Question

Why is $PbC{{l}_{4}}$ is less stable than $SnC{{l}_{4}}$, $PbC{{l}_{2}}$ is more stable than $SnC{{l}_{2}}$?

Answer 1

Hint: Both Pb and Sn are soft metals and generally form tetrachlorides in the form of $MC{{l}_{4}}$. Before we go on and answer our question, we need to know about the inert pair effect.

Complete step by step solution:

Inert pair effect: Inert pair effect is defined as the reluctance of the s-electrons of the valence shell to take part in bonding. This effect happens due to the poor shielding of the $ns^2$ electrons in the valence shell by the intervening d and/or f-electrons. Inert pair effect increases as we move down a group. So, the elements present in the lower part of the group show lower oxidation states which is 2 units less than the highest group oxidation state.

Now coming to our question both Pb and Sn belong to the Carbon family or Group 14 which have the electronic configuration of $ns^{2}np^{6}$ and show a general oxidation state of +4. As we move down the group from C to Pb, the stability of +4 oxidation state decreases while that of +2 oxidation state increases due to the same inert pair effect. Till Sn, +4 oxidation state is stable but after that it starts decreasing, making +2 the new stable oxidation state. Therefore, for Pb +2 oxidation state becomes more stable than +4 oxidation state and vice-versa for Sn.

Thus, $PbC{{l}_{4}}$is less stable than $SnC{{l}_{4}}$ whereas $PbC{{l}_{2}}$ is more stable than $SnC{{l}_{2}}$.

Note: We should take note that Sn can exhibit both +2 and +4 oxidation states but +4 oxidation states are more stable. $SnC{{l}_{4}}$ is stable and $SnC{{l}_{2}}$ is a strong reducing agent. $PbC{{l}_{4}}$ is an unstable molecule and doesn’t exist at room temperature.