A complex containing K^{+},Pt(IV) and Cl^{-} | vedclass

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(B) The van't Hoff factor $i$ is given by $i = 1 + (y-1) \alpha$,where $y$ is the number of ions produced per formula unit and $\alpha$ is the degree

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$K_2[PtCl_6]$

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(B) The van't Hoff factor $i$ is given by $i = 1 + (y-1) \alpha$,where $y$ is the number of ions produced per formula unit and $\alpha$ is the degree of ionisation.Given $\alpha = 1$ ($100\%$ ionisation) and $i = 3$,we have $i = y = 3$.This means the complex must dissociate into $3$ ions.For $K_2[PtCl_6]$,the dissociation is $K_2[PtCl_6] \rightarrow 2K^{+} + [PtCl_6]^{2-}$.Total ions $y = 2 + 1 = 3$.Also,the oxidation state of $Pt$ in $[PtCl_6]^{2-}$ is $x + 6(-1) = -2$,so $x = +4$,which matches $Pt(IV)$.Thus,$K_2[PtCl_6]$ is the correct complex.

$A$ complex containing $K^{+}$,$Pt(IV)$ and $Cl^{-}$ is $100\%$ ionised giving $i = 3$. Thus,the possible complex is

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