Calculate the oxidation number of sulphur,chromium,and nitrogen in $H_2SO_5$,$Cr_2O_7^{2-}$,and $NO_3^-$. Suggest the structures of these compounds and explain the fallacy in the calculated oxidation number for $H_2SO_5$.

(A) $(i)$ For $H_2SO_5$: Let the oxidation number of $S$ be $x$. $2(+1) + x + 5(-2) = 0$ $\Rightarrow 2 + x - 10 = 0$ $\Rightarrow x = +8$. However,the oxidation number of $S$ cannot exceed its valence electrons,which is $6$. This fallacy arises because $H_2SO_5$ (Caro's acid) contains a peroxide linkage. The correct structure is $HO-S(=O)_2-O-OH$. Assigning $-1$ to each peroxide oxygen: $2(+1) + x + 3(-2) + 2(-1) = 0 \Rightarrow x = +6$.
$(ii)$ For $Cr_2O_7^{2-}$: Let the oxidation number of $Cr$ be $x$. $2x + 7(-2) = -2$ $\Rightarrow 2x - 14 = -2$ $\Rightarrow 2x = 12$ $\Rightarrow x = +6$. The structure is $[O_3Cr-O-CrO_3]^{2-}$,where each $Cr$ is in the $+6$ oxidation state.
$(iii)$ For $NO_3^-$: Let the oxidation number of $N$ be $x$. $x + 3(-2) = -1$ $\Rightarrow x - 6 = -1$ $\Rightarrow x = +5$. The structure is a trigonal planar arrangement with $N$ at the center,exhibiting an oxidation state of $+5$.