Calculate the oxidation number of sulphur, chromium, and nitrogen in `H_(2)SO_(5), Cr_(2)O_(7)^(2-)` and `NO_(3)^(Θ)`. Suggest the structure of these compounds. Count for the fallacy.

Explanation

a. Oxidation number of S in `H_(2)SO_(5)`
By convenctional method, the oxidation number of `S` in `H_(2)SO_(5)` is
`2(+1)+x+5(-2)=0` or, `x=+8`
This is impossible because the maximum oxidation number of `S` cannot be more than six since it has only six electrons in the valence shell. This fallacy is overcome if we calculate the oxidation number of S by chemical bonding method.
`H-O-underset(O)underset(||)overset(O)overset(||)(S)-O-O-H`
`underset(("for H"))(2xx(+1))+underset(("for S"))(x)+underset("for(O-O)")(2(-1))+underset(("for other O atoms"))(3xx(-2)=0)`
or `x=+6`
b. Oxidation number of `Cr` in `CrO_(5)`
According to conventional method, oxidation number of `Cr` is: `x+5(-2)=0` or `x=+10`. Thus is impossible because the maximum oxidation number of `Cr` cannot be more than six since it has `3d^(5) 4s^(1)` outer orbital configuration. This fallacy is removed if we calculate oxidation number of `Cr` by chemical bonding method. The structure of. `CrO_(5)` is (Butterfly structure)

From the structure, the oxidation number of `Cr` can be calculated as follows:
`x+4(-1)+1(-2)=0` or `x=+6`
`("for O-O") ("for=O")`
c. Oxidation number of `N` in `NO_(3)^(Θ)`.
According to conventional method, oxidation number of `N` is `NO_(3)^(Θ)=x+3(-2)=-1` or `x=+5`
According to the chemical bonding method,

`underset(("for O"^(Θ)))(x+1(-1))+underset(("for=O"))(1(-2))+underset(("for "rarr))(1(-2)=0)`
or `x=5`
Thus, there is no fallacy about the oxidation number of `N` in `NO_(3)^(Θ)` whether one calculates by conventional methoc or by chemical bonding method.