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# The oxidation number of ${\text{Pt}}$ in ${\left[ {{\text{Pt}}{{\text{C}}_2}{{\text{H}}_4}{\text{C}}{{\text{l}}_3}} \right]^ - }$ is:

Last updated date: 07th Aug 2024
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Hint: In the complex anion, the sum of all the oxidation numbers will be equal to the charge on the anion. The oxidation number of chlorine is -1. The oxidation number of ethylene is zero as ethylene is a neutral molecule, with zero charge on it. Ethylene only donates pi electrons to the metal. Ethylene does not form a sigma bond with metal.

Let X be the oxidation number of ${\text{Pt}}$ in ${\left[ {{\text{Pt}}{{\text{C}}_2}{{\text{H}}_4}{\text{C}}{{\text{l}}_3}} \right]^ - }$.
In a neutral molecule, the sum of all the oxidation numbers is zero. In an anion, the sum of all the oxidation numbers is equal to the charge on the anion. Thus if anion has -1 charge, then the sum of all the oxidation numbers will be -1. In the ion ${\left[ {{\text{Pt}}{{\text{C}}_2}{{\text{H}}_4}{\text{C}}{{\text{l}}_3}} \right]^ - }$ the charge on the ion is -1. So the sum of all the oxidation numbers will be -1. Here, for the complex ion ${\left[ {{\text{Pt}}{{\text{C}}_2}{{\text{H}}_4}{\text{C}}{{\text{l}}_3}} \right]^ - }$, the sum of all the oxidation numbers will be the oxidation numbers of ${\text{Pt, }}{{\text{C}}_2}{{\text{H}}_4}{\text{ and Cl}}$.
Calculate the oxidation number of ${\text{Pt}}$.
$X{\text{ + 0 + 3}}\left( { - 1} \right) = - 1 \\ X - 3 = - 1 \\ X = 3 - 1 \\ X = + 2 \\$
Hence, the oxidation number of ${\text{Pt}}$ in ${\left[ {{\text{Pt}}{{\text{C}}_2}{{\text{H}}_4}{\text{C}}{{\text{l}}_3}} \right]^ - }$ is +2.
Note: Oxidation number represents general distribution of electrons. Oxidation number gives the total number of electrons gained or lost by an atom to form a chemical bond. Thus, when neutral platinum atom loses two electrons to form ${\left[ {{\text{Pt}}{{\text{C}}_2}{{\text{H}}_4}{\text{C}}{{\text{l}}_3}} \right]^ - }$ complex anion, it attains oxidation number of +2.