Animated Solution for Chemistry - s and p-Block Elements: Among the following, the number of halide(s) which is/are inert to hydrolysis is ............... . (i) ${\text{BF}}_3$ (ii) ${\text{SiCl}}_4$ (iii) ${\text{PCl}}_5$ (iv) ${\text{SF}}_6$

When we talk about the hydrolysis of covalent halides, we are essentially looking at a nucleophilic substitution reaction. The water molecule ($H_{2}O$) acts as a nucleophile, seeking a positively charged or electron-deficient center to attack.

For this attack to be successful, two golden rules must be satisfied. First, the central atom of the halide must possess vacant orbitals (such as empty $p$ or $d$ orbitals) to accept the lone pair of electrons donated by the oxygen atom in water. Second, the central atom must be sterically accessible. If the central atom is heavily guarded by bulky surrounding atoms, the water molecule simply cannot reach it, no matter how many vacant orbitals are waiting inside.

Let's evaluate the first three candidates: $B{F}_3$, $SiC{l}_4$, and $PC{l}_5$.

In Boron trifluoride ($B{F}_3$), Boron is $s{p}^2$ hybridized and has an empty $2p$ orbital ready to accept electrons. In Silicon tetrachloride ($SiC{l}_4$) and Phosphorus pentachloride ($PC{l}_5$), both Silicon and Phosphorus have vacant $3d$ orbitals. More importantly, the spatial arrangement of atoms in these molecules leaves enough room for a water molecule to approach and initiate the attack. Consequently, all three of these halides are easily hydrolysed.

Now, we turn our attention to Sulphur hexafluoride ($S{F}_6$). At first glance, Sulphur, being in the third period, definitely possesses vacant $3d$ orbitals. One might assume it should undergo hydrolysis just like $SiC{l}_4$ or $PC{l}_5$.

However, the reality is quite different. In $S{F}_6$, the central Sulphur atom is surrounded by six highly electronegative Fluorine atoms arranged in a perfect octahedral geometry. Fluorine atoms, though small individually, create a remarkably dense and tightly packed electron cloud around the Sulphur atom when six of them are present.

When a water molecule attempts to approach the Sulphur atom to donate its lone pair, it crashes into this dense wall of negative charge. The physical crowding prevents the water molecule from ever reaching the vacant $d$ orbitals. This phenomenon is known as severe steric hindrance.

Because the water molecule is sterically blocked from attacking the central Sulphur atom, the hydrolysis reaction is completely shut down. $S{F}_6$ stands as a classic example of a molecule that is kinetically inert despite being thermodynamically capable of reacting.

Therefore, out of the four given options, only one halide is inert to hydrolysis, and that is $S{F}_6$.