The most characteristic oxidation state for lead and tin is, respectively:

In Group 14 elements, the general valence shell electronic configuration is $ns^2np^2$, allowing for +2 and +4 oxidation states. As we move down the group from Carbon to Lead, the stability of the +4 oxidation state decreases, while the stability of the +2 oxidation state increases due to the inert pair effect (the reluctance of the inner $ns^2$ electrons to participate in bonding due to poor shielding by $d$ and $f$ orbitals) [Source 19].

1. Lead (Pb): For lead, the inert pair effect is very strong. Consequently, the +2 oxidation state is its most characteristic and stable state. Compounds in the +4 state (like $PbO_2$) are strong oxidising agents as they tend to revert to the +2 state.
2. Tin (Sn): For tin, the +4 oxidation state is more stable and characteristic. Sn(II) compounds act as reducing agents because they readily lose electrons to form the stable Sn(IV) species.

Therefore, the correct order is +2 for Lead and +4 for Tin.