Question

Cation and anion are called basic and acidic radicals, respectively, because during salt formation cation comes from base and anion comes from acid.

$K_4\left[Fe\right(CN{)}_6]$ can be used to detect one or more out of $F{e}^{2+}$, $F{e}^{3+}$, $Z{n}^{2+}$, $C{u}^{2+}$, $C{d}^{2+}$

• A. $F{e}^{2+}$, $F{e}^{3+}$ only
• B. $F{e}^{3+}$, $Z{n}^{2+}$, $C{u}^{2+}$ only
• C. All but not $C{d}^{2+}$
• D. All of these

Answer 1

Answer: (D)

All of these

$C{u}^{2-}$ ions chocolate-brown precipitate with $K_4\left[Fe\right(CN{)}_6]$ solution.

$2C{u}^{2+}+K_4\left[Fe\right(CN{)}_6]⟶\underset{Chocolate-brownppt}{C{u}_2\left[Fe\right(CN{)}_6]\downarrow +4K^{+}}$

With potassium hexacyanoferrate (II) solution, $F{e}^{3+}$ given intense blue precipitate of iron (III) hexacyanoferrate (Prussian blue)

$4F{e}^{3+}+3\left[Fe\right(CN{)}_6{]}^{4-}⟶\underset{Prussianblue}{F{e}_4\left[Fe\right(CN{)}_6{]}_3\downarrow }$

with potassium hexacyanoferrate (II) solution, a white precipitate of variable composition is obtained. If the reagent is added in some excess, the composition of the precipitate is $K_2Zn_3[Fe(CN)_6]_2.

$3Z{n}^{2+}+2K^{+}+2\left[Fe\right(CN{)}_6{]}^{4-}⟶\underset{Whiteppt}{K_{2}Z{n}_3\left[Fe\right(CN{)}_6{]}_2\downarrow }$

With potassium hexacyanoferrate (II) solution, $F{e}^{2+}$ forms a white precipitate of potassium iron (II) hexacyanoferrate (II) in the absence of air.

$F{e}^{2+}+2K^{+}+\left[Fe\right(CN{)}_6{]}^{4-}c⟶\underset{Whiteppt}{K_{2}Fe\left[Fe\right(CN{)}_6]\downarrow }$

All of these can detected using this as