Chemistry · Redox Equilibrium · NEET
Going down a group (Ga to Tl, Sn to Pb, As to Bi), the atom grows larger and the ns² pair sits in a level that is poorly shielded by the filled d and f electrons. This poor shielding makes the ns² pair more strongly attracted to the nucleus and held very firmly. Because these two electrons are held so tightly, the atom does not want to un-pair and use them in bonding. So the higher oxidation state (which needs those two electrons) becomes harder to reach, and the lower state (group number minus 2) becomes more stable at the bottom of the group.
Sn²⁺ is a REDUCING agent and Pb⁴⁺ is an OXIDISING agent. For tin (higher in the group), the +4 state is still the stable one, so Sn²⁺ wants to lose 2 electrons and go to Sn⁴⁺, which means it gets oxidised itself and reduces something else. For lead (lower in the group), the inert pair effect makes +2 the stable state, so Pb⁴⁺ wants to gain 2 electrons and drop to Pb²⁺, which means it gets reduced itself and oxidises something else. This exact idea was asked in NEET 2017.
Thallium is at the very bottom of Group 13, so the inert pair effect is at its strongest. The 6s² pair is held so tightly that Tl prefers to lose only its single 6p electron, giving the very stable +1 state (Tl⁺). Reaching the +3 state (Tl³⁺) would need the 6s² pair too, which costs a lot of energy. So Tl³⁺ is unstable and readily gains 2 electrons to become Tl⁺, making Tl³⁺ a strong oxidising agent (E° for Tl³⁺/Tl = +1.26 V, a high positive value).
Yes. Bismuth is the heaviest common element of Group 15, so its 6s² pair is inert. The stable state of Bi is +3 (group number 5 minus 2 = 3), and the +5 state is unstable. That is why Bi⁵⁺ compounds like sodium bismuthate (NaBiO₃) are powerful oxidising agents, easily dropping back to Bi³⁺. The same rule (lower state stable at the bottom) links Tl (+1), Pb (+2) and Bi (+3).
For a p-block group, the two common oxidation states are the group's maximum (using all valence electrons) and that maximum MINUS 2 (skipping the ns² pair). Group 13 shows +3 and +1, Group 14 shows +4 and +2, Group 15 shows +5 and +3. As you go down each group the (max − 2) state becomes the stable one. So the stable states become Tl⁺ (Group 13), Pb²⁺ (Group 14) and Bi³⁺ (Group 15).
It is because of inability of ns² electrons of the valence shell to participate in bonding that:
The E° values are: Al⁺/Al = +0.55 V, Tl⁺/Tl = −0.34 V, Al³⁺/Al = −1.66 V and Tl³⁺/Tl = +1.26 V. Identify the incorrect statement.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the reluctance of the two ns² valence electrons to take part in bonding, which increases down a p-block group and makes the lower oxidation state more stable.
Tl is stable as +1, Pb as +2 and Bi as +3. These are all (group maximum − 2) states, made stable by the inert pair effect.
Because the stable state of lead is +2. Pb⁴⁺ readily gains 2 electrons to fall to the stable Pb²⁺, and in doing so it oxidises other species, so it acts as an oxidising agent.
Yes. Tin is higher in Group 14 where +4 is still stable, so Sn²⁺ loses 2 electrons to become Sn⁴⁺, acting as a reducing agent. This was tested directly in NEET 2017.
Yes. Bismuth is stable in the +3 state, so the +5 state in NaBiO₃ is unstable and readily gains electrons to become Bi³⁺, making it a powerful oxidising agent.
