Detection of Iridium

Metallic Iridium, like rhodium, is insoluble in all acids, save that in a very finely divided condition it is slowly attacked by aqua regia. Fusion with potassium hydrogen sulphate oxidises the metal but does not effect its solution (contrast ruthenium and rhodium). When fused with a mixture of potassium nitrate and hydroxide an insoluble residue containing the sesquioxide, Ir₂O₃, with alkali is obtained.

Hydrogen sulphide in excess causes the precipitation of dark brown iridium sesquisulphide, Ir₂S₃. If the iridium is present in solution in the tetravalent condition, as for example in the form of alkali chlor-iridates, M₂IrCl₆ or 2MCl.IrCl₄, the dark solution is first decolorised and a yellowish white precipitate of sulphur forms in consequence of the reduction of the iridium salt to the trivalent condition, namely, as IrCl₃. When this has been accomplished the brown sesquisulphide is obtained.

Ammonium sulphide yields the same precipitate, soluble in excess.

Alkali hydroxides in excess give the solution a green tint and precipitate a little double chloride. On heating the solution a red tint develops, changing to a deep azure blue in consequence of the precipitation of the hydrated dioxide, Ir(OH)₄. This serves to distinguish iridium from platinum.

Potassium chloride and ammonium chloride precipitate from concentrated solutions their respective chlor-iridates, K₂IrCl₆ and (NH₄)₂IrCl₆, as brownish red deposits.

Addition of reducing agents such as ferrous sulphate or stannous chloride to solutions of the alkali chlor-iridates reduces them to double salts of iridium trichloride, namely, 3MCl.IrCl₃ or M₃IrCl₆, known generally as chlor-iridites. The solution is simultaneously decolorised, and the salts crystallise out on cooling.

Metallic zinc reduces iridium salts to the metal, which is deposited as a black powder from solution.

Alcohol reduces iridic salts to the corresponding iridous compounds.