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.

 

i H+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, the O.N. of S is +6.
ii  Cxr2O72-2-
2x+7-2=-2
2x-14=-2
x=+6
Here, there is no fallacy about the O.N. of Cr in Cr2O72-.
The structure of Cr2O72-  is shown as follows:



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

iii  NxO3-2-
1x+3-2=-1
x-6=-1
x=+5
Here, there is no fallacy about the O.N. of N in NO3-.
The structure of NO3- is shown as follows:


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