On the basis of the inert pair effect, the correct order of reducing power of the divalent ions is

A $Pb X^{2 +} > Sn X^{2 +} > Ge X^{2 +}$
B $Sn X^{2 +} > Ge X^{2 +} > Pb X^{2 +}$
C $Ge X^{2 +} = Sn X^{2 +} = Pb X^{2 +}$
D $Ge X^{2 +} > Sn X^{2 +} > Pb X^{2 +}$

Answer: D

Stability of the +2 state increases down the group (Ge < Sn < Pb). An ion that is less stable as M(II) is more easily oxidised to M(IV), so reducing power falls from Ge(II) to Pb(II).

Practice p-Block Elements: Groups 13 and 14 More on Trends in group 13 and group 14

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