Question

The oxidation number of $Pt$ in ${[Pt({C_2}{H_4})C{l_3}]^ - }$ is:
A. +1
B. +2
C. +3
D. +4

Answer 1

For solving this question, we need to understand the concept of oxidation number. We can define the oxidation number of an ion or molecule as the number of charges it would carry when the electrons are completely transferred.

Complete step by step answer:
For determining the oxidation number of a molecule or ion we need to follow the following rules:
The oxidation number of an element in its free states is zero. For example: Oxidation number of Aluminum Al.= 0.
The oxidation number of a monatomic ion will be equal to the charge on the ion or molecule. For example: Oxidation number of $N{a^ + }$ ion = 1.
H shows +1 oxidation in case of non-metals and -1 oxidation number in case of metals.
Also, O always shows -2 oxidation numbers and acts as an exception in case of peroxide. Thus, it shows -1 oxidation number.
Therefore, to determine the oxidation number of $Pt$ in ${[Pt({C_2}{H_4})C{l_3}]^ - }$
Here, let x be the oxidation number of Pt. We can see that the overall charge of ${[Pt({C_2}{H_4})C{l_3}]^ - }$is -1. Thus, the sum of oxidation states of all elements in it should be equal to −1.
So, we will place the values in terms of x.
Thus, x + 0 -3 = -1

{x{\text{ }} = {\text{ }}\left( { - 1} \right){\text{ }} + {\text{ }}3} \\\ {x{\text{ }} = {\text{ }} + 2} \end{array}$$ Therefore, the oxidation number of Pt in ${[Pt({C_2}{H_4})C{l_3}]^ - }$ is +2. **So, the correct answer is Option B .** **Note:** For solving this question, we need to know the oxidation number of carbon in ethene (${C_2}{H_4}$ ) is –2 and in ethane (${C_2}{H_6}$) it has an oxidation number of –3. Thus, the oxidation number of hydrogen in ethene and ethane is +1.