Verified Solution • 2025–26 Session moderate

QJEEM23Cha353: JEE Main 2023 Chemistry - Class 12 JEEM23Cha353 Solution: Given below are two statements : Statement I: Upon heating a…

Given below are two statements :

Statement I: Upon heating a borax bead dipped in cupric sulphate in a luminous flame, the colour of the bead becomes green

Statement II: The green colour observed is due to the formation of copper(I) metaborate

In light of the above statements, choose the most appropriate answer from the options given below :

Multiple Choice Options

  • A Both Statement I and Statement II are false
  • B Both Statement I and Statement II are true
  • C Statement I is false but Statement II is true
  • D Statement I is true but Statement II is false

Step-by-Step Verified Solution

On treatment with metal salt, boric anhydride forms metaborate of the metal which gives different colours in oxidising and reducing flame. For example, in the case of copper sulphate, following reactions occur.

CuSO4+B2O3⟶(Oxidation)Non−lumin⁡ous flameCu(BO2)2+SO3\mathrm{CuSO}_4+\mathrm{B}_2 \mathrm{O}_3 \mathrel{\mathop{\kern0pt\longrightarrow}\limits_{\left( {Oxidation} \right)}^{Non - lu\min ous\,flame}} \mathrm{Cu}\left(\mathrm{BO}_2\right)_2+\mathrm{SO}_3

Two reactions may take place in reducing flame (Luminous flame)

(i) The blue-green Cu(BO2)2\mathrm{Cu}\left(\mathrm{BO}_2\right)_2 is reduced to colourless cuprous metaborate as :

amp;2Cu(BO2)2+2NaBO2+C⟶ Luminous amp;2CuBO2+Na2 B4O7+CO\begin{aligned}& 2 \mathrm{Cu}\left(\mathrm{BO}_2\right)_2+2 \mathrm{NaBO}_2+\mathrm{C} \stackrel{\text { Luminous }}{\longrightarrow} \\\\& 2 \mathrm{CuBO}_2+\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+\mathrm{CO}\end{aligned}

(ii) Cupric metaborate may be reduced to metallic copper and bead appears red opaque.

amp;2Cu(BO2)2+4NaBO2+2C⟶ Luminous  flame amp;2Cu+2Na2 B4O7+2CO\begin{aligned}& 2 \mathrm{Cu}\left(\mathrm{BO}_2\right)_2+4 \mathrm{NaBO}_2+2 \mathrm{C} \underset{\text { flame }}{\stackrel{\text { Luminous }}{\longrightarrow}} \\\\& 2 \mathrm{Cu}+2 \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+2 \mathrm{CO}\end{aligned}

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