The Vibration-Rotation Spectrum of Si H4

The infra-red bands of Si${\mathrm{H}}_{4}$, measured originally by Steward and Nielsen, have been remeasured at considerably higher resolving power. From a rotational analysis of the bands the equilibrium value of $B$ and values of $B$ for the vibration states ${\ensuremath{\nu}}_{2}$, ${\ensuremath{\nu}}_{3}$, ${\ensuremath{\nu}}_{4}$, ${\ensuremath{\nu}}_{2}+{\ensuremath{\nu}}_{3}$, and ${\ensuremath{\nu}}_{3}+{\ensuremath{\nu}}_{4}$ have been determined, $B$ being $\frac{h}{8{\ensuremath{\pi}}^{2}\mathrm{Ic}}$. These have, respectively, the values ${B}_{0}=3.00$ ${\mathrm{cm}}^{\ensuremath{-}1}$, ${B}^{\ensuremath{'}\ensuremath{'}}=2.96$ ${\mathrm{cm}}^{\ensuremath{-}1}$, ${B}^{\ensuremath{'}}({\ensuremath{\nu}}_{3})=2.95$ ${\mathrm{cm}}^{\ensuremath{-}1}$, $B_{+}^{}{}_{}{}^{\ensuremath{'}}({\ensuremath{\nu}}_{2})=3.06$ ${\mathrm{cm}}^{\ensuremath{-}1}$, $B_{\ensuremath{-}}^{}{}_{}{}^{\ensuremath{'}}({\ensuremath{\nu}}_{2})=3.02$ ${\mathrm{cm}}^{\ensuremath{-}1}$, ${B}^{\ensuremath{'}}({\ensuremath{\nu}}_{4})=2.89$ ${\mathrm{cm}}^{\ensuremath{-}1}$, $B_{+}^{}{}_{}{}^{\ensuremath{'}}({\ensuremath{\nu}}_{3}+{\ensuremath{\nu}}_{4})=3.05$ ${\mathrm{cm}}^{\ensuremath{-}1}$, $B_{0}^{}{}_{}{}^{\ensuremath{'}}({\ensuremath{\nu}}_{3}+{\ensuremath{\nu}}_{4})=2.97$ ${\mathrm{cm}}^{\ensuremath{-}1}$, $B_{\ensuremath{-}}^{}{}_{}{}^{\ensuremath{'}}({\ensuremath{\nu}}_{3}+{\ensuremath{\nu}}_{4})=2.94$ ${\mathrm{cm}}^{\ensuremath{-}1}$, $B_{+}^{}{}_{}{}^{\ensuremath{'}}({\ensuremath{\nu}}_{2}+{\ensuremath{\nu}}_{3})=3.05$ ${\mathrm{cm}}^{\ensuremath{-}1}$, and $B_{\ensuremath{-}}^{}{}_{}{}^{\ensuremath{'}}({\ensuremath{\nu}}_{2}+{\ensuremath{\nu}}_{3})=2.98$ ${\mathrm{cm}}^{\ensuremath{-}1}$. The centers of the bands have been determined to be the following: ${\ensuremath{\nu}}_{2}=974.6$, ${\ensuremath{\nu}}_{3}=2190.6$ ${\mathrm{cm}}^{\ensuremath{-}1}$, ${\ensuremath{\nu}}_{4}=914.2$ ${\mathrm{cm}}^{\ensuremath{-}1}$, $({\ensuremath{\nu}}_{2}+{\ensuremath{\nu}}_{3})=3148.8$, $({\ensuremath{\nu}}_{3}+{\ensuremath{\nu}}_{4})=3100.2$. The value of ${\ensuremath{\zeta}}_{4}=\frac{1}{2}\ensuremath{-}{\ensuremath{\zeta}}_{3}=0.454$ is in good agreement with the value predicted by Jahn for valence force fields. From the band centers the valence force constants are calculated.