Question

Consider the following reactions: 

The oxidation states of Cu in Z and Q, respectively are:

• A. +2 and +1
• B. +1 and +2
• C. +2 and +2
• D. +1 and +1

Hint:

Always track oxidation states step by step: - Decomposition products first. - Reaction intermediates next. - Final reduction/oxidation products last.

Answer 1

The Correct Option is A

To find the oxidation states of Cu in compounds Z and Q, let's analyze the given reactions step-by-step:

The reactions provided are:

1. Decomposition of \(\text{Na}_2\text{B}_4\text{O}_7\):

\(\text{Na}_2\text{B}_4\text{O}_7 \xrightarrow{\Delta} 2\text{X} + \text{Y}\)

1. Reaction of CuSO₄ with Y:

\(\text{CuSO}_4 + \text{Y} \xrightarrow{\text{Non-Luminous Flame}} \text{Z} + \text{SO}_3\)

Here, CuSO₄ is reduced to Z. In this reaction, Cu²⁺ in CuSO₄ is reduced to Cu in compound Z, indicating that Cu in Z will have an oxidation state of +2.

1. Further reduction with Carbon:

\(2\text{Z} + 2\text{X} + \text{Carbon} \xrightarrow{\text{Luminous Flame}} 2\text{Q} + \text{Na}_2\text{B}_4\text{O}_7 + \text{CO}\)

Since Z is CuO (with Cu at +2 oxidation state), and it's reduced further with carbon, Cu²⁺ in Z undergoes reduction to form Q where Cu is reduced to Cu¹⁺. Hence, in Q, Cu will have an oxidation state of +1.

Given all the analysis, the oxidation states of Cu in compounds Z and Q are:

• Z: +2
• Q: +1

Therefore, the correct answer is:

+2 and +1