In Tollen’s test for aldehyde, the overall number of electron(s) transferred to the Tollens’ reagent formula ${\left[Ag{\left(N{H}_3\right)}_2\right]}^{+}$ per aldehyde group to form silver mirror is ______

$\begin{array}{c}+1\\ {\mathrm{CH}}_3\mathrm{CHO}+2{\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}+3{\mathrm{OH}}^{-}\end{array}$$⟶$$\begin{array}{r}+3\\ {\mathrm{CH}}_3{\mathrm{COO}}^{-}+2\mathrm{Ag}+2{\mathrm{H}}_2\mathrm{O}+4{\mathrm{NH}}_3\end{array}$

The oxidation state of carbon charges from +1 to +3 that means carbon loses 2 electrons while silver accepts one electron. So, 2 mol silver will be formed by accepting 2 mol electrons from 1 mol of carbon