2.8: Ligand Field Theory - Chemistry LibreTexts

Ligand Field Theory looks at the effect of donor atoms on the energy of d orbitals in the metal complex. The effect depends on the ... Skiptomaincontent OctahedralcaseReasonsforLow-spinvs.High-spin:TheEffectoftheMetalIonAttribution Conceptsfrommolecularorbitaltheoryareusefulinunderstandingthereactivityofcoordinationcompounds.OneofthebasicwaysofapplyingMOconceptstocoordinationchemistryisinLigandFieldTheory.LigandFieldTheorylooksattheeffectofdonoratomsontheenergyofdorbitalsinthemetalcomplex. Therearetwowaysinwhichwesometimesthinkabouttheeffectofligandsonthedelectronsonametal.Onthebasisofsimpleelectron-electronrepulsion,donationofalonepairmightraiseanoccupieddorbitalinenergy.Alternatively,wecanthinkaboutbondinginteractionsbetweenligandorbitalsanddorbitals.Thissecondwayofthinkingaboutthingsisalittlebitmoreuseful,andthat'stheapproachwe'llfocuson,here. Eitherway,thereareinteractionsbetweenligandelectronsanddelectrons,thatusuallyendupraisingthedelectronsinenergy.Theeffectdependsonthecoordinationgeometryoftheligands.Ligandsinatetrahedralcoordinationspherewillhaveadifferenteffectthanligandsinanoctahedralcoordinationsphere,becausetheywillinteractwiththedifferentdorbitalsindifferentways. LigandFieldTheorylooksattheeffectofdonoratomsontheenergyofdorbitalsinthemetalcomplex.Theeffectdependsonthecoordinationgeometrygeometryoftheligands. Octahedralcase Supposeacomplexhasanoctahedralcoordinationsphere.Assumethesixligandsallliealongthex,yandzaxes. Therearetwodorbitalsthatwillinteractverystronglywiththeseligands:thedx2-y2,whichliesdirectlyonthexandyaxes,andthedz2,whichliesdirectlyonthezaxis.Together,thesetwometalorbitalsandtheligandorbitalsthatinteractwiththemwillformnewbondingandantibondingmolecularorbitals. Thedrawingbelowissimplified.Theligandswillalsointeractwithsandporbitals,butforthemomentwe'renotgoingtoworryaboutthem.Wealsowon'tworryaboutinteractionsfromtheotherfourligandswiththedorbitals(possiblebysymmetryconsiderations,butalsoamorecomplicatedpicture). Now,rememberthatmetalsusuallyhavedelectronsthataremuchhigherinenergythanthoseontypicaldonoratoms(likeoxygen,sulfur,nitrogenorphosphorus).Thatmeanstheantibondingcombinationswillbemuchcloserinenergytotheoriginaldorbitals,becausebotharerelativelyhighinenergy.Thebondingcombinationwillbemuchcloserinenergytotheoriginalligandorbitals,becausetheseonesareallrelativelylowinenergy. Thatenergeticsimilaritygenerallytranslatesintoasimilarityinshapeandlocationaswell.Inotherwords,theantibondingcombinationbetweenadorbitalandaligandorbitalisalotliketheoriginaldorbital.Thebondingcombinationismoreliketheoriginalligandorbitalthantheoriginaldorbital.Becauseofthosesimilarities,inorganicchemistsoftenrefertothoseantibondingorbitalsasiftheywerestilltheoriginaldorbitals. Thesetwoorbitalswillberaisedrelativelyhighinenergybysigmabondinginteractionswiththedonororbitals.Ifthereareelectronsinthepicture,itmightlooksomethinglikethis: Assumethesixligandsallliealongthex,yandzaxes. Thedx2-y2andthedz2orbitalsliealongthebondaxes. Thesetwoorbitalswillberaisedrelativelyhighinenergy. Theseorbitalsarelikeantibondinglevels. Theseorbitalsaresometimescalledthe"eg"setoforbitals.Theterm"eg"comesfromthemathematicsofsymmetry. Ontheotherhand,theotherthreedorbitals,thedxy,dxzanddyz,allliebetweenthedonorligands,ratherthanhittingthemhead-on.Theseorbitalswillinteractlessstronglywiththedonorelectrons. Thedxy,dxzanddyzorbitalsallliebetweenthebondaxes. Thesethreeorbitalswillbechangedinenergyonlyalittle. Theseorbitalsaremorelikenon-bondingorbitals. Theseorbitalsaresometimescalledthe"t2g"setoforbitals. Remember,onlytheenergyoftheelectronsaffectstheoverallenergyofthesystem.Theunoccupieddorbitalsareraisedinenergy,buttheoccupiedorbitalsgodowninenergy(orelsestaythesame). Apartfromthestabilizationofthecomplex,thereisanotherconsequenceofthispicture.Whatweareleftwithistwodistinctsetsofdenergylevels,onelowerthantheother.Thatwillhaveaneffectontheelectronconfigurationatthemetalatominthecomplex.Thatmeanstherewillbecaseswhereelectronscouldbepairedorunpaired,dependingonhowtheseorbitalsareoccupied. Takethecaseofthebiologicallyimportantiron(II)ion.Ithasad6valenceelectronconfiguration.Inlessformalparlanceofinorganicchemistry,"iron(II)isd6".Inaniron(II)ionallaloneinspace,allthed-obitalswouldhavethesameenergylevel.Wewouldputoneelectronineachorbital,andhaveoneleft.Itwouldneedtopairupinoneofthedorbitals.(Noticethat,inthechemistryoftransitionmetalions,thevalencesandporbitalsarealwaysassumedtobeunoccupied). Thingsareverydifferentinanoctahedralcomplex,likeK4[Fe(CN)6].Inthatcase,thedorbitalsarenolongeratthesameenergylevel.Therearetwopossibleconfigurationstoconsider. Inonecase,oneelectronwouldgointoeachofthelowerenergydorbitals.Achoicewouldbemadeforthefourthelectron.Doesitgointothehigherenergydorbital,ordoesitpairupwithoneofthelowerenergydelectrons?Thechoicedependsonhowmuchhigherinenergytheupperdorbitalsare,comparedtohowmuchenergyitcoststoputtwoelectronsinthesamedorbital. Ifthe"dorbitalsplittingenergy"isprettylow,sothatthetwosetsofdorbitalsarestillprettysimilarinenergy,thenextelectroncangointoahigherorbital.Pairingwouldnotberequireduntilthefinalelectron.Overall,thatwouldleavefourunpairedelectrons,justlikeinthecaseofalonemetalioninspace.Thisiscalledthe"high-spin"case,becauseelectronscaneasilygointothehigherorbital. Ifthedorbitalsplittingenergyistoohigh,thenextelectronmustpairupinalowerorbital.Allthreeremainingelectronspairup,andsotherearenounpairedelectronsinthecomplex.Thisiscalledthe"low-spin"case,becauseelectronsmoreeasilypairupintheorbital. Sotheoverallruleisthatiftheenergytopairuptheelectronsisgreaterthantheenergyneededtogettothenextlevel,theelectronwillgoaheadandoccupythenextlevel. However,iftheenergyittakestogettothenextlevelismorethanitwouldcosttopairup,theelectronswilljustpairupinstead. Theelectronconfigurationcanbe"highspin"or"low-spin",dependingonhowlargetheenergysplittingisbetweenthetwosetsofdorbitals. Thedifferencebetweenthehigh-spincaseandthelow-spincaseissignificant,becauseunpairedelectronsaffectthemagneticpropertiesofamaterial.Thelow-spincasewouldbediamagnetic,resultinginnointeractionwithamagneticfield.However,thehigh-spincasewouldbeparamagnetic,andwouldbeattractedtoamagneticfield. ItturnsoutK4[Fe(CN)6]isdiamagnetic.Thus,itisprettyclearthatitisalow-spincomplex.Theenergydifferencebetweenthetwodorbitallevelsisrelativelylargeinthiscase. Inadditiontoinfluencingmagneticproperties,whetheracomplexishigh-orlow-spinalsoinfluencesreactivity.Compoundswithhigh-energydelectronsaregenerallymorelabile,meaningtheyletgoofligandsmoreeasily. electronconfigurationinfluencesmagneticproperties electronconfigurationinfluenceslability(howeasilyligandsarereleased) ReasonsforLow-spinvs.High-spin:TheEffectoftheMetalIon Thereareafewfactorsthatdeterminethemagnitudeofthedorbitalsplitting,andwhetheranelectroncanoccupythehigherenergysetoforbitals,ratherthanpairingup.Itisbasedpartlyonligandfieldstrength,whichisexploredonthenextpage.Italsodependsonthechargeonthemetalion,andwhetherthemetalisinthefirst,secondorthirdrowofthetransitionmetals. Thehigherthechargeonthemetal,thegreaterthesplittingbetweenthedorbitalenergylevels.Forexample,Fe(II)isusuallyhighspin.Ithasasmallersplittingbetweenthelowerandhigherdorbitallevels,soelectronscanmoreeasilygotothehigherlevelratherthanpairupunthelowerlevel. Ontheotherhand,Fe(III)isusuallylowspin.Ithasalargersplittingbetweenthedlevels.Inthatcase,itcostslessenergyforelectronstopairupinthelowerlevelthantogouptothehigherlevel. High-spinversuslow-spincasesinvolveatrade-offbetweenthedorbitalsplittingenergyandthepairingenergy. 2ndand3rdrowtransitionmetalsareusuallylowspin 1strowtransitionmetalsareoftenhighspin However,1strowtransitionmetalsandbelowspiniftheyareverypositive(usually3+orgreater) Thereisalotgoingoninmetalions,butwe'lltakeasimplifiedviewofthings.Thinkingonlyaboutelectrostatics,wecantrytoimaginewhathappenstothoseelectronswhenthechargeonthemetalionchanges. FirstweneedtoknowaboutCoulomb'slaw.Coulomb'slawstatesthattheforceofattractionbetweentheelectronandthenucleusdependsononlytwofactors:theamountofpositivechargeinthenucleus,andthedistancebetweenthenucleusandtheelectron. Thegreaterthechargeonthenucleus,thegreatertheattractionbetweentheelectronandthenucleus. Thefartheranelectronisfromthenucleus,theweakertheattractionbetweentheelectronandthenucleus. Coulomb'slawcanbeusedtoevaluatethepotentialenergyoftheelectron.Itisoneofthefactorsthatdetermineshowhighorlowthoseelectronicenergylevelsarethatweseeinenergyleveldiagramsforatoms,ionsandmolecules.Theenergyoftheelectronvariesinaroughlysimilarway:thegreaterthechargeonthenucleus,thelowertheenergyoftheelectron.Also,theclosertheelectronistothenucleus,theloweritsenergy. Roughlyspeaking,electronsathigherenergyarefartherfromthenucleus.Electronsatlowerenergyareclosertothenucleus. Whathappensifthechargeincreases?Maybealotmoreprotonsareaddedtothenucleus.Maybesomeelectronsarelost,sothattotheremainingelectronsitjustfeelslikethechargeofthenucleushasincreased.Thentheelectronsshouldbemoreattractedtothenucleus.Theygetalittlecloser.Theirpotentialenergydrops. Ofcourse,ifoneelectronisclosertothenucleusalready,itfeelsthatincreaseinpositivechargemorestronglythananelectronthatisfartheraway.Consequently,itdropsfurtherinenergythananelectronthatisfurtheraway. Ifwetranslatethatideaintoapictureofthedorbitalenergylevelsinanoctahedralgeometry,itlookslikethis: Whenthechargeonthemetalionisincreased,boththehigherandthelowerlevelsdropinenergy.However,thelowerleveldropsmore.Thus,thegapbetweenthelevelsgetswider. Metalsinthesecondandthirdrowoftheperiodictablealmostneverformhigh-spincomplexes.Thedorbitalenergysplittinginthesecasesislargerthanforfirstrowmetals.Fromaverysimplepointofview,thesemetalshavemanymoreprotonsintheirnucleithanthefirstrowtransitionmetals,droppingthatlowersetofdelectronslowerwithrespecttothehigherset. Thatisn'tthewholepictureforthesecondandthirdrowtransitionmetals,however.Remember,wearesimplifying,andtherearefactorswewon'tgointo.However,itisimportanttoknowthatmetal-ligandbondstrengthsaremuchgreaterinthesecondandthirdrowthaninthefirst.We'lllookatthewholeinteractiondiagramforanoctahedralcomplexnow,includingcontributionsformmetalsandporbitals. Figure\(\PageIndex{15}\): Likeallligand-metalinteractiondiagrams,theenergylevelsoftheligandsbythemselvesareshownononeside.Themetal'selectronicenergylevelsareshownontheotherside.Theresultoftheirinteraction,ametal-ligandcomplex,isshowninthemiddle.Thedorbitalsplittingdiagramisshowninabox. Supposethediagramaboveisforafirstrowtransitionmetal.Thediagramforasecondorthirdrowmetalissimilar,butwithstrongerbonds. Ifthebondinginteractionisstrongerbetweenthemetalandligand,thensoistheantibondinginteraction.Theantibondinglevelsarebumpedhigherinenergyasthebondinglevelssinklower.Generallythat'sOK,becausewhentheelectronsarefilledin,theywillbefoundpreferentiallyatthelowerlevels,notthehigherones.Therewillbeanetloweringofelectronicenergy. Whydosecondandthirdrowtransitionmetalsformsuchstrongbonds?Bondstrengthsareverycomplicated.Ingeneral,thereisgreatercovalencybetweenthesemetalsandtheirligandsbecauseofincreasedspatialandenergeticoverlap.Ratherthangointothosefactors,we'lljustthinkaboutallthoseextraprotonsinthenucleusthatareattractingtheligandelectronsmorestrongly. Thereisonemoreimportantdistinctionthatmakessecondandthirdrowtransitionmetalslowspin.Inaddition,thepairingenergyislowerinthesemetalsbecausetheorbitalsarelarger.Thereismoreroomfortwoelectronsinoneorbital,withlessrepulsion.Asaresult,electronsaremuchmorelikelytopairupthantooccupythenextenergylevel. 2ndand3rdrowtransitionmetalshavestrongerbonds,leadingtoalargergapbetweendorbitallevels 2ndand3rdrowtransitionmetalshavemorediffuseorbitals,leadingtoalowerpairingenergy Itissignificantthatmostimportanttransitionmetalionsinbiologyarefromthefirstrowofthetransitionblockandareprettylabile.Thatfactplaysanimportantroleintheeaseofformationanddeconstructionoftransition-metalcontainingproteins.Intermsofformation,ifthemetalismoreeasilyreleasedbyitspreviousligands(eitherwaterorsomecompoundthatdeliversthemetaltothesiteofproteinconstruction),itcanformthenecessaryproteinmorequickly.However,evenifametal-containingenzymeplaysausefulrole,itshouldnotbetoostable,becauseweneedtobeabletoregulatethelevelofproteinconcentrationforoptimumactivity,ordisassembleproteinifitbecomesdamaged.Thus,itisimportantthatthemetalioncanberemovedeasily. Exercise\(\PageIndex{1}\) Drawbothhighspinandlowspind-orbitalsplittingdiagramsforthefollowingionsinanoctahedralenvironmentanddeterminethenumberofunpairedelectronsineachcase. a)Mn2+b)Co2+c)Ni2+d)Cu+e)Fe3+f)Cr2+g)Zn2+ Answer Exercise\(\PageIndex{2}\) Thedorbitalsplittingdiagramforatetrahedralcoordinationenvironmentisshownbelow.Giventhisdiagram,andtheaxesintheaccompanyingpicture,identifywhichdorbitalsarefoundatwhichlevel.Inthepicture,themetalatomisatthecenterofthecube,andthecirclerepresenttheligands. Answer Thethreeorbitalsshownaboveinteractalittlemorestronglywiththeligands.Thethreeorbitalsshownbelowinteractalittlemoreweakly. Thereasonforthedifferenceintheinteractionhastodowithhowclosethenearestlobeofadorbitalcomestoaligand.Therearereallytwopossiblepositions:thefaceofacubeortheedgeofacube.Iftheligandsareatalternatingcornersofthecube,thentheorbitalspointingattheedgesarealittlecloserthanthosepointingatthefacesofthecube. Exercise\(\PageIndex{3}\) Typically,thedorbitalsplittingenergyinthetetrahedralcaseisonlyabout4/9aslargeasthesplittingenergyintheanalogousoctahedralcase.Explainwhyitissmallerforthetetrahedralcase. Answer Theligandsdonotoverlapwiththedorbitalsaswellintetrahedralcomplexesastheydoinoctahedralcomplexes.Thus,thereisaweakerbondinginteractioninthetetrahedralcase.Thatmeanstheantibondingorbitalinvolvingthedelectronsisnotraisedashighinenergy,sothesplittingbetweenthetwodlevelsissmaller. Exercise\(\PageIndex{4}\) SupposeeachoftheionsinExercise\(\PageIndex{1}\)(CC8.1)wereintetrahedral,ratherthanoctahedral,coordinationenvironments.Drawthedorbitaldiagramsforthehighspinandthelowspincaseforeachion. Answer Exercise\(\PageIndex{5}\) Usually,tetrahedralionsarehighspinratherthanlowspin.Explainwhy. Answer Becausethedorbitalsplittingismuchsmallerinthetetrahedralcase,itislikelythattheenergyrequiredtopairtwoelectronsinthesameorbitalwillbegreaterthantheenergyrequiredtopromoteanelectrontothenextenergylevel.Inmostcases,thecomplexwillbehighspin. Exercise\(\PageIndex{6}\) Thedorbitalsplittingdiagramforasquareplanarenvironmentisshownbelow.Giventhisdiagram,andtheaxesintheaccompanyingpicture,identifywhichdorbitalsarefoundatwhichlevel. Answer Theorbitalsareshowninorderofenergy. Exercise\(\PageIndex{7}\) Predictwhethereachcompoundwillbehighorlowspin. [Fe(py)6]2+ [Fe(H2O)6]2+ [FeBr6]3- [Co(NH3)6]3+ [Cu(NH3)6]2+ [Rh(CO)6]3+ [Cr(CO)6]3+ [PtCl6]2- Answera [Fe(py)6]2+3dmetal,M+2,piacceptorligand→lowspin Answerb [Fe(H2O)6]2+3dmetal,M+2,pidonorligand→highspin Answerc [FeBr6]3-3dmetal,M+3,pidonorligand→highspin Answerd [Co(NH3)6]3+3dmetal,M+3,sigmadonorligand→lowspin Answere [Cu(NH3)6]2+3dmetal,M+2,sigmadonorligand→lowspin Answerf [Rh(CO)6]3+4dmetal,M+3→lowspin Answerg [Cr(CO)6]3+3dmetal,M+3,piacceptorligand→lowspin Answerh [PtCl6]2-5dmetal,M+4→lowspin Exercise\(\PageIndex{8}\) Predictwhethereachcompoundwillbesquareplanarortetrahedral. [Zn(NH3)4]2+ [NiCl4]2+ [Ni(CN)4]2- [Ir(CO)(OH)(PCy3)2]2+;Cy=cyclohexyl [Ag(dppb)2]+;dppb=1,4-bis(diphenylphosphino)butane PtCl2(NH3)2 PdCl2(NH3)2 [CoCl4]2– Rh(PPh3)3Cl Answera [Zn(NH3)4]2+3dmetal,d10,sigmadonorligand→tetrahedral Answerb [NiCl4]2+3dmetal,d8,pidonorligand→tetrahedral Answerc [Ni(CN)4]2-3dmetal,d8,piacceptorligand→squareplanar Answerd [Ir(CO)(OH)(PCy3)2]2+5dmetal,d8→squareplanar Answere [Ag(dppb)2]1+4dmetal,d10,sigmadonorligand→tetrahedral Answerf [PtCl2(NH3)2]5dmetal,d8→squareplanar Answerg [PdCl2(NH3)2]4dmetal,d8,M+2,sigmadonorligand→squareplanar Answerh [CoCl4]2–3dmetal,d7,sigmadonorligand→tetrahedral Answeri [Rh(PPh3)3Cl]5dmetal,d8→squareplanar Attribution ChrisPSchaller,Ph.D.,(CollegeofSaintBenedict/SaintJohn'sUniversity)