8.5 Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O72– and NO3. Suggest structure of these compounds. Count for the fallacy.

 

iH+1SxO-25
2+1+1x+5-2=0
2+x-10=0
x=+8

However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6.
The structure of H2SO5 is shown as follows:

Now,2+1+1x+3-2+2-1=0
2+x-6-2=0
x=+6
Therefore,theO.N.ofSis+6.
iiCxr2O72-2-
2x+7-2=-2
2x-14=-2
x=+6
Here,thereisnofallacyabouttheO.N.ofCrinCr2O72-.
ThestructureofCr2O72-isshownasfollows:



Here, each of the two Cr atoms exhibits the O.N. of +6.

iiiNxO3-2-
1x+3-2=-1
x-6=-1
x=+5
Here,thereisnofallacyabouttheO.N.ofNinNO3-.
ThestructureofNO3-isshownasfollows:


The N atom exhibits the O.N. of +5.