Consider the following elements In, Tl, Al, Pb, Sn and Ge . The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are

Solution

Chemistry ma’am · 4 steps

Do not assume IE falls neatly down a group: Ge tops this list and In, not Tl, is lowest.

1. Across the period. Group 14 elements have one more proton than their group 13 neighbours in the same period, so they hold the outer electron more tightly. That is why Ge, Sn and Pb all have higher first ionisation enthalpies than Al, In and Tl.
2. Down the group. The trend is not smooth. The 4f electrons in Tl and Pb shield the nucleus poorly, so Tl ends up with a higher first ionisation enthalpy than In, and Pb a higher one than Sn.

   Element	Group	First IE (kJ mol⁻¹)	Most stable oxidation state
   Ge	14	762, highest	+4
   Pb	14	716	+2 (inert pair)
   Sn	14	709	+4
   Tl	13	589	+1 (inert pair)
   Al	13	577	+3
   In	13	558, lowest	+3

3. Stable states. Ge is light enough that the inert pair effect is weak, so it prefers +4. In also uses all three valence electrons, so +3 is its most stable state. The +1 state becomes the more stable one only at Tl.
4. So. The highest IE belongs to Ge (+4) and the lowest to In (+3), so the answer is +4 and +3, option (b).

The idea to keep. In groups 13 and 14, poor shielding by d and f electrons breaks the simple top-to-bottom drop in IE. Tl and Pb sit higher than In and Sn, and only these heaviest elements prefer the lower, inert-pair state.

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