The point group of Cis - [PtCl2(NH3)2] is :
C2v
Platinum(II) is d8, so [PtCl2(NH3)2] is square planar, not tetrahedral. That alone eliminates Td.
In the cis isomer the two chlorides occupy adjacent corners of the square and the two ammines occupy the other two. Now find the symmetry elements.
E — the identity, always present.
C2 — one two-fold axis, lying in the molecular plane and passing between the two Cl ligands and between the two NH3 ligands. Rotating by 180° swaps the two chlorides with each other and the two ammines with each other, restoring the molecule.
Two mirror planes containing that axis — the molecular plane itself, and the plane perpendicular to it that contains the C2 axis.
An element set of E, one C2 and two vertical mirror planes is exactly the definition of C2v.
Crucially, the cis isomer has no centre of inversion: inversion through platinum would carry a chloride onto an ammine position. That rules out both C2h and D2h, since each requires a centre of symmetry. The trans isomer, by contrast, does have a centre and belongs to D2h — which is precisely why the two isomers can be told apart by their vibrational spectra, the polar cis form showing more IR-active Pt-Cl stretches.
Hence cis-[PtCl2(NH3)2] belongs to the C2v point group.
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