- A$CH_2Cl_2$ and $CCl_4$
- B$He$ and $He$
- ✓$CHCl_3$ and $CH_2Cl_2$
- D$C_6H_6$ and $CH_4$
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$NO _{3}^{-}, H _{2} O _{2}, BF _{3}, PCl _{3}, XeF _{4},SF _{4}, XeO _{3}, PH _{4}^{+}, SO _{3},\left[ Al ( OH )_{4}\right]^{-}$
$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,} \\
{\,\,\,\,||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,} \\
{C{H_3} - C - C{H_2} - C - CN} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$

Statement $I :$ In 'Lassaigne's Test, when both nitrogen and sulphur are present in an organic compound, sodium thiocyanate is formed.
Statement $II :$ If both nitrogen and sulphur are present in an organic compound, then the excess of sodium used in sodium fusion will decompose the sodium thiocyanate formed to give $NaCN$ and $Na _{2} S$.
In the light of the above statements, choose the most appropriate answer from the options given below...
