The orders of reactivity of ligands, $\mathrm{NMe_3}$ $\mathrm{PMe_3}$, and $\mathrm{CO}$ with complexes $\mathrm{MeTiCl}$, and $\mathrm{(CO)_5Mo\,(thf)}$ are
(a)$\mathrm{CO}>\mathrm{PMe_3}>\mathrm{NMe_3}$ and $\mathrm{CO}>\mathrm{NMe_3}>\mathrm{PMe_3}$,
(b)$\mathrm{PMe_3}>\mathrm{CO}>\mathrm{NMe_3}$ and $\mathrm{NMe_3}>\mathrm{CO}>\mathrm{PMe_3}$
(c)$\mathrm{NMe_3}>\mathrm{PMe_3}>\mathrm{CO}$ and $\mathrm{CO}>\mathrm{PMe_3}>\mathrm{NMe_3}$
(d)$\mathrm{NMe_3}>\mathrm{CO}>\mathrm{PMe_3}$ and $\mathrm{PMe_3}>\mathrm{NMe_3}>\mathrm{CO}$
Answer
Answer (as printed): C
Explanation
Since $\mathrm{MeTiCl_3}$ has no electrons for back donation with $\pi$-accepting ligands like $\mathrm{PMe_3}$ and CO, therefore it reacts in the order $\mathrm{NMe_3} > \mathrm{PMe_3} >$ CO.Now, in $(\mathrm{CO})\mathrm{Mo(thf)}$, Mo has sufficient $e^{-}$s and a vacant site to react with the $\pi$-acceptor ligand. Thus, the order is CO $> \mathrm{PMe_3} > \mathrm{NMe_3}$.