14. Oxidation number

1. If the oxidation number is positive, it means that the atom loses an electron, and if it is negative, it means that the atom gains electrons.
2. If it is zero, then the atom neither gains nor loses electrons.

3. The amount of oxidation numbers of all the atoms in the formula for a neutral compound is $0$.

4. The amount of oxidation numbers of an ion is the same as the charge on that ion.

5. A negative oxidation number in a compound of two, unlike atoms, is assigned to the more electronegative atom.

  

Illustration $1$ Oxidation Number of $H$ and $O$ in $H_2O$:Let us take oxidation number of $H$ = $+1$ and oxidation number of $O$ = $-2$
$2$ × ($+1$) + $1$ × ($-2$) = $0$
($+2$) + ($-2$) = $0$
Thus, the oxidation number of $H$ is $+1$, and the oxidation number of $O$ is $-2$.

 

Let oxidation number  of $S$ be $x$ and we know that oxidation number of $H$ = $+1$ and $O$ = $-2$
$2$ × ($+1$) + $x$ + $4$ × ($-2$) = $0$
($+2$) + $x$ + ($-8$) = $0$
$x$ = $+6$ 

Therefore, oxidation number of $S$ is $+6$.

 

Let oxidation number of $Cr$ be $x$ and we know that oxidation number of $K$ = $+1$ and $O$ = $-2$
$2$ × ($+1$) + $2$ × $x$ + $7$ × ($-2$) = $0$
($+2$) + $2x$ + ($-14$) = $0$
$2x$ = $+12$
$x$ = $+6$ 

Therefore, the oxidation number of $Cr$ in $K_2Cr_2O_7$ is $+6$.

 

  

Let oxidation number of $Fe$ be $x$ and we know that oxidation number of $S$ = $+6$ and $O$ = $-2$

$x$ + $1$ × ($+6$) + $4$ × ($-2$) = $0$
$x$ + ($+6$) + ($-8$) = $0$
$x$ = $+2$

Therefore, the oxidation number of $Fe$ in $FeSO_4$ is $+2$.