Question

Assertion (A): \[S{F_6}\] cannot be hydrolyzed but \[S{F_4}\] can be.
Reason(R): Six F- atoms in \[S{F_6}\] prevent the attack of \[{H_2}O\] on the Sulphur atom of \[S{F_6}\].

Answer 1

Hint: \[S{F_6}\] is electrically neutral, and \[S{F_4}\] is surrounded by only 4 fluorine atoms. \[S{F_6}\] is very stable molecule which is \[{d^2}s{p^3}\] hybridised and its geometry is octahedral, its dipole moment is zero and its atom is highly tense that repel the incoming reacting molecules of water. \[S{F_6}\] is chemically inert. These properties are helpful in determining the changes that SF₆ and SF₄ undergo in different reactions.

Complete step by step solution:
In Sulphur hexafluoride the f-atom is connected with Sulphur atoms and act like a shield and this is the reason behind \[S{F_6}\] is chemically not reactive with water. That is why, \[S{F_6}\] cannot be hydrolyzed. In case of \[S{F_4}\], its molecule has sp3d hybridization and its shape is see-saw that provides little steric hindrance. Here the Sulphur atom is surrounded by only 4 f-atoms, which is why the water molecule is free to attack on Sulphur tetrafluoride. Water molecules are not sterically hindered.

In Sulphur tetrafluoride is a vacant orbital and this molecule is not fully coordinated. hence, \[S{F_4}\] can be
hydrolyzed.
Here, assertion is correct and reason is also correct. Reason is correctly explaining the assertion.
Additional information: Hydrolysis generally needs the use of an acid or base catalyst. Hydrolysis means breaking down with water. This process is chemical decomposition in this bond is broken by addition of water. Hydrolysis can be reduced by increasing the value of ph.

Note: Loan pair of fluorine and loan pair of oxygen in water repel each other in \[S{F_6}\]. Hydrolysis is not a spontaneous process and the rate of hydrolysis depends on pH.