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The $\pi -\pi$ phosphorescence spectrum (PS) of N-heterocyclic molecules is generally negative-polarized, i.e. the emitting oscillator is oriented perpendicularly to the molecular plane. The phosphorescence results from a transition, forbidden by the selection rule $\Delta S=0$, which obtains its intensity from first-order perturbations (direct spin-orbit coupling). The polarization of the PS of phenazine in 3-methylpentane as a solvent was measured. If the strong 0,0 band, which was negative-polarized, is disregarded: the other bands are found about equally positive- and negative-polarized. This indicates the presence of a rather strong, inplane component of the emitting oscillator. The Raman and infrared spectra, and a sharp PS of phenazine in n-heptane as a solvent, were also recorded. A vibrational analysis of these spectra was performed and the results compared with the polarization of the PS of phenazine in 3-methylpentane. All negative-polarized lines of the PS could be assigned to Raman frequencies, and all those of positive-polarized lines to infrared modes. The presence of these positive-polarized lines (out-of-plane modes) of appreciable intensity indicated rather strong, vibronic spin-orbit coupling (second-order perturbations) in phenazine. The possible coupling schemes could be: $S(\pi ,\pi \ast )⟷SOT(\pi ;\pi \ast )⟷OPviharT(\pi ,\pi \ast )$ respectively, $S(\pi ,\pi \ast )⟷OPviharS(\pi ;\pi \ast )⟷SOT(\pi ,\pi \ast )$ Both coupling schemes involve out of plane vibrations (OP vibr.).