The last drop of water can be removed from $H_2O_2$ by

Assertion: The $O-O$ bond length in $H_2O_2$ is shorter than that of $O_2F_2$.
Reason: $H_2O_2$ is an ionic compound.

$1 \ mL$ of "$x$ volume" $H_2O_2$ solution on heating gives $20 \ mL$ of oxygen gas at $STP$. The $(w/v) \%$ corresponding to "$x$ volume" of $H_2O_2$ is

Consider the following reaction:
$2 HSO_{4}^{-}(aq)$ $\xrightarrow[\text{(2) Hydrolysis}]{\text{(1) Electrolysis}} H_{2}S_{2}O_{8}(aq)$ $\xrightarrow{\text{Hydrolysis}} 2 HSO_{4}^{-}(aq) + 2 H^{+}(aq) + A(aq)$
The dihedral angle in product $A$ in its solid phase at $110 \ K$ is ....$^{\circ}$

The correct statements about $H_2O_2$ are:
$A$. Used in the treatment of effluents.
$B$. Used as both oxidising and reducing agents.
$C$. The two hydroxyl groups lie in the same plane.
$D$. Miscible with water.
Choose the correct answer from the options given below:

Which of the following yields $H_2O_2$ upon oxidation?