JEE Main 2026 — Chemistry Inorganic Chemistry

The given reactions describe the Borax Bead Test.

When borax ($Na_2B_4O_7 \cdot 10H_2O$Na2​B4​O7​⋅10H2​O) is heated, it first loses water and then melts to form a transparent glassy bead consisting of sodium metaborate ($NaBO_2$NaBO2​) and boric anhydride ($B_2O_3$B2​O3​).

$$Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 (X) + B_2O_3 (Y)$$Na2​B4​O7​Δ​2NaBO2​(X)+B2​O3​(Y)

In the non-luminous (oxidizing) flame, $B_2O_3$B2​O3​ reacts with $CuSO_4$CuSO4​ to form copper(II) metaborate ($Z$Z), which is blue in color.

$$CuSO_4 + B_2O_3 \xrightarrow{Non-Luminous} Cu(BO_2)_2 (Z) + SO_3$$CuSO4​+B2​O3​Non−Luminous​Cu(BO2​)2​(Z)+SO3​

In $Cu(BO_2)_2$Cu(BO2​)2​, the oxidation state of $Cu$Cu is $+2$+2.

In the luminous (reducing) flame, the copper(II) metaborate is reduced by carbon to copper(I) metaborate ($Q$Q), which is colorless/red.

$$2Cu(BO_2)_2 (Z) + 2NaBO_2 (X) + C \xrightarrow{Luminous} 2CuBO_2 (Q) + Na_2B_4O_7 + CO$$2Cu(BO2​)2​(Z)+2NaBO2​(X)+CLuminous​2CuBO2​(Q)+Na2​B4​O7​+CO

In $CuBO_2$CuBO2​, the oxidation state of $Cu$Cu is $+1$+1.

Therefore, the oxidation states of $Cu$Cu in $Z$Z and $Q$Q are $+2$+2 and $+1$+1 respectively.