Question

Which of the following reaction(s) is/are example(s) of intra-redox reaction?

• A. ${N{H}_4}_{2}C{O}_3\stackrel{\Delta }{\to }2N{H}_3+H_{2}C{O}_3$
• B. ${N{H}_4}_{2}S\stackrel{\Delta }{\to }2N{H}_3+H_{2}S$
• C. $(N{H}_4{)}_{2}C{r}_2{O}_7\stackrel{\Delta }{\to }2N_2+C{r}_2{O}_3+H_{2}O$
• D. $N{H}_{4}I{O}_3\stackrel{\Delta }{\to }{N}_2+I_2+4H_{2}O+O_2$

Answer 1

Answer: (C), (D)

The correct options are
C $(N{H}_4{)}_{2}C{r}_2{O}_7\stackrel{\Delta }{\to }2N_2+C{r}_2{O}_3+H_{2}O$
D $N{H}_{4}I{O}_3\stackrel{\Delta }{\to }{N}_2+I_2+4H_{2}O+O_2$

Intra - Redox reaction.: The intra- redox reactions are those reactions where there is the same compound undergo reduction and oxidation, leading to change in oxidation state of any element to its lower oxidation state and at the same time other element undergo increases in its oxidation state.

Such condition is found in the following reaction: $(N{H}_4{)}_{2}C{r}_2{O}_7\stackrel{\Delta }{\to }2N_2+C{r}_2{O}_3+H_{2}O$

Here, in $(N{H}_4{)}_{2}C{r}_2{O}_7$ the oxidation state of $N$ is $-3$ and changes to $0$ in $N_2$, whereas $Cr$ has $+6$ oxidation state in $(N{H}_4{)}_{2}C{r}_2{O}_7$ and after reaction its oxidation state changes to $+3$ in $C{r}_2{O}_3$.

$N{H}_{4}I{O}_3\stackrel{\Delta }{\to }{N}_2+I_2+4H_{2}O+O_2$

Same as above, there is an increase in $N$ oxidation state here and $I$ oxidation state also decreases to $0$.