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KCET · Chemistry · Redox Reactions

The oxidation number of nitrogen atoms in \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) are

  1. A \(+5\), \(+5\)
  2. B \(-3,+5\)
  3. C \(+3,-5\)
  4. D \(-3,-3\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(-3,+5\)

Step-by-step Solution

Detailed explanation

\(\mathrm{NH}_{4} \mathrm{NO}_{3} \longrightarrow \mathrm{NH}_{4}^{+}+\mathrm{NO}_{3}^{-}\)
\(\left(N_{1}\right) \quad\left(N_{2}\right)\)
For oxidation number of \(N_{1} \rightarrow x+4=+1\)
\(x=+1-4=-3\)
So, oxidation state of first nitrogen is \(-3\).
For oxidation state of second nitrogen \(\left(N_{2}\right) \rightarrow\)
\(\begin{aligned}
x-6 &=-1 \\
x &=-1+6=+5
\end{aligned}\)
So, the answer is \(-3\) and \(+5\).