Question

$SnCl_{ 2 }$ acts as a reducing agent because:
(A) $SnCl_{ 2 }$ can accept electrons readily
(B) $Sn^{ 2+ }$ is more stable than $Sn^{ 4+ }$
(C) $Sn^{ 4+ }$ is more stable than $Sn^{ 2+ }$
(D) $Sn^{ 2+ }$ can be easily converted to metallic tin

Answer 1

Hint: You should know that the reducing agent is an element or compound that loses or donates an electron to an electron recipient in a redox chemical reaction. Now you just need to find the correct option according to this statement.

Complete step by step answer:
We can prove that $SnCl_{ 2 }$ acts as a reducing agent by the following statements-

* Sn has electronic configuration [Kr] $4d^{ 10 }5s^{ 2 }5p^{ 2 }$
* Sn doesn't exhibit inert pair effect and thus higher oxidation state is more stable.
* $Sn^{ 4+ }$ is more stable than $Sn^{ 2+ }$.
* $SnCl_{ 2 }$ readily reacts with other compounds and gets converted to a stable $SnCl_{ 4 }$.
* During the reaction, $SnCl_{ 2 }$ gets oxidized while another compound is reduced.

Thus, we can say $SnCl_{ 2 }$ is a good reducing agent.

Therefore, we can conclude that the correct answer to this question is option C.

Additional information:
We know Sn and Pb are the members of the 4th group of the periodic table (carbon family). The valence shell electronic configuration of these elements is $ns^{ 2 }np^{ 2 }$ type. All these elements contain four electrons in the valence shell. These elements show variable oxidation states of +2 and +4.
From top to bottom, the lower oxidation state is more stable than the higher one due to the inert pair effect. Thus Pb, due to greater inert pair effect, shows +2 as a stable oxidation state rather than +4. Thus by accepting two electrons $Pb^{ 4+ }$ will get converted into $Pb^{ 2+ }$. Hence $Pb^{ 4+ }$ by undergoing self-reduction acts as an oxidizing agent.

Note: We can define the inert pair effect as the non-participation of the two s electrons in bonding due to the high energy needed for unpairing them.