(NH$_{4}$)Pb$_{2}$Br$_{5}$ ($K3_{4}$) Structure (Revised): A5BC2_tI32_140_cl_a_h-001

Picture of Structure; Click for Big Picture
Prototype Br$_{5}$(NH$_{4}$)Pb$_{2}$
AFLOW prototype label A5BC2_tI32_140_cl_a_h-001
Strukturbericht designation $K3_{4}$
ICSD 26662
CCDC 1602052
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.

First cited in

  • 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, Comp. Mat. Sci. 240, 112988 (2024). (doi=10.1016/j.commatsci.2021.110450)

Geometry files


Prototype Generator

aflow --proto=A5BC2_tI32_140_cl_a_h --params=$a,c/a,x_{3},x_{4},z_{4}$

Species:

Running:

Output: