Edgarite (FeNb$_{3}$S$_{6}$) $P6_{3}$ Structure: AB3C6_hP20_173_b_a2b_2c-001

Picture of Structure; Click for Big Picture
Prototype FeNb$_{3}$S$_{6}$
AFLOW prototype label AB3C6_hP20_173_b_a2b_2c-001
Mineral name edgarite
ICSD 230540
CCDC 2037169
Pearson symbol hP20
Space group number 173
Space group symbol $P6_3$
AFLOW prototype command aflow --proto=AB3C6_hP20_173_b_a2b_2c-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak z_{3}, \allowbreak z_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}$

Other compounds with this structure

CrNb$_{2}$S$_{3}$


  • The name Edgarite is also given to the FeNb$_{3}$S$_{6}$ structure in space group $P6_{3}22$ #182, for which we use CoNb$_{3}$S$_{6}$ as the prototype.
  • (Bindi, 2018) attributes this discrepancy to the difference between natural and synthetic edgarite, although in the original discovery paper for the mineral (Barkov, 2000) say that The space group is most probably $P6_{3}22$. The ICSD lists many more structures in the $P6_{3}22$ structure than it does in the $P6_{3}$ structure.

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

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) Nb 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) Nb I
$\mathbf{B_{3}}$ = $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ (2b) Fe I
$\mathbf{B_{4}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{2} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2b) Fe 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) Nb 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) Nb 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) Nb III
$\mathbf{B_{8}}$ = $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}+\left(z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c \left(z_{4} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (2b) Nb III
$\mathbf{B_{9}}$ = $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6c) S I
$\mathbf{B_{10}}$ = $- y_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6c) S I
$\mathbf{B_{11}}$ = $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ (6c) S I
$\mathbf{B_{12}}$ = $- x_{5} \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) S I
$\mathbf{B_{13}}$ = $y_{5} \, \mathbf{a}_{1}- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{5} + 2 y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) S I
$\mathbf{B_{14}}$ = $\left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}+\left(z_{5} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+c \left(z_{5} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) S I
$\mathbf{B_{15}}$ = $x_{6} \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (6c) S II
$\mathbf{B_{16}}$ = $- y_{6} \, \mathbf{a}_{1}+\left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(x_{6} - 2 y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (6c) S II
$\mathbf{B_{17}}$ = $- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ (6c) S II
$\mathbf{B_{18}}$ = $- x_{6} \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $- \frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) S II
$\mathbf{B_{19}}$ = $y_{6} \, \mathbf{a}_{1}- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(- x_{6} + 2 y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) S II
$\mathbf{B_{20}}$ = $\left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}+\left(z_{6} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ = $\frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}+c \left(z_{6} + \frac{1}{2}\right) \,\mathbf{\hat{z}}$ (6c) S II

References

  • L. Bindi and R. F. Martin, Edgarite, FeNb$_{3}$S$_{6}$, from the Khibina Alkaline Complex, Russia: Solution of the Crystal Structure, Can. Mineral. 56, 259–264 (2018), doi:10.3749/canmin.1800019.
  • A. Y. Barkov, R. F. Martin, Y. P. Men'shikov, Y. E. Savchenko, Y. Thibault, and K. V. O. Laajoki, Edgarite, FeNb$_{3}$S$_{6}$, first natural niobium-rich sulfide from the Khibina alkaline complex, Russian Far North: evidence for chalcophile behavior of Nb in a fenite, Contrib. Mineral. Petrol. 138, 229–236 (2000), doi:10.1007/s004100050559.

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=AB3C6_hP20_173_b_a2b_2c --params=$a,c/a,z_{1},z_{2},z_{3},z_{4},x_{5},y_{5},z_{5},x_{6},y_{6},z_{6}$

Species:

Running:

Output: