The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions. The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers +2, +4 and +6. XeF$$_4$$ reacts violently with water to give XeO$$_3$$. The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.
XeF$$_4$$ and XeF$$_6$$ are expected to be
$$ \mathrm{B}(\mathrm{OH})_3+\mathrm{NaOH} \quad \mathrm{NaBO}_2+\mathrm{Na}\left[\mathrm{~B}(\mathrm{OH})_4\right] $$$+\mathrm{H}_2 \mathrm{O}$
How can this reaction be made to proceed in forward direction?
$$ \text { Match the Column I with Column II : } $$
| Column I | Column II | ||
|---|---|---|---|
| (A) | $$ \mathrm{Bi}^{3+} \rightarrow(\mathrm{BiO})^{+} $$ |
(P) | Heat |
| (B) | $$ \left[\mathrm{AlO}_2\right]^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_3 $$ |
(Q) | Hydrolysis |
| (C) | $$ \mathrm{SiO}_4^{4-} \rightarrow \mathrm{Si}_2 \mathrm{O}_7^{6-} $$ |
(R) | Acidification |
| (D) | $$ \left(\mathrm{B}_4 \mathrm{O}_7^{2-}\right) \rightarrow\left[\mathrm{B}(\mathrm{OH})_3\right] $$ |
(S) | Dilution by water |
JEE Advanced Subjects
Browse all chapters by subject