《基础化学》英文教学课件：chapter-11

1、Chapter 11. Coordination Compounds 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 covalen

2、t bond in which both electrons come from one of the atoms is called coordinate covalent bond(配位共价键).H+:NH3+:N:HHHH+1. Basic ConceptsCuSO4NH3H2O Blue NH3H2ODark bluesolution White No BaCl2NaOHCuSO4 + 4NH3H2O = Cu(NH3)4SO4 + 4H2OCu2+ + 4NH3 = Cu(NH3)42+ExperimentBa2+ + SO42- = BaSO4 Cu2+ + 2OH- = Cu(O

3、H)2 little Cu2+1. Basic ConceptsExperiment 2Zn2+ + 2OH- = Zn(OH)2 Zinc hydroxide reacts with a strong 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-Amphoteric hydroxideZn2+NaOHZn(OH)2Zn(OH)42-1. Basic Con

4、ceptsI. Definition of coordination compoundA complex ion (molecule)(配离子或配位分子) (e.g., Cu(NH3)42+, Pt(NH3)2Cl2) is a cation (or an atom) with anion or neutral molecules attached to it through coordinate covalent bonds. A coordination compound (or complex)(配位化合物) is a compound consisting either of comp

5、lex ion and ions of opposite charge (e.g., the compound Cu(NH3)4SO4 of the complex ion Cu(NH3)42+ and SO42- ion) or of a neutral complex species (e.g., Fe(CO)5).2+1. Basic ConceptsII. Composition of coordination compoundCu(NH3)42+SO42-Ligating atom配位原子 Number of ligands配位数Central atom中心原子Ligand配体Out

6、er sphere外层Inner sphere内层Coordination compound配位化合物1. Basic Concepts1. Central atom(中心原子)A Central atom resides at the center of a coordination compound. Generally, the central atom is transition metal cation (Cu(NH3)42+， Zn(OH)42-) or neutral atom (Fe(CO)5), which have unoccupied orbital to accept

7、electron pairs from ligands.SiF62-1. Basic Concepts2. Ligand(配体)Ligands 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 ligating atom(配位原子). Cu(NH3)42+，Zn(OH)42-，Fe(CO)

8、5。thiocyanate(硫氰根)isothiocyanate(异硫氰根)Why ?1. 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 as a ligand.

9、An example is the pyrazinium ion:1. Basic ConceptsA monodentate ligand(单齿配体, meaning “one-toothed”) is a ligand that has one ligating atom (or a ligand that bonds to a central atom through one atom of the ligand). For example, NH3，H2O，X-，CN-，SCN-。1. Basic ConceptsA polydentate ligand(多齿配体, “having m

10、ore teeth”) is a ligand that has two or more ligating atoms (or a ligand that can bond with two or more atoms to a central atom).Ethylenediamine (en)1. Basic ConceptsEthylenediaminetetraacetate (edta)1. Basic Concepts1. Basic Concepts3. Coordination number(配位数)The coordination number of a central at

11、om is the total number of bonds the central atom forms with ligands.Very high coordination numbers (9 to 12) are known for some complex ions of the lanthanide elements (镧系元素).1. Basic Concepts4. Charge on a complex ionThe charge on a complex ion equals the sum of the charges on the ions (central ato

12、m and the ligands) from which it is formed.Cu(NH3)42+HgI42-Pt(NH3)4Cl2Cl2Pt(NH3)2Cl2K2PtCl61. Basic ConceptsCu(NH3)42+SO42-Ligating atom配位原子Number of ligands配位数Central atom中心原子Ligand配体Outer sphere外层Inner sphere内层Coordination compound配位化合物1. Basic ConceptsExample 11-1: For coordination compound KFe(e

13、n)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 sphere.Solution:Central atom:Ligands:Ligating atoms:Number of ligands:Coordination number:Fe(III);en, Cl

14、-, Br-;N, Cl, Br;5;6;Charge on the complex ion:-1;Outer sphere:K+.1. Basic ConceptsIII. Naming coordination compoundsCu(NH3)42+配离子及配位分子命名顺序：配位体数（中文数字）配位体名称合中心原子的名称和电荷数（罗马数字加括号）1. 五羰基合铁（0）四氨合铜（II）配离子Ag(NH3)2+二氨合银（I）配离子Cu(en)22+二（乙二胺）合铜（II）配离子Fe(CN)63-六氰合铁（III）配离子Pt(Cl)62-六氯合铂（IV）配离子Fe(CO)51. Basic Co

15、ncepts硫酸四氨合铜（II）配合物的命名是阴离子在前、阳离子在后，像一般无机化合物中的二元化合物、酸、碱和盐一样命名为“某化某”、“某酸”、“氢氧化某”和“某酸某”。2. Ag(NH3)2ClAg(NH3)2 OHH2Pt(Cl)6Cu(NH3)4SO4氯化二氨合银（I）氢氧化二氨合银（I）六氯合铂（IV）酸1. Basic Concepts3. 在配合物中若有多种配体时，不同配体之间用圆点“”分开，配体顺序为：先无机配体，后有机配体；先阴离子配体，后中性分子；若配体均为阴离子或中性分子时，按配位原子元素符号的英文字母的顺序排列。Pt(NH3)2Cl2氯化二 氨二(乙二胺)合钴（III）二

16、氯三氨水合钴（III）配离子Co(NH3)2(en)2Cl3二氯二氨合铂（II）Co(NH3)3(H2O)Cl2-IV. Geometric isomerism of coordination compoundsCoordnation Isomers(配位异构体)Linkage Isomers(键联异构体)Geometric Isomers(几何异构体)Optical Isomers(光学异构体) Constitutional isomers (结构异构) are isomers that differ in how the atoms are joined togetherthat is,

17、in the order in which the atoms are bonded to each other. Pt(NH3)4Cl2(NO2)2Pt(NH3)4(NO2)2Cl2Coordnation Isomers(配位异构体)1. Basic ConceptsCo(NH3)5(SO4)BrCo(NH3)5BrSO4Co(NH3)5(NO2)Cl2Linkage Isomers(键联异构体)1. Basic Concepts1. Basic ConceptsGeometric isomers(几何异构) are isomers in which the atoms are joined

18、 to one another in the same way but differ because some atoms occupy different relative positions in space.Co(NH3)4Cl2Cl Tetraamminedichlorocobalt(III)chlorideThe cis compound is purple;The trans compound is green.Pt(NH3)2Cl21. Basic Concepts1. In the formation of a coordinate covalent bond in a com

19、plex, a ligand orbital containing two electrons overlaps an unoccupied orbital 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 equivalent hybrid orbita

20、ls from the central atom are required.2. Valence Bond Theory of ComplexesOutline of hybrid orbitals Only atomic orbitals with approximate energy from the isolated atom can be combined to give hybrid orbitals, in which the energy and orientation of the orbitals are redistributed. The reorientation of

21、 the hybrid orbitals is favorable for the maximum overlap during covalent bond formation. The reorientation of the hybrid orbitals minimizes the repulsion between bonding electron pairs, thus the resulting covalent bonds are more stable.3. The spatial configuration of complex is determined by the nu

22、mber and type of hybrid orbitals that the central atom provided. 2. Valence Bond Theory of Complexes2. Valence Bond Theory of ComplexesZn(OH)42-Tetrahedral2. Valence Bond Theory of ComplexesCr(NH3)63+Octahedral Complexes2. Valence Bond Theory of ComplexesSquare Planar ComplexesNi(CN)42-2. Valence Bo

23、nd 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 bonds, octahedral hybrid orbitals are required. These hybrid orbitals will use two d orbitals, the 4s orbital, and the three 4p orbitals. Th

24、e 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+3d4s4pAr4d2. Valence Bond Theory of ComplexesFe(CN)63-3d4s4pAr4dd2sp33d4s4pAr4dsp3d2Fe(H2O)63+2. Valence Bond Theory of ComplexesIII. Parama

25、gnetism and magnetic momentParamagnetic(顺磁性的) substances are attracted to a strong magnetic field. Paramagnetism is due to unpaired electrons in a substance. The magnitude of the paramagnetism can be measured with a Gouy balance, in which the force of magnetic attraction is balanced with weights.Gou

26、y balance2. Valence Bond Theory of ComplexesMagnetic moment(磁矩) n: number of unpaired electrons;B: unit of magnetic moment, called Bohr magneton(玻尔磁子).2. Valence Bond Theory of ComplexesIV. Crystal field theory and the color of complexesM(H2O)6n+2. Valence Bond Theory of Complexes1. Fe(SCN)(H2O)52+2

27、. 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(克拉), this blue sapphire is the largest one on public display. The colors of many gemstones are due to transition-metal-ion

28、 impurities in the mineral (alumina, Al2O3).Sapphire(蓝宝石) is alumina with Fe3+ and Ti4+ 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 Smithsonian Museum of Natural History.

29、Ruby(红宝石) is alumina with Cr3+ 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.Topaz(黄玉) is alumina with Fe3+ i

30、mpurities.3. Complex-Ion EquilibriaCu2+ + 4NH3 = Cu(NH3)42+Cu2+ + 4NH3 Cu(NH3)42+Ks: 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(稳定常数) of a complex ion3. Complex-Ion Equil

31、ibiaWhen a large amount of ligand is added to a solution of metal ion, you expect most of the metal ion to react to form the complex ion.Note: 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.CN-Generally, Ks h

32、as a quite large value, which means that 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

33、primarily as this ion. This suggest that you do 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 A

34、g+ (aq).Method 1Problem strategy:3. Complex-Ion EquilibiaSolution: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

35、 mol NH3. You now look at the equilibrium for the dissociation of Ag(NH3)2+. 3. Complex-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+

36、and 0.98 mol NH3. The complex ion dissociates slightly, so that 1 L of solution contains x 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 e

37、quals to 5.910-8. If you assume x to be small compared with 0.010, ThenThe silver ion concentration 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

38、 0.01 1.00 0Change -x -2x xEquilibrium 0.01-x 1.00-2x xMethod 23. Complex-Ion EquilibiaII. Shift of complex-ion equilibriumAcidity of solution; Precipitate agents; Oxidizing or reducing agents; Other ligands.H. Le ChatelierLe Chateliers principle(勒夏特列原理): When a stress (concentration, pressure, temp

39、erature) is applied to a system in dynamic equilibrium, the equilibrium tends to adjust to minimize the effect of the stress.3. Complex-Ion Equilibia1. Acidity of solutionAcid effectHydrolytic effect(水解作用)3. Complex-Ion Equilibia2. Precipitate agentsKs vs. Ksp3. Complex-Ion Equilibia3. Oxidizing or

40、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 precipitate will form in the presence of the complex ionExample 11-3: a. Will silver chloride pr

41、ecipitate 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 you compare Ip with Ksp. Part b. differs only in that you need to know the Ag+ concentra

42、tion 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).This is greater than Ksp=1.810-10, so a precipitate should form.3. Complex-Ion Equi

43、libiab. 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 precipitate should form.3. Complex-Ion Equilibia Calculating the solubilit

44、y 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 the complex-ion equilibrium. As AgCl dissolves to give ions, the Ag+ ion

45、 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 the overall equation isYou obtain the overall equat

46、ion 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. (molL-1) AgCl(s) + 2NH3(aq) Ag(NH3)2+ (aq

47、) + 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 solution. Because the stabili

48、ty of Ag(NH3)2+, most of the silver in solution will be in the form of this complex ion. Therefore, the solubility of AgCl equals 0.050 molL-1.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 t

49、han in pure water.4. Chelate and Biological LigandsI. Chelate(螯合物)A complex formed by polydentate ligands is called chelate.The polydentate ligands are called chelating agents(螯合剂).The term chelate is derived from the Greek chele for “claw”, because a polydentate ligand appears to attach itself to t

50、he metal atom like a crab claws to some object.Co(en)33+4. 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 LigandsChelate is generally quite stable. As a result, chela

51、ting agents are often used to remove metal ions from a chemical system. EDTA, for example, is added to 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 a

52、toms)Ni(NH3)62+Ks = 5.5 108Ni(en)32+Ks = 2.0 1018Factors affecting the stability of chelate:4. Chelate and Biological LigandsIII. Biological ligands(生物配体)The hemoglobin molecule in red blood cells is an example of a complex with a quadridentate ligand. Hemoglobin consists of the protein globin chemi

53、cally bonded to heme. Heme is a planarmolecule consisting of iron(II) to which a quadridentate ligand is bonded through its four nitrogen atoms.Heme5. Medical Applications of ComplexesI. Maintaining Physiological functionsCOHemoglobin, the molecule in red blood cells that is responsible for the transport of oxygen, O2, fr