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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(image) $\frac{{conc.\,HN{O_3}}}{{conc.\,{H_2}S{O_4}}}X$
the structure of the major product $'X'$ is


$(1)$ In gas phase $SO_2$ molecule is $V-$ shape.
$(2)$ In gas Phase $SO_3$ molecule is planar
$(3)$ $\gamma - SO_3$ is cyclic trimer
Which oi the above statement are correct ?