Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A9B16C7_cF128_225_acd_2f_be-001

This structure originally had the label A9B16C7_cF128_225_acd_2f_be. Calls to that address will be redirected here.

If you are using this page, please cite:
M. J. Mehl, D. Hicks, C. Toher, O. Levy, R. M. Hanson, G. L. W. Hart, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 1, Comp. Mat. Sci. 136, S1-S828 (2017). (doi=10.1016/j.commatsci.2017.01.017)

Links to this page

https://aflow.org/p/FENM
or https://aflow.org/p/A9B16C7_cF128_225_acd_2f_be-001
or PDF Version

Model of Ferrite Structure (cF128): A9B16C7_cF128_225_acd_2f_be-001

Picture of Structure; Click for Big Picture
Prototype Cr$_{9}$Fe$_{16}$Ni$_{7}$
AFLOW prototype label A9B16C7_cF128_225_acd_2f_be-001
ICSD none
Pearson symbol cF128
Space group number 225
Space group symbol $Fm\overline{3}m$
AFLOW prototype command aflow --proto=A9B16C7_cF128_225_acd_2f_be-001
--params=$a, \allowbreak x_{5}, \allowbreak x_{6}, \allowbreak x_{7}$

  • Ferritic steels are alloys of iron and other metals with an averaged body-centered cubic structure. This model represents one approximation for a ferritic steel. It is not meant to represent a real steel, and the selection of atom types for each Wyckoff position is arbitrary.
  • If we set $x_{5} = 1/4, x_{6} = 1/8$, and $x_{7} = 3/8$ and replace the nickel atoms by chromium, this structure reverts to CsCl ($B2$) with $a_{B2} = 1/4 a$.
  • If we replace both the nickel and chromium atoms by iron the structure becomes a body-centered cubic lattice ($A2$) again with $a_{A2} = 1/4 a$.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $0$ = $0$ (4a) Cr I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}a \,\mathbf{\hat{z}}$ (4b) Ni I
$\mathbf{B_{3}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ (8c) Cr II
$\mathbf{B_{4}}$ = $\frac{3}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}a \,\mathbf{\hat{y}}+\frac{3}{4}a \,\mathbf{\hat{z}}$ (8c) Cr II
$\mathbf{B_{5}}$ = $\frac{1}{2} \, \mathbf{a}_{1}$ = $\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ (24d) Cr III
$\mathbf{B_{6}}$ = $\frac{1}{2} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ (24d) Cr III
$\mathbf{B_{7}}$ = $\frac{1}{2} \, \mathbf{a}_{2}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ (24d) Cr III
$\mathbf{B_{8}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ (24d) Cr III
$\mathbf{B_{9}}$ = $\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}$ (24d) Cr III
$\mathbf{B_{10}}$ = $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+\frac{1}{2}a \,\mathbf{\hat{z}}$ (24d) Cr III
$\mathbf{B_{11}}$ = $- x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+x_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}$ (24e) Ni II
$\mathbf{B_{12}}$ = $x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}- x_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}$ (24e) Ni II
$\mathbf{B_{13}}$ = $x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+x_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{y}}$ (24e) Ni II
$\mathbf{B_{14}}$ = $- x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}- x_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{y}}$ (24e) Ni II
$\mathbf{B_{15}}$ = $x_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}- x_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{z}}$ (24e) Ni II
$\mathbf{B_{16}}$ = $- x_{5} \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+x_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{z}}$ (24e) Ni II
$\mathbf{B_{17}}$ = $x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}+x_{6} \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}+a x_{6} \,\mathbf{\hat{y}}+a x_{6} \,\mathbf{\hat{z}}$ (32f) Fe I
$\mathbf{B_{18}}$ = $x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}- 3 x_{6} \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}- a x_{6} \,\mathbf{\hat{y}}+a x_{6} \,\mathbf{\hat{z}}$ (32f) Fe I
$\mathbf{B_{19}}$ = $x_{6} \, \mathbf{a}_{1}- 3 x_{6} \, \mathbf{a}_{2}+x_{6} \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}+a x_{6} \,\mathbf{\hat{y}}- a x_{6} \,\mathbf{\hat{z}}$ (32f) Fe I
$\mathbf{B_{20}}$ = $- 3 x_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}+x_{6} \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}- a x_{6} \,\mathbf{\hat{y}}- a x_{6} \,\mathbf{\hat{z}}$ (32f) Fe I
$\mathbf{B_{21}}$ = $- x_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+3 x_{6} \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}+a x_{6} \,\mathbf{\hat{y}}- a x_{6} \,\mathbf{\hat{z}}$ (32f) Fe I
$\mathbf{B_{22}}$ = $- x_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}- x_{6} \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}- a x_{6} \,\mathbf{\hat{y}}- a x_{6} \,\mathbf{\hat{z}}$ (32f) Fe I
$\mathbf{B_{23}}$ = $- x_{6} \, \mathbf{a}_{1}+3 x_{6} \, \mathbf{a}_{2}- x_{6} \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}- a x_{6} \,\mathbf{\hat{y}}+a x_{6} \,\mathbf{\hat{z}}$ (32f) Fe I
$\mathbf{B_{24}}$ = $3 x_{6} \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}- x_{6} \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}+a x_{6} \,\mathbf{\hat{y}}+a x_{6} \,\mathbf{\hat{z}}$ (32f) Fe I
$\mathbf{B_{25}}$ = $x_{7} \, \mathbf{a}_{1}+x_{7} \, \mathbf{a}_{2}+x_{7} \, \mathbf{a}_{3}$ = $a x_{7} \,\mathbf{\hat{x}}+a x_{7} \,\mathbf{\hat{y}}+a x_{7} \,\mathbf{\hat{z}}$ (32f) Fe II
$\mathbf{B_{26}}$ = $x_{7} \, \mathbf{a}_{1}+x_{7} \, \mathbf{a}_{2}- 3 x_{7} \, \mathbf{a}_{3}$ = $- a x_{7} \,\mathbf{\hat{x}}- a x_{7} \,\mathbf{\hat{y}}+a x_{7} \,\mathbf{\hat{z}}$ (32f) Fe II
$\mathbf{B_{27}}$ = $x_{7} \, \mathbf{a}_{1}- 3 x_{7} \, \mathbf{a}_{2}+x_{7} \, \mathbf{a}_{3}$ = $- a x_{7} \,\mathbf{\hat{x}}+a x_{7} \,\mathbf{\hat{y}}- a x_{7} \,\mathbf{\hat{z}}$ (32f) Fe II
$\mathbf{B_{28}}$ = $- 3 x_{7} \, \mathbf{a}_{1}+x_{7} \, \mathbf{a}_{2}+x_{7} \, \mathbf{a}_{3}$ = $a x_{7} \,\mathbf{\hat{x}}- a x_{7} \,\mathbf{\hat{y}}- a x_{7} \,\mathbf{\hat{z}}$ (32f) Fe II
$\mathbf{B_{29}}$ = $- x_{7} \, \mathbf{a}_{1}- x_{7} \, \mathbf{a}_{2}+3 x_{7} \, \mathbf{a}_{3}$ = $a x_{7} \,\mathbf{\hat{x}}+a x_{7} \,\mathbf{\hat{y}}- a x_{7} \,\mathbf{\hat{z}}$ (32f) Fe II
$\mathbf{B_{30}}$ = $- x_{7} \, \mathbf{a}_{1}- x_{7} \, \mathbf{a}_{2}- x_{7} \, \mathbf{a}_{3}$ = $- a x_{7} \,\mathbf{\hat{x}}- a x_{7} \,\mathbf{\hat{y}}- a x_{7} \,\mathbf{\hat{z}}$ (32f) Fe II
$\mathbf{B_{31}}$ = $- x_{7} \, \mathbf{a}_{1}+3 x_{7} \, \mathbf{a}_{2}- x_{7} \, \mathbf{a}_{3}$ = $a x_{7} \,\mathbf{\hat{x}}- a x_{7} \,\mathbf{\hat{y}}+a x_{7} \,\mathbf{\hat{z}}$ (32f) Fe II
$\mathbf{B_{32}}$ = $3 x_{7} \, \mathbf{a}_{1}- x_{7} \, \mathbf{a}_{2}- x_{7} \, \mathbf{a}_{3}$ = $- a x_{7} \,\mathbf{\hat{x}}+a x_{7} \,\mathbf{\hat{y}}+a x_{7} \,\mathbf{\hat{z}}$ (32f) Fe II

References

  • M. J. Mehl, D. Hicks, C. Toher, O. Levy, R. M. Hanson, G. Hart, and S. Curtarolo, The AFLOW library of crystallographic prototypes: part 1, Comput. Mater. Sci. 136, S1–S828 (2017), doi:10.1016/j.commatsci.2017.01.017.

Prototype Generator

aflow --proto=A9B16C7_cF128_225_acd_2f_be --params=$a,x_{5},x_{6},x_{7}$

Species:

Running:

Output: