The Hexagonal Diamond (Lonsdaleite) Structure
      
      
        
      
      
        You can now
      
      
      
      
        - 
          This is related to the hcp (A3)
          lattice in the same way that
          diamond (A4) is related to the
          fcc lattice (A1). It can also be
          obtained from Wurtzite (B4) by
          replacing both the Zn and S atoms by Carbon. We've
          shifted the origin compared to Wurtzite, so that the
          inversion site is between one pair of Carbon atoms.
        
 
        - 
          The ``ideal'' structure, where the nearest-neighbor
          environment of each atom is the same as in diamond, is
          achieved when we take c/a = (8/3)1/2
          and u = 1/16. Alternatively, we can take
          u = 3/16, in which case the origin is at the
          center of a C-C bond aligned in the [0001] direction.
        
 
        - 
          When u = 0 this structure becomes a set of
          graphitic sheets, but not true graphite
          (A9).
        
 
        - 
          9 March 2003: Corrected the exponents in the
          Cartesian expressions for the Basis Vectors (-1/3 ==>
          ½).
        
 
        - 
          6 June 2006: corrected sign of
          Y component of
          B4.
        
 
        - 
          10 October 2006: The text of the page previously
          mislabeled the z coordinates of some of the
          atoms. The coordinates used in generating the pictures
          were correct. We have corrected the basis vectors and
          reordered them so that they agree with the
          coordinate file.
        
 
      
      
      
        - 
          Prototype: C (hexagonal diamond)
        
 
        - 
          Pearson Symbol:
          hP4
        
 
        - 
          Space Group:
          P63/mmc
          (Cartesian and
          lattice coordinate
          listings available)
        
 
        - 
          Number: 194
        
 
        - 
          Other Compounds with this Structure: Si (Hexagonal)
        
 
        - 
          Primitive Vectors: 
          
            
              | 
                A1
               | 
              
                 = 
               | 
              
                ½ a X - ½
                3½
                a Y
               | 
            
            
              | 
                A2
               | 
              
                 = 
               | 
              
                ½ a X + ½
                3½
                a Y
               | 
            
            
              | 
                A3
               | 
              
                 = 
               | 
              
                c Z
               | 
            
          
         
        - 
          Basis Vectors: 
          
            
              | 
                B1
               | 
              
                 = 
               | 
              
                1/3 A1 + 2/3
                A2 + z
                A3
               | 
              
                 = 
               | 
              
                ½ a X +
                12-½ a Y + z c
                Z 
               | 
              
                 (C) 
               | 
              
                 (4f)
               | 
            
            
              | 
                B2
               | 
              
                 = 
               | 
              
                2/3 A1 + 1/3
                A2 + (½ + z)
                A3
               | 
              
                 = 
               | 
              
                ½ a X -
                12-½ a Y +
                (½ + z) c
                Z 
               | 
              
                 (C) 
               | 
              
                 (4f)
               | 
            
            
              | 
                B3
               | 
              
                 = 
               | 
              
                1/3 A1 + 2/3
                A2 + (½ - z)
                A3
               | 
              
                 = 
               | 
              
                ½ a X +
                12-½ a Y +
                (½ - z) c
                Z 
               | 
              
                 (C) 
               | 
              
                 (4f)
               | 
            
            
              | 
                B4
               | 
              
                 = 
               | 
              
                2/3 A1 + 1/3
                A2 - z
                A3
               | 
              
                 = 
               | 
              
                ½ a X -
                12-½ a Y - z c
                Z 
               | 
              
                 (C) 
               | 
              
                 (4f)
               | 
            
          
         
      
      
      
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