Iridium(IV) iodide is a binary chemical compound of iridium and iodide with the chemical formula IrI
4.[1][2][3]
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 |
Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa).
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Preparation
Iridium(IV) iodide can be obtained by reacting dipotassium hexachloroiridate or hexachloroiridic acid with an aqueous solution of potassium iodide.[4][5]
Properties
Iridium tetraiodide forms black crystals, does not dissolve in water and alcohol.[6][7][8] 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
Uses
Iridium(IV) iodide can be used as a catalyst in organic chemistry.[9][8]
References
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