JEE Advanced 2021 Paper 2 Q15 Chemistry Inorganic Chemistry d-Block Elements Medium

JEE Advanced 2021 Paper 2 · Q15 · d-Block Elements

Precipitate X is

  1. A. Fe$_4$[Fe(CN)$_6$]$_3$
  2. B. Fe[Fe(CN)$_6$]
  3. C. K$_2$Fe[Fe(CN)$_6$]
  4. D. KFe[Fe(CN)$_6$]
Reveal answer + step-by-step solution

Correct answer:C

Solution

K$_4$[Fe(CN)$_6$] (potassium ferrocyanide, Fe(II)) reacts with FeSO$_4$ (Fe(II)) in the complete absence of air to give a white precipitate which contains only Fe(II) and the ferrocyanide ion:

K$_4$[Fe(CN)$_6$] + FeSO$_4$ $\rightarrow$ K$_2$Fe[Fe(CN)$_6$] $\downarrow$ + K$_2$SO$_4$.

The white precipitate is K$_2$Fe$^{II}$[Fe$^{II}$(CN)$_6$], which turns blue (Prussian-blue analogue) on air oxidation of Fe(II) to Fe(III). Option (C).

(A) Fe$_4$[Fe(CN)$_6$]$_3$ is Prussian blue (Fe$^{III}_4$[Fe$^{II}$(CN)$_6$]$_3$) - not white. (B) Fe[Fe(CN)$_6$] is Turnbull's blue. (D) KFe[Fe(CN)$_6$] is the soluble Prussian blue (Berlin blue), not white. Hence X = K$_2$Fe[Fe(CN)$_6$], option (C).

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