SrCl$_{2}\cdot$(H$_{2}$O)$_{6}$ Structure: A2B12C6D_hP21_150_d_2g_ef_a-001
| Prototype | Cl$_{2}$H$_{12}$O$_{6}$Sr |
| AFLOW prototype label | A2B12C6D_hP21_150_d_2g_ef_a-001 |
| ICSD | 59143 |
| Pearson symbol | hP21 |
| Space group number | 150 |
| Space group symbol | $P321$ |
| AFLOW prototype command |
aflow --proto=A2B12C6D_hP21_150_d_2g_ef_a-001
--params=$a, \allowbreak c/a, \allowbreak z_{2}, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak x_{5}, \allowbreak y_{5}, \allowbreak z_{5}, \allowbreak x_{6}, \allowbreak y_{6}, \allowbreak z_{6}$ |
Other compounds with this structure
CaCl$_{2}\cdot$(H$_{2}$O)$_{6}$, CaBr$_{2}\cdot$(H$_{2}$O)$_{6}$
- This is a redetermination of the $I1_{3}$ structure. In addition to locating the hydrogen atoms in the water molecule, (Agron, 1986) show that the space group is $P321$ #150, not $P\overline{3}$ #147.
\[ \begin{array}{ccc}
\mathbf{a_{1}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \,\mathbf{\hat{y}}\\\mathbf{a_{2}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \,\mathbf{\hat{y}}\\\mathbf{a_{3}}&=&c \,\mathbf{\hat{z}}
\end{array}\]
Basis vectors
| Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
|---|---|---|---|---|---|---|
| $\mathbf{B_{1}}$ | = | $0$ | = | $0$ | (1a) | Sr I |
| $\mathbf{B_{2}}$ | = | $\frac{1}{3} \, \mathbf{a}_{1}+\frac{2}{3} \, \mathbf{a}_{2}+z_{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}+c z_{2} \,\mathbf{\hat{z}}$ | (2d) | Cl I |
| $\mathbf{B_{3}}$ | = | $\frac{2}{3} \, \mathbf{a}_{1}+\frac{1}{3} \, \mathbf{a}_{2}- z_{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{6}a \,\mathbf{\hat{y}}- c z_{2} \,\mathbf{\hat{z}}$ | (2d) | Cl I |
| $\mathbf{B_{4}}$ | = | $x_{3} \, \mathbf{a}_{1}$ | = | $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}$ | (3e) | O I |
| $\mathbf{B_{5}}$ | = | $x_{3} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}a x_{3} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{3} \,\mathbf{\hat{y}}$ | (3e) | O I |
| $\mathbf{B_{6}}$ | = | $- x_{3} \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}$ | = | $- a x_{3} \,\mathbf{\hat{x}}$ | (3e) | O I |
| $\mathbf{B_{7}}$ | = | $x_{4} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (3f) | O II |
| $\mathbf{B_{8}}$ | = | $x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a x_{4} \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{4} \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (3f) | O II |
| $\mathbf{B_{9}}$ | = | $- x_{4} \, \mathbf{a}_{1}- x_{4} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}+\frac{1}{2}c \,\mathbf{\hat{z}}$ | (3f) | O II |
| $\mathbf{B_{10}}$ | = | $x_{5} \, \mathbf{a}_{1}+y_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (6g) | H I |
| $\mathbf{B_{11}}$ | = | $- y_{5} \, \mathbf{a}_{1}+\left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (6g) | H I |
| $\mathbf{B_{12}}$ | = | $- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- x_{5} \, \mathbf{a}_{2}+z_{5} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}+c z_{5} \,\mathbf{\hat{z}}$ | (6g) | H I |
| $\mathbf{B_{13}}$ | = | $y_{5} \, \mathbf{a}_{1}+x_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} + y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{5} - y_{5}\right) \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (6g) | H I |
| $\mathbf{B_{14}}$ | = | $\left(x_{5} - y_{5}\right) \, \mathbf{a}_{1}- y_{5} \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{5} - 2 y_{5}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (6g) | H I |
| $\mathbf{B_{15}}$ | = | $- x_{5} \, \mathbf{a}_{1}- \left(x_{5} - y_{5}\right) \, \mathbf{a}_{2}- z_{5} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{5} - y_{5}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{5} \,\mathbf{\hat{y}}- c z_{5} \,\mathbf{\hat{z}}$ | (6g) | H I |
| $\mathbf{B_{16}}$ | = | $x_{6} \, \mathbf{a}_{1}+y_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (6g) | H II |
| $\mathbf{B_{17}}$ | = | $- y_{6} \, \mathbf{a}_{1}+\left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{6} - 2 y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (6g) | H II |
| $\mathbf{B_{18}}$ | = | $- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}- x_{6} \, \mathbf{a}_{2}+z_{6} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}+c z_{6} \,\mathbf{\hat{z}}$ | (6g) | H II |
| $\mathbf{B_{19}}$ | = | $y_{6} \, \mathbf{a}_{1}+x_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{6} + y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a \left(x_{6} - y_{6}\right) \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (6g) | H II |
| $\mathbf{B_{20}}$ | = | $\left(x_{6} - y_{6}\right) \, \mathbf{a}_{1}- y_{6} \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \left(x_{6} - 2 y_{6}\right) \,\mathbf{\hat{x}}- \frac{\sqrt{3}}{2}a x_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (6g) | H II |
| $\mathbf{B_{21}}$ | = | $- x_{6} \, \mathbf{a}_{1}- \left(x_{6} - y_{6}\right) \, \mathbf{a}_{2}- z_{6} \, \mathbf{a}_{3}$ | = | $- \frac{1}{2}a \left(2 x_{6} - y_{6}\right) \,\mathbf{\hat{x}}+\frac{\sqrt{3}}{2}a y_{6} \,\mathbf{\hat{y}}- c z_{6} \,\mathbf{\hat{z}}$ | (6g) | H II |
References
- P. A. Agron and W. R. Busing, Calcium and strontium dichloride hexahydrates by neutron diffraction, Acta Crystallogr. Sect. C 42, 141–143 (1986), doi:10.1107/S0108270186097007.
Found in
- P. Villars, PAULING FILE (2016). In: Inorganic Solid Phases, SpringerMaterials (online database), Springer, Heidelberg SpringerMaterials.
Prototype Generator
aflow --proto=A2B12C6D_hP21_150_d_2g_ef_a --params=$a,c/a,z_{2},x_{3},x_{4},x_{5},y_{5},z_{5},x_{6},y_{6},z_{6}$