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Phosphorus Pentabromide, PBr5

In the presence of excess of bromine phosphorus can form not only the pentabromide but also, according to the thermal diagram, other perbromides. It was discovered early that the presence of some iodine greatly facilitates the combination, probably by the formation of ICl3, thus:—

PCl3 + ½I + 5Br = PBr5 + ICl3

It is also produced by the decomposition of the dibromotrifluoride formed when trifluoride is passed into bromine at -10° C.:—

5PF3Br2 = 3PF5 + 2PBr5

Properties

Phosphorus pentabromide is a yellow crystalline solid which melts to a red liquid with decomposition. It fumes in the air. A red form has also been described, but this is probably PBr7.

The densities and specific volumes of the liquid under the pressure of its own vapour were determined in a sealed evacuated glass dilatometer. One set of results was as follows:—

t° C85100130165
v0.35300.36210.37550.3899


From these results and others the coefficient of expansion is calculated as vt-85 = v85{1 + 0.0019(t-85)}

and
vt-100 = v100{1 + 0.0012t}, up to 165° C.

The vapour is formed with dissociation. The pressures of the total vapour formed were determined by a static method. From the results it was calculated that the boiling- (sublimation) point was 106° C.

The heat of formation

P (solid) +2½Br2 (liq.) = PBr5 (solid) + 63.5 Cals.

was found to be greater than that of PBr3, but less than that of PCl5. The additional heat on combination with the last two atoms of bromine was found to be small.

PBr3 (liq.) + Br2 (liq.) = PBr5 (solid) +20.3 Cals.

The hydrolysis with excess of water produces hydrobromic and phosphoric acids. The heptabromide, PBr7, which was prepared by subliming PBr5 with Br2 in a sealed tube at 90° C., hydrolyses with the production of the same acids and bromine in addition. In the presence of a small amount of water the oxybromide POBr3 may be produced. Hydrogen sulphide by an analogous reaction gave the sulphobromide:—

PBr5 + H2S = PSBr3 + 2HBr

The pentabromide may also be used to replace hydroxyl groups in organic compounds by halogen. Thus:

PBr5 + CH3COOH = POBr3 + CH3COBr + HBr

An ammine, PBr5.9NH3, was formed by passing dry ammonia into a solution of PBr5 in CCl4.

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