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I was asked which geometrical isomer of $\ce{[M(NH3)2Cl2]}$, viz. cis or trans would react with silver oxalate. The possible reaction could be formation of silver chloride but since the coordination entity does not break up, any reaction thus, in my opinion can not happen. But the asker says that the answer is not none, but actually one of them.

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[…] but since the coordination entity does not break up, any reaction thus, IMO can not happen.

Your opinion contradicts experimental findings. Two chlorido ligands in metal complexes can be replaced by (other monodentate ligands or) a bidentate chelating ligand, as demonstrated for the reaction of potassium tetrachloropalladate, $\ce{K2[PdCl4]}$, with diaminosuccinic acid (and its diethylester))(DOI).

When the binding atoms of a bidentate ligand are close together, as in oxalate, the ligand can't just somehow wrap around the central metal. Instead specific geometric requiries have to be met.

For a tetra-coordinated complex, such as $\ce{[M(NH3)2Cl2]}$, two major fundamental geometries are conceivable, namely

  • tetrahedral or
  • square-planar

Only in the latter case, cis-trans isomerism can be observed!

In the case of the trans-isomer the two square-planar complexes, the substitution of two chlorido ligands that point to opposite corners of the square is not possible: oxalate would have to bind all across the complex. Consequently, it is only the cis-isomer which can react here!

$$\ce{\mathit{cis}-[M(NH3)2Cl2] + ox^{2-} -> \mathit{cis}-[M(ox)(NH3)2] +2Cl-}$$

(For the abbreviation ox for oxalate in coordination compounds, see the IUPAC Red Book, Table VII, entry 118.)

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