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JEE Advanced levelSingle correct optionDifficulty 3 of 3pt210

Which of the following correctly lists the divalent ions of group 14 in increasing order of reducing power?

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

Answer: A

The inert pair effect makes the

+ 2

state more stable and the

+ 4

state less stable down the group. Pb(II) therefore has least tendency to be oxidised, and Ge(II), the most prone to attain

+ 4

, is the strongest reductant. The increasing order of reducing power is

Pb X^{2 +} < Sn X^{2 +} < Ge X^{2 +}

.

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