Find whether the following statement is true or false.
$SnC{l_2}$ is a better reducing agent than $HgC{l_2}$ .
A) True
B) False

Hint:We will calculate the oxidation states of Tin and Mercury for the given compounds. Check for the stability of Tin and Mercury in these oxidation states and compare them to the stabilities when they are oxidized. The compound that has more stability at higher oxidation states will be a better reducing agent.


Complete solution:
An oxidizing agent is a compound that can easily take up electrons and a reducing agent is a compound that can take up electrons.Firstly, let us find the oxidation state of Tin in $SnC{l_2}$
Since the oxidation state of Chlorine is $ - 1$ and there are two Chlorines, the oxidation state of Tin would be $ + 2$ .Similarly, we can also calculate the oxidation state of Mercury in $HgC{l_2}$ which is also $ + 2$.
Since, both the oxidation states are equal, we find out the oxidation states in which Tin and Mercury can exist. Tin can exist in both $ + 2$ and $ + 4$ oxidation states whereas Mercury’s oxidation can only extend up to $ + 1$ and $ + 2$
Because Tin is more stable at higher oxidation states, it can easily be oxidized and therefore is a better reducing agent compared to $HgC{l_2}$

Therefore, from the above explanation we can say that the correct option is (A).


Note:Compounds that have better stability at higher oxidation states tend to reduce other compounds to reach that higher oxidation state. Same is the case with Tin in $SnC{l_2}$ . Tin is more stable at $ + 4$ oxidation state than it is at $ + 2$ oxidation state so the compound $SnC{l_2}$ is a stronger reducing agent.