MCQ
The correct structure of tribromooctaoxide is
- A

- B

- ✓

- D






$3 XeF _4+6 H _2 O \rightarrow 2 Xe + XeO _3+\frac{3}{2} O _2+12 HF$
$\therefore$ One mole of $XeF _4$ gives 4 moles of $HF$ on hydrolysis.
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