Question

Calculate the oxidation number of sulphur, chromium and nitrogen in $H_{2}S{O}_5,$
$C{r}_2{O}_7^{2-}$ and $N{O}_{3-}^{-}$. Suggest structure of these compounds. Count for the fallacy.

$\stackrel{x}{C{r}_2}\stackrel{2-}{O_7^{2-}}$

$\stackrel{x}{N}\stackrel{2-}{O_3^{-}}$

Answer 1

$\stackrel{+1}{H_2}\stackrel{x}{S}\stackrel{-2}{O_5}2(+1)+1\left(x\right)+5(-2)=0\Rightarrow 2+x-10=0\Rightarrow x=+8$
However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6.
The structure of $H_{2}S{O}_5$ is shown as follows:

Now, 2 (+1) + 1(x) +3 (-2) + 2(-1) = 0
$\Rightarrow 2+x-6-2=0\Rightarrow x=+6$
Therefore, the O. N. of S is + 6.

$\stackrel{x}{C{r}_2}\stackrel{2-}{O_7^{2-}}$
$2\left(x\right)+7(-2)=-2\Rightarrow 2x-14=-2\Rightarrow x=+6$
Here, there is no fallacy about the O.N. of Cr in $C{r}_2{O}_7^{2-}$
The structure of $C{r}_2{O}^{2-}7$

is shown as follows: Here, each of the two Cr atoms exhibits the O.N. of +6.

$\stackrel{x}{N}\stackrel{2-}{O_3^{-}}$
$1\left(x\right)+3(-2)=-1\Rightarrow x-6=-1\Rightarrow x=+5$
Here, there is no fallacy about the O.N. of N in $N{O}^{-}3$.
The structure of $N{O}^{-3}$

is shown as follows:
The N atom exhibits the O.N. of +5.