First slide

13th group or boron group

Question

In the reaction $B_{2} H_{6} + 2 KOH + 2 X ⟶ 2 Y + 6 H_{2}, X$ and Y are respectively

Moderate

A

$H_{2} O, {KBO}_{2}$
B

$HCl, {KBO}_{3}$
C

$H_{2} O, {KBO}_{3}$
D

$H_{2}, H_{3} {BO}_{3}$

Solution

$B_{2} H_{6} + 2 KOH + 2 H_{2} O ⟶ 2 {KBO}_{2} + 6 H_{2}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App