Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: AB3_oP24_59_ae_befg-001

If you are using this page, please cite:
H. Eckert, S. Divilov, M. J. Mehl, D. Hicks, A. C. Zettel, M. Esters. X. Campilongo and S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 4. Submitted to Computational Materials Science.

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https://aflow.org/p/5BSF
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β-NbPd$_{3}$ Structure: AB3_oP24_59_ae_befg-001

Picture of Structure; Click for Big Picture
Prototype NbPd$_{3}$
AFLOW prototype label AB3_oP24_59_ae_befg-001
ICSD 105193
Pearson symbol oP24
Space group number 59
Space group symbol $Pmmn$
AFLOW prototype command aflow --proto=AB3_oP24_59_ae_befg-001
--params=$a, \allowbreak b/a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak y_{3}, \allowbreak z_{3}, \allowbreak y_{4}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}$

  • Although $\beta$–NbPd$_{3}$ and $\alpha$–NbPd$_{3}$, which has the $D0_{22}$ structure, coexist over a wide range of temperatures, it is thought that $\alpha$–NbPd$_{3}$ is the ground state (Chandrasekharaiah, 1988).
  • (Giessen, 1964) state that their coordinates are given in the $Pmmn$ setting of space group #59, but it is apparent that they actually use the $Pmnm$ setting. We transformed this to $Pmmn$.

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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{1} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{1} \,\mathbf{\hat{z}}$ (2a) Nb I
$\mathbf{B_{2}}$ = $\frac{3}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{1} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{1} \,\mathbf{\hat{z}}$ (2a) Nb I
$\mathbf{B_{3}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (2b) Pd I
$\mathbf{B_{4}}$ = $\frac{3}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ (2b) Pd I
$\mathbf{B_{5}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+y_{3} \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+b y_{3} \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (4e) Nb II
$\mathbf{B_{6}}$ = $\frac{1}{4} \, \mathbf{a}_{1}- \left(y_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{3} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}- b \left(y_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{3} \,\mathbf{\hat{z}}$ (4e) Nb II
$\mathbf{B_{7}}$ = $\frac{3}{4} \, \mathbf{a}_{1}+\left(y_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}+b \left(y_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ (4e) Nb II
$\mathbf{B_{8}}$ = $\frac{3}{4} \, \mathbf{a}_{1}- y_{3} \, \mathbf{a}_{2}- z_{3} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}- b y_{3} \,\mathbf{\hat{y}}- c z_{3} \,\mathbf{\hat{z}}$ (4e) Nb II
$\mathbf{B_{9}}$ = $\frac{1}{4} \, \mathbf{a}_{1}+y_{4} \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}+b y_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (4e) Pd II
$\mathbf{B_{10}}$ = $\frac{1}{4} \, \mathbf{a}_{1}- \left(y_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{4} \, \mathbf{a}_{3}$ = $\frac{1}{4}a \,\mathbf{\hat{x}}- b \left(y_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ (4e) Pd II
$\mathbf{B_{11}}$ = $\frac{3}{4} \, \mathbf{a}_{1}+\left(y_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}+b \left(y_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (4e) Pd II
$\mathbf{B_{12}}$ = $\frac{3}{4} \, \mathbf{a}_{1}- y_{4} \, \mathbf{a}_{2}- z_{4} \, \mathbf{a}_{3}$ = $\frac{3}{4}a \,\mathbf{\hat{x}}- b y_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ (4e) Pd II
$\mathbf{B_{13}}$ = $x_{5} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $a x_{5} \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (4f) Pd III
$\mathbf{B_{14}}$ = $- \left(x_{5} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- a \left(x_{5} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{1}{4}b \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (4f) Pd III
$\mathbf{B_{15}}$ = $- x_{5} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $- a x_{5} \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (4f) Pd III
$\mathbf{B_{16}}$ = $\left(x_{5} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ = $a \left(x_{5} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+\frac{3}{4}b \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ (4f) Pd III
$\mathbf{B_{17}}$ = $x_{6} \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}+b y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (8g) Pd IV
$\mathbf{B_{18}}$ = $- \left(x_{6} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(y_{6} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $- a \left(x_{6} - \frac{1}{2}\right) \,\mathbf{\hat{x}}- b \left(y_{6} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (8g) Pd IV
$\mathbf{B_{19}}$ = $- x_{6} \, \mathbf{a}_{1}+\left(y_{6} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}+b \left(y_{6} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (8g) Pd IV
$\mathbf{B_{20}}$ = $\left(x_{6} + \frac{1}{2}\right) \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $a \left(x_{6} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- b y_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (8g) Pd IV
$\mathbf{B_{21}}$ = $- x_{6} \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $- a x_{6} \,\mathbf{\hat{x}}- b y_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (8g) Pd IV
$\mathbf{B_{22}}$ = $\left(x_{6} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(y_{6} + \frac{1}{2}\right) \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ = $a \left(x_{6} + \frac{1}{2}\right) \,\mathbf{\hat{x}}+b \left(y_{6} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ (8g) Pd IV
$\mathbf{B_{23}}$ = $x_{6} \, \mathbf{a}_{1}- \left(y_{6} - \frac{1}{2}\right) \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $a x_{6} \,\mathbf{\hat{x}}- b \left(y_{6} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (8g) Pd IV
$\mathbf{B_{24}}$ = $- \left(x_{6} - \frac{1}{2}\right) \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $- a \left(x_{6} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+b y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (8g) Pd IV

References

  • B. C. Giessen and N. J. Grant, New intermediate phases in system of Nb or Ta with Rh, Ir, Pd, or Pt, Acta Cryst. 17, 615–616 (1964), doi:10.1107/S0365110X64001438.
  • M. S. Chandrasekharaiah, The Nb−Pb (Niobium-Palladium) system, Bull. Alloy Phase Diag. 9, 449–452 (1988), doi:10.1007/BF02881865.

Found in


Prototype Generator

aflow --proto=AB3_oP24_59_ae_befg --params=$a,b/a,c/a,z_{1},z_{2},y_{3},z_{3},y_{4},z_{4},x_{5},z_{5},x_{6},y_{6},z_{6}$

Species:

Running:

Output: