Question

An element (A) belongs to period number 4 (four) and group number 11 (eleven) and is extracted from its pyrite ore. Element (A) reacts with oxygen at two different temperatures forming compounds (B) and (C). Element (A) also reacts with conc. $HN{O}_3$ to give compound (D) with the evolution of $N{O}_2$. Identify (A), (B), (C) and (D). Explain the reactions.

Answer 1

From the periodic table, copper $\left(Cu\right)$ belongs to period $4$ and group $1$.

$2Cu+O_2\stackrel{<1370K}{\to }\underset{(B-\text{Cupric Oxide})}{2CuO}$

$2Cu+O_2\stackrel{>1370K}{\to }\underset{(C-\text{Cuprous Oxide})}{2{Cu}_{2}O}$

$Cu\stackrel{\text{Conc.}HN{O}_3}{\to }\underset{(D-\text{Cupric Nitrate})}{Cu{\left(N{O}_3\right)}_2}+N{O}_2+H_{2}O$

Hence the compound $A,B,C$ and $D$ are copper, cupric oxide, cuprous oxide and cupric nitrate respectively.