Question

Iodine Oxidises Sodium Borohydride To Give_____.

• A. ${\mathrm{I}}_3$
• B. $\mathrm{HI}$
• C. ${\mathrm{B}}_2{\mathrm{H}}_6$
• D. Sodium Hydride

Answer 1

The correct option is C

${\mathrm{B}}_2{\mathrm{H}}_6$

Explanation for the Correct Option:

Option (C): ${\mathrm{B}}_2{\mathrm{H}}_6$

• Iodine oxidizes to give diborane.
• The chemical equation for which can be written as;

$\underset{(\mathrm{Sodium}\mathrm{borohydride})}{2{\mathrm{NaBH}}_4}+\underset{\left(\mathrm{Iodine}\right)}{{\mathrm{I}}_2}\stackrel{\mathrm{Diglyme}}{\to }\underset{\left(\mathrm{Diborane}\right)}{{\mathrm{B}}_2{\mathrm{H}}_6}+2\underset{(\mathrm{Sodium}\mathrm{iodide})}{\mathrm{NaI}}+\underset{\left(\mathrm{Hydrogen}\right)}{{\mathrm{H}}_2}$

Explanation for the incorrect Option:

Option (A): ${\mathrm{I}}_3$

• when Iodine undergoes oxidization with sodium borohydride, ${\mathrm{I}}_3$ is not formed as a product.

• Hence, the given Option is an incorrect one.

Option (B): $\mathrm{HI}$

• The oxidation of Iodine with sodium borohydride doesn’t give $\mathrm{HI}$ as its product.
• Hence, this Option is given here does not represent the correct concept.

Option (D): Sodium Hydride.

• Sodium Hydride is not formed as a product when Iodine undergoes oxidization with Sodium borohydride
• Thus the given Option doesn’t coincide with the correct concept.

Hence, Iodine Oxidises Sodium Borohydride to Give${\mathrm{B}}_2{\mathrm{H}}_6$(diborane).