some Carbonates of the Calcite Group. 341 



Six equivalent points. 



(a) u u 0, u u, u u, u to 0, u u, u u. 



(b) u u 1/2, u 1/2 u, 1/2 u u, u u 1/2, u 1/2 w, 1/2 u u. 



(c) u u v, u v u, v u u, u u V, u v ii, v u u. 



T)6 



V sd 



Two equivalent points. 



(a) 0, 1/2 1/2 1/2. 



(ft) 1/4 1/4 1/4, 3/4 3/4 3/4. 



.FWr equivalent points. 



(a) tc u n; u u u; 1/2 — u, 1/2 — u, 1/2 — u; u + 1/2, 

 ^ + 1/2,^ + 1/2. 



Six equivalent points. 



(a) 1/4 3/4 3/4; 3/4 3/4 1/4; 3/4 1/4 3/4; 

 1/4 3/4 1/4; 3/4 1/4 1/4; 1/4 1/4 3/4. 



(h) u u 0/ u uj 



1/2-n, u + 1/2, 1/2; u+1/2, 1/2, 1/2-u; 



u uj 

 1/2 1/2 -n, u + 1/2. 



(c) u u 1/2/ u 1/2 w/ 

 1/2-u, u + 1/2, 0; u + 1/2, 0, 1/2-u; 



1/2 u u; 

 0, i/2-u, u + 1/2. 



Two molecules of CaC0 3 are to be associated with the 

 simplest unit of structure— that corresponding with the 

 newest set of axes. Since in neither group are found 

 four cases of one equivalent point, the two calcium atoms 

 must be equivalent and the two carbon atoms must also be 

 equivalent one to the other. The oxygen atoms must 

 either be all equal (at six equivalent positions), or four 

 alike and two different, or three sets of two like atoms, or 

 lastly two sets of three like atoms. This last arrange- 

 ment is improbable because it is to be supposed that the 

 three oxygen atoms associated with each carbon atom 

 would be similarly related to this carbon atom — a state of 

 affairs impossible with two sets of three equivalent posi- 

 tions. The above conditions are only fulfilled by arrang- 

 ing the atoms of 2CaC0 3 in any one of the following ways. 



