题目

英语翻译
Complex 1 crystallizes in an orthorhombic lattice without
solvates.The acetylenic phenyl ring is perpendicular to the
(N^C^N)Pt plane,and the latter is coplanar with the
ab plane.Infinite PtII···PtII chains exist along the c axis,with
a uniform intermetal contact of 3.383 2 (shorter than the sum
of van der Waals radii of 3.44 2) and a Pt-Pt-Pt angle of
172.858.Neighboring molecules are arranged in a staggered
fashion with a torsion angle of 49.48 around the Pt–Pt axis
(Figure 1b).Intermolecular C2H···p interactions have been
discerned between every two adjacent molecules along the
b axis (Figure 1c),as indicated by a short contact of 2.691 2
between H15 of the (N^C^N) ligand and acetylenic C19 of
another molecule,and a C15-H15-C19 angle of 159.68.No
short intermolecular contacts could be identified along the
a axis.
Complex 4 crystallizes as thin plates with intrinsic
twinning problems.Nevertheless,we were able to

提问时间：2020-08-10

答案

络合物1在没有溶剂化物的正交晶格中结晶.乙炔基苯基环与(N∧C∧N)Pt平面垂直,后者与ab平面共平面.沿着c轴存在无限的PtII…PtII链,伴随着3.383Å（短于3.44的范德瓦尔斯半径之和）的均匀的金属间接触和172.85°的Pt-Pt-Pt角.相邻分子以一种交错形式排列,围绕Pt-Pt轴的扭转角为49.4°（图1b）.分子间C-H…π互作用在沿b轴的每两个相邻分子之间已辨别清楚（图1c）,就如由(N∧C∧N)配体的H15和另一分子的乙炔基C19之间2.691 Å的短接触,以及159.6°的C15-H15-C19角度所表明的那样.沿a轴不存在短的分子间接触可加以识别.
络合物4结晶为薄片,其存在两个固有的孪生问题.不管怎样,我们能够利用直接的方法解决络合物4的晶体结构问题,这种方法基于一组具有87％完整性的X射线衍射数据（详情见支持信息）.该晶体结构具有单斜晶的空间基团,络合物的每三个晶体学上独立的分子带半个己烷溶剂化物分子.这些分子的(N∧C∧N)Pt平面与[10-1]晶面是接近于共平面的（参见支持信息的图）.在两个相邻分子之间的金属间距离为3.229、4.879和3.710 Å.类似于晶体1,其具有2.646 Å短接触距离和163.6°接触角的分子间C-H…π（C≡C）互作用已经在沿c轴的每两个相邻的分子之间得到识别.