It is because of the inability of $ns^{2}$ electrons of the valence shell to participate in bonding that:
- A$Sn^{2+}$ is oxidising while $Pb^{4+}$ is reducing
- B$Sn^{2+}$ and $Pb^{2+}$ are both oxidising and reducing
- C$Sn^{4+}$ is reducing while $Pb^{4+}$ is oxidising
- D$Sn^{2+}$ is reducing while $Pb^{4+}$ is oxidising.
Explore More
Similar Questions
What is the correct order of acidic character for the given oxides?
Medium
View SolutionWhich gas is used in the manufacturing of nitrogenous fertilizers?
Easy
View SolutionWhich of the following is the anhydride of orthophosphoric acid $(H_3PO_4)$?
Easy
View SolutionWhich of the following are the correct orders of stability of oxidation states of cations?
$(a) \ Pb^{+2} > Pb^{+4}, Tl^{+1} < Tl^{+3}$
$(b) \ Bi^{+3} < Sb^{+3}, Sn^{+2} < Sn^{+4}$
$(c) \ Pb^{+2} > Pb^{+4}, Bi^{+3} > Bi^{+5}$
$(d) \ Tl^{+3} < In^{+3}, Sn^{+2} > Sn^{+4}$
$(e) \ Sn^{+2} < Pb^{+2}, Sn^{+4} > Pb^{+4}$
$(f) \ Sn^{+2} < Pb^{+2}, Sn^{+4} < Pb^{+4}$
$(a) \ Pb^{+2} > Pb^{+4}, Tl^{+1} < Tl^{+3}$
$(b) \ Bi^{+3} < Sb^{+3}, Sn^{+2} < Sn^{+4}$
$(c) \ Pb^{+2} > Pb^{+4}, Bi^{+3} > Bi^{+5}$
$(d) \ Tl^{+3} < In^{+3}, Sn^{+2} > Sn^{+4}$
$(e) \ Sn^{+2} < Pb^{+2}, Sn^{+4} > Pb^{+4}$
$(f) \ Sn^{+2} < Pb^{+2}, Sn^{+4} < Pb^{+4}$
Difficult
View SolutionWhich of the following is a true acidic anhydride?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo