MCQ
If $N$ and $S$ are present in an organic compound during Lassaigne test, then both changes into
- A$N{a_2}S$ and $NaCN$
- ✓$NaSCN$
- C$N{a_2}S{O_3}$ and $NaCN$
- D$N{a_2}S$ and $NaCNO$
$3NaCNS + FeC{l_3} \to \mathop {\mathop {Fe{{(CNS)}_3}}\limits_{{\rm{Ferric sulpho cyanide}}} }\limits_{{\rm{(Blood red colour)}}} + 3NaCl$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)\,\, KO_2$
$(B)\,\,$ cis $-[Co(en)_2Cl_2]^+$
$(C)\,\, K_3[Co(Ox)_3]$
$(D)\,\,[CoCl_3]^{2-}$
$\underset{({{C}_{2}}{{H}_{6}}O)}{\mathop{X}}\,\,\xrightarrow[573\,\,K]{Cu}$ $A$ $\xrightarrow[^{-}OH\,,\,\Delta ]{{{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}}$ Silver mirror
$A\,\xrightarrow{^{-}OH\,,\,\Delta }Y$
$A\,\xrightarrow{N{{H}_{2}}NHCON{{H}_{2}}}Z$
Reason : Low spin complexes have lesser number of unpaired electrons.