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# A complex containing ${{K}^{+}}$, $Pt\left( IV \right)$and $C{{l}^{-}}$is $100%$ ionised giving $i=\text{ }3$. Thus, the complex is:A.$K_2\left[ PtCl_4 \right]$B.$K_2\left[ PtCl_6 \right]$C.$K_3\left[ PtCl_5 \right]$D.$K_3\left[ PtCl_3 \right]$

Last updated date: 13th Jun 2024
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Hint:Since the complex is ionised completely, the value of degree of dissociation would be one.
And the oxidation number of Pt is given as $4$, we will use this clue to see which of these complexes has the same oxidation number.
Formula used: $i=1+\left( y-1 \right)\alpha$
Where, $y$ represents the number of ions present in the solution
$\alpha$ Represents the degree of dissociation and $i$ stands for the van’t Hoff’s factor

In this case, $i$ stands for the van’t Hoff’s factor, which is the ratio of the actual concentration of the particles which are produced when the substance is dissolved and the concentration of a substance as calculated from its mass. Here van’t Hoff’s factor will be equal to the sum of one and the degree of ionisation into subtraction of one from the number of ions present in the solution. The equation can be expressed as
$i=1+\left( y-1 \right)\alpha$
Where, $y$ represents the number of ions present in the solution
$\alpha$ Represents the degree of dissociation and $i$ stands for the van’t Hoff’s factor
As we all know that this salt will dissociate into its positive and negative ions, but the ions inside the square bracket will not dissociate as they are not free to do so. The whole coordination sphere will act as one entity and the positive ion will get dissociated from this entity.
If we consider the first option, the coordination complex is ​$K_2\left[ PtCl_4 \right]$ , so it will get dissociated to form three ions, that is , two positive and one negative ions. This can be represented by the chemical equation,
${{K}_{2}}PtC{{l}_{4}}\to 2{{K}^{+}}+{{\left[ PtCl_4 \right]}^{2-}}$
So if we calculate the van’t hoff factor, we get
$i=1+2$ Which is equal to $3$
Similarly, if we take the second option, $K_2\left[ PtCl6 \right]$ , it will get dissociated to form three ions, that is , two positive and one negative ions. This can be represented by the chemical equation,
${{K}_{2}}PtC{{l}_{6}}\to 2{{K}^{+}}+{{\left[ PtC{{l}_{6}} \right]}^{2-}}~$;
So if we calculate the van’t hoff factor, we get
$i=1+2=3$
In option A, the oxidation number of $Pt\text{ }=\text{ }2$. Whereas In option B, the oxidation number of $Pt\text{ }=\text{ }4$.Therefore, $K_2\left[ PtCl_6 \right]$ is the correct answer.
So the correct answer is option B.

Note:
-$K_2\left[ PtCl_6 \right]$ is a coordination complex whose central metal is $Pt\text{ }$, and $6$ chlorine atoms as ligands are attached to it through coordinate bonds.
-$K_2\left[ PtCl_6 \right]$ has a number of industrial, commercial and laboratory usage known to us.