Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: A2B4C5_hP22_186_ab_4b_a4b-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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Ti$_{3}$Al$_{2}$N$_{2}$ Structure: A2B4C5_hP22_186_ab_4b_a4b-001

Picture of Structure; Click for Big Picture
Prototype Al$_{2}$N$_{2}$Ti$_{2}$
AFLOW prototype label A2B4C5_hP22_186_ab_4b_a4b-001
ICSD 52643
Pearson symbol hP22
Space group number 186
Space group symbol $P6_3mc$
AFLOW prototype command aflow --proto=A2B4C5_hP22_186_ab_4b_a4b-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak z_{5}, \allowbreak z_{6}, \allowbreak z_{7}, \allowbreak z_{8}, \allowbreak z_{9}, \allowbreak z_{10}, \allowbreak z_{11}$

  • This compound only exists at temperatures near 1573K, where this data was taken.
  • (Schuster, 1984) placed this system in the trigonal space group $P31c$ #159. The (2a) and (2b) Wyckoff positions for space group #159 are identical to those in the hexagonal space group $P6_{3}mc$ #186, so we follow (Cenzual, 1991) and place this in the higher symmetry space group.
  • The origin of the $z$-axis in space group $P6_{3}mc$ is arbitrary. We use the origin specified by (Schuster, 1984).
  • Most of the Wyckoff positions in this structure are only partially occupied:
    • Sites Al-I and Ti-I are fully occupied.
    • Sites Ni-I, Ni-II, Ti-II, and Ti-III are 80% occupied.
    • Sites Ni-III, Ni-IV, Ti-IV, and Ti-V are 20% occupied.
  • This explains the short Ti-N bond lengths shown in the figure - for any given Ti-N pair, only one of the two atoms is actually present.

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

Basis vectors

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

References

Found in

  • K. Cenzual, L. M. Gelato, M. Penzo, and E. Parthé, Inorganic structure types with revised space groups. I, Acta Crystallogr. Sect. B 47, 433–439 (1991), doi:10.1107/S0108768191000903.

Prototype Generator

aflow --proto=A2B4C5_hP22_186_ab_4b_a4b --params=$a,c/a,z_{1},z_{2},z_{3},z_{4},z_{5},z_{6},z_{7},z_{8},z_{9},z_{10},z_{11}$

Species:

Running:

Output: