When borax is heated with CoO on a platinum loop, blue coloured bead formed is largely due to
\(B_2O_3\)
\(CO(BO_2)_2\)
\(COB_4O_7\)
\(CO[B_4O_5(OH)_4]\)
The Correct Option is B
Solution and Explanation
The question involves the borax bead test, a qualitative analysis technique used to identify certain metal ions based on the color of the bead formed when a metal compound is fused with borax in a flame. Let's go through the process to identify why the blue-colored bead is formed when borax is heated with cobalt oxide (CoO).
When borax \((\text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O})\) is heated, it loses water and forms sodium metaborate \((\text{NaBO}_2)\) and boric oxide \((\text{B}_2\text{O}_3)\):
\(\text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O} \rightarrow 2\text{NaBO}_2 + \text{B}_2\text{O}_3 + 10\text{H}_2\text{O}\)
The boric oxide \((\text{B}_2\text{O}_3)\) then reacts with the cobalt oxide \((\text{CoO})\) present on the platinum loop to form cobalt metaborate \((\text{Co(BO}_2)_2)\). This compound is responsible for the distinctive blue color of the bead:
\(\text{CoO} + \text{B}_2\text{O}_3 \rightarrow \text{Co(BO}_2)_2\)
Now, let's analyze why the correct answer is \(\text{Co(BO}_2)_2\):
- Formation of the compound: During the fusion in the flame, the formed cobalt metaborate \(\text{Co(BO}_2)_2\) imparts a blue color to the glassy bead as it is integrated into the structure of the glass matrix.
- Ionic influence: The presence of cobalt ion \(\text{Co}^{2+}\) specifically imparts the deep blue coloration which is typical in cobalt-containing glass and glazes.
- Elimination of other options:
- \(\text{B}_2\text{O}_3\): Boric oxide itself does not contribute to any specific coloration; it acts as a basic constituent enabling the formation of metaborate structures.
- \(\text{CoB}_4\text{O}_7\) and \(\text{Co[B}_4\text{O}_5(\text{OH})_4]\): These structures do not typically form or exist in the bead under the heated conditions from borax bead preparation.
Thus, the blue-colored bead is indeed due to the formation of cobalt metaborate, \(\text{Co(BO}_2)_2\).
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