On reaction with $\mathrm{Cl}_2$, phosphorus forms two types of halides ' A ' and ' B '. Halide ' $A$ ' is yellowish-white powder but halide ' $B$ ' is colourless oily liquid. Identify A and B and write the formulae of their hydrolysis products.
Phosphorus on reaction with $\mathrm{Cl}_2$ forms two types of halides $A$ and $B$. ' $A$ ' is $\mathrm{PCl}_5$ and ' $B$ ' is $\mathrm{PCl}_3$.
$$\begin{aligned} \mathrm{P}_4+10 \mathrm{Cl}_2 & \longrightarrow 4 \mathrm{PCl}_5 \\ \mathrm{P}_4+6 \mathrm{Cl}_2 & \longrightarrow 4 \mathrm{PCl}_3 \end{aligned}$$
When 'A' and 'B' are hydrolysed
(a) $\underset{\begin{array}{c}\text { [A] } \\ \text { Phosphorus } \\ \text { pentachloride }\end{array}}{\mathrm{PCl}_5}+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\begin{array}{c}\text { Phosphoric } \\ \text { acid }\end{array}}{\mathrm{H}_3 \mathrm{PO}_4}+5 \mathrm{HCl}$
(b) $\underset{\begin{array}{c}\text { [B] } \\ \text { Phosphorus } \\ \text { trichloride }\end{array}}{\mathrm{PCl}_3}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\begin{array}{c}\text { Phosphorus } \\ \text { acid }\end{array}}{\mathrm{H}_3 \mathrm{PO}_3}+3 \mathrm{HCl}$
In the ring test of $\mathrm{NO}_3^{-}$ion, $\mathrm{Fe}^{2+}$ ion reduces nitrate ion to nitric oxide, which combines with $\mathrm{Fe}^{2+}(\mathrm{aq})$ ion to form brown complex. Write the reactions involved in the formation of brown ring.
$\mathrm{NO}_3^{-}+3 \mathrm{Fe}^{2+}+4 \mathrm{H}^{+} \longrightarrow \mathrm{NO}+3 \mathrm{Fe}^{3+}+2 \mathrm{H}_2 \mathrm{O}$
$\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}+\mathrm{NO} \longrightarrow \underset{\text { Brown ring }}{\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{NO}\right]^{2+}}+\mathrm{H}_2 \mathrm{O}$
This test is known as brown ring test of nitrates generally used to identify the presence of nitrate ion in given solution.
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.