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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Solution:
- The correct increasing order of oxidation state of nitrogen for nitrogen oxides is
$$ \stackrel{+1}{\mathrm{~N}}{2} \mathrm{O}<\stackrel{+2}{\mathrm{~N}} \mathrm{O}<\stackrel{+3}{\mathrm{~N}} \mathrm{O}{2}<\stackrel{+4}{\mathrm{~N}} \mathrm{O}_{2} $$
- 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
$$ \begin{aligned} x+2(-2) & =0 \ x-4 & =0 \ x & =+4 \end{aligned} $$
