Identify Correct Statements About Transition Metal Compounds and Catalyst Processes

To determine the correct statements, let us evaluate each statement individually:

1. Statement A: Basic vanadium oxide is used in the manufacture of $\text{H}_2\text{SO}_4$.

   • In the Contact Process for the manufacture of sulfuric acid ($\text{H}_2\text{SO}_4$), vanadium pentoxide ($\text{V}_2\text{O}_5$) is used as a catalyst.
   • The acid-base nature of vanadium oxides changes with the oxidation state of vanadium: $\text{VO}$ and $\text{V}_2\text{O}_3$ are basic, $\text{VO}_2$ is amphoteric, and $\text{V}_2\text{O}_5$ is amphoteric (reacts with both acids and bases).
   • Since $\text{V}_2\text{O}_5$ is an amphoteric oxide rather than a basic oxide, Statement A is incorrect.

2. Statement B: The spin-only magnetic moment value of the transition metal halide employed in Ziegler-Natta polymerization is $2.84\text{ BM}$.

   • The Ziegler-Natta catalyst typically comprises a transition metal halide, titanium tetrachloride ($\text{TiCl}_4$), and a co-catalyst like triethylaluminium ($\text{Al(C}_2\text{H}_5)_3$).
   • In $\text{TiCl}_4$, titanium is in the $+4$ oxidation state ($\text{Ti}^{4+}$).
   • Ground state electronic configuration of $\text{Ti}$ ($Z=22$) is $[\text{Ar}] 3\text{d}^2 4\text{s}^2$. Therefore, for $\text{Ti}^{4+}$, the configuration is $[\text{Ar}] 3\text{d}^0$.
   • As the number of unpaired electrons $n = 0$, the spin-only magnetic moment is: $\mu_{\text{s}} = \sqrt{n(n+2)} = 0\text{ BM}$
   • Thus, Statement B is incorrect.

3. Statement C: The p-block metal compound employed in Ziegler-Natta polymerization has the metal in $+3$ oxidation state.

   • The p-block compound in the Ziegler-Natta catalyst is triethylaluminium, $\text{Al(C}_2\text{H}_5)_3$.
   • Aluminium ($\text{Al}$) is a Group 13 p-block metal, and in $\text{Al(C}_2\text{H}_5)_3$, its oxidation state is $+3$.
   • Thus, Statement C is correct.

4. Statement D: The number of electrons present in the outer most 'd' orbital of metal halide employed in Wacker process is $8$.

   • In the Wacker process, palladium(II) chloride ($\text{PdCl}_2$) is used as the primary catalyst along with $\text{CuCl}_2$.
   • In $\text{PdCl}_2$, palladium is in the $+2$ oxidation state ($\text{Pd}^{2+}$).
   • Ground state electronic configuration of $\text{Pd}$ ($Z=46$) is $[\text{Kr}] 4\text{d}^{10} 5\text{s}^0$.
   • Removal of two electrons to form $\text{Pd}^{2+}$ yields the configuration $[\text{Kr}] 4\text{d}^8$.
   • Thus, there are $8$ electrons in the $4\text{d}$ orbital, so Statement D is correct.

Hence, the correct statements are C and D Only.

Correct Option: C and D Only