8.5. Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O72- and NO3-. Suggest structure of these compounds. Count for the fallacy.

(a) H2? SO5? 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 an oxidation number more than 6. The fallacy is overcome if we calculate the oxidation number from its structure HO−S (O2)−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.

8.5. Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇^2-and NO₃^-. Suggest structure of these compounds. Count for the fallacy.

18 Views|Posted 10 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Vishal Baghel | Contributor-Level 10

10 months ago

(a) H_2?SO₅? 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 an oxidation number more than 6. The fallacy is overcome if we calculate the oxidation number from its structure HO−S (O₂)−O−O−H.
−1+X+2 (−2)+2