Note 25 bis — Chapter VI · TTHE COMPOUNDS OF NITROGEN WITH HYDROGEN AND OXYGEN

In order to illustrate the application of the law of substitution to a given case, and to show the connection between ammonia and the oxides of nitrogen, let us consider the possible products of an oxygen and hydroxyl substitution in caustic ammonia, NH4(OH). It is evident that the substitution of H by OH can give: (1) NH3(OH)2; (2) NH2(OH)3; (3) NH(OH)4; and (4) N(OH)5. They should all, like caustic ammonia itself, easily part with water and form products (hydroxylic) of the oxidation of ammonia. The first of them is the hydrate of hydroxylamine, NH2(OH) + H2O; the second, NH(OH)2 + H2O (and also the substance NH(OH)4 or NH3O2), containing, as it does, both hydrogen and oxygen, is able to part with all its hydrogen in the form of water (which could not be done by the first product, since it contained too little oxygen), forming, as the ultimate product, 2NH2(OH)3 - 5H2O = N2O—that is, it corresponds with nitrous oxide, or the lower degree of the oxidation of nitrogen. So, also, nitrous anhydride corresponds with the third of the above products, 2NH(OH)4 - 5H2O = N2O3, and nitric anhydride with the fourth, 2N(OH)5 - 5H2O = N2O5. As, in these three equations, two molecules of the substitution products (-5H2O) are taken, it is also possible to combine two different products in one equation. For instance, the third and fourth products: NH(OH)4 + N(OH)5 - 5H2O corresponds to N2O4 or 2NO2, that is, to peroxide of nitrogen. Thus all the five (see later) oxides of nitrogen, N2O, NO, N2O3, NO2, and N2O5, may be deduced from ammonia. The above may be expressed in a general form by the equation (it should be remarked that the composition of all the substitution products of caustic ammonia may be expressed by NH3O5 - a, where a varies between 0 and 4):

NH₅O_5-a + NH₅O_5-b - 5H₂O = N₂O_5-(a+b),

where a + b can evidently be not greater than 5; when a + b = 5 we have N₂—nitrogen, when = 4 we have N₂O nitrous oxide; when a + b = 3 we have N₂O₂ or NO—nitric oxide, and so on to N₂O₅, when a + b = 0. Besides which it is evident that intermediate products may correspond with (and hence also break up into) different starting points; for instance, N₂O is obtained when a + b = 2, and this may occur either when a = 0 (nitric acid), and b = 2 (hydroxylamine), or when a = b = 1 (the third of the above substitution products).