Question

Why \[{\text{NC}}{{\text{l}}_{\text{3}}}\] and \[{\text{PC}}{{\text{l}}_{\text{3}}}\] are hydrolyzed differently?

Answer 1

Hint: We know that the bond between nitrogen and chlorine in \[{\text{NC}}{{\text{l}}_{\text{3}}}\] is nonpolar and the bond between phosphorus and chlorine in \[{\text{PC}}{{\text{l}}_{\text{3}}}\] is polar. As water is the polar molecule, the hydrolysis of the polar and nonpolar compound is different in it.

Complete step by step answer:
First, we will find the hydrolysis of \[{\text{NC}}{{\text{l}}_{\text{3}}}\]. We know that electronegativity (EN) of nitrogen is approximately 3. We also know that electronegativity (EN) of chlorine is approximately 3. So, the electronegativity difference between nitrogen and chlorine is zero. Hence, the bond between chlorine and nitrogen in \[{\text{NC}}{{\text{l}}_{\text{3}}}\] is non-polar.
As nitrogen belongs to the second period of the modern periodic table, it does not have an empty d-orbital to accept electrons from the oxygen of water molecules. Instead, during hydrolysis, nitrogen donates its lone pair of electrons to the hydrogen atom of water to form \[{\text{N}}{{\text{H}}_{\text{3}}}\].
We can show the hydrolysis of \[{\text{NC}}{{\text{l}}_{\text{3}}}\] as follows.
\[{\text{NC}}{{\text{l}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{N}}{{\text{H}}_{\text{3}}}{\text{ + 3HOCl}}\]
Now, we will find out about the hydrolysis of \[{\text{PC}}{{\text{l}}_{\text{3}}}\]. We know that electronegativity (EN) of chlorine is approximately 3. It is also known that electronegativity (EN) of Phosphorus is approximately 2.1. So, the electronegativity difference between phosphorus and chlorine is not zero. Hence, the bond between chlorine and phosphorus in \[{\text{PC}}{{\text{l}}_{\text{3}}}\] is polar.
As phosphorus belongs to the third period of the modern periodic table it has the empty d-orbital to accept electrons from the oxygen of water molecules. So, we came to know that during hydrolysis phosphorous accepts lone pair of electrons from the oxygen atom of water to form \[{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}\].
We can show hydrolysis of \[{\text{PC}}{{\text{l}}_{\text{3}}}\] as follows.
\[{\text{PC}}{{\text{l}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}} \to {{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}{\text{ + HCl}}\]
Therefore, we can conclude that \[{\text{NC}}{{\text{l}}_{\text{3}}}\] and \[{\text{PC}}{{\text{l}}_{\text{3}}}\] are hydrolyzed differently.

Note:
Here, the electronegativity difference between atoms leads to the formation polar bond and if the electronegativity difference is not much then, the bond is nonpolar. Nitrogen does not have an empty d orbital. So, it does not accept the electrons whereas, phosphorus has an empty d orbital for acceptance of electrons if necessary.