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CBSEClass–12SubjectChemistry
NCERTSolutions
Chapter–09
CoordinationCompounds
In-textquestion
1.Writetheformulasforthefollowingcoordinationcompounds:
(i)Tetraamminediaquacobalt(III)chloride
(ii)Potassiumtetracyanonickelate(II)
(iii)Tris(ethane-1,2-diamine)chromium(III)chloride
(iv)Amminebromidochloridonitrito-N-platinate(II)
(v)Dichloridobis(ethane-1,2-diamine)platinum(IV)nitrate
(vi)Iron(III)hexacyanoferrate(II)
Ans.(i)
(ii)
(iii)
(vi)
(v)
(vi)
2.WritetheIUPACnamesofthefollowingcoordinationcompounds:
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(i)
(ii)
(iii)
(iv)
(v)
(vi)
Ans.(i)Hexaamminecobalt(III)chloride
(ii)Pentaamminechloridocobalt(III)chloride
(iii)Potassiumhexacyanoferrate(III)
(iv)Potassiumtrioxalatoferrate(III)
(v)Potassiumtetrachloridopalladate(II)
(vi)Diamminechlorido(methanamine)platinum(II)chloride
3.Indicatethetypesofisomerismexhibitedbythefollowingcomplexesanddrawthe
structuresfortheseisomers:
(i)
(ii)
(iii)
(iv)
Ans.(i)Bothgeometrical(cis-,trans-)isomersfor canexist.Also,
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opticalisomersforcis-isomerexist.
Trans-isomerisopticallyinactive.Ontheotherhand,cis-isomerisopticallyactive.
(ii)Twoopticalisomersfor exist.
Twoopticalisomersarepossibleforthisstructure.
(iii)
Apairofopticalisomers:
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Itcanalsoshowlinkageisomerism.
and
Itcanalsoshowionizationisomerism.
(iv)Geometrical(cis-,trans-)isomersof canexist.
4. Give evidence that and are ionization
isomers.
Ans.Whenionizationisomersaredissolvedinwater,theyionizetogivedifferentions.These
ionsthenreactdifferentlywithdifferentreagentstogivedifferentproducts.
5. Explain on the basis of valence bond theory that ion with square
planar structure is diamagnetic and the ion with tetrahedral geometry is
paramagnetic.
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Ans.Niisinthe+2oxidationstatei.e.,in configuration.
There are 4 ions. Thus, it can either have a tetrahedral geometry or square planar
geometry. Since ion is a strong field ligand, it causes the pairing of unpaired 3d
electrons.
It now undergoes hybridization. Since all electrons are paired, it forms diamagnetic
compound.
Incaseof ,Cl-ionisaweakfieldligand.Therefore,itdoesnotleadtothepairing
ofunpaired3delectrons.Therefore,itundergoes hybridization.
Since there are 2 unpaired electrons in this case, due to weak field ligand so, it is
paramagneticinnature.
6. is paramagnetic while is diamagnetic though both are
tetrahedral.Why?
Ans.Though both and are tetrahedral, their magnetic characters
aredifferent.Thisisduetoadifferenceinthenatureofligandsleadingtoparamagnetism.
isaweakfieldligandanditdoesnotcausethepairingofunpaired3delectrons.Hence,
isparamagnetic.
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In ,Niisinthezerooxidationstatei.e.,ithasaconfigurationof .
ButCOisastrongfieldligand.Therefore,itcausesthepairingofunpaired3delectrons.Also,
itcausesthe4selectronstoshift to the3dorbital, therebygivingriseto hybridization.
Sincenounpairedelectronsarepresentinthiscase, isdiamagnetic.
7. is strongly paramagnetic whereas is weakly
paramagnetic.Explain.
Ans.Inboth and ,Feexistsinthe+3oxidationstatei.e.,in
configuration.
Since is a strong field ligand, it causes the pairing of unpaired electrons. Therefore,
thereisonlyoneunpairedelectronleftinthed-orbital.
Therefore,
=1.732BM
On the other hand, is aweak field ligand. Therefore, it cannot cause the pairing of
electrons.Thismeansthatthenumberofunpairedelectronsis5.
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Therefore,
=6BM
Thus, it is evident that is strongly paramagnetic, while is
weaklyparamagnetic.
8. Explain is an inner orbital complex whereas is an
outerorbitalcomplex.
Ans.
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9.Predictthenumberofunpairedelectronsinthesquareplanar ion.
Ans.
Inthiscomplex,Pt is inthe+2state. It formsasquareplanarstructure.Thismeansthat it
undergoes hybridization.Now,theelectronicconfigurationofPd(+2)is .
beingastrongfield ligandcauses thepairingofunpairedelectrons.Hence, thereare
nounpairedelectronsin .
10. The hexaquo manganese (II) ion contains five unpaired electrons, while the
hexacyanoioncontainsonlyoneunpairedelectron.ExplainusingCrystalFieldTheory.
Ans.
Hence,hexaaquomanganese (II) ionhas fiveunpairedelectrons,whilehexacyano ionhas
onlyoneunpairedelectron.Sopairingtakesplaceandlowspincomplexesareformedwith
this1unpairedelectron.
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11. Calculate the overall complex dissociation equilibrium constant for the
ion,giventhat forthiscomplexis
Ans. =
The overall complex dissociation equilibrium constant is the reciprocal of the overall
stabilityconstant, .
ChapterEndQuestion
1.ExplainthebondingincoordinationcompoundsintermsofWerner'spostulates.
Ans.Werner'spostulatesexplainthebondingincoordinationcompoundsasfollows:
(i) A metal exhibits two types of valencies namely, primary and secondary valencies.
Primaryvalenciesaresatisfiedbynegative ionswhilesecondaryvalenciesaresatisfiedby
bothnegativeandneutralions.
(Inmodern terminology, theprimaryvalency corresponds to theoxidationnumberof the
metal ion,whereas the secondary valency refers to the coordinationnumber of themetal
ion.
(ii)Ametalionhasadefinitenumberofsecondaryvalenciesaroundthecentralatom.Also,
thesevalenciesprojectinaspecificdirectioninthespaceassignedtothedefinitegeometry
ofthecoordinationcompound.
(iii)Primaryvalenciesareusuallyionizable,whilesecondaryvalenciesarenon-ionizable.
2. solutionmixedwith solutionin1:1molarratiogivesthetestof
ionbut solutionmixedwithaqueousammoniain1:4molarratiodoesnot
givethetestof ion.Explainwhy?
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Ans.
Both the compounds i.e., and fall
underthecategoryofadditioncompoundswithonlyonemajordifferencei.e.,theformeris
anexampleofadoublesalt,whilethelatterisacoordinationcompound.
Adoublesaltisanadditioncompoundthatisstableinthesolidstatebutthatwhichbreaks
up into its constituent ions in the dissolved state. These compounds exhibit individual
properties of their constituents. For e.g. breaks into
and ions.Hence,itgivesapositivetestfor ions.
Acoordinationcompoundisanadditioncompoundwhichretainsitsidentityinthesolidas
wellasinthedissolvedstate.However,theindividualpropertiesoftheconstituentsarelost.
Thishappensbecause doesnotshowthetestfor .Theions
presentinthesolutionof are and .
3. Explain with two examples each of the following: coordination entity, ligand,
coordinationnumber,coordinationpolyhedron,homolepticandheteroleptic.
Ans.(i)Coordinationentity:
A coordination entity is an electrically charged radical or species carrying a positive or
negativecharge.Inacoordinationentity,thecentralatomorionissurroundedbyasuitable
numberofneutralmoleculesornegativeions(calledligands).Forexample:
=cationiccomplex
=anioniccomplex
=neutralcomplex
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(ii)Ligands
The neutral molecules or negatively charged ions that surround the metal atom in a
coordinationentityoracoordinalcomplexareknownasligands.Forexample, ,
Cl-, OH- . Ligands are usually polar in nature and possess at least one unshared pair of
valenceelectrons.
(iii)Coordinationnumber:
Thetotalnumberofligands(eitherneutralmoleculesornegativeions)thatgetattachedto
thecentralmetalatominthecoordinationsphereiscalledthecoordinationnumberofthe
centralmetalatom.Itisalsoreferredtoasitsligancy.
Forexample:
(a) Inthecomplex, , thereas sixchloride ionsattached toPt in thecoordinate
sphere.Therefore,thecoordinationnumberofPtis6.
(b)Similarly,inthecomplex ,thecoordinationnumberofthecentralatom
(Ni)is4.
(vi)Coordinationpolyhedron:
Coordinationpolyhedronsaboutthecentralatomcanbedefinedasthespatialarrangement
oftheligandsthataredirectlyattachedtothecentralmetalioninthecoordinationsphere.
Forexample:
(a)
(b)Tetrahedral
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(v)Homolepticcomplexes:
These are those complexes in which themetal ion is bound to only one kind of a donor
group.Foreg: etc.
(vi)Heterolepticcomplexes:
Heterolepticcomplexesare thosecomplexeswhere thecentralmetal ion isboundtomore
thanonetypeofadonorgroup.
Fore.g.:
4.Whatismeantbyunidentate,didentateandambidentateligands?Givetwoexamples
foreach.
Ans.A ligandmay contain one ormore unshared pairs of electronswhich are called the
donorsitesof ligands.Now,dependingon thenumberof thesedonorsites, ligandscanbe
classifiedasfollows:
(a)Unidentateligands:Ligandswithonlyonedonorsitesarecalledunidentateligands.For
e.g., ,Cl-etc.
(b)Didentate ligands:Ligands thathave twodonorsitesarecalleddidentate ligands.For
e.g.,
(a)Ethane-1,2-diamine
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(b)Oxalateion
(c)Ambidentateligands:
Ligands that canattach themselves to thecentralmetalatom through twodifferentatoms
arecalledambidentateligands.Forexample:
(a)
(ThedonoratomisN)
(Thedonoratomisoxygen)
(b)
(ThedonoratomisS)
(ThedonoratomisN)
5.Specifytheoxidationnumbersofthemetalsinthefollowingcoordinationentities:
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(i)
(ii)
(iii)
(iv)
(v)
Ans.(i)
LettheoxidationnumberofCobex.
Thechargeonthecomplexis+2.
(ii)
LettheoxidationnumberofPtbex.
Thechargeonthecomplexis–2.
x+4(–1)=–2
x=+2
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(iv)
6.UsingIUPACnormswritetheformulasforthefollowing:
(i)Tetrahydroxozincate(II)
(ii)Potassiumtetrachloridopalladate(II)
(iii)Diamminedichloridoplatinum(II)
(iv)Potassiumtetracyanonickelate(II)
(v)Pentaamminenitrito-O-cobalt(III)
(vi)Hexaamminecobalt(III)sulphate
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(vii)Potassiumtri(oxalato)chromate(III)
(viii)Hexaammineplatinum(IV)
(ix)Tetrabromidocuprate(II)
(x)Pentaamminenitrito-N-cobalt(III)
Ans.(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
7.UsingIUPACnormswritethesystematicnamesofthefollowing:
(i)
(ii)
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(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
Ans.(i)Hexaamminecobalt(III)chloride
(ii)Diamminechlorido(methylamine)platinum(II)chloride
(iii)Hexaaquatitanium(III)ion
(iv)Tetraamminechloridonitrito-N-Cobalt(III)chloride
(v)Hexaaquamanganese(II)ion
(vi)Tetrachloridonickelate(II)ion
(vii)Hexaamminenickel(II)chloride
(viii)Tris(ethane-1,2-diammine)cobalt(III)ion
(ix)Tetracarbonylnickel(0)
8. List various types of isomerism possible for coordination compounds, giving an
exampleofeach.
Ans.
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(a)Geometricisomerism:
This typeof isomerism is common inheteroleptic complexes. It arisesdue to thedifferent
possiblegeometricarrangementsoftheligands.Forexample:
(b)Opticalisomerism:
Thistypeof isomerismarises inchiralmolecules. Isomersaremirrorimagesofeachother
andarenon-superimposable.
(c) Linkage isomerism: This type of isomerism is found in complexes that contain
ambidentateligands.Forexample:
and
YellowformRedform
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(d)Coordinationisomerism:
This type of isomerism arises when the ligands are interchanged between cationic and
anionicentitiesofdiffernetmetalionspresentinthecomplex.
and
(e)Ionizationisomerism:
Thistypeofisomerismariseswhenacounterionreplacesaligandwithinthecoordination
sphere. Thus, complexes that have the same composition, but furnish different ionswhen
dissolvedinwaterarecalledionizationisomers.Fore.g.,[Co(NH3)5SO4]Brand
[Co(NH3)5Br]SO4.
(f)Solvateisomerism:
Solvateisomersdifferbywhetherornotthesolventmoleculeisdirectlybondedtothemetal
ionormerelypresentasafreesolventmoleculeinthecrystallattice.
VioletBlue-greenDarkgreen
9.Howmanygeometricalisomersarepossibleinthefollowingcoordinationentities?
(i)
(ii)
Ans.(i)For ,nogeometricisomerispossibleasitisabidentateligand.
(ii)
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Twogeometricalisomersarepossible.
10.Drawthestructuresofopticalisomersof:
(i)
(ii)
(iii)
Ans.(i)
(ii)
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(iii)
11.Drawalltheisomers(geometricalandoptical)of:
(i)
(ii)
(iii)
Ans.(i)
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Intotal,threeisomersarepossible.
(ii)
Trans-isomersareopticallyinactive.
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Cis-isomersareopticallyactive.
(iii)
12.Write all the geometrical isomers of andhowmanyof
thesewillexhibitopticalisomers?
Ans.
From the above isomers, none will exhibit optical isomers. Tetrahedral complexes rarely
show optical isomerization. They do so only in the presence of unsymmetrical chelating
agents.
13.Aqueouscoppersulphatesolution(blueincolour)gives:
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(i)agreenprecipitatewithaqueouspotassiumfluoride,and
(ii)abrightgreensolutionwithaqueouspotassiumchloride
Explaintheseexperimentalresults.
Ans.Aqueous existsas .Itisblueincolourduetothepresence
of
ions.
(i)WhenKFisadded:
(ii)WhenKClisadded:
Inboththesecases,theweakfieldligandwaterisreplacedbythe and ions.
14.WhatisthecoordinationentityformedwhenexcessofaqueousKCNisaddedtoan
aqueoussolutionofcoppersulphate?Whyisitthatnoprecipitateofcoppersulphideis
obtainedwhen ispassedthroughthissolution?
Ans.
i.e.,
Thus,thecoordinationentityformedintheprocessis . is
averystablecomplex,whichdoesnotionizetogive ionswhenaddedtowater.Hence,
ionsarenotprecipitatedwhen ispassedthroughthesolution.
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15.Discussthenatureofbondinginthefollowingcoordinationentitiesonthebasisof
valencebondtheory:
(i)
(ii)
(iii)
(iv)
Ans.(i) In the above coordination complex, iron exists in the +II oxidation
state.
:Electronicconfigurationis
Orbitalsof ion:
As isastrongfieldligand,itcausesthepairingoftheunpaired3delectrons.
Sincetherearesix ligandsaroundthecentralmetal ion, themostfeasiblehybridizationis
.
hybridizedorbitalsof are:
6electronpairsfrom ionsoccupythesixhybrid orbitals.
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Then,
Hence,thegeometryofthecomplexisoctahedralandthecomplexisdiamagnetic(asthere
arenounpairedelectrons).
(ii)
Inthiscomplex,theoxidationstateofFeis+3.
Orbitalsof ion:
Thereare6F-ions.Thus,itwillundergo or hybridization.As isaweakfield
ligand, it does not cause the pairing of the electrons in the 3d orbital. Hence, the most
feasiblehybridizationis .
hybridizedorbitalsofFeare:
Hence,thegeometryofthecomplexisfoundtobeoctahedral.
(iii)
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Cobaltexistsinthe+3oxidationstateinthegivencomplex.
Orbitalsof ion:
Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the 3d orbital
electrons. As there are 6 ligands, hybridization has to be either or
hybridization.
hybridizationof :
The6electronpairsfromthe3oxalateions(oxalateanionisabidentateligand)occupythese
orbitals.
Hence,thegeometryofthecomplexisfoundtobeoctahedral.
(iv)
Cobaltexistsinthe+3oxidationstate.
Orbitalsof ion:
Again,fluorideionisaweakfieldligand.Itcannotcausethepairingofthe3delectrons.Asa
result,the ionwillundergo hybridization.
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hybridizedorbitalsof ionare:
Hence,thegeometryofthecomplexisoctahedralandparamagnetic.
16.Drawfiguretoshowthesplittingofdorbitalsinanoctahedralcrystalfield.
Ans.
The splitting of the d orbitals in an octahedral field takes palce in such a way that
experience a rise in energy and form the level, while and
experienceafallinenergyandformthe level.
ChapterEndQuestion
17.Whatisspectrochemicalseries?Explainthedifferencebetweenaweakfieldligand
andastrongfieldligand.
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Ans.Aspectrochemicalseriesisthearrangementofcommonligandsintheincreasingorder
oftheirfieldstrengthandcrystal-fieldsplittingenergy(CFSE)values.Theligandspresenton
theR.H.SoftheseriesarestrongfieldligandswhilethatontheL.H.Sareweakfieldligands.
Also, strong field ligands cause higher splitting in the d orbitals than weak field ligands.
Strongligandsareabletoproducestrongfieldwhereasweakligandsproduceweakfield.
Weak ligands can not cause pairing of electrons and form high spin complexes whereas
strongfieldligandscancausepairingofelectronsandthusformlowspincomplexes.
18. What is crystal field splitting energy? How does the magnitude of decide the
actualconfigurationofd-orbitalsinacoordinationentity?
Ans.Thedifferencebetweenenergiesoftwosetsofd-orbitalsiscalledcrystalfieldsplitting
energyi.e.deltanote.Thedegenerated-orbitals(inasphericalfieldenvironment)splitinto
twolevelsi.e., and inthepresenceofligands.Thesplittingofthesedegeneratelevels
duetothepresenceofligandsiscalledthecrystal-fieldsplittingwhiletheenergydifference
betweenthetwolevels( and )iscalledthecrystal-fieldsplittingenergy.Itisdenoted
by .
Aftertheorbitalshavesplit,thefillingoftheelectronstakesplace.After1electron(each)has
beenfilledinthethree orbitals,thefillingofthefourthelectrontakesplaceintwoways.
Itcanenterthe orbital(givingriseto likeelectronicconfiguration)orthepairing
of the electrons can take place in the orbitals (giving rise to like electronic
configuration). If the value of a ligand is less than the pairing energy (P), then the
electronsenterthe orbital.Ontheotherhand,ifthe valueofaligandismorethanthe
pairingenergy(P),thentheelectronsenterthe orbital.
Although orbital splitting energies are not sufficiently large to force pairing so low spin
configurationarerarelyobserved.
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19. isparamagneticwhile isdiamagnetic.Explainwhy?
Ans.Cr is in the+3oxidation state i.e., configuration.Also, is aweak field ligand
thatdoesnotcausethepairingoftheelectronsinthe3dorbital.
Therefore, it undergoes hybridization and the electrons in the 3d orbitals remain
unpaired.Hence,itisparamagneticinnature.
In ,Niexistsinthe+2oxidationstatei.e., configuration.
:
isa strong field ligand. It causes thepairingof the3d orbitalelectrons.Then,
undergoes hybridization.
Astherearenounpairedelectrons,itisdiamagneticwithsquareplanarshape.
20.Asolutionof isgreenbutasolutionof iscolourless.
Explain.
Ans. In , is a weak field ligand. Therefore, there are unpaired
electrons in . In this complex, the d electrons from the lower energy level can be
excited to the higher energy level i.e., the possibility ofd - d transition is present. Hence,
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light is absorbed from visible region and complementary green colour is radiated. So,
isgreencoloured.
In ,theelectronsareallpairedas isastrongfieldligand.Therefore,d-
dtransitionisnotpossiblein .Hence,itiscolourless.
21. and areofdifferentcoloursindilutesolutions.Why?
Ans.The colour of a particular coordination compound depends on themagnitude of the
crystal-fieldsplittingenergy, .ThisCFSE in turndependson thenatureof the ligand. In
caseof and ,thecolourdiffersbecausethereisadifference
intheCFSE.Now, isastrongfield ligandhavingahigherCFSEvalueascomparedto
theCFSEvalueofwater.Thismeansthattheabsorptionofenergyfromvisibleregionoccurs
fortheintrad-dtransitionandcomplementarycoloursareradiated.Hence,thetransmitted
colouralsodiffers.
22.Discussthenatureofbondinginmetalcarbonyls.
Ans.Themetal-carbonbondsinmetalcarbonylshaveboth and characters. bondis
formedwhenthecarbonylcarbondonatesalonepairofelectronstothevacantorbitalofthe
metal. A bond is formed by the donation of a pair of electrons from the filledmetal d
orbitalintothevacantanti-bonding orbital(alsoknownasbackbondingofthecarbonyl
group). The bond strengthens the bond and vice-versa. Thus, a synergic effect is
createdduetothismetal-ligandbonding.Thissynergiceffectstrengthensthebondbetween
COandthemetal.
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23. Give the oxidation state, d-orbital occupation and coordination number of the
centralmetalioninthefollowingcomplexes:
(i)
(ii)cis-
(iii)
(iv)
Ans.(i)
ThecentralmetalionisCo.
Itscoordinationnumberis6.
Theoxidationstatecanbegivenas:
x–6=–3
x=+3
Thedorbitaloccupationfor is .
(ii)cis-
ThecentralmetalionisCr.
Thecoordinationnumberis6.
Theoxidationstatecanbegivenas:
x+2(0)+2(–1)=+1
x–2=+1
x=+3
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Thedorbitaloccupationfor is .
(iii)
ThecentralmetalionisCo.
Thecoordinationnumberis4.
Theoxidationstatecanbegivenas:
x–4=–2
x=+2
Thedorbitaloccupationfor is .
(iv)
ThecentralmetalionisMn.
Thecoordinationnumberis6.
Theoxidationstatecanbegivenas:
x+0=+2
x=+2
ThedorbitaloccupationforMnis .
24.WritedowntheIUPACnameforeachof thefollowingcomplexesandindicatethe
oxidation state, electronic configuration and coordination number. Also give
stereochemistryandmagneticmomentofthecomplex:
(i)
(ii)
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(iii)
(iv)
(v)
Ans.(i)Potassiumdiaquadioxalatochromate(III)trihydrate.
Oxidationstateofchromium=3
Electronicconfiguration:
Coordinationnumber=6
Shape:octahedral
Stereochemistry:
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Magneticmoment,
(ii)
IUPACname:Pentaamminechloridocobalt(III)chloride
OxidationstateofCo=+3
Coordinationnumber=6
Shape:octahedral.
Electronicconfiguration: .
Stereochemistry:
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MagneticMoment=0
(iii)
IUPACname:Trichloridotripyridinechromium(III)
Oxidationstateofchromium=+3
Electronicconfigurationfor
Coordinationnumber=6
Shape:octahedral.
Stereochemistry:
Bothisomersareopticallyactive.Therefore,atotalof4isomersexist.
Magneticmoment,
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(iv)
IUPACname:Caesiumtetrachloroferrate(III)
OxidationstateofFe=+3
Electronicconfigurationof
Coordinationnumber=4
Shape:tetrahedral
Stereochemistry:opticallyinactive
Magneticmoment:
(v)
Potassiumhexacyanomanganate(II)
Oxidationstateofmanganese=+2
Electronicconfiguration:
Coordinationnumber=6
Shape:octahedral.
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Streochemistry:opticallyinactive
Magneticmoment,
=1.732
25.Whatismeantbystabilityofacoordinationcompoundinsolution?Statethefactors
whichgovernstabilityofcomplexes.
Ans.Thestabilityofacomplexinasolutionreferstothedegreeofassociationbetweenthe
two species involved ina stateof equilibrium.Stability canbeexpressedquantitatively in
termsofstabilityconstantorformationconstant.
Forthisreaction,thegreaterthevalueofthestabilityconstant,thegreateristheproportion
of inthesolution.
Stabilitycanbeoftwotypes:
(a)Thermodynamicstability:
Theextenttowhichthecomplexwillbeformedorwillbetransformedintoanotherspecies
atthepointofequilibriumisdeterminedbythermodynamicstability.
(b)Kineticstability:
Thishelpsindeterminingthespeedwithwhichthetransformationwilloccurtoattainthe
stateofequilibrium.
Factorsthataffectthestabilityofacomplexare:
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(a)Chargeonthecentralmetalion:Thegreater thechargeon thecentralmetal ion, the
greateristhestabilityofthecomplex.
(b)Basicnatureoftheligand:Amorebasicligandwillformamorestablecomplex.
(c)Presenceofchelaterings:Chelationincreasesthestabilityofcomplexes.
26.Whatismeantbythechelateeffect?Giveanexample.
Ans.Whenaligandattachestothemetalioninamannerthatformsaring,thenthemetal-
ligand association is found to bemore stable. In otherwords,we can say that complexes
containingchelateringsaremorestablethancomplexeswithoutrings.Thisisknownasthe
chelateeffect.
Forexample:
27.Discussbrieflygivinganexampleineachcasetheroleofcoordinationcompounds
in:
(i)biologicalsystem
(ii)medicinalchemistry
(iii)analyticalchemistry
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(iv)extraction/metallurgyofmetals
Ans.(i)Roleofcoordinationcompoundsinbiologicalsystems:
Weknowthatphotosynthesisismadepossiblebythepresenceofthechlorophyllpigment.
This pigment is a coordination compound ofmagnesium. In the humanbiological system,
several coordination compounds play important roles. For example, the oxygen-carrier of
blood,i.e.,haemoglobin,isacoordinationcompoundofiron.
(ii)Roleofcoordinationcompoundsinmedicinalchemistry:
Certaincoordinationcompoundsofplatinum(forexample,cis-platin)areusedforinhibiting
thegrowthoftumours.
(iii)Roleofcoordinationcompoundsinanalyticalchemistry:
During salt analysis, a number of basic radicals are detected with the help of the colour
changes they exhibit with different reagents. These colour changes are a result of the
coordinationcompoundsorcomplexesthatthebasicradicalsformwithdifferentligands.
(iii)Roleofcoordinationcompoundsinextractionormetallurgyofmetals:
Theprocess of extraction of someof themetals from their ores involves the formation of
complexes. For example, in aqueous solution, gold combines with cyanide ions to form
.Fromthissolution,goldislaterextractedbytheadditionofzincmetal.
28.Howmanyionsareproducedfromthecomplex insolution?
(i)6
(ii)4
(iii)3
(iv)2
Ans.(iii)Thegivencomplexcanbewrittenas .
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Thus, alongwithtwoCl-ionsareproduced.
29.Amongstthefollowingionswhichonehasthehighestmagneticmomentvalue?
(i)
(ii)
(iii)
Ans.(i)No.ofunpairedelectronsin =3
Then,
(ii)No.ofunpairedelectronsin =4
Then,
(iii)No.ofunpairedelectronsin
Hence, hasthehighestmagneticmomentvalue.
30.Theoxidationnumberofcobaltin is
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(i)+1
(ii)+3
(iii)-1
(iv)-3
Ans.WeknowthatCOisaneutralligandandKcarriesachargeof+1.
Therefore, the complex can be written as . Therefore, the oxidation
numberofCointhegivencomplexis-1.Hence,option(iii)iscorrect.
31.Amongstthefollowing,themoststablecomplexis
(i)
(ii)
(iii)
(iv)
Ans.We know that the stability of a complex increases by chelation. Therefore, themost
stablecomplexis .
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32. What will be the correct order for the wavelengths of absorption in the visible
regionforthefollowing:
Ans.Thecentralmetal ioninall thethreecomplexesis thesame.Therefore,absorptionin
thevisibleregiondependsontheligands.TheorderinwhichtheCFSEvaluesoftheligands
increasesinthespectrochemicalseriesisasfollows:
Thus,theamountofcrystal-fieldsplittingobservedwillbeinthefollowingorder:
Hence,thewavelengthsofabsorptioninthevisibleregionwillbeintheorder:
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