Which of the following oxides undergoes dispr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Disproportionation is a reaction where an element in an intermediate oxidation state is simultaneously oxidized and reduced.In the case of $N_2O_4

Vedclass · Accepted Answer

$N_2O_4$

Vedclass · Answer

(B) Disproportionation is a reaction where an element in an intermediate oxidation state is simultaneously oxidized and reduced.In the case of $N_2O_4$,the oxidation state of nitrogen is $+4$.When $N_2O_4$ reacts with $NaOH$,it forms a mixture of nitrite $(NO_2^-)$ and nitrate $(NO_3^-)$ salts:$N_2O_4 + 2NaOH \rightarrow NaNO_2 + NaNO_3 + H_2O$.Here,nitrogen is reduced from $+4$ to $+3$ in $NaNO_2$ and oxidized from $+4$ to $+5$ in $NaNO_3$.

Which of the following oxides undergoes disproportionation with $NaOH$ solution?

Similar Questions

Can phosphorus with outer electronic configuration $3s^{2} 3p^{3}$ form $PH_{5}$?

$A + H_2O \to B + HCl$
$B + H_2O \to C + HCl$
Compound $(A), (B)$ and $(C)$ will be respectively

In which of the following arrangements,the order is not strictly according to the property written against it?

$A$ non-metal $M$ forms $MCl_3$,$M_2O_5$ and $Mg_3M_2$ but does not form $MI_5$. Then the incorrect statement regarding non-metal $M$ is:

Very pure $N_2$ can be obtained by

Vedclass Test Series

Exam Paper Generator

Online Exam Module