JEE Main 2023 · 31 January · evening · Single correct · +4 / −1
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 :
- (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
Answer and solution
Answer (a) Both Statement I and Statement II are false
Solution
Chemistry ma’am · 4 steps
A luminous flame is a reducing flame, so copper is pulled down from +2 and the green goes away.
- Make the bead. On heating, borax gives Na₂B₄O₇ → 2NaBO₂ + B₂O₃ and cupric sulphate gives CuO. The glassy B₂O₃ then combines with the metal oxide to form a metaborate.
- Know your flame. The non-luminous flame is oxidising, so copper stays as Cu²⁺. The luminous flame has unburnt carbon, so it is reducing and pushes copper down to +1 or to the metal.
Flame What forms Bead colour Non-luminous (oxidising) CuO + B₂O₃ → Cu(BO₂)₂, copper(II) metaborate blue-green Luminous (reducing) 2Cu(BO₂)₂ + 2NaBO₂ + C → 2CuBO₂ + Na₂B₄O₇ + CO, copper(I) metaborate colourless Luminous, further reduction 2Cu(BO₂)₂ + 4NaBO₂ + 2C → 2Cu + 2Na₂B₄O₇ + 2CO, copper metal red, opaque - Check each statement. Statement I is false: in a luminous flame the bead turns colourless or red, not green. Statement II is false: the blue-green colour comes from copper(II) metaborate, Cu(BO₂)₂, because copper(I) metaborate is colourless.
- So. Both statements are false, which is option (a).
The idea to keep. Non-luminous flame oxidises, so you see the higher oxidation state's colour. Luminous flame reduces: copper(II) goes to colourless copper(I) metaborate or red copper metal.
Answer from the NTA final answer key. Solution by PrepWiser.