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

cbse
redoxreactions

#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