Validity, and Residual Affinity. 305 



that the platinum is divalent, and yet that the peculiarity of 

 the chlorine atoms — their residual affinity, validity, or even 

 valency — enables Cl 3 Pt to play the part of an unsaturated 

 radical, but not because Pt is necessarily tetravalent. On 

 this hypothesis I would write PtCl 4 as 



ci=ci /Ci==ci ^ci— a 



Pt( | or P< | ; and possibly Pt ||. 



X C1=C1 X C1==C1 \C1— CI 



This indicates the chemical asymmetry of the two chlorine atoms. 



Either heat or free iodine removes the two remoter atoms (or, 



say, more accessible vortex-rings), giving in the latter case 



/C1=I 

 red crystals of PtCl 2 I 2 , i. e. Pt^ | . Platinum is therefore 



01=1 

 not octavalent in H 2 PtCl 6 , and probably not even tetravalent. 

 The existence of the so-called molecular compound is simply 

 due to the peculiarities of the chlorine, as suggested by 

 C1==C1==C1-H 

 \C1==C1= =C1— H. 

 This is probably the type of the characteristic compounds of 

 the platinum and palladium triplets, Rh, Ru, Pd and Ir, Os, Pt. 

 The double chlorides are asymmetrical. The same idea applies 

 to the characteristic W 2 MF e of the Ti, Zr, Th and Si, Ge, Sn 

 triplets. In cases where the chlorine atoms are symmetrical we 

 often find an equal number of molecules associated by the mys- 

 terious dot ( • ). The type is XG\ n . nM where M is == C1K, 

 == NH 3 , == CO, KCN == and such like. Thus there is not 

 only T1C1 3 and T1C1, but also T1C1 3 . 3C1K. 



Validity is inter alia conditioned by temperature, and a 

 temperature of 200° resolves either H 2 PtCl 6 or PtCl 4 into 

 PtCl 2 . That Pt is divalent in K 2 Pt(CN) 4 seems to be indi- 

 cated by its formations from either PtCl 2 or PtCl 4 . Lio-ht 

 will probably also be thrown upon the difficulties which beset 

 attempts to give constitutional formulae to the numerous iso- 

 meric or even metameric resultants between ammonia and 

 chlorides. Their existence would appear to be due to this 

 validity of the nitrogen and chlorine. For instance, of the 

 two different substances formulated, PdCl 2 N 2 H 6 , Pd beino- a 

 well-defined dyad, the red one yields 1STH 3 easily, and the 

 yellow one gives no NH 3 on treatment with KHO, but is the 

 dichloride of a divalent radical, (PdNgHg)", whose oxide and 

 whose alkaline hydrate are known. In this yellow compound 

 the chlorine is probably monovalent or the nitrogen pentavalent, 



pd/ NH s Cl 

 x NH 3 Cf 



