144 



MR J. Y. BUCHANAN ON THE 



CIO3 and Br03 change places, we get 1/64 [Rb] ; and when CIO3 further changes places 

 with I, we get 1/64 [Cs]. 



CIO, 



BrOo 



BrO, 



lOs / 1/64 \ BrO„ 



Br 



K 



IO3 ' i/g4 ^ C10„ 



Br 



Rb 



I0„ 



Br 



1/64 

 Cs 



I 

 CIO. 



CI 



CI CI 



The expressions 1/64 [Cs], 1/64 [CI], etc., are used as abbreviations to mean 

 the hexagons corresponding to the nuclei Cs, CI, etc., and the solutions containing 

 1/64 grm.-mol. salt per thousand grams of water. The general expressions 7n [R] and 

 m [M] indicate the hexagons corresponding to salts with a metalloidal or a metallic 

 nucleus respectively, the solutions of which contain m grm.-mol. of the salt per thousand 

 grams of water. 



§ 64. It has already been pointed out that the hexagonal arrangement of residues 

 in ascending order of magnitude of I'/m is the same for each nucleus at all concentrations 

 for which m>l/64, and we have shown how the arrangements expressed by 1/64 [R] 

 and 1/64 [M] are derived from that corresponding to 1/64 [CI]. We now proceed 

 to consider the case of in [R] and m [M] when m<l/64. It is only in solutions of 

 such low concentration that the phenomenon of expansion on further dilution with 

 water shows itself 



Beginning with the solutions of salts having a metalloidal nucleus, R, we arrange 

 the hexagons in three lines of four hexagons each. For the top lines R = CI, for the 

 middle line Br, and for the lowest line I. In each line the concentrations are defined 

 by m=l/64. 1/128, 1/256, and 1/512 in consecutive order. This arrangement is 

 exhibited in the group of hexagons on next page. 



Considering the hexagons with nucleus CI, we see that, for m^ 1/256, the 

 residues follow in the orders of the triads M, MO3, and in each triad in the ascending 

 order of atomic weight, K, Eb, Cs, KO3, RbOg, CsOg. This may be called the regular 

 system. For m= 1/512 this order is preserved, with the exception that place 1 is 

 taken by KO3 and we have KO3, K, Rb, Cs, RbOg, CsOg. 



When the nucleus is Br and ??i=l/64, 1/128, the order of residues is that of 

 consecutive triads, with transposition of KO3 and Cs. For m= 1/256 and 1/512, Rb 

 and KO3 are transposed, and also RbOg and Cs, so that the residues come to be 

 arranged alternately after the type M and MO3 respectively round the hexagons — 

 K, KO3, Rb, RbOg, Cs, CSO3. In the eight hexagons m [CI], m [Br], place 6 is occupied 

 by CSO3, and in seven out of the eight hexagons place 1 is occupied by K. 



When the nucleus is I, we find that place 1 is occupied in all cases by KO3, and 

 place 6 by Cs ; that is, the initial residue of the second triad takes the lowest place, 

 and the final residue of the first triad takes the highest place. Places 2 and 3 are 



