A) \[{{N}_{2}}\,to\text{ }N{{O}_{2}}\]
B) \[{{N}_{2}}{{H}_{4}}to\text{ }{{N}_{2}}{{O}_{3}}\]
C) \[NO\text{ }to\,HN{{O}_{3}}\]
D) \[N{{H}_{3\,}}to\text{ }NO\]
Correct Answer: B
Solution :
\[\overset{(+1)}{\mathop{H{{g}_{2}}}}\,\,\,\overset{(-2)}{\mathop{S}}\,\,\,\,\to \,\,H{{g}^{+\,2}}+{{\overset{(+\,6)}{\mathop{S{{O}^{-2}}}}\,}_{4}}\]\ \[n-factor\,\,\,\to \,\,H{{g}^{+2}}=\left( +1\text{ }to+2 \right)\times 2\] \[=\text{ }2e{{\,}^{-}}loss\] \[S=\left( -2\text{ }to+6 \right)=8e{{\,}^{-}}loss\] \[n-factor\text{ }=10\] Also, \[\overset{(-2)}{\mathop{{{N}_{2}}{{H}_{4}}}}\,\,\,\to \,\,\overset{(+3)}{\mathop{{{N}_{2}}{{O}_{3}}}}\,\] n-factor \[=\,\,\left( -2\text{ }to+3 \right)\times 2=10\text{ }e{{\,}^{-}}loss\]You need to login to perform this action.
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