Question

Which of the following does not hydrolyse in water,
$1.PC{l_3}$
$2.N{F_3}$
$3.NaC{l_3}$
$4.B{F_3}$

Answer 1

Hydrolysis reaction is that reaction in which water is a solvent and the compound is solvated (gets hydrolysed) or breaks down to give other new compounds. Hydrolyses have some conditions according to which we can determine whether a compound will get hydrolysed in water or it cannot be hydrolysed.

Complete answer:
Now we know that Hydrolysis reaction is that reaction in which water is a solvent and the compound is solvated (gets hydrolysed) or breaks down to give other new compounds.
In order to know that an inorganic compound will undergo hydrolysis in water or not, we need to know about the conditions of hydrolysis:

1. First condition is that there must be an electron deficient centre with low energy vacant orbitals so that it can accept the electrons and get hydrolysed.
2. Second condition is that there should not be any steric crowding in the molecule.
   Now, if we see the given molecule:
   $1.PC{l_3}$
   In $PC{l_3}$ molecule phosphorus atom is electropositive and has low lying vacant orbitals and also it does not have any steric crowding. So, it can be hydrolysed.
   $2.N{F_3}$
   In the $N{F_3}$ molecule , the Nitrogen atom is electropositive but does not have low lying vacant orbitals. So, it cannot be hydrolysed.
   $3.NaC{l_3}$
   Ionic compounds undergo hydration.
   $4.B{F_3}$
   In $B{F_3}$ molecule boron atom is electropositive and has low lying vacant orbitals and also it does not have any steric crowding. So, it can be hydrolysed.
   Hence, we can conclude that $N{F_3}$ do not get hydrolyse in water.

Note:
We should note that the electropositive element must have low lying vacant orbitals so that it can accept electron pairs and gets hydrolysed. There should not be any steric crowding in the molecule otherwise the coming nucleophile (hydroxyl ion in this case) will not get easily attached to the central atom for hydrolysis.