Question

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) & (B) are respectively

• A. +4, +1
• B. +4, +2
• C. +2, +4
• D. +1, +4

Hint:

Comparison: Zn + Dil \(\text{HNO}_3\) \(\rightarrow \text{N}_2\text{O}\) Cu + Dil \(\text{HNO}_3\) \(\rightarrow \text{NO}\) Both Zn/Cu + Conc \(\text{HNO}_3\) \(\rightarrow \text{NO}_2\)

Answer 1

The Correct Option is A

Step 1: Reaction with Concentrated \(\text{HNO}_3\):
Zinc reacts with concentrated nitric acid to evolve Nitrogen Dioxide gas. \[ \text{Zn} + 4\text{HNO}_3(\text{conc}) \rightarrow \text{Zn(NO}_3)_2 + 2\text{NO}_2 + 2\text{H}_2\text{O} \] Oxide (A) is \(\text{NO}_2\). Oxidation number of N in \(\text{NO}_2\): \(x + 2(-2) = 0 \implies x = +4\).
Step 2: Reaction with Dilute \(\text{HNO}_3\):
Zinc is a stronger reducing agent (compared to copper). With dilute nitric acid, it reduces nitrogen to a lower oxidation state, forming Nitrous Oxide (Laughing Gas). \[ 4\text{Zn} + 10\text{HNO}_3(\text{dil}) \rightarrow 4\text{Zn(NO}_3)_2 + \text{N}_2\text{O} + 5\text{H}_2\text{O} \] Oxide (B) is \(\text{N}_2\text{O}\). Oxidation number of N in \(\text{N}_2\text{O}\): \(2x + (-2) = 0 \implies 2x = 2 \implies x = +1\).
Step 3: Answer Match:
A: +4, B: +1. Final Answer:
Option (A).