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.