The number of moles of potassium dichromate needed to oxidize 3 moles…

The correct answer is Option (1) → 0.5

In acidic solution the dichromate ion is a strong oxidising agent; each chromium goes from +6 to +3, so one dichromate ion takes up six electrons. Each ferrous ion gives only one electron as it is oxidised to the ferric ion:

    \( \displaystyle \ce{Cr2O7^2- + 14H+ + 6e- -> 2Cr^3+ + 7H2O} \)

    \( \displaystyle \ce{Fe^2+ -> Fe^3+ + e-} \)

Multiplying the iron half reaction by 6, so that 6 electrons are given and taken, and adding:

    \( \displaystyle \ce{Cr2O7^2- + 14H+ + 6Fe^2+ -> 2Cr^3+ + 6Fe^3+ + 7H2O} \)

So 1 mol of dichromate oxidises 6 mol of \( \ce{Fe^2+} \). For 3 mol of \( \ce{Fe^2+} \):

    \( \displaystyle n(\ce{K2Cr2O7}) = \frac{3\ \mathrm{mol}}{6} = 0.5\ \mathrm{mol} \)

So 0.5 mol of potassium dichromate is needed.

Shortcut: Electrons lost must equal electrons gained: 3 mol of \( \ce{Fe^2+} \) give 3 mol of electrons, and each mole of dichromate takes 6, so 3/6 = 0.5 mol.

NCERT (2026-27 edition):
Oxides and Oxoanions of Metals – …