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Phosphorus Diiodide, P2I4

Phosphorus Diiodide, P2I4, was first prepared by fusing the constituents in equivalent proportions. 50 grams of iodine and 4 grams of red phosphorus are melted in a flask and, after partial cooling, 2.5 grams of white phosphorus are added in small pieces. Or, equal parts by weight of iodine and phosphorus are dissolved in carbon disulphide and the solution is cooled to 0° C., when the compound crystallises. It has also been prepared by the action of iodine on phosphorous oxide:—

5P4O6 + 8I2 = 4P2I4 + 6P2O5


Phosphorus diiodide forms orange-coloured crystals which belong to the triclinic system. The analysis gave (PI2). The vapour density, determined in the presence of nitrogen at a pressure slightly below 100 mm. and at 265° C., was between 18.0 and 20.2 (air = 1), which corresponds to a molecule P2I4. At 15 mm. and 100° to 120° C. the compound dissociates into P2I3 and P. The melting-point was given as about 110° C. The heat of formation is given as

P + I2 (solid) = PI2 + 9.88 Cals.

P2I4 can be ignited in a current of oxygen and burnt to phosphoric oxide and iodine. The hydrolysis appears somewhat complex and yields P, PH3, HI and H3PO3. The phosphine is a secondary product, since by gradual addition to water in the cold neither phosphine nor red phosphorus are formed:—

P2I4 + 5H2O = 4HI + H3PO3 + H3PO2

P2I4 is soluble in CS2 and slightly soluble in liquid H2S. At about 100° C. it reacted with H2S giving HI and a sulphoiodide P4S3I2.

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