The addition of dilute NaOH to $\mathrm{Cr}^{3+}$ salt solution will give:

(B) precipitate of $\mathrm{Cr}_{2} \mathrm{O}_{3}\left(\mathrm{H}_{2} \mathrm{O}\right)_{n}$

(A) precipitate of $\left[\mathrm{Cr}(\mathrm{OH})_{6}\right]^{3-}$

(C) precipitate of $\mathrm{Cr}(\mathrm{OH})_{3}$

(D) a solution of $\left[\mathrm{Cr}(\mathrm{OH})_{4}\right]^{-}$

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · d and f Block Elements

The addition of dilute NaOH to

Cr^{3 +}

salt solution will give:

Answer: (B)

When dilute NaOH is added to a

Cr^{3 +}

salt solution, chromium(III) hydroxide precipitates as a green gelatinous solid:

Cr^{3 +} + 3 OH^{-} \to Cr (OH)_{3} ↓

. This precipitate is amphoteric and dissolves in excess strong base to form the chromite ion

[Cr (OH)_{4}]^{-}

, but with dilute NaOH the precipitate remains. The correct answer is (C).

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option