Atomistry » Boron » Chemical properties » Borohydrates
Atomistry »
  Boron »
    Chemical properties »
      Borohydrates »

Borohydrates

When one part of boron sesqui-oxide is heated with 2¼ parts of magnesium powder for 45 minutes at a red heat in a rapid stream of hydrogen, the main products are magnesia and magnesium boride, Mg3B2:

6Mg + B2O3 = 3MgO + Mg3B2.

It appears to be essential to the success of the experiment that the mixture should gently deflagrate for about five minutes after the heating commences.

The crude mixture from the foregoing reaction evolves hydrogen when treated with water. It has been shown by Travers, Ray, and Gupta that the magnesium boride is decomposed as follows: -

Mg3B2 + 6H2O = Mg3B2(OH)6 + 3H2.

The product is an almost white powder, insoluble in water; it is the magnesium derivative of a compound B2(OH2)6, which, from analogy with the carbohydrates, may be termed a borohydrate.

The solution obtained by treating the boride with water is found to contain small quantities of substances which exhibit the remarkable property of evolving hydrogen, with brisk effervescence, when acidified. These substances are also borohydrates or allied compounds, and, like the hydroborons obtained from the magnesium boride by the action of acids, appear to be the products of certain unknown side reactions. The amounts of these substances obtained are very small in comparison with the amount of boride required for their production.

The solution is usually yellow, owing to the presence of colloidal boron. It decomposes slowly at the ordinary temperature, hydrogen being evolved. In the presence of platinum black the rate of decomposition is greatly accelerated. The solution precipitates silver and mercury from their salts immediately; with copper salts either copper hydride or amorphous boron appears to be precipitated, according to circumstances. When acidified, the solution evolves hydrogen; the liquid thus obtained decolorises iodine.

A careful, quantitative study of the properties of the solution has led Travers, Ray, and Gupta to the conclusion that the solution contains two substances of the formulae H6B2O2 and H6B2O3Mg, the latter being the magnesium derivative of a compound H8B2O3. It is suggested that in these compounds boron has a valency of five; and if, as seems highly probable, the evolution of hydrogen takes place by the elimination of pairs of hydrogen atoms attached to adjacent boron atoms, the properties of these compounds can be explained by assigning to them the formulae: -

and

The products formed on treatment with acid arise in the following manner: -

(i) = +2H2

(ii) = = = + H2

The compound ho.b: b.oh, which is analogous to hyponitrous acid in its structure, and the compound bh3: b(oh)3 are oxidised by iodine to boron dioxide, B2O2: -

(iii) + I2 = + 3HI

(iv) + 2I2 = + 4HI + H2O

When the solution obtained by the action of water on magnesium boride is treated with ammonia, magnesium hydroxide is precipitated. In the conversion of the magnesium derivative H6B2O3Mg into the ammonium compound, however, intra-molecular change apparently occurs, thus: -

HB(OH)3:HBH3H2B(OH)2:H2BH(OH)

for the new product, when acidified, evolves twice as much hydrogen as the initial: -

H2B(OH)2.H3BH(OH) = B(OH)2.BH(OH)

yielding a product which is oxidised by iodine to boron dioxide, B2O2: -

B(OH)2.BH(OH) + I2 = B:O.B:O + 2HI + H2O

The magnesium derivative Mg3B2(OH)6, already mentioned, undergoes decomposition when treated with strong ammonia in an atmosphere of hydrogen. Travers and Ray, who have investigated the reaction, conclude that the soluble product of the change is a di-ammonium derivative of the compound -

BH2(OH)2.BH(OH):BH(OH).BH2(OH)2.

When acidified, this compound loses hydrogen, thus: -

BH2(OH)2.BH(OH):BH(OH).BH2(OH)2 = 2H2 + BH(OH)2:B(OH):B(OH):BH(OH)2,

and the new product loses more hydrogen when treated with iodine: -

BH(OH)2:B(OH):B(OH):BH(OH)2 + I2 = 2HI + B(OH)2.B(OH).B(OH).B(OH)2.

By evaporating the ammoniacal solution in vacuo and heating the residue, the oxide B4O5 is obtained, thus: -

BP2(OH)2.BH(OH).BH(OH).BH2(OH2) + 2NH3 = + 5H2 + H2O + 2NH3

Last articles

Zn in 7RE3
Zn in 7RDX
Zn in 7RDZ
Zn in 7RWM
Zn in 7PGU
Zn in 7PGR
Zn in 7PGT
Zn in 7PGS
Zn in 7SQE
Zn in 7RWK
© Copyright 2008-2020 by atomistry.com
Home   |    Site Map   |    Copyright   |    Contact us   |    Privacy