Consider the following reactions
Na2B4O7→Δ2X+Y
CuSO4+Y→Non-luminous flameZ+SO3
2Z+2X+Carbon→Luminous flame2Q+Na2B4O7+CO
The oxidation states of Cu in Z and Q, respectively are: [2026]
+2 and +2
+1 and +1
+2 and +1
+1 and +2
(3)
Na2B4O7→Δ(2NaBO2)(X)+B2O3
CuSO4+B2O3→Non-luminous flameΔCu+2(BO2)2(Blue bead)(Z)+SO3
2Cu(BO2)2+2NaBO2+C→Luminous flameΔ2Cu+1BO2Colorless(Q)+Na2B4O7+CO