Question

Complete and balance the following chemical equations: \begin{center} \includegraphics[width=10cm]{31ii.png} \end{center}

Hint:

- KMnO$_4$ decomposition is a key reaction in the production of oxygen in labs. - Dichromate-Iodide reaction is a classic redox reaction where Cr$^{6+}$ is reduced, and I$^-$ is oxidized to I$_2$. - Always balance chemical equations by ensuring both mass and charge conservation.

Answer 1

CuCl$_2$ is more stable than Cu$_2$Cl$_2$ in aqueous solution because Cu$^{2+}$ is more stable than Cu$^+$ due to its higher hydration enthalpy ($\Delta_{\text{hyd}} H^\circ$). In aqueous solution, Cu$^+$ undergoes disproportionation, as shown in the equation: \[ 2Cu^+(aq) \rightarrow Cu^{2+}(aq) + Cu(s) \] (ii) Write the general electronic configuration of f-block elements. % Solution % Solution Solution: The general electronic configuration of f-block elements is: \[ (n-2)f^{1-14} (n-1)d^{0-1} ns^2 \] (iii) Predict which of the following will be coloured in aqueous solution and why? Sc$^{3+$, Fe$^{3+}$, Zn$^{2+}$} [Atomic numbers: Sc = 21, Fe = 26, Zn = 30] % Solution % Solution Solution: Among the given ions, Fe$^{3+}$ is coloured in aqueous solution because it has unpaired electrons in its d-orbital, which allows d-d transitions. On the other hand: - Sc$^{3+}$ has an empty d-orbital (d$^0$ configuration), so no d-d transition occurs → Colourless. - Zn$^{2+}$ has a fully filled d-orbital (d$^{10}$ configuration), so no d-d transition occurs → Colourless. (iv) How can you obtain potassium dichromate from sodium chromate? % Solution % Solution Solution: Potassium dichromate (K$_2$Cr$_2$O$_7$) can be obtained from sodium chromate (Na$_2$CrO$_4$) through the following reactions: \[ 2Na_2CrO_4 + 2H^+ \rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O \] \[ Na_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl \] (v) Why do transition metals and their compounds show catalytic activities? % Solution % Solution Solution: Transition metals and their compounds show catalytic activity because: - They can exhibit variable oxidation states, allowing them to form intermediate complexes during reactions. - They can adsorb reactants onto their surface, increasing reaction rates. - Their large surface area provides active sites for catalytic activity.