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

(i) `overset(+1)(H)_(2) overset(x)(S) overset(-2)(O_(5))`
`2 (+1) + 1(x) + 5(-2) = 0`
`implies 2 + x - 10 = 0`
`implies 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 `H_(2)SO_(5)` is shown as follows :

Now , `2(+1) + 1(x) + 3 (-2) + 2(-1) = 0`
`implies 2 + x - 6 - 2= 0 `
`implies x = +6`
Therefore , the O.N. of S is `+6`.
(ii) `overset(x)(Cr_(2)) overset(-2)(O_(7)^(-2))`
`2(x) + 7 (-2) = -2`
`implies 2x - 14 = -2`
`implies x = +6`
Here, there is no fallacy about the O.N. of Cr in `Cr_(2)O_(7)^(2-)`
The structure of `Cr_(2)O_(7)^(2-)` is shown as follows :

Here, each of the two Cr atoms exhibits the O.N. of +6. lt brgt (iii)
`overset(x)(N)overset(2-)O_(3)^(-)`
`1(x) + 3(-2) = -1`
`implies x - 6 = -1`
`implies x = +5`
Here , there is no fallacy about the O.N. of N in `NO_(3)^(-)`
The structure of `NO_(3)^(-)` is shown as follows :

The atom exhibits the O.N. of +5