Question:
Consider the following reactions:
The oxidation states of Cu in Z and Q, respectively are:
Consider the following reactions:
The oxidation states of Cu in Z and Q, respectively are:
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Always track oxidation states step by step:
- Decomposition products first.
- Reaction intermediates next.
- Final reduction/oxidation products last.
Updated On: Feb 24, 2026
- +2 and +1
- +1 and +2
- +2 and +2
- +1 and +1
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The Correct Option is A
Solution and Explanation
Step 1: Identify decomposition of borax.
Na2B4O7 → (heat) → NaBO2 (X) + B2O3 (Y)
Step 2: Reaction with CuSO4.
CuSO4 + B2O3 → CuO (Z) + SO3
Here, copper in CuO has oxidation state +2.
Step 3: Reaction in luminous flame.
2CuO + 2NaBO2 + C → 2Cu (Q) + Na2B4O7 + CO
In this step, copper is reduced and finally forms Cu2O in the luminous flame, where copper has oxidation state +1.
Step 4: Conclusion.
Thus, oxidation states of copper are:
• In Z (CuO): +2
• In Q (Cu2O): +1
Answer:
+2 and +1
Na2B4O7 → (heat) → NaBO2 (X) + B2O3 (Y)
Step 2: Reaction with CuSO4.
CuSO4 + B2O3 → CuO (Z) + SO3
Here, copper in CuO has oxidation state +2.
Step 3: Reaction in luminous flame.
2CuO + 2NaBO2 + C → 2Cu (Q) + Na2B4O7 + CO
In this step, copper is reduced and finally forms Cu2O in the luminous flame, where copper has oxidation state +1.
Step 4: Conclusion.
Thus, oxidation states of copper are:
• In Z (CuO): +2
• In Q (Cu2O): +1
Answer:
+2 and +1
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