Calculate the oxidation number of sulphur, chromium, and nitrogen in H2SO5, Cr2O2–7 and NO–3. Suggest structure of these compounds. Count for the fallacy.

(i) H2SO5


Let the oxidation number(O.N.) of S be ‘x’


We know that,


O.N. of H = + 1


O.N. of O = -2


Therefore, we have,


2(+ 1) + 1(x) + 5(-2) = 0


2 + x-10 = 0


X = + 8


The O.N. of S is + 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 H2SO5 is shown as follows:



Now, 2(+ 1) + 1(x) + 3(-2) + 2(-1) = 0


2 + x-6-2 = 0


X = + 6


Therefore, the O.N. of S is + 6.


(ii) Cr2O72-


Let the oxidation number (O.N.) of Cr be ‘x’


O.N. of O = -2


Therefore, we have,


2(x) + 7(-2) = -2


2x-14 = -2


X = + 6


The O.N. of Cr is + 6.


Here, there is no fallacy about the O.N. of Cr in Cr2O72-


The structure of Cr2O72- is shown as follows



Therefore, the O.N. of Cr is + 6.


(iii) NO3-


Let the oxidation number (O.N.) of N be ‘x’


O.N. of O = -2


Therefore, we have,


1(x) + 3(-2) = -1


1x-6 = -1


X = + 5


Here, there is no fallacy about the O.N. of N in NO3-.


The structure of NO3- is shown as follows:



The N atom exhibits the O.N. of + 5.


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