(NH$_{4}$)Pb$_{2}$Br$_{5}$ ($K3_{4}$) Structure (Revised): A5BC2_tI32_140_cl_a_h-001
| Prototype | Br$_{5}$(NH$_{4}$)Pb$_{2}$ |
| AFLOW prototype label | A5BC2_tI32_140_cl_a_h-001 |
| Strukturbericht designation | $K3_{4}$ |
| ICSD | 26662 |
| Pearson symbol | tI32 |
| Space group number | 140 |
| Space group symbol | $I4/mcm$ |
| AFLOW prototype command |
aflow --proto=A5BC2_tI32_140_cl_a_h-001
--params=$a, \allowbreak c/a, \allowbreak x_{3}, \allowbreak x_{4}, \allowbreak z_{4}$ |
Other compounds with this structure
CsPb$_{2}$Br$_{5}$, CsSn$_{2}$Br$_{5}$, ISe$_{2}$Tl$_{5}$, InPb$_{2}$I$_{5}$, InSn$_{2}$Br$_{5}$, InSn$_{2}$I$_{5}$, KPb$_{2}$Br$_{5}$, KSn$_{2}$Br$_{5}$, KSn$_{2}$I$_{5}$, RbPb$_{2}$Br$_{5}$, RbSn$_{2}$Br$_{5}$, TlSn$_{2}$Br$_{5}$
- (Powell, 1937) gives the first bromine site a (2b) label, but gives the position as (000), which corresponds to the (2c) Wyckoff position. In our previous attempt at presenting this structure we followed the first choice, but the second choice is the correct one. See the previous $K3_{4}$ page for more information.
- The positions of the hydrogen atoms in the NH$_{4}$ ions were not determined, so we only provide the positions of the nitrogen atoms (labeled as NH$_{4}$).
\[ \begin{array}{ccc}
\mathbf{a_{1}}&=&- \frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}\\\mathbf{a_{2}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}c \,\mathbf{\hat{z}}\\\mathbf{a_{3}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}- \frac{1}{2}c \,\mathbf{\hat{z}}
\end{array}\]
Basis vectors
| Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
|---|---|---|---|---|---|---|
| $\mathbf{B_{1}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}$ | = | $\frac{1}{4}c \,\mathbf{\hat{z}}$ | (4a) | N I |
| $\mathbf{B_{2}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}$ | = | $\frac{3}{4}c \,\mathbf{\hat{z}}$ | (4a) | N I |
| $\mathbf{B_{3}}$ | = | $0$ | = | $0$ | (4c) | Br I |
| $\mathbf{B_{4}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $\frac{1}{2}c \,\mathbf{\hat{z}}$ | (4c) | Br I |
| $\mathbf{B_{5}}$ | = | $\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{1}+x_{3} \, \mathbf{a}_{2}+\left(2 x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a x_{3} \,\mathbf{\hat{x}}+a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (8h) | Pb I |
| $\mathbf{B_{6}}$ | = | $- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{1}- x_{3} \, \mathbf{a}_{2}- \left(2 x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{3} \,\mathbf{\hat{x}}- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{y}}$ | (8h) | Pb I |
| $\mathbf{B_{7}}$ | = | $x_{3} \, \mathbf{a}_{1}- \left(x_{3} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a \left(x_{3} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{3} \,\mathbf{\hat{y}}$ | (8h) | Pb I |
| $\mathbf{B_{8}}$ | = | $- x_{3} \, \mathbf{a}_{1}+\left(x_{3} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $a \left(x_{3} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a x_{3} \,\mathbf{\hat{y}}$ | (8h) | Pb I |
| $\mathbf{B_{9}}$ | = | $\left(x_{4} + z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{4} + z_{4}\right) \, \mathbf{a}_{2}+\left(2 x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (16l) | Br II |
| $\mathbf{B_{10}}$ | = | $\left(- x_{4} + z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{4} - z_{4}\right) \, \mathbf{a}_{2}- \left(2 x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (16l) | Br II |
| $\mathbf{B_{11}}$ | = | $\left(x_{4} + z_{4}\right) \, \mathbf{a}_{1}+\left(- x_{4} + z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (16l) | Br II |
| $\mathbf{B_{12}}$ | = | $- \left(x_{4} - z_{4}\right) \, \mathbf{a}_{1}+\left(x_{4} + z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}+c z_{4} \,\mathbf{\hat{z}}$ | (16l) | Br II |
| $\mathbf{B_{13}}$ | = | $\left(x_{4} - z_{4}\right) \, \mathbf{a}_{1}- \left(x_{4} + z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (16l) | Br II |
| $\mathbf{B_{14}}$ | = | $- \left(x_{4} + z_{4}\right) \, \mathbf{a}_{1}+\left(x_{4} - z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{x}}- a x_{4} \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (16l) | Br II |
| $\mathbf{B_{15}}$ | = | $\left(x_{4} - z_{4} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{4} - z_{4}\right) \, \mathbf{a}_{2}+\left(2 x_{4} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a x_{4} \,\mathbf{\hat{x}}+a \left(x_{4} + \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (16l) | Br II |
| $\mathbf{B_{16}}$ | = | $- \left(x_{4} + z_{4} - \frac{1}{2}\right) \, \mathbf{a}_{1}- \left(x_{4} + z_{4}\right) \, \mathbf{a}_{2}- \left(2 x_{4} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{4} \,\mathbf{\hat{x}}- a \left(x_{4} - \frac{1}{2}\right) \,\mathbf{\hat{y}}- c z_{4} \,\mathbf{\hat{z}}$ | (16l) | Br II |
References
- H. M. Powell and H. S. Tasker, The valency angle of bivalent lead: the crystal structure of ammonium, rubidium, and potassium pentabromodiplumbites, J. Chem. Soc. p. 119 (1937), doi:10.1039/JR9370000119.
Prototype Generator
aflow --proto=A5BC2_tI32_140_cl_a_h --params=$a,c/a,x_{3},x_{4},z_{4}$