Question

The oxidation state of platinum in xenon hexafluoroplatinate is:
a.) +4
b.) +2
c.) +6
d.) +5

Answer 1

In order to solve the given problem first we will see the basic definition and the uses of finding the oxidation number of the compound. Further we will see the method to find the oxidation number of the compound also we will see how we break the compound on the basis of initial reaction of the formation of compound to find out the oxidation number of an element in a compound. First of all we will see the chemical formula of the compound.

Complete step by step answer:
Given that: to find the oxidation state of platinum in xenon hexafluoroplatinate.
The chemical representation of xenon hexafluoroplatinate is $Xe\left[ {Pt{F_6}} \right]$ .
In basic terms, the oxidation number can be represented as the number allocated to elements in a chemical mixture. Basically, the oxidation number is the number of electrons that atoms will share, lose or acquire in a molecule when forming chemical bonds with other atoms of another substance.
The oxidation number is also known as the state of oxidation. Often, though, these terms may have a different definition based on whether or not we are discussing the atom's electronegativity. In coordination chemistry, the oxidation number concept is commonly used.
In order to find the oxidation state of an unknown element in a compound there are a set of rules for the oxidation number. For some of the particular types of elements there are some fixed oxidation numbers and for the unknown ones we consider as an unknown and further we solve the problem like an algebraic equation to find the oxidation number of unknown.
For coordination compounds like above we can break the compounds into 2 parts on the basis of the origin of the compound and the oxidation number in both the cases will be the same.
Xenon hexafluoroplatinate is formed when the xenon metal reacts with platinum hexafluoride. So the oxidation number of platinum in xenon hexafluoroplatinate and platinum hexafluoride will be the same.
We know the fact that the oxidation number of fluorine is -1
Let us take the compound $Pt{F_6}$ . Now let us take the oxidation number of fluorine as x.
So let us solve for x.

$$\Rightarrow x + \left( {6 \times - 1} \right) = 0 \\\ \Rightarrow x + \left( { - 6} \right) = 0 \\\ \Rightarrow x = + 6 \\\$$

So, the oxidation number of platinum in platinum hexafluoride is +6. Therefore the oxidation number of fluorine in xenon hexafluoroplatinate is also +6.
Hence, the oxidation state of platinum in xenon hexafluoroplatinate is +6.
So, the correct answer is “Option C”.

Note: In order to solve the problems related to finding the oxidation number of an element in a compound, students must remember the different set of rules and also the oxidation number of some common type of elements. Students must solve the problem in the algebraic method. Also the students must keep in mind that there are some elements which have different oxidation numbers according to the coordination compound.