Reference-answers(参考答案).doc

The reference answers for the exercises of organometallics asigned已布置金属有机化学习题参考答案The First Asignment（第一次作业）1. What was the first olefin complex?（第一个烯烃络合物是什么？）Answer: The first olefin complex is Zeises salt NaPtCl3C2H4答案：第一个烯烃络合物是蔡斯盐NaPtCl3C2H4。2. In what year did P. Ehrlich won Nobel prize and what was his invention?（P额尔利屈获得诺贝尔奖？他的发明是什么？）Answer: P. Ehrlich won Nobel prize in the year of 1908 for his invention of Salvarsan, a medicin for the treatment of syphilis.答案：P额尔利屈在1908年获得诺贝尔奖，他的发明是治疗梅毒的药物洒尔佛散。3. What was K. Ziegler and G. Nattas major discovery?（K齐格勒和G纳塔的主要发现是什么？）Answer: K. Ziegler, G. Natta fund ways to manufacture polyolefins (聚烯烃) from ethylene and propylene, respectively, in a low pressure process employing mixed metal catalysts (transition-metal halide/ AlR3).答案：K. 齐格勒，G. 纳塔找到了应用混合金属催化剂（过渡-金属卤化物/AlR3）在低压过程中分别从乙烯和丙烯制造聚烯烃的途径。4. What was G. Wilkinsons discovery?（G. 威尔金森的发现是什么？）Answer: G. Wilkinson discovered that (PPh3)3RhCl acts as a homogeneous (均相的) catalyst in the hydrogenation of alkenes(烯烃).答案：G. 威尔金森发现(PPh3)3RhCl在烯烃的催化氢化中充当均相催化剂。5. What did W. N. Lipscomb win Nobel prize for?（W. N.李普斯考姆何以获得诺贝尔奖？）Answer: W. N. Lipscomb won Nobel prize for the theoretical and experimental clarification of structure and bonding in boranes.答案：W. N. 李普斯考姆获得诺贝尔奖是因为对硼烷结构和成键所作的理论和试验澄清。6. For what did G. Wittig win Nobel prize?（G. 维悌希何以获得诺贝尔奖？）Answer: H. C. Brown and G. Wittig won Nobel prize for the application of organoboranes and methylenephosporanes, respectively, in organic synthesis.答案：H. C. 布朗和G. 维悌希获得诺贝尔奖是由于分别将有机硼烷和甲烯磷烷应用于有机合成之中。7. What were R. Hoffman and K. Fukui awarded Nobel prize for?（R. 霍夫曼和K. 福井谦一何以获得诺贝尔奖？）Answer: R. Hoffman and K. Fukui were awarded Nobel Prize for the semiempirical MO-concept in a unified discussion of structure and reactivity of inorganic, organic and organometallic molecules, isolobal analogies.答案：R. 霍夫曼和K. 福井谦一获得诺贝尔奖是由于半经验分子轨道概念对无机、有机、金属有机分子的结构和反应性的统一讨论，即等叶类比。8. What project R. G. Bergman and W. A. G. Graham work for? （R. G. 波戈曼和W. A. G. 格雷姆从事什么项目？）Answer: R. G. Bergman and W. A. G. Graham work for intermolecular reactions of organotransition-metal compounds with alkanes (C-H activation)答案：R. G. 波戈曼和W. A. G. 格雷姆从事有机过渡-金属化合物与烷烃的分子间反应工作（即碳-氢键活化）。9. What have William S. Knowles and RyoJi Noyori accomplished for their Nobel prize?（威廉姆斯 诺里和野依良治因何成就得到诺贝尔奖？）Answer：William S. Knowles and RyoJi Noyori have accomplished for chirally catalyzed hydrogenation reactions.答案：威廉姆斯 诺里和野依良治的成就是手性催化氢化反应工作。10. What kind of achievement does the Nobel Prize award for the year 2010?（2010年的诺贝尔奖奖励何种成就？）Answer：The Nobel prize is awarded jointly to Richard F. Heck, Ei-Ichi Negishi, and Akira Suzuki for palladium-catalyzed cross couplings in organic synthesis this year. 答案：今年的诺贝尔奖授予理查德 骇客、根岸英一、铃木章三人，奖励他们在有机合成中的钯催化交联中的研究成就。The Second Asignment(第二次作业)1 What are organometallic compounds?Answer：Organometallic compounds are defined as materials which possess direct, more or less polar bonds Md+ - Cd- between metal and carbon atoms.（金属有机化合物定义为在金属和碳原子之间具有直接的，或多或少极性Md+ - Cd-键的材料）2 How to distinguish s-. p-, d- bond types?Answer：s-bond is formed when the electron clouds of the bonding atoms are overlapping head to head along an axis, the number of nodal planes is zero; p-bond is formed when the electron clouds of the bonding atoms are overlapping shoulder to shoulder on a plane, the number of nodal planes is one; d-bond is formed when the electron clouds are facing each other across a wider space, the number of nodal plane is two.（s-键是由两个成键原子的电子云沿轴向重叠而形成，其界面数为零；p-键是由两个成键原子的电子云在一个平面上肩并肩重叠而形成，其界面数为壹；d-键是由两个成键原子的电子云跨越一个大的空间相向而成，其界面数为贰。）3 Generally speaking, what is the relationship between M-C bond energy and the atomic number of M in main-group organometallic compounds?Answer: The mean bond energy E(M-C) with a main-group decreases with the increasing atomic number（对于主族来说平均键能E（M-C）随着原子序数的增加而下降）.4 Zn(C2H5)2 is pyrophoric while Sn(CH3)4 is airstable. Explain their differences in terms of thermodynamics and kinetics.Answer：Thermodynamically both Zn(C2H5)2 and Sn(CH3)4 are not very stable. The drastic difference between these two compounds lies on kinetics. Zn(C2H5)2 is labile and react readily while Sn(CH3)4 is inert and somehow less reactive.（就热力学而言Zn(C2H5)2 和 Sn(CH3)4都不够稳定。这两个化学物戏剧性的不同在于动力学方面。Zn(C2H5)2具有活性容易反应，Sn(CH3)4而具有惰性，反应性较低。）5 Describe the properties of Me3In, Me4Sn and Me3Sb and explain in relevant factors.Answer：The properties of these compounds and relevant factors attributed to these properties are summarized in the following table.（这些化合物的性质及相归因的相对因素总结在下表当中。）In air在空气中In water在水中Relavant factors相关因素Me3InPyrophoric起火Hydrolysed水解Electron gap at In, highly bond polarity.铟有电子缺口，高度键极性Me4SnInert 惰性inertSn shielded well, low bond polarity.锡得到很好屏蔽，低键极性。Me3SbpyrophoricinertFree electron pair on Sb.在锑上有一对自由电子6 Write out the molecular structures for the following organic compounds（写出下述有机化合物的分子结构式）1） 1-bromo-3-ethyl-5-methylcyclohexaneAnswer: 2） cyclopentadieneAnswer:3） propyneAnswer: 4） ethoxyethaneAnswer: 5） ethanolAnswer: 6） ethanalAnswer:7） propanoneAnswer:8） butanoic acidAnswer:9） ethanamideAnswer:10）para-bromoclorobenzeneAnswer:11）5-hydroxy-3-hexanoneAnswer:7. Name the following organic compoundsAnswer:(1) 1-chloro-1-methylcyclohexane（1-氯-1-甲基环己烷）(2) 2, 2-dimethylpropane8. Wright out the English name and chemical symbols for the following elements, and briefly describe their original meaning and sources：（写出下列化学元素英文名称和元素符号，并简单说明原意和出处）1） 钠2）磷3）钾4）钛5）铬6）氪7）钯8）银9）钷10）钽Answer:1） 钠，英文名称 Sodium；元素符号Na。钠的元素符号来自碳酸钠的拉丁语Natron，英文名称sodium来源于阿拉伯语suda2） 磷，英文名称Phosphurus，元素符号P。磷在黑暗中可以发光，英文名称phosphurus来源于希腊语“带来光明的东西”之意。3） 钾，英文名称Potassium，元素符号K。钾的英文名称是根据potash（草木灰）加-ium组合而成，potash是pot（锅）和ash（灰）的合成词。钾的元素符号取自拉丁文名Kalium。4） 钛，英文名称Titanium，元素符号Ti。钛的英文名称来源于希腊神话故事中由天神（Uranus）和地母（Gaea）所生的巨人儿子Titan的名字命名。5） 铬，英文名称Chromium，元素符号Cr。由于铬在不同氧化状态下变化为紫、红、黄、绿等颜色，它的英文名称取自希腊语中表示颜色的词“chroma”6） 氪，英文名称Krypton，元素符号Kr。1898年拉姆齐和特拉弗斯将1L液体空气慢慢蒸发得到 1 mL液体，分析表明除大部分是氩外，还有其他比氩分子质量更大的气体，颇有隐士风范，用希腊语中“隐藏”（kryptos）命名。7） 钯，英文名称Palladium，元素符号Pa。1803年由英国人沃拉斯顿发现。以一年前发现的小行星 Pallas (古希腊女神雅典娜的别名Pallas Athena) 命名。8） 银，英文名称Silver，元素符号Ag。元素符号源于拉丁语 argentum，是光辉的意思。9） 钷，英文名称Promethium，元素符号Pm。1947年美国人科里尔等人从核分裂物质中分离了半衰期仅有2.6年的物质，借用古希腊神话中给人类盗取火种的普罗米修斯（Prometheus）的名字命名。10）钽，英文名称Tantalum，元素符号Ta。瑞典人厄克贝里1802年发现钽元素，为比喻在发现这种元素过程中的种种困难，以希腊神话中被主神宙斯以吊胃口之刑惩罚的Tantalus 的名字为其命名。9. Write the formula for each of the following compounds:（写出下列每个化合物的分子式：）(1) dibromotetraammineruthenium(III)nitrate. 硝酸二溴四氨合镣(III)。Answer: RuBr2(NH3)4NO3(2) chloroaquobis(ethylenediamine) rhodium(III)chloride. 盐酸氯水双(乙二胺)合铑(III)。Answer: RhCl(H2O)(en)2Cl2(3) calcium dioxalatodiamminecobaltate(III). 二草酸根二氨合钴(III)酸钙。Answer: CaCo(C2O4)2(NH3)22(4) octaammine-m-amido-m-hydroxodicobalt(III)sulfate. 硫酸八氨-m-氨基-m-羟基合二钴(III)。Answer:(5) sodium dithiosulfato-S-argentate(I). 二硫代硫酸根-S-合银(I)酸钠。Answer: NaAg(S2O3)210. Name each of the following compounds:命名下列化合物：(1) Co(NH3)62(SO3)3Answer: hexaamminecolbalt(III)sulfate. 硫酸六氨合钴(III)。(2) Pt(en)Cl4Answer: tetrachloroethylenediamineplatinum(IV). 四氯乙二胺合铂(IV)。(3) NH4Cr(NH3)2(NCS)4Answer: ammonium diamminetetrathiocyanato-N-chromate(III). 二氨四硫氰酸根-N-合铬(III)酸铵。(4) K4Ni(CN)4Answer: potassium tetracyanonickelate(0). 四氰合镍(0)酸钾。(5)(6)Answer: (5) 1-bromo-3-chloroamminemethylamineplatinum(II). 1-溴-3-氯氨甲胺合铂(II)。(6) potassium tetraoxalato-m-dihydroxodicobaltate(III). 四草酸根-m-二羟基合二钴(III)酸钾。11. White out the names for the following organometallic compounds（写出下列金属有机化合物的名称）(1)Answer: Decacarbonyldimanganese(0)(2)Answer: di(h5-cyclopentadienyl)iron(II)(3)Answer: Hexacarbonyltri(m-carbonyl)diiron(0)The Third Asignment（第三次作业）1. Describe the packing structure of solid methyllithium. What is its molecular bonding?（描述固态甲基锂的堆积结构。它的分子键如何？）Answer: solid methyllithium is best described as cubic body-centered packing of (LiCH3)4 units, consisting of Li4-tetrahedra with methyl groups caping the triangular faces.答案：固态甲基锂最好可描述为(LiCH3)4单位的立方体心堆积，由Li4-四面体与甲基基团加盖三角面组成。2. In order to yield a favorable NMR signal what spin quantum number a nucleus must have? （为了得到有利的核磁共振信号，一个核必须具有什么自选量子数？）Answer: For small molecules, I = usually yield sharp resonance lines.答案：对于小分子而言，I = 1/2 通常产生敏锐的共振线条。3. How does strong donor ligand TMEDA affect the reactivity of n-BuLi?（配体TMEDA怎样影响正丁基锂的反应性？）Answer: TMEDA effects both the cleavage of n-BuLi oligomers and, through complexation of the Li+ cation, the polarization of the Li-C bond.答案TMEDA影响正丁基锂寡聚物的裂解，并且通过与Li正离子的络合，影响Li-C键的极性。4. How is alkyllithium reagent used in Wittig reaction?（烷基锂在威悌希反应中怎样使用？）Answer: Alkyllithium reagent is used in Wittig reaction to deprotonate organophosphonium ion to produce an Ylid. Ph3PCH3+ + RLi Ph3P+-CH2- + RH答案：在威悌希反应中烷基锂试剂被用来对有机磷鎓离子去质子化从而产生伊利德。Ph3PCH3+ + RLi Ph3P+-CH2- + RHThe Fourth Asignment（第四次作业）1. Describe structures of alkylberrylium R2Be in both solid and gaseous phase, what are their valence electron hybridizations respectively?（描述烷基铍R2Be在固相核气相的结构，它们分别有怎样的价电子杂化？）Answer: Compound BeR2 is a polymeric solid. Be atom adopts sp3 hybridization.答案：化合物R2Be是一个聚合态固体。铍原子采用sp3杂化。In the gas phase, BeR2 is monomeric and linear and therefore is to be described by Be (sp) hybridization. 在气相，R2Be为单体并且呈线性分子，因此可以将铍原子描述为sp杂化。2. Describe the structure of berryllocene. Does the compound possess a dipole moment?（描述茂铍的结构。该化合物具有偶极矩吗?）Answer: Berryllocene (C5H5)2Be adopts a “slipped sandwich” shape with a symmetry Cs and h3-, h5-structure. The molecule possesses a dipole moment (m = 2.24 Debye).答案：茂铍(C5H5)2Be采用了“滑落三明治”形状，Cs对称群和h3-, h5-结构。该分子拥有偶极矩（m = 2.24德拜）3. What is the situation of Grignard reagent in etherate solution? (hint: Schlenk equilibrium)格氏试剂在乙醚化溶液中的情况如何？（暗示：舒兰克平衡）Answer: A concise description of the situation in solution is given by the Schlenk equilibrium.答案：在溶液中精确的描述是由舒兰克平衡给出。4. Explain why is RMgX in Et2O conductive and what evidence point to the involvement of radical R.?（解释为什么RMgX在Et2O中有导电性？什么证据指向自由基R.的参与？）Answer: RMgX in Et2O undergoes heterocleavage to produce RMg+ and RMgX2- ions, which can serve as electric current carriers. During electrolysis radicals R. are generated at both electrodes and dimerization may occur. Detection of R-R can be regarded as an evidence of the involvement of radical R.答案：RMgX在Et2O中发生异裂产生RMg+ 与 RMgX2- 离子，它们可充当电流载体。电解期间在两电极都有自由基R.产生，双聚可能发生。探测到R-R可以被看作是自由基R.参与的证据。5. What are the general structures of di(cyclopentadienyl)metal complexes of the heavier alkaline earth Ca, Sr, and Ba?（重碱土金属钙、锶及其钡的双环戊二烯基络合物的一般结构是什么？）Answer: The di(cyclopentadienyl)metal complexes of the heavier alkaline earth (Ca, Sr, and Ba) possess bent sandwich structure.答案：重碱土金属（钙、锶及其钡）的双环戊二烯基络合物具有弯折三明治结构。The Fifth Asignment（第五次作业）1. What structure does an organoboron hydride usually assume? How to interpret it with infrared spectroscopy?（有机硼氢化物通常采用什么结构？怎样用光外波谱解读它？）Answer: Organoboron hydrides R2BH and RBH2 form dimers which always display hydride bridges rather than alkyl bridges. The steching frequencies for the terminal and bridging B-H bonds are different enough to distiguish their structure by IR spectroscopy. For example:答案：有机硼氢化物R2BH 和 RBH2形成二聚物，并且氢桥基总是优先于烷基桥基。而端位和桥位BH键的伸缩频率的差别大的足以用红外波谱来区别它们的结构。例如：2. Why is cyclotrisborazine环硼氮烷 (B3N3H6) called “inorganic benzene”? （为什么环硼氮烷(B3N3H6)被称为“无机苯”？）Answer: In cyclotrisborazine B3N3H6 the isoelectronic nature of the units -C=C- and B=N- suggests mutual replacement and makes C6H6 and B3N3H6 electronicly equivalent. The pi electron counts of the both: 4n + 2 = 4 x 1 + 2 = 6 (Huckels Rule) make them aromatic. Since this “benzene” does not contain carbon atoms, it is best be called “inorganic benzene”.答案：在环硼氮烷B3N3H6中CC和BN单元等电子的意味着它们的相互置换并且使得C6H6 和 B3N3H6电子等当量。两者的p电子计数为：4n + 2 = 4 x 1 + 2 = 6（休克尔规则）使得它们都具有芳香性。鉴于该“苯”不含碳原子，它最好被称为“无机苯”。3. What structural category does tetraborane B4H10 belong according to Wades rules? （根据Wade规则四硼烷B4H10属于哪类结构？）Answer: The skeletal electron count according to Wades rule答案：根据Wade规则骨架电子计数Electrons from B-H units 4 x 2 = 8来自B-H单元的电子Electrons from additional H6 x 1 = 6来自额外H电子Total skeletal electrons14, which makes 7 pairs. Since there are totally 4 boron atomsin this molecule. The electron pairs are fomulated as 7 = 4 + 3 = n + 3总骨架电子14，它配成7对。鉴于该分子中共有4个硼原子。将电子对数公式化为7 = 4 + 3 = n + 3The structure of the molecule should be arachno or cobweb or spiderweb.该分子的结构应为蜘蛛网4. What structural category does dodecaborane ion B12H122- belong according to Wades rules? （ 根据Wade规则十二硼烷B12H122-属于哪类结构？）Answer: The skeletal electron count according to Wades rule答案：根据Wade规则骨架电子计数Electrons from B-H units 12 x 2 = 24来自B-H单元的电子Electrons from the charge2来自电荷的电子Total skeletal electrons26, which makes 13 pairs. Since there are totally 12 boron atoms in this molecule. The electron pairs are fomulated as 13 = 12 + 1 = n + 1总骨架电子26，它配成13对。鉴于该分子中共有12个硼原子。将电子对数公式化为13 = 12 + 1 = n + 1The structure of the molecule should be closo or closed or cage.该分子的结构应为封闭式或鸟笼The Sixth Asignment（第六次作业）1. Describe the structure and bonding of binary aluminum organyl Al2Me6.（描述二元有机铝Al2Me6的结构和成键）。Answer: The structure of the compound is as follows：答案：化合物的结构如下：The bonding of the compound can be described either as a or as b alternatively.化合物的成键可以描绘成a或者b两者选一。2. Complete following reactions:（完成下列反应：）Answer:a) (CH3)3Ga(g) + AsH3(g) GaAs(s) + 3CH4(g)b) Me2GaCl + 2NH3 Me2Ga(NH3)2+Cl-c) InCl3 + 3Na(C5H5) In(C5H5)3 + 3NaCld) InCl + Na(C5H5) In(C5H5) + NaCl3. Describe the structure and bonding of In(I) (p-complex) (C5H5)In.（描述In(I)p络合物(C5H5)In的结构和成键。）Answer: (p-complex) (C5H5)In adopts a pentagonal pyramidal structure, atom In has a lone pair pointing opposite to the cyclopetadienyl ring.答案：(p-络合物)(C5H5)In采取五角锥结构，原子In有一对孤电子指向环戊二烯基环的反面。The bonding of this complex resides on the symmetry based atomic orbital interactions between metal In and ligand Cp.这个络合物的成键基于金属In和配体环戊二烯基对称原子轨道的相互作用。The Seventh Asignment（第七次作业）1. Complete the following reactions:（完成下列反应：）Answer:a) HSiCl3 + RCH=CH2 RCH2CH2SiCl3(anti-Markovnikov direction)b) Me2GeCl2 + 2RMgX Me2GeR2 + 2MgXClc) R3SnCl + NaMn(CO)5 R3SnMn(CO)5 + NaCl2. Write following molecular formula:（写出下列分子式;）a)Stannocene（茂锡） Answer: (C5H5)2Snb)Decamethylgermanocene（十甲基茂锗）Answer: (C5Me5)Gec)Tetraphenylsilane（四苯基硅）Answer: (C6H5)4Si3. Describe Sn = Sn bonding situation in stannylene dimmer R2Sn=SnR2（描述在锡烯二元体R2Sn=SnR2中的SnSn成键情况）Answer: Unlike C=C double bond, which consists an s bond formed by overlap of two sp2 hybridized orbiats and a p-bond formed by overlap of two p orbitals perpedicular to the sp2 plane. Both bonds in Sn=Sn were formed by overlap of a sp2 lone pair orbital and an empty p orbital. Therefore stannylene R2Sn=SnR2 is not a planar compound.答案：CC双键是由两个sp2杂化轨道重叠形成的一个s键和与sp2平面垂直的两个p轨道重叠形成的一个p键组成。而SnSn的两个键均由一个sp2孤对电子轨道和一个p空轨道的重叠而形成。因此R2Sn=SnR2锡烯不是一个平面化合物。The Eighth Asignment（第八次作业）1. What molecular structures do pentaphenyl arsorane and pentaphenylstiborane adopt?（五苯基砷烷和五苯基碲烷采取什么样的分子结构？）Answer: Pentaphenylarsorane adopts trignoal bipyramidal structure and pentaphenylstiborane adopts a square pyramidal structure.答案：五苯基砷烷采用三角双锥结构，而五苯基碲烷采取四方锥结构。2. Why are trisorganoarsane, trisorganostibane, and trisorganobismuthane MeEtPhE are chiral?（为什么三有机砷烷、三有机碲烷和三有机铋烷MeEtPhE是手性的？）Answer: Trisorganoarsanes, trisorganostibanes, and trisorganobismuthanes (R3E) have pyramidal structures with one lone pair of occupied orbital. High inversion barrier prevents the molecules RRRE from configuration change and chirality is resulted.答案：三有机砷烷、三有机碲烷和三有机铋烷（R3E）具有锥形结构，有一占据孤电子对。高翻转能障阻止了RRRE的构形改变，其结果为手性的产生。翻转能障：177千焦/摩尔3. Complete the syntheses of compounds Ph5As, Ph5Sb, and Ph5bi.（完成化合物Ph5As, Ph5Sb, 和 Ph5bi的合成）。Answer: The Nineth Asignment（第九次作业）1. Complete following reactions:完成下列反应：）Answer: (a) RCCH + Cu(NH3)2+ RCCCu + NH3 + NH4+(b) 2 LiAr + AgX LiAgAr2 + LiX(c) Et3PAuCl + 2 LiMe LiAuMe2 + LiCl + Et3P(d) Me3PAuMe + Me3P=CH2 Me3P+-CH2-AuMe + Me3P(e) LiAuMe2 + MeI + Ph3P Ph3PAuMe3 + LiIThe Tenth Asignment（第十次作业）1. Draw molecular structures of following compounds and calculate corresponding total valence electrons:（画出下列化合物的分子结构并计算相应的总价电子数：）(a) Di(cyclopentadienyl)iron Fe(C5H5)2 双(环戊二烯基)铁Fe(C5H5)2Answer: (b) Decacarbonyldimanganese Mn2(CO)10 十羰基二锰Mn2(CO)10Answer: (c) Noncarbonyldiiron Fe2(CO)9 九羰基二铁Fe2(CO)9Answer: (d) Cyclopentadienyltricarbonylmanganese (C5H5)Mn(CO)3 环戊二烯基三羰基锰(C5H5)Mn(CO)3Answer:(e) Di(cyclopentadienyl)hexacarbonyldichromium (C5H5)Cr(CO)32 双环戊二烯基六羰基二铬(C5H5)Cr(CO)32Answer:2. Draw molecular orbital diagram for an octahedral complex (simplified), s-bonding only. Indicate orbital hybridization and frontier orbitals in such complex.（画八面体络合物（简化）分子轨道图，只考虑s-键。指出这种络合物中的轨道杂化和前线轨道。0Answer: The general form of molecular diagram in an octahedral complex (simplified), s-bonding only is as following:答案：（简化）八面体络合物分子轨道的一般形式, 只考虑s-成键, 如下：The hybridization of metal bonding orbitals is d2sp3 (1s, 3p, and dx2-y2, dz2) and the remaining metal orbitals are nonbonding (dxy,dxz,dyz).金属成键轨道的杂化是d2sp3 (1s, 3p, and dx2-y2, dz2)，剩余的金属轨道不成键(dxy,dxz,dyz)。The frontier orbitals of the molecule are: HOMO t2g (dxy,dxz,dyz); LUMO eg* (dx2-y2,dz2).分子的前线轨道是：最高占据分子轨道t2g (dxy,dxz,dyz);最低非占据分子轨道eg* (dx2-y2,dz2)。3. One of the major causes for decomposition of a transition-metal complex is b-elimination. To prevent this from occur what measures should be taken? Give one example to support each of your argument.（过渡金属络合物分解的一个主要原因