The change in the oxidation number of the underlined nitrogen atom in the following chemical reactions are :
a. Nitrous acid $(H N - - O_{2})$ reduces $M n O_{4}^{⊖}$ in acid solution.
b. Nitrous acid $(H N - - O_{2})$ oxidises $I^{⊖}$ to $I_{2}$ in acid solution.

(a) +3 to +5
(b) +3 to +2

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

(a) +3 to +6
(b) +3 to +2

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

(a) +3 to +5
(b) +3 to +1

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

(a) +3 to +2
(b) +3 to +5

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

Open in App

Solution

The correct option is B (a) +3 to +5
(b) +3 to +2
The balanced reactions are as follows:
i. $2 M n O_{4}^{⊖} + 6 H^{+} + 5 N O_{2}^{⊖} ⟶ 5 N O_{3}^{⊖} + 3 H_{2} O + 2 {M n}^{2 +}$
ii. $2 H N O_{2} + 2 H^{\oplus} + 2 I^{⊖} ⟶ 2 N O + 2 H_{2} O + I_{2}$
Oxidation number of $N$ changes from +3 to +5 in (i).
Oxidation number of $N$ changes from +3 to +2 in (ii).

Suggest Corrections

Similar questions

H N O_{2}

H N O_{2}

(H N O_{2})

Nitrous acid $H N O_{2}$ can decompose to $N O$ and $N O_{2}$ .

Nitrous acid $H N O_{2}$ decomposes to both $N O$ and $N O_{2}$ .

2 a^{3} + 4 a b^{2} - 5 b^{3} - 6 a^{2} b

3 a^{3} - 4 a^{2} b + 11 a b^{2} + b^{3}

Join BYJU'S Learning Program