《基础化学》英文教学课件：chapter-11.ppt

1、Chapter 11.11-1.Basic Concepts 11-2.Valence Bond Theory of Complexes 11-3.Complex-Ion Equilibrium 11-4.Chelate and Biological Ligands 11-5.Medical Applications of ComplexesCo(NH3)5H2OCl3Co(NH3)5ClCl2CoCl2(H2O)2Co(H2O)6Cl21.Basic ConceptsCoordinate covalent bondA covalent bond in which both electrons

2、 come from one of the atoms is called.H+:NH3+:N:HHHH+A+BABA+BAB1.Basic ConceptsCuSO4NH3H2O Blue NH3H2ODark bluesolution White No BaCl2NaOHCuSO4+4NH3H2O=Cu(NH3)4SO4+4H2OCu2+4NH3=Cu(NH3)42+Ba2+SO42-=BaSO4 Cu2+2OH-=Cu(OH)2 little Cu2+1.Basic ConceptsZn2+2OH-=Zn(OH)2 Zinc hydroxide reacts with a strong

3、acid and the metal hydroxide dissolves:Zn(OH)2+2H+=Zn2+2H2OWith a base,Zn(OH)2 reacts to form a complex ion:Zn(OH)2+2OH-=Zn(OH)42-Zn2+NaOHZn(OH)2Zn(OH)42-1.Basic ConceptsI.Definition of coordination compound(e.g.,)is a cation(or an atom)with anion or neutral molecules attached to it through coordina

4、te covalent bonds.A is a compound consisting either of complex ion and ions of opposite charge(e.g.,the compound of the complex ion Cu(NH3)42+and SO42-ion)or of a neutral complex species(e.g.,Fe(CO)5).Cu NH3NH3NH3H3N2+1.Basic ConceptsII.Composition of coordination compoundCu(NH3)42+SO42-Ligating ato

5、m Number of ligandsCentral atomLigandOuter sphereInner sphereCoordination compound1.Basic Concepts1.Central atom()resides at the center of a coordination compound.Generally,the central atom is transition metal cation(NH3)42+，(OH)42-)or neutral atom(CO)5),which have unoccupied orbital to accept elect

6、ron pairs from ligands.Cu NH3NH3NH3H3NF62-1.Basic Concepts2.Ligand()are neutral molecules or anions that attached to the central atom via coordinate covalent bond,they are electron-pair donors.The electron-pair donating atom in a ligand is called().(H3)42+，(H)42-，(O)5。C:.O:.:+C:O:C:.C:.O:.:O:.:+C:O:

7、C:O:O:thiocyanate(硫氰根硫氰根)isothiocyanate(异硫氰根异硫氰根):SCN:CN:S1.Basic ConceptsCations only rarely function as ligands because an electron pair on a cation is held securely by the positive charge.A cation in which the positive charge is far removed from an electron pair that could be donated can function

8、 as a ligand.An example is the pyrazinium ion:1.Basic Concepts is a ligand that has ligating atom(or a ligand that bonds to a central atom through atom of the ligand).For example,NH3，H2O，X-，N-，SCN-。Cu NH3NH3NH3H3N1.Basic Concepts is a ligand that has ligating atoms(or a ligand that can bond with two

9、 or more atoms to a central atom).Ethylenediamine(en)1.Basic ConceptsEthylenediaminetetraacetate(edta)1.Basic Concepts1.Basic Concepts3.Coordination number()The coordination number of a central atom is the the central atom forms with ligands.Very high coordination numbers(9 to 12)are known for some

10、complex ions of the.1.Basic Concepts4.Charge on a complex ionThe charge on a complex ion equals the of the charges on the ions(central atom and the ligands)from which it is formed.Cu(NH3)42+HgI42-Pt(NH3)4Cl2Cl2Pt(NH3)2Cl2K2PtCl61.Basic ConceptsCu(NH3)42+SO42-Ligating atomNumber of ligandsCentral ato

11、mLigandOuter sphereInner sphereCoordination compound1.Basic ConceptsExample 11-1:For coordination compound KFe(en)Cl2Br2,please point out the central atom and its oxidation number,the ligands,the ligating atoms,the number of ligands,the coordination number,the charge on the complex ion,and the outer

12、 sphere.Solution:Central atom:Ligands:Ligating atoms:Number of ligands:Coordination number:Fe(III);en,Cl-,Br-;N,Cl,Br;5;6;Charge on the complex ion:-1;Outer sphere:K+.1.Basic ConceptsIII.Naming coordination compoundsCu(NH3)42+命名顺序：命名顺序：1.五羰基合铁（五羰基合铁（0）四氨合铜（四氨合铜（II）配离子配离子Ag(NH3)2+二氨合银（二氨合银（I）配离子配离子Cu

13、(en)22+二（乙二胺）合铜（二（乙二胺）合铜（II）配离子配离子Fe(CN)63-六氰合铁（六氰合铁（III）配离子配离子Pt(Cl)62-六氯合铂（六氯合铂（IV）配离子配离子Fe(CO)51.Basic Concepts硫酸四氨合铜（硫酸四氨合铜（II）是阴离子在前、阳离子在后，像是阴离子在前、阳离子在后，像一般无机化合物中的二元化合物、酸、碱和盐一般无机化合物中的二元化合物、酸、碱和盐一样命名为一样命名为“”、“”、“”和和“”。2.Ag(NH3)2ClAg(NH3)2 OHH2Pt(Cl)6Cu(NH3)4SO4氯化二氨合银（氯化二氨合银（I）氢氧化二氨合银（氢氧化二氨合银（I）六

14、氯合铂（六氯合铂（IV）酸酸1.Basic Concepts3.在配合物中若有在配合物中若有时，不同配体之间用圆点时，不同配体之间用圆点“”分开，配体顺序为：分开，配体顺序为：Pt(NH3)2Cl2氯化二氯化二 氨氨二二(乙二胺乙二胺)合钴（合钴（III）二氯二氯三氨三氨水合钴（水合钴（III）配离子配离子Co(NH3)2(en)2Cl3二氯二氯二氨合铂（二氨合铂（II）Co(NH3)3(H2O)Cl2-IV.Geometric isomerism of coordination compoundsCoordnation IsomersLinkage Isomers(键联异构体键联异构体)Ge

15、ometric Isomers(几何异构体几何异构体)Optical Isomers(光学异构体光学异构体)are isomers that differ in how the atoms are joined togetherthat is,in the order in which the atoms are bonded to each other.Pt(NH3)4Cl2(NO2)2Pt(NH3)4(NO2)2Cl2Coordnation Isomers1.Basic ConceptsCo(NH3)5(SO4)BrCo(NH3)5BrSO4Co(NH3)5(NO2)Cl2Linkage

16、Isomers(键联异构体键联异构体)1.Basic Concepts1.Basic Concepts are isomers in which the atoms are joined to one another in the same way but differ because some atoms occupy different relative positions in space.Co(NH3)4Cl2Cl The cis compound is purple;The trans compound is green.Pt(NH3)2Cl21.Basic Concepts1.In

17、 the formation of a coordinate covalent bond in a complex,a ligand orbital containing overlaps an on the central atom.2.Valence Bond Theory of ComplexesI.Valence bond theory of complex2.To bond electron pairs from several ligating atoms,forming several equivalent bonds,the same number of from the ce

18、ntral atom are required.2.Valence Bond Theory of Complexes Only atomic orbitals with from the isolated atom can be combined to give hybrid orbitals,in which the.The reorientation of the hybrid orbitals is favorable for the during covalent bond formation.The reorientation of the hybrid orbitals minim

19、izes the repulsion between bonding electron pairs,thus the resulting covalent bonds are.3.The spatial configuration of complex is determined by the number and type of hybrid orbitals that the central atom provided.2.Valence Bond Theory of Complexes2.Valence Bond Theory of ComplexesZn(OH)42-Tetrahedr

20、al2.Valence Bond Theory of ComplexesCr(NH3)63+Octahedral Complexes2.Valence Bond Theory of ComplexesSquare Planar ComplexesNi(CN)42-2.Valence Bond Theory of Complexes2.Valence Bond Theory of ComplexesII.ExamplesFe(H2O)63+Fe(CN)63-To bond electron pairs from six ligands to Fe3+,forming six equivalent

21、 bonds,octahedral hybrid orbitals are required.These hybrid orbitals will use two d orbitals,the 4s orbital,and the three 4p orbitals.The d orbitals could be either 3d or 4d,depending on the valence electron configuration of the central atom and the nature of ligands.Fe:Ar3d64s2Fe3+:Ar3d5Fe3+3d4s4pA

22、r4d2.Valence Bond Theory of ComplexesFe(CN)63-3d4s4pAr4dd2sp33d4s4pAr4dsp3d2Fe(H2O)63+2.Valence Bond Theory of ComplexesIII.Paramagnetism and magnetic moment substances are attracted to a strong magnetic field.Paramagnetism is due to in a substance.The magnitude of the paramagnetism can be measured

23、with a,in which the force of magnetic attraction is balanced with weights.Gouy balance2.Valence Bond Theory of ComplexesB2)n(nMagnetic moment()n:number of unpaired electrons;B:unit of magnetic moment,called Bohr magneton().2.Valence Bond Theory of ComplexesIV.Crystal field theory and the color of co

24、mplexesM(H2O)6n+2.Valence Bond Theory of Complexes1.Fe(SCN)(H2O)52+2.Co(SCN)4(H2O)22-3.Cu(NH3)4(H2O)22+4.CuBr42-12342.Valence Bond Theory of ComplexesThe Logan Sapphire is displayed in the Smithsonian Museum of Natural History.At 423 carats(),this blue sapphire is the largest one on public display.T

25、he colors of many gemstones are due to transition-metal-ion impurities in the mineral(alumina,Al2O3).is alumina with and impurities.2.Valence Bond Theory of ComplexesThis ruby and diamond bracelet contains 31 matched Burmese rubies with a total of 60 carats.The bracelet is on display at the Smithson

26、ian Museum of Natural History.is alumina with in place of some Al3+ions.2.Valence Bond Theory of ComplexesThis extraordinary topaz gem is labeled 22,892.5 carats.It is displayed in the Smithsonian Museum of Natural History.It has a mass of 4.6 kg and measures about 15 cm across.is alumina with impur

27、ities.3.Complex-Ion EquilibriaCu2+4NH3 Cu(NH3)42+Cu2+4NH3 Cu(NH3)42+stability constant of a complex ion,which is the equilibrium constant for the formation of the complex ion from the aqueous metal ion and the ligands.I.Stability constant()of a complex ion3.Complex-Ion EquilibiaWhen a large amount o

28、f ligand is added to a solution of metal ion,you expect most of the metal ion to react to form the complex ion.The stability constant can be used to compare the stability of complex ion only if the number of ligands is the same for both complex ions.Generally,Kshas a quite large value,which means th

29、at the complex ion is quite stable.3.Complex-Ion EquilibiaExample 11-2:What is the concentration of Ag+(aq)ion in 0.010 molL-1 AgNO3 that is also 1.00 molL-1 NH3?Ks for Ag(NH3)2+ion is 1.7107.The stability constant for Ag(NH3)2+is large,so silver exists primarily as this ion.This suggest that you do

30、 this problem in two parts.First you do the stoichiometry calculation,in which you assume that Ag+(aq)reacts completely to form Ag(NH3)2+(aq).Then you do the equilibrium calculation,in which this complex ion dissociate to give a small amount of Ag+(aq).Problem strategy:3.Complex-Ion EquilibiaSolutio

31、n:Ag+2NH3 Ag(NH3)2+0.010 1.00Ag(NH3)2+Ag+2NH30.010 0.98Stoichiometry calculation:In 1 L of solution,you initially have 0.010 mol Ag+(aq).This reacts to give 0.010 mol Ag(NH3)2+,leaving(1.00-20.010)mol NH3,which equals 0.98 mol NH3.You now look at the equilibrium for the dissociation of Ag(NH3)2+.3.C

32、omplex-Ion EquilibiaEquilibrium calculation:Conc.(molL-1)Ag(NH3)2+(aq)Ag+(aq)+2NH3(aq)Starting 0.0100 0.98Change -x +x +2xEquilibrium 0.010-xx 0.98+2xSTEP 1 One liter of the solution contains 0.010 mol Ag(NH3)2+and 0.98 mol NH3.The complex ion dissociates slightly,so that 1 L of solution contains x

33、mol Ag+.These data are summarized in the table:3.Complex-Ion EquilibiaSTEP 2 The stability constant isSubstituting into this equation gives3.Complex-Ion EquilibiaSTEP 3 The right-hand side of the equation equals to 5.910-8.If you assume x to be small compared with 0.010,ThenThe silver ion concentrat

34、ion is 6.110-10 molL-1(over 10 million times smaller than its value in 0.010 molL-1 AgNO3 that does not contain ammonia).3.Complex-Ion EquilibiaConc.(molL-1)Ag+(aq)+2NH3(aq)Ag(NH3)2+(aq)Starting 0.01 1.00 0Change -x -2x xEquilibrium 0.01-x 1.00-2x x3.Complex-Ion EquilibiaII.Shift of complex-ion equi

35、libriumAcidity of solution;Precipitate agents Oxidizing or reducing agents;Other ligands.H.Le ChatelierLe Chateliers principle():When a stress(concentration,pressure,temperature)is applied to a system in dynamic equilibrium,the equilibrium tends to adjust to minimize the effect of the stress.3.Compl

36、ex-Ion Equilibia1.Acidity of solution3.Complex-Ion Equilibia2.Precipitate agents3.Complex-Ion Equilibia3.Oxidizing or reducing agents4.Other ligandsAg(NH3)2+2CN-Ag(CN)2+2NH33.Complex-Ion EquilibiaAg(NH3)2+2CN-Ag(CN)2+2NH3Ag+2CN-Ag(CN)2+Ag+2NH3 Ag(NH3)2+3.Complex-Ion Equilibia Predicting whether a pr

37、ecipitate will form in the presence of the complex ionExample 11-3:a.Will silver chloride precipitate from a solution that is 0.010 molL-1 AgNO3 and 0.010 molL-1 NaCl?b.Will silver chloride precipitate from this solution if it is also 1.00 molL-1 NH3?Part a.is a simple precipitation problem in which

38、 you compare Ip with Ksp.Part b.differs only in that you need to know the Ag+concentration in equilibrium with a complex ion.Problem strategy:3.Complex-Ion EquilibiaSolution:a.To determine whether a precipitate should form,you calculate the ion product Ip and compare it with Ksp for AgCl(1.810-10).T

39、his is greater than Ksp=1.810-10,so a precipitate should form.3.Complex-Ion Equilibiab.You first need to calculate the concentration of Ag+(aq)in a solution containing 1.00 molL-1 NH3.We did this in Example 11-2 and found that Ag+equals 6.110-10.Hence,Because the ion product is smaller than Ksp,no p

40、recipitate should form.3.Complex-Ion Equilibia Calculating the solubility of a slightly soluble ionic compound in a solution of the complex ion.Example 11-4:Calculate the solubility of AgCl in 1.0 molL-1 NH3.Problem strategy:This problem involves the solubility equilibrium for AgCl,in addition to th

41、e complex-ion equilibrium.As AgCl dissolves to give ions,the Ag+ion reacts with NH3 to give the complex ion Ag(NH3)2+.The equilibria are:AgCl(s)Ag+(aq)+Cl-(aq)Ag+(aq)+2NH3(aq)Ag(NH3)2+(aq)3.Complex-Ion EquilibiaSolution:AgCl(s)Ag+(aq)+Cl-(aq)Ag+(aq)+2NH3(aq)Ag(NH3)2+(aq)The equilibrium constant for

42、the overall equation isYou obtain the overall equation for the process by adding the solubility and complex-ion equilibria.AgCl(s)+2NH3(aq)Ag(NH3)2+(aq)+Cl-(aq)3.Complex-Ion Equilibia3.Complex-Ion EquilibiaYou now solve the equilibrium problem.Substituting into the equilibrium-constant equationConc.

43、(molL-1)AgCl(s)+2NH3(aq)Ag(NH3)2+(aq)+Cl-(aq)Starting1.00 0Change-2x+x +xEquilibrium1.0-2xx xgives3.Complex-Ion EquilibiaYou solve this equation by taking the square root of both sides,Rearranging,Hence,3.Complex-Ion EquilibiaNote that the solubility of AgCl equals the concentration of silver in the

44、 solution.Because the stability of Ag(NH3)2+,most of the silver in solution will be in the form of this complex ion.Therefore,.When Ag+(aq)reacts to give the complex ion,more AgCl dissolves to partially replenish the Ag+(aq)ion.Therefore,AgCl is more soluble in aqueous ammonia than in pure water.4.C

45、helate and Biological LigandsI.Chelate()A complex formed by polydentate ligands is called.The polydentate ligands are called.The term chelate is derived from the Greek chele for“claw”,because a polydentate ligand appears to attach itself to the metal atom like a crab claws to some object.Co(en)33+4.

46、Chelate and Biological LigandsChelating ligand EDTA is a hexadentate ligand,it forms 5 chelating cycles with metal ion,and the coordination number is 6.4.Chelate and Biological Ligands As a result,chelating agents are often used to remove metal ions from a chemical system.EDTA,for example,is added t

47、o certain canned food to remove transition-metal ions that can catalyze the deterioration of the food.II.Stability of ChelateSize of chelating cyclesNumber of chelating cycles(number of ligating atoms)Ni(NH3)62+Ks=5.5 108Ni(en)32+Ks=2.0 1018Factors affecting the stability of chelate:4.Chelate and Bi

48、ological LigandsIII.Biological ligands()The hemoglobin molecule in red blood cells is an example of a complex with a quadridentate ligand.Hemoglobin consists of the protein globin chemically bonded to heme.Heme is a planarmolecule consisting of iron(II)to which a quadridentate ligand is bonded throu

49、gh its four nitrogen atoms.Heme5.Medical Applications of ComplexesI.Maintaining Physiological functionsHemoglobin,the molecule in red blood cells that is responsible for the transport of oxygen,O2,from the lungs to other body tissues.5.Medical Applications of ComplexesII.Antidotes()has been used to

50、treat lead poisoning because it binds Pb2+ions as the chelate,forming a substance that can then be excreted by the kidneys.HCCOHHSHHSH5.Medical Applications of ComplexesIII.Complexes as anticancer drugsThe platinum-containing substance,development by Barnett Rosenberg at MSU,is a leading anticancer