Question

Ammonium hydroxide is added first in a small quantity and then in excess to a solution of copper sulphate. Which of the following observations can be made?
(A) Pale blue / bluish white precipitate is formed which is soluble in excess of $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ and deep blue / inky blue solution is formed
(B) Pale blue / bluish white precipitate is formed which is soluble in excess of $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ and deep black solution is formed
(C) Pale red / reddish white precipitate is formed which is soluble in excess of $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ and deep blue / inky blue solution is formed
(D) Pale red / reddish white precipitate is formed which is soluble in excess of $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ and deep red / inky red solution is formed

Answer 1

The copper sulphate $\text{ CuS}{{\text{O}}_{\text{4}}}\text{ }$is a $\text{ C}{{\text{u}}^{\text{2+}}}\text{ }$salt. The $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ when added to the copper sulphate solution, the precipitate of copper hydroxide is formed. When treated with an excess of $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ the copper forms a complex with 4 molecule of ammonia to give tetraamminecopper (II) complex which enhances the colour of the solution.

Complete step by step solution:
When an aqueous solution of ammonia is added to the solution of copper (II) sulphate, a pale blue coloured precipitate of copper hydroxide is formed .this dissolves in the excess of the aqueous solution to give the deep blue colour solution containing the complex ion, tetraamminecopper (II).
The reaction is given as follows,
The ammonium hydroxide $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ dissociates into the solution as ammonia $\text{ N}{{\text{H}}_{\text{3}}}\text{ }$ and water molecule. The dissociation of ammonium hydroxide is as follows,
$\text{ N}{{\text{H}}_{\text{4}}}\text{OH}\left( \text{aq} \right)\text{ }\rightleftharpoons \text{ N}{{\text{H}}_{\text{3}}}\left( \text{aq} \right)\text{ + }{{\text{H}}_{\text{2}}}\text{O}\left( \text{l} \right)\text{ }$
The water-soluble copper sulphate $\text{ CuS}{{\text{O}}_{\text{4}}}\text{ }$salt produces the $\text{ C}{{\text{u}}^{\text{2+}}}\text{ }$ or the $\text{Cu (II)}$ and the $\text{ SO}_{4}^{2-}\text{ }$ ions as follows,
$\text{ CuS}{{\text{O}}_{\text{4}}}\text{(s) }\to \text{ C}{{\text{u}}^{\text{2+}}}(l)\,\text{ + SO}_{4}^{2-}(l)\text{ }$
The $\text{ C}{{\text{u}}^{\text{2+}}}\text{ }$ ions in the solution react with the ammonium hydroxide or with the ammonia in the solution. When ammonium hydroxide $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$ is added to the dropwise in the sulphate solution the copper ion $\text{ C}{{\text{u}}^{\text{2+}}}\text{ }$ reacts with the 2 molecules of ammonia. The copper reacts to form copper hydroxide.
$\text{ }\begin{matrix} \text{CuS}{{\text{O}}_{\text{4}}} & \+ & \text{2N}{{\text{H}}_{\text{4}}}\text{OH} & \to & \text{Cu}{{\left( \text{OH} \right)}_{\text{2}}}\text{ (}\downarrow \text{)} & \+ & {{\left( \text{N}{{\text{H}}_{\text{4}}} \right)}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} \\\ (\text{colourless}) & {} & {} & {} & (\text{Pale blue colour)} & {} & {} \\\ \end{matrix}\text{ }$
Copper hydroxide is a hydroxide of copper and a pale blue-greenish precipitate is formed.
On the further addition of ammonium hydroxide or ammonia molecules to the copper sulphate solution, the insoluble precipitate of copper hydroxides dissolves in the excess of $\text{N}{{\text{H}}_{\text{4}}}\text{OH}$. The copper forms a complex with four ammonia molecules. The complex formation is as shown below,

\text{C}{{\text{u}}^{\text{2+}}}\left( \text{aq} \right) & \+ & \text{4 N}{{\text{H}}_{\text{3}}}\left( \text{aq} \right) & \to & {{\left[ \text{Cu}{{\left( \text{N}{{\text{H}}_{\text{3}}} \right)}_{\text{4}}} \right]}^{\text{2+}}}\left( \text{aq} \right) \\\ {} & {} & {} & {} & (\text{Deep blue colour complex)} \\\ \end{matrix}$$ Or $$\text{ Cu}{{\left( \text{OH} \right)}_{\text{2}}}\text{ + 4N}{{\text{H}}_{\text{4}}}\text{OH }\to \text{ Cu}{{\left( \text{N}{{\text{H}}_{\text{3}}} \right)}_{\text{4}}}{{\left( \text{OH} \right)}_{\text{2}}}\text{ + 4}{{\text{H}}_{\text{2}}}\text{O }$$ The copper complex with the ammonia to form the tetraamminecopper (II) complex $${{\left[ \text{Cu}{{\left( \text{N}{{\text{H}}_{\text{3}}} \right)}_{\text{4}}} \right]}^{\text{2+}}}$$. This amine complex of copper is a deep blue colour complex. The overall reaction between the ammonium hydroxide and the copper sulphate is given as follows, $$\text{ CuS}{{\text{O}}_{\text{4}}}\left( \text{aq} \right)\text{ + 4 N}{{\text{H}}_{\text{4}}}\text{OH}\left( \text{aq} \right)\text{ }\to \text{ Cu}{{\left( \text{N}{{\text{H}}_{\text{3}}} \right)}_{\text{4}}}\text{S}{{\text{O}}_{\text{4}}}\left( \text{s} \right)\text{ + 4 }{{\text{H}}_{\text{2}}}\text{O}\left( \text{l} \right)\text{ }$$ The reaction proceeds with the formation of a pale blue coloured precipitate when treated with the excess of ammonium hydroxide the deep blue coloured solution is formed. **Hence (A) is the correct option.** **Note:** The copper has an ability to form a complex. Most of the complex compounds are coloured. This is because the tetraamminecopper (II) has the unpaired electron which is responsible for the absorption of radiation. The tetraamminecopper (II) sulphate is complex. Here, the copper (Ii) is bound to the four unidentate ammonia molecules. The coordination number of copper (II) is 4. The complex is square planar and has one unpaired electron.