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If in a test, when sodium fusion extract of an organic compound is boiled with \mathrm{FeSO_4} then treated with \mathrm{H_2SO_4}, a Prussian blue colouration is observed. This colour is due to the presence of

Option: 1

\mathrm{CuSO _{4}}


Option: 2

\mathrm{NO _{3}^{-}}


Option: 3

\mathrm{FeSO _{4}}


Option: 4

\mathrm{F e_{4}\left[F e(C N)_{6}\right]_{3}}


Answers (1)

best_answer

As we have learnt,

Test for Nitrogen :

The sodium fusion extract is boiled with Ferrous Sulphate and further, the solution is acidified and heated with concentrated \mathrm{H_2SO_4}. Upon reaction, the formation of prussian blue colour confirm the presence of nitrogen.

\begin{aligned} &6 \mathrm{CN}^{-}+\mathrm{Fe}^{2+} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \\ &3\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+4 \mathrm{Fe}^{3+} \stackrel{\mathrm{xH}_{2} \mathrm{O}}{\longrightarrow} \mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3} \cdot \mathrm{xH}_{2} \mathrm{O} \\ \end{aligned}

It is to be noted that some Ferrous ions \mathrm{(Fe^{2+})} are oxidised to Ferric ions \mathrm{(Fe^{3+})} upon reaction with conc. \mathrm{H_2SO_4} which act as the counter ion outside the complex.

Hence, the correct answer is Option (4)

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