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Consider the following reaction :
$2 {HSO}_{4}^{-} (aq) \frac{(1) E l e c t r o l y s i s}{(2) Hydrolysis} 2 {HSO}_{4}^{-} + 2 H^{+} + A$
The dihedral angle in product $A$ in its solid phase at $110 K$ is :

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$111.5 °$

$111.0 °$

$90.2 °$

$104 °$

answer is C.

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Detailed Solution

$2 {HSO}_{4}^{-} (a q .) \to_{(2) H y d r o l y s i s}^{(1) E l e c t r o l y s i s} 2 H {SO}_{4}^{-} + 2 H^{+} + \underset{(A)}{H_{2} O_{2}}$

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