p Block Elements Part 3

Concept:Thermal decomposition of borax and reactions in oxidizing and reducing flames determine the oxidation states of copper in the products.Explanation:Step 1: $\mathrm{Na}_2\mathrm{B}_4\mathrm{O}_7 \xrightarrow{\Delta} 2\mathrm{NaBO}_2 + \mathrm{B}_2\mathrm{O}_3$. Thus $\mathrm{X} = \mathrm{NaBO}_2$ and $\mathrm{Y} = \mathrm{B}_2\mathrm{O}_3$.Step 2: $\mathrm{CuSO}_4 + \mathrm{B}_2\mathrm{O}_3$ (non‑luminous flame, oxidizing) gives $\mathrm{Cu}(\mathrm{BO}_2)_2 + \mathrm{SO}_3$. So $\mathrm{Z} = \mathrm{Cu}(\mathrm{BO}_2)_2$ and copper is in $+2$ oxidation state.Step 3: $2\mathrm{Cu}(\mathrm{BO}_2)_2 + 2\mathrm{NaBO}_2 + \mathrm{C}$ (luminous flame, reducing) yields $2\mathrm{CuBO}_2 + \mathrm{Na}_2\mathrm{B}_4\mathrm{O}_7 + \mathrm{CO}$. Thus $\mathrm{Q} = \mathrm{CuBO}_2$ and copper is reduced to $+1$ state.Answer:Oxidation states of Cu in Z and Q are $+2$ and $+1$ respectively. Option A is correct.