Question

$\text{SO}_{2}$ reduces \text{Cr}_{2}\text{O}_{7}^{2-}$$$ to $$$\text{Cr}^{2+} . The change in oxidation number of $\text{Cr}$ is:

Answer 1

We can see that it is given in the reduction of $\text{Cr}_{2}\text{O}_{7}^{2-}$ takes place due to the presence of $\text{SO}_{2}$ and leads to the formation of $\text{Cr}^{2+}$ . Therefore, we can say that in this reaction, $\text{SO}_{2}$ act as a reducing agent.

Complete step by step answer:
We can assume that the oxidation number of $\text{Cr}$ in $\text{Cr}_{2}\text{O}_{7}^{2-}$ be $x$. The value of $x$ is calculated as shown below.
$2 x+7(-2) =-2$
x =$\dfrac{-2+14}{2}$
=+6
We can say that in the case of ${Cr}^{2+}$ the oxidation number of ${Cr}$ in $Cr^{2+}$ is equal to the charge on ${Cr}$ atom that is $+2$.
We can write the reduction reaction in which $t{Cr}_{2}{O}_{7}^{2-}$ reduces to ${Cr}^{2+}$ can be written as follows.
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \rightarrow 2 \mathrm{Cr}^{2+}$
We can see that in the above reaction the change in the oxidation number per $\text{Cr}$ atom is $-4$. Thus we can conclude that the oxidation number of $\text{Cr}$ decreases by four.

Note:
We know that the decrease in the oxidation number of species in a redox reaction is regarded as reduction. However, the increase in the oxidation number is considered as oxidation.