Among the following, the number of halide(s) which is/are inert to hydrolysis is ______
(A) $$BF_3$$
(B) $$SiCl_4$$
(C) $$PCl_5$$
(D) $$SF_6$$

We need to determine which of the given halides is/are inert (resistant) to hydrolysis.

(A) $$BF_3$$: Boron trifluoride is readily hydrolysed by water. Boron, being electron-deficient, acts as a Lewis acid and reacts with water. So $$BF_3$$ is not inert to hydrolysis.

(B) $$SiCl_4$$: Silicon tetrachloride is easily hydrolysed because silicon has vacant 3d orbitals that can accommodate the lone pair from water, facilitating the hydrolysis reaction. So $$SiCl_4$$ is not inert to hydrolysis.

(C) $$PCl_5$$: Phosphorus pentachloride is vigorously hydrolysed by water. Phosphorus has vacant d-orbitals and the molecule reacts readily with water. So $$PCl_5$$ is not inert to hydrolysis.

(D) $$SF_6$$: Sulfur hexafluoride is kinetically inert to hydrolysis despite the thermodynamic feasibility of the reaction. This is because the six fluorine atoms sterically shield the sulfur atom so completely that water molecules cannot access it. The compact size of fluorine atoms and the octahedral arrangement create an effective steric barrier. So $$SF_6$$ is inert to hydrolysis.

Therefore, the number of halides inert to hydrolysis is $$\mathbf{1}$$ (only $$SF_6$$).