First slide

13th group or boron group

Question

Diborane $(B_{2} H_{6})$ reacts independently with $O_{2}$ and $H_{2} O$ to produce, respectively.

Moderate

A

$B_{2} O_{3}$ and $H_{3} B O_{3}$
B

$B_{2} O_{3}$ and ${[B H_{4}]}^{-}$
C

$H_{3} B O_{3}$ and $B_{2} O_{3}$
D

$H B O_{2}$ and $H_{3} B O_{3}$

Solution

Following reactions take place
$B_{2} H_{6} + 3 O_{2} \to B_{2} O_{3} + 3 H_{2} O$
$B_{2} H_{6} + 6 H_{2} O \to H_{3} B O_{3} + 6 H_{2}$

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