Which of the following complex is formed when A reacts with $K_{4} [F e (C N)_{6}]$ ?

Prussian blue

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Turnbull's blue

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Brown ring complex

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Sodium nitroprusside

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B Prussian blue
Option (A) is correct.
Compound A will be $F e_{2} (S O_{4})_{3}$ .
When A reacts with $K_{4} [F e (C N)_{6}]$ gives prussian blue color.
$\begin{matrix} F e_{2} (S O_{4})_{3} & Δ - \to & F e_{2} O_{3} + 3 S O_{3} (A) \end{matrix}$
Reddish brown precipitate of ferric hydroxide dissolves in hydrochloric acid and ferric chloride is formed.
$F e (O H)_{3} + 3 H C l ⟶ F e C l_{3} + 3 H_{2} O$
When the solution containing ferric chloride is treated with potassium ferrocyanide solution a blue precipitate/colouration is obtained. The colour of the precipitate is Prussian blue. It is ferric ferro-cyanide.
$4 F e C l_{3} + 3 K_{4} [F e (C N)_{6}] ⟶ F e_{4} [F e (C N)_{6}]_{3} + 12 K C l$

Suggest Corrections

Similar questions

F e C l_{3}

K_{4} [F e (C N)_{6}]

C u S O_{4}

K_{4} [F e (C N)_{6}]

K_{4} [F e (C N)_{6}]

[C u (N H_{3})_{4}]^{2 +}

[C u (H_{2} O)_{6}]^{2 +}

K_{4} [F e (C N)_{6}]

[C u (N H_{3})_{4}]^{2 +}

[C u (H_{2} O)_{4}]^{2 +}

K_{4} [F e (C N)_{6}]

Join BYJU'S Learning Program