Redox Reactions And Electrochemistry Ques 747

Question: The order of increasing O.N. of S in $ S_8,S_2O_8^{-2},S_2O_3^{-2},S_4O_6^{-2} $ is:

Options:

A) $ S_8<S_2O_8^{-2}<S_2O_3^{-2}<S_4O_6^{-2} $

B) $ S_2O_8^{-2}<S_2O_3^{-2}<S_4O_6^{-2}<S_8 $

C) $ S_2O_8^{-2}<S_8<S_4O_6^{-2}<S_2O_3^{-2} $

D) $ S_8<S_2O_3^{-2}<S_4O_6^{-2}<S_2O_8^{-2} $

Show Answer

Answer:

Correct Answer: D

Solution:

[d] The oxidation number of S are shown below along with the compounds $ \underset{0}{\mathop{S_8}},,\underset{+6}{\mathop{\text{ }S_2O_8^{-2}}},,\underset{+2}{\mathop{\text{ }S_2O_3^{-2},}},\underset{+2.5}{\mathop{\text{ }S_4O_6^{-2}}}, $ Hence the order of increasing O.N. of S is $ S_8<S_2O_3^{-2}<S_4O_6^{-2}<S_2O_8^{-2} $



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