Question

The oxidation number of Pt in ${{\text{ }\!\![\!\!\text{ Pt(}{{\text{C}}_{2}}{{\text{H}}_{4}}\text{)C}{{\text{l}}_{3}}\text{ }\!\!]\!\!\text{ }}^{-}}$is:
(A) +1
(B) +2
(C) +3
(D) +4

Answer 1

Hint: Oxidation state refers to the charge which an atom carries after being oxidized (i.e. loss of electrons) or reduced (i.e. gain of electrons). Oxidation results into positive and reduction results into negative oxidation states and knowing the oxidation number of ${{\text{C}}_{2}}{{\text{H}}_{4}}$ and Cl in ${{\text{ }\!\![\!\!\text{ Pt(}{{\text{C}}_{2}}{{\text{H}}_{4}}\text{)C}{{\text{l}}_{3}}\text{ }\!\!]\!\!\text{ }}^{-}}$, you can easily find the oxidation number of Pt.


Complete step by step answer:
Platinum is a silvery metal and it comes under the category of transition metals having the atomic number as 78 and it is represented by the symbol Pt. It is mostly used as a catalyst in a large number of chemical reactions.
First, we should know what the oxidation number is. By the oxidation number we simply mean the total number of the electrons an element gains or loses during a chemical reaction as a result of the bond which it forms with the other atom.
It is also called as the oxidation state because it tells us about the state of that particular atom whether it has undergone oxidation (i.e. lose electrons) and it always gives positive oxidation state or reduction (i.e. gain electrons) in a chemical reaction and it always gives negative oxidation state. Oxidation increases the oxidation state whereas on the other hand, reduction decreases the oxidation state.
Now, calculating the oxidation number of Pt in ${{\text{ }\!\![\!\!\text{ Pt(}{{\text{C}}_{2}}{{\text{H}}_{4}}\text{)C}{{\text{l}}_{3}}\text{ }\!\!]\!\!\text{ }}^{-}}$,
Oxidation number of ${{\text{C}} _ {2}} {{\text{H}} _ {4}} $=0
Oxidation number of Cl= -1
Then, we get;
Oxidation number of Pt = x+0+3(-1) = -1
= x -3 = -1
=x = -1 +3
Oxidation number of Pt = x= +2
Therefore, the oxidation number of Pt in ${{\text{ }\!\![\!\!\text{ Pt(}{{\text{C}}_{2}}{{\text{H}}_{4}}\text{)C}{{\text{l}}_{3}}\text{ }\!\!]\!\!\text{ }}^{-}}$is +2.


Hence, option(b) is correct.


Note: In ${{\text{C}} _ {2}} {{\text{H}} _ {4}} $, the oxidation number is zero because here the carbon neither gains nor loses any electrons and forms covalent bond with the hydrogen atoms i.e. there is no charge separation involved and hence its oxidation state is zero.