Discuss the pattern of variation in the oxidation states of (a)`B to TI` and (b) `C to Pb`.

Explanation

(i) B and Al have no d-or f-electrons. Therefore, they do not exhibit inert pair effect. Consequently, they slow show an oxidation state of +3 only due to th persence of two electrons in the s-and one electron in the p-orbital of the valence shell. In contrast, all other element from Ga to Tl contains either only d-or d- and f-electrons and hence show two oxidation states of +1 and +3 due to inert pair effect. Further, as the number of d- and f-electrons increases down te group , the inert pair effect becomes more and more pronounced. In other words, as we move down the group from Ga to Tl, the stability of +1 oxidation state increases (i.e, `Ga lt ln lt Tl`) while that of +3 oxidation state decreases (i.e., `Ga lt ln gt Tl`). Thus, `+I` oxidation state of Tl is more stable than its +3 oxidation state.
(ii) Carbon and silicon have no d-or f-electrons.Therefore , they do not exhibit inert pair effect.
Consequently they show an oxidation state of +4 due to the presence of two electrons in the s- and two electrons in the p-orbital of the valence shell. In contrast , all other elemen from Ge to Pb contain either only d- or f-electron and hence show two oxidation states of +2 and +4 due to inert pair effect.
Further , as the number of d- and f-electrons increases , the inert pair effect becomes more and more pronounced. In other words, as we more down the group from Ge to Pb, the stability of +2 oxidation state increases (i.e., `Ge lt Sn lt Pb`) while that of `+4` oxidation state decreases (i.e., `Ge gt Sn gt Pb`).
Thus, +2 oxidation state of Pb is more stable than its +4 oxidation state.