Question

# Find the oxidation number of $Pt$ in ${\left[ {PtC{l_6}} \right]^{2 - }}$ ion.

Hint: Oxidation number is defined as the charge which is present on an atom due to electronegativity difference of other atoms present in the same compound. According to Werner's theory there are two types of valency of a central metal atom: primary and secondary.

According to Werner's theory there are two types of valency of an atom. Primary and secondary. Primary valency represents oxidation state (charge) and secondary valency represents coordinate number (molecules to which central metal atom is linked). For example in ${K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]$ central atom is iron. It is linked with $6$ $CN$ molecules which represent its secondary valency i.e. secondary valency of iron is $6$. As each $CN$ molecule possess $- 1$ charge, on whole coordinate sphere charge is $- 4$ (each potassium atom has charge of $- 1$ and as a whole compound is neutral so there will be $- 4$charge on anionic part) so charge on iron will be $+ 2$, which represents primary valency of iron.
In this question we have to find primary valency (oxidation number) of ${\left[ {PtC{l_6}} \right]^{2 - }}$. In this compound $Pt$ is linked with $6$ $Cl$ atoms. We know each chlorine atom possesses $- 1$ charge. So total charge due to $Cl$ atoms will be $- 6$ (there are six chlorine atoms). There is a charge of $- 2$ on the ion as a whole. Therefore the sum of charge of platinum $\left( {Pt} \right)$ and chlorine must be equal to $- 2$. Let charge on platinum be $x$. Therefore:
$\ x - 6 = - 2 \\ x = - 2 + 6 \\ x = 4 \\ \$
Therefore charge on platinum $\left( {Pt} \right)$ is $+ 4$.