Step-by-step analysis:

1. Analyze Statement 1: Pb ${}^{4+}$ compounds are stronger oxidising agents than Sn ${}^{4+}$ compounds.

• An oxidising agent is a substance that causes oxidation in another substance by accepting electrons and getting reduced itself. A stronger oxidising agent has a greater tendency to accept electrons.
• Both tin (Sn) and lead (Pb) are elements of Group 14 of the periodic table. The general electronic configuration for this group is $ns^2np^2$. They can exhibit oxidation states of +2 (by losing two p-electrons) and +4 (by losing both s and p valence electrons).
• Due to the inert pair effect, the stability of the lower oxidation state (+2) increases, and the stability of the higher oxidation state (+4) decreases as we move down the group. This effect is particularly significant for the heavier elements like Sn and Pb.
• For tin (Sn), the +4 oxidation state is more stable than the +2 oxidation state. Therefore, Sn${}^{4+}$ has a relatively low tendency to gain two electrons to form Sn${}^{2+}$. $Sn${}^{4+}$+ 2e${}^-$→ Sn${}^{2+}$$ (This reaction is not highly favorable as Sn${}^{4+}$ is stable).
• For lead (Pb), the heavier element in the group, the inert pair effect is very pronounced. The +2 oxidation state is much more stable than the +4 oxidation state.
• Consequently, Pb${}^{4+}$ has a very strong tendency to accept two electrons and get reduced to the more stable Pb${}^{2+}$ state. $Pb${}^{4+}$+ 2e${}^-$→ Pb${}^{2+}$$ (This reaction is highly favorable as Pb${}^{2+}$ is much more stable than Pb${}^{4+}$).
• Since Pb${}^{4+}$ has a much greater tendency to be reduced than Sn${}^{4+}$, Pb${}^{4+}$ compounds are stronger oxidising agents than Sn${}^{4+}$ compounds.
• Therefore, Statement 1 is True.

2. Analyze Statement 2: The higher oxidation states for the group 14 elements are more stable for the heavier members of the group due to 'inert pair effect'.

• This statement describes the trend of stability for the higher oxidation state (+4) in Group 14.
• The inert pair effect is the reluctance of the outermost s-electrons (the 'inert pair') to participate in chemical bonding for heavier p-block elements. This is due to the poor shielding of the nuclear charge by the intervening d- and f-orbitals.
• As a result of the inert pair effect, as we descend Group 14, the stability of the lower oxidation state (+2) increases, while the stability of the higher oxidation state (+4) decreases.
• The stability order for the +4 oxidation state is: C${}^{4+}$ > Si${}^{4+}$ > Ge${}^{4+}$ > Sn${}^{4+}$ > Pb${}^{4+}$.
• The stability order for the +2 oxidation state is: Ge${}^{2+}$ < Sn${}^{2+}$ < Pb${}^{2+}$.
• Statement 2 claims that the higher oxidation state (+4) becomes more stable for heavier members. This is the exact opposite of the actual trend caused by the inert pair effect.
• Therefore, Statement 2 is False.

3. Conclusion:

• Statement 1 is True.
• Statement 2 is False.
• Based on this analysis, the correct option is C.