A) \[{{N}_{3}}H\]
B) \[N{{H}_{2}}OH\]
C) \[{{N}_{2}}{{H}_{4}}\]
D) \[N{{H}_{3}}\]
Correct Answer: A
Solution :
\[3\times x+1\,(1)=0\] \[3x+1=0\] \[3x=-1,\,\Rightarrow x=-\frac{1}{3}\,\,\text{in}\,\,{{N}_{3}}H\] \[x+2\,(+1)+1\,(-2)+1(1)=0\] \[x=-1\,\,\text{in}\,\,N{{H}_{2}}OH\] \[x\times 2+4\,(1)=0\]\[x=-\frac{4}{2}=-2\,\text{in}\,\,{{N}_{2}}{{H}_{4}}\] \[x+3\,(1)=0\]\[x=-3\,\,in\,\,N{{H}_{3}}\] Hence, highest in \[{{N}_{3}}H\].
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