Select correct statement

All statements are correct

Explanation:

• (a) In the decomposition of an oxide into oxygen and solid/liquid metal, entropy increases: When a compound like an oxide decomposes, the solid or liquid oxide breaks down into its metal component and oxygen gas. This process leads to an increase in entropy (∆S), as the gaseous state of oxygen has significantly more disorder compared to the solid or liquid state. This aligns with the thermodynamic principle that gaseous products contribute to a higher entropy in a chemical reaction.
• (b) Decomposition of an oxide is an endothermic change: The decomposition of oxides typically requires the input of heat energy, making it an endothermic process. For example, in the decomposition of mercury(II) oxide (HgO): HgO → Hg (liquid) + ½ O₂ (gas) Heat energy is absorbed to break the bonds in HgO, demonstrating the endothermic nature of the reaction.
• (c) To make ∆G° negative, the temperature should be high enough so that T∆S° > ∆H°: According to Gibbs free energy equation, ∆G° = ∆H° - T∆S°, where: - ∆G°: Free energy change - ∆H°: Enthalpy change - T∆S°: Temperature times the entropy change For a process to be spontaneous, ∆G° must be negative. In the case of oxide decomposition, increasing the temperature raises the T∆S° term, eventually making it greater than ∆H°, ensuring ∆G° becomes negative.

Thus, all the given statements are correct, supporting the selection of option (D).