Describe what is seen when excess ammonia solution is added gradually to copper(II)chloride solution. Give the formula of the complex ion formed.

A copper(II) salt like CuCl2 will dissolve in water and
form a complex hexaaquacopper ion as shown below:

CuCl2 +
6H2O  --> [Cu(H2O)6]2+ + 2Cl-

The hexaaquacopper ion
is a complex ion meaning that it is an ion with a metal atom in the center (in this case
copper) surrounded by several coordinating ligands (in this case 6 water molecules). 
The entire complex has a 2+ charge and is highly soluble in water, thus giving rise to
the light blue color associated with copper salts in water.  When ammonia (NH3) is added
gradually, the initial reaction that occurs is an acid/base reaction where the ammonia
acts as a base and converts 2 of the water molecules to hydroxide ions.  This is shown
in the equation below:

[Cu(H2O)6]2+ + 2NH3  --> 
[Cu(H2O)4(OH)2] + 2(NH4)+

As you can see, 2 of the water
molecules have been deprotonated by the ammonia to form 2 hydroxide (OH) ions.  This
[Cu(H2O)4(OH)2] is a neutral complex ion meaning it has no charge.  It will precipitate
from the solution as a light blue solid.

As excess ammonia
is added to the solution, a new reaction will occur where the ammonia acts as a ligand
and will replace water in the complex ion as shown
below:

[Cu(H2O)6]2+ + 4NH3  -->  [Cu(NH3)4(H2O)2]2+
+ 4H2O

This tetraammoniadiaquacopper complex ion is highly
soluble in water and also a very deep blue color.  So as more ammonia is added to the
solution the inital light blue precipitate [Cu(H2O)4(OH)2] will dissolve as
[Cu(NH3)4(H2O)2]2+ becomes the only complex ion present and the solution turns a very
deep blue.