中级无机化学翻译(1)

1、第一章1.4指出下列物质在液氨中的酸碱行为（写出反应方程式）：CH3CO2H , NH2CONH2 , CH3CH2OH , NaH1.4 Pointed out that the following substances in acid-base behavior in liquid ammonia(write reaction equation):CH3CO2H , NH2CONH2 , CH3CH2OH , NaH参考答案：CH3CO2H为强酸 CH3COOH+NH3CH3COO-+NH4+ NH2CONH2为弱酸 NH2CONH2+NH3NH4+ +NH2CONH- CH3CH2OH为

2、弱酸 CH3CH2OH+NH3NH4+CH3CH2O- NaH为强碱 NaH+NH3NH2-+H21.4 指出下列物质在液氨中的酸碱行为（写出反应方程式）：1.4 Points out the following material in the soda acid liquid ammonia behavior (write a response equation)CH3COOH,H2NCONH2,CH3CH2OH,NaH解：CH3COOH strong acid CH3COOH+NH3=CH3COO+NH4H2NCONH2 weak acid H2NCONH2+NH3=NH4+NH2CONH

3、CH3CH2OH weak acid CH3CH2OH+NH3=NH4+CH3CH2ONaH strong soda H+NH3=NH2+H21.5指出下列物质在100%H2SO4中的酸碱行为（写出反应方程式）：HNO3 , CH3CO2H , HClO4 , C2H5OH , H2NCONH2 , SO31.5 Pointed out that the following substances in acid-base behavior in 100%H2SO4(write reaction equation):HNO3 , CH3CO2H , HClO4 , C2H5OH , H2NCON

4、H2 , SO3参考答案：HNO3为碱 HNO3 + 2H2SO42HSO4-+NO2+ H3O+ CH3CO2H为碱 CH3COOH + H2SO4HSO4-+ CH3COOH2+ HClO4为酸 HClO4+ H2SO4H3SO4+ CH3COOH2+ C2H5OH为碱 C2H5OH+ 2H2SO4CH3CH2HSO4+ HSO4-+ H3O+ H2NCONH2为碱 H2NCONH2+ H2SO4HSO4-+ H2NCONH3+ SO3为碱 SO3+ H2SO4H2S2O7 H2SO4 +H2S2O7 H3SO4+ HS2O7-1.8 Please write down the chemi

5、cal equation of the following reactions.(1) Potassium amide and aniline have reactions in liquid ammonia.(2) Inject hydrochloric acid into concentrated sulfuric acid and concentrated nitric acid solution to get NO2Cl.(3) Boric acid is dissolved in sulfuric acid.1.9 In liquid ammonia,what is the conc

6、entration of ammonium ion?1.9 in the liquid ammonia, what is the concentration of ammonium ion ?Answer: 2NH3 = NH4 + NH2- K=1.0*10-33NH4 +=K1/2=3.16*10-7 ( MOl/L)1.12 estimate rough the value of PKa about each of theFollowing acids?Answer: 1.phosphite pka1=2 pka2=72. nitric acid pka1=-3 3. Percholor

7、ic acid pka=84. Periodate pka1=2 pka2=71.12 粗略估计下述各酸的pKa值1.12 A rough estimate the following the sour pKa value解：(1)H3PO3 m=1 pK1=2 pK2=pK1+5=7(2)HNO3 m=2 pK=-3(3)HClO4 m=3 pK=-8(4)H5IO6 m=1 pK1=2 pK2=7第二章2.1Determine the following molecules or objects belong point group(1)SiHDBr2 Tetrahedral Cs poi

8、nt group(2)SiFClBrI Tetrahedral C1 point group(3)SiH2Br2 Tetrahedral C2v point group(4)PCl3 Three sides of vertebral C3v point group(5)OPCl3 Three sides of vertebral C3v point group(6)CO2 Linear Dh point group(7)P4O6 Tetrahedral Td point group(8)Mn(CO)5I Octahedral C4v point group(9) Letter T and ZT

9、 C2v point groupZ C2h point group(10)A piece of chalk Cv point group2.1确定下列分子或物体所属的点群（Sure the following molecular or object points of subordinate）   2.3指出下列分子或离子所属的点群，并写出他们所含有的对称元素（Point out the following molecular or ion points of subordinate ,and write out their symmetry element2.8 Which poi

10、nt group PtCl42 belong to ? Draw its structure , mark the elements of these operations(each operation mark one).2.9 Known following molecular point group , draw their structures. 2.9 Given the following molecules(ions) belong to which point group,draw their structure.(1)B(OH)3 C3h(2)Cr(en)3 D3(3)Co(

11、gly)3 C3(4)Mn2(CO)10 D4d(5)Allene D2d第三章3.1 Try to estimate the geometries and electronic structures of these following coordinative ions.Co（CN）63（diamagnetism）：octahedron，Oh strong field，low spin， t2g6eg0；NiF64（2 single electron）：octahedron，Oh weak field，high spin， t2g6eg2；CrF64（4 single electron）：

12、octahedron，Oh weak field，high spin， t2g3eg1；AuCl4（diamagnetism）：plane square，D4h field， eg4a1g2b2g2b1g0；FeCl4（5 single electron）：tetrahedron，Td field， e2t23；NiF62（diamagnetism）：octahedron，Oh strong field，low spin，t2g6 eg0 。3.1、Describe the structure and electron structure of the following coordinati

13、on ion.Co(CN)63- (diamagnetism) NiF64- (two unpaired electron)CrF64- (four unpaired electron) AuCl4- (diamagnetism) FeCl4- (five unpaired electron) NiF62- (diamagnetism) 3.5、Which ones are high-spin state in the following coordination ion. Mn(H2O)2+6 Fe(H2O)3+6 Co(NH3)3+6 Co(H2O)2+6 CoCl2- Fe(CN)4-6

14、3.5：The following complex ion in those belonging to a high spin configuration? (a): Mn(H2O)62+ ; (b): Fe(H2O)63+ ; (c):Co(NH3)63+(d):Co(H2O)62+ ; (e):CoCl42- ； (f):Fe(CN)64-Solution: H2O is weak filed, so Mn(H2O)62+ 、Fe(H2O)63+ 、Co(H2O)62+ are high spin. NH3 is strong filed , so Co(NH3)63+ is low sp

15、in. Cl- is weak filed,so CoCl42- is high spin. CN- is strong filed, so Fe(CN)64- is low spin.3.5 The following cations which are high spin configuration ？(a)Six hydrated manganese ()cation；（b）Six hydrated iron () cation ; (c)Six ammonia cobalt()cation；(d) Six hydrated cobalt()cation ；（e）Four chlorid

16、e cobalt( II )anion; (f) Six-iron cyanide ()anion.The answer: Water is the weak-field ligands，its splitting can be low，that (a),(b),(d) are high-spin configuration. Cyanide is the strong-field ligands, its splitting can be high, that (f) is low-spin configuration. Amine is a medium strength ligand,h

17、ave a greater split coordinated with cobalt ions, that (c) is low-spin configuration.Four cobalt II chloride anion is Tetrahedral geometry, because of the regular tetrahedron split in smaller, so (e) is high spin configuration3.8 The following cations, which may produce Jahn-Teller effect ?(1) Six-i

18、ron four cyanide anion (2) Six hydrated iron Divalent Cation(3) Hexahydrate cadmium trivalent positive ions(4) Six ammonia cobalt trivalent positive ionsThe answer: (1) Cyanide is the strong field ligand, low-spin,its electronic structure is t62g eg0 ,t2g is full filled, it doesnt produce John-Telle

19、r effect.(2)Water is the weak-field ligand,high-spin, its electronic structure is t42ge2g, its t2g track is asymmetric occupy, it do produce Jahn-Teller effect.(3) Water is the weak-field ligand , its electronic structure is t32geg0, its t2g track is symmetric occupy, it doesnt produce John-Teller e

20、ffect.(4) Ammonia is a strong field ligand, low-spin, its electronic structure is t62geg0, its t2g track is symmetric occupy, it doesnt produce John-Teller effect.3.8下列离子中，哪一种可能产生john-teller效应？The follering with ion in,which one may produce the john-teller effect?3.8: The following complex ion in wh

21、ich a possible Jahn-Teller effects?(a): Fe(CN)64- ; (b): Fe(H2O)62 + ; (c): Cr(H2O)63+ ; (d): Co(NH3)63+Solution: (a):CN- is strong filed ligands ,Fe(CN)64- is low spin, and electronic structure is t2g6eg0 ,so it is not Jahn-Teller effects (b): H2O is weak filed, Fe(H2O)62 + is high spin,and electro

22、nic structure is t2g4eg2,so it is Jahn-Teller effects.(c): H2O is weak filed, Cr(H2O)63+ is high spin,and electronic structure is t2g3eg0,so it is not Jahn-Teller effects.(d): NH3 is strong filed, Co(NH3)63+ is low spin,and electronic structure is t2g6eg0, so it is not Jahn-Teller effects.3.14 Co2+的

23、四面体配合物比Ni2+的相应的四面体稳定，但是Ni2+的弱场正八面体配合物都比Co2+形影的正八面体配合物稳定。请通过计算说明原因。Co(2)ion tetrahedron complexes is stableer than Ni(2) ion,but Ni(2) ion weak field are octagon complexes is stabler than Co(2)ion.please Through the calculation of the reason why.P1373.14:Co2+ tetrahedral complexes than the Ni2+ corre

24、sponding tetrahedral complexes stability, but Ni2+ weak filed are eight surface complexes are better than those of Co2+ eight surface corresponding is stable ,please by calculation.Solution: in tetrahedral Co2+ complexes as high spin,and the electronic structure is e4t(2) 3.CFSE=-0.6t*4+0.4t*3=-1.2t

25、.in tetrahedral Ni2+ complexes as high spin,and the electronic structure ise4t(2)4. CFSE=-0.6t*4+0.4t*4=-0.8t.So Co2+ complexes are stable.In eight bodies Co2+ complexes as high spin,and the electronic structure ist2g5eg2. CFSE=-0.4o*5+0.6o*2=-0.8o.In eight bodies, Ni2+ complexes as high spin,and th

26、e electronic structure t2g6eg2. CFSE=-0.4o*6+0.6o*2=-1.2o.So Ni2+ complexes are stable.3.20 According to the Crystals field theory and the nature of the list ions, Write their d electron configuration structure and calculate its magnetic strength .coordination ionpairing energyP /cm /cmd electron co

27、nfigurationmagnetic strength Co(NH3)63+1780023000(1)(2)Fe(H2O)63+2650013700(3)(4)Key: (1) t2g6eg0 (2) 0 (3) t2g3eg2 (4) 5.923.22 Contrast complexes Fe(H2O)6SO4和K4Fe(CN)6:(a) Separately them is low or high spin complexes；(b) Draw their center of metal ions d electron configuration ;(c) Clculate their

28、 LESE respectively ;Key: (a) Fe(H2O)6SO4 is high spin complexes；K4Fe(CN)6 is low spin complexes；(b) The Fe2+ of Fe(H2O)6SO4 electron configuration is t2g4eg2 ; The Fe2+ of K4Fe(CN)6 electron configuration is t2g6eg0 ;(c) Fe(H2O)6SO4 : LFSE = -0.4 * 4+ 0.6 * 2 =-0.4 K4Fe(CN)6 : LFSE = -0.4 * 6 + 2P =

29、 -2.4 + 2P3.22 Compare the complexes Fe（H2O）6SO4 against K4Fe（CN）6（a）Separately declare whether they are low spin or high spin complexes.Fe（H2O）6SO4 is high spin complexes；K4Fe（CN）6 is low spin complexes 。（b）Draw out their center metal ions d electron configuration.Fe（H2O）6SO4 Fe2 t2g4 eg2 ；K4Fe（CN）

30、6 Fe2 t2g6eg0。（c）Separately calculate their LFSEFe（H2O）6SO4： CFSE0.4o×4 0.6o×2 0.4oK4Fe（CN）6： CFSE0.4o×62P2 .4o2P第四章4.2 Why excessive metal Mn, Tc, Re and Co, Rh, Ir  easy to form more nuclear carbony

31、l compounds4.2 Why transitionmetal Mn, Tc, Re and Co, Rh, Ir  easy to form multinuclear carbonyl compounds?Answer:This metals atomic number is odd,So they can not form dualistic mononuclear carbonyl compounds.If not

32、,the valence electron of the metals are inconformity rule EAN. 4.5 Try to calculate X value the following stable compounds (1)X=(18-(7+1)/2=5(2) X= (18-9-3)/2=3(3) 18-6-6-7=-1 X=1(4) 8+8=16 X=18-16=24.5 For the following stable compounds in the patent X&#

33、160;valueMn(CO)xI   Co(CO)x(NO)   (5-C5H5)Fe(6-C6H6)x+  Fe(CO)4 x-4.6 Try to explain the following experiment facts:(1) Ferrocene Weng ion (5-C5H5)2Fe+ is a strong oxidizer;(2) In the appropriate solvent cobaltocene Co (5-C5H5)2 is a strong reducing agent

34、;(3) Nickelocene reacts with nitric oxide gas to (5-C5H5)Ni(NO).4.6 Try to explain following experiment results （1）Two ferrocene onium ion is a kind of strong oxidant The compound owns 17 Electronics. If this compound gets one Electronic, it conforms to the EAN rule.(2) Two cyclopentadienyl cobalt i

35、s a kind of Reducing agent in the suitable solution. The compound owns 19 Electronics. If this compound loses one Electronic, it conforms to the EAN rule.(3) Two cyclopentadienyl nickel can react with nitric oxide to product The product conforms to EAN rule.4.13 Write and balancing the following res

36、ponse:(1) Mn2(CO)10 and I2 total heat;(2) In the THF Mn(CO)6 with KI together backflow;(3) Fe(CO)5 and KOH solution reaction;(4) Ni(CO)4 and PCl3 reaction;4.13 Write and balance the following reaction (1)Mn(CO)10 and I2 total heat(2)Mo(CO)6 and KI refluxing in THF(3)Reation of Fe(CO)5 and KOH soluti

37、on(4)Reation of Ni(CO)4 and PCl34.18 Discuss the following experimental observations(1)、The Infrared stretching vibration frequencies of CO、Mo(CO)6、Mo(CO)3(NH3)3 and Mo(CO)3(PPh3)3 were 2143、2004、1855 and 1950 cm-1(2)、The symmetric stretching vibration frequencies of CO in Fe(CO)42-、Co(CO)4- and Ni(

38、CO)4 were 1788、1918 and 2121 cm-1(3)、The bond lengths of V(CO)6 and V(CO)6 are 193 pm and 200 pm.4.18、Discuss the following experimental observation results（1）CO, Mo (CO) 6, Mo (CO) 3 (NH3) 3 and Mo (CO) 3 (PPh) 3 infrared stretching vibration frequency is respectively 2143, 2004, 1855 and 1950cm-1（

39、2）In the Fe (CO) 4 2 -, Co (CO) 4 - and Ni (CO) 4 CO symmetrical stretching vibration frequency is respectively 1788, 1918 and 2121cm-1（3）V(CO)6 - and V(CO)6 in V-C bond lengths are respectively 193pm and 200pmSolution: (1) Free CO VCO2143cm-1, Mo ( CO )6of all the CO are end type combination, VCO f

40、or2004cm-1, because the N atom d track, cannot accept feedback Pie, here CO accept more feedback Pie, so VCO1855cm-1, P d track is empty, can accept Pie, so VCO1950.(2) For the three electronic system, but the center of atomic electron cloud density decreased, so the formation of feedback Pie keys a

41、re weaken, VCO increased in turn.（3）The matching key is similar, but the former feedback Pie bond strength on the latter, namely the V-C key strength of the former is larger than the latter, so bond length is smaller than the latter.4.24、How to prepare two ferrocene? Sketch the crystal two ferrocene

42、 structural features, belongs to which point group; qualitative description of its chemical bonding modes.Solution: 2NaC5H5+FeCl2Fe(C5H5)2+2NaClTwo ferrocene structural features into two parallel cyclopentadienyl ring, metal atoms are located in the center of the middle point between the CP ring.Two CP ring with staggered and overlapping two typical configuration, molecular belong to D5d, D5h point group.Two ferrocene, the two Mau anionic conjugated to Pie electronic