Cr_2O_7^2- dissolves in aqueous NaOH to give:

The correct answer is Option (3) → \({CrO_4}^{2-}\)

In alkali, the orange dichromate ion is converted into the yellow chromate ion.

This is not a redox reaction. Chromium is +6 in both ions:

    \( \displaystyle \ce{Cr2O7^2-}:\ 2x + 7(-2) = -2 \Rightarrow x = +6 \)

    \( \displaystyle \ce{CrO4^2-}:\ x + 4(-2) = -2 \Rightarrow x = +6 \)

So in aqueous NaOH, dichromate gives the yellow chromate ion.

Why the other options are incorrect

• Option (1) \(Cr_2O_3\) — Incorrect. Chromium is +3 here; forming it needs a reducing agent, which NaOH is not.

• Option (2) \(Cr(OH)_3\) — Incorrect. Also a chromium(III) compound; it is precipitated when \( \ce{Cr^3+} \) salts, not dichromate, react with alkali.

• Option (4) \(CrO_3\) — Incorrect. Chromium(VI) oxide separates when dichromate is treated with concentrated sulphuric acid; alkali moves the equilibrium the other way.

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.

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

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