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When borax is heated with CoO on a platinum loop, blue colour bead formed is largely due to

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$B_{2} O_{2}$

$C o {(B O_{2})}_{2}$

$C O B_{4} O_{7}$

$C O [B_{4} O_{5} {(O H)}_{4}]$

answer is B.

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Detailed Solution

$N a_{2} B_{4} O_{7} .10 H_{2} O \overset{Δ}{\to} N a_{2} B_{4} O_{7} \overset{Δ}{\to} 2 N a B O_{2} + B_{2} O_{3}$

$B_{2} O_{3} + C o O \to C o {(B O_{2})}_{2}$

Bead(blue colored)

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