Question

The compound $\mathit{W}{\mathit{H}}_{\mathbf{2}}{\left({\mathit{C}}_5{\mathit{H}}_5\right)}_{\mathbf{2}}$acts as a base, but $\mathit{T}\mathit{a}{\mathit{H}}_{\mathbf{3}}{\left({\mathit{C}}_5{\mathit{H}}_5\right)}_{\mathbf{2}}$ does not. Explain.

Short answer

The compound $W{H}_2{\left(C_5{H}_5\right)}_2$acts as a base, but $Ta{H}_3{\left(C_5{H}_5\right)}_2$does not.

Step-by-step solution

Take the WH2(C5H5)2 compound

Let consider first,$W{H}_2{\left(C_5{H}_5\right)}_2$ compound, the tungsten has donated two electrons to form a pair of ${\left(C_5{H}_5\right)}^{-}$ions; it becomes a $d^4$ compound (ion). While in the case of $Ta{H}_3{\left(C_5{H}_5\right)}_2$tantalum becomes compound (ion) when donating two electrons to form a pair of ${\left(C_5{H}_5\right)}^{-}$ions.

Take the TaH3(C5H5)2 compound

Furthermore, when tungsten shares two electrons with bonded H atoms to form a single bonds, it has two electrons left to form a single bond with an approaching $H^{+}$ion; it is thus a good base. By contrast, tantalum has already shared its three valence electrons with three H atoms and has little affinity for an additional$H^{+}$ ion.

Hence, the compound $W{H}_2{\left(C_5{H}_5\right)}_2$acts as a base, but $Ta{H}_3{\left(C_5{H}_5\right)}_2$does not.