White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of HCl obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water.
Equations for the reactions
$\because \quad 124 \mathrm{~g}$ of white phosphorus produces $\mathrm{HCl}=438 \mathrm{~g}$
$\therefore 62 \mathrm{~g}$ of white phosphorus will produces
$$\mathrm{HCl}=\frac{438}{124} \times 62=219.0 \mathrm{~g} \mathrm{~HCl}$$
Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.
Three oxoacids of nitrogen having oxidation state +3 are
(a) $\mathrm{HNO}_2$, nitrous acid
(b) $\mathrm{HNO}_3$, nitric acid
(c) Hyponitrous acid, $\mathrm{H}_2 \mathrm{~N}_2 \mathrm{O}_2$
In $\mathrm{HNO}_2, \mathrm{~N}$ is in +3 oxidation state
Disproportionation reaction
Nitric acid forms an oxide of nitrogen on reaction with $\mathrm{P}_4 \mathrm{O}_{10}$. Write the reaction involved. Also write the resonating structures of the oxide of nitrogen formed.
$\mathrm{P}_4 \mathrm{O}_{10}$ being a dehydrating agent, on reaction with $\mathrm{HNO}_3$ removes a molecule of water and forms anhydride of $\mathrm{HNO}_3$.
$$4 \mathrm{HNO}_3+\mathrm{P}_4 \mathrm{O}_{10} \longrightarrow 4 \mathrm{HPO}_3+2 \mathrm{~N}_2 \mathrm{O}_5$$
Resonating structures of $\mathrm{N}_2 \mathrm{O}_5$ are
(i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white red and black phosphorus on the basis of their structure and reactivity.
Phosphorus has three allotropic forms -
| White phosphorus | Red phosphorus | Black phosphorus | |
|---|---|---|---|
| 1. | It is less stable form of P | More stable than white P. | It is most stable form of P |
| 2. | It is highly reactive. |
Less reactive than white P. |
It is least reactive. |
| 3. | It has regular tetrahedron structure. | It has polymeric structure. | It has a layered structure. |
Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.
Effect of concentration of nitric acid on the formation of oxidation product can be understood by its reaction with conc $\mathrm{HNO}_3$. Dilute and concentrated nitric acid give different oxidation products on reaction with copper metal.
$$\begin{gathered} 3 \mathrm{Cu}+8 \mathrm{HNO}_3 \text { (Dil.) } \longrightarrow 3 \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}+4 \mathrm{H}_2 \mathrm{O} \\ \mathrm{Cu}+4 \mathrm{HNO}_3 \text { (Conc.) } \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{NO}_2+2 \mathrm{H}_2 \mathrm{O} \end{gathered}$$