The correct option is D 6 mole of Fe2+
Oxidation half reaction 6Fe+2→6Fe+3+6e−
Reduction half reaction Cr2O2−7+14H++6e−→2Cr+3+7H2O
The full ionic equation may be obtained by adding the half-reaction for potassium dichromate to the half-reaction for the reducing agent
6Fe2++Cr2O2−7+14H+→6Fe+3+2Cr+3+7H2O
Thus, One mole of Cr2O2−7 oxidises 6 mole of Fe2+