A `0.001` molal solution of a complex represented as `Pt(NH_(3))_(4)Cl_(4)` in water had freezing point depression of `0.0054^(@)C`. Given `K_(f)` for `H_(2)O=1.86 K m^(-1)`. Assuming `100%` ionization of the complex, write the ionization nature and formula or complex.

Explanation

Let `n` atoms of `Cl` be the acting as ligand. Then formula of complex and its ionisation is:
`{:([Pt(NH_(3))_(4)CI_(n)]CI_((4-n))rarr,[Pt(NH_(3))_(4)CI_(n)]^(+(4-n))+,(4-n)CI^(-)),(" 1"," 0"," 0"),(" "0," "1," "(4-n)):}`
Thus particles after dissocation `=4-n+1=5-n`
and therefore, van't Holf factor (i)`=5-n`
Now `DeltaT_(f)=K'_(f)xx"molality"xx"van't Holf factor"`
`0.0054=1.86xx0.001xx(5-n)`
`therefore n=2.1approx2` (integer value)
Thus complex and its ionisation is:
`[Pt(NH_(3))_(4)Cl_(2)]Cl_(2)rarr[Pt(NH_(3))_(4)Cl_(2)]^(2+)+2Cl^(-)`