SrAu$_{3}$Al$_{2}$ Structure: A2B3C_oP24_62_d_3c

Basis vectors

		Lattice coordinates		Cartesian coordinates	Wyckoff position	Atom type
$B_{1}$	=	$x_{1} a_{1} + \frac{1}{4} a_{2} + z_{1} a_{3}$	=	$a x_{1} \hat{x} + \frac{1}{4} b \hat{y} + c z_{1} \hat{z}$	(4c)	Au I
$B_{2}$	=	$- (x_{1} - \frac{1}{2}) a_{1} + \frac{3}{4} a_{2} + (z_{1} + \frac{1}{2}) a_{3}$	=	$- a (x_{1} - \frac{1}{2}) \hat{x} + \frac{3}{4} b \hat{y} + c (z_{1} + \frac{1}{2}) \hat{z}$	(4c)	Au I
$B_{3}$	=	$- x_{1} a_{1} + \frac{3}{4} a_{2} - z_{1} a_{3}$	=	$- a x_{1} \hat{x} + \frac{3}{4} b \hat{y} - c z_{1} \hat{z}$	(4c)	Au I
$B_{4}$	=	$(x_{1} + \frac{1}{2}) a_{1} + \frac{1}{4} a_{2} - (z_{1} - \frac{1}{2}) a_{3}$	=	$a (x_{1} + \frac{1}{2}) \hat{x} + \frac{1}{4} b \hat{y} - c (z_{1} - \frac{1}{2}) \hat{z}$	(4c)	Au I
$B_{5}$	=	$x_{2} a_{1} + \frac{1}{4} a_{2} + z_{2} a_{3}$	=	$a x_{2} \hat{x} + \frac{1}{4} b \hat{y} + c z_{2} \hat{z}$	(4c)	Au II
$B_{6}$	=	$- (x_{2} - \frac{1}{2}) a_{1} + \frac{3}{4} a_{2} + (z_{2} + \frac{1}{2}) a_{3}$	=	$- a (x_{2} - \frac{1}{2}) \hat{x} + \frac{3}{4} b \hat{y} + c (z_{2} + \frac{1}{2}) \hat{z}$	(4c)	Au II
$B_{7}$	=	$- x_{2} a_{1} + \frac{3}{4} a_{2} - z_{2} a_{3}$	=	$- a x_{2} \hat{x} + \frac{3}{4} b \hat{y} - c z_{2} \hat{z}$	(4c)	Au II
$B_{8}$	=	$(x_{2} + \frac{1}{2}) a_{1} + \frac{1}{4} a_{2} - (z_{2} - \frac{1}{2}) a_{3}$	=	$a (x_{2} + \frac{1}{2}) \hat{x} + \frac{1}{4} b \hat{y} - c (z_{2} - \frac{1}{2}) \hat{z}$	(4c)	Au II
$B_{9}$	=	$x_{3} a_{1} + \frac{1}{4} a_{2} + z_{3} a_{3}$	=	$a x_{3} \hat{x} + \frac{1}{4} b \hat{y} + c z_{3} \hat{z}$	(4c)	Au III
$B_{10}$	=	$- (x_{3} - \frac{1}{2}) a_{1} + \frac{3}{4} a_{2} + (z_{3} + \frac{1}{2}) a_{3}$	=	$- a (x_{3} - \frac{1}{2}) \hat{x} + \frac{3}{4} b \hat{y} + c (z_{3} + \frac{1}{2}) \hat{z}$	(4c)	Au III
$B_{11}$	=	$- x_{3} a_{1} + \frac{3}{4} a_{2} - z_{3} a_{3}$	=	$- a x_{3} \hat{x} + \frac{3}{4} b \hat{y} - c z_{3} \hat{z}$	(4c)	Au III
$B_{12}$	=	$(x_{3} + \frac{1}{2}) a_{1} + \frac{1}{4} a_{2} - (z_{3} - \frac{1}{2}) a_{3}$	=	$a (x_{3} + \frac{1}{2}) \hat{x} + \frac{1}{4} b \hat{y} - c (z_{3} - \frac{1}{2}) \hat{z}$	(4c)	Au III
$B_{13}$	=	$x_{4} a_{1} + \frac{1}{4} a_{2} + z_{4} a_{3}$	=	$a x_{4} \hat{x} + \frac{1}{4} b \hat{y} + c z_{4} \hat{z}$	(4c)	Sr I
$B_{14}$	=	$- (x_{4} - \frac{1}{2}) a_{1} + \frac{3}{4} a_{2} + (z_{4} + \frac{1}{2}) a_{3}$	=	$- a (x_{4} - \frac{1}{2}) \hat{x} + \frac{3}{4} b \hat{y} + c (z_{4} + \frac{1}{2}) \hat{z}$	(4c)	Sr I
$B_{15}$	=	$- x_{4} a_{1} + \frac{3}{4} a_{2} - z_{4} a_{3}$	=	$- a x_{4} \hat{x} + \frac{3}{4} b \hat{y} - c z_{4} \hat{z}$	(4c)	Sr I
$B_{16}$	=	$(x_{4} + \frac{1}{2}) a_{1} + \frac{1}{4} a_{2} - (z_{4} - \frac{1}{2}) a_{3}$	=	$a (x_{4} + \frac{1}{2}) \hat{x} + \frac{1}{4} b \hat{y} - c (z_{4} - \frac{1}{2}) \hat{z}$	(4c)	Sr I
$B_{17}$	=	$x_{5} a_{1} + y_{5} a_{2} + z_{5} a_{3}$	=	$a x_{5} \hat{x} + b y_{5} \hat{y} + c z_{5} \hat{z}$	(8d)	Al I
$B_{18}$	=	$- (x_{5} - \frac{1}{2}) a_{1} - y_{5} a_{2} + (z_{5} + \frac{1}{2}) a_{3}$	=	$- a (x_{5} - \frac{1}{2}) \hat{x} - b y_{5} \hat{y} + c (z_{5} + \frac{1}{2}) \hat{z}$	(8d)	Al I
$B_{19}$	=	$- x_{5} a_{1} + (y_{5} + \frac{1}{2}) a_{2} - z_{5} a_{3}$	=	$- a x_{5} \hat{x} + b (y_{5} + \frac{1}{2}) \hat{y} - c z_{5} \hat{z}$	(8d)	Al I
$B_{20}$	=	$(x_{5} + \frac{1}{2}) a_{1} - (y_{5} - \frac{1}{2}) a_{2} - (z_{5} - \frac{1}{2}) a_{3}$	=	$a (x_{5} + \frac{1}{2}) \hat{x} - b (y_{5} - \frac{1}{2}) \hat{y} - c (z_{5} - \frac{1}{2}) \hat{z}$	(8d)	Al I
$B_{21}$	=	$- x_{5} a_{1} - y_{5} a_{2} - z_{5} a_{3}$	=	$- a x_{5} \hat{x} - b y_{5} \hat{y} - c z_{5} \hat{z}$	(8d)	Al I
$B_{22}$	=	$(x_{5} + \frac{1}{2}) a_{1} + y_{5} a_{2} - (z_{5} - \frac{1}{2}) a_{3}$	=	$a (x_{5} + \frac{1}{2}) \hat{x} + b y_{5} \hat{y} - c (z_{5} - \frac{1}{2}) \hat{z}$	(8d)	Al I
$B_{23}$	=	$x_{5} a_{1} - (y_{5} - \frac{1}{2}) a_{2} + z_{5} a_{3}$	=	$a x_{5} \hat{x} - b (y_{5} - \frac{1}{2}) \hat{y} + c z_{5} \hat{z}$	(8d)	Al I
$B_{24}$	=	$- (x_{5} - \frac{1}{2}) a_{1} + (y_{5} + \frac{1}{2}) a_{2} + (z_{5} + \frac{1}{2}) a_{3}$	=	$- a (x_{5} - \frac{1}{2}) \hat{x} + b (y_{5} + \frac{1}{2}) \hat{y} + c (z_{5} + \frac{1}{2}) \hat{z}$	(8d)	Al I

Prototype	Al $_{2}$ Au $_{3}$ Sr
AFLOW prototype label	A2B3C_oP24_62_d_3c_c-001
ICSD	429904
Pearson symbol	oP24
Space group number	62
Space group symbol	$P nma$
AFLOW prototype command	aflow --proto=A2B3C_oP24_62_d_3c_c-001 --params= $a, b / a, c / a, x_{1}, z_{1}, x_{2}, z_{2}, x_{3}, z_{3}, x_{4}, z_{4}, x_{5}, y_{5}, z_{5}$

Encyclopedia of Crystallographic Prototypes

SrAu $_{3}$ Al $_{2}$ Structure: A2B3C_oP24_62_d_3c_c-001

Simple Orthorhombic primitive vectors

References

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Encyclopedia of Crystallographic Prototypes

SrAu3​Al2​ Structure: A2B3C_oP24_62_d_3c_c-001

Simple Orthorhombic primitive vectors

References

Found in

Running:

Output:

SrAu $_{3}$ Al $_{2}$ Structure: A2B3C_oP24_62_d_3c_c-001