【摘要】 合成了苯并 15 冠 5与Na2 [M(SCN) 4](M =Pd ,Pt)生成的配合物 :[Na(B15 C 5 ) ]2 [Pd(SCN) 4](1) ,{ [Na(B15 C 5 ) ][Na(B15 C 5 ) (H2 O) ]} [Pt(SCN) 4](2 ) .1为单斜晶系 ,空间群P2 1/n ,a =1.0 16 4(6 )nm ,b =1.3743(3)nm ,c=1.4987(7)nm ,β =95 .2 48(6 )° ,V =2 .0 847nm3 ,Z =2 ,Dcalcd=1.47g/cm3 ,F(0 0 0 ) =944 ,R =0 .0 5 3,wR =0 .0 72 .2为三斜晶系 ,空间群P1,a=1.1484(2 )nm ,b=1.42 10 (3)nm ,c=1.5 0 2 6 (3)nm ,α =6 2 .5 0 0 (3)°,β =72 .393(3)°,γ =73.10 6 (4 )° ,V =2 .0 398(7)nm3 ,Z =2 ,Dcalcd=1.6 74g/cm3 ,F(0 0 0 ) =10 2 8,R1=0 .0 32 7,wR2 =0 .0 885 .1由两个[Na(B15 C 5 ) ]+ 配阳离子和一个 [Pd(SCN) 4]2 -配阴离子组成 ,两者通过Na—N键形成中性配合物 ,[Na(B15 C 5 ) ]+相对钯原子呈反式排列 .2由 [Na(B15 C 5 ) ]+ 和 [Na(B15 C 5 ) (H2 O) ]+ 配阳离子和一个 [Pt(SCN) 4]2 -配阴离子组成 ,它们也通过Na—N键形成中性配合物 ,配阳离子相对铂原子呈顺式排列 .2的两个分子通过氢键形成二聚结构 .更多还原

【Abstract】 Two benzo-15-crown-5 （B15-C-5） complexes: [Na（B15-C-5）] 2[Pd（SCN） 4] （1） and {[Na（B15-C-5）][Na（B15-C-5）（H 2O）]}[Pt（SCN） 4] （2） have been synthesized and characterized by IR, elemental analysis, X-ray diffraction analysis. 1 belongs to monoclinic, space group P2 1/n with cell dimensions: a= 1.0164（6） nm, b=1.3743（3） nm, c=1.4987（7） nm, β=95.248（6）°, V=2.0847 nm 3, Z=2, D calcd= 1.47 g/cm 3, F（000）=944, R=0.053, wR=0.072. 2 is triclinic crystal system, space group P1 with cell dimensions: a=1.1484（2） nm, b=1.4210（3） nm, c=1.5026（3） nm, α=62.500（3）°, β=72.393（3）°, γ= 73.106（4）°, V= 2.0398（7） nm 3, Z=2, D calcd=1.674 g/cm 3, F（000）=1028, R 1= 0.0327, wR 2= 0.0885. 1 is composed of two [Na（B15-C-5）] + complex cations and one [Pd（SCN） 4] 2- complex anion. Complex cations and complex anion form neutral complex through Na-N interactions. Two [Na（B15-C-5）] + are trans-orientation with respect to Pd atom. 2 is made up of [Na（B15-C-5）] + and [Na（B15-C-5）^{H 2O}] + complex cations and one [Pt（SCN） 4] 2- complex anion. They also form neutral complex through Na-N interactions. [Na（B15-C-5）] + and [Na（B15-C-5）^{H 2O}] + are cis-orientation with respect to Pt atom. Two molecules of 2 form dimer structure though hydrogen bonds of H 2O and N atom of SCN group.更多还原