Identify the oxidizing agent and the reducing agent in the following redox reaction.
\[7C{N^ - } + 2O{H^ - } + 2Cu{(N{H_3})_4}^{2 + } \to 2Cu{(CN)_3}^{2 - } + 8N{H_3} + CN{O^ - } + {H_2}O\]
(A) Oxidising agent: $Cu{(N{H_3})_4}^{2 + }$ ; Reducing agent: $CN{O^ - }$
(B) Oxidising agent: $Cu{(N{H_3})_4}^{2 + }$ ; Reducing agent: $C{N^ - }$
(C) Oxidising agent: $CN{O^ - }$ ; Reducing agent: $C{N^ - }$
(D) None of these

Hint: The compound that undergoes oxidation in the reaction is called the reducing agent and the compound that undergoes reduction during the reaction is called an oxidizing agent. Calculate the oxidation state of all atoms before and after the reaction in order to identify the oxidising and reducing agent.

Complete step by step solution:
We know that in a redox reaction, oxidation and reduction reaction occurs simultaneously. That means some compounds are reduced and some are oxidized. We will check which reagents are oxidising agents and which are reducing ones.
-Oxidising agents are the compounds that undergo reduction during the reaction. Reducing agents are the compounds that undergo oxidation during the reaction.
-To find the oxidation number of an atom in the molecule, we will add the oxidation numbers of all the atoms which are there in the molecule and that sum will be equal to the overall charge of the molecule.
So, for $C{N^ - }$,
Overall charge on $C{N^ - }$ = Oxidation number of C + Oxidation number of N
-1 = Oxidation number of C + (-3)
Oxidation number of C = +2
Now, in $CN{O^ - }$,
Overall charge on $CN{O^ - }$ = Oxidation number of C + Oxidation number of N + Oxidation number of O
-1 = Oxidation number of C + (-3) + (-2)
Oxidation number of C = +4
-So, we can see that carbon is oxidized from +2 to +4 oxidation state. So, we can say that oxidation of $C{N^ - }$ has occurred here in the reaction. So, $C{N^ - }$ is the reducing agent.
In $Cu{(N{H_3})_4}^{2 + }$,
Overall charge on $Cu{(N{H_3})_4}^{2 + }$= Oxidation number of Cu + 4(Overall charge on $N{H_3}$)
+2 = Oxidation number of Cu + 4(0)
The oxidation number of Cu = +2
Now, in $Cu{(CN)_3}^{2 - }$,
Overall charge on $Cu{(CN)_3}^{2 - }$= Oxidation number of Cu + 3(Overall charge on $C{N^ - }$)
-2 = Oxidation number of Cu + 3(-1)
The oxidation number of Cu = +1
-So here we can see that reduction of copper occurs here from +2 to +1 state. Therefore there is a reduction of $Cu{(N{H_3})_4}^{2 + }$ during the reaction. So, $Cu{(N{H_3})_4}^{2 + }$ is called the oxidizing agent.
Thus, we can say that the oxidizing agent is $Cu{(N{H_3})_4}^{2 + }$ and the reducing agent is $C{N^ - }$.

Therefore, the correct answer is (B).

Note: Remember that the process of losing electrons is called oxidation and the process of gain of the electrons is called reduction. Also, note that $N{H_3}$ is a neutral molecule and it does not have any charge on it. It forms bonds with the metal by the lone pair of nitrogen atoms.