A) \[N{{a}_{2}}S\]and\[NaCN\]
B) \[NaSCN\]
C) \[N{{a}_{2}}S{{O}_{4}}\]and\[NaCN\]
D) \[N{{a}_{2}}S\]and\[NaCN\]
Correct Answer: A
Solution :
If both N and S are present, it may result in the formation of\[NaCNS\]during fusion with sodium metal\[Na+C+N+S\xrightarrow[{}]{{}}NaCN\] If sodium metal is present in excess, the sulphocyanide formed decomposes into sulphide and cyanide. \[NaCNS+2Na\xrightarrow[{}]{{}}NaCN+N{{a}_{2}}S\]
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