MCQ
In $Xe{F_2}$ hybridisation of $Xe$ is
- A$s{p^2}$
- ✓$s{p^3}d$
- C$s{p^3}$
- D$s{p^3}{d^2}$
Hybridisation of a atom with $3$ lone pairs and $2$ bond pairs is $sp ^3 d ^2$.
So, the hybridisation of $Xe$ is $sp ^3 d$.
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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)
Product $(A)$ is