The correct code for stability, of oxidation stares for given cations is :
(i) ${Pb}^{2 +} > {Pb}^{4 +}, {Tl}^{+} < {Tl}^{3 +}$ $(ii) {Bi}^{3 +} < {Sb}^{3 +}, {Sn}^{2 +} < {Sn}^{4 +}$
$(iii) {Pb}^{2 +} > {Pb}^{4 +}, {Bi}^{3 +} > {Bi}^{5 +}$ $(iv) {Tl}^{3 +} < {In}^{3 +}, {Sn}^{2 +} > {Sn}^{4 +}$
$(v) {Sn}^{2 +} < {Pb}^{2 +}, {Sn}^{4 +} > {Pb}^{4 +}$ $(vi) {Sn}^{2 +} < {Pb}^{2 +}, {Sn}^{4 +} < {Pb}^{4 +}$

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

(i), (iii) and (vi)

(v) and (vi)

(ii) and (iv)

(iii) and (v)

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Due to inert effect the stability of lower oxidation state gradually increases while stability of higher oxidation state gradually decreases down the group in 13th to 15th. So correct orders are :
$(iii) {Pb}^{2 +} > {Pb}^{4 +}, {Bi}^{3 +} > {Bi}^{5 +}$
$(iv) {Sn}^{2 +} < {Pb}^{2 +}, {Sn}^{4 +} > {Pb}^{4 +}$