The Chalcopyrite (E1₁) Structure

Last modified 5 April 2001

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When c = 2a and x = 1/8 the atoms are on the sites of the diamond (A4) structure. In this case, if we replace the Fe atoms by Cu, we get the zincblende (B3) structure.

• Prototype: CuFeS₂
• Pearson Symbol: tI16
• Strukturbericht Designation: E1₁
• Space Group: I(-4)2d (Cartesian and lattice coordinate listings available)
• Number: 122
• Primitive Vectors:
  • A₁ = a X
  • A₂ = a Y
  • A₃ = ½ a X + ½ a Y + ½ c Z
• Basis Vectors:

  The "x" parameter here is related to the one in Pearson by xPearson = x + 1/8

  B1		=	0		(Cu)	(4a)
  B2	=	- ¼ A1 + ¼ A2 + ½ A3	=	½ a Y + ¼ c Z	(Cu)	(4a)
  B3	=	+ ½ A1 + ½ A2	=	½ a X + ½ a Y	(Fe)	(4b)
  B4	=	+ ¼ A1 - ¼ A2 + ½ A3	=	½ a X + ¼ c Z	(Fe)	(4b)
  B5	=	+ x A1 + 1/8 A2 + ¼ A3	=	+ (1/8+x) a X + ¼ a Y + 1/8 c Z	(S)	(8d)
  B6	=	- (¼ + x) A1 - 3/8 A2 + ¼ A3	=	- (1/8+x) a X - ¼ a Y + 1/8 c Z	(S)	(8d)
  B7	=	- 1/8 A1 + (¼ + x) A2 - ¼ A3	=	-¼ a X + (1/8+x) a Y - 1/8 c Z	(S)	(8d)
  B8	=	+ 3/8 A1 - x A2 - ¼ A3	=	+ ¼ a X - (1/8+x) a Y - 1/8 c Z	(S)	(8d)

See these vectors in a better notation.

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