Kurnakov test addition of thiourea (Th) to cis-Platin yields [Pt(Th)4]Cl2. In a similar process, trans-Platin will react with thiourea to yield :
Trans-[Pt(NH3)2(Th)2]Cl2
The Kurnakov test distinguishes cis- from trans-platin, and the whole difference comes from the trans effect — the ability of a ligand to labilise whatever sits opposite it. Thiourea is a soft, strongly π-accepting sulphur donor with a very high trans effect, far greater than that of ammonia or chloride.
In cis-platin, the two chlorides are adjacent and each is trans to an ammonia. Thiourea first replaces the two chlorides. Now each incoming thiourea is trans to an NH3, and its powerful trans effect labilises that ammonia, so both ammines are displaced in turn. The result is complete substitution to [Pt(Th)4]2+, as the stem states.
In trans-platin, the geometry changes the outcome. The two chlorides are trans to each other, and the two ammines are likewise mutually trans. Thiourea again displaces the two chlorides — but now each thiourea finds itself trans to the other thiourea, not to an ammonia. Its trans effect is therefore directed at another thiourea, which is itself firmly bound, and the ammines are left untouched because nothing with a strong trans effect is opposite them.
Substitution therefore stops after two ligands, giving trans-[Pt(NH3)2(Th)2]Cl2, with the original trans arrangement of the ammines preserved.
That is exactly what makes the test diagnostic: the cis isomer loses all four original ligands while the trans isomer retains its two ammines, and the two products are readily told apart.
Hence trans-platin yields trans-[Pt(NH3)2(Th)2]Cl2.
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