In acidic solution, potassium dichromate converts tin(II) ions into t…

The correct answer is Option (3) → 0.30 mol

In acid solution the dichromate ion takes six electrons: each of its two chromium atoms goes from +6 to +3. Each tin(II) ion gives two electrons when it becomes tin(IV). So one dichromate ion can oxidise three Sn²⁺ ions.

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

    \( \displaystyle \ce{3Sn^2+ -> 3Sn^4+ + 6e^-} \)

Adding the two half-reactions gives the full ionic equation:

    \( \displaystyle \ce{Cr2O7^2- + 14H+ + 3Sn^2+ -> 2Cr^3+ + 3Sn^4+ + 7H2O} \)

So 1 mol of \( \ce{K2Cr2O7} \) oxidises 3 mol of \( \ce{Sn^2+} \).

    \( \displaystyle n(\ce{K2Cr2O7}) = \frac{0.90\ \mathrm{mol}}{3} = 0.30\ \mathrm{mol} \)

NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “…In acidic solution, its oxidising action can be represented as follows: Cr₂O₇^2– + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O (E^o = 1.33V) Thus, acidified potassium dichromate will oxidise iodides to iodine, sulphides to sulphur, tin(II) to tin(IV) and iron(II) salts to iron(III).”

“3 Sn²⁺ → 3Sn⁴⁺ + 6 e^–” … “The full ionic equation may be obtained by adding the half-reaction for potassium dichromate to the half-reaction for the reducing agent, for e.g., Cr₂O₇^2– + 14 H⁺ + 6 Fe²⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O”