When borax is heated with CoO on a platinum loop, blue coloured bead formed is

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

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

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B_{2} O_{3}

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C o B_{4} O_{7}

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Solution

The correct option is B $C o (B O_{2})_{2}$
$N a_{2} B_{4} O_{7} \cdot 10 H_{2} O Δ - - \to N a_{2} B_{4} O_{7} + 10 H_{2} O$

$N a_{2} B_{4} O_{7} Δ - - \to N a B O_{2} (s o d i u m m e t a b o r a t e) + B_{2} O_{3}$

$B_{2} O_{3} + C o O ⟶ C o (B O_{2})_{2} B l u e B e a d (c o b a l t (I I) m e t a b o r a t e)$

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