1. Analyze transformation (A): $CO_2(s) \to CO_2(g)$

This is sublimation, i.e. solid directly to gas.

• Hence it is a phase transition $\Rightarrow (p)$
• Sublimation requires heat, so $\Delta H > 0$ $\Rightarrow (r)$
• Disorder increases from solid to gas, so $\Delta S > 0$ $\Rightarrow (s)$

So, $A \to p,r,s$

2. Analyze transformation (B): $CaCO_3(s) \to CaO(s) + CO_2(g)$

This is a decomposition reaction, not a phase transition.

• Thermal decomposition requires heat, so $\Delta H > 0$ $\Rightarrow (r)$
• A gas is produced from solids, so randomness increases, hence $\Delta S > 0$ $\Rightarrow (s)$

So, $B \to r,s$

3. Analyze transformation (C): $2H \to H_2(g)$

Two hydrogen atoms combine to form one molecule.

• Number of gaseous particles decreases, so entropy decreases.
• Therefore $\Delta S < 0$ $\Rightarrow (t)$

Also, bond formation is exothermic, so $\Delta H < 0$; hence $(r)$ is not correct.

So, $C \to t$

4. Analyze transformation (D): $P(\text{white, solid}) \to P(\text{red, solid})$

This is conversion of one allotrope of phosphorus into another.

• Hence it is an allotropic change $\Rightarrow (q)$
• It is not a phase transition because both are solids of different structures, not different physical states.
• White phosphorus is more ordered molecular solid? Red phosphorus is polymeric and thermodynamically more stable; entropy decreases in going to the more ordered network form, so $\Delta S < 0$ $\Rightarrow (t)$

So, $D \to q,t$

5. Match with options

We obtained:

• $A \to p,r,s$
• $B \to r,s$
• $C \to t$
• $D \to q,t$

This matches Option D.

6. Compare with stored correct answer

Stored correct answer = A

But option A says for $D$: $p,q,t$, which incorrectly includes phase transition $(p)$. Allotropy is not a phase transition.

Therefore, the stored answer appears incorrect.