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

  • A
    $K_2[PtCl_4]$
  • B
    $K_2[PtCl_6]$
  • C
    $K_3[PtCl_5]$
  • D
    $K[PtCl_3]$

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