Iridium(IV) iodide
Chemical compound
From Wikipedia, the free encyclopedia
Iridium(IV) iodide is a binary inorganic compound of iridium and iodide with the chemical formula IrI
4.[2]
| Names | |
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| Other names
Iridium(IV) iodide, tetraiodoiridium | |
| Identifiers | |
3D model (JSmol) |
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| ChemSpider | |
| ECHA InfoCard | 100.029.279 |
| EC Number |
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PubChem CID |
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CompTox Dashboard (EPA) |
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| Properties | |
| I4Ir | |
| Molar mass | 699.835 g·mol−1 |
| Appearance | Black powder |
| Melting point | 100 °C (212 °F; 373 K) |
| insoluble | |
| Structure | |
| hexagonal | |
| Related compounds | |
Related compounds |
Iridium triiodide, platinum tetraiodide |
| Hazards | |
| GHS labelling:[1] | |
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| Warning | |
| H315, H319, H335 | |
| P261, P264, P264+P265, P271, P280, P302+P352, P304+P340, P305+P351+P338, P319, P321, P332+P317, P337+P317, P362+P364, P403+P233, P405, P501 | |
Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa).
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Properties
Iridium tetraiodide forms black crystals, does not dissolve in water and alcohol.[3][4] In alkali metal iodide solutions, the compound dissolves easily to give a ruby red solution, forming complex salts.[5]
The compound decomposes when heated:[citation needed]
- IrI4 → Ir + 2I2
Preparation
Iridium(IV) iodide can be obtained by reacting dipotassium hexachloroiridate or hexachloroiridic acid with an aqueous solution of potassium iodide.[6][5]
