Crystal Lattice Structures:	Creation Date: 18 Dec 2000
	Last Modified:  7 Jun 2006

The CaTiO₃ (Pnma) Structure

You can now

• see the structure from several perspectives;
• examine the structure with an external viewer; or
• download the coordinates of the atoms in these pictures in XYZ format.
• visualize the structure (Uses the JMOL Applet)

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• This is the true ground state of the prototype Perovskite crystal CaTiO₃. Roberto Colella of Purdue suggested that we look at another substance, LaMnO₃, so the pictures here refer to this compound. See Douglas du Boulay's Thesis for more details about perovskites, and P. Norby et al. for more information about LaMnO₃.
• This is a small distortion of the Perovskite unit cell. One can see this by looking at the icon above, or look at the XYZ file representation of one-fourth of the LaMnO₃ unit cell.
• Reference: (LaMnO₃) P. Norby et al. Solid State Chem. 119, 191 (1995).
• 7 June 2006: As pointed out to us over a year ago, the (Ca,La) atoms occupy the (4c) sites while the (Ti,Mn) atoms occupy the (4a) sites. The files used to generate the pictures have always been correct.

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• Prototype: CaTiO₃
• Pearson Symbol: oP20
• Space Group: Pnma (Cartesian and lattice coordinate listings available)
• Number: 62
• Other Compounds with this Structure: LaMnO₃ (shown), YAlO₃,RFeO₃ (R = La, Pr, Nd, Sm, Eu, Y).
• Primitive Vectors:

  A1	=	a X
  A2	=	b Y
  A3	=	c Z

• Basis Vectors:

  B1	=		0		(Ti, Mn)	(4a)
  B2	=	½ A2	=	½ b Y	(Ti, Mn)	(4a)
  B3	=	½ A1 + ½ A3	=	½ a X + ½ c Z	(Ti, Mn)	(4a)
  B4	=	½ A1 + ½ A2 + ½ A3	=	½ a X + ½ b Y + ½ c Z	(Ti, Mn)	(4a)
  B5	=	+ x1 A1 + ¼ A2 + z1 A3	=	+ x1 a X + ¼ b Y + z1 c Z	(Ca, La)	(4c)
  B6	=	- x1 A1 + ¾ A2 - z1 A3	=	- x1 a X + ¾ b Y - z1 c Z	(Ca, La)	(4c)
  B7	=	(½ - x1) A1 + ¾ A2 + (½ + z1) A3	=	(½ - x1) a X + ¾ b Y + (½ + z1) c Z	(Ca, La)	(4c)
  B8	=	(½ + x1) A1 + ¼ A2 + (½ - z1) A3	=	(½ + x1) a X + ¼ b Y + (½ - z1) c Z	(Ca, La)	(4c)
  B9	=	+ x2 A1 + ¼ A2 + z2 A3	=	+ x2 a X + ¼ b Y + z2 c Z	(O-I)	(4c)
  B10	=	- x2 A1 + ¾ A2 - z2 A3	=	- x2 a X + ¾ b Y - z2 c Z	(O-I)	(4c)
  B11	=	(½ - x2) A1 + ¾ A2 + (½ + z2) A3	=	(½ - x2) a X + ¾ b Y + (½ + z2) c Z	(O-I)	(4c)
  B12	=	(½ + x2) A1 + ¼ A2 + (½ - z2) A3	=	(½ + x2) a X + ¼ b Y + (½ - z2) c Z	(O-I)	(4c)
  B13	=	+ x3 A1 + y3 A2 + z3 A3	=	+ x3 a X + y3 b Y + z3 c Z	(O-II)	(8d)
  B14	=	- x3 A1 - y3 A2 - z3 A3	=	- x3 a X - y3 b Y - z3 c Z	(O-II)	(8d)
  B15	=	(½ - x3) A1 - y3 A2 + (½ + z3) A3	=	(½ - x3) a X - y3 b Y + (½ + z3) c Z	(O-II)	(8d)
  B16	=	(½ + x3) A1 + y3 A2 + (½ - z3) A3	=	(½ + x3) a X + y3 b Y + (½ - z3) c Z	(O-II)	(8d)
  B17	=	+ x3 A1 + (½ - y3) A2 + z3 A3	=	+ x3 a X + (½ - y3) b Y + z3 c Z	(O-II)	(8d)
  B18	=	- x3 A1 + (½ + y3) A2 - z3 A3	=	- x3 a X + (½ + y3) b Y - z3 c Z	(O-II)	(8d)
  B19	=	(½ + x3) A1 + (½ - y3) A2 + (½ - z3) A3	=	(½ + x3) a X + (½ - y3) b Y + (½ - z3) c Z	(O-II)	(8d)
  B20	=	(½ - x3) A1 + (½ + y3) A2 + (½ + z3) A3	=	(½ - x3) a X + (½ + y3) b Y + (½ + z3) c Z	(O-II)	(8d)

See these vectors in LaTeX output format.

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