In which of the following cases there is a change in oxidation number?

The correct answer is Option (2) → \(SO_2\) gas is passed through acidified \(Cr_2O_7^{2-}\) solution

An oxidation number changes only in a redox reaction. Acidified dichromate is a strong oxidising agent and oxidises \( \ce{SO2} \) to sulphate, being itself reduced to \( \ce{Cr^{3+}} \):

    \( \ce{Cr2O7^{2-}} \) + \( \ce{3SO2} \) + \( \ce{2H^{+}} \) → \( \ce{2Cr^{3+}} \) + \( \ce{3SO4^{2-}} \) + \( \ce{H2O} \)

In \( \ce{Cr2O7^{2-}} \): 2x + 7(−2) = −2, so x = +6; in \( \ce{Cr^{3+}} \) it is +3. Sulphur goes from +4 (\( \ce{SO2} \)) to +6 (\( \ce{SO4^{2-}} \)). The orange solution turns green.

Why the other options are incorrect

• Option (1) Aqueous solution of \(CrO_4^{2-}\) is acidified — Incorrect. \( \ce{2CrO4^{2-}} \) + \( \ce{2H^{+}} \) → \( \ce{Cr2O7^{2-}} \) + \( \ce{H2O} \). In \( \ce{CrO4^{2-}} \), x + 4(−2) = −2 gives x = +6, the same as in \( \ce{Cr2O7^{2-}} \); only the colour changes (yellow to orange).

• Option (3) \(Cr_2O_7^{2-}\) solution is made alkaline — Incorrect. \( \ce{Cr2O7^{2-}} \) + \( \ce{2OH^{-}} \) → \( \ce{2CrO4^{2-}} \) + \( \ce{H2O} \); chromium stays +6 (the reverse of Option (1)).

• Option (4) \(CrO_2Cl_2\) is dissolved in \(NaOH\) — Incorrect. \( \ce{CrO2Cl2} \) + 4NaOH → \( \ce{Na2CrO4} \) + 2NaCl + \( \ce{2H2O} \). In \( \ce{CrO2Cl2} \), x + 2(−2) + 2(−1) = 0 gives x = +6, and chromium is +6 in \( \ce{Na2CrO4} \) too.

NCERT (2026-27 edition):
Oxides and Oxoanions of Metals – “… The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. The oxidation state of chromium in chromate and dichromate is the same.”

    \( \ce{2CrO4^2- + 2H+ -> Cr2O7^2- + H2O} \)
    \( \ce{Cr2O7^2- + 2OH- -> 2CrO4^2- + H2O} \)

Note: The same/similar question has been asked in different years by NTA as per the details given below: