Questions Answers Coordination PDF
| Title | Questions Answers Coordination |
|---|---|
| Author | Elihle Yiliwe |
| Course | Analytical Chemistry |
| Institution | Walter Sisulu University |
| Pages | 15 |
| File Size | 622.7 KB |
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Questions&AnswersonCoordinationChemistry(DRay) 1.Whatarethegeometriesofthefollowingtwocomplexes? (i)[AlCl4]‐ (ii)[Ag(NH3)2]+ Ans.tetrahedral,linear 2.Whatistherespectivecentral‐metaloxidationstate,coordinationnumber,andthe overallchargeonthecomplexioninNH4[Cr(NH3)2(NCS)4]? Ans.+3;6;1‐ 3. The formula for lithium iodotris(trifluorophosphine)nickelate(0) is: (Note: trifluorophosphineisaneutralligandwiththeformulaPF3.) Ans.Li[Ni(PF3)3I3] 4.Whichismorelikelytoformahigh‐spincomplex—en,F‐,orCN‐? Ans.F‐
5.Namethefollowingcompound:K2[CrCO(CN)5]. Ans.potassiumcarbonylpentacyanochromium(III) 6.Whatismacrocycliceffect? Ans. Macrocyclic ligands of appropriate size form more stable complexes than chelate ligands. A competition between a noncyclic chelating ligand and a macrocyclic (chelating)ligandhavingthesamenumberandtypeof donoratomswillgenerallylead to complex formation predominantly by the macrocyclic ligand. This is known as the macrocycliceffect.
Theentropy termdrivesthistypeofreactiontotheright.Thelongnoncyclicligandis more flexible than the corresponding macrocyclic ligand and can adopt many more conformationsthanthemacrocyclicligandwhenitisnotcoordinated. 7. Predict the no. of unpaired electrons, the spin‐only magnetic moments at 25 C for
eachofthefollowing. a)[Fe(CN)6]4‐ b)[Ru(NH3)6]3+ c)[Cr(NH3)6]2+ d)[EuCl6]4‐ 2+ 6 Ans. a) Fe is 3d . Has 0 unpaired electron in l.s. complexes and thus the magnetic momentwouldbecloseto0µB. b)Ru3+is4d5.Has1unpairedelectroninl.s.complexesandthusthemagneticmoment wouldbecloseto1.73µB. c) Cr2+ is 4d4. Have 4 and 2 unpaired electrons in h.s. and l.s. complexesand thus the magneticmomentwouldbecloseto4.90and2.83µB,respectively. d)Eu2+is4f7.Has7unpairedelectronsinh.s.complexesandthusthemagneticmoment wouldbecloseto7.94µB. Themagneticmomentsdeviateconsiderably fromthe spin‐ only values because of strong spin‐orbit coupling. The f‐orbitals have so little overlap andinteractionswithligandorbitals. 8. With reference to the 3d elements in the periodic table identify the elements and associated oxidation numbers that form square planar complexes. Give formulas for threeexamplesofsuchcomplexes. Ans. The square‐planar geometry is primarily associated with the d8 electron configuration, and the following elements and oxidation states commonly form the d8 configuration: 3456789101112 Ni2+ Rh+Pd2+ Ir+Pt2+Au3+ Examplesofsuchcomplexes:RhCl(PPh3)3,IrCl(CO)(PPh3)2,Ni(CN)42–, PdCl42‐,cis‐and trans‐PtCl2(NH3)2 and AuCl4–. Note however that NiCl42– is not square planar, but is insteadtetrahedral.Ingeneral,thepreferenceforsquareplanarvs.tetrahedralincreases withincreasingligandfieldstrength;theligandfieldsforthe2ndand3rdtransitionseries are intrinsically larger than for the first series metals. Note however, that Ni(CO)4 is alsotetrahedral,notsquareplanar,butthatisbecausethiscomplexisd10,ratherthan d8! 9. Sketch the two structures that describe most six‐coordinate complexes. Which is morecommon? Ans.OctahedralTrigonalprismatic
Hexamminecobalt(III)chloride Trigonalprismaticisanextremelyraregeometry,butoctahedralisextremelycommon. 10. The compound Na2IrCl6 reacts with triphenylphosphine in diethylene glycol under an atmosphere of CO to give trans‐[IrCl(CO)(PPh3)2], known as Vaska’s compound. Excess CO produces a five‐coordinate species and treatment with NaBH4 in ethanol gives[IrH(CO)2(PPh3)2].Drawandnamethethreecomplexes. Ans.
Names,inoder:Carbonylchlorobis(triphenylphosphine)iridium(I), Dicarbonylchlorobis(triphenylphosphine)iridium(I),and Carbonylhydridobis(triphenylphosphine)iridium(I). 11. Which of the following complexes are chiral? (a) [Cr(ox)3]3–; (b) cis‐[PtCl2(en)]; (c) cis‐[RhCl2(NH3)4]+; (d) [Ru(bipy)3]4+; (e) [Co(edta)]–; (f) fac‐[Co(NO2)3(dien)]; (g) mer‐ [Co(NO2)3(dien)].Identifytheenantiomersaschiralandachiralcomplexes. Ans. (a) [Cr(ox)3]3– It is definitely chiral; two chelating ligands on an octahedral framework are sufficient do impart chirality to the complex. The two enantiomers providetheleftandrightofaplaneofreflection. (b)cis‐[PtCl2(en)]Thismoleculewouldbechiralbecauseofthetwistin theenligand. However, ring inversion in chelate complexes of this type (as in cyclopentane) is very fast at room temperature and well below, so that enantiomers cannot be isolated, and thecomplexisnormallyconsideredtobeachiral. (c)cis‐[RhCl2(NH3)4]+This complexisachiral.It hasC2v symmetry,andthushastwo internal mirror planes, which can be clearly seen in the picture of the molecule, one virticalandonehorizontaltothepage.Notethatthesymmetry requiresthe ammine ligands to be in a specific orientation. In practice, the Rh—N bonds rotate sufficiently fast that the ammine ligands are effectively spherical, so the molecule definitelyactsasaC2vspecies.
(d)[Ru(bipy)3]4+ Thisisanotherexampleofatrischelatecomplex,andthereforewillbe both chiral and easily resolvable. The stucture is shown with its non‐superimposible mirrorimagebelow(Hatomsomittedforclarity). (e) [Co(edta)]– When EDTA chelates a metal such that all six donor atoms coordinate (themostcommon,butnottheonlywaythatthisligandattaches tometals) thenthe resultingcomplexesarechiral.Thenon‐superimposablemirrorimagesimpartchirality inthesystem. (f)fac‐[Co(NO2)3(dien)](notetypoinquestion!)and(g)mer‐[Co(NO2)3(dien)].Ifone ignores the conformations of the chelate rings, the complexes are not chiral. The chelates are ethylenediamine linkages, and as discussed for (b) above, such rings interconverttoorapidlytoallowtheseparationofenantiomersofthesecomplexes.The facialisomerhasaninternalplaneofsymmetrythat includesthecentral Nofdien, the metalandoneofthenitrogroups.Themeridionalisomerhasaplanethroughthe dien ligand and the metal (as well as one through the metal and the three nitro N atoms.) Theseplanesareverticaltothepaperthroughthemiddleofthemolecules. 12.WhatisthecoordinationnumberoftheFeatomisK3[Fe(C2O4)3]? Ans.Thecoordinationnumberis6.Therearethreebidentateligandsattachedto thecentralmetal,3x2=6. 13.WhatisthecoordinationnumberoftheAuatominK[Au(CN)2(SCN)2]? Ans.Thecoordinationnumberis4.Therearefourmonodentateligandsattached tothecentralmetal. 14.Whichofthefollowingcanfunctionasabidentateligand? NH3,C2O42‐,CO,OH‐ 2‐ Ans.OnlyC2O4 ,oxalate,isabidentateligand.Theothersareallmonodentate. 15.Ethylenediaminetetraacetateion(EDTA4‐)iscommonlyreferredtoasa ___________ligand. Ans.hexadentate 16.Calculatetheoxidationstateofthemetalandthenumberofdelectronsinthe followingcoordinationcomplexes: a)[CoCl4]2‐;b)[Fe(bpy)3]3+;c)[Cu(ox)2]2‐;d)[Cr(CO)6] Ans.a)EachClligandhasachargeof‐1,so4x‐1=‐4 Overallchargeonthecomplexis‐2,sotheoxidationstateofCo=+2. GroundstateconfigurationforCo=[Ar]3d74s2
Onlossof2e‐,Co2+hasconfiguration[Ar]3d7,sosevendelectrons. b)bpy(2,2’‐Bipirydyl)isuncharged=neutral OxidationstateofFe=+3.GroundstateconfigurationforFe=[Ar]3d64s2 Onlossof3e‐Fe3+hasconfiguration[Ar]3d5,sofivedelectrons. c)ox(oxalate,C2O42‐)hascharge‐2peroxalate,sototal=2x‐2=‐4 Overallchargeoncomplex=‐2,sotheoxidationstateofCu=+2. GroundstateconfigurationforCu=[Ar]3d104s1 Onlossof2e‐,Cu2+hasconfiguration[Ar]3d9,soninedelectrons. d)COisuncharged=neutral. Oxidation stateof Cr = 0. In thiscase, all electrons are in 3dorbitals which are now of lowerenergy(becausefilled)than4s orbitals.Ground stateconfigurationfor Cr=[Ar] 3d54s1.ConfigurationforCr0=[Ar]3d6,sosixdelectrons. 17.Drawthestructureofthefollowingcomplexes: a)trans‐diaquadichloroplatinum(II) b)diamminetetra(isothiocyanato)chromate(III) Ans.a)
H2O(aqua)isneutral.Diaquaindicatesthattherearetwoofthem. EachClligand=‐1. Dichloroindicatestherearetwochlorineligands.Platiniumisinthe+2oxidationstate sothecomplexisuncharged.TransindicatedthattheClandH2O ligands arelocated oppositeeachother. b)
AmmineisNH3,whichisuncharged.Thereare2ofthem.Isothiocyanato=SCN‐,which attachesthroughtheS.Eachhasachargeof‐1andthereare 4.Chromate (III)indicates
that this is an anion with Cr in the 3+ oxidation state. Charge on SCN‐= 4 x‐1 =‐4. ChargeonCr=+3,sooverallcharge=‐1. 18.Fora nickel(II)complexexplain the following electronic spectrumwiththehelp of theadjacentTanabe‐Suganodiagram.
Ans.Tanabe‐Suganodiagramsareusedincoordinationchemistrytopredictabsorptions intheUVandvisibleelectromagneticspectrumofcoordinationcompounds. It can be used to assign transitions for the absorptions. The first peak is due to the 4 A2g(F)4T2g(F)transitionandhasanenergyequalto∆o.Thesecondpeakisduetothe 4 A2g(F)4T1g(F)transition.Thethirdpeakisduetothe4A2g(F)4T1g(P)transition.
19.ExplianHund’srules. Ans. In atomic physics, Hund's rules refer to a set of rules formulated by German physicist Friedrich Hund around 1927, which are used to determine the term symbol thatcorrespondsto thegroundstateofamulti‐electronatom. Inchemistry, rule one is especiallyimportantandisoftenreferredtoassimplyHund'srules. Thethreerulesare: 1) For a given electron configuration, the term with maximum multiplicity has the lowestenergy.Sincemultiplicityis2S+1equalto,thisisalsothetermwithmaximumS. Sisthespinangularmomentum. 2) For a given multiplicity, the term with the largest value of L has the lowest energy, whereListheorbitalangularmomentum. 3)Foragiventerm,inanatomwithoutermostsub‐shellhalf‐filledorless,thelevelwith thelowestvalueofJlieslowest inenergy. If theoutermost shellismorethan half‐filled, thelevelwithhighestvalueofJislowestinenergy.Jisthetotalangularmomentum,J=
L+S. 20.ExplaintheJahn‐Tellerdistortionin[Cu(H2O)6]2+. Ans.TheJahn‐Teller(J‐T)theoremstatesthatinmolecules/ionsthathave adegenerate ground‐state, the molecule/ion will distort to remove the degeneracy. This is a fancy wayofsayingthatwhenorbitalsinthesamelevelareoccupied bydifferentnumbersof electrons,thiswill leadtodistortionofthemolecule.Forus,whatisimportant is thatif thetwoorbitalsoftheeglevelhavedifferentnumbersofelectrons,thiswillleadtoJ‐T distortion.Cu(II)withitsd9configurationisdegenerateandhasJ‐Tdistortion.
21. Why the magnetic moment of an octahedral monothiocarbamate complex of Iron (III)droppedfrom5.8μBat300K,to5.8μBat150K,to4μBat78K? Ans:InthiscomplexthecrystalfieldsplittingΔ O isalmost exactlyequalto thepairing energy (P). A change in temperature can cause a partial spin crossover from high‐ to low‐spin behavior, with unusual magnetic behavior. The magnetic moment value changes as the proportion of molecules in the high‐ and low‐spin states changes with temperature. 22.Howcomplexanionsareseparated frombyproductsandisolatedincrystallineform? Ans:Thecomplexanionsuchas[Ni(CN)5]3‐canbecrystallizedusingsolubilitytendency by adding a large counterion of opposite charge. Five coordinate [Ni(CN)5]3‐ ion prevented isolation from solution until the matched counterion 3+ ) ] was added, whereupon [Cr(en) ][Ni(CN) ].15H O crystallized [Cr(NH2CH2CH2NH2 3 3 5 2 out.Evaporatingthesolventoraddinganothersolventthatmixeswiththefirstbutdoes not dissolve the product (e.g., adding EtOH to the aqueous solution) can help to crystallizeasaltoftheproduct.
23.HowwillyouprepareK3[Rh(ox)3]fromkineticallyinertK3[RhCl6]? 100C K+ 3‐ 2‐ [Rh(ox)3]3‐ Ans: [RhCl6] +3ox Winered2hYellowEvaporation With kinetically inert starting complex, the reaction requires boiling in some high boiling solvent for some hours followed by the supply of appropriate (matching) counterion. 24.Howwillyouprepareinert[Cr(en)3]Cl3from inert[Cr(H2O)6]Cl3? Ans:[Cr(H2O)6]3++3en [Cr(OH)3(H2O)3]∙(s)+3enH+ Iftheligandisbasicinwater,andthemetalionisappreciablyacidicas well asbeinga hardacid, the metal ion may prefer to react with the hydroxide ions generated by the ligandinwater,ratherthantheliganditself. d3 Cr3+ ion is inert and the water molecules do not dissociate as quickly as the basic liganddeprotonatestheacidiccoordinatedwatermolecules. dryEt2O CrCl3(anhydrous)+3en [Cr(en)3]Cl3 Yellow 25.Giveexampleforsubstitutionofvolatileligandsinmetalcomplexes. Ans: Volatile ligands such as CO, H2O or NH3 are driven off by heating and then are replacedbyadesiredligandinsolutionorbythecounterionsofthestartingcomplex. 250C [Pt(NH3)4]Cl2(s) trans‐[PtCl2(NH3)2]+2NH3(g) White hυPPh3 [W(CO)6] CO(g)+W(CO)5 [W(CO)5(PPh3)] reactiveunsaturated complex 26.Whatismacrocycliceffect? Ans: Macrocyclic ligands of appropriate size form more stable complexes than chelate ligands. A competition between a non‐cyclic chelating ligand and a macrocyclic (chelating) ligand having the same no. and type of donor atoms will generally lead to
complex formation predominantly by the macrocyclic ligand. This is known as the macrocycliceffect.
Theentropytermderivesthistypeofreactionto theright.Thelong noncyclicligandis more flexible than the corresponding macrocyclic ligand and can adopt many more conformationsthanthemacrocyclicwhenitisnotcoordinated. 27.Predictthenumberofunpairedelectrons, themagneticmomentsat25◦Cforeacho f thefollowing (a)[Fe(CN)6]4‐,(b)[Ru(NH3)6]3+,(c)[Cr(NH3)6]2+,(d)[EuCl6]4‐. 4‐ 2+ 6 Ans:(a)[Fe(CN)6] →Fe →3d . Lowspincomplex→0(zero)unpairedelectron→μB=0 (b)[Ru(NH3)6]3+→Ru3+→4d5. Lowspincomplex→oneunpairedelectron→μB=1.73BM. (c)[Cr(NH3)6]2+→Cr2+→3d4. Highspincomplex→4unpairedelectrons→μB=4.90BM. Lowspincomplex→2unpairedelectrons→μB=2.83BM. (d)[EuCl6]4‐→Eu2+→4f7. Highspincomplex→7unpairedelectrons→μB=7.9BM. Theforbitalshavesolittleoverlapandinteractionwithligandorbitals. 28.Amongthefollowing3dtransitionmetalionswhichoneiskineticallyinert? Cr2+,C03+,C02+,Fe3+ Ans.C03+
29.Amongthefollowingwhichstatementaboutthetrans‐effectandthetrans‐influence ofanysquare‐planarcomplexiscorrect? A.Thetrans‐influenceisaground‐stateeffect,whereasthetrans‐effecthasakinetic origin B.Thetrans‐effectisaground‐stateeffect,whereasthetrans‐influencehasakinetic origin
C. Boththetrans‐effectandtrans‐influence areground‐stateeffects D.Ratesofsubstitutionareaffectedbythetrans‐effectbuthavenothingtodowiththe Ans.[A] 30.Marcus‐Hushtheoryappliesto: A.anyelectrontransferreactionofconsideration B.outer‐sphereelectrontransferinredoxreactions C.inner‐sphereelectrontransferinredoxreactions D.allbiologicalredoxreactions Ans.ThecorrectanswerisB. 31.Showthepossiblestereoisomersofoctahedral[Mn(H2O)2(ox)2]2‐. Ans.H2Oismonodentateandoxisbidentateoxalateionandrepresentedby
32.Sketchthetwostructuresthatdescribemostsix‐coordinatecomplexes.Whichis morecommon? Ans.
Incoordinationchemistrytrigonalprismaticisararegeometry,butoctahedralis extremelycommon. 33.Giveformulaforthreedifferentcomplexesthathavethemorecommonsix‐ coordinatestructure. Ans.Somecommonexamplesofsix‐coordinatecomplexesare[Cr(NH3)6]3+,[Cr(CO)6], [Cr(OH2)6]2+,[Fe(CN)6]3–,[Fe(CN)6]4–,and[RhCl6]3–.Inaddition,almostallmetal aquaions(saveonlyCu2inthefirstTMseries)haveoctahedralaquacomplexesinwater ordiluteacid. 34.Nameanddrawstructuresofthefollowingcomplexes:(a)[Ni(CO)4];(b)[Ni(CN)4]2–; (c)[CoCl4]2–;(d)[Ni(NH3)6]2+. Ans.
35. Determine the configuration (in the form t2gmegn or emt2n, as appropriate), the numberofunpairedelectrons,andtheligandfieldstabilizationenergyasamultipleof ∆oor∆T foreach of thefollowingcomplexesusing thespectrochemicalseriestodecide, where relevant, which are likely to be strong‐field and which weak‐field. (a)[Co(NH3)6]3+; (b) [Fe(OH2)6]2+; (c) [Fe(CN)6]3–; (d) [Cr(NH3)6]3+; (e) [W(CO)6]; (f) [FeCl4]2–and(g)[Ni(CO)4]. 3+ Ans.(a)[Co(NH3)6] Since the NH3 ligands are neutral, this is a Co3+ complex, so we are dealing with a d6 metalion.Althoughammoniaisinthemiddleof thespectrochemicalseries,ametal in the3+oxidationstateispast themiddleofthe metalseries,sothecombinationisvery likely to be a low‐spin configuration, t2g6eg0. Indeed, octahedral d6 has S = 0 (no unpairedelectrons)andisdiamagnetic.TheLFSEis6×0.4∆o=2.4∆o. Notethatthis is thelargestpossibleLFSEforanoctahedralcomplex. (b)[Fe(OH2)6]2+ The iron ion in this octahedral complex with six neutral water ligands is in the 2+ oxidationstate,andwaterislowonthespectrochemicalseries.Wecertainlyexpectthis d6iontobeinahighspinconfiguration,t2g4eg2,withS=2.Thatis, ithas fourunpaired electronsandisstronglyparamagnetic.TheLFSEis4×0.4∆o–2×0.6∆o=0.4∆o. NotethemuchsmallerLFSEcomparedtothelowspind6casediscussedin(a). (c)[Fe(CN)6]3– The iron in this octahedral complex with six 1– cyanide ligands is in the 3+ ...
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