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Valency and Oxidation State

Consider the ...

Question

Consider the following unbalanced reactions:
$I : Z n + d i l . H_{2} S O_{4} ⟶ Z n S O_{4} + H_{2}$
$I I : Z n + c o n c . H_{2} S O_{4} ⟶ Z n S O_{4} + S O_{2} + H_{2} O$
$I I I : Z n + H N O_{3} ⟶ Z n {({N O}_{3})}_{2} + {N H}_{4} {N O}_{3} + H_{2} O$
Oxidation number of hydrogen changes in :

I, I I, I I I

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I, I I

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I I, I I I

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I

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Solution

The correct option is C $I$
Change in oxidation state of H is shown below:
$I : Z n + 2 H^{+} + 1 ⟶ H_{2} 0$ oxidation state changes (reduction)
$I I : Z n + H^{+} + 1 + S O_{4}^{2 -} ⟶ {Z n}^{2 +} + S O_{2} + H_{2} + 1 O$
no change in oxidation state of H
$I I I : Z n + H^{+} + 1 + N O_{3}^{-} ⟶ {Z n}^{2 +} + {N H}_{4} ↑ + 1 H_{2} ↑ + 1 O$ no change
Therefore, from the above reactions, we can see oxidation number of $H$ changes in $I$ .

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