A) \[{{\text{N}}_{\text{2}}}{{\text{H}}_{\text{4}}}\]
B) \[\text{N}{{\text{H}}_{3}}\]
C) \[{{\text{N}}_{\text{3}}}\text{H}\]
D) \[\text{N}{{\text{H}}_{\text{2}}}\text{OH}\]
Correct Answer: C
Solution :
Let the oxidation state of nitrogen in the given compounds be x. \[{{\overset{x}{\mathop{N}}\,}_{2}}{{\overset{x1}{\mathop{H}}\,}_{4}}\] \[2(x)+(+1)4=0\] \[2x=-4\] \[\therefore \] \[x=-2\] \[\overset{x}{\mathop{N}}\,{{\overset{+1}{\mathop{H}}\,}_{3}}\] \[x+(+1)3=0\] \[\therefore \] \[x=-3\] \[{{\overset{x}{\mathop{N}}\,}_{3}}\overset{+1}{\mathop{H}}\,\] \[(x)3+(+1)=0\] \[3x=-1\] \[\therefore \] \[x=-\frac{1}{3}\] \[\overset{x}{\mathop{N}}\,\overset{+1}{\mathop{{{H}_{2}}}}\,\overset{-2+1}{\mathop{OH}}\,\] \[x+(+1)2+(-2)+(+1)=0\] \[x+2-2+1=0\] \[x+1=0\] \[x=-1\] Thus, oxidation state of nitrogen is highest in\[{{N}_{3}}H.\]
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