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Question: In the reaction of \({ KCN }\) and \({ CuSO }_{ 4 }\) solution: (a) \({ KCN }\) acts as a reduc...

In the reaction of KCN{ KCN } and CuSO4{ CuSO }_{ 4 } solution:

(a) KCN{ KCN } acts as a reducing agent
(b) KCN{ KCN } acts as a complexing agent
(c) K3[Cu(CN)4]{ K }_{ 3 }{ [Cu }{ (CN) }_{ 4 }{ ] } is formed
(d) All of the above are correct

Explanation

Solution

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{ KCN } reacts with CuSO4{ CuSO }_{ 4 };
2KCN+CuSO4→Cu(CN)2+K2SO4{ 2 } { KCN } + { CuSO }_{ 4 } → { Cu }{ (CN) }_{ 2 }+{ K }_{ 2 }{ SO }_{ 4 }
2Cu(CN)2→Cu2(CN)2+(CN)2{ 2 } { Cu }{ (CN) }_{ 2 }→ { { Cu }_{ 2 } }{ (CN) }_{ 2 } + { (CN) }_{ 2 }
Cu2(CN)2+6KCN→2K3[Cu(CN)4]{ { 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{ KCN } acts as a reducing agent as it causes reduction in copper sulphate. Also, KCN{ 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{ KCN } reacts with CuSO4{ CuSO }_{ 4 } reacts then tri potassium tetracyanocuprate(I), a stable complex will be formed.
KCN+CuSO4→K3[Cu(CN)4]{ 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{ Cu } in K3[Cu(CN)4]{ K }_{ 3 }{ [Cu }{ (CN) }_{ 4 }{ ] } present in +1 oxidation state with electronic configuration of 3d10{ 3d }^{ 10 }.
The hybridization will be sp3{ sp }^{ 3 } as Cu forms 4{ 4 } coordinate bonds with 4{ 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.