Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: AB3C_oP20_62_c_3c_c-001

This structure originally had the label AB3C_oP20_62_c_3c_c. Calls to that address will be redirected here.

If you are using this page, please cite:
D. Hicks, M.J. Mehl, M. Esters, C. Oses, O. Levy, G.L.W. Hart, C. Toher, and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 3, Comp. Mat. Sci. 199, 110450 (2021). (doi=10.1016/j.commatsci.2021.110450)

Links to this page

https://aflow.org/p/Z6H8
or https://aflow.org/p/AB3C_oP20_62_c_3c_c-001
or PDF Version

(NH$_{4}$)CdCl$_{3}$ ($E2_{4}$) Structure: AB3C_oP20_62_c_3c_c-001

Picture of Structure; Click for Big Picture
Prototype CdCl$_{3}$(NH$_{4}$)
AFLOW prototype label AB3C_oP20_62_c_3c_c-001
Strukturbericht designation $E2_{4}$
ICSD 28829
Pearson symbol oP20
Space group number 62
Space group symbol $Pnma$
AFLOW prototype command aflow --proto=AB3C_oP20_62_c_3c_c-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak x_{1}, \allowbreak z_{1}, \allowbreak x_{2}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak z_{3}, \allowbreak x_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}$

Other compounds with this structure

CeCrS$_{3}$,  CeCrSe$_{3}$,  LaCrS$_{3}$,  NdCrS$_{3}$,  PbSnS$_{3}$,  RbCdBr$_{3}$,  RbCdCl$_{3}$,  SmCrS$_{3}$,  (Ti,  Sn)$_{2}$S$_{3}$


  • (Brasseur, 1938) gave the Wyckoff positions in the $Pnam$ setting of space group #62. We used FINDSYM to rotate the structure and put it in the standard $Pnma$ setting.
  • This structure has the same AFLOW label as CuTaS$_{3}$, AB3C_oP20_62_c_3c_c. The structures are generated by the same symmetry operations with different sets of parameters (--params) specified in their corresponding CIF files.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $x_{1} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ = $a x_{1} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{1} \,\mathbf{\hat{z}}$ (4c) Cd I
$\mathbf{B_{2}}$ = $- \left(x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{1} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cd I
$\mathbf{B_{3}}$ = $- x_{1} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{1} \, \mathbf{a}_{3}$ = $- a x_{1} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{1} \,\mathbf{\hat{z}}$ (4c) Cd I
$\mathbf{B_{4}}$ = $\left(x_{1} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{1} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{1} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{1} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cd I
$\mathbf{B_{5}}$ = $x_{2} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $a x_{2} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{6}}$ = $- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{2} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{7}}$ = $- x_{2} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ = $- a x_{2} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{8}}$ = $\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{2} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{2} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl I
$\mathbf{B_{9}}$ = $x_{3} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{10}}$ = $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{3} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{11}}$ = $- x_{3} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{12}}$ = $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{3} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl II
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (4c) Cl III
$\mathbf{B_{14}}$ = $- \left(x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl III
$\mathbf{B_{15}}$ = $- x_{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (4c) Cl III
$\mathbf{B_{16}}$ = $\left(x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{4} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) Cl III
$\mathbf{B_{17}}$ = $x_{5} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (4c) NH I
$\mathbf{B_{18}}$ = $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) NH I
$\mathbf{B_{19}}$ = $- x_{5} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (4c) NH I
$\mathbf{B_{20}}$ = $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- \left(z_{5} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c \left(z_{5} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (4c) NH I

References

  • H. Brasseur and L. Pauling, The Crystal Structure of Ammonium Cadmium Chloride, NH$_{4}$CdCl$_{3}$, J. Am. Chem. Soc. 60, 2886–2890 (1938), doi:10.1021/ja01279a016.

Prototype Generator

aflow --proto=AB3C_oP20_62_c_3c_c --params=$a,b/a,c/a,x_{1},z_{1},x_{2},z_{2},x_{3},z_{3},x_{4},z_{4},x_{5},z_{5}$

Species:

Running:

Output: