Among the following, the most characteristic oxidation states for lead and tin are, respectively:

Group 14 elements exhibit oxidation states of +2 and +4. As we move down the group, the stability of the higher oxidation state (+4) decreases and the stability of the lower oxidation state (+2) increases due to the inert pair effect (reluctance of s-electrons to participate in bonding).

1. Lead (Pb): Being at the bottom of the group, the inert pair effect is most pronounced. Consequently, the +2 oxidation state is more stable than the +4 state. This is supported by the fact that $Pb^{4+}$ is a strong oxidising agent ($E^o = +1.67 V$ for $Pb^{4+} \rightarrow Pb^{2+}$), indicating it readily reduces to the stable $Pb^{2+}$ form .
2. Tin (Sn): Tin lies above Lead. While it shows both +2 and +4 states, the +4 state is more characteristic and stable. Tin(II) acts as a reducing agent ($Sn^{2+}$ oxidises to $Sn^{4+}$), whereas Lead(II) is stable.

Therefore, the characteristic oxidation states are +2 for Lead and +4 for Tin.