Zinc on reaction with concentrated nitric aci | vedclass

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(A) The reaction of zinc with concentrated nitric acid is given by: $Zn + 4HNO_3 (	ext{conc.}) ightarrow Zn(NO_3)_2 + 2NO_2 + 2H_2O$. Here, the oxi

Vedclass · Accepted Answer

$+4, +1$

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(A) The reaction of zinc with concentrated nitric acid is given by: $Zn + 4HNO_3 (	ext{conc.}) ightarrow Zn(NO_3)_2 + 2NO_2 + 2H_2O$. Here, the oxide of nitrogen $(A)$ is $NO_2$. The oxidation number of $N$ in $NO_2$ is $x + 2(-2) = 0$, so $x = +4$.The reaction of zinc with dilute nitric acid is given by: $4Zn + 10HNO_3 (	ext{dil.}) ightarrow 4Zn(NO_3)_2 + N_2O + 5H_2O$. Here, the oxide of nitrogen $(B)$ is $N_2O$. The oxidation number of $N$ in $N_2O$ is $2x + (-2) = 0$, so $2x = +2$, $x = +1$.Therefore, the oxidation numbers of nitrogen in $(A)$ and $(B)$ are $+4$ and $+1$ respectively.

Zinc on reaction with concentrated nitric acid gives an oxide of nitrogen $(A)$. Zinc with dilute nitric acid gives another oxide of nitrogen $(B)$. Oxidation numbers of nitrogen in $(A)$ and $(B)$ are respectively:

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