Calculate the O.N of sulphur, chromium and nitrogen in `H_(2)SO_(5), CrO_(5) "and NO"^(-)3` ion Suggest structure of these compounds.Account for the fallacy if any.

Explanation

(i) Oxidation number of S in `H_(2)SO_(5)`
By conventional method ,O.N of suphur may be calculated as:
`2xx1+x+5xx(-2)=0 or 2+x-10=0 or x=+8 (wrong)` But this cannot be true as maximum oxidation number for suphur cannot exceed +6 as fit has only six valence electrons .The oxidation number of the element can be calculted from the structure of the persulphuric acid. The suphur atom is linked one OH group (O.N=-1) two oxygen atoms (O.N=-2) and one peroxide (-O-O-) linkage (O.N=-1)
O.N of `S=1(-1)+x+2(-2)+1(-1)=0` or (ii) oxidation number of Cr in `CrO_(5)`
By conventional method O.N of chromimum may be calculted as:

`x+5(-2)=0 or x=+10` (wrong).
(iii)Oxidation number N in `NO_(3)^(-)` ion
By conventional method, O.N of N atom in the ion mayt be calculated as : `x+3(-2) =-1 or x=+5`.
This can be further made clear form the structure of the ion In it the nitrogen atom is linked to one single bonded oxygen atom (O.N=-2), one double bonded oxygen atom (O.N=-2) and one oxygen atom by dtive bond (O.N =-2)
`therefore` O.N of `N = 1(-2)+x+1(-2)+1(-2)=-1 "or" x=+5`