Out of $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}_2 \mathrm{S}$, which one has higher bond angle and why?
Bond angle of $\mathrm{H}_2 \mathrm{O}\left(\mathrm{H}-\mathrm{O}-\mathrm{H}=104.5^{\circ}\right)$ is larger than that of $\mathrm{H}_2 \mathrm{S}\left(\mathrm{H}-\mathrm{S}-\mathrm{H}=92^{\circ}\right)$ because oxygen is more electronegative than sulphur therefore, bond pair electron of $\mathrm{O}-\mathrm{H}$ bond will be closer to oxygen and there will be more bond pair-bond pair repulsion between bond pairs of two $\mathrm{O}-\mathrm{H}$ bonds.
$\mathrm{SF}_6$ is known but $\mathrm{SCl}_6$ is not. Why?
Fluorine atom is smaller in size so, sixF- ions can surround a sulphur atom. The case is not so with chlorine atom due to its large size. $\mathrm{So}^{-} \mathrm{SF}_6$ is known but $\mathrm{SCl}_6$ is not known due to interionic repulsion between larger $\mathrm{Cl}^{-}$ions.
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
$$\mathrm{HClO}<\mathrm{HClO}_2<\mathrm{HClO}_3<\mathrm{HClO}_4$$