1. Cu(NO₃)₂ and NO₂ 2. Cu(NO₃)₂ and NO 3. Cu(NO₃)₂ , NO and NO₂ 4. Cu(NO₃)₂ and N₂O

Answer: 1. Cu(NO₃)₂ and NO₂ Key idea: the reduction product of nitric acid depends on its concentration. Cu+4HNO₃(conc.)→ Cu(NO₃)₂+2NO₂+2H₂O Concentrated acid gives NO₂ . Dilute acid (about 3:8 with Cu) gives NO, and very dilute acid can give N₂O .

Topic: General trends, Group 15 and 16· p-Block Elements

When copper is heated with conc. HNO_3 it produces :

1. $Cu(NO_3)_2$ and $NO_2$

2. $Cu(NO_3)_2$ and $NO$

3. $Cu(NO_3)_2$, $NO$ and $NO_2$

4. $Cu(NO_3)_2$ and $N_2O$

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Answer: 1. $Cu(NO_3)_2$ and $NO_2$

Key idea: the reduction product of nitric acid depends on its concentration.

$$\mathrm{Cu+4HNO_3(conc.)\to Cu(NO_3)_2+2NO_2+2H_2O}$$

Concentrated acid gives $\mathrm{NO_2}$. Dilute acid (about 3:8 with Cu) gives NO, and very dilute acid can give $\mathrm{N_2O}$.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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When copper is heated with conc. HNO_3 it produces :

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