AFLOW Prototype: AB3C_hP20_190_ac_i_f
Prototype | : | CsO3S |
AFLOW prototype label | : | AB3C_hP20_190_ac_i_f |
Strukturbericht designation | : | $K1_{2}$ |
Pearson symbol | : | hP20 |
Space group number | : | 190 |
Space group symbol | : | $P\bar{6}2c$ |
AFLOW prototype command | : | aflow --proto=AB3C_hP20_190_ac_i_f --params=$a$,$c/a$,$z_{3}$,$x_{4}$,$y_{4}$,$z_{4}$ |
which gives the S2O6 group exactly the same shape it has in K2S2O6 [ Strukturbericht $K11$].
Basis vectors:
\[ \begin{array}{ccccccc} & & \text{Lattice Coordinates} & & \text{Cartesian Coordinates} &\text{Wyckoff Position} & \text{Atom Type} \\ \mathbf{B}_{1} & = & 0 \, \mathbf{a}_{1} + 0 \, \mathbf{a}_{2} + 0 \, \mathbf{a}_{3} & = & 0 \, \mathbf{\hat{x}} + 0 \, \mathbf{\hat{y}} + 0 \, \mathbf{\hat{z}} & \left(2a\right) & \text{Cs I} \\ \mathbf{B}_{2} & = & \frac{1}{2} \, \mathbf{a}_{3} & = & \frac{1}{2}c \, \mathbf{\hat{z}} & \left(2a\right) & \text{Cs I} \\ \mathbf{B}_{3} & = & \frac{1}{3} \, \mathbf{a}_{1} + \frac{2}{3} \, \mathbf{a}_{2} + \frac{1}{4} \, \mathbf{a}_{3} & = & \frac{1}{2}a \, \mathbf{\hat{x}} + \frac{1}{2\sqrt{3}}a \, \mathbf{\hat{y}} + \frac{1}{4}c \, \mathbf{\hat{z}} & \left(2c\right) & \text{Cs II} \\ \mathbf{B}_{4} & = & \frac{2}{3} \, \mathbf{a}_{1} + \frac{1}{3} \, \mathbf{a}_{2} + \frac{3}{4} \, \mathbf{a}_{3} & = & \frac{1}{2}a \, \mathbf{\hat{x}}- \frac{1}{2\sqrt{3}}a \, \mathbf{\hat{y}} + \frac{3}{4}c \, \mathbf{\hat{z}} & \left(2c\right) & \text{Cs II} \\ \mathbf{B}_{5} & = & \frac{1}{3} \, \mathbf{a}_{1} + \frac{2}{3} \, \mathbf{a}_{2} + z_{3} \, \mathbf{a}_{3} & = & \frac{1}{2}a \, \mathbf{\hat{x}} + \frac{1}{2\sqrt{3}}a \, \mathbf{\hat{y}} + z_{3}c \, \mathbf{\hat{z}} & \left(4f\right) & \text{S} \\ \mathbf{B}_{6} & = & \frac{1}{3} \, \mathbf{a}_{1} + \frac{2}{3} \, \mathbf{a}_{2} + \left(\frac{1}{2} - z_{3}\right) \, \mathbf{a}_{3} & = & \frac{1}{2}a \, \mathbf{\hat{x}} + \frac{1}{2\sqrt{3}}a \, \mathbf{\hat{y}} + \left(\frac{1}{2}-z_{3}\right)c \, \mathbf{\hat{z}} & \left(4f\right) & \text{S} \\ \mathbf{B}_{7} & = & \frac{2}{3} \, \mathbf{a}_{1} + \frac{1}{3} \, \mathbf{a}_{2}-z_{3} \, \mathbf{a}_{3} & = & \frac{1}{2}a \, \mathbf{\hat{x}}- \frac{1}{2\sqrt{3}}a \, \mathbf{\hat{y}}-z_{3}c \, \mathbf{\hat{z}} & \left(4f\right) & \text{S} \\ \mathbf{B}_{8} & = & \frac{2}{3} \, \mathbf{a}_{1} + \frac{1}{3} \, \mathbf{a}_{2} + \left(\frac{1}{2} +z_{3}\right) \, \mathbf{a}_{3} & = & \frac{1}{2}a \, \mathbf{\hat{x}}- \frac{1}{2\sqrt{3}}a \, \mathbf{\hat{y}} + \left(\frac{1}{2} +z_{3}\right)c \, \mathbf{\hat{z}} & \left(4f\right) & \text{S} \\ \mathbf{B}_{9} & = & x_{4} \, \mathbf{a}_{1} + y_{4} \, \mathbf{a}_{2} + z_{4} \, \mathbf{a}_{3} & = & \frac{1}{2}\left(x_{4}+y_{4}\right)a \, \mathbf{\hat{x}} + \frac{\sqrt{3}}{2}\left(-x_{4}+y_{4}\right)a \, \mathbf{\hat{y}} + z_{4}c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \mathbf{B}_{10} & = & -y_{4} \, \mathbf{a}_{1} + \left(x_{4}-y_{4}\right) \, \mathbf{a}_{2} + z_{4} \, \mathbf{a}_{3} & = & \left(\frac{1}{2}x_{4}-y_{4}\right)a \, \mathbf{\hat{x}} + \frac{\sqrt{3}}{2}x_{4}a \, \mathbf{\hat{y}} + z_{4}c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \mathbf{B}_{11} & = & \left(-x_{4}+y_{4}\right) \, \mathbf{a}_{1}-x_{4} \, \mathbf{a}_{2} + z_{4} \, \mathbf{a}_{3} & = & \left(-x_{4}+\frac{1}{2}y_{4}\right)a \, \mathbf{\hat{x}}-\frac{\sqrt{3}}{2}y_{4}a \, \mathbf{\hat{y}} + z_{4}c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \mathbf{B}_{12} & = & x_{4} \, \mathbf{a}_{1} + y_{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} - z_{4}\right) \, \mathbf{a}_{3} & = & \frac{1}{2}\left(x_{4}+y_{4}\right)a \, \mathbf{\hat{x}} + \frac{\sqrt{3}}{2}\left(-x_{4}+y_{4}\right)a \, \mathbf{\hat{y}} + \left(\frac{1}{2}-z_{4}\right)c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \mathbf{B}_{13} & = & -y_{4} \, \mathbf{a}_{1} + \left(x_{4}-y_{4}\right) \, \mathbf{a}_{2} + \left(\frac{1}{2} - z_{4}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2}x_{4}-y_{4}\right)a \, \mathbf{\hat{x}} + \frac{\sqrt{3}}{2}x_{4}a \, \mathbf{\hat{y}} + \left(\frac{1}{2}-z_{4}\right)c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \mathbf{B}_{14} & = & \left(-x_{4}+y_{4}\right) \, \mathbf{a}_{1}-x_{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} - z_{4}\right) \, \mathbf{a}_{3} & = & \left(-x_{4}+\frac{1}{2}y_{4}\right)a \, \mathbf{\hat{x}}-\frac{\sqrt{3}}{2}y_{4}a \, \mathbf{\hat{y}} + \left(\frac{1}{2}-z_{4}\right)c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \mathbf{B}_{15} & = & y_{4} \, \mathbf{a}_{1} + x_{4} \, \mathbf{a}_{2}-z_{4} \, \mathbf{a}_{3} & = & \frac{1}{2}\left(x_{4}+y_{4}\right)a \, \mathbf{\hat{x}} + \frac{\sqrt{3}}{2}\left(x_{4}-y_{4}\right)a \, \mathbf{\hat{y}}-z_{4}c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \mathbf{B}_{16} & = & \left(x_{4}-y_{4}\right) \, \mathbf{a}_{1}-y_{4} \, \mathbf{a}_{2}-z_{4} \, \mathbf{a}_{3} & = & \left(\frac{1}{2}x_{4}-y_{4}\right)a \, \mathbf{\hat{x}}-\frac{\sqrt{3}}{2}x_{4}a \, \mathbf{\hat{y}}-z_{4}c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \mathbf{B}_{17} & = & -x_{4} \, \mathbf{a}_{1} + \left(-x_{4}+y_{4}\right) \, \mathbf{a}_{2}-z_{4} \, \mathbf{a}_{3} & = & \left(-x_{4}+\frac{1}{2}y_{4}\right)a \, \mathbf{\hat{x}} + \frac{\sqrt{3}}{2}y_{4}a \, \mathbf{\hat{y}}-z_{4}c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \mathbf{B}_{18} & = & y_{4} \, \mathbf{a}_{1} + x_{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} +z_{4}\right) \, \mathbf{a}_{3} & = & \frac{1}{2}\left(x_{4}+y_{4}\right)a \, \mathbf{\hat{x}} + \frac{\sqrt{3}}{2}\left(x_{4}-y_{4}\right)a \, \mathbf{\hat{y}} + \left(\frac{1}{2} +z_{4}\right)c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \mathbf{B}_{19} & = & \left(x_{4}-y_{4}\right) \, \mathbf{a}_{1}-y_{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} +z_{4}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2}x_{4}-y_{4}\right)a \, \mathbf{\hat{x}}-\frac{\sqrt{3}}{2}x_{4}a \, \mathbf{\hat{y}} + \left(\frac{1}{2} +z_{4}\right)c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \mathbf{B}_{20} & = & -x_{4} \, \mathbf{a}_{1} + \left(-x_{4}+y_{4}\right) \, \mathbf{a}_{2} + \left(\frac{1}{2} +z_{4}\right) \, \mathbf{a}_{3} & = & \left(-x_{4}+\frac{1}{2}y_{4}\right)a \, \mathbf{\hat{x}} + \frac{\sqrt{3}}{2}y_{4}a \, \mathbf{\hat{y}} + \left(\frac{1}{2} +z_{4}\right)c \, \mathbf{\hat{z}} & \left(12i\right) & \text{O} \\ \end{array} \]