Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A7B2C_tP40_128_egi_h_e-001

This structure originally had the label A7B2C_tP40_128_egi_h_e. Calls to that address will be redirected here.

If you are using this page, please cite:
D. Hicks, M. J. Mehl, E. Gossett, C. Toher, O. Levy, R. M. Hanson, G. L. W. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 2, Comp. Mat. Sci. 161, S1-S1011 (2019). (doi=10.1016/j.commatsci.2018.10.043)

Links to this page

https://aflow.org/p/YX16
or https://aflow.org/p/A7B2C_tP40_128_egi_h_e-001
or PDF Version

FeCu$_{2}$Al$_{7}$ ($E9_{a}$) Structure: A7B2C_tP40_128_egi_h_e-001

Picture of Structure; Click for Big Picture
Prototype Al$_{7}$Cu$_{2}$Fe
AFLOW prototype label A7B2C_tP40_128_egi_h_e-001
Strukturbericht designation $E9_{a}$
ICSD 57677
Pearson symbol tP40
Space group number 128
Space group symbol $P4/mnc$
AFLOW prototype command aflow --proto=A7B2C_tP40_128_egi_h_e-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak y_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}$

Other compounds with this structure

CoCu$_{2}$Al$_{7}$,  NiCu$_{3}$Al$_{6}$,  T(CoCuAl) -- an alloy with approximate composition Co$_{2}$Cu$_{4.9}$Al$_{17.7}$


  • Our original (Hicks, 2019) CIF for this structure has incorrect $z$-coordinates for the Al I and Fe I atoms. We have corrected the errors with this release.

\[ \begin{array}{ccc} \mathbf{a_{1}}&=&a \,\mathbf{\hat{x}}\\\mathbf{a_{2}}&=&a \,\mathbf{\hat{y}}\\\mathbf{a_{3}}&=&c \,\mathbf{\hat{z}} \end{array}\]

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $z_{1} \, \mathbf{a}_{3}$ = $c z_{1} \,\mathbf{\hat{z}}$ (4e) Al I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}- \left(z_{1} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}- c \left(z_{1} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4e) Al I
$\mathbf{B_{3}}$ = $- z_{1} \, \mathbf{a}_{3}$ = $- c z_{1} \,\mathbf{\hat{z}}$ (4e) Al I
$\mathbf{B_{4}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4e) Al I
$\mathbf{B_{5}}$ = $z_{2} \, \mathbf{a}_{3}$ = $c z_{2} \,\mathbf{\hat{z}}$ (4e) Fe I
$\mathbf{B_{6}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4e) Fe I
$\mathbf{B_{7}}$ = $- z_{2} \, \mathbf{a}_{3}$ = $- c z_{2} \,\mathbf{\hat{z}}$ (4e) Fe I
$\mathbf{B_{8}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4e) Fe I
$\mathbf{B_{9}}$ = $x_{3} \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (8g) Al II
$\mathbf{B_{10}}$ = $- x_{3} \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (8g) Al II
$\mathbf{B_{11}}$ = $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (8g) Al II
$\mathbf{B_{12}}$ = $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (8g) Al II
$\mathbf{B_{13}}$ = $- x_{3} \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (8g) Al II
$\mathbf{B_{14}}$ = $x_{3} \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (8g) Al II
$\mathbf{B_{15}}$ = $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (8g) Al II
$\mathbf{B_{16}}$ = $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (8g) Al II
$\mathbf{B_{17}}$ = $x_{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}$ = $a x_{4} \,\mathbf{\hat{x}}+a y_{4} \,\mathbf{\hat{y}}$ (8h) Cu I
$\mathbf{B_{18}}$ = $- x_{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}$ = $- a x_{4} \,\mathbf{\hat{x}}- a y_{4} \,\mathbf{\hat{y}}$ (8h) Cu I
$\mathbf{B_{19}}$ = $- y_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}$ = $- a y_{4} \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}$ (8h) Cu I
$\mathbf{B_{20}}$ = $y_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}$ = $a y_{4} \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}$ (8h) Cu I
$\mathbf{B_{21}}$ = $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (8h) Cu I
$\mathbf{B_{22}}$ = $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (8h) Cu I
$\mathbf{B_{23}}$ = $\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $a \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (8h) Cu I
$\mathbf{B_{24}}$ = $- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $- a \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (8h) Cu I
$\mathbf{B_{25}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{26}}$ = $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}- a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{27}}$ = $- y_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- a y_{5} \,\mathbf{\hat{x}}+a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{28}}$ = $y_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a y_{5} \,\mathbf{\hat{x}}- a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{29}}$ = $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{30}}$ = $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{31}}$ = $\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{32}}$ = $- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{33}}$ = $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}- a y_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{34}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+a y_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{35}}$ = $y_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $a y_{5} \,\mathbf{\hat{x}}- a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{36}}$ = $- y_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a y_{5} \,\mathbf{\hat{x}}+a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{37}}$ = $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{38}}$ = $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{39}}$ = $- \left(y_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(y_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (16i) Al III
$\mathbf{B_{40}}$ = $\left(y_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(y_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (16i) Al III

References

  • M. G. Bown and P. J. Brown, The structure of FeCu$_{2}$Al$_{7}$ and T(CoCuAl), Acta Cryst. 9, 911–914 (1956), doi:10.1107/S0365110X56002576.
  • D. Hicks, M. J. Mehl, E. Gossett, C. Toher, O. Levy, R. M. Hanson, G. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 2, Comput. Mater. Sci. 161, S1–S1011 (2019), doi:10.1016/j.commatsci.2018.10.043.

Prototype Generator

aflow --proto=A7B2C_tP40_128_egi_h_e --params=$a,c/a,z_{1},z_{2},x_{3},x_{4},y_{4},x_{5},y_{5},z_{5}$

Species:

Running:

Output: