AFLOW Prototype: A2B3_oP20_62_2c_3c
Prototype | : | Cr3C2 |
AFLOW prototype label | : | A2B3_oP20_62_2c_3c |
Strukturbericht designation | : | $D5_{10}$ |
Pearson symbol | : | oP20 |
Space group number | : | 62 |
Space group symbol | : | $Pnma$ |
AFLOW prototype command | : | aflow --proto=A2B3_oP20_62_2c_3c --params=$a$,$b/a$,$c/a$,$x_{1}$,$z_{1}$,$x_{2}$,$z_{2}$,$x_{3}$,$z_{3}$,$x_{4}$,$z_{4}$,$x_{5}$,$z_{5}$ |
Basis vectors:
\[ \begin{array}{ccccccc} & & \text{Lattice Coordinates} & & \text{Cartesian Coordinates} &\text{Wyckoff Position} & \text{Atom Type} \\ \mathbf{B}_{1} & = & x_{1} \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + z_{1} \, \mathbf{a}_{3} & = & x_{1}a \, \mathbf{\hat{x}} + \frac{1}{4}b \, \mathbf{\hat{y}} + z_{1}c \, \mathbf{\hat{z}} & \left(4c\right) & \text{C I} \\ \mathbf{B}_{2} & = & \left(\frac{1}{2} - x_{1}\right) \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} +z_{1}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2} - x_{1}\right)a \, \mathbf{\hat{x}} + \frac{3}{4}b \, \mathbf{\hat{y}} + \left(\frac{1}{2} +z_{1}\right)c \, \mathbf{\hat{z}} & \left(4c\right) & \text{C I} \\ \mathbf{B}_{3} & = & -x_{1} \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2}-z_{1} \, \mathbf{a}_{3} & = & -x_{1}a \, \mathbf{\hat{x}} + \frac{3}{4}b \, \mathbf{\hat{y}}-z_{1}c \, \mathbf{\hat{z}} & \left(4c\right) & \text{C I} \\ \mathbf{B}_{4} & = & \left(\frac{1}{2} +x_{1}\right) \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} - z_{1}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2} +x_{1}\right)a \, \mathbf{\hat{x}} + \frac{1}{4}b \, \mathbf{\hat{y}} + \left(\frac{1}{2} - z_{1}\right)c \, \mathbf{\hat{z}} & \left(4c\right) & \text{C I} \\ \mathbf{B}_{5} & = & x_{2} \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + z_{2} \, \mathbf{a}_{3} & = & x_{2}a \, \mathbf{\hat{x}} + \frac{1}{4}b \, \mathbf{\hat{y}} + z_{2}c \, \mathbf{\hat{z}} & \left(4c\right) & \text{C II} \\ \mathbf{B}_{6} & = & \left(\frac{1}{2} - x_{2}\right) \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} +z_{2}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2} - x_{2}\right)a \, \mathbf{\hat{x}} + \frac{3}{4}b \, \mathbf{\hat{y}} + \left(\frac{1}{2} +z_{2}\right)c \, \mathbf{\hat{z}} & \left(4c\right) & \text{C II} \\ \mathbf{B}_{7} & = & -x_{2} \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2}-z_{2} \, \mathbf{a}_{3} & = & -x_{2}a \, \mathbf{\hat{x}} + \frac{3}{4}b \, \mathbf{\hat{y}}-z_{2}c \, \mathbf{\hat{z}} & \left(4c\right) & \text{C II} \\ \mathbf{B}_{8} & = & \left(\frac{1}{2} +x_{2}\right) \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} - z_{2}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2} +x_{2}\right)a \, \mathbf{\hat{x}} + \frac{1}{4}b \, \mathbf{\hat{y}} + \left(\frac{1}{2} - z_{2}\right)c \, \mathbf{\hat{z}} & \left(4c\right) & \text{C II} \\ \mathbf{B}_{9} & = & x_{3} \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + z_{3} \, \mathbf{a}_{3} & = & x_{3}a \, \mathbf{\hat{x}} + \frac{1}{4}b \, \mathbf{\hat{y}} + z_{3}c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr I} \\ \mathbf{B}_{10} & = & \left(\frac{1}{2} - x_{3}\right) \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} +z_{3}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2} - x_{3}\right)a \, \mathbf{\hat{x}} + \frac{3}{4}b \, \mathbf{\hat{y}} + \left(\frac{1}{2} +z_{3}\right)c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr I} \\ \mathbf{B}_{11} & = & -x_{3} \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2}-z_{3} \, \mathbf{a}_{3} & = & -x_{3}a \, \mathbf{\hat{x}} + \frac{3}{4}b \, \mathbf{\hat{y}}-z_{3}c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr I} \\ \mathbf{B}_{12} & = & \left(\frac{1}{2} +x_{3}\right) \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} - z_{3}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2} +x_{3}\right)a \, \mathbf{\hat{x}} + \frac{1}{4}b \, \mathbf{\hat{y}} + \left(\frac{1}{2} - z_{3}\right)c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr I} \\ \mathbf{B}_{13} & = & x_{4} \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + z_{4} \, \mathbf{a}_{3} & = & x_{4}a \, \mathbf{\hat{x}} + \frac{1}{4}b \, \mathbf{\hat{y}} + z_{4}c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr II} \\ \mathbf{B}_{14} & = & \left(\frac{1}{2} - x_{4}\right) \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} +z_{4}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2} - x_{4}\right)a \, \mathbf{\hat{x}} + \frac{3}{4}b \, \mathbf{\hat{y}} + \left(\frac{1}{2} +z_{4}\right)c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr II} \\ \mathbf{B}_{15} & = & -x_{4} \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2}-z_{4} \, \mathbf{a}_{3} & = & -x_{4}a \, \mathbf{\hat{x}} + \frac{3}{4}b \, \mathbf{\hat{y}}-z_{4}c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr II} \\ \mathbf{B}_{16} & = & \left(\frac{1}{2} +x_{4}\right) \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} - z_{4}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2} +x_{4}\right)a \, \mathbf{\hat{x}} + \frac{1}{4}b \, \mathbf{\hat{y}} + \left(\frac{1}{2} - z_{4}\right)c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr II} \\ \mathbf{B}_{17} & = & x_{5} \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + z_{5} \, \mathbf{a}_{3} & = & x_{5}a \, \mathbf{\hat{x}} + \frac{1}{4}b \, \mathbf{\hat{y}} + z_{5}c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr III} \\ \mathbf{B}_{18} & = & \left(\frac{1}{2} - x_{5}\right) \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} +z_{5}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2} - x_{5}\right)a \, \mathbf{\hat{x}} + \frac{3}{4}b \, \mathbf{\hat{y}} + \left(\frac{1}{2} +z_{5}\right)c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr III} \\ \mathbf{B}_{19} & = & -x_{5} \, \mathbf{a}_{1} + \frac{3}{4} \, \mathbf{a}_{2}-z_{5} \, \mathbf{a}_{3} & = & -x_{5}a \, \mathbf{\hat{x}} + \frac{3}{4}b \, \mathbf{\hat{y}}-z_{5}c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr III} \\ \mathbf{B}_{20} & = & \left(\frac{1}{2} +x_{5}\right) \, \mathbf{a}_{1} + \frac{1}{4} \, \mathbf{a}_{2} + \left(\frac{1}{2} - z_{5}\right) \, \mathbf{a}_{3} & = & \left(\frac{1}{2} +x_{5}\right)a \, \mathbf{\hat{x}} + \frac{1}{4}b \, \mathbf{\hat{y}} + \left(\frac{1}{2} - z_{5}\right)c \, \mathbf{\hat{z}} & \left(4c\right) & \text{Cr III} \\ \end{array} \]