‘Objective questions - 3 Fill in the blank ~ 1 Long answer type Questions ~1 C Quick Review) When two simple salts containing same — : © Weoichiomethic ratio, we get double sate Some important double salts are— MMohr's salt ~ FeSO, (NH,),SO,,.6H,0 Porash alum ~ K,SO,, AlxSO,4,, 24H,0 Camalite— KCL MgCi,6H,0 2 Ferric alum ~ (NH,),S0,, Fe,(SO,),.24H,0 Co-ordination compounds are formed from two appar- ently saturated molecules capable of independent existance, Co-ordination compounds in aqueous solution, does not iveall simple ions but also gives complex ion/ions. © Theproperties of co-ordination compounds are completely different from their own ions. +The co-ordination compounds are of three types. 1. Simple cation and complex anion. KyIFe(CN)g] =24K* +[Fe(CN) 6] simple cation complexanion 2. Simple anion and complex cation [Ci0NH)elp S003 —cr(Niy)¢f* +3807 ‘complex calton simple anion Both are complex ions [CrINH)g] [CO(CN)g] =S{Cr(NH)g}* +1C0(CN6}° ‘complex cation complex anion © Theatoms or group of atoms having a tendency to donate electron pair are known as ligand. % The atom of ligand which donates electron pair is known as donar atom. © On the basis of number of donor atoms, ligands are clas- sified into six. (8) Monodentate ligands-Having one donor atom. 3 2 3 8 & NO,” F, Cl, SO,>-, H,0. NH, Pyridine etc © HG is named as agua, NH, > is named as Ammine (b) Bidentate-Having two donor atoms CUlgNlly — Fuiptene COP” oat; CHyNH (en) coo~ (OX) CHyNHy | Glyeinato(Gly) coo aso) Dipyridine (dipy) [This long type question is asked with or 3 x a= 1% Marks ‘A Marks 3 Marks Total Marks =5 (©) Tridentate-Having three donor atoms © Dicthylene triamine (dien) © Terpyridine (Terpy) (d) Tetradentate-Having four donor atoms. © Triethylene tetramine (Trien) (c) Pentadentate-Having five donor atoms CH,COO® CHW. | H,Coo® CH,NH —CH,COO® () Hexadentate- Having six donor atoms CH, i Cit,COO™ | ‘CH,COO™ City - wt ‘CHCOO™ tepray- CH,COO™ tepray~ + In some unidentate ligands, there are two types of donor atoms which can contribute with the central atom through. ‘one site at a time, such ligands are called ambidenate ligands. ( NO," can co-ordinate with the central metal atom/ion through N or O as— MeO-N=0 cyanide or ~ N =C isocyanide © When a di or polydentate ligand uses its two or more donor atoms to bind the same central atonvion, itis called Chelation. Due to this formation the complex has a ring structure and the ligand is co-ordinating through two or ‘more donor atoms. © Such types of compounds (Chelating compounds) are more stable. © Co-ordination Number-The total number of ligands [or donor atoms} attached to a central atonv/ion is called the co-ordination number of that metal ion. @ scanned with OKEN Scanner$= It is not a truc isomerism because isomers “ e jon number * ont cmanie ss [Co(NH,)6) Cls co-ordination te rated by following © Effective atomic number-It 1s y mul. Eteane Atomic ~ Oxidation atomic mumber of state hd iber metal ‘metal sine offective atomic number of Co in [CO(CN), 27-3+216] 27-341 Homolytic and Heterolytic complexes~ + The complexes in which only one kind of groups is at- tached to central metal/ion are called as homolytic com- plexes : " {Ni(H,O),P*: [Co(CN),; [Fe(NH)¢)° + The complexes in which more than one kind of donor ‘groups attached to central atom ion are called hetetrotytic complexes. [NiC1(H0),1: + Nomenclature— = K [Co(CN),NO} Potassium pentacyanonitrosy/ cobaltate ai) [Co Cl; (NH),}5 [CCNY] ‘Tetramminedichloridocobalt III. hexacyano chromate IIT = [Cofen)s]Cl, ‘Tris (ethylenediamine) cobalt (IMT) Chloride + Isomerism- (a) Lonisation Isomerism-Those compounds which give dif- ferent ions in their aqueous solution, ICo(NH,),Cl] SO, & [Co(NH,);SO,]CI (b) Co-ordination Isomerism-Compounds containing both anionic and cationic entities and the isomers differ in the distribution of ligands in the co-ordination sphere. [CotNHs)g] [Cr(CN)g] & [Cr(NH.)] [Co(CN),] (©) Linkage Isomerism-Compounds differ in the mode of attachement of aligand to the metal atomvion, [CoNO-(NH,)s] Cl, & [Co(ONO) (NH,),]Cl, (d) Hydrate Isomerism-Compounds differ by whether or not a solvent molecule i.e, HO is directly attached to the central metal atomv/ion or present as free solvent mol- ceules in the crystal laltice, [Cr(#4,0),] Cl; & [Cr(H,0)Cl] C1H,0 (e) Co-ordination position isomerism=This type of isom- crism is exibited by bridged complexes, differ in different placement of ligands. + 2 [coordination] Number pe 2 [CoCl,(NH),]° Alli 10NH3)g Oy CNHs )2Clz|S0q or OH, ICKNH), COX SOON )3CISOy (9) Polymerisation Isomerism same molecular formula. ners do not hay [PUCL(NHL)3] & [Pt (NH)4][PICI,) Geometrical Isomerism- : © When two same ligands occupy postions either aj tocone another or ic. (90°), known as cis-isomer, a ligands ocecupy opposite position ic. 180", knoya”® trans-isomers. Compounds of co-ordination number—4 © ‘There are two types of geometry in case of 4 co-onding, tion number, tetrahedral & square planar + Geometrical isomerism is not possible in tetrahedral metry, because all the four positions are adjacent oe another or have same bond angle ic.—109°28" © ‘Square planar complexes show geometrical isomerism The positon 1,2; 2, 3; 3, 4; & 1,4. are adjacent to cay other (at 90°) known as cis isomers. angle known as trans-isomer, + Square planar complexes of type MAX; MABX, & M(AB)X, can exist. + [Pt(gly),] shows geometrical isomerism. ps NH Gh Nou I ANG re CL ¢ Wier 07 N, oO oO © The shape of complexes having co-ordination number 6 is octahedral geometry. ‘> In Octahedral complexes, the Positions 1, E 2,3; 3,45 4,5; 4,6 are adjacent & at 90° cis isomer & the Position 1,3; 2, 4 and 5,6 are opposite positions i.e. 180° angle- trans isomers, 13:16: * The octahedral complexes of the ty pe MAX, MAX MA,X,j, MA3B3. M(AA)X5, M(AA),X, ‘Show Pn ‘metrical isomerism, 7 Optical isomerism- © Those complex compounds which rotate plane polarised light cither to left or right are known as cae Somers > These isomers are mirror images to each other. A @ scanned with OKEN Scannervalence Bond Theory 2 This theory was given by Linus Pauling, > First the central metal atom climinates the cloctrons fe ing a cation on the basis of its oxidation state Je The number of empty orbitals is equal to en -ondinal number of the central metal ion which take pata Iubridisation to form hybrid orbitals * bar tn > Fach ligand has atleast one orbi taining a lone pair of electron, The empty hybrid orbital of with the filled orbitals of the fi co-ordinate covalent bonds, |, If ligand is stronger then it has a tendency to pair the unpaired electrons present in central metal Ton terete inthis case inner d-orbitals may take part in hybridisation us called low spin complex, + If ligand is weaker than it has no tendeney to pair the unpaired electrons. Therefore in this case outer-d-orbit. als will take part in hybridisation and it is called high spin, complex. 2 Strong ligands~ NCS [Hel,|” > sp” hybridisation Trigonal planar shape. = [NWCO),], [ZnCl >, PNiCl,]>-sp* Tetrahedral, 109°28" Dia-magnetic © [NUCN),F>. [Cu(NH,)4)*, dsp? square planar, 90° dia- magnetic. * {CoF}* - sp?4? hybridisation, octahedral, 90° paramag- tal of donor atom con- ntral metal ion overlaps igand to form metal ligand netic, + {Co(H,0),|>* - sp3d? hybridisation, octahedral, 90° para- magnetic (Fe(CN), > 4?sp? hybridisation, octahedral, 90° para~ magnetic Crystal field Theory— =" According to this theory, two forces acting between metal ion & ligand known as attraction & repulsion forces. © The attraction forces carried between nucleus of metal cation and electrons of ligand. While the repulsion forces carried between electrons of metal ion & electron of ligands. : The orbitals lying in the direction of ligands will experi- ‘nce greater repulsion & their energies will be raised as ‘compared to their positions in a symmetrical field If the orbitals are lying away from the approach of the ligands will have lesser repulsion therefore their energies Will b lower than they would be in a spherical field. This conversion of five degenerate d-orbitals of the metal ion into different sets of orbitals of- different energies in the presence of electrical field of ligands is known as crys ‘al field splitting, 4, tay Gt spitting oh @orieae “Sot ows ow ‘The crystal field spitting, val be diferent sx different structurs with different co-ordination seatoers © In case of octahedral complexes, the fr split up into two sets, one oct consestang of two 1 dx? ~ y2 & dz? of higher energy called ¢, oronals am another set consisting of throe orbitals 4,4, 204 bf lower energy called tp, orbitals + Incase of tetrahedral comple y? & dz ic. c, orbital become more nergy and the three orbitals i 1, arbmals boone le stable and of higher encgy © dl, d? & d electronic configurations by. By thy > d*Seclectronic configuration, there are tac conditions— (For weak ligands~ 2 eg? 13637. yee! Byes i) For strong ligands— thee. thee hy ee" ther! 1” electronic configuration will be same m both the cases. © shyea*tigee* Stability of complexes— + Ifthe charge density on metal ion increases then stability of complex increases. * ‘The charge density on Fe~ is more than Fe>~ Therefore we can say that [Fe(CN),}> is more stable thar [Fe(CN),1* > In case of Cu2*, NP, Co? & Fe ions. The charge ty on Cu?” is maximum therefore the complex of is more stable, © If the tendeney of donating electron pair of ligand im creases, the stability will increase, CN™ ioaiis more basic than NH. Therefore [Cu(CN),P> is more stable than ICn(NH,),P ion. + -Ifmetal ion forms chelate with ligand than it will become more stable [Ni(cn);)°* is more stable than (Ni(NH, ion @ scanned with OKEN ScannerQ.1.(a) Give the IUPAC names of following compounds (@_[Cofen)s]>* 7 Gi) Ky [Zn(0H),|——_(CRAFABSATERTSApHNZOT) (b) Give one example on each (i) Homolytic complex a Ambidentate ligand [[RAje/BOHFA SiIppA2023] (iii) Chelate ligand iv) Which complex is used in eanc ae ; aa Bowed supp 2012/2019) Ans.(i) [Co(en)3]>* — Tris-(ethylenediamine) cobalt IIT ion a) K [Zn(OH),] — Potassium tetrahydroxo zincate a (b) (i) [Co(NH3)g}3*, [Fe(CN)g}= (i)-Cen-8 Ct ~NH, (iii) Ethylene diamine CH, - Sn, (iv) Cis Diamminedichlorido platinium (II) is used. Q.2.(a) [NiCly]?- ion is paramagnetic while INi(CN)4]- ion is diamagnetic. Explain it with the help of VBT. (b) Write IUPAC names of the follwoing compounds, (@ [CoCl{en),}C1 ans Gii) Ks[Fe(CN)g) ‘(Ra Board 2012)! Ans.(a) In both the complex ions, Ni has +2 oxidation state, Ni2* [Ar] 3d8490 4a 4s ate [t [4 3a’ 4s) 4p aTa[utiyi due to the presence of strong CN= ligand. The unpaired d electrons are paired, 3d 4s) 4p veya la dsp" In [Ni(CN)4]?~ ion there is no unpaired electron therefore [Ni(CN)4]?- ion is diamagn« (b) (i) [CoCla(en),}Cl Dichloridobis (ethylene diamine) cobalt (III) chloride (ii) K3[Fe(CN)g] Potassium hexacyanoferrate (III) Q.3.(a) Write down the structure of cis [CoClp(en)2]* and [Co(NH3)3 (NO2)3] (©) Give one example of ambidentateligang ,, explain why it is ambidentate ligand. (©) Give oxidation number and c number of central metal ion in K3[Fe(Cz04)3} com compound. Ans,(a) cis-[CoCl2(en)2]* f cl [bo s+ en cis-[Co(NH3)3(NO>)3] ' on, NE yn, 0. 9 No, NH (b) Those ligands which have two donor atoms but at time only one donor atom takes part in complesation are cle ambidentate ligands, G=N cyanide ligand contain C donor atom while in isocyanide ligand contains N-donor atom. S CN & NC is an ambidentate ligand, (©) Ks[Fe(C20,)3] 3¢+1) +1) +3 C2) 3+x-6 case of N= 3 ‘Thus we can say that the oxidati rh neon oa SY i oxidation state of Fe in 5 ae co-ordination number of Fe in complex compourtd is3x2= Q.4.(a) Write down th {) KateyOr ICo(NH3)4(H20)C)] Cl, Gil) [Ag(NH53)3] [Ag(CN).] 7 i (b) In purfication of metals, ordination compounds, give one example of co (©) Give differences in double salt example. and ns Ans (a) (i) Ka[Zn(OH),) Potassium tetrahydroxo zincate (I Gil) [Co(NH3)4(20)Cl] Cl, 2 tetrammine aquachlorido cobal i Gi) (ABONH)a] [Ag(CN),f (UD chloride diammine silver (I) dicyanoargentae (1) the TUPAC name of following- @ scanned with OKEN Scanner(b) Purficati done by sycbuconnestsomed Whichon fain ge Ni+4CO-> NI(CO), ~» Ni¥4CO i tp vowt t's ny ed by two saturated compounds Is which are ‘ A ae ound ut they lov thrid same Mohr's salt EeSOy.(NHy)> SO4 , 6Hy Potash al K 804, We soe wo Co-ordination compounds are a spe ee ee os or molecules beyond their normal valency. Examples (i) K3 [Fe(CN)] a Gii) [Co (NH3)g] [CrClg] Q.5a) Write the formula of ionisation isomer of [Cr (H:0),Br21Cl (b) Write the formula of co-ordination compound of percury etrathiocyanato cobaltate (111). (o) Explain on the basis of valence bond theory that |Ni(CN),2- is low spin complex ion, (RjeBGaia 2015) "Ans. (a) The ionisation isomer is [Cx(H;0)4BrCI[ Br (by Hg [Co(CNS)4] (The oxidation state of Ni in [Ni(CN)4J?-ion is +2. In {P+ ion, the co-ordination number is 4. Electronic configuration of Ni is 3d" As 4p Tees (1) [ev ] Electronic configuration of Ni?" ions is 3d" As 4p. tte titt] CI According fo valence bond theory, cf strong liganed (CN-), the unpaired electrons J trials are paired resulting in one empty d-orbital. 3d° As 4p toa dae to the presence ns present in d= i cp Therefore a low spin complex is formed and dsp? ‘nbridisation takes place. Inthis way four tmpty dsp? hybrid orbitals of Ni form fur co-ordinate bonds with CN ion. dsp" dsp’ dsp" dsp” toa: 4 4 -CN _CN -CN-CN tendency sig Explain colour and complex form “to The ural in central. Metal ion s. The unpaired electron present 18 a rt ste Agency from oe and excited £0 higher Kay orbs veo d-orbital, therefore duc 0 ded cle ‘transition, they show colour. © complex formation of a metal following factors i) The size of cation should be smallest, “The charge on cation should be maximum. ‘The cation should have vaccant d-orbitals. “lnssify the following into homoleptic and heteroleptic complexes. (i) KglFe(CN)g1 ) [Co(NH3)<(CO3)|C1 Gil) KzIZn( 1 [PU(NH)2C(NO2)] (Raj Board 2006) Jomoleptic complex-(i) KalFe( (ii) Kp{Zn(OH)4), Hederoleptic-(i) [Co(NHs)s(CO)sCliv) [PUONH)2 C1(NO2)| Q.8. Explain the violet colour of [Ti(H,0)6]>* com- plex on basis of the erystal field theory? ‘Ans, The complex cation [Ti(H0),)>* absorbs the Ao ‘500A. From visible region orbitals to higher eg orbitals. energy which is equivalent of and electron is excited from tag ‘Thus due to d-d electronic transition. Itshows Violet-colour— “8 t { - Ao Y te { ] & Q.9.(a) Write chemical formula and TUPAC name of cis-platin used effectively to inhibit the growth of tumours. ‘(b) Write two limitations of crystal field theory. OR (a) Write one example of bidentate ligand. (b) Write chemical formula and IUPAC name of wilkinson catalyst used for the hydrogenation of alkenes. imitations of valence bond theo Ans. (a)Cis-Platin ch NH, Pr oo < i, ] Cis-diamminedichloridoplatinum (11) (b)(i) CFT does take care of covalent character between metal and ligands. It only took the ionic character into con- sideration, (ii) This thery gives no significance to the orbits of the ligands. So it cannot explain any properties related to ligand orbitals and their interaction with metal orbitals. OR CH,NH, 1 © Cun, (b) Wilkinson Catalyst PhP: PPh, Rh ><, PhyP- Ethylenediamine (en) @ scanned with OKEN ScannerChloridotris (tripheny! phosphane) rhodium (I) Tonisation Isomerism ‘ty (©) Gi) This could not explain inner auter orbital entities or complexes (ii) This could not explain the colour & spectrum of omrgin, i) Define primary and secondary valency of alli ion with the help of werner's principle. eG) Write down the primary and secondary valency F Co in [Co(NH3)6 C13. ; - (i) Wrke the structural formula of Ni (CO), and CHCO),. be (iv) Define Co-reactivity and valence bond principle in carbonyl complex. |(Raj. Board 2017) Or Draw the synergic bonding present in corbonyl comple. (Raj: Board 2023] Ans (i) According to wemer’s theory, prim. valency is ionic and itis balanced by ions. while see, valency is unionic and balanced by neutral molecules or anions and itis equivalent to coordination number. {In sec. valency metal ions and molecules or anions are placed in square bracket while prim valency, anion are placed ut side the square bracket (ii) In [CofNH3)c] Cl, prim, valeney valency is 6. (Gi Structural formula of Ni(CO); and Cr(CO), Ni co7 | Sco co Tetrahedral structure is 3 white sec Octahedral structure 2B Dp ” SSE S—G Q.11.(i) Define functional and Ionisation Isomerism, ii) Which type of structural iso merism is shown by ICo(NHs)] [Cr(CN)g] and [Cr(NH)g] [Co(CN)¢l ii) Write the figure showing electronic transition in ITi(H,0)¢]3* ion. (iv) What effect takes place when it ITi(H,0)¢]3* ion. - Ans.(i) Functional Isomerism-The ‘dinate ‘compounds having same molecular formula but differ in the mode of attachment of ligand to the metal atomvion are called funetional isomers and the phenomenon is known as functional isomerism z+ Ex- [Co(NH3)s (NOz)] Cl showing two functional isomers. A Ve Co(NH,); [N z (NH3)s (NZ and [Co(NH,),(0-N =0)]Cl, excited in ‘The co-ordinate compounds having, same Molecy formula but giving different ions, known as ionisation ison, and the phenomenon in known as ionisation isomerism, Ex, = {CoCl(NH3)4] NO2 and {CoCI(NH3)4NO2|CI Ae ionisation isorsers. Gi) [Co(NHy)g] {Cr(CN)g] and [Cr(NH3)6] and [Co(CN)g] are co-ordination isomets OO, OO. togleg? ta,°. eg! (iv) When ¢ is excited from tog! ¢2° to tag, eg! i, [Ti(H,0), 5 the colour appears purple. Q.12. Write IUPAC name of the following comple, compound K3[Fe(C204)3] Ans. Ka[Fe(C04)3] potassium trioxalatoferrate (III) Q.13. On the basis of valence bond theory explain thy ation state, hybridisation, geometry and magnetic nature of metal in complex [CoF¢|>-. Ans. [CoF¢]3- ‘The oxidation state of Co is x. X-6=-3 4s 4p 4a Co Se pe F-ion isa weak ligand therefore, the d electrons preset in Co” ion is not paired thus in hybridisation the outer orbitals take part. pe A411) EEE Hybridisation sates spa ~ Presence of 4 unpaired elect a Paramagnetic and coloured jon, "8 (COFe)™ ion ~ Shape is octahedral, cs FOF og Nol Q.14. Write IUPAC name of i coca gaal: of the following complex KuIFe(CN)c] (Raf BoRealSupp20I8) Ans. K,[Fe(CN), ]= Potassium hexacynoferrate (I!) QS. (a) Why Tetrahedral complexes are always high spin complex? Explain on the basis of erystal reld ther — @ scanned with OKEN Scanner(A) According CFT, Ligands ogbitals than 6g 50 energy Of ty increases Mabe ctrahedral complex ‘ Intetr P ways A, (B) Kgl Fe(CN)g] Ans. (A) [Co(en)3}'3 Suppose Ovidation no. of Co = ve MOE +3 vet3 Here en (ethylene diamine) is neutral ligand and todentate ligand, Coordination no, 2» 3 = 6 {BIKy [Fe(CN)o] KyIFe(CN)g] > [Fe(CN} ‘dation no. of Fe = x and ind =— 1 444K dation no. (Charge) X=-446=42 Cynide is monodentate ligand so coordination no. CN=6*1=6 Q.17.Homolytic and heteroly! Ans, Homolytic and heterolytic coordination compounds When the metal atom / ion is bonded by only one kind of ligands ae called as homolytic complexes For Example : [Ni (NH3)5]?* [Co(H,0)¢)>* [Co(en)3* ‘when 2 metal atom / ion is bonded by more than one lind of ligand known as heterolytic complexes. For Ex. [NiClo(H30)4], [CrCly(NHs)4]* Q18. Define ambidenate and chelating ligands and sive one example of each, Ans. Ambidentate ligand—The ligand in which more then caedonor atoms are present but at a time only one atom donate lene pair to metal ion to form co-ordinate bond, is called ‘bidantate ligand e.g M-~CN-= Cyno ligand (Donor atom is carbon) M.«NC- = Iso cyno ligand (Donor atom is Nitrogen) Chlating ligand-Bidantate and polydentate ligands are ‘town as chelating ligans as they form ring by making co- ‘dinate bond with metal ion. This ring is called chelate and ‘sch ype of ligands are called chelating ligand. eg Oxalato (0X2) ligand cook >M boo” (C,0,)2 or OX? 19. Write oxidation state of Ni in [Ni(CO)4} Bitrm ‘Ans. Onidation state of Ni 0, x + 4[O] © 0,%= 0 Q.20. Write the geometry and magnetic nature of the complex ion [NiCI4[2- on the basis of valence bond theory. ‘Ans, ‘The oxidation state of Niin [NiClg} 2 15 42; bx 4=-2 aut 493 dp N APTA a" 4s 4 Ni? aya TATA I Chlorido (CI-) ligand is a weak ligand so one 4s orbital and three 4p orbitals are used for sp? hybridization, Hence geometry will be tetrahedral a | Ni? ay mK a ‘As we observe, two unpaired electrones are present in 3d orbitals so it will be paramagnetic and the magnetic ‘momentum will be a n= Jala 2) Here n= no. of = J22+2) unpaird n= J§ =28BM Q.21. The Co-ordination compounds having tetra hedral geometry is [Raje Board 2022] (a) IN(CN)1? (b) [Ni(CO)4] (c) [Fe(CO)s] (d) [Cr(CO)g] Ans, (b) [Ni(CO)4] Q. 22. The name of metal present in haemoglobin ion polyhadron. [Raj. Board 2022] ‘Ans, The spatial arrangement of ligands that are directly attached to the central atom or ion, Q.24, Write the IUPAC name of K3[A(C30,)31- + [Raj: Board 2022) ‘Ans, Potassium trioxalatoaluminate (IID. 4 Q.25. Write the type of isomerism exhibited by [Co(NH)e] [Cr(CN)oI and [(Cr(NH3)g] [Co(CN)¢]. Q.23. Define co-ordi Ans. Co-ordination isomerism, @ scanned with OKEN Scanner0.26. Bidenate ligand is- @) 0,7 ont, 0,2 in Dave the gemoetry of geomet 0), it “ir configuration. poy weet conpraon Ans (Pt(3,0),Br,] square planer comples with two gemetrical isomers- (i) cis isomer (i) transisomer H,0 Br 1,0 Bu Sac Br~ H,0 HO“ Br Q.28. Write the IUPAC name of following com- pounds— (Raj. Board 2023) (@ [CuH,0),)"? (ii [Co(NH,),IBr, Gi etc) 1 Ans. (i) Tetra aqua copper (II) ion (ii) Hexa ammine cobalt (II1) bromide. (ii) Hexa eyno ferrate (ID ion. Q.29. The chemical formula of brown ring formed in the brown ring test for nitrates is ((RajeBoard Supp.2023) (a) [Fe(H1,0),NO}? __(b) [Fe(H,0),NOy'> (b) [Fe(H,0) (NO),]*? _(d) [Fe(H,0) (NO),I"? Ans. (a) [Fe(H,0), NO}? Q.30. Draw the figure of crystall field spl octahedral coordination entity. ((Raj2/Board'Supp!2023) ‘Ans. Inan octahedral coordination entity the central metal atom or ion is surrounded by six ligends, In the presence of ligads all five d-orbitals of metal ion are devided in two parts. Three lower energy d-orbitals called, (dd, and d..) and two higher energy d-orbitals called eg (dx! yand dz’). It is called crystal field splitting. The energy of eg orbitals is raised 3 by 540 and energy oft, orbitals is lowered by 24, Average energy of ™ d-orbitals ~* Q.31. @ Define coordination sphere. Write the difference between d- and I- isomers of [PtCI,(en),]** on the basis of structure and optical behaviour. Gii) Draw the figure of crystal field splitting in a tet- rahedral coordination entity. ((Raj/Board)Suppy2023) per of molecules or ions any cootinen icket. It is called coordi ne bracket is called ionic sp, Ans. (i) A definite num! with central metal atom or io! which is written into big bra sphere, The species out side tl [Co(NH,),] Cl; corrdination Ionic sphere Gi) sphere Mirror d-form The both forms of [PtCl,(en),}"? are cnantiomers a cannot be overlapped on each other. d-form rotates the play polarised light in right side (clock wise) and I-form rotatssi in left side (anti clockwise). Only cis-form of [PtCL(en),}: show the optical isomerism. (iii) In the tetrahedral coordination entity the splitting of d-orbitals occurs opposite to octahedral entity. Here splitting energy Ay = Say is not so more that the pairing of electroos is not occur so generally low orbital complex or spin pire complex are not formed ‘Average energy of d-orbitals Q. 32. 7 . 32. Number of moles of precipi : Numbe Precipitated AgCl on adé- ing excess silver nitrate solution in the solution of one nk of CoCl, SNH, is- (1 3 os Ans. (b) 2 oss Ambidentate ligand ig oo (b) H,0 c) NH, @NO- Ans. (d) NO, of oh ee ", Seometry and the magnetic natut® UeP* on the basis of valence bot! @ scanned with OKEN Scanner{Raj Bonra 2025) 0" ‘Ans. Co is preset in COON), At] 3d? dg? ap A As ay? M$ a tp [Nic © INWCN), P The coonliation mumber of (EDTA) 4 is @ (6 wa 1 The number of geometrical isomers of [P{(NH,)xCls] (b) 2 4 ligands and with AgNO gives two moles ppt of AgCl. ia will be {a) [COUNH,)sNO3ICL (0) [Co(NH,),CIJNO;CI (c) Co(NH)sCHNO,—(@) None of these Which of the following compound gives optical Sm (a) [Co(CN),F* (b) [ZnCl {6) [Cofen),Cly} (@) [Cu(NF),)?* ‘The hybridisation present in [Ni(CO)4] is~ (sp (b) SP? (o) dSP? (@) sP> Which metal present in chlorophyll (Cobalt (b) Magnetism (6) ron (d) Nickle Double salt is— (@) Mohr salt (b) Prusian blue {c) Neslar reagent (d) Frankland reagent Inthe following bidentate neutral ligand is @) aly (bX (en (4) EDTA KuIFe(CN),] the oxidation number of central metal Fe is fa) 43 (b) +2 (hg (+l The colour of a complex compound is explained by- @vBT (b) d-d- transition {©) Pairing enemy (qd) None of these IFe(H130)sNO}'2 the coordination number of Fe in Complex is eons NH, is a strong figand so pairing of d-clectrons takes place. sp we 1] { pty orbitals hybridised so hybridisation wall be dsp", geometry is octahedral and due to all es paired it will be di aaa Important Questions M4, 15, 16. 17. 18, 19. 20. 21. 22. 2. 24, 25. lestions and|/Answers: (ays (b) 4 ()6 (a) 8 The oxidation number of Nickle in Ni(CO)s is- (a)4 (b)2 (c)0 (d) None of these In KalFe(CNYo) the hybridisation of Fe is~ (a) spi (b) dsp? (©) sp¥d (d) sp? By which combination a complex compound is formed~ (a) Metal and Non metal (b) Two non metals (©) Metal and ligand (d) metal and metaloid Haemoglobin is a complex compound of- (a) Co (b) Mg (Fe (a) Cu ‘The species written in big bracket is collectively called— (@) Ligand (b) Coordination entity (©) Anti ion (@) Central atom From the following which is a chelate ligand (a) Oxlate (b) Cyanide (©) Nitrate (@ Ammonia From the following which have tetrahedral geometry (a) [Ni(CN) 4]? (b) [NiClg}-? (©) [PtCly-? (@ [Pa(CN),} 2 ‘On mixing excess AgNO3 solution in PdCIy.4NHy solution it's one mole precipitates 2 mole AgCl. The secondary valency of Pd in PdCl; 4NHs wall bo- (a)2 (b)4 6 @o Geometrical isomerism is found in~ (a) Homoleptic complex (b) heteroleptic complex (©)Aand B both (@) None of these Gemetrical isomerism is not shown by~ (a) Tetrahedral complex. (b) Square planer complex (©) Octahedral complex (4) Alll of these From the following which is not a bidentate ligand~ (a) oxlato (b) Ethylene diamine (©) Nitro (-NO>) (@) All of the above In dsp? hybridisation the geometry of complex will bo- (a) Tetrahedral (b) square planer (c) Octahedral (d) None of these 1.) 2.00) 3.0) 4.(b) 5.(a) 6. (0) 7.(0) 8. (b) @ scanned with OKEN Scanner7 11. (b) 12.(b) 8. tetrahedral, square planer “ 16 17. 18, 19, 20. 9 (a) 10. (e) B@ Ho 15. (0) 16. (©) 17.) 18.0) 19. (a) 20.(b) 2) 22.0) 23. (a) mo 25.(b) Vitamin B,, 1s a coontination compound made by metal first of all told about the structure of coordination compounds Jn coordination compounds the of nictal is ally tonic and satisfied by negative ions, 1sthe total namber of donar atoms of ligands attached with central metal atom oF ion, The ions or molecules attached with central metal atom ‘oF jon are called Such compley in which central metal atom is attached with only one type of ligands is called complex. ‘Such compounds which have same chemical formula but different arrangement of atoms are called There are two orientations possible for coordination 10. 4, and Inthe presence of ligands, the splitting of equal energy orbitals into two groups is called Exhylene diamine (en) and oxalato (0x) ate. ligands ‘The species written in big brackets called outside the bracket is . Such ligands which use two or more donar atom with ‘one metal ion are called ligands, ‘Such complex which have two or more types of ligands are called... complex. In the name of ligands the last latter is ~O. isomerism is of two types geometrical and optical isomerism, sss: Somers are mirror images of each other. Jon or molecule which can not super imposed on each other are called. ion or molecule. bonding isomerism is found in coordination compound and which have ligands. hybridisation of gentral atom is present in tetrahedral geometry. is used in the treatment of cancer. Answers 1, Cobalt 2. Alfred wemer 3. Primary valency 4, Coordination number 5, Ligand 6, homoleptic 7. isomers 9. crystal field splitting 10. bidentate 11, Complex, antiion 12, Chelate 13, hetroleptic 14. Anionic creo 16, Optical 18, Ambidentate 7 20. Cis platin Q. in I an, iation compounds. tetramminenquachloridocobalt (111) chloride i incate 11 (i) {Co (NH), HpOCI] Cly (ii) Ky [Zn (OH)41 (iii) Ky [Al (C204)3) (iv) [CoC (eng]* (W) [Ni(CO)s] Q.2. Give IUPAC names of following : [Pt Cl (NO2) (NH3)2] Kg [Cr(C204)3] ii) [CoCl, (en)y] Cl (iv)[Co(CO3)(NH5)s] Cl (v) Hg{Co (SCN)4] ‘Ans, (i)diamminechloridonitrito-O-platinum (II) (ii) potassium trioxalatochromate (III) (iii) dichloridobis-(cthane-1,2-diamine) chromium (il) chloride (iv) pentaamminecarbonatocobalt(IIl) chloride (v)_ mercurytetrathiocyanatocobaltate [II Q.3. Write down the structures of following Co-or dination compounds 1. tetramminediaquacobali (III) chloride Ans. {Co (NH); (H30).)>* Cl [Co (NH3); (H20)>] Cl 2. potassium tetracyanonickelate (II) K* INi(CN),P Ka [Ni(CN),] 3. tris-(ethane-1,2 chloride {Cr(en)3* CL «geen amminebromidochlorido nitrito-N platis ion {Pt Br (Cl (NO) (NH) ca aa 5. dichloridobis (ethane-1, (LY) nitrate [Pt Cy (ens? NOs [PtCh (en)a] (NO3)s 6. iron (11) hexacyanoferrate 11 mine) chromium (III) diamine) platinum @ scanned with OKEN ScannerQ.4. Write the TUPAC names of 1. [Co(NH3)6ICly hexamminecobalt (I11) chloride 2. (CoCHNHsIC pentamminechloridocobalt (III) chloride name (11) chlor sium hesacyanoferrate (Il) 4. KylFe(C0)31 potassium triosalatoferrate (III) KjIPd Cul potassium tetrachloridopalladimate (II) 1. [PCCL(NH3)2 (NH{CH3)|CI, iammineehloridomethylamineplatinum (11) chloride Q.5. When two different unidentate ligands are at- yched co central metal ion in tetrahedral complex, it does tox show geometrical isomers, explain, ‘Ans, In case of tetrahedral compl following structures I the four posi- ons are adjacent to each other. Therefore they do not show gometrcal isomerism Q.6. Draw the geometrical isomers of the H)(CN)4- complex ion, cis-isomer trans-isomer Q.7. Which of the following two co-ordination entities ischiral (optically active)? (0) Cis [Cr Cl(OX) 23 () trans [CrCly (OX) 2° ; Ans. ()Cis [Cr Cl(OX)3)> shows optical a a 3 a 2 ol a | oS De Os UV Le & Cis (Cis) {) trans {CrCly (OX)3|>- does not show. Q8. FeSO, solution mixed with (NH4): molar ratio gives the test of Fe2* io ‘tition mixed with aqueous ammonia i “ss nat give the test of Cu2* ion. Explain why? say FeSO, solution on mixing with (NHa04 nto” in 1:1 molar ratio forms double salt FeSO. )2804.6H,0. $04 solution but CuSO. 4 molar ratio “The double salt dissociates in axpuecns soi aad wes tosts for all the ions FeSO4.(NH));504 games Pe?" 4 INH! + 380y? © When CuSO, solution iv mised with aqueous amonenis in 1-4 molar ratin, it forms a coordination compound {Cutty 4180, © ‘Tho complex entity (Cu(NH )4P?* does nut dissociate in aqueous solution and hence does not wive lets for 02" jon Q.9. Specify the oxidatio the following coordination (a) [CoO (CN) (eng? (by [UCP (c) (CAN) Ch (a) [CoBrsfen)p}" (ce) KafPe(CN),) Ans. (a) [Co(H,0) (CN) (en)? XH OF ED FM) = 12 pers of he metals in sities cea t3 tb) [PCP x rac lye 2 x=t2 ©) (CAN) CII x30) 13-1) =0 cKeHS (d) © Be (IR OEN AO OK ea8 Q.10, Aqueous copper sulphate solu colour) gives: (a) @ green precipitate with aqueous potassium fluoride and (b) a bright green solution with aqueous potassium chloride, Explain these experimental results ‘Ans, Aqueous copper sulphate solution exists as {Cu(F,0),]$O,4 which has blue colour due to {CulH0)4} ions, (a) When KF is added, the weak water ligands are roplaced by F ligands and forms [Cuf?4[2- ions which 1s a green precipitate, [Cu(Hy0)4]$Oq + 4F- > [CuPyP (green ppt ALLO (b) When KCI is added, the weak water ligands a replaced by Cl- ions forming [CuCl > which has bright yi colour. (blue in [CuH,0),BO, ACL — {Cue VAH,O. Q.11, What in spectrochemical series ? Kaptain the difference between a wenk field ligand and # strong field @ scanned with OKEN Scanneroe “Tho arrangement of ligands in the increasing order of'erystal field splitting is called spectrochemical series, This is shown below P P Q.12, What is crystal field splitting energy ? How oes the magnitude of Ap decide the actual configuration, of d-orbitals in a coordination entity? Ans. See text 13. How many ions are produced from the com- plex Co(NH3}sCls in solution? (a) 6 (b) 4 (6) 3,@) 2 Ans, [Co(NH)|Cls [Co(NH,)6ICly ——» [Co(NH3))3* + 3C- “. 4 ions (b) is correct. __. Q.14. Amongst the following ions which one has the highest magnetic moment value? (a) [Cr(H20), 5 ‘or example, Ans, 343: 3 unpaired electrons \d® : 4 unpaired electrons d!0 : Diamagnetic Jaluc HF etH20)c}* ion show highest magnetic moment, value Q.15, What will be the correct order for the wave- Jengths of absorption in the visible region for the follow. ing: INi(NO2)6I+, [Ni(NH3)6]2*, [Ni(F20),]2+ Ans. The order of ligands in the spectrochemical series H0 [Ni(NH;)6]?* > [Ni(NO3)¢}*-. Q.16.(a) Write down the TUPAC name of the following complex. [Co(NHs)sCl]2* (b) Write the formula for the following complex ive, Potassium tetrachlorido nickelate I, ‘Ans.(a) Pentaammine chloridocobalt Ill ion (b) KofNi(Cl),} Q.17.(a) What type of isomerism is shown by the complex. [Cr(Hz0),IClz (b) On the basis of crystal field theory, write the electronic configuration for d8 jon i.f, Ap > P. (©) Write the hybridisation and shape of [CoF)~ [atomic number of Co = 27] is ) It shows hydration (b)_ If dp > P In such condition, first t3y- orbitals, paired. The electronic configuration of dis t,6 ex2. (©). The hybridsation state of [CoFg} is spi because here weak ligand (Fis present. Q.18.(a) Which of the following is a more Stab complex and why, [Co(NH3)«[3* and [Co(en)3]3* ‘Ans. Chelating ligands form more stable complexes compared to non chelating ligands ‘Thus ethylenediamine fen} isa bidentate ligand forms stable chelate i.e. why [Co(en)3]3* is more Stable thay {Co(NH3)6I3* (b) Write the IUPAC name of the complex (Cr(NH)Cly|*. What type of isomerism does 4 it? Ans. Tetrammine dichlorido chromium (III) ion. ‘This complex exhibits geometrical isomerism, al li ch] fis Hx nn, || 4 is trans, . Q.19. Write the IUPAC name of the following co- ordination compounds. (@) [Cr(NH3)3Cl3} (iii) [CoBr2(en)2}* (i) [Cr(NH3)3Cl3] ‘Triamminetrichlorido chromium (III) (i) Ks[Fe(CN)g] Pot. hexacyanoferrate IIT Gil) (CoBr2(en)3]* Dibromido bis (cthanediamine) cobalt III ion. Q.20. Give the form ordination entities, (i) Co ion is bidentate ethylenedi (ii) Ka[Fe(CN)g] Ans. wula of each of the following co- bound to one Cl- one NHg and two ane iamine (en) molecules. oxaigl) N%* fonis bound to two HO molecules, and two Write the name and magnetic behavi ich of the above co-ordination entities. At No of eo 27 Megs. nen eat ona lo of Co 27, Ni28. imine chlorido bis (ethylene di: i Here hybridisation is a2. nd an ™ (i) INiGH-O C2031 = Diaquadioxalato nickelate I ion Hybridisation is spd? paramagnetic, Q.21.Indicate common salt, double salt and coordination compounds from following. @ Nacl : @) KCL Gil) KCL MgCl, 6,0 - @ scanned with OKEN ScannerPg a oo 61)__ee (0), FeSO, (NH,),S04.6H,0 (si) Al(CO3)3 (vii) CSOs (vii)K2S04 Al; (S0,)3.24 11,0 (és) Ky [Fe(CN)g] fs) Kj [Fe(CN)g] (ai) K3 [Cu(CN) 4] Ans. (i) NaCl Common salt. (i) KCL Common salt. (ii) KC MgCly.6H,0 Double salt (iv) [Cu(NH3)4]SOq Coordination ‘compound (v) FeSO4(NH4)2 SOq . 60 Double Sat (vi) Alp (CO3)3, Common Salt (vii) CaSO4 Common Salt (Vii) K2S04.Al ($0,)3. 240 Double Salt (ix) Ky [Fe(CN)g] Coordination compound (x) K3 [Fe (CN)g] Coordination compound (xi) Ks [Cu(CN)4] Coordination compound Q.22.Write the formula of the following compounds: (Potash alum: (ii) Mobr's salt. (ii) Carnallite (jv) Potassium ferrocyanide (») Potassium ferricyanide. Ans. (i) K380q. Alp (SO,)3.24H20 (i) FeSO4(NH,)S0,.6H,0 Gii) KCI MgCly.6H20 iv) KylFe (CN)e} (vy) Ka[Fe (CN)g] Q.23.Which types of ions are produced per mole of the following double salts in their aqueous solutions = @ KCLMgCh, 6H,0 Gi) FeSO, (NH,)2S04.6H20 {ii) KS804.Alg[SO4]3.24H30 (iv) (NH,)2504. Fez[SOy]3-24H20 6) KySO4.Crp[SO4]3.24H20 Ans. ()) KCL MgCly 6H0 It prdouces K*, Mg?* and Cl- ions (il) FeSO,.(NH,)2804.6H20 It produces Fe2*, NHy* and SO,? ions. (ili) K2S04.Aly(SO,)3.24H20 It produces K+, AF* and S04?" ions (iv) (NH,)2S04.Fes(SO,)3 24H20 It produces NH4*, Fe3* and SO,?- fons: () Ky804. Cra($Oq)3.24H20 Itproduces K+, Cr3* and $04?" ions ions are produced per mole of solution @_ {Ag(NHy)IC} Gi) [Cu(H,0)4] SO, (Gi) Ky [PUR (iv) Kg [AI (C,04)3] (%) [Co (NH3)6] [Cr(CN)o] (vi) [Pt(NHG)4] [PtCla] i) (Ag(NH3)2/C1 It gives [Ag(NH3)2]* and Cl" ions Gi) [Cu (H20),}S0, It gives [Cu(H0)q]?* and SO,?> ions ii) KafPt Fe} Ttgives K* and [Pt F]?- ions (iv) K [Al (C204)3] It gives K* and [Al (C2q)3]>- ions (v)_ [Co(NFis)6] [Cr(CN)6] It gives [Co (NH3)s}>* and [Cr(CN)g>- ions (vi) [PtQNH3)4] [PtCla] It gives [Pt (NH3)4]?* and [PtCl,]?- ions Q.25.Define ligands. Ans. Ligand is an atom or group of atoms having a tendency to donate electron pair. © Itmay be -ve,neutral or +ve ion © On the basis of number of donor atoms present in ligand, they are classified as (i) Monodentate, bidentate,” tridentate, tetradentate, pentadentate & hexadentate ligands. Q.26.Define unidentate ligand, give examples. Ans. Unidentate / Monodentate ligand ‘© Those ligands, which have only one donor atom are known as unidentate ligand. * For Ans, 9 7 example-:Ci? Chlorido NH ammine «= @ : NH,-NH,, hydrazonium cation Q.27.Define tridentate ligand, give examples. Ans. Tridentate ligand «© Those ligands which have three donor atoms are known as tridentate ligands. # Forex: M2 NH, by. Diethylenetriamine H CH, 2 [Three N atoms} Lo I ote Q.28. Define tetradentate ligand, give examples. Ans, Tetradentate ligands ‘© Those ligands which have four donor atoms are known, as (etradentate ligands. @ scanned with OKEN Scanner4 © Forex: (iii) Am NH3 % My Non, Na Ny ay ny NIL Biethylenetoramine ‘ hy (Four Natoms ony guy S suey Q.29.Give the names of following ligands in their coordination compounds » 2 ee 0 (i) No 7 SH Gi crscoo? Ro wi :Si, (it) CJ0}- & 8,03 Cyano Nirito (uy cn,coo: Acetato 0) Reo Cyanato w § Koo Isocyanato wy Si, Amido (vit) NOs Nitrato (sin) C03> Oxalato (x) co Carbonato @) $,02- Thiosulphato Q.30.Write down the formula of the following ligands @Carbonato (ii) Imido Gii) Ammine () Aqua () Pyridine (vi) Nitronium ion (ii) Hydrazonium cation (viii) Ethylenediamine (ix) Oxalato (8) Glycinato (xi) Diethylene triamine (xii) Triethylene tetramine (xiii) Ethylene diamine triacetato (xiv) [EDTA] Ans, Write down the formula of following ligands () Carbonato CO? (ii). Imido -NH-2 (iv) Aqua 1,0 (v) Pyridine CsHsN (vi) Nitronium ion NO*> (vii) hydrazonium cation NH ~N®H3 (ol) Ethylenediamine —_'NH3CH~CHy-NH (ix) Oxalato C2042 (x) Glycinato Niy-CH,-COO ( C.N. of Co¥ = 6 (iv) Co-ordination Polyhedron * Coordination polyhedron is the spatial arrangement of the ligand atoms which are directly attached to the central atom / ion, ; « The shape of common’ coordination entities are ‘Tetrahedral, square planar, octahedral, square pyramidal and trigonal bipyramidal ete 1 1 u ‘ ; CK ‘ bX 1 Cu a L cab cdral Tetrahedral ny Square planar entity entity L Squarepyramidal cennty (v) _ Homoleptic and heteroleptic coordination compounds ~~ When the metal atom / ion is bonded by only one kind of ligands are called as homolytic complexes = For Example: [Ni (NH3)g}?* [Co(H20)¢I>* [Co(en)3>* when a metal atom / ion is bonded by more than one kind of ligand known as heterolytic complexes. For Ex. [NiCl,(H20),], [CrCla(NH3)4]" Q.36.What is meant by unidentate, bidentate and ambidentate ligands ? Give two examples for each, ‘Ans. Unidentate ligands are those which bind to the metal ion through a single donor atom. c.g., C1-, H;0. Bidentate ligands are those which bind to the metal ion through two donor atoms. e.g., ethylene diamine (H:NCH,CHNH2), oxalate (C20,?>) ion. ‘Ambidentate ligands are those which can bind to metal ion through two different donor atoms. e.g., NO~ and SCN ions. Q.37.Specify the oxidation numbers of the metals in the following coordination enti (a) {Co(H30)|(CN) (en)2)* (b) IPtCLP- @ scanned with OKEN Scanner(© [Ce(NH)3Ch1 () [N(CO)s) (a) [CoBra(en)aI* Ans, (a) hexaamminecobalt(II)chloride (©) KslFe(CN)¢) (b)_ tetraaminechloridonitrito-N- “hae! cobalt(Il})chloride (a) [Co(H;0)|(CN) (en)>? (c)_hexaammincnickel(I}}chloride X+OF EN + 20)" 42 (@) diamine chloridomethylamine platinum (11) axet3 chloride (b) PCLP. (©) hexaaquamanganese(II)ion xtdcl)=-2 (6) tris (cthan-1,2-diamine)cobalt(IH)ion weed (g)_hexaaquatitanium(Il)ion (©) [CrtNH)3Ch) (h)_tetrachloridonickelate(I)ion N+ 30) #31) =0 (i) _tetracarbonylnickel(0) Q.40. How many geometrical isomers are possible in the following coordination entities? (a) [Cr(C204)3)- (b) [Co(NH3)3Cls| Ans. (a) does not show geometircal isomers (b) two geometrical isomers : fac - and mer- Q.41. Draw the structures of optical isomers of : (a) [Cr(C204)3)?- (© KylFeiCNYel 3x GI +X+6E=0 Bxet3 Q.38. Using IUPAC norms write the formula for the followin b) [PtChjfen)s|2* )) tetrahydroxozincate(I1) a Ce tenr )hexamminecobalt(IIT)sulphate Ans. (a) Gii)_potasium tetrachloridopalladimate({1) (iv) potassiumtri (oxalato) chromate( III) (%) diamminedichloridoplatinum(t1) (vi) hexaammineplatinum(IV) (vi) potassium tetracyanonickelate(II) (vili)tetrabromidocuprate({I) (&) pentaammin ()_ pentaammineni Ans. (i) [Zn(OH)4P- (ii) [Co(NHs)s]2(SO4)3, ii) Ko{PaCl,] (iv) K3[Cr(C204)3] (vy) [Pt(NH3)2Cly] (vi) [Pt(NH3)6]#* (vii) Ko[Ni(CN),] (viii) [CuBr4)?~ (ix) [Co(NH3)s(ONO)>* (x) [Co(NHs)s(NO.))?* Q.39, Using IUPAC norms write the systematic names of the following : (a) [Co(NH3)¢ICls (b) [Co(NH3)4C(NO,)|CI (c) [Ni(NH3)6]Cl, a {2 NH CINHE,CHICI og, OAD raW all the isomers (geometrical and opi Ce + (a) [CoCh(« I* pr to mane" (3 catinehemys © merrier (by [NCI = t = @ scanned with OKEN Scanner