AFLOW Prototype: A6BC3D_hR22_167_f_b_e_a
Prototype | : | Cl6FeK3Na |
AFLOW prototype label | : | A6BC3D_hR22_167_f_b_e_a |
Strukturbericht designation | : | None |
Pearson symbol | : | hR22 |
Space group number | : | 167 |
Space group symbol | : | $R\bar{3}c$ |
AFLOW prototype command | : | aflow --proto=A6BC3D_hR22_167_f_b_e_a --params=$a$,$c/a$,$x_{3}$,$x_{4}$,$y_{4}$,$z_{4}$ |
Basis vectors:
\[ \begin{array}{ccccccc} & & \text{Lattice Coordinates} & & \text{Cartesian Coordinates} &\text{Wyckoff Position} & \text{Atom Type} \\ \mathbf{B}_{1} & = & \frac{1}{4} \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + \frac{1}{4} \, \mathbf{a}_{3} & = & \frac{1}{4}c \, \mathbf{\hat{z}} & \left(2a\right) & \text{Na} \\ \mathbf{B}_{2} & = & \frac{3}{4} \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2} + \frac{3}{4} \, \mathbf{a}_{3} & = & \frac{3}{4}c \, \mathbf{\hat{z}} & \left(2a\right) & \text{Na} \\ \mathbf{B}_{3} & = & 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(2b\right) & \text{Fe} \\ \mathbf{B}_{4} & = & \frac{1}{2} \, \mathbf{a}_{1} + \frac{1}{2} \, \mathbf{a}_{2} + \frac{1}{2} \, \mathbf{a}_{3} & = & \frac{1}{2}c \, \mathbf{\hat{z}} & \left(2b\right) & \text{Fe} \\ \mathbf{B}_{5} & = & x_{3} \, \mathbf{a}_{1} + \left(\frac{1}{2} - x_{3}\right) \, \mathbf{a}_{2} + \frac{1}{4} \, \mathbf{a}_{3} & = & \left(- \frac{1}{8} +\frac{1}{2}x_{3}\right)a \, \mathbf{\hat{x}} + \left(\frac{\sqrt{3}}{8}-\frac{\sqrt{3}}{2}x_{3}\right)a \, \mathbf{\hat{y}} + \frac{1}{4}c \, \mathbf{\hat{z}} & \left(6e\right) & \text{K} \\ \mathbf{B}_{6} & = & \frac{1}{4} \, \mathbf{a}_{1} + x_{3} \, \mathbf{a}_{2} + \left(\frac{1}{2} - x_{3}\right) \, \mathbf{a}_{3} & = & \left(- \frac{1}{8} +\frac{1}{2}x_{3}\right)a \, \mathbf{\hat{x}} + \left(- \frac{\sqrt{3}}{8} +\frac{\sqrt{3}}{2}x_{3}\right)a \, \mathbf{\hat{y}} + \frac{1}{4}c \, \mathbf{\hat{z}} & \left(6e\right) & \text{K} \\ \mathbf{B}_{7} & = & \left(\frac{1}{2} - x_{3}\right) \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + x_{3} \, \mathbf{a}_{3} & = & \left(\frac{1}{4}-x_{3}\right)a \, \mathbf{\hat{x}} + \frac{1}{4}c \, \mathbf{\hat{z}} & \left(6e\right) & \text{K} \\ \mathbf{B}_{8} & = & -x_{3} \, \mathbf{a}_{1} + \left(\frac{1}{2} +x_{3}\right) \, \mathbf{a}_{2} + \frac{3}{4} \, \mathbf{a}_{3} & = & -\left(\frac{1}{2}x_{3}+\frac{3}{8}\right)a \, \mathbf{\hat{x}} + \left(\frac{1}{8\sqrt{3}} +\frac{\sqrt{3}}{2}x_{3}\right)a \, \mathbf{\hat{y}} + \frac{5}{12}c \, \mathbf{\hat{z}} & \left(6e\right) & \text{K} \\ \mathbf{B}_{9} & = & \frac{3}{4} \, \mathbf{a}_{1}-x_{3} \, \mathbf{a}_{2} + \left(\frac{1}{2} +x_{3}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{8}-\frac{1}{2}x_{3}\right)a \, \mathbf{\hat{x}}-\left(\frac{\sqrt{3}}{2}x_{3}+\frac{5}{8\sqrt{3}}\right)a \, \mathbf{\hat{y}} + \frac{5}{12}c \, \mathbf{\hat{z}} & \left(6e\right) & \text{K} \\ \mathbf{B}_{10} & = & \left(\frac{1}{2} +x_{3}\right) \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2}-x_{3} \, \mathbf{a}_{3} & = & \left(\frac{1}{4} +x_{3}\right)a \, \mathbf{\hat{x}} + \frac{1}{2\sqrt{3}}a \, \mathbf{\hat{y}} + \frac{5}{12}c \, \mathbf{\hat{z}} & \left(6e\right) & \text{K} \\ \mathbf{B}_{11} & = & x_{4} \, \mathbf{a}_{1} + y_{4} \, \mathbf{a}_{2} + z_{4} \, \mathbf{a}_{3} & = & \frac{1}{2}\left(x_{4}-z_{4}\right)a \, \mathbf{\hat{x}} + \left(-\frac{1}{2\sqrt{3}}x_{4}+\frac{1}{\sqrt{3}}y_{4}-\frac{1}{2\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}} + \frac{1}{3}\left(x_{4}+y_{4}+z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \mathbf{B}_{12} & = & z_{4} \, \mathbf{a}_{1} + x_{4} \, \mathbf{a}_{2} + y_{4} \, \mathbf{a}_{3} & = & \frac{1}{2}\left(-y_{4}+z_{4}\right)a \, \mathbf{\hat{x}} + \left(\frac{1}{\sqrt{3}}x_{4}-\frac{1}{2\sqrt{3}}y_{4}-\frac{1}{2\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}} + \frac{1}{3}\left(x_{4}+y_{4}+z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \mathbf{B}_{13} & = & y_{4} \, \mathbf{a}_{1} + z_{4} \, \mathbf{a}_{2} + x_{4} \, \mathbf{a}_{3} & = & \frac{1}{2}\left(-x_{4}+y_{4}\right)a \, \mathbf{\hat{x}} + \left(-\frac{1}{2\sqrt{3}}x_{4}-\frac{1}{2\sqrt{3}}y_{4}+\frac{1}{\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}} + \frac{1}{3}\left(x_{4}+y_{4}+z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \mathbf{B}_{14} & = & \left(\frac{1}{2} - z_{4}\right) \, \mathbf{a}_{1} + \left(\frac{1}{2} - y_{4}\right) \, \mathbf{a}_{2} + \left(\frac{1}{2} - x_{4}\right) \, \mathbf{a}_{3} & = & \frac{1}{2}\left(x_{4}-z_{4}\right)a \, \mathbf{\hat{x}} + \left(\frac{1}{2\sqrt{3}}x_{4}-\frac{1}{\sqrt{3}}y_{4}+\frac{1}{2\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}} + \left(\frac{1}{2} - \frac{1}{3}x_{4} - \frac{1}{3}y_{4} - \frac{1}{3}z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \mathbf{B}_{15} & = & \left(\frac{1}{2} - y_{4}\right) \, \mathbf{a}_{1} + \left(\frac{1}{2} - x_{4}\right) \, \mathbf{a}_{2} + \left(\frac{1}{2} - z_{4}\right) \, \mathbf{a}_{3} & = & \frac{1}{2}\left(-y_{4}+z_{4}\right)a \, \mathbf{\hat{x}} + \left(-\frac{1}{\sqrt{3}}x_{4}+\frac{1}{2\sqrt{3}}y_{4}+\frac{1}{2\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}} + \left(\frac{1}{2} - \frac{1}{3}x_{4} - \frac{1}{3}y_{4} - \frac{1}{3}z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \mathbf{B}_{16} & = & \left(\frac{1}{2} - x_{4}\right) \, \mathbf{a}_{1} + \left(\frac{1}{2} - z_{4}\right) \, \mathbf{a}_{2} + \left(\frac{1}{2} - y_{4}\right) \, \mathbf{a}_{3} & = & \frac{1}{2}\left(-x_{4}+y_{4}\right)a \, \mathbf{\hat{x}} + \left(\frac{1}{2\sqrt{3}}x_{4}+\frac{1}{2\sqrt{3}}y_{4}-\frac{1}{\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}} + \left(\frac{1}{2} - \frac{1}{3}x_{4} - \frac{1}{3}y_{4} - \frac{1}{3}z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \mathbf{B}_{17} & = & -x_{4} \, \mathbf{a}_{1}-y_{4} \, \mathbf{a}_{2}-z_{4} \, \mathbf{a}_{3} & = & \frac{1}{2}\left(-x_{4}+z_{4}\right)a \, \mathbf{\hat{x}} + \left(\frac{1}{2\sqrt{3}}x_{4}-\frac{1}{\sqrt{3}}y_{4}+\frac{1}{2\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}}-\frac{1}{3}\left(x_{4}+y_{4}+z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \mathbf{B}_{18} & = & -z_{4} \, \mathbf{a}_{1}-x_{4} \, \mathbf{a}_{2}-y_{4} \, \mathbf{a}_{3} & = & \frac{1}{2}\left(y_{4}-z_{4}\right)a \, \mathbf{\hat{x}} + \left(-\frac{1}{\sqrt{3}}x_{4}+\frac{1}{2\sqrt{3}}y_{4}+\frac{1}{2\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}}-\frac{1}{3}\left(x_{4}+y_{4}+z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \mathbf{B}_{19} & = & -y_{4} \, \mathbf{a}_{1}-z_{4} \, \mathbf{a}_{2}-x_{4} \, \mathbf{a}_{3} & = & \frac{1}{2}\left(x_{4}-y_{4}\right)a \, \mathbf{\hat{x}} + \left(\frac{1}{2\sqrt{3}}x_{4}+\frac{1}{2\sqrt{3}}y_{4}-\frac{1}{\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}}-\frac{1}{3}\left(x_{4}+y_{4}+z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \mathbf{B}_{20} & = & \left(\frac{1}{2} +z_{4}\right) \, \mathbf{a}_{1} + \left(\frac{1}{2} +y_{4}\right) \, \mathbf{a}_{2} + \left(\frac{1}{2} +x_{4}\right) \, \mathbf{a}_{3} & = & \frac{1}{2}\left(-x_{4}+z_{4}\right)a \, \mathbf{\hat{x}} + \left(-\frac{1}{2\sqrt{3}}x_{4}+\frac{1}{\sqrt{3}}y_{4}-\frac{1}{2\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}} + \left(\frac{1}{2} +\frac{1}{3}x_{4} + \frac{1}{3}y_{4} + \frac{1}{3}z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \mathbf{B}_{21} & = & \left(\frac{1}{2} +y_{4}\right) \, \mathbf{a}_{1} + \left(\frac{1}{2} +x_{4}\right) \, \mathbf{a}_{2} + \left(\frac{1}{2} +z_{4}\right) \, \mathbf{a}_{3} & = & \frac{1}{2}\left(y_{4}-z_{4}\right)a \, \mathbf{\hat{x}} + \left(\frac{1}{\sqrt{3}}x_{4}-\frac{1}{2\sqrt{3}}y_{4}-\frac{1}{2\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}} + \left(\frac{1}{2} +\frac{1}{3}x_{4} + \frac{1}{3}y_{4} + \frac{1}{3}z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \mathbf{B}_{22} & = & \left(\frac{1}{2} +x_{4}\right) \, \mathbf{a}_{1} + \left(\frac{1}{2} +z_{4}\right) \, \mathbf{a}_{2} + \left(\frac{1}{2} +y_{4}\right) \, \mathbf{a}_{3} & = & \frac{1}{2}\left(x_{4}-y_{4}\right)a \, \mathbf{\hat{x}} + \left(-\frac{1}{2\sqrt{3}}x_{4}-\frac{1}{2\sqrt{3}}y_{4}+\frac{1}{\sqrt{3}}z_{4}\right)a \, \mathbf{\hat{y}} + \left(\frac{1}{2} +\frac{1}{3}x_{4} + \frac{1}{3}y_{4} + \frac{1}{3}z_{4}\right)c \, \mathbf{\hat{z}} & \left(12f\right) & \text{Cl} \\ \end{array} \]