On the basis of dielectric studies on polycrystalline specimens, ${\mathrm{Bi}}_4$${\mathrm{Ti}}_3$${\mathrm{O}}_{12}$ has been established as a ferroelectric with a Curie temperature of 675°C. The symmetry is orthorhombic with $a=5.411\mathrm{}$ A, $c=32.83\mathrm{}$ A, and $\frac{b}{a}=1.007\mathrm{}$ at 25°C. A high-temperature x-ray study revealed a symmetry change to tetragonal at 675°C. The polar axis is probably the orthorhombic $b$ axis. ${\mathrm{Bi}}_4$${\mathrm{Ti}}_3$${\mathrm{O}}_{12}$ is a member of oxides with the general formula ${\left({\mathrm{Bi}}_2{\mathrm{O}}_2\right)}^{2+}{\left({\mathrm{Me}}_{x-1\mathrm{}}{R}_x{\mathrm{O}}_{3x+1}\right)}^{2-}$. According to Aurivillius, the crystal structure of ${\mathrm{Bi}}_4$${\mathrm{Ti}}_3$${\mathrm{O}}_{12}$ comprises a stacking of ${\mathrm{Bi}}_2$${\mathrm{O}}_2$ and perovskite-like ${\mathrm{Bi}}_2$${\mathrm{Ti}}_3$${\mathrm{O}}_{10}$ layers along the pseudotetragonal $c$ axis. Multiple ion substitutions of (${\mathrm{Bi}}^{3+}$${\mathrm{Ti}}^{4+}$) in ${\left({\mathrm{Bi}}_2{\mathrm{O}}_2\right)}^{2+}$${\left({\mathrm{Bi}}_2{\mathrm{Ti}}_3{\mathrm{O}}_{10}\right)}^{2-}$ by (${\mathrm{Me}}^{2+}{\mathrm{Nb}}^{5+}$) where ${\mathrm{Me}}^{2+}=\mathrm{B}\mathrm{a},\mathrm{P}\mathrm{b},\mathrm{o}\mathrm{r}\mathrm{S}\mathrm{r}$ lead to a steep decrease of the Curie temperature.

• Received 21 December 1960

DOI:https://doi.org/10.1103/PhysRev.122.804