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

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° C	85	100	130	165
v	0.3530	0.3621	0.3755	0.3899

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

and
v_t-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.