Consider the following reactions. $\mathrm{Na}_{2}\mathrm{B}_{4}\mathrm{O}_{7}\xrightarrow{\Delta}2\mathrm{X}+\mathrm{Y}$ $\mathrm{CuSO}_{4}+\mathrm{Y}\xrightarrow{\text{Non-Luminous flame}}\mathrm{Z}+\mathrm{SO}_{3}$ $2\mathrm{Z}+2\mathrm{X}+$ Carbon $\xrightarrow{\text{Luminous flame}}2\mathrm{Q}+\mathrm{Na}_{2}\mathrm{B}_{4}\mathrm{O}_{7}+\mathrm{CO}$ The oxidation states of Cu in Z and Q, respectively are:

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · p-Block Elements

Consider the following reactions.

Na_{2} B_{4} O_{7} Δ 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:

(A)+2 and +2
(B)+2 and +1✓ Correct
(C)+1 and +2
(D)+1 and +1

Answer: (B)

Solution

Heating borax gives

N a_{2} B_{4} O_{7} Δ 2NaB O_{2} + B_{2} O_{3}

, so X is

NaB O_{2}

and Y is

B_{2} O_{3}

. In the oxidising (non-luminous) flame

CuS O_{4} + B_{2} O_{3} \to Cu (B O_{2})_{2} + S O_{3}

gives the blue bead Z,

Cu (B O_{2})_{2}

, in which copper is

+ 2

. In the reducing (luminous) flame the carbon reduces this copper metaborate, lowering copper to the

+ 1

state in Q. The oxidation states of copper in Z and Q are

+ 2

and

+ 1

respectively.

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option

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