Pu$_{2}$C$_{3}$ ($D5_{c}$) Structure: A3B2_cI40_220_d_c-001
| Prototype | C$_{3}$Pu$_{2}$ |
| AFLOW prototype label | A3B2_cI40_220_d_c-001 |
| Strukturbericht designation | $D5_{c}$ |
| ICSD | 16511 |
| Pearson symbol | cI40 |
| Space group number | 220 |
| Space group symbol | $I\overline{4}3d$ |
| AFLOW prototype command |
aflow --proto=A3B2_cI40_220_d_c-001
--params=$a, \allowbreak x_{1}, \allowbreak x_{2}$ |
Other compounds with this structure
Am$_{2}$C$_{3}$, Ce$_{2}$C$_{3}$, Dy$_{2}$C$_{3}$, Ho$_{2}$C$_{3}$, Hf$_{2}$C$_{3}$, La$_{2}$C$_{3}$, Nd$_{2}$C$_{3}$, Np$_{2}$C$_{3}$, Pr$_{2}$C$_{3}$, Pu$_{2}$C$_{3}$, Sm$_{2}$C$_{3}$, Th$_{2}$C$_{3}$, U$_{2}$C$_{3}$, Y$_{2}$C$_{3}$, Cs$_{2}$O$_{3}$, Ru$_{2}$Er$_{3}$, Rb$_{2}$O$_{3}$, Ru$_{2}$Y$_{3}$
- We use the data for $^{240}$Pu.
\[ \begin{array}{ccc}
\mathbf{a_{1}}&=&- \frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}a \,\mathbf{\hat{z}}\\\mathbf{a_{2}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{2}a \,\mathbf{\hat{z}}\\\mathbf{a_{3}}&=&\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}- \frac{1}{2}a \,\mathbf{\hat{z}}
\end{array}\]
Basis vectors
| Lattice coordinates | Cartesian coordinates | Wyckoff position | Atom type | |||
|---|---|---|---|---|---|---|
| $\mathbf{B_{1}}$ | = | $2 x_{1} \, \mathbf{a}_{1}+2 x_{1} \, \mathbf{a}_{2}+2 x_{1} \, \mathbf{a}_{3}$ | = | $a x_{1} \,\mathbf{\hat{x}}+a x_{1} \,\mathbf{\hat{y}}+a x_{1} \,\mathbf{\hat{z}}$ | (16c) | Pu I |
| $\mathbf{B_{2}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}- \left(2 x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{1} \,\mathbf{\hat{x}}- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{y}}+a x_{1} \,\mathbf{\hat{z}}$ | (16c) | Pu I |
| $\mathbf{B_{3}}$ | = | $- \left(2 x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{x}}+a x_{1} \,\mathbf{\hat{y}}- a x_{1} \,\mathbf{\hat{z}}$ | (16c) | Pu I |
| $\mathbf{B_{4}}$ | = | $- \left(2 x_{1} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $a x_{1} \,\mathbf{\hat{x}}- a x_{1} \,\mathbf{\hat{y}}- a \left(x_{1} - \frac{1}{2}\right) \,\mathbf{\hat{z}}$ | (16c) | Pu I |
| $\mathbf{B_{5}}$ | = | $\left(2 x_{1} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(2 x_{1} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(2 x_{1} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{1} + \frac{1}{4}\right) \,\mathbf{\hat{x}}+a \left(x_{1} + \frac{1}{4}\right) \,\mathbf{\hat{y}}+a \left(x_{1} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (16c) | Pu I |
| $\mathbf{B_{6}}$ | = | $\frac{1}{2} \, \mathbf{a}_{1}- 2 x_{1} \, \mathbf{a}_{3}$ | = | $- a \left(x_{1} + \frac{1}{4}\right) \,\mathbf{\hat{x}}- a \left(x_{1} - \frac{1}{4}\right) \,\mathbf{\hat{y}}+a \left(x_{1} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (16c) | Pu I |
| $\mathbf{B_{7}}$ | = | $- 2 x_{1} \, \mathbf{a}_{1}+\frac{1}{2} \, \mathbf{a}_{2}$ | = | $a \left(x_{1} + \frac{1}{4}\right) \,\mathbf{\hat{x}}- a \left(x_{1} + \frac{1}{4}\right) \,\mathbf{\hat{y}}- a \left(x_{1} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (16c) | Pu I |
| $\mathbf{B_{8}}$ | = | $- 2 x_{1} \, \mathbf{a}_{2}+\frac{1}{2} \, \mathbf{a}_{3}$ | = | $- a \left(x_{1} - \frac{1}{4}\right) \,\mathbf{\hat{x}}+a \left(x_{1} + \frac{1}{4}\right) \,\mathbf{\hat{y}}- a \left(x_{1} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (16c) | Pu I |
| $\mathbf{B_{9}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}+\left(x_{2} + \frac{1}{4}\right) \, \mathbf{a}_{2}+x_{2} \, \mathbf{a}_{3}$ | = | $a x_{2} \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ | (24d) | C I |
| $\mathbf{B_{10}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}- \left(x_{2} - \frac{1}{4}\right) \, \mathbf{a}_{2}- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $- a x_{2} \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ | (24d) | C I |
| $\mathbf{B_{11}}$ | = | $x_{2} \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}+\left(x_{2} + \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+a x_{2} \,\mathbf{\hat{y}}$ | (24d) | C I |
| $\mathbf{B_{12}}$ | = | $- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}- \left(x_{2} - \frac{1}{4}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}- a x_{2} \,\mathbf{\hat{y}}+\frac{1}{2}a \,\mathbf{\hat{z}}$ | (24d) | C I |
| $\mathbf{B_{13}}$ | = | $\left(x_{2} + \frac{1}{4}\right) \, \mathbf{a}_{1}+x_{2} \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{y}}+a x_{2} \,\mathbf{\hat{z}}$ | (24d) | C I |
| $\mathbf{B_{14}}$ | = | $- \left(x_{2} - \frac{1}{4}\right) \, \mathbf{a}_{1}- \left(x_{2} - \frac{1}{2}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}- a x_{2} \,\mathbf{\hat{z}}$ | (24d) | C I |
| $\mathbf{B_{15}}$ | = | $\left(x_{2} + \frac{3}{4}\right) \, \mathbf{a}_{1}+\frac{3}{4} \, \mathbf{a}_{2}+\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{3}$ | = | $\frac{1}{4}a \,\mathbf{\hat{x}}+a \left(x_{2} + \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{1}{2}a \,\mathbf{\hat{z}}$ | (24d) | C I |
| $\mathbf{B_{16}}$ | = | $- \left(x_{2} - \frac{3}{4}\right) \, \mathbf{a}_{1}+\frac{1}{4} \, \mathbf{a}_{2}- x_{2} \, \mathbf{a}_{3}$ | = | $- \frac{1}{4}a \,\mathbf{\hat{x}}- a \left(x_{2} - \frac{1}{4}\right) \,\mathbf{\hat{y}}+\frac{1}{2}a \,\mathbf{\hat{z}}$ | (24d) | C I |
| $\mathbf{B_{17}}$ | = | $\frac{3}{4} \, \mathbf{a}_{1}+\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{2}+\left(x_{2} + \frac{3}{4}\right) \, \mathbf{a}_{3}$ | = | $a \left(x_{2} + \frac{1}{4}\right) \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}+\frac{1}{4}a \,\mathbf{\hat{z}}$ | (24d) | C I |
| $\mathbf{B_{18}}$ | = | $\frac{1}{4} \, \mathbf{a}_{1}- x_{2} \, \mathbf{a}_{2}- \left(x_{2} - \frac{3}{4}\right) \, \mathbf{a}_{3}$ | = | $- a \left(x_{2} - \frac{1}{4}\right) \,\mathbf{\hat{x}}+\frac{1}{2}a \,\mathbf{\hat{y}}- \frac{1}{4}a \,\mathbf{\hat{z}}$ | (24d) | C I |
| $\mathbf{B_{19}}$ | = | $\left(x_{2} + \frac{1}{2}\right) \, \mathbf{a}_{1}+\left(x_{2} + \frac{3}{4}\right) \, \mathbf{a}_{2}+\frac{3}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}+\frac{1}{4}a \,\mathbf{\hat{y}}+a \left(x_{2} + \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (24d) | C I |
| $\mathbf{B_{20}}$ | = | $- x_{2} \, \mathbf{a}_{1}- \left(x_{2} - \frac{3}{4}\right) \, \mathbf{a}_{2}+\frac{1}{4} \, \mathbf{a}_{3}$ | = | $\frac{1}{2}a \,\mathbf{\hat{x}}- \frac{1}{4}a \,\mathbf{\hat{y}}- a \left(x_{2} - \frac{1}{4}\right) \,\mathbf{\hat{z}}$ | (24d) | C I |
References
- J. L. Green, G. P. Arnold, J. A. Leary, and N. G. Nereson, Crystallographic and magnetic ordering studies of plutonium carbides using neutron diffraction, J. Nucl. Mater. 34, 281–289 (1970), doi:10.1016/0022-3115(70)90194-7.
Found in
- P. Villars and L. Calvert, Pearson's Handbook of Crystallographic Data for Intermetallic Phases (ASM International, Materials Park, OH, 1991), 2nd edn.
Prototype Generator
aflow --proto=A3B2_cI40_220_d_c --params=$a,x_{1},x_{2}$