The value of $n$ in the following equation if balanced

${Cr}_2{O}_7{}^{2-}+14H^{+}+{nFe}^{2+}⟶2{Cr}^{3+}+{nFe}^{3+}+7H_{2}O$

For any given reaction, the charges of reactants and products must be same

Upon calculation of charges on reactant and products, we get:

$-2+14\left(1\right)+2n=2(+3)+3n$

$-2+14+2n=6+3n$

$-2+14-6=3n-2n$

$n=6$

Hence the correct option is D.