Optical and Zeeman studies of the first excited singlet state of zinc porphin in a single crystal of n -octane: Evidence for Jahn-Teller instability

Author： Canters G.W. van Egmond J. Schaafsma T.J. van der Waals J.H.

Publisher： Taylor & Francis Ltd

ISSN： 1362-3028

Source： Molecular Physics, Vol.24, Iss.6, 1972-12, pp. : 1203-1215

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Abstract

The absorption and fluorescence spectra of Zn porphin in an n -octane single crystal at 4·2 K are reported in the region between 17 400 and 18 500 cm -1 . A strong peak appears in both spectra at 17 961 cm -1 and is assigned to the origin of one component (| x , 0>) of the nearly degenerate Q -band. In absorption a second strong line occurs at a frequency δ = 109 cm -1 above the first; a corresponding line is almost totally absent in the emission spectrum at 4·2 K, but it appears as a hot band of appreciable intensity when the temperature is raised to 80 K. This feature is assigned to the origin of the other component (| y , 0>) of the Q -band. The lifting of the degeneracy of the Q -band is interpreted as a crystal field splitting of the Jahn-Teller unstable 1 E u state. The Zeeman effect is investigated for the 0–0 transition of the phosphorescence spectrum and the | x , 0> and | y , 0> components of the Q -band absorption spectrum. From the phosphorescence experiment it is concluded that the great majority of the ZnP guests are oriented in the host with an angle of about 25° between the out-of-plane molecular axes and the crystal a -axis. The analysis of the Zeeman effect in absorption (H//crystal a -axis) is complicated by the Jahn-Teller instability which causes two additional unknowns to appear in the problem: the frequency &ngr; and the nuclear displacement parameter &agr; of the active mode. When not making an assumption about these parameters one can only derive a lower limit for the matrix element of the orbital angular momentum between the two electronic components: Λ > 4·6. If is identified with the low-frequency mode of 180 cm -1 appearing in the absorption spectrum, then it follows that Λ = 6·1 ± 0·6 with &agr; = 1·4 ± 0·1.