The reducing power of divalent species decreases in the order

G e^{2 +} > S n^{2 +} > P b^{2 +}

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S n^{2 +} > G e^{2 +} > P b^{2 +}

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P b^{2 +} > S n^{2 +} > G e^{2 +}

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None of these

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Solution

The correct option is A $G e^{2 +} > S n^{2 +} > P b^{2 +}$
Inert pair effect may be defined as inertness of inner ns sub-shell towards chemical reaction. It is seen in group 3A and 4A group especially (5s and 6s electrons). Therefore, Pb shows mainly +2 oxidation state frequently as compared to +4 oxidation state. Carbon and silicon mostly show +4 oxidation state. Germanium forms stable compounds in +4 state and only few compounds in +2 state. Tin forms compounds in both oxidation states.
So as we move down the group, due to inert pair effect lower oxidation state become more stable.
So,
The stability of + 2 O.S. follows the order
$P b^{2 +} > S n^{2 +} > G e^{2 +}$
Higher the stability of +2 oxidation states, lesser will be its reducing power.
Therefore, order of reducing power is
$G e^{+ 2} > S n^{+ 2} > P b^{+ 2}$

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Similar questions

Stability of ions of $G e, S n, a n d P b$ will be in the order:

[1] P b^{2 +} > P b^{4}, S n^{2 +} > S n^{4 +}

[2] P b^{2 +} < P b^{4}, S n^{2 +} < S n^{4 +}

[3] P b^{2 +} > P b^{4}, S n^{2 +} < S n^{4 +}

[4] P b^{2 +} < P b^{4}, S n^{2 +} > S n^{4 +}

[5] S n^{2 +} < P b^{2 +}, S n^{4 +} > P b^{4 +}

[6] S n^{2 +} < P b^{2 +}, S n^{4 +} < P b^{4 +}

P b^{4 +} + S n^{2 +} \to S n^{4 +} + P b^{2 +}

S n^{4 +} + P b^{2 +} \to P b^{4 +} + S n^{2 +}

E_{P b^{4 +} / P b^{2 +}}^{o} = 1.69 V, E_{S n^{4 +} / S n^{2 +}}^{o} = 0.15 V

{S n}^{2 +} + {2 H g}^{2 +} ⟶ {H g}_{2}^{2 +} + {S n}^{4 +}

{G e}^{2 +} + {P b}^{4 +} ⟶ {p b}^{2 +} + {G e}^{4 +}

{T 1}^{+} + {G a}^{3 +} ⟶ {G a}^{+} + {T 1}^{3 +}

P b + S n^{2 +} \to P b^{2 +} + S n

\frac{[P b^{2 +}]}{[S n^{2 +}]}

E_{c e l l} = 0

E_{P b / P b^{2 +}}^{\circ} = 0.13 v o l t

E_{S n^{2 +} / S n}^{\circ} = - 0.14 v o l t

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