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*Is it necessary?*Is it possible?1.Our considerations(1)AtomsFragments Structures Molecules,(2)How to characterize a structure?N=No of fragments B=No.of interfragment bondsNB=4,3 4,4 4,5 4,6 R.Hoffmann first proposed“molecular fragment”in his“principle of isolobal analogy”Periodical Table of Molecules and Molecular FragmentsFrom topology,only 6 ways to connect 4 positions(1)NB=4,3,3 (2)NB=4,3,3 (3)NB=4,4,4 (4)NB=4,4,4 (5)NB=4,5,5 (6)NB=4,6,6Add bonds,with F1,F2,F3,F4,we have 28 more structures(7)NB=4,4,3 (8)NB=4,4,3 (9)NB=4,5,3 (10)NB=4,5,3(11)NB=4,5,3 (12)NB=4,5,3 (13)NB=4,6,3 (14)NB=4,6,3 (15)NB=4,7,3 (16)NB=4,4,3 (17)NB=4,5,4 (18)NB=4,6,4 (19)NB=4,6,4 (20)NB=4,7,4 (21)NB=4,8,4 (22)NB=4,8,4(24)NB=4,5,4 (25)NB=4,5,4 (26)NB=4,5,4 (27)NB=4,6,4 (32)NB=4,7,5 (33)NB=4,7,6 (34)NB=4,8,6(38)NB=4,6,4 (29)NB=4,6,5 (30)NB=4,6,5 (31)NB=4,7,5N B=2N,2N-1,2N-2，N-1 1 1 02 2 4 3 2 13 3 6 5 4 3 24 4 8 7 6 5 4 35 5 10 9 8 7 6 5 46 6 12 11 10 9 8 7 6 57 7 14 13 12 11 10 9 8 7 6Each block in the above table contains a vast no.of molecules as shown below Table 1 Periodical Table of Molecules2Definition of FragmentsWe define a fragment as composed of a central atom with some ligands L linked to it by covalent bonds.Fragments=Ak+Ligands L1 Fi=Mj=AkLl,j=k+l;(1)I=2s-j,s=No.of Val.Orb.(2)for diamagnetic molecules without three-center-two electron bondsMiExampleVal.OrbitalssM2LiHs1M6B(C2H5)3s p23M8CH4NH3,H2O,HF,Nes p34M10PCl5,SF4,ClF3,XeF4s p3d5M12SF6,MoF6,PF6-s p3 d26M14IF7s p3d37M18Ni(CO)4,Fe(CO)5,Cr(C6H6)2,Cp2Fe,Hg(CN)4=,Cd(NH3)4+d5s p39FiMiExampleVal.OrbitalsF1M7CH3,NH2,OH,F,Cl,Br,I(as in CH3I)s p34M17Mn(CO)5,Re(PR3)5,Co(CO)4d5s p39F2M6O,NH,NR,PR,CH2,CR2,SiR2,S(in H2S)s p34M16Fe(CO)4,Ru(PR3)4,W(CO)5d5s p39F3M5N,P(as in PR3),CH,CR,SiR,BH2s p34M15NiCp,Co(CO)3,d5s p39F4M4C,BH,AlRs p34M14Fe(CO)3,CoCp,Os(PR3)3,d5s p39M12MoCl4-,ReCl4-,Re(SCN)4-,Cr(CH3)4-,Mo(SO4)2-,Mo(CH3COO)2d5s p281A2A3B4B5B6B7B8B1B2B3A4A5A6A7A8AM1M2M3M4M5M6M7M8M9M10M1M2M3M4M5M6M7M8HHeM11M12M13M14M15M16M17M18LiBeBCNOFNeBeLlBLlCLlNLlOLlFLlB-B L2CL2NL2OL2NaMgAlSiPSClAMgLlAlLlSiLlPLlSLlSL2Al-AlL2SiL2PL2ClLlKCaScTiVCrMnFeCoNiCuZnGaGeAsSeBrKrRbSrYZrNbMoTcRuRhPdAgCdInSnSbTeIXeCsBaLaHfTaWReOsIrPtAuHgTlPbBiPoAtRnA6L1A7 L1A8 L1A9 L1A7 L4A8L4A9 L4A10 L4A11L4A11L5A11L6A10 L8A5L2A6L2A7 L2A8 L2A6 L5A7 L5A8 L5A9 L5A10 L5A10 L6A10 L7A9 L9A4L3A5 L3A6 L3A7 L3A5 L6A6L6A7 L6A8 L6A9 L6A9 L7A9 L8A8 L10A3L4A4 L4A5 L4A6 L4A4 L7A5 L7A6 L7A7 L7A8 L7A8 L8A8 L9A2L5A3 L5A4 L5A5 L5A3 L8A4L8A5 L8A6 L8A7 L8A7 L9A7 L10A1A2A3A4A5A6A7A8A9A10A11A12A13A14A15A16A17A18应用应用IOs5(CO)19=Os(CO)44Os(CO)3=M164M14=F24F4,B=(1/2)(22+4)=6Only 2 Possible structure =6,=0 =5,=1The experimentally det.structure isThe nature prefers more symmetrical structurePlanar OsskeletonNB=5,6,65.From molecular formula to structure Ex(2)Fe5(CO)15C=Fe(CO)35C=M145M4=F64Only this structure satisfies the requirements of tetravalency of 6fragments.It was verified by expt.NB=6,12,125 Ex(3)Co6(CO)18C2=Co(CO)36C=M156M42=F36F42Put the fragment of lighest valency at the center of the moleculebond is stronger than bondBN=8,13,135 Ex(4)From topology,only 6 ways to connect 4 positions(1)NB=4,3,3 (2)NB=4,3,3 (3)NB=4,4,4 (4)NB=4,4,4 (5)NB=4,5,5 (6)NB=4,6,6应用应用II 决定文献已测分子结构的成键情况决定文献已测分子结构的成键情况Ex.4a NB=4,8,4 Cu4(CH2SiMe3)4,J.A.J.Jarvis,ets.J.Chem.Soc.Chem.Comm.475(1993)M12 M12s=8(d5sp2)B=1/2(4+4)=8NB=4,8,4 =8-4=4 3c-2e bondd(Cu=Cu)=242pmEx.4b NB=4,4,4Ex.4c F2=M16 Pt4(CH3COO)8=A104L38=M164=F24 d5sp3,s=9,i=2s-j=18-16=2Others 4 CH3COOgroups plane 2C6H6=A112L34L62=M174=F14 B=1/2(14)=2N-1 4c-4e bond Bond order=1/2P.F.Rodesiler and E.L.AmmaJ.Chem.Soc.Chem.Comm.599(1974).CuCu 280pm Cu-Cu 266pm,Cu=Cu 242pmShort Cu Cu diagonal 310 pm2 C6H6 located above and below the Cu4 plane.Ex.4e NB=4,2,2
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