Question

Both $N{F}_3$ and $NC{l}_3$ are covalent but they differ in the extent of hydrolysis because :

• A. $N{F}_3$ is more stable than $NC{l}_3$ and hydrolysis product of $N{F}_3,HOF$, does not exist.
• B. dipole moment of $N{F}_3$ is greater than that of $NC{l}_3$.
• C. electrongativity of F is greater than that of $Cl$.
• D. $Cl$ can expand its octet by using d-orbitals.

Answer 1

The correct option is A $N{F}_3$ is more stable than $NC{l}_3$ and hydrolysis product of $N{F}_3,HOF$, does not exist.

In contrast to $N{F}_3,NC{l}_3$ and $NB{r}_3$ are highly unstable.

Expect $N{F}_3$ and $P{F}_3$, trihailds are hydrolysed by water and the ease of hydrolysis decreases down the group, i.e., $NC{l}_3$ is easily hydrolysed and $SbC{l}_3$ and $BiC{l}_3$ are partly hydrolysed.

$NC{l}_3+3H_{2}O\to 3HOCl+N{H}_3$

$PC{l}_3+3H_{2}O\to H_{3}P{O}_3+3HCl$