Encyclopedia of Crystallographic Prototypes

AFLOW Prototype: AB2_tI48_141_h_2eg-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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Orange (I) HgI$_{2}$ Structure: AB2_tI48_141_h_2eg-001

Picture of Structure; Click for Big Picture
Prototype HgI$_{2}$
AFLOW prototype label AB2_tI48_141_h_2eg-001
ICSD 18126
Pearson symbol tI48
Space group number 141
Space group symbol $I4_1/amd$
AFLOW prototype command aflow --proto=AB2_tI48_141_h_2eg-001
--params=$a, \allowbreak c/a, \allowbreak z_{1}, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak y_{4}, \allowbreak z_{4}$

  • HgI$_{2}$ can be found in a variety of forms (Gumiński, 1997):
    • The ground state, coccinite, also known as red or $\alpha$–HgI$_{2}$ and given the Strukturbericht designation $C13$. It is stable up to 135$^\circ$C.
    • At higher temperatures this transforms into yellow or $\beta$–HgI$_{2}$ in the HgBr$_{2}$ ($C24$) structure. This is stable up to the melting point at 258$^\circ$C.
    • (Schwarzenbach, 1969) studied the metastable orange HgI$_{2}$ (this structure) body-centered tetragonal ($I4_{1}/amd$ #141) phase. This structure was refined by (Hostettler, 2002).
    • (Hostettler, 2002) also found a second orange HgI$_{2}$ phase in a simple tetragonal ($P4_{2}/nmc$ #137) cell.
    • The last two structures differ by stacking order. (Hostettler, 2002) used them to produce an averaged orange HgI$_{2}$ structure, space group $P\overline{4}m2$ #115.
  • The ICSD entry for this structure is from the earlier work of (Schwarzenbach, 1969).

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

Basis vectors

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

References

  • M. Hostettler, H. Birkedal, and D. Schwarzenbach, The structure of orange HgI$_{2}$. I. Polytypic layer structure, Acta Crystallogr. Sect. B 58, 903–913 (2002), doi:10.1107/S010876810201618X.
  • D. Schwarzenbach, The crystal structure and one-dimensional disorder of the orange modification of HgI$_{2}$, Z. Kristallogr. 128, 97–114 (1969), doi:10.1524/zkri.1969.128.1-2.97.
  • D. Schwarzenbach, H. Birkedal, M. Hostettler, and P. Fischer, Neutron diffraction investigation of the temperature dependence of crystal structure and thermal motions of red HgI$_{2}$, Acta Crystallogr. Sect. B 63, 826–835 (2007), doi:10.1107/S0108768107043327.

Prototype Generator

aflow --proto=AB2_tI48_141_h_2eg --params=$a,c/a,z_{1},z_{2},x_{3},y_{4},z_{4}$

Species:

Running:

Output: