Question

Which of the following statement(s) is/(are) correct with reference to the ferric ions?

• A. $F{e}^{3+}$ gives brown colour with potassium ferricyanide.
• B. $F{e}^{2+}$ gives blue precipitate with potassium ferricyanide.
• C. $F{e}^{3+}$ gives red colour with potassium thiocyanate,
• D. $F{e}^{2+}$ gives brown colour with ammonium thiocyanate.

Answer 1

Answer: (B), (C)

The correct options are
B $F{e}^{2+}$ gives blue precipitate with potassium ferricyanide.
C $F{e}^{3+}$ gives red colour with potassium thiocyanate,

$F{e}^{2+}$ ions give blue precipitate with $K_3\left[Fe\right(CN{)}_6].$
$KFe\left[Fe\right(CN{)}_6]\to$This precipitate is known as Turnbull's blue and it is identical to Prussian blue.

$F{e}^{2+}+K_3\left[Fe\right(CN{)}_6]\to \underset{\underset{\underset{\left(\text{Turnbull's blue}\right)}{}}{\text{Potassium ferroferricyanide}}}{KFe\left[Fe\right(CN{)}_6]}+2K^{+}$

The red colouration that appearts in case of $F{e}^{3+}$ ions on reaction with potassium thiocyanate $\left(KSCN\right)$ is due to formation of $\left[Fe\right(SCN{)}_3].$

$F{e}^{3+}+3KSCN\to \underset{\underset{\left(\text{red}\right)}{\text{Ferric thiocyanate}}}{\left[Fe\right(SCN{)}_3]}+3K^{+}$
Options (b) and (c) are correct.