Mention the type of compounds formed when small atoms like $\mathrm{H}, \mathrm{C}$ and $\mathrm{N}$ get trapped inside the crystal lattice of transition metals. Also give physical and chemical characteristics of these compounds.
When small atoms like $\mathrm{H}, \mathrm{C}$ and $\mathrm{N}$ get trapped inside the crystal lattice of transition metals.
(a) Such compounds are called interstitial compounds.
(b) Their characteristic properties are;
(i) They have high melting points, higher than those of pure metals.
(ii) They are very hard.
(iii) They retain metallic conductivity.
(iv) They are chemically inert.
(a) Transition metals can act as catalysts because these can change their oxidation state. How does Fe (III) catalyse the reaction between iodide and persulphate ions?
(b) Mention any three processes where transition metals act as catalysts.
(a) Reaction between iodide and persulphate ions is
$2 \mathrm{I}^{-}+\mathrm{S}_2 \mathrm{O}_8^{2-} \xrightarrow{\mathrm{Fe}(\mathrm{III})} \mathrm{I}_2+2 \mathrm{SO}_4^{2-}$
$$\begin{aligned} &\text { Role of } \mathrm{Fe} \text { (III) ions }\\ &\begin{aligned} 2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} & \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2 \\ 2 \mathrm{Fe}^{2+}+\mathrm{S}_2 \mathrm{O}_8^{2-} & \longrightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{SO}_4^{2-} \end{aligned} \end{aligned}$$
(b) (i) Vanadium $(\mathrm{V})$ oxide used in contact process for oxidation of $\mathrm{SO}_2$ to $\mathrm{SO}_3$.
(ii) Finely divided iron in Haber's process in conversion of $\mathrm{N}_2$ and $\mathrm{H}_2$ to $\mathrm{NH}_3$.
(iii) $\mathrm{MnO}_2$ in preparation of oxygen from $\mathrm{KClO}_3$.
A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. $\mathrm{H}_2 \mathrm{SO}_4$ and NaCl , chlorine gas is liberated and a compound (D) of manganese alongwith other products is formed. Identify compounds A to D and also explain the reactions involved.
Since, compound $(\mathrm{C})$ on treating with conc. $\mathrm{H}_2 \mathrm{SO}_4$ and NaCl gives $\mathrm{Cl}_2$ gas, so it is manganese dioxide $\left(\mathrm{MnO}_2\right)$. It is obtained alongwith $\mathrm{MnO}_4^{2-}$ when $\mathrm{KMnO}_4$ (violet) is heated.
Thus,
$$\begin{array}{ll} (A)=\mathrm{KMnO}_4 & (B)=\mathrm{K}_2 \mathrm{MnO}_4 \\ (C)=\mathrm{MnO}_2 & (D)=\mathrm{MnCl}_2 \end{array}$$
$\underset{[A]}{\mathrm{KMnO}_4} \xrightarrow{\Delta} \underset{[B]}{\mathrm{K}_2 \mathrm{MnO}_4}+\underset{[C]}{\mathrm{MnO}_2}+\mathrm{O}_2$
$2 \mathrm{MnO}_2+4 \mathrm{KOH}+\mathrm{O}_2 \longrightarrow 2 \mathrm{~K}_2 \mathrm{MnO}_4+2 \mathrm{H}_2 \mathrm{O}$
$$\mathrm{MnO}_2+4 \mathrm{NaCl}+4 \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \underset{[D]}{\mathrm{MnCl}_2}+4 \mathrm{NaHSO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{Cl}_2$$