78 Dr. L. Vegard on 



There are two different types of arrangements according 



Tto the view we take with regard to the chemical nature of 



the substances. The one view is expressed in the formulae : 



Zr (Si0 4 ), Ti(Ti0 4 ), Sn(Sn0 4 ) ; and if these were the right 



•expressions, the structure — even in the case of rutile and 



kassiterite — should distinguish between two sorts of metallic 



-atoms. 



According to the other view, zircon is considered as a kind 

 of addition product of two dioxides Zr0 2 , Si0 2 ; and if we 

 substitute Zr and Si with either Ti or Sn, we should expect 

 .'to get a lattice where all Ti or Sn atoms are equal. 



In accordance with the first view, we should arrange four 

 atoms of oxygen round each of the Si atoms in accordance 

 with the symmetry of the crystal. 



We have to consider the following three arrangements : — 



(1) The four O-atoms are placed along the tetragonal 

 :.axis through the Si atom, and with the O-atoms symme- 

 trically arranged on both sides of the Si atom. 

 ^(2) The four O-atoms are arranged in a plane through 

 the Si atom perpendicular to the C-axis with tetragonal 

 symmetry with regard to this axis. There are two different 

 arrangements. If we take the Si atom as origin the O-atoms 

 will be situated in the (xy) plane and will either have the 

 coordinates : 



(1,1), (-l,D, (1,-1), (-1,-1), or (1,0), (0,1), (0,-0, (-1,0), 



where I is a parameter. 



(3) The oxygen atoms are arranged on the diagonals AG, 

 BH, &c, fig. 3. Let the construction-points of the two face- 

 centred Si lattices be (000) and (a/4, a/4,*cM > ), then the con- 

 struction-points of the oxygen lattices would be : 



(e : a, eja, e x a), ( -€ x a, — €l a, e^), (-e x a, e x a, — € X c) 



(€j«, — €]«, —e^), and 



/a a c \ /a a c \ 



'{l + f2(7 ' 4 + e * a > 4 + € - c )' VI "^ 4 ~~ e ' 2a > I + €2a )' 



/a a 6 \ (a a c \ 



I j — €sfl 9 ^ + e 2 a, j — Wj, ( ^ + e 2 a, ^ — e 2 a, ^ _ e 2 cj. 



To get the right spacing for the face (100), e x and e 2 must 

 have the same numerical value ; but still we have to distin- 

 guish between the two cases : 



(3 a) 6! = e 2 , 

 (3 6) ex =-e 2 . 



