Consider the following reaction.

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

The dihedral angle in product A in its solid phase at 110 K is:

Option 1 -

104°

Option 2 -

111.5°

Option 3 -

90.2°

Option 4 -

111.0°

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 3

Detailed Solution:

During the electrolysis of dilute H₂SO₄

$2 H_{2} S O_{4} \overset{(1) e l e c t r o l y s i s}{\to} 2 H S O_{4}^{-} + 2 H^{+}$

$H_{2} S_{2} O_{8} + 2 H_{2} O \overset{h y d r o l y s i s}{\to} 2 H_{2} S O_{4} + \underset{(A)}{H_{2} O_{2}}$

$2 H^{+} + 2 e \to H_{2} ↑$

In the solid form of $H_{2} O_{2}$ dihedral angle is equal to 90.2°.

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Consider the following reaction.

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

The dihedral angle in product A in its solid phase at 110 K is:

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Consider the following reaction. 2 H S O 4 − ( a q ) → ( 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 S O 4 − + 2 H + + A The dihedral angle in product A in its solid phase at 110 K is:

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Consider the following reaction.

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

The dihedral angle in product A in its solid phase at 110 K is: