Consider the following reaction at $1123 K$ :

$B (s) + N a_{2} C O_{3} (s) + N a N O_{3} (s) ⟶$

The oxidation state of boron in the product formed is:

+ 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 3

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

+ 6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $+ 3$
$2 B + 3 N a_{2} C O_{3} + 3 N a N O_{3} \to 2 N a_{3} B O_{3} + 3 N a N O_{2} + 3 C O_{2}$

Now in $N a_{3} B O_{3}$ , let x be the oxidation state of B.

$3 \times (+ 1) + x + 3 \times (- 2) = 0$

$∴ x = + 3$

The correct answer is C option.

Suggest Corrections

Similar questions

1123 K

B (s) + N a_{2} C O_{3} (s) + N a N O_{3} (s) ⟶

N a {N O}_{3}

N a_{2} C O_{3} + H C l \to

K I

Join BYJU'S Learning Program