Question

Calculate the oxidation number of Sulphur, chromium, and nitrogen in
$H_2SO_5$, $Cr_2O_7^{2−}$ and $NO_3^−$. Suggest structure of these compounds. Count for the fallacy.

Answer 1

(i) $H_2SO_5$

$2(+1) + 1(x) + 5(-2) = 0$
$2 + x – 10 = 0$
$x = +8$
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 $H_2SO_5$ is shown as follows:

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

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(ii) $\overset{x}C{r}_2\overset{2-}O_7^{2−}$
$2(x) + 7(-2) = -2$
$\Rightarrow$ $2x -14 = -2$
$\Rightarrow$ $x = +6$
Here, there is no fallacy about the O.N. of $Cr$ in $Cr_2O_7^{2−}$.
The structure of is shown as follows:

Here, each of the two $Cr$ atoms exhibits the O.N. of $+6$.

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(iii) $\overset{x}N\overset{2}O_3^−$

$1(x) + 3(-2) = -1$
$\Rightarrow$ $x – 6 = -1$
$\Rightarrow$ $x = +5$
Here, there is no fallacy about the O.N. of $N$ in $NO_3^−$ .
The structure of is shown as follows:

The $N$ atom exhibits the O.N. of $+5$.