g) Formation of complexes.7 .VARIABLE VALENCY OR VARIABLE OXIDATION STATES(a)They exhibit variable valency due to involvement of (ns) and (n-1)d electrons. Due to less energy differencebetween these electrons.(b)The oxidation states of all transiition elements of '3d' series are as follows - Element Conf.Outer electronic configurationOxidation statesSc3d 4s12+ 3 3d 4sTi3d 4s22+ 2+ 3 + 4V3d 4s32+ 2+ 3 + 4+ 5Cr3d 4s51+ 1+ 2 + 3+ 4 +5 +6Mn3d 4s52+ 2+ 3 + 4+ 5 +6 +7Fe3d 4s62+ 2+ 3 + 4+ 6Co3d 4s72+ 2+ 3 + 4Ni3d 4s82+ 2+ 3 + 4Cu3d104s1+ 1+ 2Zn3d104s2+ 2

(c)Highest oxidation state of transition elements can be calculated by n + 2 where (n = number ofunpaired electrons) It is not applied for Cr and Cu.(d)The transition metal ions having stable configuration like d d or d0 510 are more stable.Ex.Sc , Ti , V+3+4+5, Fe+3 , Mn , Zn+2+2 etc.(e)In aqueous medium Cr is stable.+3(f)Co and Ni are stable in complexes..+3+2(g)In aqeous medium due to disproportionation Cu+1 is less stable than Cu+2 while its configuration is 3d10(h)Most common oxidation state among the transition elements is +2.(i)Highest oxidation state shown by transition elements of '4d' and '5d' series is +8 by Ru (44) and Os (76).(j)The common oxidation state shown by elements of IIIB i.e., Sc, Y, La and Ac is +3 as their divalentcompounds are highly unstable.(k)In lower oxidation state transition elements form ionic compounds and in higher oxidation state theircompounds are covalent.(l)They also shows zero oxidation state in their carbonyl compounds like Ni(CO) .4(m)Usually transition metal ions in their lower oxidation state act as reducing agents and in higher oxidationstate they are oxidising agents.Ex.Sc , Ti , V , Fe , Co etc are reducing agents+2+2+2+2+2Cr , Mn , Mn , Mn , Mn etc are oxidising agents.+6+7+6+5+4The relative stability of various oxidation states(a)The relative stabilities of various oxidation states of 3d-series element can be correlated with the extrastability of 3d°,3d & 3 d5 10 configuration to some extent.Ex.Stability of Ti4+ (3d ) > Ti03 + (3d )Mn1 2+ (3d ) > Mn53+ (3d )4(b)The higher oxidation state of 4d and 5d series element are generally more stable than the elements of 3dseries. Ex.(i)vi-24 Mo O(oxidation state of Mo is +6), Mo Ovi-24(4d series) & vi24 W O, vii4 Re O (5d series) are morestable due to their maximum oxidation state.(ii)vi24 Cr O & vii4 Mn O(3d-series) are strong oxidizing agents.(c)Strongly reducing states probably do not form fluorides or oxides, but may well form the heavier halidesConversely, strong oxidizing state form oxides & fluoride, but not Bromide and lodide. Ex.(i)V (Vanadium) react with halogens to form VF VCl , VBr ,but doesn' t form VBr or VI because in + 55 5 35 5 oxidation state Vanadium is strong oxidizing agent thus convert Br & I to Br & I respectively, So VBr– – 2 2 3&VI are formed but not VBr & VI3 5 5.(ii)On the other hand VF is formed because V5 5+ ion unable to oxidize highly electronegative & small anion F–(iii)Similarly highly electronegative and small O2 – ion formed oxides Ex. VO4 3 – , CrO42– & MnO4– etc.

Diffrent oxidation state of chloride & oxides compound+2TiCl2TiCl3VCl2VCl3(Ionic, basic)Less ionic (Amphoteric)TiCl4VCl4VOCl3Covalent and Acidic (Strong lewis acid)+3+4+5+6+7TiOVOCrOMnOTi O23VO23Cr O23Mn O23TiO2MnO2 VO25CrO3MnO3 Mn O27 Less Ionic (Amphoteric) Ionic, basicAcidic, covalent(d)Such compounds are expected to be unstable except in case where vacant d-orbitals are used for acceptinglone-pair from -bonding ligand.Ex.[Ni(CO) ], [Ag(CN) ] ,[Ag)(NH ) ]42–3 2+8 .COLOUR PROPERTY(a)Most of the transition metal ions exhibit colour property.(b)This is due to d-d transition of unpaired electrons in their t2g and e sets of 'd' orbitals.g(c)They require less amount of energy to undergo excitation of electrons. Hence they absorb visible regionof light exhibiting colour.Ex.Sc+2 : [Ar]3d , Ti1+2 : [Ar]3d , V2+2 : [Ar]3d3(d)Transition metal ions which do not have any unpaired elctrons in their 'd' orbitals like 3d and 3d010configurations, do not exhibit any colour property.Ex.Sc : [Ar]3d , Cu+30+1 : [Ar]3d , Ti10+4 : [Ar]3d etc are colourless ions.0(e)A transition metal ion absorbs a part of visible region of light and emmits rest of the colours, the combinationof which, is the colour of emitted light. The colour of metal ion is the colour of the emitted light.(f)In transition metal ion the 'd' orbitals split into lower energy set t orbitals and higher energy set e orbitals.2ggThe electrons from t set get excited to higher energy set i.e., e set. This excitation of electrons is called2ggas 'd-d' transition. Due to this 'd -d' transition the transition metal ions exhibit colour property.dx y 22dz 2-d yzd xyd yzd-orbitals (degenerate)lightLower energy set = t2gHigher energy set = egPresence of ligandsd-d Trnsition(t )2g(e)gFactors affecting the colour of complexThe colour of a transition metal complex depends on-(a)The magnitude of energy difference between the two d-levels (0 ,)

(b)An increase in the magnitude of  0 decreases the wave length () of the light absorbed by the complexes.01(Wavelength of light absorb) (c)Thus with a decrease in the the colour of complex changes from Red to Violet.Ex.Complex ions[Co(H O) ]263+[Co(NH ) ]3 63+[Co(CN) ]63–Ligand field strengthH O <2NH3 <CN–Magnitude of  0 0(H O) <2 0(NH ) <3 0(CN )–Magnitude of (H O) <2(NH ) <3(CN )–Colour of the transmittedorangeGreen-bluevioletColour of absorbed light Green-blueOrangeYellow-greenLight (I.e. colour of thecomplex(d)KMnO (dark pink), K Cr O (orange) having d° configuration but they are coloured due to charge trans422 7fer spectrum and charge is transfered from anion to cation.Example of Some coloured metal ions :Ti+3 PurpleCr+3GreenMn+2Light pinkFe+2GreenFe+3 YellowCo+3PinkNi+2GreenCu+2BlueSc3+ ColourlessTi4+ColourlessTi3+PurpleV4 +BlueV3 + GreenV2 +VioletCr2+BlueCr3+GreenMn3+ VioletMn2+PinkFe2+Green(light)Fe3+YellowCo2+ PinkNi2+BlueZn2+Colourless9 .MAGNETIC PROPERTIES(a)Generally transition elements exhibits the magnetic property. A paramagnetic substance is one whichis attracted into a magnetic field. Paramagnetism is mainly due to the presence of unpaired electrons inatoms or ions or molecules. It varies inversely with temperatures.(b)Diamagnetic substance is one which is slightly repelled by a magnetic field. It's independent of temperature.(c)As is evident most of the transition metal ions have unpaired electrons in their 'd' orbitals. Hence most of thetransition metal ions are paramagnetic in nature. Ex.Ti [Ar]3d , Ti [Ar]3d . V [Ar]3d , Cr [Ar]3d+22+31+23+33(d)Transition metal ions having 3d and 3d configuration exhibit diamagnetic nature.010(e)The total magnetic moment of a substance is the resultant of the magnetic moments of all the individualelectrons.(f)The magnetic moment ( ) created due to spinning of unpaired electrons can be calculated by usingn(n2) Where - 'n' is the number of unpaired electrons in the metal ion. = Magnetic moment in Bohr Magnetons (B.M.)(g)The magnetic moment of diamagnetic substances will be zero.(h)Transition metal ions having d configuration will have maximum number of unpaired electrons therefore5they will be maximum paramagnetic in nature.

1 0 .CATALYTIC PROPERTY(a)Transition elements and their compounds exhibit catalytic properties. This is due to their variable valencyas well as due to the free valencies on their surface.(b)When transition elements and their compounds are in powdered state, their catalytic properties exhibitedwill be to a greater extent. This is due to greater surface area available in the powdered state.(c)Transition metals and their compounds exhibiting catalytic properties in various processes are-CatalystUsedTiCl3Used as the Ziegler-Natta catalyst in the production of polythene.V O25Convert SO to SO in the contact process for making H SO2324MnO2Used as a catalyst to decompose KClO to give O32FePromoted iron is used in the Haber-Bosch process for making NH3FeCl3Used in the production of CCl from CS and Cl422FeSO and H O42 2Used as Fenton's reagent for oxidizing alcohols to aldehydes.PdCl2Wacker process for converting C H + H O + PdCl to2 422CH CHO + 2HCl + Pd.3PdUsed for hydrogenation (e.g. phenol to cyclohexanone).Pt/PtOAdams catalyst, used for reductions.PtFormerly used for SO 2SO in in the contace process for making H SO324Pt/RhFormerly used in the ostwald process for making HNO to oxidize NH to NO33CuIs used in the direct process for manufacture of (CH ) SiCl used to make silicones.3 22Cu/VOxidation of cyclohexanol/cyclohexanone mixture to adipic acid which is usedto make nylone-66CuCl2Decon process of making Cl from HCl2NiRaney nickel, numerous reduction processes (e.g. manufacture ofhexamethylenediamine, productiomn of H from NH , reducing anthraquinone23to anthraquinol in the production of H O2 21 1 .FORMATION OF ALLOY(a)Transition elements have maximum tendency to form alloys.(b)The reactivity of transition elements is very less and their sizes are almost similar. Due to this a transitionmetal atom in the lattice can be easily replaced by other transition metal atom and hence they havemaximum tendency to form alloys.(c)In the alloys, ratio of component metals is fixed.(d)These are extremly hard and have high melting point.

SOME IMPORTANT ALLOY(a)BronzeCu (75 - 90 %) +Sn ( 10 - 25 %)(b)BrassCu ( 60 - 80 %) +Zn (20 - 40 %)(c)Gun metal(Cu + Zn + Sn) (87 : 3 : 10)(d)German SilverCu + Zn + Ni ( 2 : 1 : 1)(e)Bell metalCu (80 %) + Sn (20 %)(f)Nichrome(Ni + Cr + Fe)(g)Alnico(Al, Ni, Co)(h)Type MetalPb + Sn + Sb(i)Alloys of steel Vanadium steelV (0.2 - 1 %) Chromium steelCr (2 - 4 %) Nickel steelNi (3 -5 %) Manganese steelMn (10 - 18 %) Stainless steelCr (12 - 14 %) & Ni (2 - 4 %) Tunguston steelW (10 - 20 %) InvarNi (36 %)(j)14 Carat Gold54 % Au + Ag (14 to 30 %) + Cu (12 - 28 %)(k)24 Carat Gold100 % Au(l)SolderPb + Sn(m)MagnelliumMg (10%) + Al (90%)(n)Duralumin(Al + Mn + Cu)(o)Artificial GoldCu (90 %) + Al (10%)(p)ConstantanCu(60%) + Ni (40%)% of Carbon in different type of IronN am e% of C(a)Wrought Iron0.1 to 0.25(b)Steel0.25 to 2.0(c)Cast Iron2.6 to 4.3(d)Pig Iron2.3 to 4.61 2 .FORMATION OF INTERSTITIAL COMPOUNDS(a)Transition elements form interstitial compounds with smaller sized non metal elements like hydrogen,carbon, boron, nitrogen etc.(b)The smaller sized atoms get entrapped in between the interstitial spaces of the metal lattices.These interstitial compounds are nonstoichiometric in nature and hence cannot be given any definiteformula.(c)The smaller sized elements are held in interstitial spaces of transition elements by weak Vander Waalsforces of attractions.(d)The interstitial compounds have essentially the same chemical properties as the parent metals but theydiffer in physical properties such as density and hardness.The process of adsorption of excess of H atom by the transition metals like Pd, Pt etc is called occlusion.

1 3 .NONSTOICHIOMETRY(a)The transition elements sometimes form nonstoichiometric compounds due to variable valency.(b)These are the compounds of indefinite structure & proportion.(c)For example, lron (II) Oxide FeO should be written as a bar over the formula FeO to indicate the ratio ofFe & O atom is not exactly 1:1 (Fe .0 94 O & Fe0.84 O), VSe (VSe0.98VSe1.2 ,)(d)Non stoichiometry is shown particularly among transition metal compounds of the group 16 elements(O,S,Se,Te).(e)Some times nonstoichiometry is caused by defect in the solid structure.1 4 .POTASSIUM DICHROMATE (K Cr O )227PreparationIt is prepared from Chromite ore or Ferrochrome or Chrome iron. (FeO.Cr O or FeCr O ). The various steps2 32 4involved are.(a)Preparation of sodium chromate (Na CrO ) :24The powdered chromite ore in fused with sodium hydroxide or sodium carbonate in the presence of air ina reverberatory furnace.4FeCr O + 16NaOH + 7O 2 428Na CrO + 2Fe O + 8H O242 32or4FeCr O + 8Na CO + 7O 2 42328Na CrO + 2Fe O + 8CO242 32After the reaction the roasted mass is extracted with water. So sodium chromate is completely dissolvedwhile ferric oxide is left behind.(b)Formation of sodium dichromate (Na Cr O ) from sodium chromate (Na CrO ) :22724The solution of sodium chromate is filtered and acidified with dil./con. H SO acid giving its dichromate.242Na CrO + H SO 2424Na Cr O + Na SO + H O22 7242On cooling, sodium sulphate being less soluble crystallizes out as Na SO .10H O and is removed. The242resulting solution contains sodium dichromate (Na Cr O ).22 7(c)Formation of potassium dichromate from sodium dichromate :The hot concentrate solution of sodium dichromate is heated with calculated amount of KCl.Na Cr O + 2KCl 22 7K Cr O + 2NaCl22 7Sodium chloride, being the least soluble precipitates out from the hot solution and is removed by filtration.Orange red crystals of potassium dichromate separate out from mother liquid on cooling.Properties(a)Colour and Melting Point :- Orange red crystals. 670 K(b)Solubility :- Moderately soluble is cold water but readily soluble in hot water.(c)Action of Heat :- It decompose on heating to give potassium chromate, chromic oxide and oxygen.4K Cr O 22 7Heat 4K CrO + 2Cr O + 3O242 32 Potassium Chromicchromate oxide

(d)Action of Alkalies :- On heating with alkalies the orange colour of dichromate solution changes toyellow due to the formation of chromate ions.K Cr O + 2KOH 22 72K CrO + H O242orCr OOHCrOH O27 24 2222This chromate on acidifying reconverts into dichromate.2K CrO + H SO 2424K Cr O + K SO + H O22 7242or2CrO2H 4 227 22Cr OH OThe interconversion is explained by the fact that dichromate ion and chromate ion exist in equilibrium ata pH of about 4.Cr OH O27 222CrO2H 4 2 When alkali added, H consumed so forward direction. When acid added, H increases so backward++direction.(e)Chromyl chloride Test :- When potassium dichromate is heated with conc. H SO acid and a soluble24metal chloride (ex. NaCl) orange red vapours of chromyl chloride (CrO Cl ) are formed.22K Cr O + 4NaCl + 6H SO 22 7242KHSO + 4NaHSO + 2CrO Cl + 3H O4 4222(f)Reaction with H O :- Acidified solution of dichromate ions give deep blue colour solution with H O2 22 2due to the formation of [CrO(O ) ] or CrO . The blue colour fades away gradually due to the decomposition2 25of CrO into Cr ions and oxygen.5+3CrOOOOOCr OH OHCrOH O27 222524225 (Butterfly stucture)(g)Action with HCl :- Potassium dichromate reacts with hydrochloric acid and evolves chlorine.K Cr O + 14HCl 22 7 2KCl + 2CrCl + 7H O + 3Cl322(h)Action of con. H SO24(i)In cold, red crystals of chromic anhydride are formed.K Cr O + 2H SO 22 724 2CrO + 2KHSO + H O342(ii)On heating the mixture oxygen is evolved.2K Cr O + 8H SO 22 724 2K SO + 2Cr (SO ) + 8H O + 3O2424 322(i)Oxidising propertiesThe dichromates act as powerful oxidising agent in acidic medium. In presence of dil H SO , K Cr O2422 7liberates Nascent oxygen and therefore acts as an oxidising agent.K Cr O + 4H SO 22 724 K SO + Cr (SO ) + 4H O + 3[O]2424 32In terms of electronic concept the Cr O27 2  ion takes up electrons in the acidic medium and hence acts asan oxidising agent.Cr OHeCrH O27 23214627(i)It oxidises iodides to iodine :-Cr O27 2  + 14H + 6e +– 2Cr + 7H O+32 [2I – I + 2e ] × 32–Cr O27 2  + 14H + 6I +– 2Cr + 3I + 7H O+322orK Cr O + 7H SO + 6KI 22 7244K SO + Cr (SO ) + 7H O + 3I2424 322

(ii)Acidified ferrous sulphate to ferric sulphateCr O27 2  + 14H + 6e +– 2Cr + 7H O+32 Fe +2 Fe + e ] × 63+–Cr O27 2  + 14H + 6Fe ++2 2Cr + 6Fe + 7H O+3+32orK Cr O + 6FeSO + 7H SO 22 7424Cr (SO ) + 3Fe (SO ) + 7H O + K SO24 324 3224(iii)Oxidises H S to sulphur2Cr O27 2  + 14H + 6e +– 2Cr + 7H O+32H S 2 S + 2H + 2e ] × 3+–Cr O27 2  + 3H S + 8H 2+ 2Cr + 3S + 7H O+32orK Cr O + 3H S + 4H SO 22 7224Cr (SO ) + 3S + 7H O + K SO24 3224Similarly, it oxidises sulphites to sulphates, chlorides to chlorine, nitrites to nitrates, thiosulphates tosulphates and sulphur and stannous (Sn ) salts to stannic (Sn ) salts.+2+43SO8H3SO2Cr4H O3227 24 232Cr O3NO8H3NO2Cr4H O227 2332Cr O 3S O8H3SO3S2Cr4H O23 227 24 232Cr O 6Cl14H3Cl2Cr7H O27 2232Cr O 3Sn14H3Sn2Cr7H O227 2432Cr OIt oxidises SO to sulphuric acid.2K Cr O + 4H SO 22 724 K SO + Cr (SO ) + 4H O + 3O2424 32SO + O + H O 22 H SO24Us es(a)For volumetric estimation of ferrous salts, iodides and sulphites.(b)For preparation of other chromium compounds such as chrome alum (K SO , Cr (SO ) .24H O), chrome2424 32yellow (PbCrO ) and chrome red (PbCrO .PbO).44(c)Used in photography for hardening of gelatin film.(d)It is used in leather industry (chrome tanning)(e)Chromic acid mixture is used for cleaning glassware, consist of K Cr O and Con. H SO .22 724(f)In organic chemistry, it is used as an oxidising agent.(g)In dyeing and calico printing.StructureThe chromate ion has tetrahedral structure in which four atoms around chromium atom are oriented in atetrahedral arrangement.

The structure of dichromate ion consist of two tetrahedra sharing an oxygen atom at the common corner.1 5 .POTASSIUM PERMANGANATE (KMnO )4PreparationPotassium permanganate is prepared from mineral pyrolusite (MnO ). The preparation involves the following steps.2(a)Conversion of pyrolusite ore to potassium manganateThe pyrolusite MnO is fused with caustic potash (KOH) or potassium carbonate in the presence of air or2oxidising agents, such as KNO or KClO to give a green mass due to the formation of potassium33manganate (K MnO ).242MnO + 4KOH + O 222K MnO + 2H O2422MnO + 2K CO + O 22322K MnO + 2CO242(b)Oxidation of potassium manganate to potassium permanganateThe green mass is extracted with water resulting is green solution of potassium manganate. The solutionis then treated with a current of Cl or ozone or CO to oxidise K MnO to KMnO . The solution is22244concentrated and dark purple crystals of KMnO seperate out.42K MnO + Cl 2422KCl + 2KMnO42K MnO + O + H O 24322KMnO + 2KOH + O423K MnO + 2CO 2422K CO + MnO232+ 2KMnO4Alternatively, alkaline potassium manganate is electrolytically oxidised.Electrolytic method :- The potassium manganate solution is taken in an electrolytic cell which contains ironcathode and nickel anode. When current is passed the manganate ion in oxidised to permanganate ion at anodeand hydrogen is liberated at cathode.K MnO 24  2K + MnO+42–At anode :MnO4–2MnO + e4 ––Green PurpleAt cathode : 2H + 2e +–2H2H H 2

Properties(a)Colour and M.P. :- Dark violet crystalline solid, M.P. 523 K(b)Solubility :- Moderately soluble is room temperature, but fairly soluble in hot water giving purple solution.(c)Heating :- When heated strongly it decomposes at 746 K to give K MnO and O .2422KMnO 4746 K K MnO + MnO + O2422Solid KMnO gives KOH, MnO and water vapours, when heated in current of hydrogen.42KMnO + 5H42 2KOH + 2MnO + 4H O2(d)Action of alkali :- On heating with alkali, potassium permanganate changes into potassium manganateand oxygen gas is evolved.4KMnO + 4KOH 44K MnO + 2H O + O2422(e)Action of con. H SO :- With cold H SO , it gives Mn O which on heating decomposes into MnO .24242 722KMnO + 2H SO 424Mn O + 2KHSO + H O2 7422Mn O 2 74MnO + 3O22(f)Oxidising character :- KMnO acts as powerful oxidising agent in neutral, alkaline or acidic solution4because it liberates nascent oxygen as :-Acidic solution :- Mn ions are formed+22KMnO + 3H SO 424K SO + 2MnSO + 3H O + 5[O]2442orMnO + 8H + 5e 4 –+– Mn + 4H O +22equal wt.M  5Neutral solution :- MnO is formed22KMnO + H O 422KOH + 2MnO + 3[O]2orMnO + 2H O + 3e 4 –2– MnO + 4OH2–equal wt.M  3During the reaction the alkali produced generates the alkaline medium even if we start from neutral medium.Alkaline medium :- Manganate ions are formed.2KMnO + 2KOH 42K MnO + H O + [O]242Reactions in Acidic Medium : In acidic medium KMnO oxidizes –4(a)Ferrous salts to feric saltsMnO + 8H + 5e 4 –+– Mn + 4H O+22 Fe +2 Fe + e ] × 5+3–MnO + 5Fe + 8H 4 –+2+ Mn + 5Fe + 4H O2++32(b)Oxalates to CO :2MnO + 8H + 5e 4 –+– Mn + 4H O] × 2+22 C O2 42– 2CO + 2e ] × 52–2MnO + 5C O4 –2 42– + 16H + 2Mn + 10CO + 8H O+222

(c)Iodides to IodineMnO + 8H + 5e 4 –+–Mn + 4H O] × 2+222I –I + 2e ] × 52–10I + 2MnO + 16H –4 –+ 2Mn + 5I + 8H O+222(d)Sulphites to sulphatesMnO + 8H + 5e 4 –+–Mn + 4H O] × 2+22 SO32– + H O 2SO42– + 2H + 2e ] × 5+–5SO32– + 2MnO + 6H 4 –+ 2Mn + 5SO+242– + 3H O2(e)It oxidizes H S to S2MnO + 8H + 5e 4 –+–Mn + 4H O] × 2+22 S 2–S + 2e ] × 5–2MnO + 16H + 5S 4 –+–22Mn + 5S + 8H O+22(f)It oxidizes SO to sulphuric acid22KMnO + 3H SO 424K SO + 2MnSO + 3H O + 5[O]2442 SO + H O + [O] 22H SO ] × 5242KMnO + 5SO + 2H O 422K SO + 2MnSO + 2H SO24424(g)It oxidizes Nitrites to nitrates2KMnO + 3H SO 424K SO + 2MnSO + 3H O + 5[O]2442 KNO + O 2KNO ] × 532KMnO + 5KNO + 3H SO 4224K SO + 2MnSO + 5KNO + 3H O24432Reactions in Neutral Medium :(a)It oxidizes H S to sulphur :22KMnO + H O 422KOH + 2MnO + 3 [O]2 H S + O 2H O + S] × 322KMnO + 3H S 422KOH + 2MnO + 2H O + 3S22(b)It oxidizes Manganese sulphate (MnSO to MnO ) manganese dioxide :422KMnO + H O 422KOH + 2MnO + 3 [O]2MnSO + H O + O 42MnO + H SO ] × 32242KOH + H SO 24 K SO + 2H O2422KMnO + 3MnSO + 2H O 4425MnO + K SO + 2H SO22424

(c)It oxidizes Sodium thiosulphate to sulphate and sulphur :2KMnO + H O 422KOH + 2MnO + 3 [O]2Na S O + O 2 2 3Na SO + S] × 3242KMnO + 3Na S O + H O 42 2 322MnO + 3Na SO + 2KOH + 3S224Reactions in alkaline Medium(a)It oxidizes Iodides to Iodates in alkaline medium :2KMnO + H O 422KOH + 2MnO + 3 [O]2 KI +3O KIO32KMnO + KI + H O 422MnO + 2KOH + KIO23(b)Alkaline KMnO (Baeyers reagent) oxidizes ethylene to ethylene glycol.4CH 2CH 2+ H O + [O] 2CH—OH2CH—OH2StructureMnO4 –Us es(a)Used in volumetric analysis for estimation of ferrous salts, oxalates, and other reducing. agents. It is notused as primary standard because it is difficult to obtain it in the pure state.(b)It is used as strong oxidising agent in the laboratory as well as industry.(c)As disinfectant and germicide.(d)In dry cells.(e)A very dilute solution of KMnO is used for washing wounds.41 6 .COMPOUNDS OF IRONFERROUS SULPHATE (GREEN VITRIOL), FeSO ·7H O : This is the best known ferrous salt. It occurs in42nature as copper and is formed by the oxidation of pyrites under the action of water and atmospheric air.2FeS + 7O + 2H O 2222FeSO + 2H SO4 24It is commonly known as harakasis.

PreparationIt is obtained by dissolving scrap iron in dilute sulphuric acid.Fe + H SO 242FeSO + H42The solution is crystallised by the addition of alcohol as ferrous sulphate is sparingly soluble in it.Properties(a)Action of heat : At 300°C, it becomes anhydrous. The anhydrous ferrous sulphate is colourless. Theanhydrous salt when strongly heated, breaks up to form ferric oxide with the evolution of SO and SO .23FeSO 7HO42·Green300°C– 7HO22FeSO4WhiteHightemperature Fe O + SO + SO2 323(b)The aqueous solution of ferrous sulphate is slightly acidic due to its hydrolysis.FeSO + 2H O 42 Fe(OH) + H SO224 Weak base Strong acid(c)It reduces gold chloride to gold.AuCl + 3FeSO 34Au + Fe (SO ) + FeCl24 33(d)It reduces mercuric chloride to mercurous chloride.[2HgCl 2Hg Cl + 2Cl] × 322[3FeSO + 3Cl 4Fe (SO ) + FeCl ] × 224 336HgCl + 6FeSO 43Hg Cl + 2Fe2(SO ) + 2FeCl224 33(e)A cold solution of ferrous sulphate absorbs nitric oxide forming dark brown addition compound, nitrosoferrous sulphate.FeSO + NO 4FeSO · NO4Nitroso ferrous sulphate (Brown)The NO gas is evolved when the solution is heated.Us es(a)Ferrous sulphate is used for making blue black ink.(b)It is used as a mordant in dyeing.(c)It is also used as an insecticide in agriculture.(d)It is employed as a laboratory reagent and in the preparation of Mohr's salt.Ferrous-oxide FeO (Black)Preparation : FeC O 2 4In absence of air FeO + CO + CO2Properties : It is stable at high temperature and on cooling slowly disproportionates into Fe O and iron3 4Ferrous chloride (FeCl )2Preparation : Fe + 2HCl a current of HClheated in FeCl + H22Properties : 2FeCl + H 32 2FeCl + 2HCl2(a)It is deliquescent in air like FeCl3(b)It is soluble in water, alcohol and ether also because it is sufficiently covalent in nature

(c)It volatilises at about 1000°C and vapour density indicates the presence of Fe Cl . Above 1300°C24density becomes normal(d)It oxidises on heating in air12FeCl + 3O 22 2Fe O + 8FeCl2 33(e)H evolves on heating in steam23FeCl + 4H O 22 Fe O + 6HCl + H3 42(f)It can exist as different hydrated formFeCl · 2H O 22 colourlessFeCl · 4H O 22 pale greenFeCl · 6H O 22 green1 7 .COMPOUND OF ZINCZinc oxide (ZnO) zinc whitePreparation(a)ZnO is formed when ZnS is oxidised2ZnS + 3O 2 2ZnO + 2SO2(b)Zn(OH) on strongly heating gives ZnO2Zn(OH) 2ZnO + H O2(c)Zinc on burning in air gives ZnO (commercial method)2Zn + O 22ZnOProperties(a)ZnO is white when it is cold, a property that has given it a use as a pigment in paints. However, it changescolour, when hot, to a pale yellow. This is due to change in the structure of lattice.(b)ZnO is soluble both in acid and alkali and is thus amphoteric in nature.ZnO + 2H +Zn + H O2+2ZnO + 2OH + H O –2[Zn(OH) ] or 42–22 ZnO zincate ionZnO + 2HCl ZnCl + H O22ZnO + 2NaOH Na ZnO + H O222sodium zincate(c)ZnO + C >1000°CZn + COZnO + COZn + CO2It is preferred to white lead as it is not blackened by H S. It is also used in medicine and in the perparation of2Rinman's green (ZnCo O )2 4Zinc Sulphate (ZnSO )4Preparation(a)ZnSO · 7H O (also called white vitriol) is formed by decomposing ZnCO with dil. H SO42324ZnCO + H SO 324ZnSO + H O + CO422

(b)By heating ZnS (zinc blende) in air at lower temperature and dissolving the product in dil. H SO242ZnS + 3.5O 2ZnO + ZnSO + SO42ZnO + H SO 24ZnSO + H O42Properties(a)Highly soluble in water and solution is acidic in nature due to hydrolysisZnSO + 2H O 42Zn(OH) + H SO2 24(b)ZnSO · 7H O 42100°C ZnSO · 6H O 42280°C ZnSO4 T > 760°C ZnO + SO3It slowly effloresces when exposed to air.(c)It is isomorphous with Epsom salt and used in the manufacture of lithophone (which is a mixture ofBaS + ZnSO and is used as white pigment).4Zinc chloride (ZnCl2 )PreparationZnO + 2HCl ZnCl + H O22ZnCO + 2HCl 3ZnCl + H O + CO 222It crystallises as ZnCl 2HO 22·Zn(OH) + 2HCl 2ZnCl + 2H O22Anhydrous ZnCl cannot be made by heating ZnCl · 2H O because222ZnCl · 2H O 22 Zn(OH)Cl + HCl + H O2Zn(OH)Cl  ZnO + HClTo get anhydrous ZnCl2Zn + Cl2 ZnCl2Zn + 2HCl(dry) ZnCl + H2Zn + HgCl 2ZnCl + Hg2Properties(a)It is deliquescent white solid (when anhydrous)(b)ZnCl + H S 22ZnSZnCl + NaOH 2Zn(OH) 2excess Na [Zn(OH) ]24ZnCl + NH OH 24Zn(OH) 2excess [Zn(NH ) ]3 42+Us es(a)Used for impregnating timber to prevent destruction by insects(b)As dehydrating agent when anhydrous(c)ZnO · ZnCl used in dental filling2

1 8 .COMPOUND OF SILVERSilver Nitrate (Lunar Caustic) AgNO3Preparation(a)When Ag is heated with dil HNO , AgNO is formed. Crystals separate out on cooling the concentrated33solution of AgNO33Ag + 4HNO 3 3 AgNO + NO + 2H O32Colourless crystalline compound soluble in H O and alcohol ; m.p. 212°C2(b)When exposed to light, it decomposes hence, stored in a brown coloured bottle:2Ag + 2NO + O 22, red hot2AgNO 3, T > 212°C2AgNO + O22Properties(a)It is reduced to metallic Ag by more electropositive metals like Cu, Zn, Mg and also by PH .32AgNO + Cu 3Cu(NO ) + 2Ag3 26AgNO + PH + 3H O 3326Ag + 6HNO + H PO333(b)It dissolves in excess of KCN:AgNO 3 KCN AgCN  KCN K[Ag(CN) ]2 white pptsoluble potassium argentocyanideAgNO gives white precipitate with Na S O ; white precipitate changes to black.32 2 32AgNO + Na S O 32 23Ag S O + 2NaNO2 233 white pptAg S O + H O 2 2 32Ag S + H SO224 black(c)Ammoniacal AgNO is called Tollen's reagent and is used to identify reducing sugars (including aldehydes):3RCHO + 2Ag + 3OH +– RCOO + 2Ag + 2H O–2It is called 'silver mirror test' of aldehydes and reducing sugar (like glucose, fructose).Some important reaction of AgNO3Ag ,T>212°CAgNO3AgKCNHNO3RCHO,OH—Na S O223PH 3CuCu + Ag2+Ag + H PO + HNO333O + Ag + NO22AgK[Ag(CN) ]2Ag S O223whiteH O2Ag Sblack2

1 9 .COMPOUND OF COPPERCupric oxide (CuO)It is called black oxide of copper and is found in nature as tenorite.Preparation(a)By heating Cu O in air or by heating copper for a long time in air (the temperature should not exceed2above 1100°C)Cu O + 212O 2 2CuO2Cu + O 2 2CuO(b)By heating cupric hydroxide,Cu(OH) 2 CuO + H O2(c)By heating copper nitrate,2Cu(NO ) 3 2 2CuO + 4NO + O22(d)On a commercial scale, it is obtained by heating molachite which is found in nature.CuCO · Cu(OH) 32 2CuO + CO + H O22Properties(a)It is black powder and stable to moderate heating.(b)The oxide is insoluble in water but dissolves in acids forming corresponding salts.CuO + 2HCl  CuCl + H O22CuO + H SO 24 CuSO + H O42CuO + 2HNO 3 Cu(NO ) + H O3 22(c)When heated to 1100 – 1200°C, it is converted into cuprous oxide with evolution of oxygen.4CuO 2Cu O + O22(d)It is reduced to metallic copper by reducing agents like hydrogen, carbon and carbon monoxide.CuO + H 2 Cu + H O2CuO + C  Cu + COCuO + CO  Cu + CO2Us esIt is used to impart green to blue colour to glazes and glass.Cupric Chloride, (CuCl · 2H O)22Preparation(a)2Cu + 4HCl + O 22CuCl + 2H O22CuO + 2HCl CuCl + H O22Cu(OH) CuCO + 4HCl 232CuCl + 3H O + CO222 (b)Cu + Cl 2CuCl2CuCl · 2H O 2 2150°CHCl gas CuCl + 2H O22

Properties(a)The aqueous solution is acidic due to its hydrolysis.CuCl2 + 2H O 2Cu(OH) + 2HCl2(b)The anhydrous salt on heating forms Cu Cl and Cl2222CuCl 2Cu Cl + Cl222(c)It is readily reduced to Cu Cl by copper turnings or SO gas, or hydrogen (Nascent–obtained by the222action of HCl on Zn) or SnCl .2CuCl + Cu 2Cu Cl222CuCl + SO + 2H O 222Cu Cl + 2HCl + H SO22242CuCl + 2H 2Cu Cl + 2HCl222CuCl + SnCl 22Cu Cl + SnCl224(d)A pale blue precipitate of basic cupric chloride, CuCl · 3Cu(OH) is obtained when NaOH is added.22CuCl + 2NaOH 2Cu(OH) + 2NaCl2CuCl + 3Cu(OH) 22CuCl · 3Cu(OH)22It dissolves in ammonium hydroxide forming a deep blue solution. On evaporating of this solution deepblue crystals of tetrammine cupric chloride are obtained.CuCl + 4NH OH 24Cu(NH ) Cl · H O + 3H O3 4222Us esIt is used as a catalyst in Deacon's proces. It is also used in medicines and as an oxygen carrier in the preparationof organic dyestuffs.Copper Sulphate (Blue Vitriol), CuSO · 5H O42Copper sulphate is the most common compound of copper. It is called as blue vitriol or nila thotha.Preparation(a)CuO + H SO 24 CuSO + H O42Cu(OH) + H SO 224CuSO + 2H O42Cu(OH) CuCO + 2H SO 23242CuSO + 3H O + CO422(b)On commercial scale : it is prepared from scrap copper. The scrap copper is placed in a perforatedlead bucket which the dipped into hot dilute sulphuric acid. Air is blown through the acid. Coppersulphate is crystallised from the solution.Cu + H SO + 2412 O (air)2CuSO + H O42Properties(a)It is a blue crystalline compound and is fairly soluble in water.(b)Heating effectCuSO · 5H O 42Exposure CuSO · 3H O 42100°CCuSO · H O 42230°C CuSO4Blue Pale blue Bluish white WhiteCuSO 4720°CCuO + SO3SO + 2 12 O2

(c)Action of NH OH : With ammonia solution, it forms the soluble blue complex. First it forms a precipitate4of Cu(OH) which dissolves in excess of ammonia solution2CuSO + 2NH OH44Cu(OH) + (NH )SO244Cu(OH) + 2NH OH + (NH ) SO244 24Cu(NH ) SO + 4H O3 442 Tetrammine cupric sulphateThe complex is known as Schwitzer's reagent which is used for dissolving cellulose in the manufacture ofartificial silk.(d)Action of alkalies : Alkalies form a pale bule precipitate of copper hydroxide.CuSO + 2NaOH 4Cu(OH) + Na SO224(e)Action of potassium iodide : First cupric iodide is formed which decomposes to give white cuprousiodide and iodine.[CuSO + 2KI 4CuI + K SO ] × 22242CuI 2Cu I + I2 222CuSO + 4KI 4Cu I + 2K SO + I2 2242(f)Action of H S : When H S is pased through copper sulphate solution, a black precipitate of copper22sulphide is formed.CuSO + H S 42CuS + H SO24The black precipitate dissolves in conc. HNO33CuS + 8HNO 33Cu(NO ) + 2NO + 3S + 4H O3 22(g)Action of potassium sulphocyanide : Cupric sulphocyanide is formed.CuSO + 2KCNS 4Cu(CNS) + K SO224If SO is passed through the solution, a white precipitate of cuprous sulphocyanide is formed.22CuSO + 2KCNS + SO + 2H O 422Cu (CNS) + K SO + 2H SO222424[This is the general method for obtaining curprous compounds.](h)Action of sodium thiosulphate etc.CuSO + Na S O 42 23CuS O + Na SO23242CuS O + Na S O 232 23Cu S O + Na S O2 232 463Cu S O + 2Na S O 2 232 23Na [Cu (S O ) ]4623 5 Sodium cuprothiosulphateUs es(a)Copper sulphate is used for the preparation of other copper compounds.(b)It is used in agriculture as a fungicide and germicide.(c)It is extensively used in electric batteries.