454 Dr. S. H. C. Briggs : The Elements 



we have a nitrogen kernel at the centre of the complex with 

 two electrons at each corner of a regular tetrahedron sur- 

 rounding the kernel, and a hydrogen nucleus united to each 

 pair of electrons (see Langmuir, J. Amer. Chem. Soc. xlii. 

 p. 274 (1920)). The secondary negative affinity of the 

 hydrogen nuclei is, however, sufficient to pull the electron 

 into the complex with the formation of hydrogen and 

 ammonia, and ammonium is consequently unknown in the 

 free state. We might expect, however, from the example 

 of cobaltic chloride and hexammine cobaltic chloride, that 

 the ammonium would be stabilized by partially saturating 

 the secondary negative affinity of the hydrogen nuclei by 

 the free affinity of the electrons of nercury atoms to give an 

 amalgam (NH^rHg^E. This is actually the case, ammonium 

 amalgam being a definite compound, capable of independent 

 existence, although easily decomposed (Smith, Ber. der 

 deutsch. chem. Gesellschaft, xl. p. 2941 (1907)). 



It is clear from the researches of Kraus (J. Amer. Chem. 

 Soc. xxx. p. 1323 (1908); xxxvi. p. 864 (1914)) that -the 

 alkali and alkaline earth metals undergo electrolytic dis- 

 sociation in liquid ammonia into kernel and electrons. The 

 compound Oa6NH 3 , for example, must therefore have the 

 structure (Ca*6NH 3 )E 2 , in which we have a complex ion 

 with a calcium kernel at the centre of a regular octahedron, 

 united to the electrons in the nitrogen atoms of six molecules 

 of ammonia situated at the corners of the octahedron, the 

 hydrogen atoms of the ammonia forming an outside layer of 

 the complex with unsaturated secondary negative affinity. 

 The electrons without the complex will therefore be in 

 proximity to the hydrogen nuclei, as seen from the extended 

 formula (Oa6E 2 N A 'E 6 H A ' 3 )E 2 , and with increasing thermal 

 agitation will tend to unite with the hydrogen nuclei giving 

 atomic hydrogen. This explains the decomposition 



M (NH 3 ) 6 = M (NH 2 ) 2 + 4NH 3 + fl 2 . 



(See also Biltz & Hiittig, Zeitsch. fur anorg. Chem. cxiv. 

 p. 241 (1920).) 



(3) Co-ordination compounds with a poly-nuclear complex. 

 The sharing of electrons by atoms with the production of 

 octets, corresponds to the sharing of atoms or radicles in the 

 formation of poly-nuclear complexes. 



As an example we may take Langmuir's formula for 

 hydrazoic acid H(N = X = N). Writing the formula as a 

 compound of kernels and electrons, we obtain (3), in which 

 we have a complex made up of three cubes in a line, 



