To determine which metal is the weakest reducing agent among the given options ($Rb$, $Na$, $Li$, $K$), we need to understand the concept of reducing agents and their tendencies in terms of electrochemical series.
In the electrochemical series, metals are arranged according to their standard electrode potentials. The metal with a higher negative standard electrode potential is a stronger reducing agent as it readily loses electrons to form cations. Conversely, the metal with a less negative or more positive standard electrode potential is a weaker reducing agent.
| Metal | Standard Electrode Potential (V) |
|---|---|
| \(Li\) | -3.04 |
| \(Rb\) | -2.92 |
| \(K\) | -2.92 |
| \(Na\) | -2.71 |
From the table above, we see that the standard electrode potential of $Na$ is less negative compared to the other metals listed. This indicates that $Na$ is less likely to lose electrons and hence, is the weakest reducing agent among the given options.
Conclusion: The weakest reducing agent among $Rb$, $Na$, $Li$, and $K$ is \(Na\).
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 