Question

On heating, lead(II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless solid/liquid (B). (B) on heating with NO changes to a blue solid (C). The oxidation number of nitrogen in solid (C) is:

• A. +3
• B. +4
• C. +2
• D. +5

Answer 1

The correct option is A

+3

Explanation for correct answer:-

Option (A) +3

Step 1: Heating o Lead nitrate

When Lead nitrate$\left({\mathrm{PbNO}}_3\right)$ is heated, it produces Lead oxide$\left(\mathrm{PbO}\right)$, Nitrogen dioxide, and Oxygen.

This nitrogen dioxide is the compound ‘A’

$\underset{\mathrm{Lead}\mathrm{nitrate}}{\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)}\stackrel{∆}{\to }\underset{\mathrm{Lead}\mathrm{oxide}}{\mathrm{PbO}\left(\mathrm{s}\right)}+\underset{\mathrm{Nitrogen}\mathrm{dioxide}}{2{\mathrm{NO}}_2\left(\mathrm{g}\right)(\mathrm{brown}\mathrm{gas})}+\underset{\mathrm{Oxygen}}{\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})}$

Step 2: Cooling of Nitrogen dioxide

Now this compound ‘A’ that is Nitrogen dioxide on cooling forms Dinitrogen tetroxide, which is the compound ‘B’

$\underset{\mathrm{Nirogen}\mathrm{dioxide}}{2{\mathrm{NO}}_2\left(\mathrm{g}\right)}\stackrel{\mathrm{cooling}}{\to }\underset{\mathrm{Dinitrogen}\mathrm{tetroxide}}{{\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)}$

Step 3: Heating of Dinitrogen tetroxide

This compound ‘B’ which is the Dinitrogen tetroxide is now finally heated with Nitric oxide$\left(\mathrm{NO}\right)$.

This produces a blue solid Dinitrogen trioxide$\left({\mathrm{N}}_2{\mathrm{O}}_3\right)$ that is our required compound C.

$\underset{\mathrm{Dinitrogen}\mathrm{tetroxide}}{{\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)}\stackrel{∆}{\to }\underset{\mathrm{Dinitrogen}\mathrm{trioxide}}{{\mathrm{N}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)}$

Step 4: Calculating the oxidation number of Nitrogen

Let us assume the oxidation number of Nitrogen be $\mathrm{x}$

We know that the oxidation number of oxygen is $-2$.

$$\begin{gathered} \therefore 2\mathrm{x}+(3\times -2)=0 \\ \Rightarrow 2\mathrm{x}-6=0 \\ \Rightarrow 2\mathrm{x}=6 \\ \Rightarrow \mathrm{x}=+3 \end{gathered}$$

Therefore, the oxidation number of Nitrogen in compound ‘C’ which is Dinitrogen trioxide$\left({\mathrm{N}}_2{\mathrm{O}}_3\right)$ is $+3$