JEE Main 2026 — p Block Elements (Group 13 & 14) Question with Solution

The given reactions describe the Borax Bead Test.
When borax ($N{a}_2{B}_4{O}_7\cdot 10H_{2}O$) is heated, it first loses water and then melts to form a transparent glassy bead consisting of sodium metaborate ($NaB{O}_2$) and boric anhydride ($B_2{O}_3$).
$N{a}_2{B}_4{O}_7\stackrel{\Delta }{\to }2NaB{O}_2(X)+B_2{O}_3(Y)$
In the non-luminous (oxidizing) flame, $B_2{O}_3$ reacts with $CuS{O}_4$ to form copper(II) metaborate ($Z$), which is blue in color.
$CuS{O}_4+B_2{O}_3\stackrel{Non-Luminous}{\to }Cu(B{O}_2{)}_2(Z)+S{O}_3$
In $Cu(B{O}_2{)}_2$, the oxidation state of $Cu$ is $+2$.
In the luminous (reducing) flame, the copper(II) metaborate is reduced by carbon to copper(I) metaborate ($Q$), which is colorless/red.
$2Cu(B{O}_2{)}_2(Z)+2NaB{O}_2(X)+C\stackrel{Luminous}{\to }2CuB{O}_2(Q)+N{a}_2{B}_4{O}_7+CO$
In $CuB{O}_2$, the oxidation state of $Cu$ is $+1$.
Therefore, the oxidation states of $Cu$ in $Z$ and $Q$ are $+2$ and $+1$ respectively.