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相互作用的v6 = 2级D3Si35Cl:一种从440 - 590和900 - 1100 cm-1区域的高分辨率光谱确定基态常数的工具。

The Interacting v6 = 2 Levels of D3Si35Cl: A Tool to Determine Ground State Constants from High-Resolution Spectra in the 440-590 and 900-1100 cm-1 Regions.

作者信息

Ceausu A, Graner G, Bürger H, Mkadmi EB, Pracna P

机构信息

Laboratoire de Physique Moléculaire et Applications, CNRS, Laboratoire Associé aux Universités Pierre et Marie Curie et Paris-Sud, Bât. 350, Campus d'Orsay, Orsay Cedex, F-91405, France

出版信息

J Mol Spectrosc. 1999 Mar;194(1):128-141. doi: 10.1006/jmsp.1998.7768.

DOI:10.1006/jmsp.1998.7768
PMID:9986782
Abstract

High-resolution FTIR spectra of monoisotopic D3Si35Cl have been recorded in the regions 440-590 cm-1 (nu3/nu6) and 900-1100 cm-1 (2nu6/nu3 + nu6/2nu3). A detailed rovibrational study was done for the 2nu06, 2nu-/+26, and 2nu3 overtone bands and for the 2nu+/-26-nu+/-16, 2nu06-nu-/+16, and (nu3 + nu+/-16)-nu+/-16 hot bands. For the first time the interactions occurring between the v6 = 2, v3 = v6 = 1, and v3 = 2 levels of any trideuterated silyl halide have been analyzed. The (nu3 + nu+/-16)-nu+/-16 hot band served to obtain accurate energies of the v3 = v6 = 1 level, the nu3 + nu6 combination band not being detectable on our spectra. The first experimental determination of A0 and D0K for this molecule was undertaken by a well-established method, using the nu6, 2nu+/-26-nu+/-16, and 2nu-/+26 bands. Ground state energy differences DeltaK(J) = E0(J, K) - E0(J, K - 3) were calculated for K values from 2 to 16. By a least-squares fit of 163 such differences, the A0 and D0K values thus obtained were (in cm-1): A0 = 1.4278230(8) and D0K = 5.3916(31) x 10(-6). Copyright 1999 Academic Press.

摘要

已记录了单同位素D3Si35Cl在440 - 590厘米-1(ν3/ν6)和900 - 1100厘米-1(2ν6/ν3 + ν6/2ν3)区域的高分辨率傅里叶变换红外光谱。对2ν06、2ν±26和2ν3泛音带以及2ν±26 - ν±16、2ν06 - ν±16和(ν3 + ν±16) - ν±16热带进行了详细的振转研究。首次分析了任何三氘代甲硅烷基卤化物的v6 = 2、v3 = v6 = 1和v3 = 2能级之间发生的相互作用。(ν3 + ν±16) - ν±16热带用于获得v3 = v6 = 1能级的精确能量,我们的光谱上未检测到ν3 + ν6组合带。通过一种成熟的方法,利用ν6、2ν±26 - ν±16和2ν±26带对该分子的A0和D0K进行了首次实验测定。计算了K值从2到16时的基态能量差ΔK(J) = E0(J, K) - E0(J, K - 3)。通过对163个这样的差值进行最小二乘拟合,得到的A0和D0K值(单位为厘米-1)为:A0 = 1.4278230(8),D0K = 5.3916(31)×10(-6)。版权所有1999年学术出版社。

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