Practical Chemistry Question 4
Question 4 - 31 January - Shift 2
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 the light of the above statements, choose the most appropriate answer from the options given below :
(1) Both Statement I and Statement II are true
(2) Statement I is true but Statement II is false
(3) Both Statement I and Statement II are false
(4) Statement I is false but Statement II is true
Show Answer
Answer: (3)
Solution:
Formula: Borax Bead Test
(Borax Bead Test)
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.
$CuSO_4+B_2 O_3 \xrightarrow[\text{ (Oxidising ) }]{\text{ Non -luminos flame }} Cu(BO_2)_2+SO_3$
Cupric metaborate
blue-green
Two reactions may take place in reducing flame (Luminous flame)
(i) The blue-green $Cu(BO_2)_2$ is reduced to colourless cuprous metaborate as :
$2 Cu(BO_2)_2+2 NaBO_2+C \xrightarrow[\text{ flame }]{\text{ Luminous }}$
$2 CuBO_2+Na_2 B_4 O_7+CO$
(ii) Cupric metaborate may be reduced to metallic copper and bead appears red opaque.
$2 Cu(BO_2)_2+4 NaBO_2+2 C \xrightarrow[\text{ flame }]{\text{ Luminous }}$
$2 Cu+2 Na_2 B_4 O_7+2 CO$
