pBlock ElementsII 1 Question 4
4. The correct order of the oxidation states of nitrogen in NO, $\mathrm{NO} _{2}, \mathrm{NO} _{2}$ and $\mathrm{N} _{2} \mathrm{O} _{3}$ is
(2019 Main, 9 April I)
(a) $\mathrm{NO} _{2}<\mathrm{NO}<\mathrm{N} _{2} \mathrm{O} _{3}<\mathrm{N} _{2} \mathrm{O}$
(b) $\mathrm{N} _{2} \mathrm{O}<\mathrm{NO}<\mathrm{N} _{2} \mathrm{O} _{3}<\mathrm{NO} _{2}$
(c) $\mathrm{O} _{2}<\mathrm{N} _{2} \mathrm{O} _{3}<\mathrm{NO}<\mathrm{N} _{2} \mathrm{O}$
(d) $\mathrm{N} _{2} \mathrm{O}<\mathrm{N} _{2} \mathrm{O} _{3}<\mathrm{NO}<\mathrm{NO} _{2}$
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Answer:
Correct Answer: 4. (b)
Solution:
- The correct increasing order of oxidation state of nitrogen for nitrogen oxides is
$ \stackrel{+1}{\mathrm{N}} _{2} \mathrm{O}<\stackrel{+2}{\mathrm{N}} \mathrm{O}<\stackrel{+3}{\mathrm{N}} _{2} \mathrm{O} _{3}<\stackrel{+4}{\mathrm{N}} \mathrm{O} _{2} $
- Oxidation state of $\mathrm{N}$ in $\mathrm{N} _{2} \mathrm{O}$ is
$ \begin{aligned} & 2(x)-2=0 \\ & x=+\frac{2}{2}=+1 \end{aligned} $
- Oxidation state of $\mathrm{N}$ in $\mathrm{NO}$ is
$ \begin{aligned} x-2 & =0 \\ x & =+2 \end{aligned} $
- Oxidation state of $\mathrm{N}$ in $\mathrm{N} _{2} \mathrm{O} _{3}$ is
$ \begin{array}{r} 2 x+3(-2)=0 \\ x=\frac{6}{2}=3 \end{array} $
- Oxidation state of $\mathrm{N}$ in $\mathrm{NO} _{2}$ is
$ \begin{aligned} x+2(-2) & =0 \\ x-4 & =0 \\ x & =+4 \end{aligned} $
