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When borax is heated, the reaction that takes place is :

$$\ce{Na_2B_4O_7.10H_2O \ce{->} 2NaBO_2 + B_2O_3 + 10H_2O}$$ And the compounds that are responsible for the colors are $\ce{M_x(BO_2)_y}$ salts, like $\ce{Co(BO_2)_2}$. My question is, does this $\ce{BO_2}$ come from $\ce{NaBO_2}$ or from $\ce{B_2O_3}$ or do both of them react?

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A little chemical detective work gives a plausible answer.

Say the metal is cobalt and it's oxidation state is +2. Then, with the cobalt calcined to the oxide, you have two possible acid-base reactions:

$\ce{CoO + 2 NaBO2 -> Co(BO2)2 + Na2O}$

$\ce{CoO + B2O3 -> Co(BO2)2}$

Both reactions look good, but upon further review the first one gives something we recognize as a powerful base, more likely to be a reactant than a product. Moreover, a quick check here reveals that boric oxide has a low enough melting point to liquefy in the flame, favoring the second reaction kinetically.

So the second reaction above is really the only plausible one.

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    $\begingroup$ Does the NaBO2 have no use then? Is It just present due to the decomposition of borax into NaBO2 and B2O3 and because NaBO2 cannot be separated from B2O3 in the glass bead? $\endgroup$
    – user600016
    Feb 15, 2019 at 2:39
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    $\begingroup$ In the case of $\ce Cu(BO2)2 $ when this compound is heated in a reducing flame, the $\ce NaBO2$ present in the bead reacts with it forming $\ce CuBO2$ and $\ce Na2B4O7$ OR it may form metallic Copper and $\ce Na2B4O7$ $\endgroup$
    – McSuperbX1
    Jan 17, 2020 at 12:49

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