Question

In the reaction of ${ KCN }$ and ${ CuSO }_{ 4 }$ solution:

(a) ${ KCN }$ acts as a reducing agent
(b) ${ KCN }$ acts as a complexing agent
(c) ${ K }_{ 3 }{ [Cu }{ (CN) }_{ 4 }{ ] }$ is formed
(d) All of the above are correct

Answer 1

Hint: A reducing agent is a substance that reduces other substances, especially by donating an electron or electrons is known as reducing agents.
Complexing agents:- A coordination complex, consists of an atom or an ion, and a surrounding array of bound molecules or anions which is called as complexing agents or ligands.


Complete answer step-by-step:

The following reactions will takes place when ${ KCN }$ reacts with ${ CuSO }_{ 4 }$;
${ 2 } { KCN } + { CuSO }_{ 4 } → { Cu }{ (CN) }_{ 2 }+{ K }_{ 2 }{ SO }_{ 4 }$
${ 2 } { Cu }{ (CN) }_{ 2 }→ { { Cu }_{ 2 } }{ (CN) }_{ 2 } + { (CN) }_{ 2 }$
${ { Cu }_{ 2 } }{ (CN) }_{ 2 }+{ 6 } { KCN } → { 2 }{ K }_{ 3 }{ [Cu }{ (CN) }_{ 4 }{ ] }$
                                                                                             (Stable complex)
It is a three step reaction:
Firstly, cupric cyanide will be formed.
Secondly, cupric cyanide will decompose to form cuprous cyanide.
Thirdly, when cuprous cyanide reacts with potassium cyanide, a stable complex will be formed.

Here, ${ KCN }$ acts as a reducing agent as it causes reduction in copper sulphate. Also, ${ KCN }$ acts as a complexing agent (A molecule or ion of non-metal forms a coordinate bond with a metal atom or ion).
When ${ KCN }$ reacts with ${ CuSO }_{ 4 }$ reacts then tri potassium tetracyanocuprate(I), a stable complex will be formed.
${ KCN+CuSO_{ 4 } }\rightarrow { K }_{ 3 }{ [Cu(CN) }_{ 4 }{ ] }$

Additional Information:

Complexing agent:- These are the substances that are used to form a complex compound with another material in the solution.
The ${ Cu }$ in ${ K }_{ 3 }{ [Cu }{ (CN) }_{ 4 }{ ] }$ present in +1 oxidation state with electronic configuration of ${ 3d }^{ 10 }$.
The hybridization will be ${ sp }^{ 3 }$ as Cu forms ${ 4 }$ coordinate bonds with ${ 4 }$ cyanide ligands. Therefore, the geometry will be tetrahedral and it will be diamagnetic due to the absence of paired electrons.

Note:
The possibility of making a mistake is that you may confuse the oxidation state. Here, the reduction of Cu takes place and oxidation states change from +2 to 0.