MCQ
Conc. $H_2SO_4$ cannot be used to prepare $HBr$ from $NaBr$ because it :
- ✓reacts slowly with $NaBr$
- Boxidises $HBr$ into $Br_2$
- Creduces $HBr$
- Ddisproportionates $HBr$
Acidic nature $ \propto EN \propto + O.S.$
For compounds of an element
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Reason : Oxidation state of $Cl$ in $HClO_4$ is $+VII$ and in $HClO_3$ $+V$.
$\mathrm{C}\mathrm{H}_{3} -\mathrm{C} \equiv \mathrm{CH} \xrightarrow[873\;K]{\text { red hot iron tube}} \mathrm{A}$
the number of sigma$(\sigma)$ bonds present in the product $A$ is
