Atomic Symmetries of Ammonium Chloride. 179 



immediate method of selecting between values of n = 

 1, 2, 3, or 4, with corresponding values of m = 1, 8, 27, and 

 64, and it is natural to look to a more complicated struc- 

 ture arising from one of these for the apparently 

 necessary reconciliation between the symmetry of atom 

 arrangement and face development. 



It has, however, already been intimated that in this par- 

 ticular instance no such more complicated arrangement is 

 available. This can be shown as follows. There is a 

 maximum of 96 equivalent positions in the unit for any 

 space group having an enantiomorphic cubic hemihedry. 

 This means that if we make the apparently chemically 

 necessary assumption that all of the nitrogen and all of 

 the chlorine atoms in ammonium chloride are alike, there 

 may be as many as 96 nitrogen and 96 chlorine atoms in 

 the unit. If furthermore these atoms lie upon elements of 

 symmetry their number within the unit will be some sub- 



Table 1. 

 Special cases of the enantiomorphic cubic space groups. 



Equivalent positions in unit cell 

 Group 



Sixteen 



KD 



s — — 2 2 1(1) 1(1) — 



Explanatory Note: If a number in the preceding table is not followed 

 by a parenthesis, the positions in the corresponding arrangement are com- 

 pletely determined by the symmetry considerations; the parenthesized 

 values give the number of variable parameters possessed by the various 

 arrangements to which they refer. 



structure for ammonium chloride. Thus these calculations yield for the 

 strongest diffractions that are recorded the following calculated intensities: 



Intensities 

 Indices of Plane Order Estimated Calculated 



110 1 10 3,050 arbitrarv units 



111 1 1 220 

 100 2 2 679 



210 1 1.5 359 



211 1 3 1,690 





One 



Four 



Eight 



Twelve 



1 

 2 

 3 



2 



2 

 2 



1(1) 

 1(1) 

 1 



1(1) 

 KD 



4 



— 



— 



2 







5 

 6 



— 



2 



2 



1 



1(1) 

 1(1) 



1;1(D 



KD 



KD 



Twenty- 



Thirty- 



four 



two 



1(3) 



— 



1(3) 



— 



l;i(D 



1(D 







KD 



KD 



— 



1(3) 



— 



1(3) 



— 



