Question

“\[{\rm{AgCl}}\] dissolve in an ammonia solution due to the formation of a complex”.
Answer whether the above statement is true or false.
If true enter 1, else enter 0.

Answer 1

Hint:
As we know that the compounds formed have bonding strength, there are some forces which bind the atoms or molecules together. When the lattice energy is greater than the hydration energy, compounds do not break into its ions when they dissolve in water.

Complete step by step solution
Silver chloride is a salt of silver nitrate and hydrochloric acid. This salt is a white precipitate which does not dissolve in water and get settled down. This is due to silver being a second- row transition metal and it is small in size and the silver cation has great polarization effect which polarises chloride ion and form covalent bonding. Therefore, the force of attraction between silver ion and chloride ion is very strong and the compound is a solid precipitate.

Now, when \[{\rm{AgCl}}\] dissolves in ammonia it forms complex as \[\left[ {{\rm{Ag(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}} \right]\]shown below.
\[{\rm{AgCl(s) + }}\,{\rm{2N}}{{\rm{H}}_{\rm{3}}}{\rm{(l)}} \to {\left[ {{\rm{Ag(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}} \right]^{\rm{ + }}}{\rm{ + C}}{{\rm{l}}^{\rm{ - }}}\]

The formation of \[\left[ {{\rm{Ag(N}}{{\rm{H}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}} \right]\] complex is the spontaneous process in which the entropy of the above reaction decreases and the enthalpy of formation of complex is negative thus, the Gibbs free energy is negative and hence, the silver chloride is dissolved in ammonia.

Hence, the correct answer is true that is 1.

Note:
According to Fajan’s rule the smaller the cation and bigger the anion, more will be the polarization effect and there will be covalent bonding.
If cation and anion both are bigger than ionic bonding will be there.