Hafnium tetraiodide
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(Redirected from Hafnium(IV) iodide)
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Identifiers | |
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13777-23-6 | |
Properties | |
HfI4 | |
Molar mass | 686.11[1] |
Appearance | red-orange[1] |
Density | 5.60 g/cm3[1] |
Melting point | 449 °C (840 °F; 722 K)[1] |
Boiling point | 394 °C (741 °F; 667 K)[1] (sublimes) |
Structure | |
Monoclinic, mS40 | |
C2/c, No. 15[2] | |
a = 1.1787 nm, b = 1.1801 nm, c = 1.2905 nm
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Vapor pressure | {{{value}}} |
Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa).
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Infobox references | |
Hafnium tetraiodide is the inorganic compound with the formula HfI4. It is a red-orange, moisture sensitive, sublimable solid that is produced by heating a mixture of hafnium with excess iodine.[3] It is an intermediate in the crystal bar process for producing hafnium metal.
In this compound, the hafnium centers adopt octahedral coordination geometry. Like most binary metal halides, the compound is a polymeric. It is one-dimensional polymer consisting of chains of edge-shared bioctahedral Hf2I8 subunits, similar to the motif adopted by HfCl4. The nonbridging iodide ligands have shorter bonds to Hf than the bridging iodide ligands.[3]
References
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- Iodides
- Hafnium compounds
- Metal halides