A) -3 to +3
B) -3 to zero
C) -3 to + 5
D) 0 to -3
Correct Answer: A
Solution :
\[\underset{\begin{smallmatrix} manganese \\ oxide(black)I \end{smallmatrix}}{\mathop{\operatorname{Mn}{{O}_{2}}}}\, + {{N}_{2}}S{{O}_{4}} + 2HCl\,\,\xrightarrow{{}}\,\,MnS{{O}_{4}}\] \[+2{{H}_{2}}O\underset{\begin{smallmatrix} chlorine\,\,gas \\ (greenish\,\,yellow) \end{smallmatrix}}{\mathop{+C{{l}_{2}}}}\,\] \[\underset{ammonia}{\mathop{N{{H}_{3}}}}\,+3C{{l}_{2}}\xrightarrow{{}}\underset{\begin{smallmatrix} Nitrogen\,\,trichloride \\ (unstable\,\,and\,\,explosive) \end{smallmatrix}}{\mathop{NC{{l}_{3}}+3HCl}}\,\] The oxidation state of N in \[N{{H}_{3}}\] is -3 while in \[NC{{l}_{3}}\] it is +3. Hence, in given process, oxidation state of nitrogen changed from -3 to +3.
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