Question

Q.Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O2 and NOT.

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Answer 1

Answer:

(a) H

2

SO

5

by conventional method.

Let x be the oxidation number of S

2(+1)+x+5(−2)=0

x=+8

+8 Oxidation state of S is not possible as S cannot have oxidation number more than 6. The fallacy is overcomed if we calculate the oxidation number from its structure HO−S(O

2

)−O−O−H.

−1+X+2(−2)+2(−1)+1=0

x=+6(b) Dichromate ion

Let x be the oxidation number of Cr in dichromate ion

2x+7(−2)=−2

x=+6

Hence the oxidation number of Cr in dichromate ion is +6. This is correct and there is no fallacy.

(c) Nitrate ion, by conventional method

Let x be the oxidation number of N in nitrate ion.

x+3(−2)=−1

From the structure O

−

−N

+

(O)−O

−

x+1(−1)+1(−2)+1(−2)=0

x=+5

Thus there is no fallacy

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