Calculate the oxidation number of sulphur, chromium and nitrogen ion `H_(2)SO_(5), Cr_(2)O_(7)^(2-)andNO_(3)^(-)`. Suggest structure of those compounds.

Explanation

`H_(2)SO_(5)`
Oxidation number of H = +1
Oxidation number of O = -2
Oxidation number of S = x
`(2xx1)+x+3(-2)+2(-1)=0`
2 + x - 6 - 2 = 0
x - 6 = 0
x = +6
(i) The oxidation number of sulphur is exceeding its group number which cannot exist. So it should contain peroxy bond.
Structure of `H_(2)SO_(5)`
`HO-underset(O)underset(||)overset(O)overset(||)(S)-O-OH`
(ii) `Cr_(2)O_(7)^(2-)`
Oxidation state of chromium = x
Oxidation state of oxygen = -2
`(2x)+(-2xx7)=-2`
2x = +12
`thereforex=(+12)/2=+6`
The oxidation state Cr = +6
Structure of `Cr_(2)O_(7)^(2-)`

(iii) `NO_(3)^(-)`
Oxidation state of N = x
Oxidation state of O = -2
`x+(-2xx3)=-1`
x = +5
Structure of `NO_(3)^(-)`