Calculate the oxidation number of sulphur chromium and nitrogen in \${{H}_{2}}\$S\${{O}_{5}}\$, \${{Cr}_{2}}\$\${{O}_{7}}^{2-}\$, N\${{O}_{3}}^{-}\$. Suggest structure of these compounds. Count for the fallacy

#1

Oxidation No. of S in \${{H}{2}}\$S\${{O}{5}}\$ can be calculated byunderstanding its structure.
(i) By conventional method
\${{H}{2}}\$S\${{O}{5}}\$ or 2(+ 1) + x + 5(- 2) = 0
or x = + 8 (wrong)
But this is wrong as the maximum oxidation no. of sulphur cannot be more than + 6, since it has only
six valence electrons.

in which two oxygen atoms are joined by peroxide linkage.
The oxidation no. of S can be calculated as
2(+ 1) + x + 3(- 2) + 2(-1) =0
(for H) (for O-O)
+2+x-6-2 =0
or x = + 6
Thus, oxidation no. of S in \${{H}{2}}\$S\${{O}{5}}\$ = + 6.
(ii) Oxidation no. of Cr in \${{Cr}{2}}\$\${{O}{7}}^{2-}\$
By conventional method
\${{Cr}{2}}\$\${{O}{7}}^{2-}\$ or 2x + 7(- 2)=-2 or x=+6
In this case there is no fallacy.
(iii) Oxidation no. of N in N\${{O}{3}}^{-}\$
Let Oxidation no. of N in N\${{O}
{3}}^{-}\$ = x
x + 3(-2) =-l
x = + 5
Oxidation no. of N in N\${{O}{3}}^{-}\$ = + 5
Structure of N\${{O}
{3}}^{-}\$ ion is