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All the products formed in the oxidation of $\mathrm{NaBH}_{4}$ by $\mathrm{I}_{2}$, are
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$\mathrm{B}_{2} \mathrm{H}_{6}, \mathrm{H}_{2}$ and $\mathrm{NaI}$
$\mathrm{Na} \mathrm{BH}_{4}+\mathrm{I}_{2} \rightarrow \mathrm{Nal}+\mathrm{B}_{2} \mathrm{H}_{6}+\mathrm{H}_{2}$
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