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} $
