The reducing power of divalent species decreases in the order

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

Suggest Corrections

Similar questions

K, R b

C s

The size of species $I$ , $I^{\oplus}$ and $I^{⊖}$ decreases in the order

Z n^{2 +} + 2 e^{-} \to Z n; (- 0.762 V

C r^{3 +} + 3 e^{-} \to C r; (- 0.74 V

2 H^{+} + 2 e^{-} \to H_{2}; (0.0 V

F e^{3 +} + e^{-} \to F e^{2 +}; (0.77 V

Z n

P b

A l

E_{Z n^{2 +} | Z n}^{0} = - 0.76 V, E_{P b^{2 +} | P b}^{0} = - 0.13 V

E_{A l^{3 +} | A l}^{0} = - 1.66 V

M^{2 +}

Z n^{2 +} > S c^{2 +} = C u^{2 +} > T i^{2 +} = N i^{2 +} > V^{2 +} = C o^{2 +} > C r^{2 +} = F e^{2 +} > M n^{2 +}

Join BYJU'S Learning Program