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MCQEasyJEE 2026Oxidation Number & Redox Reactions

JEE Chemistry 2026 Question with Solution

Consider the following reactions:

Reaction scheme showing Na2B4O7 on heating giving 2X and Y; CuSO4 plus Y in non-luminous flame giving Z and SO3; and 2Z plus 2X plus carbon in luminous flame giving 2Q, Na2B4O7, and CO.

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

  • A

    +2+2 and +1+1

  • B

    +1+1 and +2+2

  • C

    +2+2 and +2+2

  • D

    +1+1 and +1+1

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The reaction sequence is

Na2B4O7→Δ2X+Y\mathrm{Na_2B_4O_7 \xrightarrow{\Delta} 2X + Y} CuSO4+Y→Non-Luminous flameZ+SO3\mathrm{CuSO_4 + Y \xrightarrow{\text{Non\text{-}Luminous\ flame}} Z + SO_3} 2Z+2X+C→Luminous flame2Q+Na2B4O7+CO\mathrm{2Z + 2X + C \xrightarrow{\text{Luminous\ flame}} 2Q + Na_2B_4O_7 + CO}

Find: The oxidation states of Cu in Z and Q.

From the decomposition of borax:

Na2B4O7→heatNaBO2 (X)+B2O3 (Y)\mathrm{Na_2B_4O_7 \xrightarrow{heat} NaBO_2\,(X) + B_2O_3\,(Y)}

Now use the second reaction:

CuSO4+B2O3→CuO (Z)+SO3\mathrm{CuSO_4 + B_2O_3 \rightarrow CuO\,(Z) + SO_3}

In CuOCuO, oxygen is −2-2, so copper must be +2+2.

In the luminous flame step:

2CuO+2NaBO2+C→2Q+Na2B4O7+CO\mathrm{2CuO + 2NaBO_2 + C \rightarrow 2Q + Na_2B_4O_7 + CO}

The solution identifies the final copper product as Cu2OCu_2O in the luminous flame. In Cu2OCu_2O, oxygen is −2-2, so the two copper atoms together contribute +2+2. Hence each copper atom has oxidation state +1+1.

Therefore, the oxidation states of Cu in Z and Q, respectively, are +2+2 and +1+1. The correct option is A.

Oxidation State Tracking

Given: Track the copper species formed in the second and third reactions. Find: Oxidation states of copper in Z and Q.

  1. Identify YY from borax decomposition:
Na2B4O7→NaBO2+B2O3\mathrm{Na_2B_4O_7 \rightarrow NaBO_2 + B_2O_3}

So, Y=B2O3Y = B_2O_3.

  1. Identify ZZ from reaction with copper sulfate:
CuSO4+B2O3→CuO+SO3\mathrm{CuSO_4 + B_2O_3 \rightarrow CuO + SO_3}

Thus Z=CuOZ = CuO.

  1. Find oxidation state in ZZ: For CuOCuO,
Oxidation state of Cu+(−2)=0\text{Oxidation state of Cu} + (-2) = 0 Oxidation state of Cu=+2\text{Oxidation state of Cu} = +2
  1. Identify QQ from the luminous flame step. The extracted solution states that copper is reduced and finally forms Cu2OCu_2O in the luminous flame. Therefore Q=Cu2OQ = Cu_2O.
  1. Find oxidation state in QQ: For Cu2OCu_2O,
2x+(−2)=02x + (-2) = 0 x=+1x = +1

So, copper has oxidation states +2+2 in Z and +1+1 in Q. Hence the correct option is A.

Common mistakes

  • Mistake: Assuming QQ is metallic CuCu because reduction occurs in the last step. Why it is wrong: the extracted solution explicitly states that the luminous flame product is Cu2OCu_2O, not elemental copper. What to do instead: identify the product named in the solution first, then calculate its oxidation state.

  • Mistake: Taking the oxidation state of copper in Cu2OCu_2O as +2+2 by copying the value from CuOCuO. Why it is wrong: oxidation state depends on the compound, not only the element. What to do instead: write the charge-balance equation separately for each compound.

  • Mistake: Not identifying ZZ correctly from the second reaction. Why it is wrong: if ZZ is misread, the oxidation state question becomes incorrect from the start. What to do instead: first map XX, YY, ZZ, and QQ from the reaction sequence, then evaluate oxidation states.

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