To determine in which of the given equilibria $\mathrm{K}_p$ and $$\mathrm{K}_{\mathrm{c}}$$ are not equal, it's important to understand the relationship between these two equilibrium constants. This relationship is expressed by the equation:

$$\mathrm{K}_p = \mathrm{K}_c (RT)^{\Delta n}$$

where $R$ is the gas constant, $T$ is the temperature in Kelvin, and $\Delta n$ is the change in the number of moles of gas (number of moles of gaseous products minus number of moles of gaseous reactants).

If $\Delta n = 0$, then $\mathrm{K}_p$ and $$\mathrm{K}_{\mathrm{c}}$$ are equal because $(RT)^0 = 1$. However, if $\Delta n \neq 0$, the constants will not be the same, and the degree to which they differ will depend on the temperature and the value of $\Delta n$.

Now, let's analyze each option:

• Option A: $$\mathrm{PCl}_{5(\mathrm{~g})} \rightleftharpoons \mathrm{PCl}_{3(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}$$

Reactant side moles = 1, Product side moles = 2; $\Delta n = 2 - 1 = 1$.

• Option B: $$\mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})}$$

Reactant side moles = 2, Product side moles = 2; $\Delta n = 2 - 2 = 0$.

• Option C: $$\mathrm{CO}_{(\mathrm{g})}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2(\mathrm{~g})}$$

Reactant side moles = 2, Product side moles = 2; $\Delta n = 2 - 2 = 0$.

• Option D: $$2 \mathrm{BrCl}_{(\mathrm{g})} \rightleftharpoons \mathrm{Br}_{2(\mathrm{~g})}+\mathrm{Cl}_{2(\mathrm{~g})}$$

Reactant side moles = 2, Product side moles = 2; $\Delta n = 2 - 2 = 0$.

From this analysis, it is evident that $\mathrm{K}_p$ and $$\mathrm{K}_{\mathrm{c}}$$ are not equal in Option A where $\Delta n = 1$. In all other options, since $\Delta n = 0$, $\mathrm{K}_p$ is equal to $$\mathrm{K}_{\mathrm{c}}$$. Thus, the correct answer is Option A.