Answer
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Hint: $AgCl$ is an ionic compound but it is insoluble in water. It defies the aqueous law of solubility. $AgCl$ has a very strong bond between silver and chlorine ions. They have a very strong attraction between them. $AgCl$ precipitate is white in colour.
Complete answer:
$CrCl_3.5NH_3$ is a coordinate compound. now a coordinate compound formed by d block elements. What happens is that the d block element loses their electrons. The ligands, hugely negative electrons, attack the empty orbitals. These are highly electronegative and distort the orbital. This is called hybridisation.
The chromium has a total twenty-four number of electrons. Then it has six electrons in its last subshell, which are divided in the d orbital and the s orbital. There are only six electrons and these all occupy the six orbitals and the all are alone, unpaired. As we all know that the most stable electronic configuration of an element is when the electrons are all paired or all are unpaired.
$CrCl_3.5NH_3$ has eight ligands and we know that the chromium can lose two, three or six electrons, so there will be only six ligands, the formula would be like $[Cr{(N{H_3})_5}Cl]C{l_2}$
Here ammonia is present in five coordinate spheres and one chlorine in coordinate spheres. The other two chlorine ions are present in the ionization sphere.
The reaction of the two compounds would go as the follows so as to form the precipitate of silver chloride. Silver chloride forms readily due to high bond strength between silver and chlorine.
$[Cr{(N{H_3})_5}Cl]C{l_2} + 2AgN{O_3} \to [Cr{(N{H_3})_5}Cl](N{O_3}) + 2AgCl$
On balancing the equation, we get the above reaction. The nitrate ion of silver nitrate only occupies the position of the chlorine present in the ionisation sphere.
There are two moles of silver chloride precipitate present in the solution.
Note:
The three silver halides which form precipitates in the water solution, all have different colour precipitate. The silver bromide forms a creamy coloured precipitate while the silver iodide forms a very unique yellow coloured precipitate.
Complete answer:
$CrCl_3.5NH_3$ is a coordinate compound. now a coordinate compound formed by d block elements. What happens is that the d block element loses their electrons. The ligands, hugely negative electrons, attack the empty orbitals. These are highly electronegative and distort the orbital. This is called hybridisation.
The chromium has a total twenty-four number of electrons. Then it has six electrons in its last subshell, which are divided in the d orbital and the s orbital. There are only six electrons and these all occupy the six orbitals and the all are alone, unpaired. As we all know that the most stable electronic configuration of an element is when the electrons are all paired or all are unpaired.
$CrCl_3.5NH_3$ has eight ligands and we know that the chromium can lose two, three or six electrons, so there will be only six ligands, the formula would be like $[Cr{(N{H_3})_5}Cl]C{l_2}$
Here ammonia is present in five coordinate spheres and one chlorine in coordinate spheres. The other two chlorine ions are present in the ionization sphere.
The reaction of the two compounds would go as the follows so as to form the precipitate of silver chloride. Silver chloride forms readily due to high bond strength between silver and chlorine.
$[Cr{(N{H_3})_5}Cl]C{l_2} + 2AgN{O_3} \to [Cr{(N{H_3})_5}Cl](N{O_3}) + 2AgCl$
On balancing the equation, we get the above reaction. The nitrate ion of silver nitrate only occupies the position of the chlorine present in the ionisation sphere.
There are two moles of silver chloride precipitate present in the solution.
Note:
The three silver halides which form precipitates in the water solution, all have different colour precipitate. The silver bromide forms a creamy coloured precipitate while the silver iodide forms a very unique yellow coloured precipitate.
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