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

Explanation

(1) Oxidation number of sulphur in `H_(2)SO_(5)`,
`2 xx (+1) + x + 5 xx (-2) = 0`
`x= 10 -2 =+8`
But the oxidation number + 8 for sulphur is not possible as it has only 6 electrons in its valence shell. It can exhibit maximum oxidation state= +6. Hence, in `H_(2)SO_(5)`, two oxygen atoms must be linked together. Considering this fallacy can be removed. Therefore, the structure of `H_(2)SO_(5)` should be:
`overset(+1)(H) - overset(-2)(O) - underset(underset(underset(-2)(O))(|))overset(overset(overset(-2)(O))(|))(S)- overset(-1)(O)-overset(-1)(O) - overset(+1)(H)`
Based on the above structure, we have, `(+1) + (-2) + x+ (-2) 2 + (-1)2 + (+1)= 0`
`-1 + xx -4-2 +1= 0`
`x= +6`
(2) Oxidation number of Cr in `CrO_(5)`. `x + 5xx (-2)= 0`
`x= +10`
But the oxidation number +10for chromium is not possible as it has only 6 electrons in its valence shell. It can exhibit maximum oxidation state =+6. Fallacy can be removed by considering the structure of `CrO_(5)` as given below:

Based on the above structure, we have,
`(-1) xx 4 + x + (-2) = 0`
`-4 + x -2= 0`
x= +6
(3) The oxidation number of N in `NO_(3)^(-), x+ (-2)3= -1`
(as `NO_(3)^(-)` bears charge-1) x= +5
The structure of `NO_(3)` is

Based on the above structures, `(-2) xx 2 + x + (-1) = 0`
`x= +5`
Therefore, this structure gives the same oxidation number for N in `NO_(3)`. Hence there is no fallacy.