JEE Main 2026 — Jan 28, Shift 2 — Chemistry solution

JEE Main2026Jan 28, Shift 2Chemistry

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 $+ 1$
C
$+ 1$ and $+ 2$
D
$+ 1$ and $+ 1$

Solution

This is the classical borax bead test.

Step 1: $Na_{2} B_{4} O_{7} Δ 2 NaBO_{2} + B_{2} O_{3}$ . So $X = NaBO_{2}$ , $Y = B_{2} O_{3}$ .

Step 2 (oxidising/non-luminous flame): $CuSO_{4} + B_{2} O_{3} \to Cu(BO_{2})_{2} + SO_{3}$ . Here Cu is in the $+ 2$ state, giving the characteristic blue bead. $Z = Cu(BO_{2})_{2}$ , Cu = $+ 2$ .

Step 3 (reducing/luminous flame): carbon reduces Cu(II) to Cu(I): $2 Cu(BO_{2})_{2} + 2 NaBO_{2} + C \to 2 CuBO_{2} + Na_{2} B_{4} O_{7} + CO$ . So $Q = CuBO_{2}$ , Cu = $+ 1$ (colourless/red bead).

$∴ + 2$ and $+ 1$ .

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