A) \[{{N}_{2}}\]and \[HCl\]
B) \[N{{H}_{4}}Cl\] and \[NC{{l}_{3}}\]
C) \[NC{{l}_{3}}\] and \[HCl\]
D) \[{{N}_{2}}\] and \[N{{H}_{4}}Cl\]
Correct Answer: C
Solution :
\[N{{H}_{3}}+3C{{l}_{2}}\to NC{{l}_{3}}+3HCl\]
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