Question
Consider the following reactions Na _ 2 ~B _ 4 O _ 7 2 X + Y CuSO _ 4 + Y Non-Luminous flame Z + SO _ 3 2 Z +2 X + Carbon Luminous flame 2 Q + Na _ 2 ~B _ 4 O _ 7 + CO The oxidation states of Cu in Z and Q, respectively are :
JEE Main
Chemistry
p Block Elements (Group 13 & 14)
2026
JEE Main 2026 (28 January Shift 2)
JEE Main Chemistry Question (2026) — Solution
Options
- A. +1 and +1
- B. +2 and +2
- C. +1 and +2
- D. +2 and +1
Answer
D. +2 and +1
Step-by-step solution
The given reactions describe the Borax Bead Test. When borax (Na_2B_4O_7 10H_2O) is heated, it first loses water and then melts to form a transparent glassy bead consisting of sodium metaborate (NaBO_2) and boric anhydride (B_2O_3). Na_2B_4O_7 2NaBO_2 (X) + B_2O_3 (Y) In the non-luminous (oxidizing) flame, B_2O_3 reacts with CuSO_4 to form copper(II) metaborate (Z), which is blue in color. CuSO_4 + B_2O_3 Non-Luminous Cu(BO_2)_2 (Z) + SO_3 In Cu(BO_2)_2, the oxidation state of Cu is +2. In the luminous (reducing) flame, the copper(II) metaborate is reduced by carbon to copper(I) metaborate (Q), which is colorless/red. 2Cu(BO_2)_2 (Z) + 2NaBO_2 (X) + C Luminous 2CuBO_2 (Q) + Na_2B_4O_7 + CO In CuBO_2, the oxidation state of Cu is +1. Therefore, the oxidation states of Cu in Z and Q are +2 and +1 respectively.
Related: Chemistry — p Block Elements (Group 13 & 14) · All PYQ Banks