Given below are two statements :
Statement (I) : The first ionization energy of Pb is greater than that of Sn .
Statement (II) : The first ionization energy of Ge is greater than that of Si .
In the light of the above statements, choose the correct answer from the options given below :

Solution

Chemistry ma’am · 4 steps

Group 14 IE falls from Si to Ge, but Pb climbs back above Sn because its 4f electrons shield poorly.

1. Usual trend. Down a group, atoms get bigger, so the outer electron is easier to remove and ionisation energy usually falls. Group 14 mostly follows this, with one exception at the bottom.

   Element	First IE (kJ mol⁻¹)	What decides it
   Si	786	small atom, outer electrons held tightly
   Ge	762	the 3d¹⁰ shell shields poorly, but the larger size wins, so Ge < Si
   Sn	709	larger still, so IE falls further
   Pb	716	4f¹⁴ and 5d¹⁰ shield very poorly, so Pb > Sn

2. Statement I. In Pb, 14 new protons are added across the 4f series, and 4f electrons shield the outer 6p electrons very poorly. The effective nuclear charge rises enough that Pb's first IE ends up slightly above Sn's. True.
3. Statement II. Ge does have a poorly shielding 3d¹⁰ shell. That only slows the fall in IE from Si to Ge. It does not reverse it, so Ge's first IE is lower than Si's. False.
4. So. Statement I is true and Statement II is false, which is option (d).

The idea to keep. In group 14 the first IE order is Si > Ge > Pb > Sn. Poorly shielding d and f electrons can slow the usual fall down a group, and only at Pb, with its 4f shell, do they reverse it.

Answer from the NTA final answer key. Solution by PrepWiser.