The number of moles of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ reduced by one mole of $S{n}^{2+}$ ion is

It is given to find out the number of moles of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ reduced by one mole of ${\mathrm{Sn}}^{2+}$ ion. 

Explanation

The balanced chemical processes (ionic reactions) for K₂Cr₂O₇ reducing by Sn²⁺ are as follows:

${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$

$({\mathrm{Sn}}^{2+}\to {\mathrm{Sn}}^{4+}+2{\mathrm{e}}^{-})\times 3$

Balanced reaction $3{\mathrm{Sn}}^{2+}+{\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}+14{\mathrm{H}}^{+}\to 3{\mathrm{Sn}}^{4+}+2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$

As a result of the balanced reaction: 1 mole of ${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}$will be decreased by 3 moles of ${\mathrm{Sn}}^{2+}$. 
Thus, one mole of ${\mathrm{Sn}}^{2+}$ will decrease to three moles of ${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}$.

Hence, the correct option is  A)  $\frac{1}{3}$.