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I have a small doub there and pls clear that .............. How do u find the oxidation state of Manganese(Mn) in this equation. 3MnO 4 2- + 4H + -------> 2MnO 4 - + MnO 2 + 2H 2 o Please find the oxidation states for all the Manganese (Mn) ions present in this equation !!! Please tell me how to find the oxidation states for problems like this !! I will gladly approve ur aanswer if it is right !! Thank You Geoffrey Richards

I have a small doub there and pls clear that ..............


 


How do u find the oxidation state of Manganese(Mn) in this equation.


3MnO42- + 4H+ -------> 2MnO4- + MnO2 + 2H2o


 


Please find the oxidation states for all the Manganese (Mn) ions present in this equation !!!


Please tell me how to find the oxidation states for problems like this !!


I will gladly approve ur aanswer if it is right !!


Thank You


Geoffrey Richards

Grade:12

2 Answers

SAGAR SINGH - IIT DELHI
879 Points
10 years ago
Dear student, The iodoform test or reaction is a chemical reaction where iodoform (CHI3) is produced by the multiple halogenation of a methyl ketone (a molecule containing the R-CO-CH3 group) in the presence of a base . R may be H, alkyl or aryl. The reaction can be used to produce CHCl3, CHBr3 or CHI3. In analytical chemistry, this reaction was traditionally used to determine the presence of a methyl ketone, or a secondary alcohol oxidizable to a methyl ketone through the the iodoform test. Nowadays, spectroscopic techniques such as NMR and infrared spectroscopy are preferred because they require small samples, may be non-destructive (for NMR) and are easy and quick to perform. Formerly, it was used to produce iodoform and bromoform and even chloroform industrially. In organic chemistry, this reaction may be used to convert a terminal methyl ketone into the appropriate carboxylic acid.
Abhi Pandey
10 Points
10 years ago

taking MnO42-  first assume OS of Mn to be x and os state of oxygen is -2 except in superoxides n cmpds of fluorine since any molecule is electrically neutral but here the overall charge is given so its ionised 1x+(-2)4=-2 we get x=6 similary forMnO4-  x=7 and for MnO2 x=4

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