Consider the following reactions.Na_{2}B_{4}O | vedclass

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(B) The reaction of borax on heating is: $Na_2 B_4 O_7 \xrightarrow{\Delta} 2 NaBO_2 (X) + B_2 O_3 (Y)$.In the non-luminous flame, $CuSO_4$ reacts wit

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$+2$ and $+1$

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(B) The reaction of borax on heating is: $Na_2 B_4 O_7 \xrightarrow{\Delta} 2 NaBO_2 (X) + B_2 O_3 (Y)$.In the non-luminous flame, $CuSO_4$ reacts with $B_2 O_3$ to form copper$(II)$ metaborate: $CuSO_4 + B_2 O_3 \xrightarrow{\Delta} Cu(BO_2)_2 (Z) + SO_3$. Here, the oxidation state of $Cu$ in $Z$ is $+2$.In the luminous flame, $Cu(BO_2)_2$ is reduced by carbon in the presence of $NaBO_2$ to form copper$(I)$ metaborate: $2 Cu(BO_2)_2 + 2 NaBO_2 + C \xrightarrow{\Delta} 2 CuBO_2 (Q) + Na_2 B_4 O_7 + CO$. Here, the oxidation state of $Cu$ in $Q$ is $+1$.Thus, the oxidation states of $Cu$ in $Z$ and $Q$ are $+2$ and $+1$ respectively.

Consider the following reactions.
$Na_{2}B_{4}O_{7} \xrightarrow{\Delta} 2 X + Y$
$CuSO_{4} + Y \xrightarrow{\text{Non-Luminous flame}} Z + SO_{3}$
$2Z + 2X + \text{Carbon} \xrightarrow{\text{Luminous flame}} 2Q + Na_{2}B_{4}O_{7} + CO$
The oxidation states of $Cu$ in $Z$ and $Q$, respectively are:

Similar Questions

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$b$) The correct formula of Borax is $Na_2[B_4O_5(OH)_4] \cdot 8H_2O$.
$c$) Trihalides of boron are Lewis bases.

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Consider the following reactions.$Na_{2}B_{4}O_{7} \xrightarrow{\Delta} 2 X + Y$$CuSO_{4} + Y \xrightarrow{\text{Non-Luminous flame}} Z + SO_{3}$$2Z + 2X + \text{Carbon} \xrightarrow{\text{Luminous flame}} 2Q + Na_{2}B_{4}O_{7} + CO$The oxidation states of $Cu$ in $Z$ and $Q$, respectively are: