Write balanced net ionic equation for the following reactions in acidic solution.

$S_{2}O{{}_6}^{2-}\left(aq\right)+C{r}_{2}O{{}_7}^{2-}\left(aq\right)\to S_{4}O{{}_6}^{2-}\left(aq\right)+C{r}^{3+}\left(aq\right)$

The correct option is A $6S_{2}O{{}_6}^{2-}\left(aq\right)+C{r}_{2}O{{}_7}^{2-}\left(aq\right)+14H^{+}\to 3S_{4}O{{}_6}^{2-}\left(aq\right)+2C{r}^{3+}\left(aq\right)+7H_{2}O$

The unbalanced net ionic equation is $S_{2}O{{}_6}^{2-}\left(aq\right)+C{r}_{2}O{{}_7}^{2-}\left(aq\right)\to S_{4}O{{}_6}^{2-}\left(aq\right)+C{r}^{3+}\left(aq\right)$

Balance $S$ atoms and $Cr$ atoms

$2S_{2}O{{}_6}^{2-}\left(aq\right)+C{r}_{2}O{{}_7}^{2-}\left(aq\right)\to S_{4}O{{}_6}^{2-}\left(aq\right)+2C{r}^{3+}\left(aq\right)$

The oxidation number of $Cr$ decreases from $+6$ to $+3$. Total decrease in the oxidation number for $2Cr$ atoms is $6$.

To balance the decrease in the oxidation number with increase in the oxidation number, multiply sulphur species with $3$.

$6S_{2}O{{}_6}^{2-}\left(aq\right)+C{r}_{2}O{{}_7}^{2-}\left(aq\right)\to 3S_{4}O{{}_6}^{2-}\left(aq\right)+2C{r}^{3+}\left(aq\right)$

To balance $O$ atoms, add $7$ water molecules on RHS

$6S_{2}O{{}_6}^{2-}\left(aq\right)+C{r}_{2}O{{}_7}^{2-}\left(aq\right)\to 3S_{4}O{{}_6}^{2-}\left(aq\right)+2C{r}^{3+}\left(aq\right)+7H_{2}O$

To balance hydrogen atoms, add $14$ protons on the LHS.

$6S_{2}O{{}_6}^{2-}\left(aq\right)+C{r}_{2}O{{}_7}^{2-}\left(aq\right)+14H^{+}\to 3S_{4}O{{}_6}^{2-}\left(aq\right)+2C{r}^{3+}\left(aq\right)+7H_{2}O$

This is balanced chemical equation