Question

An organic compound was used with sodium metal and extracted with distilled water. On adding freshly prepared $FeS{O}_4$ solution, followed by the addition of $FeC{l}_3$ and dil. $HCl$ produced greenish blue solution. $0.30$ g of the organic compound after kjeldahlisation evolved a gas (X) which was passed on $100$ ml of $0.1$ M $H_{2}S{O}_4$. The excess of acid required is $20$ ml of $0.5$ M $NaOH$ for neutralisation.
The blue colour of solution is due to:

• A. $K_4\left[Fe\right(CN{)}_6]$
• B. $N{a}_3\left[Fe\right(CN{)}_6]$
• C. $N{a}_2\left[Fe\right(CN{)}_{5}NO]$
• D. $F{e}_4\left[Fe\right(CN{)}_6{]}_3$

Answer 1

The correct option is D $F{e}_4\left[Fe\right(CN{)}_6{]}_3$

The sodium fusion test or Lassaigne's test is used in the elemental analysis for the qualitative determination of the presence of foreign elements namely halogens, nitrogen and sulphur, in an organic compound.
In this test, when a small amount of organic substance is fused with a small quantity of sodium metal in a fusion tube and which is then plunged into the distilled water and the contents are boiled for a few minutes then cooled and filtered, an extract named sodium fusion extract (SFE) is obtained.
The above particular question is an example of a test for Nitrogen.
The obtained SFE is used to detect the presence of elements $N,S,Cl,Br$ & $I$.
For nitrogen,
$Na+C+N⟶NaCN$
With addition of $FeS{O}_4$ freshly prepared solution and $Fe{Cl}_3$ solution with conc. $HCl$, the presence of a prussian blue colour indicates nitrogen presence.
$FeS{O}_4+NaOH⟶Fe{\left(OH\right)}_2+{Na}_{2}S{O}_4$
$6NaCN+Fe{\left(OH\right)}_2⟶{Na}_4\left[Fe{\left(CN\right)}_6\right]$
$3{Na}_4\left[Fe{\left(CN\right)}_6\right]+4Fe{Cl}_3⟶{Fe}_4{\left[Fe{\left(CN\right)}_6\right]}_3$(greenish blue (or) prussian blue colour) $+12NaCl$