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Iodine oxidises sodium borohydride to give
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The correct answer is:
$\mathrm{B}_2 \mathrm{H}_6$
The oxidation of sodium borohydride with iodine in diglyme gives diborane.
$$
2 \mathrm{NaBH}_4+\mathrm{I}_2 \stackrel{\text { Diglyme }}{\longrightarrow} \mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NaI}+\mathrm{H}_2
$$
$$
2 \mathrm{NaBH}_4+\mathrm{I}_2 \stackrel{\text { Diglyme }}{\longrightarrow} \mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NaI}+\mathrm{H}_2
$$
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