MCQ
The structure of the noble gas compound $XeF_4$ is
- ✓square planar
- Bdistoned tetrahedral
- Ctetrahedral
- Doctahedral
so, it has square planar structure
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$(A)\,{{C}_{8}}{{H}_{10}}\xrightarrow{KMn{{O}_{4}}}(B){{C}_{8}}{{H}_{6}}{{O}_{4}}\xrightarrow[Fe]{B{{r}_{2}}}{{C}_{8}}{{H}_{5}}Br{{O}_{4}}(C)$ (one-product only)
$S{O_2} + \frac{1}{2}{O_2} \to S{O_3}$
