Answer:
In principle, if the organic compound contains both N and S, sodium thiocyanate should be formed in Lassaigne's test and this should give blood red colouration with \[FeC{{l}_{3}}\]
\[Na\,\,+\,\,\,\,\,\,\,\underset{\text{(From}\,\text{organic}\,\text{compound)}}{\mathop{C\,\,\,+\,\,\,N+\,\,\,\,S}}\,\,\,\,\,\,\,\xrightarrow{Fusion}\,\underset{\text{Sod}\text{.}\,\,\text{thiocyanate}}{\mathop{NaCNS}}\,\]
\[3\,NaCNS+FeC{{l}_{3}}\,\,\xrightarrow{\,}\,\,\,\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\text{Ferric}\,\text{thiocyanate} \\ \text{(Blood}\,\,\text{red}\,\,\text{colouration)} \end{smallmatrix}}{\mathop{Fe{{(SCN)}_{3}}}}\,\,\,\,\,\,+3\,NaCl\]
However, if blood red colouration is not obtained, it does not necessarily mean that S is absent. This is because in presence of excess of sodium metal, sodium thiocyanate initially formed, decomposes to form sodium cyanide and sodium sulphide.
\[2Na+2NaCNS\,\,\xrightarrow{\Delta }\,NaCN+N{{a}_{2}}S\]
As a result, blood red colouration is not obtained.
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