A) \[\text{N}{{\text{H}}_{\text{4}}}\text{N}{{\text{O}}_{\text{2}}}\]
B) \[{{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{2}}\text{S}{{\text{O}}_{4}}\]
C) \[\text{Ba(}{{\text{N}}_{\text{3}}}{{\text{)}}_{\text{2}}}\]
D) \[{{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{C}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}\]
Correct Answer: B
Solution :
\[{{\text{(N}{{\text{H}}_{\text{4}}}\text{)}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\xrightarrow{\Delta }\text{N}{{\text{H}}_{\text{3}}}\text{+}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\] \[Ba{{({{N}_{3}})}_{2}}\xrightarrow{\Delta }Ba+3{{N}_{2}}\uparrow \] \[{{(N{{H}_{4}})}_{2}}C{{r}_{2}}{{O}_{7}}\xrightarrow{\Delta }{{N}_{2}}\uparrow +\underset{(Green)}{\mathop{C{{r}_{2}}{{O}_{3}}}}\,+4{{H}_{2}}O\] \[N{{H}_{4}}N{{O}_{2}}\xrightarrow{\Delta }{{N}_{2}}\uparrow +2{{H}_{2}}O\]
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