Question:
Ferric ion forms a Prussian blue precipitate due to the formation of
Ferric ion forms a Prussian blue precipitate due to the formation of
Updated On: Apr 27, 2024
- $K_4[Fe(CN)_6]$
- $K_3[Fe(CN)_6]$
- $Fe(CNS)_3$
- $Fe_4[Fe(CN)_6]_3$
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The Correct Option is D
Solution and Explanation
Ferric ion forms a prussian blue precipitate due to the formation of $Fe _{4}\left[ Fe ( CN )_{6}\right]_{3}$. This complex is formed during the determine action of presence of nitrogen in the given sample. In this method, to portion of sodium fusion extract, freshly prepared ferrous sulphate, $FeSO _{4}$ solution is added and warmed. Then about 2 to 3 drops of $FeCl _{3}$ solution are added and acidified with conc. HCl. The appearance of a prussian blue colour indicate the presence of nitrogen.
$FeSO _{4}+2 NaOH \longrightarrow Fe ( OH )_{2}+ Na _{2} SO _{4}$
$6 NaCN + Fe ( OH )_{2} \longrightarrow \underset{\text{Sodium ferrocyanide}}{{Na _{4}\left[ Fe ( CN )_{6}\right]}}+2 NaOH$
$3 Na _{4}\left[ Fe ( CN )_{6}\right]+4 FeCl _{3} \longrightarrow \underset{\text{Ferric ferrocyanide (prussian blue)}}{{Fe _{4}\left[ Fe ( CN )_{6}\right]_{3}}}+12 NaCl$
$FeSO _{4}+2 NaOH \longrightarrow Fe ( OH )_{2}+ Na _{2} SO _{4}$
$6 NaCN + Fe ( OH )_{2} \longrightarrow \underset{\text{Sodium ferrocyanide}}{{Na _{4}\left[ Fe ( CN )_{6}\right]}}+2 NaOH$
$3 Na _{4}\left[ Fe ( CN )_{6}\right]+4 FeCl _{3} \longrightarrow \underset{\text{Ferric ferrocyanide (prussian blue)}}{{Fe _{4}\left[ Fe ( CN )_{6}\right]_{3}}}+12 NaCl$
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