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Question
How are xenon fluorides XeF2, XeF4 and XeF6 obtained?
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Solution
Xenon fluorides are obtained by the direct reaction of xenon (Xe) with fluorine gas (F2). The product formed depends on the reaction conditions, such as temperature, pressure, and the ratio of Xe to F2:
1. Formation of XeF2:
\[\ce{Xe + F2 -> XeF2}\] (673 K, 1 bar, excess Xe)
2. Formation of XeF4:
\[\ce{Xe + 2F2 -> XeF4}\] (873 K, 7 bar, 1 : 5 ratio)
3. Formation of XeF6:
\[\ce{Xe + 3F2 -> XeF6}\] (573 K, 60–70 bar, 1:20 ratio)
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| Column I | Column II |
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| (B) XeO3 | (2) sp3d2 – square planar |
| (C) XeOF4 | (3) sp3 – pyramidal |
| (D) XeF4 | (4) sp3d2 – square pyramidal |
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\[\ce{XeF6 + H2O ->[Partial][Hydrolysis] \underline{}\underline{}\underline{}\underline{} + \underline{}\underline{}\underline{}\underline{}}\]
\[\ce{XeF4 + H2O - \underline{}\underline{}\underline{}\underline{} + \underline{}\underline{}\underline{}\underline{}}\]
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