Consider the following reaction :
$$2HSO_4^-(aq) \xrightarrow[(2) Hydrolysis]{(1) Electrolysis \atop } 2HSO_4^- + 2H^+ + A$$
The dihedral angle in product $$A$$ in its solid phase at $$110$$ K is

We need to identify product A from the electrolysis and hydrolysis of bisulfate ions ($$HSO_4^-$$).

The electrolysis of bisulfate ions ($$HSO_4^-$$) produces peroxodisulfuric acid ($$H_2S_2O_8$$), also known as Marshall's acid, according to the reaction: $$2HSO_4^- \xrightarrow{Electrolysis} S_2O_8^{2-} + 2H^+ + 2e^-$$.

Subsequent hydrolysis of the peroxodisulfate ion yields hydrogen peroxide ($$H_2O_2$$): $$S_2O_8^{2-} + 2H_2O \to 2HSO_4^- + H_2O_2$$.

Combining these steps gives the overall reaction: $$2HSO_4^- \xrightarrow{(1) Electrolysis \atop (2) Hydrolysis} 2HSO_4^- + 2H^+ + H_2O_2$$, which shows that product A is $$H_2O_2$$ (hydrogen peroxide).

Hydrogen peroxide ($$H_2O_2$$) adopts a non-planar, open book-like geometry, and its dihedral angle between the two H-O-O planes varies with phase: it is approximately $$111.5°$$ in the gas phase and about $$90.2°$$ in the solid phase at 110 K.

Since the question specifies the solid phase at 110 K, the dihedral angle is $$90.2°$$. Therefore, the correct answer is Option C: $$90.2°$$.