Question

Ammonia is used in the detection of $C{u^{2 + }}$ ion because:
A.) aqueous solution of $N{H_3}$ reacts with $C{u^{2 + }}$ ion to form a deep blue coloured complex.
B.) $N{H_3}$ reacts with $C{u^{2 + }}$ ion to give blue precipitate of $CuO$.
C.) aqueous solution of $N{H_3}$ reacts with $C{u^{2 + }}$ ion to form white coloured complex.
D.) $N{H_3}$ reacts with $C{u^{2 + }}$ ion to give green precipitate.

Answer 1

Hint: To solve this question we need to know that ammonia that is $(N{H_3})$ is a lewis base and when ammonia reacts with the copper it forms a deep blue colored complex because of the lone pair of ammonia.

Complete step by step answer:
As given in question, ammonia has formula $N{H_3}$. Also we know that ammonia acts as a lewis base because it can donate an electron pair which is present on nitrogen. This is because nitrogen has a valency of five but only three hydrogens are attached to it and two remaining electrons are shown as lone pairs. Due to the presence of lone pair ammonia has the ability to form complex compounds.
The $C{u^{2 + }}$ can be in any form of solution. We will take it in the form of copper$(II)$ chloride that can be written as $CuC{l_2}$. When the ammonia solution is added to the light blue solution of copper chloride, a light blue precipitate is obtained. This reaction can be shown as:
$CuC{l_2} + 2N{H_3} + 3{H_2}O \to Cu{(OH)_2}(s) + 2NH_4^ + Cl$
When the obtained copper$(II)$ hydroxide reacts with excess ammonia, the copper ions again go back into the solution and form a deep blue ammonia complex. This explained reaction can be shown as:
$Cu{(OH)_2} + 4N{H_3}(aq) \to {[Cu{(N{H_3})_4}]^{2 + }}(aq) + 2O{H^ - }$
  Blue Deep blue complex
Here, in the above given reaction ${[Cu{(N{H_3})_4}]^{2 + }}$ is the deep blue coloured complex. This deep blue colouration in this reaction indicates the presence of $C{u^{2 + }}$ ions. Hence, ammonia is used for the detection of $C{u^{2 + }}$.
Thus, we can conclude that option A is the correct answer.


Note:
In case of ammonia we need to remember that it acts as a lewis acid due to the presence of lone pairs of electrons on it. And thus it has the ability to form coordinate bonds with electron deficient molecules or with transition metal cations to form complex compounds.