Explain why the stability of oxoacids of chlorine increases in the order given below.
$$\mathrm{HClO}<\mathrm{HClO}_2<\mathrm{HClO}_3<\mathrm{HClO}_4$$
Oxygen is more electronegative than chlorine, therefore dispersal of negative charge present on chlorine increases from $\mathrm{ClO}^{-}$ to $\mathrm{ClO}_4^{-}$, ion because number of oxygen atoms attached to chlorine is increasing. Therefore, stability of ions will increase in the order given below
$$\mathrm{ClO}^{-}<\mathrm{ClO}_2^{-}<\mathrm{ClO}_3^{-}<\mathrm{ClO}_4^{-}$$
Due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the same order
$$\mathrm{HClO}<\mathrm{HClO}_2<\mathrm{HClO}_3<\mathrm{HClO}_4$$
Explain why ozone is thermodynamically less stable than oxygen?
Ozone is thermodynamically less stable than oxygen because its decomposition into oxygen results in the liberation of heat ( $\Delta H$ is negative) and an increase in entropy ( $\Delta S$ is positive). These two effects reinforce each other, resulting in large negative Gibbs energy change $(\Delta G)$ for its conversion into oxygen.
$\mathrm{P}_4 \mathrm{O}_6$ reacts with water according to equation $\mathrm{P}_4 \mathrm{O}_6+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{H}_3 \mathrm{PO}_3$. Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of $\mathrm{P}_4 \mathrm{O}_6$ in $\mathrm{H}_2 \mathrm{O}$.
$\mathrm{P}_4 \mathrm{O}_6+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 4 \mathrm{H}_3 \mathrm{PO}_3\quad\text{.... (i)}$
Neutralisation $$ \left.\mathrm{H}_3 \mathrm{PO}_3+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_2 \mathrm{HPO}_3+2 \mathrm{H}_2 \mathrm{O}\right] \times 4\quad\text{.... (ii)}$$
Adding Eqs. (i) and (ii)
$$\mathrm{\mathop {{P_4}{O_6}}\limits_{1\,mol} + \mathop {8NaOH}\limits_{8\,mol}}\longrightarrow 4 \mathrm{Na}_2 \mathrm{HPO}_3+2 \mathrm{H}_2 \mathrm{O}\quad\text{.... (iii)}$$
Number of moles of $\mathrm{P}_4 \mathrm{O}_6$,
$$n=\frac{m}{M}=\frac{1.1}{220}=\frac{1}{200} \mathrm{~mol}$$
$\left(\right.$ Molar mass of $\mathrm{P}_4 \mathrm{O}_6=(4 \times 31)+(6 \times 16)=220$
$\because$ Product formed by 1 mole of $\mathrm{P}_4 \mathrm{O}_6$ is neutralised by 8 moles NaOH
$\therefore$ Product formed by $\frac{1}{200}$ moles of $\mathrm{P}_4 \mathrm{O}_6$ will be neutralised by NaOH
$$=8 \times \frac{1}{200}=\frac{8}{200} \text { mole } \mathrm{NaOH}$$
Given, $\quad$ Molarity of $\mathrm{NaOH}=0.1 \mathrm{M}=0.1 \mathrm{~mol} / \mathrm{L}$
$$\begin{aligned} & \text { Molarity }=\frac{\text { Number of moles }}{\text { Volume in litres }} \\ & \text { Volume }=\frac{\text { Number of moles }}{\text { Molarity }}=\frac{8}{200} \times \frac{1}{0.1}=0.4 \mathrm{~L} \text { or } 400 \mathrm{~mL} \end{aligned}$$
$\therefore \quad 400 \mathrm{~mL} \mathrm{~NaOH}$ is required.
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
