Elements involved belong to Group 14

Group 14 elements are: $\text{C, Si, Ge, Sn, Pb}$

We need to compare first ionization energies.

Relevant periodic trend

Down a group, first ionization energy generally decreases because:

• atomic size increases,
• shielding increases,
• valence electron is farther from the nucleus.

However, due to poor shielding by d- and f-electrons and inert pair effect, some irregularities occur in heavier p-block elements.

Check Statement (I):

Statement (I): The first ionization energy of Pb is greater than that of Sn.

Compare: $\text{Sn} : [Kr]4d^{10}5s^25p^2$ $\text{Pb} : [Xe]4f^{14}5d^{10}6s^26p^2$

Normally, going down the group from Sn to Pb, ionization energy should decrease. But in Pb, the poor shielding of intervening $d$ and especially $f$ electrons increases effective nuclear charge felt by the outer electrons.

Hence, Pb has slightly higher first ionization enthalpy than Sn.

So, Statement (I) is true.

Check Statement (II):

Statement (II): The first ionization energy of Ge is greater than that of Si.

Compare: $\text{Si} : [Ne]3s^23p^2$ $\text{Ge} : [Ar]3d^{10}4s^24p^2$

Going down the group from Si to Ge, size increases and ionization energy generally decreases. Although $d$-electron shielding causes some irregularity later, for Si to Ge the first ionization energy of Ge is slightly lower than that of Si.

Therefore, Statement (II) is false.

Final conclusion

• Statement I: True
• Statement II: False

Therefore, the correct option is: $\boxed{\text{D}}$