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-252+1+1x+5-2=02+x-10=0x=+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=02+x-6-2=0x=+6Therefore, the O.N. of S is +6.ii  Cxr2O72-2-2x+7-2=-22x-14=-2x=+6Here, 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=-1x-6=-1x=+5Here, 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.