Redox Reactions And Electrochemistry Ques 660
Question: Oxidation state of sulphur in anions $ SO_3^{2-},,S_2O_4^{2-} $ and $ S_2O_6^{2-} $ increases in the orders:
Options:
A) $ S_2O_6^{2-}<S_2O_4^{2-}<SO_3^{2-}~ $
B) $ SO_6^{2-}<S_2O_4^{2-}<S_2O_6^{2-}~ $
C) $ S_2O_4^{2-}<SO_3^{2-}<S_2O_6^{2-}~ $
D) $ S_2O_4^{2-}<S_2O_6^{2-}<SO_3^{2-}~ $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] In $ SO_3^{^{2-}} $ $ x+3( -2 )=-2;x=+4 $ In $ S_2O_6^{2-} $ $ 2x+4( -2 )=-2 $ $ 2x-8=-2 $ $ 2x=6;x=+3 $ In $ S_2O_6^{2-} $ $ 2x+6( -2 )=-2 $ $ 2x=10;x=+5 $ hence the correct order is $ S_2O_4^{2-}<SO_3^{2-}<S_2O_6^{2-} $
