La$_{3}$TiSb$_{5}$: A3B5C_hP18_193_g_dg_b-001

Picture of Structure; Click for Big Picture
Prototype La$_{3}$Sb$_{5}$Ti
AFLOW prototype label A3B5C_hP18_193_g_dg_b-001
ICSD 80907
CCDC 1641084
Pearson symbol hP18
Space group number 193
Space group symbol $P6_3/mcm$
AFLOW prototype command aflow --proto=A3B5C_hP18_193_g_dg_b-001
--params=$a, \allowbreak c/a, \allowbreak x_{3}, \allowbreak x_{4}$

Other compounds with this structure

Al$_{3}$NHf$_{5}$,  Ba$_{3}$ScTe$_{5}$,  Bi$_{3}$BrBa$_{5}$,  Bi$_{3}$BrCa$_{5}$,  Bi$_{3}$BrLa$_{5}$,  Bi$_{3}$ClBa$_{5}$,  Bi$_{3}$ClCa$_{5}$,  Bi$_{3}$CuCe$_{5}$,  Bi$_{3}$CuGd$_{5}$,  Bi$_{3}$CuLa$_{5}$,  Bi$_{3}$CuNd$_{5}$,  Bi$_{3}$CuPr$_{5}$,  Bi$_{3}$CuTb$_{5}$,  Bi$_{3}$CuZr$_{5}$,  Ce$_{3}$HfSb$_{5}$,  Ce$_{3}$MnBi$_{5}$,  Ce$_{3}$NbSb$_{5}$,  Ce$_{3}$TiBi$_{5}$,  Ce$_{3}$TiSb$_{5}$,  Ce$_{3}$ZrSb$_{5}$,  Ge$_{3}$CrLa$_{5}$,  In$_{3}$AgZr$_{5}$,  In$_{3}$BrLa$_{5}$,  In$_{3}$CuSc$_{5}$,  In$_{3}$CuZr$_{5}$,  La$_{3}$CrAs$_{5}$,  La$_{3}$HfBi$_{5}$,  La$_{3}$HfSb$_{5}$,  La$_{3}$MgBi$_{5}$,  La$_{3}$MnBi$_{5}$,  La$_{3}$NbSb$_{5}$,  La$_{3}$ScBi$_{5}$,  La$_{3}$TiAs$_{5}$,  La$_{3}$TiBi$_{5}$,  La$_{3}$TiP$_{5}$,  La$_{3}$TiSb$_{5}$,  La$_{3}$ZrBi$_{5}$,  La$_{3}$ZrSb$_{5}$,  Nd$_{3}$HfSb$_{5}$,  Nd$_{3}$MnBi$_{5}$,  Nd$_{3}$NbSb$_{5}$,  Nd$_{3}$TiSb$_{5}$,  Nd$_{3}$ZrSb$_{5}$,  P$_{3}$NV$_{5}$,  Pb$_{3}$AgCe$_{5}$,  Pb$_{3}$AgLa$_{5}$,  Pb$_{3}$AgZr$_{5}$,  Pb$_{3}$AlZr$_{5}$,  Pb$_{3}$AsLa$_{5}$,  Pb$_{3}$AsZr$_{5}$,  Pb$_{3}$CdZr$_{5}$,  Pb$_{3}$CeLa$_{5}$,  Pb$_{3}$CoLa$_{5}$,  Pb$_{3}$CoZr$_{5}$,  Pb$_{3}$CrLa$_{5}$,  Pb$_{3}$CuCe$_{5}$,  Pb$_{3}$CuHf$_{5}$,  Pb$_{3}$CuLa$_{5}$,  Pb$_{3}$CuY$_{5}$,  Pb$_{3}$CuZr$_{5}$,  Pb$_{3}$FeCa$_{5}$,  Pb$_{3}$FeLa$_{5}$,  Pb$_{3}$FeZr$_{5}$,  Pb$_{3}$GaZr$_{5}$,  Pb$_{3}$GeZr$_{5}$,  Pb$_{3}$ILa$_{5}$,  Pb$_{3}$InZr$_{5}$,  Pb$_{3}$MnLa$_{5}$,  Pb$_{3}$NiDy$_{5}$,  Pb$_{3}$NiEr$_{5}$,  Pb$_{3}$NiHo$_{5}$,  Pb$_{3}$NiLa$_{5}$,  Pb$_{3}$NiLu$_{5}$,  Pb$_{3}$NiPr$_{5}$,  Pb$_{3}$NiTb$_{5}$,  Pb$_{3}$NiTm$_{5}$,  Pb$_{3}$NiZr$_{5}$,  Pb$_{3}$PLa$_{5}$,  Pb$_{3}$PZr$_{5}$,  Pb$_{3}$RuLa$_{5}$,  Pb$_{3}$SLa$_{5}$,  Pb$_{3}$SZr$_{5}$,  Pb$_{3}$SbLa$_{5}$,  Pb$_{3}$SbZr$_{5}$,  Pb$_{3}$SeLa$_{5}$,  Pb$_{3}$SeZr$_{5}$,  Pb$_{3}$SiZr$_{5}$,  Pb$_{3}$SnZr$_{5}$,  Pb$_{3}$TeZr$_{5}$,  Pb$_{3}$ZnLa$_{5}$,  Pb$_{3}$ZnZr$_{5}$,  Pr$_{3}$HfSb$_{5}$,  Pr$_{3}$MnBi$_{5}$,  Pr$_{3}$NbSb$_{5}$,  Pr$_{3}$TiSb$_{5}$,  Pr$_{3}$ZrSb$_{5}$,  Sb$_{3}$AgZr$_{5}$,  Sb$_{3}$AlZr$_{5}$,  Sb$_{3}$AsZr$_{5}$,  Sb$_{3}$BrBa$_{5}$,  Sb$_{3}$BrLa$_{5}$,  Sb$_{3}$ClBa$_{5}$,  Sb$_{3}$ClCa$_{5}$,  Sb$_{3}$ClSr$_{5}$,  Sb$_{3}$CoZr$_{5}$,  Sb$_{3}$CuHf$_{5}$,  Sb$_{3}$CuTi$_{5}$,  Sb$_{3}$CuZr$_{5}$,  Sb$_{3}$FeZr$_{5}$,  Sb$_{3}$GeZr$_{5}$,  Sb$_{3}$NiHf$_{5}$,  Sb$_{3}$NiZr$_{5}$,  Sb$_{3}$PZr$_{5}$,  Sb$_{3}$RuZr$_{5}$,  Sb$_{3}$SZr$_{5}$,  Sb$_{3}$SeZr$_{5}$,  Sb$_{3}$SiZr$_{5}$,  Sb$_{3}$ZnHf$_{5}$,  Sb$_{3}$ZnZr$_{5}$,  Si$_{3}$DTi$_{5}$,  Si$_{3}$PNb$_{5}$,  Sm$_{3}$HfSb$_{5}$,  Sm$_{3}$NbSb$_{5}$,  Sm$_{3}$TiSb$_{5}$,  Sm$_{3}$ZrBi$_{5}$,  Sm$_{3}$ZrSb$_{5}$,  Sn$_{3}$AgCe$_{5}$,  Sn$_{3}$AlZr$_{5}$,  Sn$_{3}$AsZr$_{5}$,  Sn$_{3}$BrLa$_{5}$,  Sn$_{3}$ClLa$_{5}$,  Sn$_{3}$CuCe$_{5}$,  Sn$_{3}$CuYb$_{5}$,  Sn$_{3}$CuZr$_{5}$,  Sn$_{3}$GaZr$_{5}$,  Sn$_{3}$GeZr$_{5}$,  Sn$_{3}$HfGa$_{5}$,  Sn$_{3}$ILa$_{5}$,  Sn$_{3}$LiTb$_{5}$,  Sn$_{3}$NZr$_{5}$,  Sn$_{3}$NiHf$_{5}$,  Sn$_{3}$PZr$_{5}$,  Sn$_{3}$SZr$_{5}$,  Sn$_{3}$SeZr$_{5}$,  Sn$_{3}$SiZr$_{5}$,  Sn$_{3}$ZnZr$_{5}$,  U$_{3}$CrSb$_{5}$,  U$_{3}$HfSb$_{5}$,  U$_{3}$MnSb$_{5}$,  U$_{3}$NbSb$_{5}$,  U$_{3}$ScSb$_{5}$,  U$_{3}$TiGe$_{5}$,  U$_{3}$TiSb$_{5}$,  U$_{3}$VSb$_{5}$,  U$_{3}$ZrSb$_{5}$



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

Basis vectors

Lattice coordinates Cartesian coordinates Wyckoff position Atom type
$\mathbf{B_{1}}$ = $0$ = $0$ (2b) Ti I
$\mathbf{B_{2}}$ = $\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}c \,\mathbf{\hat{z}}$ (2b) Ti I
$\mathbf{B_{3}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}$ (4d) Sb I
$\mathbf{B_{4}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (4d) Sb I
$\mathbf{B_{5}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}$ (4d) Sb I
$\mathbf{B_{6}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ (4d) Sb I
$\mathbf{B_{7}}$ = $x_{3} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6g) La I
$\mathbf{B_{8}}$ = $x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6g) La I
$\mathbf{B_{9}}$ = $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- a x_{3} \,\mathbf{\hat{x}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6g) La I
$\mathbf{B_{10}}$ = $- x_{3} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6g) La I
$\mathbf{B_{11}}$ = $- x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6g) La I
$\mathbf{B_{12}}$ = $x_{3} \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $a x_{3} \,\mathbf{\hat{x}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6g) La I
$\mathbf{B_{13}}$ = $x_{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6g) Sb II
$\mathbf{B_{14}}$ = $x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6g) Sb II
$\mathbf{B_{15}}$ = $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ = $- a x_{4} \,\mathbf{\hat{x}}+\frac{1}{4}c \,\mathbf{\hat{z}}$ (6g) Sb II
$\mathbf{B_{16}}$ = $- x_{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6g) Sb II
$\mathbf{B_{17}}$ = $- x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6g) Sb II
$\mathbf{B_{18}}$ = $x_{4} \, \mathbf{a}_{1}+x_{4} \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ = $a x_{4} \,\mathbf{\hat{x}}+\frac{3}{4}c \,\mathbf{\hat{z}}$ (6g) Sb II

References

  • G. Bolloré, M. J. Ferguson, R. W. Hushagen, and A. Mar, New Ternary Rare-Earth Transition-Metal Antimonides RE$_{3}$MSb$_{5}$ (RE = La, Ce, Pr, Nd, Sm; M = Ti, Zr, Hf, Nb), Chem. Mater. 7, 2229–2231 (1995), doi:10.1021/cm00060a005.

Found in

  • M. Matin, R. Kulkarni, A. Thamizhavel, S. K. Dhar, A. Provino, and P. Manfrinetti, Probing the magnetic ground state of single crystalline Ce$_{3}$TiSb$_{5}$, J. Phys.: Condens. Matter 29, 145601 (2017), doi:10.1088/1361-648X/aa57c0.

First cited in

  • N. Anderson, M. J. Mehl, H. Eckert, S. Divilov, X. Campilongo, S. Curtarolo, The AFLOW Library of Crystallographic Prototypes: Part 5. Submitted to Computational Materials Science (2026).

Geometry files


Prototype Generator

aflow --proto=A3B5C_hP18_193_g_dg_b --params=$a,c/a,x_{3},x_{4}$

Species:

Running:

Output: