Question

In which case oxidation number of Cr has been affected?

• A. $\begin{array}{c}(N{H}_4{)}_{2}C{r}_2{O}_7\\ \downarrow \\ N_2+C{r}_2{O}_3+4H_{2}O\end{array}$
• B. $\begin{array}{c}C{r}_2{O}_7^{2-}+2O{H}^{-}\\ \downarrow \\ 2Cr{O}_4^{2-}+H_{2}O\end{array}$
• C. $\begin{array}{c}2Cr{O}_4^{2-}+2H^{+}\\ \downarrow \\ C{r}_2{O}_7^{2-}+H_{2}O\end{array}$
• D. $\begin{array}{c}Cr{O}_{2}C{l}_2+2O{H}^{-}\\ \downarrow \\ Cr{O}_4^{2-}+2HCl\end{array}$

Answer 1

The correct option is A $\begin{array}{c}(N{H}_4{)}_{2}C{r}_2{O}_7\\ \downarrow \\ N_2+C{r}_2{O}_3+4H_{2}O\end{array}$

Analysing the options
Option (A):
$2Cr{O}_4^{2-}+2H^{+}\to C{r}_2{O}_7^{2-}+H_{2}O$
$Cr=+6Cr=+6$

In this reaction , no change in oxidation state of Cr is taking place.

Option (B):
$C{r}_2{O}_7^{2-}+2O{H}^{-}\to 2Cr{O}_4^{2-}+H_{2}O$
$Cr=+6Cr=+6$
In this reaction, no change in oxidation state is taking place as oxidation state of Cr in both $C{r}_2{O}_7^{2-}$ and $Cr{O}_4^{2-}$ is the same (+6).

Option (C):
$(N{H}_4{)}_{2}C{r}_2{O}_7\to N_2+C{r}_2{O}_3+4H_{2}O$
$Cr=+6Cr=+3$
In this reaction, the oxidation state of Cr is decreasing from +6 to +3, this implies that $(N{H}_4{)}_{2}C{r}_2{O}_7$ is getting reduced.
Thus, the oxidation state of Cr gets affected.

Option (D):
$Cr{O}_{2}C{l}_2+2O{H}^{-}\to Cr{O}_4^{2-}+2HCl$
$Cr=+6Cr=+6$
In this reaction, no change in oxidation state is taking place as oxidation state of Cr in $Cr{O}_{2}C{l}_{2}andCr{O}_4^{2-}$ is (+6).
So, the correct answer is option (C)